simplified mechanics & strength of materials for architects and builders

SIMPLIFIED MECHANICS AND STRENGTH OF MATERIALS

Sixth Edition

JAMES AMBROSE

Formerly Professor of ArchitectureUniversity of Southern California

Los Angeles California

based on the work of

THE LATE HARRY PARKERFormerly Professor of Architectural Construction

University of Pennsylvania

JOHN WILEY amp SONS INC

3751 P- FM 111301 1214 PM Page iii

Innodata
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3751 P- FM 111301 1214 PM Page xii

SIMPLIFIED MECHANICS

AND STRENGTH OF MATERIALS

3751 P- FM 111301 1214 PM Page i

Other titles in thePARKER-AMBROSE SERIES OF SIMPLIED DESIGN GUIDES

Harry Parker John W MacGuire and James AmbroseSimplified Site Engineering 2nd Edition

James AmbroseSimplied Design of Building Foundations 2nd Edition

James Ambrose and Dimitry VergunSimplified Building Design for Wind and Earthquake Forces 3rd Edition

James AmbroseSimplied Design of Masonry Structures

James Ambrose and Peter D BrandowSimplified Site Design

Harry Parker and James AmbroseSimplied Mechanics and Strength of Materials 5th Edition

Marc SchilerSimplied Design of Building Lighting

James PattersonSimplified Design for Building Fire Safety

William BobenhausenSimplied Design of HVAC Systems

James AmbroseSimplified Design of Wood Structures 5th Edition

James Ambrose and Jeffrey E OllswangSimplified Design for Building Sound Control

James AmbroseSimplified Design of Building Structures 3rd Edition

James Ambrose and Harry ParkerSimplified Design of Concrete Structures 7th Edition

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James AmbroseSimplified Engineering for Architects and Builders 9th Edition

3751 P- FM 111301 1214 PM Page ii


Sixth Edition

JAMES AMBROSE







3751 P- FM 111301 1214 PM Page iii

Copyright copy 2002 by John Wiley amp Sons New York All rights reserved

No part of this publication may be reproduced stored in a retrieval system or transmittedin any form or by any means electronic mechanical photocopying recording scanningor otherwise except as permitted under Sections 107 or 108 of the 1976 United StatesCopyright Act without either the prior written permission of the Publisher orauthorization through payment of the appropriate per-copy fee to the CopyrightClearance Center 222 Rosewood Drive Danvers MA 01923 (978) 750-8400 fax (978) 750-4744 Requests to the Publisher for permission should be addressed to thePermissions Department John Wiley amp Sons inc 605 Third Avenue New York NY10158-0012 (22) 850-6011 fax (212) 850-6008 E-Mail PERMREQ WILEYCOM

This publication is designed to provide accurate and authoritative information in regard to the subject matter covered It is sold with the understanding that the publisher is notengaged in rendering professional services If professional advice or other expertassistance is required the services of a competent professional person should be sought

This title is also available in print as ISBN 0-471-40052-1 [print version ISBNs--includecloth and paper ISBNs if both are available] Some content that appears in the printversion of this book may not be available in this electronic edition

For more information about Wiley products visit our web site at wwwWileycom

fcopyebkqxd 11702 943 AM

v

CONTENTS

Preface to the Sixth Edition ix

Preface to the First Edition xiii

Introduction 1Structural Mechanics 2

Units of Measurement 2

Accuracy of Computations 3

Symbols 7

Nomenclature 7

1 Structures Purpose and Function 911 Loads 11

12 Special Considerations for Loads 13

13 Generation of Structures 21

14 Reactions 24

15 Internal Forces 28

16 Functional Requirements of Structures 30

3751 P- FM 111301 1214 PM Page v

17 Types of Internal Force 39

18 Stress and Strain 46

19 Dynamic Effects 61

110 Design for Structural Response 64

2 Forces and Force Actions 6921 Loads and Resistance 69

22 Forces and Stresses 71

23 Types of Forces 73

24 Vectors 73

25 Properties of Forces 74

26 Motion 76

27 Force Components and Combinations 78

28 Graphical Analysis of Forces 83

29 Investigation of Force Actions 87

210 Friction 91

211 Moments 97

212 Forces on a Beam 102

3 Analysis of Trusses 11131 Graphical Analysis of Trusses 111

32 Algebraic Analysis of Trusses 120

33 The Method of Sections 127

4 Analysis of Beams 13241 Types of Beams 133

42 Loads and Reactions 134

43 Shear in Beams 135

44 Bending Moments in Beams 140

45 Sense of Bending in Beams 147

46 Cantilever Beams 151

47 Tabulated Values for Beam Behavior 155

5 Continuous and Restrained Beams 16051 Bending Moments for Continuous Beams 160

52 Restrained Beams 172

vi CONTENTS

3751 P- FM 111301 1214 PM Page vi

53 Beams with Internal Pins 17654 Approximate Analysis of Continuous Beams 181

6 Retaining Walls 18361 Horizontal Earth Pressure 18462 Stability of Retaining Walls 18663 Vertical Soil Pressure 188

7 Rigid Frames 19271 Cantilever Frames 19372 Single-Span Frames 199

8 Noncoplanar Force Systems 20281 Concurrent Systems 20382 Parallel Systems 20983 General Noncoplanar Systems 213

9 Properties of Sections 21491 Centroids 21592 Moment of Inertia 21893 Transferring Moments of Inertia 22394 Miscellaneous Properties 22895 Tables of Properties of Sections 229

10 Stress and Deformation 239101 Mechanical Properties of Materials 241102 Design Use of Direct Stress 243103 Deformation and Stress Relations and Issues 246104 Inelastic and Nonlinear Behavior 251

11 Stress and Strain in Beams 254111 Development of Bending Resistance 255112 Investigation of Beams 259113 Computation of Safe Loads 261114 Design of Beams for Flexure 263115 Shear Stress in Beams 265116 Shear in Steel Beams 270

CONTENTS vii

3751 P- FM 111301 1214 PM Page vii

117 Flitched Beams 272

118 Deflection of Beams 275

119 Deflection Computations 279

1110 Plastic Behavior in Steel Beams 283

12 Compression Members 293121 Slenderness Effects 293

122 Wood Columns 297

123 Steel Columns 301

13 Combined Forces and Stresses 309131 Combined Action Tension Plus Bending 309

132 Combined Action Compression Plus Bending 312

133 Development of Shear Stress 318

134 Stress on an Oblique Section 319

135 Combined Direct and Shear Stresses 321

14 Connections for Structural Steel 324141 Bolted Connections 324

142 Design of a Bolted Connection 337

143 Welded Connections 343

15 Reinforced Concrete Beams 353151 General Considerations 353

152 Flexure Stress Method 363

153 General Application of Strength Methods 375

154 Flexure Strength Method 376

155 T-Beams 382

156 Shear in Concrete Beams 387

157 Design for Shear in Concrete Beams 394

References 402

Answers to Selected Exercise Problems 403

Index 409

viii CONTENTS

3751 P- FM 111301 1214 PM Page viii

ix

PREFACE TO THE SIXTH EDITION

Publication of this book presents the opportunity for yet another newgeneration of readers to pursue a study of the fundamental topics that un-derlie the work of design of building structures In particular the workhere is developed in a form to ensure its accessibility to persons with lim-ited backgrounds in engineering That purpose and the general rationalefor the book are well presented in Professor Parkerrsquos preface to the firstedition excerpts from which follow

The fundamental materials presented here derive from two generalareas of study The first area is that of applied mechanics and most prin-cipally applications of the field of statics This study deals primarilywith the nature of forces and their effects when applied to objects Thesecond area of study is that of strength of materials which deals gener-ally with the behavior of particular forms of objects of specific structuralmaterials when subjected to actions of forces Fundamental relation-ships and evaluations derived from these basic fields provide the tools forinvestigation of structures relating to their effectiveness and safety forusage in building construction No structural design work can be satis-factorily achieved without this investigation

3751 P- FM 111301 1214 PM Page ix

In keeping with the previously stated special purpose of this book thework here is relatively uncomplicated and uses quite simple mathemat-ics A first course in algebra plus some very elementary geometry andtrigonometry will suffice for the reader to follow any derivations pre-sented here In fact the mathematical operations in applications to actualproblem solving involve mostly only simple arithmetic and elementaryalgebra

More important to the study here than mechanical mathematical op-erations is the conceptual visualization of the work being performed Tofoster this achievement extensive use is made of graphic images to en-courage the reader to literally see what is going on The ultimate exten-sion of this approach is embodied in the first chapter which presents theentire scope of topics in the book without mathematics This chapter isnew to this edition and is intended both to provide a comprehensive graspof the bookrsquos scope and to condition the reader to emphasize the need forvisualization preceding any analytical investigation

Mastery of the work in this book is essentially preparatory in natureleading to a next step that develops the topic of structural design Thisstep may be taken quite effectively through the use of the book that is es-sentially a companion to this work Simplified Engineering for Architectsand Builders That book picks up the fundamental materials presentedhere adds to them various pragmatic considerations for use of specificmaterials and systems and engages the work of creating solutions tostructural design problems

For highly motivated readers this book may function as a self-studyreference Its more practical application however is as a text for a coursein which case readers will have the advantage of guidance prodding andcounsel from a teacher For teachers accepting such a challenge aTeacherrsquos Manual is available from the publisher

While the work here is mostly quite theoretical in nature some use ofdata and criteria derived from sources of real materials and products isnecessary Those sources consist primarily of industry organizations andI am grateful for the permissions granted for such use Primary sourcesused here include the American Concrete Institute the American Institute for Steel Construction and the American Forest and PaperAssociation

A practical context for this theoretical work is presented through sev-eral illustrations taken from books that more thoroughly develop thetopic of building construction I am grateful to John Wiley amp Sons for

x PREFACE TO THE SIXTH EDITION

3751 P- FM 111301 1214 PM Page x

permission to use these illustrations from several of its publications bothcurrent and vintage works

Bringing any work to actual publication requires enormous effort andcontributions by highly competent and experienced people who cantransform the authorrsquos raw materials into intelligible and presentableform Through many engagements I continue to be amazed at the levelof quality and the skill of the editors and production staff at John Wileyamp Sons who achieve this effort

This work is the sixtieth publication that I have brought forth over thepast 35 years all of which were conceived and produced in my home of-fice None of themmdashfirst to lastmdashwould have happened there withoutthe support encouragement and lately the direct assistance of my wifePeggy I am grateful to her for that contribution and hope she will sus-tain it through the next work

JAMES AMBROSE

2002

PREFACE TO THE SIXTH EDITION xi

3751 P- FM 111301 1214 PM Page xi

3751 P- FM 111301 1214 PM Page xii

xiii

PREFACE TO THE FIRST EDITION

The following are excerpts from the preface to the first edition of thisbook written by Professor Parker at the time of publication in 1951

Since engineering design is based on the science of mechanics it is im-possible to overemphasize the importance of a thorough knowledge ofthis basic subject Regardless of the particular field of engineering inwhich a student is interested it is essential that he understand fully thefundamental principles that deal with the actions of forces on bodies andthe resulting stresses

This is an elementary treatment written for those who have had lim-ited preparation The best books on the subject of mechanics and strengthof materials make use of physics calculus and trigonometry Such booksare useless for many ambitious men Consequently this book has beenprepared for the student who has not obtained a practical appreciation ofmechanics or advanced mathematics A working knowledge of algebraand arithmetic is sufficient to enable him to comprehend the mathemat-ics involved in this volume

3751 P- FM 111301 1214 PM Page xiii

This book has been written for use as a textbook in courses in me-chanics and strength of materials and for use by practical men interestedin mechanics and construction Because it is elementary the material hasbeen arranged so that it may be used for home study For those who havehad previous training it will serve as a refresher course in reviewing themost important of the basic principles of structural design

One of the most important features of this book is a detailed explana-tion of numerous illustrative examples In so far as possible the exam-ples relate to problems encountered in practice The explanations arefollowed by problems to be solved by the student

This book presents no short-cuts to a knowledge of the fundamentalprinciples of mechanics and strength of materials There is nothingunique in the presentation for the discussions follow accepted present-day design procedure It is the belief of the author however that a thor-ough understanding of the material contained herein will afford afoundation of practical information and serve as a step to further study

HARRY PARKER

High HollowSouthamptonBucks County PennsylvaniaMay 1951

xiv PREFACE TO THE FIRST EDITION

3751 P- FM 111301 1214 PM Page xiv

1

INTRODUCTION

The principal purpose of this book is to develop the topic of structural in-vestigation also sometimes described as structural analysis To the ex-tent possible the focus of this study is on a consideration of the analyticalstudy as a background for work in structural design The work of struc-tural investigation consists of the consideration of the tasks required of astructure and the evaluation of the responses of the structure in perform-ing these tasks Investigation may be performed in various ways theprincipal ones being either the use of mathematical modeling or the con-struction of physical models

For the designer a major first step in any investigation is the visual-ization of the structure and the force actions to which it must respond Inthis book extensive use is made of graphic illustrations in order to en-courage the reader to develop the habit of first clearly seeing what is hap-pening before proceeding with the essentially abstract procedures ofmathematical investigation To further emphasize the need for visualiza-tion and the degree to which it can be carried out without any mathe-matical computations the first chapter of the book presents the wholerange of book topics in this manner The reader is encouraged to read

3751 P-00 (intro) 111301 1217 PM Page 1

Chapter 1 completely and to study the many graphic illustrations This ini-tial study should help greatly in giving the reader a grasp for the many con-cepts to be presented later and for the whole body of the bookrsquos topic scope

STRUCTURAL MECHANICS

The branch of physics called mechanics concerns the actions of forces onphysical bodies Most of engineering design and investigation is based onapplications of the science of mechanics Statics is the branch of me-chanics that deals with bodies held in a state of unchanging motion by thebalanced nature (called static equilibrium) of the forces acting on themDynamics is the branch of mechanics that concerns bodies in motion orin a process of change of shape due to actions of forces A static condi-tion is essentially unchanging with regard to time a dynamic conditionimplies a time-dependent action and response

When external forces act on a body two things happen First internalforces that resist the actions of the external forces are set up in the bodyThese internal forces produce stresses in the material of the body Secondthe external forces produce deformations or changes in shape of thebody Strength of materials or mechanics of materials is the study of the properties of material bodies that enable them to resist the actions of external forces of the stresses within the bodies and of the deforma-tions of bodies that result from external forces

Taken together the topics of applied mechanics and strength of mate-rials are often given the overall designation of structural mechanics orstructural analysis This is the fundamental basis for structural investiga-tion which is essentially an analytical process On the other hand designis a progressive refining process in which a structure is first generally vi-sualized then it is investigated for required force responses and its perfor-mance is evaluated finallymdashpossibly after several cycles of investigationand modificationmdashan acceptable form is derived for the structure

UNITS OF MEASUREMENT

Early editions of this book have used US units (feet inches poundsetc) for the basic presentation In this edition the basic work is devel-oped with US units with equivalent metric unit values in brackets [thus]

2 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 2

While the building industry in the United States is now in the process ofchanging over to the use of metric units our decision for the presentationhere is a pragmatic one Most of the references used for this book are stilldeveloped primarily in US units and most readers educated in theUnited States will have acquired use of US units as their ldquofirst lan-guagerdquo even if they now also use metric units

Table 1 lists the standard units of measurement in the US systemwith the abbreviations used in this work and a description of commonusage in structural design work In similar form Table 2 gives the corre-sponding units in the metric system (or Systegraveme International SI) Con-version factors to be used for shifting from one unit system to the otherare given in Table 3 Direct use of the conversion factors will producewhat is called a hard conversion of a reasonably precise form

In the work in this book many of the unit conversions presented aresoft conversions meaning one in which the converted value is roundedoff to produce an approximate equivalent value of some slightly morerelevant numerical significance to the unit system Thus a wood 2 times 4(actually 15 times 35 inches in the US system) is precisely 381 times 889 mmin the metric system However the metric equivalent of a 2 by 4 ismore likely to be made 40 times 90 mm close enough for most purposes inconstruction work

For some of the work in this book the units of measurement are notsignificant What is required in such cases is simply to find a numericalanswer The visualization of the problem the manipulation of the math-ematical processes for the solution and the quantification of the answerare not related to specific unitsmdashonly to their relative values In such sit-uations the use of dual units in the presentation is omitted in order to re-duce the potential for confusion for the reader

ACCURACY OF COMPUTATIONS

Structures for buildings are seldom produced with a high degree of di-mensional precision Exact dimensions are difficult to achieve even forthe most diligent of workers and builders Add this to considerations forthe lack of precision in predicting loads for any structure and the signif-icance of highly precise structural computations becomes moot This isnot to be used as an argument to justify sloppy mathematical workoverly sloppy construction or use of vague theories of investigation of

ACCURACY OF COMPUTATIONS 3

3751 P-00 (intro) 111301 1217 PM Page 3

4 INTRODUCTION

TABLE 1 Units of Measurement US System

Name of Unit Abbreviation Use in Building Design

LengthFoot ft Large dimensions building plans

beam spansInch in Small dimensions size of member

cross sections

AreaSquare feet ft2 Large areasSquare inches in2 Small areas properties of cross

sections

VolumeCubic yards yd3 Large volumes of soil or concrete

(commonly called simply ldquoyardsrdquo)Cubic feet ft3 Quantities of materialsCubic inches in3 Small volumes

Force MassPound lb Specific weight force loadKip kip k 1000 poundsTon ton 2000 poundsPounds per foot lbft plf Linear load (as on a beam)Kips per foot kipsft klf Linear load (as on a beam)Pounds per square foot lbft2 psf Distributed load on a surface

pressureKips per square foot kft2 ksf Distributed load on a surface

pressurePounds per cubic foot lbft3 Relative density unit weight

MomentFoot-pounds ft-lb Rotational or bending momentInch-pounds in-lb Rotational or bending momentKip-feet kip-ft Rotational or bending momentKip-inches kip-in Rotational or bending moment

StressPounds per square foot lbft2 psf Soil pressurePounds per square inch lbin2 psi Stresses in structuresKips per square foot kipsft2 ksf Soil pressureKips per square inch kipsin2 ksi Stresses in structures

TemperatureDegree Fahrenheit degF Temperature

3751 P-00 (intro) 111301 1217 PM Page 4


TABLE 2 Units of Measurement SI System


LengthMeter m Large dimensions building plans

beam spansMillimeter mm Small dimensions size of member

cross sections

AreaSquare meters m2 Large areasSquare millimeters mm2 Small areas properties of member

cross sections

VolumeCubic meters m3 Large volumesCubic millimeters mm3 Small volumes

MassKilogram kg Mass of material (equivalent to

weight in US units)Kilograms per cubic meter kgm3 Density (unit weight)

Force LoadNewton N Force or load on structureKilonewton kN 1000 newtons

MomentNewton-meters N-m Rotational or bending momentKilonewton-meters kN-m Rotational or bending moment

StressPascal Pa Stress or pressure (1 pascal =

1 Nm2)Kilopascal kPa 1000 pascalsMegapascal MPa 1000000 pascalsGigapascal GPa 1000000000 pascals

TemperatureDegree Celsius degC Temperature

3751 P-00 (intro) 111301 1217 PM Page 5

6 INTRODUCTION

TABLE 3 Factors for Conversion of Units

To convert from To convert fromUS Units to SI SI Units to US

Units Multiply by US Unit SI Unit Units Multiply by

254 in mm 00393703048 ft m 3281

6452 in2 mm2 1550 times 10-3

1639 times 103 in3 mm3 6102 times 10-6

4162 times 103 in4 mm4 2403 times 10-6

009290 ft2 m2 1076002832 ft3 m3 353104536 lb (mass) kg 22054448 lb (force) N 022484448 kip (force) kN 022481356 ft-lb (moment) N-m 073761356 kip-ft (moment) kN-m 07376

160185 lbft3 (density) kgm3 0062431459 lbft (load) Nm 0068531459 kipft (load) kNm 0068536895 psi (stress) kPa 014506895 ksi (stress) MPa 01450004788 psf (load or kPa 2093

pressure)4788 ksf (load or pressure) kPa 002093

0566 times (oF ndash 32) oF oC (18 times oC) + 32

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction This table is a sample from an extensive set oftables in the reference document

behaviors Nevertheless it makes a case for not being highly concernedwith any numbers beyond about the second digit

While most professional design work these days is likely to be donewith computer support most of the work illustrated here is quite simpleand was actually performed with a hand calculator (the eight-digit sci-entific type is adequate) Rounding off of these primitive computations isdone with no apologies

With the use of the computer accuracy of computational work is asomewhat different matter Still it is the designer (a person) who makesjudgements based on the computations and who knows how good the

3751 P-00 (intro) 111301 1217 PM Page 6

input to the computer was and what the real significance of the degree ofaccuracy of an answer is

SYMBOLS

The following shorthand symbols are frequently used

Symbol Reading

gt is greater thanlt is less thange is equal to or greater thanle is equal to or less than6 6 feet6 6 inchessum the sum of∆L change in L

NOMENCLATURE

Notation used in this book complies generally with that used in the build-ing design field A general attempt has been made to conform to usage inthe 1997 edition of the Uniform Building Code UBC for short (Ref 1)The following list includes all of the notation used in this book that isgeneral and is related to the topic of the book Specialized notation isused by various groups especially as related to individual materialswood steel masonry concrete and so on The reader is referred to basicreferences for notation in special fields Some of this notation is ex-plained in later parts of this book

Building codes including the UBC use special notation that is usuallycarefully defined by the code and the reader is referred to the source forinterpretation of these definitions When used in demonstrations of com-putations such notation is explained in the text of this book

Ag = gross (total) area of a section defined by the outer dimensions

An = net area

C = compressive force

NOMENCLATURE 7

3751 P-00 (intro) 111301 1217 PM Page 7

E = modulus of elasticity (general)

F = (1) force (2) a specified limit for stress

I = moment of inertia

L = length (usually of a span)

M = bending moment

P = concentrated load

S = section modulus

T = tension force

W = (1) total gravity load (2) weight or dead load of an object (3) total wind load force (4) total of a uniformly distributedload or pressure due to gravity

a = unit area

e = (1) total dimensional change of length of an object caused bystress or thermal change (2) eccentricity of a nonaxial load frompoint of application of the load to the centroid of the section

f = computed direct stress

h = effective height (usually meaning unbraced height) of a wall orcolumn

l = length usually of a span

s = spacing center to center

v = computed shear stress

8 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 8

9

1STRUCTURES PURPOSE

AND FUNCTION

This book deals with the behavior of structures in particular with struc-tures for buildings The behavior referred to is that which occurs whenthe structures respond to various force actions produced by natural andusage-generated effects Investigation of structural behaviors has the di-rect purpose of supporting an informed design of the structures and an as-surance as to the safety of the construction with regard to the buildingoccupants

Structural behaviors may be simple or complex This quality may de-rive from the nature of the loads on the structuremdashfrom simple gravity tothe dynamic effects of earthquakes It may also derive from the nature ofthe structure itself For example the simple structure shown in Figure 11has basic elements that yield to quite elementary investigation for be-havior This book provides a starting point for the most elementary in-vestigations of structures It can be the beginning of a long course ofstudy for persons interested in the investigation and design of highlycomplex structures

3751 P-01 111301 1217 PM Page 9

10 STRUCTURES PURPOSE AND FUNCTION

Figure 11 An All-American classic structure the light wood frame achieved al-most entirely with ldquo2 timesrdquo dimension lumber Wall studs serve as columns to supporthorizontal members in the time-honored post and beam system with its roots in an-tiquity While systems of much greater sophistication have been developed this isstill the single most widely used structure in the United States today

3751 P-01 111301 1217 PM Page 10

Consider the problems of the structure that derive from its perfor-mance of various load resisting functions The basic issues to be dealtwith are

The load sources and their effects

What the structure accomplishes in terms of its performance as a sup-porting spanning or bracing element

What happens to the structure internally as it performs its varioustasks

What is involved in determining the necessary structural elements andsystems for specific structural tasks

We begin this study with a consideration of the loads that affect build-ing structures

11 LOADS

Used in its general sense the term load refers to any effect that results ina need for some resistive response on the part of the structure There aremany different sources for loads and many ways in which they can beclassified The principal kinds and sources of loads on building structuresare the following

Gravity

Source The weight of the structure and of other parts of the con-struction the weight of building occupants and contents theweight of snow ice or water on the roof

Computation By determination of the volume density and type ofdispersion of items

Application Vertically downward and constant in magnitude

Wind

Source Moving air

Computation From anticipated wind velocities established by localweather history

LOADS 11

3751 P-01 111301 1217 PM Page 11

Application As pressure perpendicular to exterior surfaces or asshearing drag parallel to exterior surfaces Primarily considered asa horizontal force from any compass point but also with a verticalcomponent on sloping surfaces and vertical uplift on flat roofs

Earthquake (Seismic Shock)

Source Vibration of the ground as a result of a subterranean shock

Computation By prediction of the probability of occurrence basedon local history of seismic activity

Application Back-and-forth up-and-down movement of the groundon which a building sits resulting in forces induced by the inertialeffect of the buildingrsquos weight

Blast

Source Explosion of bomb projectile or volatile materials

Computation As pressure depending on the magnitude of the ex-plosion and its proximity to the structure

Application Slamming force on surfaces surrounding the explosion

Hydraulic Pressure

Source Principally from groundwater levels above the bottom of thebasement floor

Computation As fluid pressure proportional to the depth below thewater top surface

Application As horizontal pressure on basement walls and upwardpressure on basement floors

Thermal Change

Source Temperature changes in the building materials caused byfluctuations of outdoor temperature

Computation From weather histories coefficient of expansion ofmaterials and amount of exposure of the individual parts of theconstruction


3751 P-01 111301 1217 PM Page 12

Application Forces exerted when parts are restrained from expand-ing or contracting distortions of building if connected parts differ in temperature or have significantly different coefficients ofexpansion

Shrinkage

Natural volume reduction occurs in concrete in the mortar joints of ma-sonry in green wood and in wet clay soils These can induce forces in amanner similar to thermal change

Vibration

In addition to earthquake effects vibration of the structure may be causedby heavy machinery moving vehicles or high intensity sounds Thesemay not be a critical force issue but can be a major concern for sensationby occupants

Internal Actions

Forces may be generated within a structure by settlement of supportsslippage or loosening of connections or by shape changes due to sagwarping shrinkage and so on

Handling

Forces may be exerted on elements of the structure during productiontransportation erection storage and so on These may not be evidentwhen considering only the normal use of the building but must be con-sidered for the life of the structure

12 SPECIAL CONSIDERATIONS FOR LOADS

In addition to identifying load sources it is necessary to classify loads invarious ways The following are some such classifications

SPECIAL CONSIDERATIONS FOR LOADS 13

3751 P-01 111301 1217 PM Page 13

Live and Dead Loads

For design a distinction is made between so-called live and dead loadsA dead load is essentially a permanent load such as the weight of thestructure itself and the weight of other permanent elements of the build-ing construction supported by the structure A live load is technicallyanything that is not permanently applied as a force on the structure How-ever the specific term ldquolive loadrdquo is typically used in building codes torefer to the assumed design loads in the form of dispersed load on theroof and floor surfaces that derive from the building location and itsusage

Static versus Dynamic Forces

This distinction has to do essentially with the time-dependent characterof the force Thus the weight of the structure produces a static effect un-less the structure is suddenly moved or stopped from moving at whichtime a dynamic effect occurs due to the inertia or momentum of the massof the structure (see Figure 12a) The more sudden the stop or start thegreater the dynamic effect

Other dynamic effects are caused by ocean waves earthquakes blastssonic booms vibration of heavy machinery and the bouncing effect ofpeople walking or of moving vehicles Dynamic effects are different innature from static effects A light steel-framed building for instancemay be very strong in resisting static forces but a dynamic force maycause large distortions or vibrations resulting in cracking of plasterbreaking of window glass loosening of structural connections and so onA heavy masonry structure although possibly not as strong as the steelframe for static load has considerable stiffness and dead weight andmay thus absorb the energy of the dynamic force without perceptiblemovement

In the example just cited the effect of the force on the function of thestructure was described This may be distinct from any potential damag-ing effect on the structure The steel frame is flexible and may respondwith a degree of movement that is objectionable However from a struc-tural point of view it is probably more resistive to dynamic force than themasonry structure Steel is strong in tension and tends to dissipate someof the dynamic force through movement similar to a boxer rolling with


3751 P-01 111301 1217 PM Page 14

a punch Masonry in contrast is brittle and stiff and absorbs the energyalmost entirely in the form of shock to the material

In evaluating dynamic force effects and the response of structures tothem both the effect on the structure and the effect on its performancemust be considered (see Figure 12b) Success for the structure must bemeasured in both ways


Figure 12 (a) Static versus dynamicforce effects (b) Effects of vibration on occupantrsquos sense of the buildingrsquossolidity

3751 P-01 111301 1217 PM Page 15

Load Dispersion

Forces are also distinguished by the manner of their dispersion Gasunder pressure in a container exerts a pressure that is uniformly dispersedin all directions at all points The dead load of roofing the weight ofsnow on a roof and the weight of water on the bottom of a tank are allloads that are uniformly distributed on a surface The weight of a beamor a suspended cable is a load that is uniformly distributed in a linearmanner On the other hand the foot of a column or the end of a beam represent loads that are concentrated at a relatively small location (seeFigure 13)

Randomly dispersed live loads may result in unbalanced conditions orin reversals of internal forces in the structure (see Figure 14) Since liveloads are generally variable in occurrence it may be necessary to con-sider various arrangements and combinations of them in order to deter-mine the worst effects on the structure


Figure 13 Dispersion of loads

3751 P-01 111301 1217 PM Page 16

Wind

Wind is moving air and thus it has an impact on any static object in itspath just as water flowing in a stream has an impact on a large rock or abridge pier The fluid flow of the air also produces various other effectssuch as those shown in Figure 15 The form surface texture and size ofthe building as well as the sheltering effect of ground forms large treesor other nearby buildings may modify the effects of wind

While gravity is a constant magnitude single direction force wind isvariable in both magnitude and direction Although usually directed par-allel to the ground surface wind can cause aerodynamic effects in otherorientations resulting in both inward and outward pressures on individ-ual surfaces of a building Violent winds are usually accompanied bygusts which are brief surges in the wind velocity Gusts produce impactson surfaces and may result in jerking or rocking of small buildings

Wind magnitude is measured in terms of velocity (wind speed) Theeffect on buildings is translated into force in terms of pressures on the ex-terior building surfaces measured in pounds per square foot (psf) Fromphysics this pressure varies with the square of the velocity For the case


Figure 14 Unbalanced loads

3751 P-01 111301 1217 PM Page 17

of small to medium size buildings with flat sides sitting on the groundan approximation of the total force from these pressures is visualized inthe form of a single pressure on the building windward side of

p = 0003V 2

in which

p = pressure on the vertical surface in units of psf

V = wind velocity in units of miles per hour (mph)

A plot of this equation is shown in Figure 16 Local weather histories areused to establish the maximum anticipated wind speeds for a given loca-


Figure 15 Wind loads on buildings

3751 P-01 111301 1217 PM Page 18

tion which are then used to establish the code-required design pressuresused for design of structures in that region

Earthquakes

Earthquakes can have various disastrous effects on buildings The pri-mary direct effect is the shaking of the ground produced by the shockwaves that emanate from the center of the earthquake The rapidity du-ration and magnitude of this shaking depend on the intensity of theearthquake on the geological nature of the earth between the earth-quake and the building site and on the dynamic response character of thesite itself


Figure 16 Relation of wind velocity (speed) to surface pressure on buildings Re-produced from Simplified Building Design for Wind and Earthquake Forces 3rdedition by J Ambrose and D Vergun 1995 with permission of the publisherJohn Wiley amp Sons New York

3751 P-01 111301 1217 PM Page 19

The shaking effect of an earthquake may be a source of serious dis-tress to the building or its occupants The force effect on the structure isdirectly related to the weight of the building and is modified by variousdynamic properties of the structure As the base of a building is suddenlymoved the upper part of the building at first resists moving This resultsin a distortion of the structure with the base laterally displaced while theupper part momentarily remains stationary Then as the upper part fi-nally moves the base suddenly reverses direction which produces aforce due to the momentum of the upper part This action can producesliding toppling or total collapse of the building Repeated severaldozen times during an earthquake it can also produce progressive failureof the structure and a fun ride for the building occupants

If a structure is large tall and flexible its relatively slow response canset up whiplashlike effects as shown in Figure 17 If a structure is smallshort and stiff its motion will be essentially the same as that of theground In addition to the direct shaking action there are other potentialdestructive effects from earthquakes including

Settling cracking or lateral shifting of the ground surface

Landslides avalanches rock falls or glacial faults

Tidal waves that can travel long distances and cause damage to coastalareas

Surging of water in lakes reservoirs and large water tanks

Explosions and fires resulting from broken gas or oil pipelines

Major interruption of community services for power water supply orcommunication due to damage to buried utilities to transmissiontowers to electrical transformers and so on

The potential for disaster is enormous but the reality is tempered by theinfrequent occurrence of major earthquakes their highly localized na-ture and our steady development of more resistive structures Sadly butbeneficially each major earthquake works to reduce the inventory ofvulnerable structures for the next earthquake

Load Combinations

A difficult judgement for the designer is that of the likelihood of simul-taneous occurrence of forces from various sources Potential combina-


3751 P-01 111301 1217 PM Page 20

tions must be studied carefully to determine those that cause critical sit-uations and that have some reasonable possibility of actual simultaneousoccurrence For example it is not reasonable to design for the simul-taneous occurrence of a major wind storm and a major earthquake Noris it possible for the wind to blow simultaneously from more than onedirection

13 GENERATION OF STRUCTURES

The making of buildings involves a number of situations that generate aneed for structures

GENERATION OF STRUCTURES 21

Figure 17 Earthquake effects on tall structures Reproduced from SimplifiedBuilding Design for Wind and Earthquake Forces 3rd edition by J Ambrose andD Vergun 1995 with permission of the publisher John Wiley amp Sons New York

3751 P-01 111301 1217 PM Page 21

Need for Unobstructed Interior Space

Housing of activities creates the need for producing unobstructed interiorspaces that are free of vertical elements of the building structure Thesespaces may be very small (closets and bathrooms) or very large (sportsarenas) Generating open enclosed interior space involves the basicstructural task of spanning as shown in Figure 18 The magnitude of thespanning task is determined by the length of the span and the loads on thespanning structure As the span increases the required structural effortincreases rapidly and feasible options for the spanning structure narrowto a few choices


Figure 18 The structural task of generating unobstructed interior space

3751 P-01 111301 1218 PM Page 22

Architectural Elements

Most buildings consist of combinations of three basic elements wallsfloors and roofs These elements are arranged to create both space divi-sion and clear-spanned unobstructed interior spaces

Walls Walls are usually vertical and potentially lend themselves tothe task of supporting roofs and floors Even when they do not serve assupports they often incorporate the columns that do serve this purposeThus the design development of spanning roof and floor systems beginswith the planning of the wall systems over which they span Walls maybe classified on the basis of their architectural functions and their struc-tural tasks and this classification affects judgements about their form in terms of thickness and of stiffness in their own planes as shown inFigure 19

Floors Floor structures are often dual in function providing for afloor surface above and a ceiling surface below The floor function usu-ally dictates the need for a flat horizontal geometry thus most floorstructures are of the flat-spanning category (not arches catenary cablesetc) Most floor structures are relatively short in span owing to the highloadings and the inefficiency of the flat-spanning structure

Roofs Roofs have two primary functions to act as skin elements forthe building and to drain away water from rain and melting snowWhereas floors must usually be flat roofs must usually not be as somesloped form is required for water drainage Thus even so-called flat roofshave some minimum slope for draining the roof surface to designatedcollector elements (gutters downspouts gargoyles etc) Floors alsoneed some rigidity for a solid feeling when walked on Because of theirfreedom from requirements for horizontal flatness and solidity roofshave a great range of possibilities for geometry and nonflat structurethus most really long spans and exotic structural geometries are achievedwith roof structures


3751 P-01 111301 1218 PM Page 23

14 REACTIONS

Successful functioning of the structure in resisting loads involves twofundamental considerations First the structure must have sufficient in-ternal strength and stiffness to redirect the loads to its supports withoutdeveloping undue stress on its materials or an undesirable amount of de-formation (sag etc) Second the supports for the structure must keep the


Figure 19 Structural functions of walls

3751 P-01 111301 1218 PM Page 24

structure from collapsing The required forces developed by the supportsare called reactions

Figure 110 shows a column supporting a load that generates a linearcompressive effect The reaction generated by the columnrsquos support mustbe equal in magnitude and opposite in sense (up versus down) to the col-umn load The balancing of the active force (column load) and reactiveforce (support reaction) produces the necessary state of static equilib-rium thus no movement occurs

Figure 111 shows the reaction forces required for various structuresThe simple spanning beam requires only two vertical forces for supportHowever the gable frame arch and draped cable also require horizontalrestraint at their supports Structural behavior of the elements is differentin each of the four types of spanning structures shown in Figure 111 asis the required effort by the supports These differences are due to the dif-fering forms of the structures even though all four basically perform thesame spanning task

There is another type of reaction effort that can be visualized by con-sidering the situation of the cantilever beam as shown in Figure 112Since there is no support at the free end of the beam the support at theother end must develop a resistance to rotation of the beam end as wellas resistance to the vertical load The rotational effect is called momentand it has a unit that is different from that of direct force Force is measured

REACTIONS 25

Figure 110 Applied and reactive forces on a column

3751 P-01 111301 1218 PM Page 25


Figure 111 Reactions R for various spanning structures

Figure 112 Reactions for a cantilever beam

3751 P-01 111301 1218 PM Page 26

in weight units pounds tons and so on Moment is a product of forceand distance resulting in a compound unit of pound-feet or some othercombination of force and length units The total support reaction for thecantilever therefore consists of a combination of the vertical force (Rv)and the resisting moment (Rm)

For the rigid frame shown in Figure 113 there are three possiblecomponents of the reactions If vertical force alone is resisted at the sup-ports the bottoms of the columns will move outward and rotate as

REACTIONS 27

Figure 113 Reactions for a rigid frame

3751 P-01 111301 1218 PM Page 27

shown in Figure 113a If horizontal resistance is developed as shownfor the gable arch and cable in Figure 111 the column bottoms can bepushed back to their unloaded positions but they will still rotate asshown in Figure 113b Finally if a moment resistance is developed bythe supports the column bottoms can be held completely in their originalpositions as shown in Figure 113c

The combination of loads and support reactions constitutes the totalexternal effort on a structure This system is in some ways independentof the structure that is the external forces must be in equilibrium re-gardless of the materials strength and so on of the structure For exam-ple the task for a beam can be totally defined in terms of effort withoutreference to what the beam actually consists of

With its tasks defined however it becomes necessary to consider theresponse developed by the structure This means moving on to considerwhat happens inside the structure in terms of internal force effects

15 INTERNAL FORCES

In response to the external effects of loads and reactions internal forcesare developed within a structure as the material of the structure strives toresist the deformations caused by the external effects These internalforce effects are generated by stresses in the material of the structure Thestresses are actually incremental forces within the material and they re-sult in incremental deformations called strains

Cause and Effect External versus Internal Force

When subjected to external forces a structure twists sags stretchesshortens and so on To be more technical it stresses and strains thus as-suming some new shape as the incremental strains accumulate into over-all dimensional changes While stresses are not visually apparent theiraccompanying strains are thus it is possible to infer a stress conditionfrom observation of structural deformations

As shown in Figure 114 a person standing on a wooden plank thatspans between two supports will cause the plank to sag downward andassume a curved profile The sag may be visualized as the manifestationof a strain phenomenon accompanied by a stress phenomenon In this ex-


3751 P-01 111301 1218 PM Page 28

ample the principal cause of the structurersquos deformation is bending re-sistance called internal resistive bending moment

The stresses associated with the internal force action of bending mo-ment are horizontally directed compression in the upper portion of theplank and horizontally directed tension in the lower portion Anyonecould have predicted that the plank would sag when the person steppedon it But we can also predict the deformation as an accumulation ofstrains resulting in the shortening of the upper portion and the lengthen-ing of the lower portion of the plank Thus the stress condition can be in-ferred from observed deformation but likewise the deformation can bepredicted from known stress conditions

For the relatively thin wooden plank the bending action and strain ef-fects are quite apparent If the plank is replaced by a thick wooden beamthe sag will not be visually apparent with a light load and a short spanHowever the internal bending still occurs and the sagmdashhowever slightmdashdoes exist For the investigation of structural behaviors visualization ofinternal forces is aided by considering an exaggerated deformation of thestructure assuming it to be much more flexible than it really is

INTERNAL FORCES 29

Figure 114 Internal bending

3751 P-01 111301 1218 PM Page 29

16 FUNCTIONAL REQUIREMENTS OF STRUCTURES

Any structure subjected to loads must have certain characteristics inorder to function For purposes of structural resistance it must be inher-ently stable must have adequate strength for an acceptable margin ofsafety and must have a reasonable stiffness for resistance to deformationThese three basic characteristicsmdashstability strength and stiffnessmdasharethe principal functional requirements of structures

Stability

Stability has both simple and complex connotations In the case of thewooden plank it is essential that there be two supports and that the per-son stand between the supports As shown in Figure 115 if the plank ex-tends over one support and a person stands on the extended end disasterwill certainly occur unless a counterweight is placed on the plank or theplank is anchored to the opposite support In this case either the coun-terweight or the anchorage is necessary for the stability of the structuremdashunrelated to the strength or stiffness of the plank

A slightly different problem of stability is illustrated by another ex-ample Suppose you have a sore foot and want to use a walking stick toassist your travel You are offered a 3fraslfrasl4-in round wooden stick and a 1fraslfrasl4-in round steel rod each 3 ft long After handling both you would prob-ably choose the wooden stick since the steel rod would buckle underyour weight This buckling action can be visualized demonstrated andmeasured The essential property of a structure that determines its buck-ling potential is its slenderness

In engineering analysis the geometric property of slenderness used toestablish the likelihood of buckling is the slenderness ratio also calledthe relative slenderness expressed as

Lr

in which

L = length of the compression member over which there is nolateral bracing to prevent buckling

r = a geometric property of the member cross section called theradius of gyration


3751 P-01 111301 1218 PM Page 30

The geometric property r can be expressed as

In this formula

A = the member cross-sectional area

I = a property called the second moment of the area or themoment of inertia

rI

A=

1 2

FUNCTIONAL REQUIREMENTS OF STRUCTURES 31

Figure 115 Developing stability

3751 P-01 111301 1218 PM Page 31

While A is a direct measure of the amount of material in the member I isa measure of the memberrsquos stiffness in resisting bendingmdashwhich is whatbuckling becomes once it is initiated

In the example of the walking stick the 3fraslfrasl4-in diameter wooden stickhas an L r of 192 while the 1fraslfrasl4-in steel rod has an L r of 576 If we takethe steel and flatten it out and roll it up to produce a cylinder with a 3fraslfrasl4 indiameter the area remains the same but the I value is significantly in-creased Furthermore the r value is thus also increased so that the L rnow becomes 136 As long as the cylinder wall is not made too thin thepipe-shaped stick represents a major improvement in buckling resistanceFigure 116 shows the three cross sections and the corresponding L rvalues

Bending and buckling stiffness are also affected by the stiffness of thematerial Thus a 1frasl4 in rod of wood would be even less stiff than the oneof steel since wood is considerably less stiff than steel For a single veryslender compression member the compression force required to producebuckling is expressed by the Euler formula shown in the plot of com-pression failure versus length in Figure 117 As the member is short-ened buckling becomes less critical and the limiting effect becomessimple compressive crushing of the material At very short lengths there-fore the compression limit is determined by the stress resistance of the


Figure 116 Relative Lr values

3751 P-01 111301 1218 PM Page 32

material At the other end of the graph the curve becomes that of theEuler formula in which the index of the member resistance is stiffnessmdashof both the member cross section (I ) and the material (E which is thestiffness modulus of the material) Between the limits the curve slowlychanges from one form to the other and the buckling phenomenon con-tains some aspect of both types of failure

Stability can be a problem for a single structural member such as asingle column or it can be a problem for a whole structural assemblageThe eight-element framework shown in Figure 118 may be stable in re-sisting vertical gravity loads but it must be braced in some manneragainst any horizontal forces such as those caused by wind or earth-quakes The illustrations in Figure 118 show the three principal means


Figure 117 Compression load limit versus member slenderness E is a factorthat indicates the stiffness of the material

3751 P-01 111301 1218 PM Page 33

for achieving this stability by using rigid joints between members byusing truss bracing in the wall planes or by using rigid panels in the wallplanes called infilling

Strength

Strength is probably the most obvious requirement for a structure Eventhough it is stable the plank in Figure 114 is not strong enough to hold theweight of ten people This has to do partly with the materialmdashif the plankwere made of steel it might do the job It also has to do with the form andorientation of the plank cross sectionmdashif the wood plank were turned on itsedge like a floor joist it would probably also support ten people

Material strength often depends on the type of stress that the materialmust sustain Steel is adaptable and capable of major resistance to tensioncompression shearing twisting and bending with equal dexterity Woodhowever has different strengths depending on the direction of the stresswith reference to the wood grain As shown in Figure 119 the develop-


Figure 118 Means of stabilizing a frame structure

3751 P-01 111301 1218 PM Page 34

ment of major stresses perpendicular to the wood grain direction cancause the wood to fail easily Reforming the wood either by glue lamina-tion or by pulverising the wood and using the wood fiber to produce com-pressed fiber panels is a way of overcoming the grain limitation

Stone concrete and fired clay are examples of materials that havevarying strengths for different stresses All are relatively strong in resist-ing compression but are much less strong in resisting tension or shearThis requires caution in their use in structures to avoid these stresses orto compensate for themmdashsuch as by using steel reinforcement in con-crete structures

Attention must be given both to the form and nature of elements andto their uses A cable assembled from thin steel wires has little resistanceto compression or bending or to anything but the single task for which itis formedmdashresisting tension This is so despite the fact that the steel asa material has other stress potentials

A stack of bricks with no bonding in the joints has the capability of sup-porting a compressive load applied directly downward on the top of thestack Picking the unbonded stack up by lifting the top brick or turning thestack sideways to create a spanning structure as shown in Figure 120 isobviously not possible Thus joint formation of elements in an assembledstructure is also a concern for strength


Figure 119 Effect of orientation to load


3751 P-01 111301 1218 PM Page 35

Stiffness

All structures change shape and move when subjected to forces (seeFigure 121) The relative magnitude of these changes determines a qual-ity of the structure called rigidity or stiffness The degree of stiffness de-pends on the material of the of the structure on the configuration of itsparts andmdashfor assemblagesmdashon the arrangement of the assembledmembers It may also depend on the connections between parts and onthe type of restraint offered by supports The presence or absence ofbracing may also be a factor

Although stiffness is usually not as critical to the safety of a structureas strength and stability it is frequently important for use of the structureIf a slammed door rocks the whole building or if floors bounce whenwalked on the users of the building will probably not be satisfied withthe structure

Equilibrium of Structures

Most structures act as transfer elements receiving certain forces andtransferring them to other points This transfer capability is dependent onthe internal strength and stability of the structure As shown in Figure122 a thin sheet of aluminum may be easily buckled a block of woodmay be easily split along its grain and a rectangular framework withloose single-pin joints may be easily collapsed sideways All of thesestructures fail because of an inability to maintain internal equilibriumthrough lack of strength or because of the lack of some inherent stabil-ity or for both reasons

The complete static equilibrium of a structure requires two separatebalances that of the external forces and that of the internal forces Ex-ternally sufficient reaction components must be developed by the sup-ports Internally there must be an inherent capability for stability and


Figure 121 Deformation of structures under load

3751 P-01 111301 1218 PM Page 36

sufficient strength to do the work of transferring the applied loads to the supports

As shown in Figure 123 there are three possible conditions for exter-nal stability If support conditions are insufficient in type or number thestructure is externally unstable If support conditions are just adequate thestructure is stable If the supports provide an excess of the necessary con-ditions the structure is probably stable but may be indeterminatemdashnotnecessarily a bad quality just a problem for achieving a simple investiga-tion of structural behavior

For internal stability the structure must be formed arranged and fas-tened together to develop the necessary resistance In the examplesshown in Figure 122 the aluminum sheet was too thin for its size thewood block had weak shear planes and the frame lacked the necessaryarrangement of members or type of joints All three could be altered tomake them more functional As shown in Figure 124 the aluminumsheet can be braced with stiffening ribs the solid-sawn wood block canbe replaced with a laminated piece with alternate plies having their grain


Figure 122 Lack of internal resistance

3751 P-01 111301 1218 PM Page 37


Figure 123 Stability analysis

Figure 124 Alteration of internal conditions to improve structural resistance

3751 P-01 111301 1218 PM Page 38

directions perpendicular to each other and the frame can be stabilized byadding a diagonal member

17 TYPES OF INTERNAL FORCE

Complex actions and effects consist of combinations of the followingbasic types of internal force The simplest types to visualize are tensionand compression both of which produce simple stress and strain condi-tions as shown in Figure 125

Tension

The ability to withstand tension requires certain materials stone con-crete sandy soil and wood perpendicular to its grain all have low resis-tance to tension Stresses can become critical at abrupt changes in thecross section of a member such as at a hole or a notch Tension mayserve to straighten members or to align connected members Connectionsfor transfer of tension are often more difficult to achieve than those forcompression requiring not simply contact (as with the stack of bricks)but some form of engagement or anchorage (see Figure 126)

Compression

Compression usually causes one of two types of failure crushing orbuckling As discussed previously buckling has to do with the relativestiffness of elements while crushing is essentially a simple stress resistance

TYPES OF INTERNAL FORCE 39

Figure 125 (a) Effects of tension (b) Effects of compression

(a)

(b)

3751 P-01 111301 1218 PM Page 39

by the material Actually however most building compression elementsfall between a very slender (pure buckling) form and a very squat (purecrushing) form and their behavior thus has some aspects of both formsof response (See Figure 117 and consider the middle portion of thegraph) Compression can be transferred between elements by simple con-tact as in the case of a footing resting on soil (see Figure 126) Howeverif the contact surface is not perpendicular to the compressive force aside-slip failure might occur Some form of engagement or restraint isthus usually desirable

Shear

In its simplest form shear is the tendency for slipping of adjacent objectsThis may occur at the joint between elements or within a material suchas a grain split in wood (see Figure 127) If two wooden boards in a floorare connected at their edges by a tongue-and-groove joint shear stress isdeveloped at the root of the tongue when one board is stepped on and theother is not This type of shear also develops in bolts and hinge pins

A more complex form of shear is that developed in beams This can bevisualized by considering the beam to consist of a stack of loose boardsThe horizontal slipping that would occur between the boards in such astructure is similar to the internal shear that occurs in a solid beam If theboards are glued together to form a solid beam the horizontal slipping ef-fectmdashbeam shearmdashis what must be resisted at the glue joints


Figure 126 Considerations of tension and compression actions

3751 P-01 111301 1218 PM Page 40

Bending

Tension compression and shear are all produced by some direct forceeffect Actions that cause rotation or curvature are of a different sort Ifthe action tends to cause straight elements to curve it is called bendingIf it tends to twist elements it is called torsion (see Figure 128) When a


Figure 127 Effects of shear

Figure 128 Effect of torsion

3751 P-01 111301 1218 PM Page 41

wrench is used to turn a bolt bending is developed in the handle of thewrench and torsion is developed in the bolt shaft

Bending can be produced in a number of ways A common situationoccurs when a flat spanning structure is subjected to loads that act per-pendicular to it This is the basic condition of an ordinary beam Asshown in Figure 129 the internal force acting in the beam is a combi-nation of bending and shear Both of these internal stress effects pro-duce lateral deformation of the straight unloaded beam called sag ordeflection

Bending involves a combination of force and distance most simplyvisualized in terms of a single force and an operating moment arm (seeFigure 130) It may also be developed by a pair of opposed forces suchas two hands on a steering wheel The latter effect is similar to how abeam develops an internal bending resistancemdashby the opposing of com-pressive stresses in the top part of the beam to tension stresses in thebottom part


Figure 129 Internal effects inbeams

3751 P-01 111301 1218 PM Page 42

Since the development of moment is a product of force times dis-tance a given magnitude of force can produce more moment if the mo-ment arm is increased The larger the diameter of a steering wheel theless force required to turn itmdashor with a given limited force the moremoment it can develop This is why a plank can resist more bending if itis turned on its edge as a joist Figure 131 shows the effect of formchange on a constant amount of material used for the cross section of abeam For each shape the numbers indicate the relative resistance tobending in terms of strength (as a stress limit) and stiffness (as a strainlimit producing deflection)

In addition to the bending created when flat spanning members aretransversely loaded there are other situations in buildings that can pro-duce bending effects Two of these are shown in Figure 132 In the upperfigures bending is produced by a compression load not in line with theaxis of the member or by a combination of compressive and lateral load-ing In the lower figure bending is transmitted to the columns throughthe rigid joints of the frame


Figure 130 Development of moments

3751 P-01 111301 1218 PM Page 43


Figure 131 Relation of cross-sectional geometry to bending resistance

Figure 132 Conditions resulting in internal bending

3751 P-01 111301 1218 PM Page 44

Torsion

Torsion is similar to bending in that it is a product of force and distanceAs with bending the form of the cross section of the member resisting thetorsion is a critical factor in establishing its strength and stiffness A roundhollow cylinder (pipe shape) is one of the most efficient forms for resis-tance to torsion However if the cylinder wall is slit lengthwise its resis-tance is drastically reduced being approximately the same as that for a flatplate made by flattening out the slit cylinder Figure 133 shows the effecton torsional resistance of variations in the cross-sectional shape of a lin-ear member with the same amount of material (area) in the cross section

Often in designing structures it is a wiser choice to develop resistanceto torsion by bracing members against the twisting effect Thus the tor-sion is absorbed by the bracing rather than by stresses in the member

Combinations of Internal Forces

The individual actions of tension compression shear bending and tor-sion can occur in various combinations and in several directions at a sin-gle point in a structure For example as illustrated previously beamsordinarily sustain a combination of bending and shear In the columns ofthe frame shown in the lower part of Figure 132 the loading on the beamwill produce a combination of compression bending and shear In the ex-ample shown in Figure 134 the loading will produce a combination of in-ternal compression shear torsion and bending in two directions

Structures must be analyzed carefully for the various internal forcecombinations that can occur and for the critical situations that may


Figure 133 Relation of cross-sectional geometry to torsional resistance

3751 P-01 111301 1218 PM Page 45

produce maximum stress conditions and maximum deformations In ad-dition the external loads often occur in different combinations with eachcombination producing different internal force effects This frequentlymakes the analysis of structural behaviors for design a quite laboriousprocess making us now very grateful for the ability to utilize computer-aided procedures in design work

18 STRESS AND STRAIN

Internal force actions are resisted by stresses in the material of the struc-ture There are three basic types of stress tension compression andshear Tension and compression are similar in nature although oppositein sign or sense Both tension and compression produce a linear type ofstrain (shape change) and can be visualized as pressure effects perpen-dicular to the surface of a stressed cross section as shown in Figure 135Because of these similarities both tension and compression are referredto as direct stresses one considered positive and the other negative

Shear stress occurs in the plane of a cross section and is similar to asliding friction effect As shown in Figure 136 strain due to shear stressis of a different form from that due to direct stress it consists of an an-gular change rather than a linear shortening or lengthening

Stress-Strain Relations

Stress and strain are related not only in the basic forms they take but intheir actual magnitudes Figure 137 shows the relation between stress and


Figure 134 Combined internal force effects

3751 P-01 111301 1218 PM Page 46

strain for a number of different materials The form of such a graph illus-trates various aspects of the nature of structural behavior of the materials

Curves 1 and 2 represent materials with a constant proportionality ofthe stress and strain magnitudes For these materials a quantified rela-tionship between stress and strain can be described simply in terms of theslope or angle of the straight line graph This relationship is commonlyexpressed as the tangent of the angle of the graph and is called themodulus of elasticity of the material The higher the value of this modu-lusmdashthat is the steeper the slope of the graphmdashthe stiffer the materialThus the material represented by curve 1 in the illustration is stiffer thanthe material represented by curve 2

STRESS AND STRAIN 47

Figure 135 Direct stress and strain

Figure 136 Shear stress and strain

3751 P-01 111301 1218 PM Page 47

For direct stress of tension or compression the strain is measured as alinear change and the modulus is called the direct stress modulus of elas-ticity For shear stress the strain is measured as an angular change andthe resulting modulus is called the shear modulus of elasticity

Some materials such as glass and very high-strength steel have aconstant modulus of elasticity for just about the full range of stress up tofailure of the material Other materials such as wood concrete and plas-tic have a curved form for the stress-strain graph (curve 3 in Figure137) The curved graph indicates that the value for the modulus of elas-ticity varies continuously for the full range of stress

The complex shape of curve 4 in Figure 137 is the characteristic formfor a so-called ductile material such as low-grade steel of the type ordi-narily used for beams and columns in buildings This material respondselastically at a low level of stress but suddenly deforms excessively at alevel of stress described as its yield point However fracture does notusually occur at this level of stress but rather at a higher level after thematerial reaches a certain limiting magnitude of yielding strain This pre-dictable yield phenomenon and the secondary reserve strength are usedto predict ultimate load capacities for steel frames as well as for concretestructures that are reinforced with ductile steel rods


Figure 137 Stress and strain relationships

3751 P-01 111301 1218 PM Page 48

Stress Combinations

Stress and strain are three-dimensional phenomena but for simplicitythey are often visualized in linear or planar form As shown in Figure135 direct stress of compression in a single direction results in strain ofshortening of the material in that direction However if the volume of thematerial remains essentially unchangedmdashwhich it usually doesmdashtherewill be a resulting effect of lengthening (or pushing out) at right anglesto the compression stress This implies the development of a tension ef-fect at right angles to the compression which in some materials may bethe real source of failure as is the case for tension-weak concrete andplaster Thus a common form of failure for concrete in compression is bylateral bursting at right angles to the compression load

If direct stress is developed in a linear member as shown in Figure138 the pure direct stress occurs only on sections at right angles to thedirect force loading called cross sections If stress is considered on a sec-tion at some other angle (called an oblique section) there will be a com-ponent of shear on the section If the material is weak in shear (such aswood parallel to its grain) this angular shear stress effect may be morecritical than the direct stress effect

Although simple linear tension and compression forces produce di-rect linear stresses shear stress is essentially two-dimensional as shownin Figure 139 The direct effect of a shear force is to produce shearstresses that are parallel to the force (on faces a and b in Figure 139a)These opposed stresses in the material produce a rotational effect whichmust be balanced by other opposed stresses (at faces c and d in Figure139b) Thus whenever shear stress exists within a structure there is al-ways an equal magnitude of shear stress at right angles to it An example


Figure 138 Stress on a cross section not at right angles to the active force

3751 P-01 111301 1218 PM Page 49

of this is the stack of loose boards used as a beam as shown in Figure127 The shear failure in this case is a horizontal slipping between theboards even though the shear force is induced by vertical loading

As shown in Figures 139c and d the combination of the mutually per-pendicular shear stresses produces a lengthening of the material on onediagonal and a shortening on the other diagonal This implies the devel-opment of tension on one diagonal and compression on the other diago-nal at right angles to the tension In some cases these diagonal stressesmay be more critical than the shear stresses that produce them In con-crete for example failure due to shear stress is usually actually a diago-nal tension stress failure as this is the weakest property of the material



3751 P-01 111301 1218 PM Page 50

On the other hand high shear in the web of a steel beam may result in di-agonal compression buckling of the thin web

Separately produced direct stresses in a single direction may besummed algebraically at a given point in a structure In the case of thecolumn shown in Figure 140 the compression load produces a directcompression stress on a cross section as shown at Figure 140a if theload is placed so as not to produce bending If the load is off-center on thecolumn the stress conditions will be modified by the addition of bendingstresses on the cross section as shown in Figure 140b The true netstress condition at any point on the cross section will thus be the simpleaddition of the two stress effects with a combined stress distributionpossible as shown in Figure 140c

A more complex situation is the combination of direct stresses and shearstresses Figure 141a shows the general condition at a point in the crosssection of a beam where the net stress consists of a combination of the di-rect stress due to bending (tension or compression) and shear stress Thesestresses cannot simply be added as they were for the column What can becombined are the direct stress due to bending and the direct diagonal stressdue to shear as shown in Figure 141b Actually because there are two di-agonal stress conditions there will be two combinationsmdashone producing amaximum effect and the other a minimum effect as shown in Figure 141cThese two stress limits will occur in mutually perpendicular directions

There is also a net combined shear stress as shown in Figure 141dThis is the combination of the direct shear stress and the diagonal shearstress due to the direct stress Since the direct shear stress is at right an-gles (vertically and horizontally) and the shear stress due to direct stressis on a 45deg plane the net maximum shear will be at some angle betweenthese two This angle will be closer to a right angle when the direct shearis larger and closer to a 45deg position when the direct stress is larger

Another stress combination is that produced by triaxial stress condi-tions An example of this is a confined material subjected to compressionsuch as air or liquid in a piston chamber as shown in Figure 142 In addi-tion to being compressed by the active compressing force (the piston) thematerial is squeezed laterally by the other material around it The net effecton the confined material is a three-way push or triaxial compression Formaterials with little or no tension resistance such as air water or dry sandthis is the only situation in which they can resist compression Thus asandy soil beneath a footing can develop resistance in the form of verticalsoil pressure because of the confinement of the soil around it and above it


3751 P-01 111301 1218 PM Page 51

For visualization purposes it is common to reduce complex structuralactions to their component effects These simpler individual effects canthus be analyzed more clearly and simply and the results combined withthe effects of the other components In the end however care must betaken to include all the components for a given situation


Figure 140 Combined direct stresses

3751 P-01 111301 1218 PM Page 52

Thermal Stress

The volumes of materials change with temperature variation increasingas temperatures rise and decreasing when they fall This phenomenoncreates a number of problems that must be dealt with in building design

The form of objects determines the basic nature of significant di-mensional changes As shown in Figure 143 the critical directions of


Figure 141 Combined shear stress and direct stress

3751 P-01 111301 1218 PM Page 53


Figure 142 Development of stress in a confined material

Figure 143 Effects of thermal change on solid objects

3751 P-01 111301 1218 PM Page 54

movement depend on whether the object is essentially linear planar(two-dimensional) or three-dimensional For a linear object (beam col-umn etc) the significant change is in its length significant concerns arethose for very long objects especially in climates with a considerabletemperature range

Planar objects such as wall panels and large sheets of glass expand ina two-dimensional manner Attachments and constraints by other con-struction must allow for thermal movements Three-dimensional move-ments are mostly dealt with by providing for component movements ofa linear or two-dimensional nature

If thermal expansion or contraction is resisted stresses are producedFigure 144 shows a linear structural member in which length change is


Figure 144 Effect of thermal change on a constrained element

3751 P-01 111301 1218 PM Page 55

constrained If the temperature is raised the member will push outwardagainst the restraints developing internal compression as the constraintspush back This results in an external compression force on the memberin the same manner as a load applied to a column With quantified val-ues known for the thermal expansion coefficient and the stress-strain re-lationship for the material the compressive stress developed in themember can be determined

Another type of thermal problem is that involving differential move-ment of attached parts of the construction Figure 145 shows a commonsituation in which a cast concrete structure consists of elements of dif-ferent mass or thickness If exposed to temperature change the thinnerparts will cool down or warm up more quickly than the thicker parts towhich they are attached by the continuous casting process The result isthat the thinner parts are restrained in their movements by the thickerparts which induces stresses in all the parts These stresses are most crit-ical for the thinner parts and at the joints between the parts

Another problem of differential thermal movements occurs betweenthe exterior surface and the interior mass of a building As shown inFigure 146 the exposed skinmdashas well as any exposed structural mem-bersmdashwill tend to move in response to the changes in outdoor tempera-tures while the interior elements of the construction tend to remain at arelatively constant comfort-level temperature For a multistory build-ing this effect accumulates toward the top of the building and can resultin considerable distortions in the upper levels of the structure

A similar problem occurs with long buildings in which the part aboveground is exposed to the weather while that buried in the ground remainsat a relatively constant temperature throughout the year (see Figure 147)


Figure 145 Critical stress effects resulting from differential thermal movements

3751 P-01 111301 1218 PM Page 56


Figure 146 Effect of exposure conditions of the structure on development ofthermally induced stress and strain (a) Conditions resulting in major exposure ofthe exterior wall structure but enclosure of the interior structure (b) In the winter(outside at 0degF interior at 70degF differential of 70degF) the exterior columns becomeshorter than the interior resulting in the deformations shown (c) In the summer(outside at 100degF inside at 75degF differential of 25degF) the exterior columns becomelonger than the interior resulting in the deformations shown

Figure 147 Thermal effects in partly underground buildings

3751 P-01 111301 1218 PM Page 57

The simple solution here is to provide construction joints periodically inthe building length that literally create separated masses of the buildingeach of a controlled shorter length

Composite Structures

When structural elements of different stiffness share a load they developresistance in proportion to their individual stiffnesses As shown in Fig-ure 148a if a group of springs share a load that shortens all of thesprings the same amount the portion of the load resisted by the stiffersprings will be greater since it takes a greater effort to shorten them

Another common type of composite structure occurs when concrete isreinforced with steel rods as shown in Figure 148b When a load is ap-plied to such an element (called a composite structure) the stiffer mate-rial (steel in this case) will carry a higher portion of the load In this


Figure 148 Load sharing in compositestructures (a) A group of springs of varyingstiffness (b) Steel-reinforced concrete

3751 P-01 111301 1218 PM Page 58

manner a relatively small percentage of steel in a reinforced concretemember can be made to carry a major part of the load since steel has onaverage around 10 times the stiffness of structural grade concrete

A situation somewhat similar to this occurs when the building as awhole is distorted by loads such as the horizontal effects of wind andearthquakes Figure 149 shows two examples of this the first being abuilding with solid walls of masonry and wood frame construction in thesame exterior surface As a bracing wall for horizontal loads the muchstiffer masonry will tend to take most of the load In this case the woodframed wall may be virtually ignored for its structural resistance al-though any effects of the lateral distortion must be considered

The second example in Figure 149 involves a steel frame in the sameplane as relatively stiff walls Even though the framed walls may be lessstrong than the steel frame they will likely be much stiffer thus theywill tend to absorb a major portion of the lateral load The solution in thiscase is to either make the walls strong enough for the bracing work or tomake the steel frame stiff enough to protect the walls and actually do thebracing work

Time-Related Stress and Strain

Some stress and strain phenomena are time related Concrete is subject toan effect called creep (see Figure 150) in which the material sustains aprogressive deformation when held at a constant stress over a long timeThese deformations are added to those produced normally by the initial


Figure 149 Load sharing by elements of different construction

3751 P-01 111301 1218 PM Page 59

loading Additionally unlike the initial deformations they remain per-manent similar to the long-term sag of wood beams

Creep does not affect the stress resistance of concrete but does resultin some redistribution of stresses between the concrete and its steel rein-forcing Since the steel does not creep it effectively becomes increas-ingly stiffer in relation to the progressively softening concrete Thismakes the steel even greater in its capability of carrying a major part ofthe load in the composite structure

Soft wet clay soils are subject to a time-related flow effect similar tothe slow oozing of toothpaste from a tube as it is squeezed If the soilmass is well constrained (similar to putting the cap back on the toothpastetube) this effect can be arrested However as long as there is some-where for the clay to ooze toward and the pressure on it is maintainedthe flow will continue Instances of buildings that continue to settle overmany years have occurred with this soil condition (see Figure 151)

Another time-related stress problem occurs when structures are re-peatedly loaded and unloaded The effect of people walking of windand earthquakes and of machinery rocking on its supports are cases ofthis loading condition in buildings Some materials may fail from the fa-tigue effects of such loadings However a more common problem is thatof loosening of connections or the progressive development of cracksthat were initially created by other effects


Figure 150 Effect of creep

3751 P-01 111301 1218 PM Page 60

19 DYNAMIC EFFECTS

Vibrations moving loads and sudden changes in the state of motionsuch as the jolt of rapid braking or acceleration cause forces that resultin stresses and strains in structures The study of dynamic forces and theireffects is very complex although a few of the basic concepts can be il-lustrated simply

For structural investigation and design a significant distinction be-tween static and dynamic effects has to do with the response of the struc-ture to the loading If the principal response of the structure can beeffectively evaluated in static terms (force stress linear deformationetc) the effect on the structure is essentially static even though the loadmay be time-dependent in nature If however the structurersquos responsecan be effectively evaluated only in terms of energy capacity work doneor cyclic movement the effect of the load is truly dynamic in character

A critical factor in the evaluation of dynamic response is the funda-mental period of the structure This is the time required for one full cycleof motion in the form of a bounce or a continuing vibration The relationof this time to the time of buildup of the load is a major factor in deter-mining that a structure experiences a true dynamic response The time of

DYNAMIC EFFECTS 61

Figure 151 Time-related settlement

3751 P-01 111301 1218 PM Page 61

the period of a structure may vary from a small fraction of a second toseveral seconds depending on the structurersquos size mass (weight) andstiffness as well as on support constraints and the presence of dampingeffects

In the example in Figure 152 a single blow from the hammer causesthe board to bounce in a vibratory manner described by the time-motiongraph The elapsed time for one full cycle of this motion is the funda-mental period of the board If a 100-lb load is applied to the end of theboard by slowly stacking bricks on it the load effect on the board is sta-tic However if a 100-lb boy jumps on the end of the board he willcause both an increase in deflection and a continued bouncing of theboard both of which are dynamic effects If the boy bounces on the endof the board with a particular rhythm he can cause an extreme up anddown motion of the board He can easily find the rate of bouncing


Figure 152 Dynamic effects on elastic structures

3751 P-01 111301 1218 PM Page 62

required to do this by experimenting with different rhythms He mayalso find the exact variation in his bouncing that will result in an almostcomplete instantaneous stop of the boardrsquos motion As shown in thegraph in Figure 153 the reinforcing bouncing that generates increasingmotion of the board corresponds to the fundamental period of the boardTo stop the board the boy merely cuts the time of his bounce in half thusmeeting the board on its way up

If the boy bounces on the board once and then jumps off the boardwill continue to bounce in ever-decreasing magnitudes of displacementuntil it finally comes to rest The cause of this deterioration of the boardrsquosmotion is called damping It occurs because of energy dissipated in theboardrsquos spring mounting and in air friction as well as because of anygeneral inefficiencies in the movement of the board If no damping werepresent the boyrsquos sympathetic bouncing could eventually cause damageto the board

Dynamic forces on structures result from a variety of sources and cancreate problems in terms of the total energy delivered to the structure orin the form of the movements of the structure Excessive energy loadingcan cause structural damage or total collapse Movements may result inloosening of connections toppling of vertical elements or simply in highlyundesirable experiences for building occupants

Design for dynamic response usually begins with an evaluation of po-tential dynamic load sources and their ability to generate true dynamic ef-fects on the structure Once the full nature of the dynamic behavior isunderstood measures can be taken to manipulate the structurersquos dynamiccharacter or to find ways to reduce the actual effects of the dynamicloading itself Thus it may be possible to brace a structure more securelyagainst movements due to an earthquake but it may also be possible to

DYNAMIC EFFECTS 63

Figure 153 Motion of the diving board

3751 P-01 111301 1218 PM Page 63

dissipate some of the actual movement by placing a motion-absorbingseparator between the building and the ground

110 DESIGN FOR STRUCTURAL RESPONSE

In the practice of structural design the investigation of structural re-sponse to loads is an important part of the design process To incorporatethis investigation into the design work the designer needs to develop anumber of capabilities including the following

1 The ability to visualize and evaluate the sources that produceloads on structures

2 The ability to quantify the loads and the effects they have onstructures

3 The ability to analyze a structurersquos response to the loads in termsof internal forces and stresses and strains

4 The ability to evaluate the structurersquos safe limits for load-carryingcapacity

5 The ability to manipulate the variables of material form dimen-sions and construction details for the structure in order to maxi-mize its structural response

For any structure it is necessary to perform some computations inorder to demonstrate the existence of an adequate margin of safety for agiven loading However the complete design of a structure must also in-corporate many other considerations in addition to structural perfor-mance A successful structure must be structurally adequate but it mustalso be economical feasible for construction and must generally facili-tate the overall task it serves as part of the building construction It mustalso be fire-resistant time-enduring maybe weather-resistant and what-ever else it takes to be a working part of the building throughout the lifeof the building

Aspects of Structural Investigation

The professional designer or investigator uses all the practical meansavailable for accomplishment of the work In this age mathematical


3751 P-01 111301 1218 PM Page 64

modeling is greatly aided by the use of computers However routineproblems (that is 98 of all problems) are still often treated by use ofsimple hand computations or by reference to data in handbook tables or graphs

The purpose of this book is essentially educational so the emphasishere is on visualization and understanding not necessarily on efficiencyof computational means Major use is made of graphical visualizationand readers are strongly encouraged to develop the habit of using such vi-sualization The use of sketches as learning and problem-solving aidscannot be overemphasized Four types of graphical devices are espe-cially useful the free-body diagram the cut section the exaggerated pro-file of the load-deformed structure and the graphical plot of criticalequations

A free-body diagram consists of a picture of any isolated physical el-ement that shows the full set of external forces that operate on that ele-ment The isolated element may be a whole structure or any fractionalpart of it Consider the structure shown in Figure 154 Figure 154ashows the entire structure consisting of attached horizontal and verticalelements (beams and columns) that produce a planar rigid frame bentThis may be one of a set of such frames comprising a building structureThe free-body diagram in Figure 154a represents the entire structurewith forces external to it represented by arrows The arrows indicate thelocation sense and direction of each external force At some stage of in-vestigation numbers may be added indicating the magnitude of theseforces The forces shown include the weight of the structure the hori-zontal force of wind and the net forces acting at the points of support forthe frame

Shown in Figure 154b is a free-body diagram of a single beam fromthe framed bent Operating externally on the beam are its own weightplus the effects of interaction between the beam and the columns towhich it is attached These interactions are not visible in the free-body di-agram of the full frame so one purpose for the diagram of the singlebeam is simply the visualization of the nature of these interactions Itmay now be observed that the columns transmit to the ends of the beamsa combination of vertical and horizontal forces plus rotational bendingactions The observation of the form of these interactions is a necessaryfirst step in a full investigation of this beam

Figure 154c shows an isolated portion of the beam length producedby slicing vertical planes a short distance apart and removing the portion

DESIGN FOR STRUCTURAL RESPONSE 65

3751 P-01 111301 1218 PM Page 65

between them Operating on this free body are its own weight and the ac-tions of the of the beam segments on the opposite sides of the slicingplanes that is the effects that hold this segment in place in the uncutbeam This slicing device called a cut section is used to visualize the in-ternal force actions in the beam and is a first step in the investigation ofthe stresses that relate to the internal forces


Figure 154 Free-body diagrams

3751 P-01 111301 1218 PM Page 66

Finally in Figure 154d is shown a tiny particle of the material of thebeam on which the external effects are those of the adjacent particlesThis is the basic device for visualization of stress In the example theparticle is seen to be operated on by a combination of vertical shear (andits horizontal complement) and horizontally directed compression

Figure 155a shows the exaggerated deformed profile of the samebent under wind loading The overall form of lateral deflection of thebent and the character of bending in each member can be visualized fromthis figure As shown in Figure 155b the character of deformation ofsegments and particles can also be visualized These diagrams are veryhelpful in establishing the qualitative nature of the relationships betweenforce actions and overall shape changes or between stresses and strainsQuantitative computations often become considerably abstract in theiroperation but these diagrams are real exercises in direct visualization of behavior


Figure 155 Visualization of structural deformations

3751 P-01 111301 1218 PM Page 67

For both visualization and quantification considerable use is made ofgraphical plots of mathematical expressions in this book Figure 156shows the form of damped vibration of an elastic spring The graph con-sists of a plot of the variation of displacement (+ or ndashs) of the spring fromits neutral position as a function of elapsed time t This is a plot of the equation

which describes the function mathematically but not visually The graphhelps us to literally see the rate of decline of the vibration (damping ef-fect) and the specific location of the spring at any given point in timeOnly mathematicians can see these things from an equation for the restof us the graph is a big help

se

P Qt Rt

=

+[ ]1

sin( )


Figure 156 Displacement versus elapsed time plot of a cyclic (harmonic) motion

3751 P-01 111301 1218 PM Page 68

69

2FORCES AND

FORCE ACTIONS

The preceding chapter provided an overview of the world of structuralanalysis as an activity for the support of design of building structuresThis chapter begins a more deliberate study of the basic applications ofphysics and mathematics to the real work of structural analysis Thisstudy begins with a consideration of forces and their actions

21 LOADS AND RESISTANCE

Loads deriving from the tasks of a structure produce forces The tasks ofthe structure involve the transmission of the load forces to the supportsfor the structure Applied to the structure these external load and supportforces produce a resistance from the structure in terms of internal forcesthat resist changes in the shape of the structure In building structural sys-tems such as that shown in Figure 21 load forces are passed from ele-ment to element here from deck to rafter to purlin to truss to column tocolumn support

3751 P-02 111301 1219 PM Page 69

A first task for investigation of structural behavior is the considerationof the nature of individual forces of the combinations they occur in andof the equilibrium (balance) of all the forces that affect an individualstructure Equilibrium is an assumed condition based on not wanting thestructure go anywhere That is it may deform slightly but it is supposedto stay in place Thus when we add up all the operating forces on a struc-ture we should get a net total of zero force

The field of mechanics in the basic science of physics provides thefundamental relationships for dealing with forces and their actions Usingthose relationships to solve practical problems involves some applica-tions of mathematicsmdashfrom simple addition to advanced calculus de-pending on the complexity of the problems Here we assume the readerhas some familiarity with basic physics and a reasonable understanding

70 FORCES AND FORCE ACTIONS

Figure 21 Developed system for generation of a roof structure Columns supportspanning trusses that in turn support a combination of purlins rafters and deckingto define the roof surface Forces flow through the system passing from the deckto the columns

3751 P-02 111301 1219 PM Page 70

of arithmetic geometry elementary algebra and the first week or so ofa trigonometry course Having more background in mathematics will be useful for advanced study beyond this book but will not really helpmuch here

As the reader has already noticed we use illustrations considerably inthis book In the work that follows these are also used as part of the illu-mination of the ideas and the steps for analysis procedures There arethus three components of study literal (text description) visual (thebookrsquos or the readerrsquos sketches) and mathematical (demonstrations ofcomputations) It will work best for the reader to be fluent in all threecomponents of the study but some shortcomings in the mathematicalarea may be compensated for if the words and pictures are fully under-stood first

22 FORCES AND STRESSES

The idea of force is one of the fundamental concepts of mechanics anddoes not yield to simple precise definition An accepted definition offorce is that which produces or tends to produce motion or a change inthe state of motion of objects A type of force is the effect of gravity bywhich all objects are attracted toward the center of the earth

What causes the force of gravity on an object is the mass of the objectand in US units this force is quantified as the weight of the body Grav-ity forces are thus measured in pounds (lb) or in some other unit such astons (T) or kips (one kilopound or 1000 pounds) In the metric (or SI)system force is measured in a more purely scientific manner as directlyrelated to the mass of objects the mass of an object is a constant whereasweight is proportional to the precise value of the acceleration of gravitywhich varies from place to place Force in metric units is measured innewtons (N) kilonewtons (kN) or meganewtons (mN) whereas weightis measured in grams (g) or kilograms (kg)

Figure 22a represents a block of metal weighing 6400 lb supported ona wooden post having an 8 times 8 in cross section The wooden post is inturn supported on a base of masonry The gravity force of the metal blockexerted on the wood is 6400 lb or 64 kips Ignoring its own weight thewooden post in turn transmits a force of equal magnitude to the masonrybase If there is no motion (a state described as equilibrium) there must be

FORCES AND STRESSES 71

3751 P-02 111301 1219 PM Page 71

an equal upward force developed by the supporting masonry Thus thewooden post is acted on by a set of balanced forces consisting of the ap-plied (or active) downward load of 6400 lb and the resisting (called reac-tive) upward force of 6400 lb

To resist being crushed the wooden post develops an internal force ofcompression through stress in the material stress being defined as inter-nal force per unit area of the postrsquos cross section For the situation showneach square inch of the postrsquos cross section must develop a stress equalto 640064 = 100 lbsq in (psi) See Figure 22b


Figure 22 Direct force action and stress

3751 P-02 111301 1219 PM Page 72

23 TYPES OF FORCES

External forces may result from a number of sources as described inSection 11 For the moment we are treating only static forces and thusonly static force effects on responding objects Direct action of staticforces produces internal force responses of compression tension orshear The metal weight in Figure 22 represents a compressive forceand the resulting stresses in the wooden post are compressive stresses

Figure 22c represents a 05-in diameter steel rod suspended from anoverhead support A weight of 1500 lb is attached to the lower end of therod exerting an external tensile force on the rod The cross-sectionalarea of the rod is pR2 = 031416(025)2 = 0196 in2 where R is the radiusHence the tensile stress in the rod is 15000196 = 7653 psi

Now consider the two steel bars held together by a 075-in diameterbolt as shown in Figure 22d and subjected to a tension force of 5000lb The tension force in the bars becomes a shear force on the bolt de-scribed as a direct shear force There are many results created by theforce in Figure 22d including tensile stress in the bars and bearing onthe sides of the hole by the bolt For now we are concerned with theslicing action on the bolt (Figure 22e) described as direct shear stressThe bolt cross section has an area of 31416(0375)2 = 04418 in2 andthe shear stress in the bolt is thus equal to 500004418 = 11317 psiNote that this type of stress is visualized as acting in the plane of the boltcross section as a slicing or sliding effect while both compressive and tensile stresses are visualized as acting perpendicular to a stressedcross section

24 VECTORS

A quantity that involves magnitude direction (vertical eg) and sense(up down etc) is a vector quantity whereas a scalar quantity involvesonly magnitude and sense Force velocity and acceleration are vectorquantities while energy time and temperature are scalar quantities Avector can be represented by a straight line leading to the possibility ofconstructed graphical solutions in some cases a situation that will bedemonstrated later Mathematically a scalar quantity can be representedcompletely as +50 or ndash50 while a vector must somehow have its direc-tion represented as well (50 vertical horizontal etc)

VECTORS 73

3751 P-02 111301 1219 PM Page 73

25 PROPERTIES OF FORCES

As stated previously in order to completely identify a force it is neces-sary to establish the following

Magnitude of the Force This is the amount of the force which ismeasured in weight units such as pounds or tons

Direction of the Force This refers to the orientation of its path calledits line of action Direction is usually described by the angle that the line of action makes with some reference such as thehorizontal

Sense of the Force This refers to the manner in which the force actsalong its line of action (up or down right or left etc) Sense is usu-ally expressed algebraically in terms of the sign of the force eitherplus or minus

Forces can be represented graphically in terms of these three properties bythe use of an arrow as shown in Figure 23a Drawn to some scale thelength of the arrow represents the magnitude of the force The angle of in-clination of the arrow represents the direction of the force The location ofthe arrowhead represents the sense of the force This form of representa-tion can be more than merely symbolic since actual mathematical ma-nipulations may be performed using the vector representation that theforce arrows constitute In the work in this book arrows are used in a sym-bolic way for visual reference when performing algebraic computationsand in a truly representative way when performing graphical analyses

In addition to the basic properties of magnitude direction and sensesome other concerns that may be significant for certain investigationsare

Position of the Line of Action of the Force This is considered withrespect to the lines of action of other forces or to some object onwhich the force operates as shown in Figure 23b For the beamshifting of the location of the load (active force) effects changes inthe forces at the supports (reactions)

Point of Application of the Force Exactly where along its line of ac-tion the force is applied may be of concern in analyzing for the spe-cific effect of the force on an object as shown in Figure 23c


3751 P-02 111301 1219 PM Page 74

When forces are not resisted they tend to produce motion An inher-ent aspect of static forces is that they exist in a state of static equilibriumthat is with no motion occurring In order for static equilibrium to existit is necessary to have a balanced system of forces An important consid-eration in the analysis of static forces is the nature of the geometricarrangement of forces in a given set of forces that constitute a single sys-tem The usual technique for classifying force systems involves consid-eration of whether the forces in the system are

PROPERTIES OF FORCES 75

Figure 23 Properties of forces (a) Graphical representation of a force (b) Re-active forces (c) Effect of point of application of a force

3751 P-02 111301 1219 PM Page 75

Coplanar All acting in a single plane such as the plane of a verticalwall

Parallel All having the same direction

Concurrent All having their lines of action intersect at a commonpoint

Using these three considerations the possible variations are given inTable 21 and illustrated in Figure 24 Note that variation 5 in the tableis really not possible since a set of coacting forces that is parallel andconcurrent cannot be noncoplanar in fact the forces all fall on a singleline of action and are called collinear

It is necessary to qualify a set of forces in the manner just illustratedbefore proceeding with any analysis whether it is to be performed alge-braically or graphically

26 MOTION

A force was defined earlier as that which produces or tends to producemotion or a change of motion of bodies Motion is a change of positionwith respect to some object regarded as having a fixed position Whenthe path of a moving point is a straight line the point has motion oftranslation When the path of a point is curved the point has curvilinear


TABLE 21 Classification of Force Systemsa

Qualifications

System Variation Coplanar Parallel Concurrent

1 Yes Yes Yes2 Yes Yes No3 Yes No Yes4 Yes No No5 Nob Yes Yes6 No Yes No7 No No Yes8 No No No

aSee Figure 24bNot possiblemdashparallel concurrent forces are automatically coplanar

3751 P-02 111301 1219 PM Page 76

motion or motion of rotation When the path of a point lies in a plane thepoint has plane motion Other motions are space motions

Mostly in the design of structures a basic goal is to prevent motionHowever for visualization of potential force actions and the actual de-formation of force resisting structures it is very useful to both graphi-cally and mathematically identify the nature of motion implied by theactive forces Ultimately of course the desired state for the structure isa final condition described as one of static equilibrium with the externalforces balanced by the internal forces and with no movement except forsmall deformations

Static Equilibrium

As stated previously an object is in equilibrium when it is either at restor has uniform motion When a system of forces acting on an object pro-duces no motion the system of forces is said to be in static equilibrium

A simple example of equilibrium is illustrated in Figure 25a Twoequal opposite and parallel forces P1 and P2 have the same line of ac-tion and act on a body If the two forces balance each other the body

MOTION 77

Figure 24 Types of force systems

3751 P-02 111301 1219 PM Page 77

does not move and the system of forces is in equilibrium These twoforces are concurrent If the lines of action of a system of forces have apoint in common the forces are concurrent

Another example of forces in equilibrium is illustrated in Figure 25bA vertical downward force of 300 lb acts at the midpoint in the length ofa beam The two upward vertical forces of 150 lb each (the reactions) actat the ends of the beam The system of three forces is in equilibrium Theforces are parallel and not having a point in common are nonconcurrent

27 FORCE COMPONENTS AND COMBINATIONS

Individual forces may interact and be combined with other forces in var-ious situations The net effect of such action produces a singular actionthat is sometimes required to be observed Conversely a single forcemay have more than one effect on an object such as a vertical action anda horizontal action simultaneously This section considers both of theseissues the adding up of single forces (combination) and breaking downof single forces (resolution)

Resultant of Forces

The resultant of a system of forces is the simplest system (usually a sin-gle force) that has the same effect as the various forces in the system act-ing simultaneously The lines of action of any system of two nonparallelforces must have a point in common and the resultant of the two forceswill pass through this common point The resultant of two coplanar


Figure 25 Equilibrium of forces

3751 P-02 111301 1219 PM Page 78

nonparallel forces may be found graphically by constructing a parallel-ogram of forces

This graphical construction is based on the parallelogram law whichmay be stated thus two nonparallel forces are laid off at any scale (of somany pounds to the inch) with both forces pointing toward or bothforces pointing away from the point of intersection of their lines of ac-tion A parallelogram is then constructed with the two forces as adjacentsides The diagonal of the parallelogram passing through the commonpoint is the resultant in magnitude direction and line of action the di-rection of the resultant being similar to that of the given forces towardor away from the point in common In Figure 26a P1 and P2 representtwo nonparallel forces whose lines of action intersect at point O Theparallelogram is drawn and the diagonal R is the resultant of the givensystem In this illustration note that the two forces point away from thepoint in common hence the resultant also has its direction away frompoint O It is a force upward to the right Notice that the resultant offorces P1 and P2 shown in Figure 26b is R its direction is toward thepoint in common

Forces may be considered to act at any points on their lines of actionIn Figure 26c the lines of action of the two forces P1 and P2 are ex-tended until they meet at point O At this point the parallelogram offorces is constructed and R the diagonal is the resultant of forces P1 andP2 In determining the magnitude of the resultant the scale used is ofcourse the same scale used in laying off the given system of forces

Example 1 A vertical force of 50 lb and a horizontal force of 100 lb asshown in Figure 27a have an angle of 90deg between their lines of actionDetermine the resultant

FORCE COMPONENTS AND COMBINATIONS 79

Figure 26 Parallelogram of forces

3751 P-02 111301 1219 PM Page 79

Solution The two forces are laid off from their point of intersection at ascale of 1 in = 80 lb The parallelogram is drawn and the diagonal is theresultant Its magnitude scales approximately 112 lb its direction is up-ward to the right and its line of action passes through the point of inter-section of the lines of action of the two given forces By use of aprotractor it is found that the angle between the resultant and the force of100 lb is approximately 265deg

Example 2 The angle between two forces of 40 and 90 lb as shown inFigure 27b is 60deg Determine the resultant

Solution The forces are laid off from their point of intersection at ascale of 1 in = 80 lb The parallelogram of forces is constructed and theresultant is found to be a force of approximately 115 lb its direction isupward to the right and its line of action passes through the commonpoint of the two given forces The angle between the resultant and theforce of 90 lb is approximately 175deg

Attention is called to the fact that these two problems have beensolved graphically by the construction of diagrams Mathematics mighthave been employed For many practical problems graphical solutionsgive sufficiently accurate answers and frequently require far less timeDo not make diagrams too small Remember that greater accuracy is ob-tained by using larger parallelograms of forces

Problems 27AndashFBy constructing the parallelogram of forces determine the resultants forthe pairs of forces shown in Figures 28andashf


Figure 27 Examples 1 and 2

3751 P-02 111301 1219 PM Page 80

Components of a Force

In addition to combining forces to obtain their resultant it is often neces-sary to replace a single force by its components The components of aforce are the two or more forces that acting together have the same effectas the given force In Figure 27a if we are given the force of 112 lb itsvertical component is 50 lb and its horizontal component is 100 lb Thatis the 112-lb force has been resolved into its vertical and horizontal com-ponents Any force may be considered as the resultant of its components

Combined Resultants

The resultant of more than two nonparallel forces may be obtained byfinding the resultants of pairs of forces and finally the resultant of theresultants

Example 3 Let it be required to find the resultant of the concurrentforces P1 P2 P3 and P4 shown in Figure 29

Solution By constructing a parallelogram of forces the resultant of P1

and P2 is found to be R1 Similarly the resultant of P3 and P4 is R2 Fi-nally the resultant of R1 and R2 is R the resultant of the four given forces


Figure 28 Problems 28AndashF

3751 P-02 111301 1219 PM Page 81

Problems 27GndashIUsing graphical methods find the resultants of the systems of concurrentforces shown in Figures 210(g)ndash(i)

Equilibrant

The force required to maintain a system of forces in equilibrium is calledthe equilibrant of the system Suppose that we are required to investigate


Figure 29 Finding a resultant by pairs

Figure 210 Problems 27GndashI

3751 P-02 111301 1219 PM Page 82

the system of two forces P1 and P2 as shown in Figure 211 The paral-lelogram of forces is constructed and the resultant is found to be R Thesystem is not in equilibrium The force required to maintain equilibriumis force E shown by the dotted line E the equilibrant is the same as theresultant in magnitude and direction but is opposite in sense The threeforces P1 P2 and E constitute a system in equilibrium

If two forces are in equilibrium they must be equal in magnitude op-posite in sense and have the same direction and line of action Either ofthe two forces may be said to be the equilibrant of the other The resul-tant of a system of forces in equilibrium is zero

28 GRAPHICAL ANALYSIS OF FORCES

Force Polygon

The resultant of a system of concurrent forces may be found by con-structing a force polygon To draw the force polygon begin with a pointand lay off at a convenient scale a line parallel to one of the forces withits length equal to the force in magnitude and having the same senseFrom the termination of this line draw similarly another line corre-sponding to one of the remaining forces and continue in the same man-ner until all the forces in the given system are accounted for If thepolygon does not close the system of forces is not in equilibrium and theline required to close the polygon drawn from the starting point is the re-sultant in magnitude and direction If the forces in the given system areconcurrent the line of action of the resultant passes through the pointthey have in common

GRAPHICAL ANALYSIS OF FORCES 83

Figure 211 Resultant and equilibrant

3751 P-02 111301 1219 PM Page 83

If the force polygon for a system of concurrent forces closes the sys-tem is in equilibrium and the resultant is zero

Example 4 Let it be required to find the resultant of the four concurrentforces P1 P2 P3 and P4 shown in Figure 212a This diagram is calledthe space diagram it shows the relative positions of the forces in a givensystem

Solution Beginning with some point such as O shown in Figure 212bdraw the upward force P1 At the upper extremity of the line representingP1 draw P2 continuing in a like manner with P3 and P4 The polygondoes not close therefore the system is not in equilibrium The resultantR shown by the dot-and-dash line is the resultant of the given systemNote that its direction is from the starting point O downward to the right The line of action of the resultant of the given system shown inFigure 212a has its line of action passing through the point they have in common its magnitude and direction having been found in the forcepolygon

In drawing the force polygon the forces may be taken in any se-quence In Figure 212c a different sequence is taken but the resultant Ris found to have the same magnitude and direction as previously found inFigure 212b


Figure 212 Force polygon for a set of concurrent forces

3751 P-02 111301 1219 PM Page 84

Bowrsquos Notation

Thus far forces have been identified by the symbols P1 P2 and so on Asystem of identifying forces known as Bowrsquos notation affords many ad-vantages In this system letters are placed in the space diagram on eachside of a force and a force is identified by two letters The sequence inwhich the letters are read is important Figure 213a shows the space di-agram of five concurrent forces Reading about the point in common ina clockwise manner the forces are AB BC CD DE and EA When aforce in the force polygon is represented by a line a letter is placed ateach end of the line As an example the vertical upward force in Figure213a is read AB (note that this is read clockwise about the commonpoint) in the force polygon (Figure 213b) the letter a is placed at thebottom of the line representing the force AB and the letter b is at the topUse capital letters to identify the forces in the space diagrams and low-ercase letters in the force polygon From point b in the force polygondraw force bc then cd and continue with de and ea Since the forcepolygon closes the five concurrent forces are in equilibrium

In reading forces a clockwise manner is used in all the following dis-cussions It is important that this method of identifying forces be thor-oughly understood To make this clear suppose that a force polygon isdrawn for the five forces shown in Figure 213a reading the forces insequence in a counterclockwise manner This will produce the forcepolygon shown in Figure 213c Either method may be used but for con-sistency the method of reading clockwise is used here


Figure 213 Use of Bowrsquos notation

3751 P-02 111301 1219 PM Page 85

Use of the Force Polygon

Two ropes are attached to a ceiling and their lower ends are connected toa ring making the arrangement shown in Figure 214a A weight of 100lb is suspended from the ring Obviously the force in the rope AB is 100lb but the magnitudes of the forces in ropes BC and CA are unknown

The forces in the ropes AB BC and CA constitute a concurrent forcesystem in equilibrium The magnitude of only one of the forces is knownmdashit is 100 lb in rope AB Since the three concurrent forces are in equi-librium their force polygon must close and this fact makes it possible tofind the magnitudes of the BC and CA Now at a convenient scale drawthe line ab (Figure 214c) representing the downward force AB 100 lbThe line ab is one side of the force polygon From point b draw a lineparallel to rope BC point c will be at some location on this line Nextdraw a line through point a parallel to rope CA point c will be at someposition on this line Since point c is also on the line though b parallel toBC the intersection of the two lines determines point c The force poly-gon for the three forces is now completed it is abc and the lengths of thesides of the polygon represent the magnitudes of the forces in ropes BCand CA 866 lb and 50 lb respectively

Particular attention is called to the fact that the lengths of the ropes inFigure 214a are not an indication of magnitude of the forces within theropes the magnitudes are determined by the lengths of the correspond-


Figure 214 Solution of a problem with concurrent forces

3751 P-02 111301 1219 PM Page 86

ing sides of the force polygon (Figure 214c) Figure 214a merely deter-mines the geometric layout for the structure

Problems 28AndashDFind the sense (tension or compression) and magnitude of the internalforces in the members indicated by question marks in Figures 215andashdusing graphical methods

29 INVESTIGATION OF FORCE ACTIONS

A convenient way to determine the unknown forces acting on a body orthe unknown internal forces in a structure is to construct a free-body di-agram This may be for a whole structure or a part of a structure The

INVESTIGATION OF FORCE ACTIONS 87

Figure 215 Problems 28AndashD

3751 P-02 111301 1219 PM Page 87

usual procedure is to imagine the defined element (body) to be cut awayfrom adjoining parts and moved to a free position in space See the dis-cussion in Section 110

Graphical Solution of Forces

Consider Figure 216a which represents two members framing into awall the upper member being horizontal and the angle between the mem-bers being 30deg A weight of 200 lb is placed at the point where the mem-bers meet Figure 216b is a diagram showing the block as a free bodywith the forces acting on it consisting of its own weight and the two un-known internal forces in the members This concurrent force system isrepresented in Figure 216c with letters placed on the figure to utilizeBowrsquos notation Thus the forces acting on the body are AB (the force dueto gravity) and the unknowns BC and CA The arrows placed on the un-known forces indicating their sense would seem to be evident althoughthey have not actually been determined at this point

To determine the unknown internal forces in the frame members aforce polygon of this concurrent set of forces may be constructed Startby drawing the vector ab downward to a convenient scale measured at200 as shown in Figure 216d On this diagram through point a draw a


Figure 216 Use of the free-body diagram

3751 P-02 111301 1219 PM Page 88

horizontal line representing force ca Then through point b draw a lineat 30deg representing the force bc The intersection of these two lines lo-cates the point c on the diagram and completes the force polygon Byusing the scale that was used to lay out force ab the lengths of the othertwo sides of the polygon can be measured these are the magnitudes ofthe unknown forces Accuracy in this case will depend on how large afigure is drawn and how carefully it is constructed The sense of theforces can be determined by following the sequence of force flow on thepolygon from a to b to c to a Thus the assumed senses are shown to be correct

Algebraic Solution

The preceding problem obviously also lends itself to a mathematical so-lution Consider the free-body diagram of the forces as shown in Figure216e On this figure the force BC is shown both as a single force and asa combination of its horizontal and vertical components either represen-tation can be used for this force The relationship of force BC to its com-ponents is shown in Figure 216f The purpose for consideration of thecomponents of BC is demonstrated in the following work

The forces in the free-body diagram in this example are constituted asa concentric coplanar force system (see Section 25) For such a systemthe algebraic conditions for static equilibrium may be stated as follows

ΣFH = 0 and ΣFV = 0

That is to say the summation of the horizontal force components of allthe forces is zero and the summation of the vertical components of all the forces is zero Referring to Figure 216e and applying these con-ditions to the example

ΣFH = 0 = CA + BCH

ΣFV = 0 = AB + BCV

To implement these algebraically a sign convention must be assumedAssume the following

For vertical forces + is up ndash is down

For horizontal forces + is to the right ndash is to the left


3751 P-02 111301 1219 PM Page 89

Thus from the summation of the vertical forces using the known valueof AB

ΣFV = 0 = (ndash200) + BCV

from which

BCV = +200 or 200 lb up

If this component is up then the force BC as indicated in Figure 216 iscorrectly shown as a compression force To obtain the value for BC con-sider the relation of the force to its components as shown in Figure 216fThus

Then using the summation of horizontal forces

ΣFH = 0 = CA + BCH = CA + (+400 times cos 30deg)

from which CA is obtained as ndash346 lb the minus sign indicates the cor-rectness of the assumption shown in Figure 216e namely that CA is intension

Two-Force Members

When a member in equilibrium is acted on by forces at only two pointsit is known as a two-force member The resultant of all the forces at one point must be equal opposite in sense and have the same directionand line of action as the resultant of the forces at the other point The internal force in a linear two-force member is either tension or compression

In Figure 216a each of the two members in the frame is a two-forcemember A free-body diagram of either member will show only oneforce at an end equal and opposite in sense to the force at the other endThe members of planar trusses are assumed to be of this form so that theanalysis of the truss may be achieved by a solution of the concentricforces at the joints of the truss This is demonstrated in Chapter 3

BCBCV=

deg= =

sin lb

30

200

0 5400


3751 P-02 111301 1219 PM Page 90

210 FRICTION

Friction is a force of resistance to movement that is developed at the con-tact face between objects when the objects are made to slide with respectto each other For the object shown in Figure 217a being acted on by itsown weight and the inclined force F the impending motion is that of theblock toward the right along the supporting surface The force tending tocause the motion is the horizontal component of F that is the componentparallel to the sliding surface The vertical component of F combineswith the weight of the block W to produce a force pressing the blockagainst the plane This pressure-generating force called the normal forceis what produces friction

A free-body diagram of the forces is shown in Figure 217b For equi-librium of the block two components of resistance must be developedFor equilibrium in a direction normal to the plane of friction (verticalhere) the reactive force N is required being equal and opposite in senseto the normal force on the plane For equilibrium in a direction parallelto the plane (horizontal here) a frictional resistance Fcent must be developedthat is at least as great as the force tending to cause sliding For this situ-ation there are three possibilities as follows

1 The block does not move because the potential friction resistanceis greater than the impelling force that is

Fcent is greater than F cos Q

FRICTION 91

Figure 217 Development of sliding friction

3751 P-02 111301 1219 PM Page 91

2 The block moves because the friction is not of sufficient magni-tude that is

F cent is less than F cos Q

3 The block is in equilibrium but just on the verge of moving be-cause the potential friction force is exactly equal to the forcetending to induce sliding that is

F cent = F cos Q

From observations and experimentation the following deductionshave been made about friction

1 The friction-resisting force (F cent in Figure 217) always acts in a di-rection to oppose motion that is it acts opposite to the slide-inducing force

2 For dry smooth surfaces the frictional resistance developed up tothe moment of sliding is directly proportional to the normal pres-sure between the surfaces This limiting value for the force is ex-pressed as

F cent = mN

in which m (Greek lowercase mu) is called the coefficient offriction

3 The frictional resistance is independent of the amount of contactarea

4 The coefficient of static friction (before motion occurs) is greaterthan the coefficient of kinetic friction (during actual sliding) Thatis for the same amount of normal pressure the frictional resis-tance is reduced once motion actually occurs

Frictional resistance is ordinarily expressed in terms of its maximumpotential value Coefficients for static friction are determined by findingthe ratio between the slide-inducing force and the normal force that cre-ates pressure just at the point of sliding A simple experiment consists ofplacing a block on an inclined surface and steadily increasing the angle


3751 P-02 111301 1219 PM Page 92

of inclination until sliding occurs (see Figure 218a) Referring to thefree-body diagram of the block in Figure 218b we note

Fcent = mN = W sin fN = W cos f

and as previously noted the coefficient of friction is expressed as theratio of Fcent to N or

Approximate values for the coefficient of static friction for various com-binations of objects in contact are given in Table 22

Problems involving friction are usually one of two types The first in-volves situations in which friction is one of the forces in a system and theproblem is to determine whether the frictional resistance is sufficient tomaintain the equilibrium of the system For this type of problem the

micro φφ

φ= prime = =F

N

W

W

sin

cos tan

FRICTION 93

Figure 218 Derivation of the coefficient of friction

TABLE 22 Range of Values for Coefficient of Static Friction

Contact Surfaces Coefficient m

Wood on wood 040ndash070Metal on wood 020ndash065Metal on metal 015ndash030Metal on stone masonry concrete 030ndash070

3751 P-02 111301 1219 PM Page 93

solution consists of writing the equations for equilibrium including themaximum potential friction and interpreting the results If the frictionalresistance is not large enough sliding will occur if it is just large enoughor excessive sliding will not occur

The second type of problem involves situations in which the force required to overcome friction must be found In this case the slide-inducing force is simply equated to the maximum potential friction re-sistance and the required force is determined

Example 5 A block is placed on an inclined plane whose angle is slowlyincreased until sliding occurs (see Figure 219) If the angle of the planewith the horizontal is 35deg when sliding begins what is the coefficient forsliding friction between the block and the plane

Solution As previously derived the coefficient of friction may be statedas the tangent of the angle of inclination of the plane thus

m = tan f = tan 35deg = 070

Example 6 Find the horizontal force P required to slide a blockweighing 100 lb if the coefficient of static friction is 030 (see Figure220)

Solution For sliding to occur the slide-inducing force P must beslightly larger than the frictional resistance Fcent Thus

P = Fcent = mN = 030(100) = 30 lb

The force must be slightly larger than 30 lb


Figure 219 Use of the inclinedplane to determine the coefficient ofstatic friction

3751 P-02 111301 1219 PM Page 94

Example 7 A block is pressed against a vertical wall with a 20-lb forcethat acts upward at an angle of 30deg with the horizontal (see Figure 221a)

(a) Express the frictional resistance to motion in terms of the avail-able pressure

(b) If the block weighs 15 lb and the coefficient of static friction is040 will the block slide

FRICTION 95

Figure 220 Example 6


3751 P-02 111301 1219 PM Page 95

(c) At what angle must the 20-lb force act to cause the 15-lb block toslide upward if the coefficient of static friction is 040

Solution For (a)

F cent = mN = m(20 cos 30deg) = 1732m lb

For (b) the sliding resistance must equal the net slide-inducing force or

required F cent = [W ndash (20 sin 30deg)] = W ndash 10 = 15 ndash 10 = 5 lb

From (a) the available resistance is

F cent = 1732(040) = 693 lb

Therefore the block will not slide

For (c)

F cent = (20 sin f) ndash 15

or

040(20 cos f) = (20 sin f) ndash 15

from which f = 811deg

Problem 210AFind the angle at which the block shown in Figure 218 will slip if the co-efficient of static friction is 035

Problem 210BFor the block shown in Figure 222 find the value of P required to keepthe block from slipping if f = 10deg and W = 10 lb

Problem 210CFor the block shown in Figure 222 find the weight for the block that willresult in slipping if f = 15deg and P = 10 lb


3751 P-02 111301 1219 PM Page 96

211 MOMENTS

The term moment is commonly used to designate the tendency of a forceto cause rotation about a given point or axis The unit of measurement formoments is a compound produced by the multiplication of the force (inpounds tons etc) times a distance (in feet inches etc) A moment isthus said to consist of so many ft-lb kip-in and so on The point or axisabout which rotation is induced is called the center of moments The per-pendicular distance between the line of action of the force and the centerof moments is called the lever arm or moment arm Thus a moment hasa magnitude that is determined as

moment = (magnitude of force) times (length of moment arm)

Consider the horizontal force of 100 lb shown in Figure 223 If pointA is the center of moments the lever arm of the force is 5 ft Then themoment of the 100-lb force with respect to point A is 100 times 5 = 500

MOMENTS 97

Figure 222 Problems 210B C

Figure 223 Moment of a forceabout a point

3751 P-02 111301 1219 PM Page 97

ft-lb In this illustration the force tends to cause a clockwise rotationabout point A which is the sense or sign of the moment Ordinarilyclockwise rotation is considered to be positive and counterclockwisemoment to be negative Thus the complete designation of the momentis +500 ft-lb

In Figure 223 the 100-lb force has a moment arm of 3 ft with respectto point B With respect to point B the force has a counterclockwise mo-ment determined to be 100 times 3 = ndash300 ft-lb

Increasing Moments

A moment may be increased by increasing the magnitude of the force orby increasing the distance of the moment arm For the wrench in Figure224 the limit for rotational effort in terms of moment on the bolt head islimited by the effective wrench length and the force exerted on the han-dle Additional twisting moment on the bolt can be developed by in-creasing the force However for a limited force the wrench length mightbe extended by slipping a pipe over the wrench handle thus producing alarger moment with the same force

If a given moment is required various combinations of force andmoment arm may be used to produce the moment For example if thecombination of the given force of 50 lb was found to be just sufficient totwist the nut in Figure 224 with the pipe over the wrench handle whatforce would have been required if the pipe was not used With the pipethe moment is 50 times 25 = 1250 in-lb If the pipe is not used the requiredforce is thus found as 1250 10 = 125 lb


Figure 224 Effect of change in the moment arm

3751 P-02 111301 1219 PM Page 98

Moment of a Mechanical Couple

A mechanical couple is a means for visualization of a pure rotational ef-fect As produced by a couple it takes a form as shown in Figure 225with two parallel forces (the couple) acting in opposite directions at somedistance apart If the two forces are equal in magnitude the resultant ofthe forces is zero as a force magnitude However the resultant effect of the forces produces a moment which is the true resultant of the forcesystem a mechanical couple The magnitude of the moment is simply theproduct of one of the forces times the distance between the separatedlines of action of the parallel forces In the illustration the sense of themoment is counterclockwise

An example of a mechanical couple is that produced when a personuses two hands to turn a steering wheel The result of this push-pull ef-fort is neither a net push or a net pull on the wheel but rather a pure ro-tation of the steering column This is directly analogous to thedevelopment of internal bending resistance in structural members whereopposed tension and compressive stresses produce pure rotational effortThis phenomenon is discussed for beams in Chapter 11

Force Required to Produce Motion

Figure 226a shows a wheel under the action of a horizontal force that isattempting to roll the wheel over a fixed block In order to produce mo-tion the force must be slightly greater than that required for equilibriumPushing on the wheel produces a set of forces consisting of the weight ofthe wheel the pushing force and the force of the corner of the fixedblock that pushes back on the wheel The combination of these three

MOMENTS 99

Figure 225 A mechanical couple

3751 P-02 111301 1219 PM Page 99

forces is shown in the free-body diagram of the wheel in Figure 226bThey constitute a concentric force system for which a force polygon isshown in Figure 226c

If the wheel weighs 400 lb and the vector for this force is drawn to ascale in proportion to the 400-lb magnitude (ca on the force polygon) theforce required for equilibrium may be found by measuring the vector bcon the polygon A graphic solution that begins with the scaled layout ofthe wheel the block and the pushing force (Figure 226a) to determinethe angle of force CA will determine that the pushing force at the pointof motion must exceed a value of approximately 330 lb An algebraic so-lution can also be performed for example a summation of momentsabout the contact point between the wheel and the fixed block

Example 8 Figure 227a shows a masonry pier that weighs 10000 lbDetermine the magnitude of the horizontal force applied at the upper leftcorner that will be required to overturn the pier

Solution Tipping of the pier will occur with rotation about the lowerright corner of the pier The forces on the pier at the point of tipping willconsist of the pier weight the horizontal push at the top and the force ex-erted by the ground at the bottom right corner A free-body diagram ofthe pier under the action of these three forces is shown in Figure 227bFigure 227c shows a force polygon for these forces that includes a mag-nitude for the pushing force at the moment of the beginning of tipping Aslight increase in the tipping force above this value will produce tipping(more often described as overturning in engineering)

As with the wheel in the preceding illustration a scaled layout may beused to determine the magnitude of the pushing force However a sim-


Figure 226 Force required to produce motion graphical solution

3751 P-02 111301 1219 PM Page 100

ple algebraic solution may be performed using a summation of momentsabout the lower right corner (point O in Figure 227b) As the line of action of the force at this point has no moment in this summation theequation for moments is reduced to that involving only the pushing forceand the weight of the pier Thus

ΣMo = +(BC times 8) ndash(AB times 2)

Entering the known value of 10000 lb for AB in this equation will pro-duce an answer of 2500 lb for the pushing force Any force exceeding2500 lb will tend to tip the pier

Problem 211AUsing a graphical solution find the horizontal force P required to roll thecylinder in Figure 228a over the fixed block The cylinder is 20 in in di-ameter and weighs 500 lb

MOMENTS 101


Figure 228 Problems 211AndashC

3751 P-02 111301 1219 PM Page 101

Problem 211BThe masonry pier in Figure 228b weighs 3600 lb If the force P as shownis 800 lb will the pier tip about its lower right corner

Problem 211CIf the pier in Figure 211b weighs 5000 lb find the magnitude requiredfor force P to cause overturning

212 FORCES ON A BEAM

Figure 229a shows a cantilever beam with a single concentrated load of100 lb placed 4 ft from the face of the supporting wall In this positionthe moment of the force about point A (the face of the support) is 100 times4 = 400 ft-lb If the load is moved 2 ft farther to the right the momentabout point A is 600 ft-lb When the load is moved to the end of the beamthe moment at point A is 800 ft-lb

Figure 229b shows a cantilever beam with a uniformly distributedload over part of its length For finding moments due to distributed loadsa procedure commonly used is to find the total of the distributed load andto consider it to be a single concentrated load placed at the center of thedistributed load In this case the total load is 200 times 6 = 1200 lb and itseffective location is at a point 3 ft from the end of the beam Thus themoment of the load about point A is 1200 times 7 = 8400 ft-lb

Equilibrium of Coplanar Forces

For a general coplanar force system equilibrium can be established withthe satisfying of three equations as follows


Figure 229 Forces on cantilever beams

3751 P-02 111301 1219 PM Page 102

1 The algebraic sum of the horizontal forces is zero

2 The algebraic sum of the vertical forces is zero

3 The algebraic sum of the moments of all the forces about anypoint in the plane is zero

These summations can be made for any coplanar system of forcesHowever any additional qualifications of the forces may result in sim-plification of the algebraic conditions For example when the forces areconcurrent (all meeting at a single point) they have no moments with re-spect to each other and the condition for equilibrium of moments can beeliminated leaving only the two force equations This was the case forthe system shown in Figure 226 An even simpler qualification is that ofcolinear forces all acting on a single line of action such as the systemshown in Figure 230a Such a system if in equilibrium consists of twoequal forces of opposite sense

Beams are generally operated on by parallel coplanar forces Thiseliminates one of the force summations from the condition for generalcoplanar systems since all the forces are in a single direction There arethus only two equations of equilibrium necessary for the parallel systemand consequently only two available for solution of the system Elimi-nating one force equation from the general set leaves

1 The sum of the vertical forces equals zero

2 The sum of the moments about any point equals zero

FORCES ON A BEAM 103

Figure 230 Moment effects on a beam

3751 P-02 111301 1219 PM Page 103

However another possibility for establishing equilibrium is to satisfythe condition that the sum of the moments of the forces about two sepa-rate points is zero Thus another set of equations that may be used for thebeam is

1 The sum of the moments about point A is zero

2 The sum of the moments about point B is zeroWhere point A is a different point in the plane than point B

Consider the simple beam in Figure 230b Four vertical forces act onthis beam and are in equilibrium The two downward forces or loads are4 kips and 8 kips Opposing these are the support reaction forces at theends of the beam 44 kips and 76 kips If these parallel forces are indeedin equilibrium they should satisfy the equilibrium equations for a paral-lel system Thus

ΣFv = 0 = +44 ndash 4 ndash 8 + 76 = (+12) + (ndash12)

and the forces are in balance

ΣMA = 0 = +(44 times 20) ndash (4 times 14) ndash (8 times 4) = (+88) + (ndash88)

and the sum of the moments about point A is indeed zeroTo further demonstrate the equilibrium of the force values moments

may be taken about any other point in the plane For example for pointB which is the location of the 4-kip load

ΣMB = +(44 times 6) + (8 times 10) ndash (76 times 14) = +(1064) ndash (1064)

which verifies the balance of moments about point BAnother type of problem involves the finding of some unknown

forces in a parallel system Remember that the two conditions of equi-librium for the parallel system provide two algebraic equations whichpotentially may be used to find two unknown forces in the system Con-sider the beam shown in Figure 231 with a single support and a load of800 lb at one end The problem is to determine the required value for aload at the other end of the beam that will maintain equilibrium and thevalue for the single support reaction A summation of vertical forces willproduce an equation with two unknowns Indeed the two unknown


3751 P-02 111301 1219 PM Page 104

forces could be solved using two equations in two unknowns Howevera simpler procedure frequently used is to write equations involving onlyone unknown in a single equation at a time if possible For example anequation for the sum of moments about either the right end or the supportwill produce such an equation Thus for moments about the supportcalling the unknown load x

ΣM = 0 = ndash(800 times 6) + (x times 3) thus x = 1600 lb

Then from a summation of vertical forces calling the reaction force R

ΣF = 0 = ndash800 +R ndash1600 thus R = 2400 lb

This form of solution is frequently used to find reactions for ordinarybeams with two supports which is discussed next

Problem 212AWrite the two equations for moments for the four forces in Figure 230btaking points C and D as the centers of moments to verify the equilib-rium of the system

Determination of Reactions for Beams

As noted earlier reactions are the forces at the supports of beams thathold the loads in equilibrium A single-span beam is shown in Figure232 with two supports one at each end of the beam As these supportsare not shown to have resistance to rotation (called fixed supports) theyare assumed to be resistant only to the necessary vertical forces and de-scribed as simple supports This common beam with a single span and


Figure 231 Beam with asingle support

3751 P-02 111301 1219 PM Page 105

two simple supports is referred to as a simple beam The computationsthat follow will demonstrate the common procedure for finding the val-ues for the magnitudes of the two support reactions for a simple beamNote that the two reactions in Figure 232 are designated R1 and R2 forthe left and right reactions respectively This is a common practice thatis followed throughout the work in this book

Example 9 Compute the reactions for the beam in Figure 232

Solution Taking the right reaction as the center of moments

Taking the left reaction as the center of moments

To see whether a mistake has been made the three forces (load and tworeactions) may be checked for equilibrium of the vertical forces thus

ΣF = 0 = +450 ndash1800 +1350

and the net force is indeed zero

Example 10 Compute the reactions for the simple beam in Figure 233with three concentrated loads

Σ = = + times minus times = =M R R0 1800 9 1216 200

1213502 2( ) ( )

thus lb

Σ = = + times minus times = =M R R0 12 1800 35400

124501 1( ) ( ) thus lb



3751 P-02 111301 1219 PM Page 106

Solution Regardless of the type or number of loads the procedure is thesame Thus considering the right reaction as the center of moments

ΣM = 0 = +(R1 times 15) ndash(400 times 12) ndash(1000 times 10) ndash(600 times 4)

Thus

Using the same procedure with the left reaction as the center of moments

And for a check the summation of vertical forces is

ΣF = +11467 ndash 400 ndash 1000 ndash 600 + 8533 = 0

For any beam with two simple supports the procedure is the sameCare must be taken however to note carefully the sign of the momentsthat is plus for clockwise moments and minus for counterclockwise mo-ments about the selected center of moments The following example hasits supports drawn in from the ends of the beam producing cantileveredor overhanging ends

Example 11 Compute the reactions for the beam in Figure 234 withoverhanging ends

R2400 3 1000 5 600 11

15

12 800

15853 3= times + times + times = =( ) ( ) ( )

lb

R14800 10 000 2400

15

17 200

151146 7= + + = =

lb



3751 P-02 111301 1219 PM Page 107

Solution Using the same procedure as in the preceding two examplesfirst take moments about the right reaction thus

ΣM = 0 = ndash(200 times 22) + (R1 times 18) ndash (1000 times 10) ndash (800 times 4) + (600 times 2)

from which

Then with a summation of moments about the left reaction

ΣM = 0 = ndash(200 times 4) + (1000 times 8) + (800 times 14) ndash (R2 times 18) + (600 times 20)

Thus

A summation of vertical forces can be used to verify the answers

Example 12 The simple beam shown in Figure 235a has a single con-centrated load and a uniformly distributed load over a portion of thespan Compute the reactions

R230 400

181688 9= =

lb

R116 400

18911 1= =

lb



3751 P-02 111301 1219 PM Page 108

Solution For a simplification in finding the reactions it is common toconsider the uniformly distributed load to be replaced by its resultant inthe form of a single concentrated load at the center of the distributed loadThe total of the uniform load is 200 times 8 = 1600 lb and the beam is thusconsidered to be as shown in Figure 235b With the modified beam asummation of moments about the right reaction is

A summation of moments about the left reaction will determine a valueof 1940 lb for R2 and a summation of vertical forces may be used to ver-ify the answers

This shortcut consisting of replacing the distributed load by its resul-tant is acceptable for finding the reactions but the real nature of the dis-tributed load must be considered for other investigations of the beam aswill be demonstrated in some of the later chapters

Problems 212BndashGCompute the reactions for the beams shown in Figures 236bndashg

Σ = = + times minus times minus times = =M R R0 20 2200 14 1600 437 200

2018601 1( ) ( ) ( )

lb



3751 P-02 111301 1219 PM Page 109


Figure 236 Problems212BndashG

3751 P-02 111301 1219 PM Page 110

111

3ANALYSIS OF TRUSSES

Planar trusses comprised of linear elements assembled in triangulatedframeworks have been used for spanning structures in buildings formany centuries Figure 31 shows a form of construction used for such atruss in the early twentieth century While construction materials detailsand processes have changed considerably this classic form of truss isstill widely used Investigation for internal forces in such trusses is typi-cally performed by simple analytical procedures using the basic methodsillustrated in the preceding chapters In this chapter these procedures aredemonstrated using both graphical and algebraic methods of solution

31 GRAPHICAL ANALYSIS OF TRUSSES

When the so-called method of joints is used finding the internal forces inthe members of a planar truss consists of solving a series of concurrentforce systems Figure 32 at the top shows a truss with the truss formthe loads and the reactions displayed in a space diagram Below thespace diagram is a figure consisting of the free-body diagrams of the

3751 P-03 111301 1221 PM Page 111

individual joints of the truss These are arranged in the same manner asthey are in the truss in order to show their interrelationships Howevereach joint constitutes a complete concurrent planar force system thatmust have its independent equilibrium ldquoSolvingrdquo the problem consists ofdetermining the equilibrium conditions for all of the joints The proce-dures used for this solution are now illustrated

Figure 33 shows a single-span planar truss that is subjected to verti-cal gravity loads This example will be used to illustrate the proceduresfor determining the internal forces in the truss that is the tension andcompression forces in the individual members of the truss The space di-agram in the figure shows the truss form and dimensions the supportconditions and the loads The letters on the space diagram identify indi-vidual forces at the truss joints as discussed in Section 28 The sequenceof placement of the letters is arbitrary the only necessary considerationbeing to place a letter in each space between the loads and the individual

112 ANALYSIS OF TRUSSES

Figure 31 Details of an early twentieth century timber truss Reproduced fromMaterials and Methods of Construction by C Gay and H Parker 1932 with per-mission of the publisher John Wiley amp Sons New York This is a classic truss pat-tern still in frequent use although neither the forms of the membersmdashsteel rodsand solid timbersmdashnor any of the joint details are likely to be used today

3751 P-03 111301 1221 PM Page 112

truss members so that each force at a joint can be identified by a two-letter symbol

The separated joint diagram in the figure provides a useful means forvisualization of the complete force system at each joint as well as the in-terrelation of the joints through the truss members The individual forcesat each joint are designated by two-letter symbols that are obtained bysimply reading around the joint in the space diagram in a clockwise di-rection Note that the two-letter symbols are reversed at the oppositeends of each of the truss members Thus the top chord member at the leftend of the truss is designated as BI when shown in the joint at the leftsupport (joint 1) and is designated as IB when shown in the first interiorupper chord joint (joint 2) The purpose of this procedure will be demon-strated in the following explanation of the graphical analysis

The third diagram in Figure 33 is a composite force polygon for theexternal and internal forces in the truss It is called a Maxwell diagram

GRAPHICAL ANALYSIS OF TRUSSES 113

Figure 32 Examples of diagrams used to represent trusses and their actions

3751 P-03 111301 1221 PM Page 113

after one of its early promoters James Maxwell a British engineer Theconstruction of this diagram constitutes a complete solution for the mag-nitudes and senses of the internal forces in the truss The procedure forthis construction is as follows

1 Construct the force polygon for the external forces Before thiscan be done the values for the reactions must be found There aregraphic techniques for finding the reactions but it is usuallymuch simpler and faster to find them with an algebraic solution


Figure 33 Examples of graphic diagrams for a planar truss

3751 P-03 111301 1221 PM Page 114

In this example although the truss is not symmetrical the load-ing is and it may simply be observed that the reactions are eachequal to one-half of the total load on the truss or 5000 divide 2 = 2500lb Since the external forces in this case are all in a single direc-tion the force polygon for the external forces is actually a straightline Using the two-letter symbols for the forces and starting withthe letter A at the left end we read the force sequence by movingin a clockwise direction around the outside of the truss The loadsare thus read as AB BC CD DE EF and FG and the two reac-tions are read as GH and HA Beginning at A on the Maxwell di-agram the force vector sequence for the external forces is readfrom A to B B to C C to D and so on ending back at A whichshows that the force polygon closes and the external forces are inthe necessary state of static equilibrium Note that we have pulledthe vectors for the reactions off to the side in the diagram to indi-cate them more clearly Note also that we have used lowercaseletters for the vector ends in the Maxwell diagram whereas up-percase letters are used on the space diagram The alphabetic cor-relation is thus retained (A to a) while any possible confusionbetween the two diagrams is prevented The letters on the spacediagram designate open spaces while the letters on the Maxwelldiagram designate points of intersection of lines

2 Construct the force polygons for the individual joints Thegraphic procedure for this consists of locating the points on theMaxwell diagram that correspond to the remaining letters Ithrough P on the space diagram When all the lettered points onthe diagram are located the complete force polygon for each jointmay be read on the diagram In order to locate these points weuse two relationships The first is that the truss members can re-sist only forces that are parallel to the membersrsquo positioned di-rections Thus we know the directions of all the internal forcesThe second relationship is a simple one from plane geometry apoint may be located at the intersection of two lines Consider theforces at joint 1 as shown in the separated joint diagram in Fig-ure 33 Note that there are four forces and that two of them areknown (the load and the reaction) and two are unknown (the in-ternal forces in the truss members) The force polygon for thisjoint as shown on the Maxwell diagram is read as ABIHA ABrepresents the load BI the force in the upper chord member IH


3751 P-03 111301 1221 PM Page 115

the force in the lower chord member and HA the reaction Thusthe location of point i on the Maxwell diagram is determined bynoting that i must be in a horizontal direction from h (corre-sponding to the horizontal position of the lower chord) and in adirection from b that is parallel to the position of the upper chord

The remaining points on the Maxwell diagram are found by the sameprocess using two known points on the diagram to project lines ofknown direction whose intersection will determine the location of an un-known point Once all the points are located the diagram is complete andcan be used to find the magnitude and sense of each internal force Theprocess for construction of the Maxwell diagram typically consists ofmoving from joint to joint along the truss Once one of the letters for aninternal space is determined on the Maxwell diagram it may be used asa known point for finding the letter for an adjacent space on the space di-agram The only limitation of the process is that it is not possible to findmore than one unknown point on the Maxwell diagram for any singlejoint Consider joint 7 on the separated joint diagram in Figure 33 Tosolve this joint first knowing only the locations of letters a through h onthe Maxwell diagram it is necessary to locate four unknown points l mn and o This is three more unknowns than can be determined in a singlestep so three of the unknowns must be found by using other joints

Solving for a single unknown point on the Maxwell diagram corre-sponds to finding two unknown forces at a joint since each letter on thespace diagram is used twice in the force identification for the internalforces Thus for joint 1 in the previous example the letter I is part of theidentity of forces BI and IH as shown on the separated joint diagramThe graphic determination of single points on the Maxwell diagramtherefore is analogous to finding two unknown quantities in an algebraicsolution As discussed previously two unknowns are the maximum thatcan be solved for in equilibrium of a coplanar concurrent force systemwhich is the condition of the individual joints in the truss

When the Maxwell diagram is completed the internal forces can beread from the diagram as follows

1 The magnitude is determined by measuring the length of the linein the diagram using the scale that was used to plot the vectorsfor the external forces


3751 P-03 111301 1221 PM Page 116

2 The sense of individual forces is determined by reading the forcesin clockwise sequence around a single joint in the space diagramand tracing the same letter sequences on the Maxwell diagram

Figure 34a shows the force system at joint 1 and the force polygonfor these forces as taken from the Maxwell diagram The forces knowninitially are shown as solid lines on the force polygon and the unknownforces are shown as dashed lines Starting with letter A on the force sys-tem we read the forces in a clockwise sequence as AB BI IH and HANote that on the Maxwell diagram moving from a to b is moving in theorder of the sense of the force that is from tail to end of the force vectorthat represents the external load on the joint Using this sequence on theMaxwell diagram this force sense flow will be a continuous one Thusreading from b to i on the Maxwell diagram is reading from tail to headof the force vector which indicates that force BI has its head at the leftend Transferring this sense indication from the Maxwell diagram to thejoint diagram indicates that force BI is in compression that is it is push-ing rather than pulling on the joint Reading from i to h on the Maxwelldiagram shows that the arrowhead for this vector is on the right whichtranslates to a tension effect on the joint diagram

Having solved for the forces at joint 1 as described the fact that theforces in truss members BI and IH are known can be used to consider theadjacent joints 2 and 3 However it should be noted that the sense re-verses at the opposite ends of the members in the joint diagrams Refer-ring to the separated joint diagram in Figure 33 if the upper chordmember shown as force BI in joint 1 is in compression its arrowhead isat the lower left end in the diagram for joint 1 as shown in Figure 34aHowever when the same force is shown as IB at joint 2 its pushing ef-fect on the joint will be indicated by having the arrowhead at the upperright end in the diagram for joint 2 Similarly the tension effect of thelower chord is shown in joint 1 by placing the arrowhead on the right endof the force IH but the same tension force will be indicated in joint 3 byplacing the arrowhead on the left end of the vector for force HI

If the solution sequence of solving joint 1 and then joint 2 is chosenit is now possible to transfer the known force in the upper chord to joint2 Thus the solution for the five forces at joint 2 is reduced to findingthree unknowns since the load BC and the chord force IB are nowknown However it is still not possible to solve joint 2 since there aretwo unknown points on the Maxwell diagram (k and j) corresponding to


3751 P-03 111301 1221 PM Page 117


Figure 34 Graphic solutions for joints 1 2 and 3 (a) Joint 1 (b) Joint 3 (c) Joint 2

3751 P-03 111301 1221 PM Page 118

the three unknown forces An option therefore is to proceed from joint1 to joint 3 at which there are now only two unknown forces On theMaxwell diagram the single unknown point j can be found by projectingvector IJ vertically from i and projecting vector JH horizontally frompoint h Since point i is also located horizontally from point h this showsthat the vector IJ has zero magnitude since both i and j must be on a hor-izontal line from h in the Maxwell diagram This indicates that there isactually no stress in this truss member for this loading condition and thatpoints i and j are coincident on the Maxwell diagram The joint force di-agram and the force polygon for joint 3 are as shown in Figure 34b Inthe joint force diagram place a zero rather than an arrowhead on thevector line for IJ to indicate the zero stress condition In the force poly-gon in Figure 34b the two force vectors are slightly separated for clar-ity although they are actually coincident on the same line

Having solved for the forces at joint 3 proceed to joint 2 since thereremain only two unknown forces at this joint The forces at the joint and the force polygon for joint 2 are shown in Figure 34c As for joint 1read the force polygon in a sequence determined by reading clockwisearound the joint BCKJIB Following the continuous direction of theforce arrows on the force polygon in this sequence it is possible to es-tablish the sense for the two forces CK and KJ

It is possible to proceed from one end and to work continuously acrossthe truss from joint to joint to construct the Maxwell diagram in this ex-ample The sequence in terms of locating points on the Maxwell diagramwould be i-j-k-l-m-n-o-p which would be accomplished by solving thejoints in the following sequence 1 3 2 5 4 6 7 9 8 However it is ad-visable to minimize the error in graphic construction by working fromboth ends of the truss Thus a better procedure would be to find points i-j-k-l-m working from the left end of the truss and then to find points p-o-n-m working from the right end This would result in finding twolocations for the point m whose separation constitutes the error in draft-ing accuracy

Problems 31A BUsing a Maxwell diagram find the internal forces in the trusses in Figure 35


3751 P-03 111301 1221 PM Page 119

32 ALGEBRAIC ANALYSIS OF TRUSSES

Graphical solution for the internal forces in a truss using the Maxwell di-agram corresponds essentially to an algebraic solution by the method ofjoints This method consists of solving the concentric force systems at theindividual joints using simple force equilibrium equations The processwill be illustrated using the previous example

As with the graphic solution first determine the external forces consisting of the loads and the reactions Then proceed to consider theequilibrium of the individual joints following a sequence as in the graphicsolution The limitation of this sequence corresponding to the limit of


Figure 35 Problems 31A B

3751 P-03 111301 1221 PM Page 120

finding only one unknown point in the Maxwell diagram is that only twounknown forces at any single joint can be found in a single step (Twoconditions of equilibrium produce two equations) Referring to Figure36 the solution for joint 1 is as follows

The force system for the joint is drawn with the sense and magnitudeof the known forces shown but with the unknown internal forces repre-sented by lines without arrowheads since their senses and magnitudesinitially are unknown (Figure 36a) For forces that are not vertical orhorizontal replace the forces with their horizontal and vertical compo-nents Then consider the two conditions necessary for the equilibrium ofthe system the sum of the vertical forces is zero and the sum of the hor-izontal forces is zero

ALGEBRAIC ANALYSIS OF TRUSSES 121

Figure 36 Algebraic solution for joint 1 (a) The initial condition (b) Unknownsreduced to components (c) Solution of vertical equilibrium (d ) Solution of hori-zontal equilibrium (e) Final answer

3751 P-03 111301 1221 PM Page 121

If the algebraic solution is performed carefully the sense of the forceswill be determined automatically However it is recommended thatwhenever possible the sense be predetermined by simple observations ofthe joint conditions as will be illustrated in the solutions

The problem to be solved at joint 1 is as shown in Figure 36a In Figure 36b the system is shown with all forces expressed as vertical andhorizontal components Note that although this now increases the num-ber of unknowns to three (IH BIv and BIh) there is a numeric relation-ship between the two components of BI When this condition is added tothe two algebraic conditions for equilibrium the number of usable re-lationships totals three so that the necessary conditions to solve for thethree unknowns are present

The condition for vertical equilibrium is shown in Figure 36c Sincethe horizontal forces do not affect the vertical equilibrium the balance isbetween the load the reaction and the vertical component of the force inthe upper chord Simple observation of the forces and the known magni-tudes makes it obvious that force BIv must act downward indicating thatBI is a compression force Thus the sense of BI is established by simplevisual inspection of the joint and the algebraic equation for vertical equi-librium (with upward force considered positive) is

ΣFv = 0 = +2500 ndash 500 ndash BIv

From this equation BIv is determined to have a magnitude of 2000 lbUsing the known relationships between BI BIv and BIh the values ofthese three quantities can be determined if any one of them is knownThus

from which

and

BI = =1 000

0 5552000 3606

( ) lb

BIh = =0 832

0 5552000 3000

( ) lb

BI BI BIv h

1 000 0 555 0 832 = =


3751 P-03 111301 1221 PM Page 122

The results of the analysis to this point are shown in Figure 36d fromwhich it may be observed that the conditions for equilibrium of the hor-izontal forces can be expressed Stated algebraically (with force sense to-ward the right considered positive) the condition is

ΣFh = 0 = IH ndash 3000

from which it is established that the force in IH is 3000 lbThe final solution for the joint is then as shown in Figure 36e On this

diagram the internal forces are identified as to sense by using C to indi-cate compression and T to indicate tension

As with the graphic solution proceed to consider the forces at joint 3The initial condition at this joint is as shown in Figure 37a with the sin-gle known force in member HI and the two unknown forces in IJ and JHSince the forces at this joint are all vertical and horizontal there is noneed to use components Consideration of vertical equilibrium makes itobvious that it is not possible to have a force in member IJ Stated alge-braically the condition for vertical equilibrium is

ΣFv = 0 = IJ (since IJ is the only force)

It is equally obvious that the force in JH must be equal and oppositeto that in HI since they are the only two horizontal forces That is statedalgebraically

ΣFv = 0 = JH ndash 3000

The final answer for the forces at joint 3 is as shown in Figure 37bNote the convention for indicating a truss member with no internal force


Figure 37 Algebraic solution for joint 3 (a) The initial condition (b) The solution

3751 P-03 111301 1221 PM Page 123

Now proceed to consider joint 2 the initial condition is as shown inFigure 38a Of the five forces at the joint only two remain unknownFollowing the procedure for joint 1 first resolve the forces into their ver-tical and horizontal components as shown in Figure 38b

Since the sense of forces CK and KJ is unknown use the procedure ofconsidering them to be positive until proven otherwise That is if theyare entered into the algebraic equations with an assumed sense and thesolution produces a negative answer then the assumption was wrongHowever be careful to be consistent with the sense of the force vectorsas the following solution will illustrate

Arbitrarily assume that force CK is in compression and force KJ is intension If this is so the forces and their components will be as shown inFigure 38c Then consider the conditions for vertical equilibrium theforces involved will be those shown in Figure 38d and the equation forvertical equilibrium will be

ΣFv = 0 = ndash 1000 + 2000 ndash CKv ndash KJv

or

0 = + 1000 ndash 0555CK ndash 0555KJ (321)

Now consider the conditions for horizontal equilibrium the forceswill be as shown in Figure 38e and the equation will be

ΣFh = 0 = + 3000 ndash CKh + KJh

or

0 = + 3000 ndash 0832CK + 0832KJ (322)

Note the consistency of the algebraic signs and the sense of the forcevectors with positive forces considered as upward and toward the rightNow solve these two equations simultaneously for the two unknownforces as follows

1 Multiply equation (321) by 08320555

00 832

0 5551000

0 832

0 5550 555

0 832

0 5550 555= + + minus + minus

( )

( )

( )CK KJ


3751 P-03 111301 1221 PM Page 124


Figure 38 Algebraic solution for joint 2 (a) The initial condition (b) Unknownsreduced to components (c) Assumed sense of the unknowns for the algebraic solution (d ) Solution of vertical equilibrium (e) Solution of horizontal equilibrium(f ) Final answer in components (g) Final answer in true forces

3751 P-03 111301 1221 PM Page 125

or

0 = + 1500 ndash 0832CK ndash 0832KJ

2 Add this equation to equation (322) and solve for CK

Note that the assumed sense of compression in CK is correct since the al-gebraic solution produces a positive answer Substituting this value forCK in equation (321)

0 = + 1000 ndash 0555(2704) ndash 0555(KJ)

and

Since the algebraic solution produces a negative quantity for KJ the assumed sense for KJ is wrong and the member is actually in compression

The final answers for the forces at joint 2 are as shown in Figure 38gIn order to verify that equilibrium exists however the forces are shown in the form of their vertical and horizontal components in Figure38f

When all of the internal forces have been determined for the truss theresults may be recorded or displayed in a number of ways The most di-rect way is to display them on a scaled diagram of the truss as shown inFigure 39a The force magnitudes are recorded next to each memberwith the sense shown as T for tension or C for compression Zero stressmembers are indicated by the conventional symbol consisting of a zeroplaced directly on the member

When solving by the algebraic method of joints the results may berecorded on a separated joint diagram as shown in Figure 39b If thevalues for the vertical and horizontal components of force in slopingmembers are shown it is a simple matter to verify the equilibrium of theindividual joints

KJ = = minus500

0 555901

lb

0 4500 1 6644500

1 6642704= + minus = =

CK CK lb


3751 P-03 111301 1221 PM Page 126

Problems 32A BUsing the algebraic method of joints find the internal forces in thetrusses in Figure 35

33 THE METHOD OF SECTIONS

Figure 310 shows a simple-span flat-chorded truss with a vertical load-ing on the top chord joints The Maxwell diagram for this loading and theanswers for the internal forces are also shown in the figure This solution

THE METHOD OF SECTIONS 127

Figure 39 Presentation of the internal forces in the truss (a) Member forces (b)Separated joint diagram

3751 P-03 111301 1221 PM Page 127

is provided as a reference for comparison with the results that will be ob-tained by the method of sections

In Figure 311 the truss is shown with a cut plane passing verticallythrough the third panel The free-body diagram of the portion of the trussto the left of this cut plane is shown in Figure 311a The internal forces


Figure 310 Graphic solution for the flat-chorded truss

3751 P-03 111301 1221 PM Page 128

in the three cut members become external forces on this free body andtheir values may be found using the following analysis of the static equi-librium of the free body

In Figure 311b we observe the condition for vertical equilibriumSince ON is the only cut member with a vertical force component it


Figure 311 Investigation of the truss by the method of sections

3751 P-03 111301 1221 PM Page 129

must be used to balance the other external forces resulting in the valuefor ONv of 500 lb acting downward With the angle of inclination of thismember known the horizontal component and the true force in the mem-ber can now be found

We next consider a condition of equilibrium of moments selecting acenter of moments as a point that will eliminate all but one of the un-known forces thus producing a single algebraic equation with only oneunknown Selecting the top chord joint as shown in Figure 311c boththe force in the top chord and in member ON are eliminated Then theonly remaining unknown force is that in the bottom chord (member NI)and the summation is

ΣM = 0 = +(3000 times 24) ndash(500 times 24) ndash(1000 times 12) ndash(NI times 10)

or

Note that the sense of the force in NI was assumed to be tension and thesign used for NI in the moment summation was based on this assumption

One way to find the force in the top chord is to do a summation of hor-izontal forces since the horizontal component of ON and the force in NIare now known An alternative would be to use another moment sum-mation this time selecting the bottom chord joint shown in Figure 311din order to eliminate IN and ON from the summation equation

ΣM2 = 0 = +(3000 times 36) ndash(500 times 36) ndash(1000 times 24) ndash(1000 times 12) ndash(DO times 10)

Thus

The forces in all of the horizontal and diagonal members of the trussmay be found by cutting sections and writing equilibrium equations sim-ilar to the process just illustrated In order to find the forces in the verti-cal members it is possible to cut the truss with an angled plane as shown

DO = =54 000

105400

lb

10 72 000 12 000 12 000 48 000

48 000

104800

( )

NI

NI

= + minus minus = +

= = lb


3751 P-03 111301 1221 PM Page 130

in Figure 312 A summation of vertical forces on this free body willyield the internal force of 1500 lb in compression in member MN

The method of sections is sometimes useful when it is desired to findthe internal force in individual members of a truss without doing a com-plete analysis for all of the members

Problems 33A BFind the internal forces in the members of the trusses in Figure 313using (1) a Maxwell diagram (2) the algebraic method of sections


Figure 312 Cut section used to find theforce in the vertical members


3751 P-03 111301 1221 PM Page 131

132

4ANALYSIS OF BEAMS

A beam is a structural member that resists transverse loads The supportsfor beams are usually at or near the ends and the supporting upwardforces are called reactions The loads acting on a beam tend to bend itrather than shorten or lengthen it Girder is the name given to a beam thatsupports smaller beams all girders are beams insofar as their structuralaction is concerned For construction usage beams carry various namesdepending on the form of construction these include purlin joist rafterlintel header and girt Figure 41 shows a floor structure achieved withclosely spaced wood beams (called joists when occurring in this situa-tion) that are supported by larger wood beams which are in turn sup-ported by masonry bearing walls or wood columns This classic systemis extensively used although the materials and elements utilized and thedetails of the construction all change over time

3751 P-04 111301 1221 PM Page 132

41 TYPES OF BEAMS

There are in general five types of beams which are identified by thenumber kind and position of the supports Figure 42 shows diagram-matically the different types and also the shape each beam tends to as-sume as it bends (deforms) under the loading In ordinary steel orreinforced concrete beams these deformations are not usually visible tothe eye but some deformation is always present

A simple beam rests on a support at each end the ends of the beambeing free to rotate (Figure 42a)

A cantilever beam is supported at one end only A beam embedded ina wall and projecting beyond the face of the wall is a typical ex-ample (Figure 42b)

An overhanging beam is a beam whose end or ends project beyond itssupports Figure 42c indicates a beam overhanging one support only

TYPES OF BEAMS 133

Figure 41 Beams were the earliest elements used to achieve spanning struc-turesmdashfirst in the form of untreated cut tree trunks and then as tools were devel-oped in more useful shaped forms Large beams used for long spans usually carrypoint loadings from other structural elements such as the joists shown here hungfrom the timber beam Lighter beams such as the joists typically carry a uniformlydistributed load from a directly attached deck Although developed in wood thisclassic system is emulated in steel and concrete Reproduced from Architects andBuilders Handbook by H Parker and F Kidder 1931 with permission of the pub-lisher John Wiley amp Sons New York

3751 P-04 111301 1221 PM Page 133

A continuous beam rests on more than two supports (Figure 42d )Continuous beams are commonly used in reinforced concrete andwelded steel construction

A restrained beam has one or both ends restrained or fixed against ro-tation (Figure 42e)

42 LOADS AND REACTIONS

Beams are acted on by external forces that consist of the loads and the re-action forces developed by the beamrsquos supports The two types of loadsthat commonly occur on beams are called concentrated and distributedA concentrated load is assumed to act at a definite point such a load isthat caused when one beam supports another beam A distributed load isone that acts over a considerable length of the beam such a load is onecaused by a floor deck supported directly by a beam If the distributedload exerts a force of equal magnitude for each unit of length of thebeam it is known as a uniformly distributed load The weight of a beam

134 ANALYSIS OF BEAMS

Figure 42 Types of beams (a) Simple (b) Cantilever (c) Overhanging (d ) Con-tinuous (e) Restrained

3751 P-04 111301 1221 PM Page 134

is a uniformly distributed load that extends over the entire length of thebeam However some uniformly distributed loadings supported by thebeam may extend over only a portion of the beam length

Reactions are the upward forces acting at the supports that hold inequilibrium the downward forces or loads The left and right reactions ofa simple beam are usually called R1 and R2 respectively Determinationof reactions for simple beams is achieved with the use of equilibriumconditions for parallel force systems as demonstrated in Section 212

Figure 43a shows a portion of a floor framing plan The diagonalcrosshatching represents the area supported by one of the beams Thisarea is 8 times 20 ft the dimensions of the beam spacing and the beam spanThe beam is supported at each end by girders that span between the sup-porting columns If the total load on the crosshatched area is 100 psf thenthe total load on the beam is determined as

W = 8 times 20 times 100 = 16000 lb or 16 kips

It is common to designate this total load as W using the capital form ofthe letter However for a uniformly distributed load the loading mayalso be expressed in the form of a unit load per unit of length of the beamThis unit load is designated by w using the lowercase form Thus forthis beam

For the beam in Figure 43 the load is symmetrically placed and the tworeactions will thus each be one-half of the total load The reactions aredeveloped as concentrated loads on the girders The loading diagrams forthe beam and girder are as shown in Figures 43b and c

For unsymmetrical beam loadings the reaction forces can be deter-mined by the procedures demonstrated in Section 212

43 SHEAR IN BEAMS

Figure 44a represents a simple beam with a uniformly distributed loadover its entire length Examination of an actual beam so loaded would

w = =16 000

20800

lb ft or 800 plf (pounds per lineal foot)

SHEAR IN BEAMS 135

3751 P-04 111301 1221 PM Page 135


Figure 43 Determination of beam loads and display of the loaded beams for aframing system (a) Plan (b) Loading diagram for the beam (c) Loading diagramfor the girder

probably not reveal any effects of the loading on the beam Howeverthere are three distinct major tendencies for the beam to fail Figures44bndashd illustrate the three phenomena

First there is a tendency for the beam to fail by dropping between thesupports (Figure 44b) This is called vertical shear Second the beammay fail by bending (Figure 44c) Third there is a tendency in woodbeams for the fibers of the beam to slide past each other in a horizontaldirection (Figure 44d ) an action described as horizontal shear Natu-rally a beam properly designed does not fail in any of the ways justmentioned but these tendencies to fail are always present and must beconsidered in structural design

3751 P-04 111301 1221 PM Page 136

Vertical Shear

Vertical shear is the tendency for one part of a beam to move verticallywith respect to an adjacent part The magnitude of the shear force at anysection in the length of a beam is equal to the algebraic sum of the verti-cal forces on either side of the section Vertical shear is usually repre-sented by the letter V In computing its values in the examples andproblems consider the forces to the left of the section but keep in mindthat the same resulting force magnitude will be obtained with the forceson the right To find the magnitude of the vertical shear at any section inthe length of a beam simply add up the forces to the right or the left ofthe section It follows from this procedure that the maximum value of theshear for simple beams is equal to the greater reaction

Example 1 Figure 45a illustrates a simple beam with concentratedloads of 600 lb and 1000 lb The problem is to find the value of the ver-tical shear at various points along the length of the beam Although theweight of the beam constitutes a uniformly distributed load it is ne-glected in this example

Solution The reactions are computed as previously described and arefound to be R1 = 1000 lb and R2 = 600 lb

Consider next the value of the vertical shear V at an infinitely shortdistance to the right of R1 Applying the rule that the shear is equal to thereaction minus the loads to the left of the section we write

V = R1 ndash 0 or V = 1000 lb

SHEAR IN BEAMS 137

Figure 44 Characteristic forms of failure for a simple beam (a) Beam withuniformly distributed load (b) Vertical shear (c) Bending (d ) Horizontal shear

3751 P-04 111301 1221 PM Page 137

The zero represents the value of the loads to the left of the section whichof course is zero Now take a section 1 ft to the right of R1 again

V(x = 1) = R1 ndash 0 or V(x = 1) = 1000 lb

The subscript (x = 1) indicates the position of the section at which theshear is taken the distance of the section from R1 At this section theshear is still 1000 lb and has the same magnitude up to the 600-lb load

The next section to consider is a very short distance to the right of the600-lb load At this section

V(x = 2+) = 1000 ndash 600 = 400 lb

Because there are no loads intervening the shear continues to be thesame magnitude up to the 1000-lb load At a section a short distance tothe right of the 1000-lb load

V(x = 6+) = 1000 ndash (600 + 1000) = ndash600 lb

This magnitude continues up to the right-hand reaction R2



3751 P-04 111301 1221 PM Page 138

Example 2 The beam shown in Figure 45b supports a concentratedload of 12000 lb located 6 ft from R2 and a uniformly distributed load of800 pounds per linear foot (lbft) over its entire length Compute thevalue of vertical shear at various sections along the span

Solution By use of the equations of equilibrium the reactions are deter-mined to be R1 = 10900 lb and R2 = 13900 lb Note that the total distrib-uted load is 800 times 16 = 12800 lb Now consider the vertical shear forceat the following sections at a distance measured from the left support

V(x = 0) = 10900 ndash 0 = 10900 lbV(x = 1) = 10900 ndash (800 times 1) = 10100 lbV(x = 5) = 10900 ndash (800 times 5) = 6900 lbV(x = 10ndash) = 10900 ndash (800 times 10) = 2900 lbV(x = 10+) = 10900 ndash (800 times 10) + 12000) = ndash9100 lbV(x = 16) = 10900 ndash (800 times 16) + 12000) = ndash13900

Shear Diagrams

In the two preceding examples the value of the shear at several sectionsalong the length of the beams was computed In order to visualize the re-sults it is common practice to plot these values on a diagram called theshear diagram which is constructed as explained below

To make such a diagram first draw the beam to scale and locate theloads This has been done in Figures 46a and b by repeating the load di-agrams of Figures 45a and b respectively Beneath the beam draw ahorizontal baseline representing zero shear Above and below this lineplot at any convenient scale the values of the shear at the various sec-tions the positive or plus values are placed above the line and the neg-ative or minus values below In Figure 46a for instance the value ofthe shear at R1 is +1000 lb The shear continues to have the same valueup to the load of 600 lb at which point it drops to 400 lb The samevalue continues up to the next load 1000 lb where it drops to ndash600 lband continues to the right-hand reaction Obviously to draw a shear di-agram it is necessary to compute the values at significant points onlyHaving made the diagram we may readily find the value of the shear at any section of the beam by scaling the vertical distance in the dia-gram The shear diagram for the beam in Figure 46b is made in the same manner

SHEAR IN BEAMS 139

3751 P-04 111301 1221 PM Page 139

There are two important facts to note concerning the vertical shearThe first is the maximum value The diagrams in each case confirm theearlier observation that the maximum shear is at the reaction having thegreater value and its magnitude is equal to that of the greater reaction InFigure 46a the maximum shear is 1000 lb and in Figure 46b it is13900 lb We disregard the positive or negative signs in reading themaximum values of the shear for the diagrams are merely conventionalmethods of representing the absolute numerical values

Another important fact to note is the point at which the shear changesfrom a plus to a minus quantity We call this the point at which the shearpasses through zero In Figure 46a it is under the 1000-lb load 6 ft fromR1 In Figure 46b it is under the 12000-lb load 10 ft from R1 A major con-cern for noting this point is that it indicates the location of the maximumvalue of bending moment in the beam as discussed in the next section

Problems 43AndashFFor the beams shown in Figures 47andashf draw the shear diagrams and noteall critical values for shear Note particularly the maximum value forshear and the point at which the shear passes through zero

44 BENDING MOMENTS IN BEAMS

The forces that tend to cause bending in a beam are the reactions and theloads Consider the section X-X 6 ft from R1 (Figure 48) The force R1


Figure 46 Construction of shear diagrams

3751 P-04 111301 1221 PM Page 140

141

Fig

ure

47

Pro

blem

s 4

3A

ndashF

3751 P-04 111301 1221 PM Page 141

or 2000 lb tends to cause a clockwise rotation about this point Because theforce is 2000 lb and the lever arm is 6 ft the moment of the force is 2000times 6 = 12000 ft-lb This same value may be found by considering the forcesto the right of the section X-X R2 which is 6000 lb and the load 8000 lbwith lever arms of 10 and 6 ft respectively The moment of the reaction is6000 times 10 = 60000 ft-lb and its direction is counterclockwise with respectto the section X-X The moment of force 8000 lb is 8000 times 6 = 48000 ft-lb and its direction is clockwise Then 60000 ft-lb ndash 48000 ft-lb =12000 ft-lb the resultant moment tending to cause counterclockwise rota-tion about the section X-X This is the same magnitude as the moment ofthe forces on the left which tends to cause a clockwise rotation

Thus it makes no difference whether use is made of the forces to theright of the section or the left the magnitude of the moment is the sameIt is called the bending moment (or the internal bending moment) becauseit is the moment of the forces that causes bending stresses in the beam Itsmagnitude varies throughout the length of the beam For instance at 4 ftfrom R1 it is only 2000 times 4 or 8000 ft-lb The bending moment is the al-gebraic sum of the moments of the forces on either side of the sectionFor simplicity take the forces on the left then the bending moment atany section of a beam is equal to the moments of the reactions minus themoments of the loads to the left of the section Because the bending mo-ment is the result of multiplying forces by distances the denominationsare foot-pounds or kip-feet

Bending Moment Diagrams

The construction of bending moment diagrams follows the procedureused for shear diagrams The beam span is drawn to scale showing the


Figure 48 Development of bending at a selected cross section

3751 P-04 111301 1221 PM Page 142

locations of the loads Below this and usually below the shear diagrama horizontal baseline is drawn representing zero bending moment Thenthe bending moments are computed at various sections along the beamspan and the values are plotted vertically to any convenient scale In sim-ple beams all bending moments are positive and therefore are plottedabove the baseline In overhanging or continuous beams there are alsonegative moments and these are plotted below the baseline

Example 3 The load diagram in Figure 49 shows a simple beam withtwo concentrated loads Draw the shear and bending moment diagrams

Solution R1 and R2 are computed first and are found to be 16000 lb and14000 lb respectively These values are recorded on the load diagram

The shear diagram is drawn as described in Section 43 Note that inthis instance it is only necessary to compute the shear at one section(between the concentrated loads) because there is no distributed loadand we know that the shear at the supports is equal in magnitude to thereactions

Because the value of the bending moment at any section of the beamis equal to the moments of the reactions minus the moments of the loadsto the left of the section the moment at R1 must be zero for there are noforces to the left Other values in the length of the beam are computed asfollows The subscripts (x = 1 etc) show the distance from R1 at whichthe bending moment is computed

M(x = 1)2 = (16000 times 1) = 16000 ft-lbM(x = 2)2 = (16000 times 2) = 32000 ft-lbM(x = 5)2 = (16000 times 5) ndash (12000 times 3) = 44000 ft-lbM(x = 8)2 = (16000 times 8) ndash (12000 times 6) = 56000 ft-lbM(x = 10) = (16000 times 10) ndash (12000 times 8) + (18000 times 2) = 28000 ft-lbM(x = 12) = (16000 times 12) ndash (12000 times 10) + (18000 times 4) = 0

The result of plotting these values is shown in the bending moment di-agram of Figure 49 More moments were computed than were necessaryWe know that the bending moments at the supports of simple beams arezero and in this instance only the bending moments directly under theloads were needed

BENDING MOMENTS IN BEAMS 143

3751 P-04 111301 1221 PM Page 143

Relations Between Shear and Bending Moment

In simple beams the shear diagram passes through zero at some point be-tween the supports As stated earlier an important principle in this re-spect is that the bending moment has a maximum magnitude whereverthe shear passes through zero In Figure 49 the shear passes throughzero under the 18000-lb load that is at x = 8 ft Note that the bendingmoment has its greatest value at this same point 56000 ft-lb

Example 4 Draw the shear and bending moment diagrams for the beamshown in Figure 410 which carries a uniformly distributed load of 400lbft and a concentrated load of 21000 lb located 4 ft from R1

Solution Computing the reactions we find R1 = 17800 lb and R2 =8800 lb By use of the process described in Section 43 the critical shearvalues are determined and the shear diagram is drawn as shown in thefigure


Figure 49 Example 3

3751 P-04 111301 1221 PM Page 144

Although the only value of bending moment that must be computed isthat where the shear passes through zero some additional values are de-termined in order to plot the true form of the moment diagram Thus

M(x = 2)2 = (17800 times 2) ndash (400 times 2 times 1) = 34800 ft-lbM(x = 4)2 = (17800 times 4) ndash (400 times 4 times 2) = 68000 ft-lbM(x = 8)2 = (17800 times 8) ndash (400 times 8 times 4) + (21000 times 4) = 45600 ft-lbM(x = 12) = (17800 times 12) ndash (400 times 12 times 6) + (21000 times 8) = 16800 ft-lb

From the two preceding examples (Figures 49 and 410) it will be ob-served that the shear diagram for the parts of the beam on which no loadsoccur is represented by horizontal lines For the parts of the beam onwhich a uniformly distributed load occurs the shear diagram consists ofstraight inclined lines The bending moment diagram is represented bystraight inclined lines when only concentrated loads occur and by acurved line if the load is distributed



3751 P-04 111301 1221 PM Page 145

Occasionally when a beam has both concentrated and uniformly dis-tributed loads the shear does not pass through zero under one of the con-centrated loads This frequently occurs when the distributed load isrelatively large compared with the concentrated loads Since it is neces-sary in designing beams to find the maximum bending moment we mustknow the point at which it occurs This of course is the point where theshear passes through zero and its location is readily determined by theprocedure illustrated in the following example

Example 5 The load diagram in Figure 411 shows a beam with a con-centrated load of 7000 lb applied 4 ft from the left reaction and a uni-formly distributed load of 800 lbft extending over the full spanCompute the maximum bending moment on the beam

Solution The values of the reactions are found to be R1 = 10600 lb andR2 = 7600 lb and are recorded on the load diagram

The shear diagram is constructed and it is observed that the shearpasses through zero at some point between the concentrated load of 7000lb and the right reaction Call this distance x ft from R2 The value of theshear at this section is zero therefore an expression for the shear for thispoint using the reaction and loads is equal to zero This equation con-tains the distance x

V x xx( ) at ft= minus + = = =7600 800 07600

8009 5



3751 P-04 111301 1221 PM Page 146

The zero shear point is thus at 95 ft from the right support and (as shownin the diagram) at 45 ft from the left support This location can also bedetermined by writing an equation for the summation of shear from theleft of the point which should produce the answer of 45 ft

Following the convention of summing up the moments from the leftof the section the maximum moment is determined as

Problems 44AndashFDraw the shear and bending moment diagrams for the beams in Figure47 indicating all critical values for shear and moment and all significantdimensions (Note These are the beams for Problem 43 for which theshear diagrams were constructed)

45 SENSE OF BENDING IN BEAMS

When a simple beam bends it has a tendency to assume the shape shownin Figure 412a In this case the fibers in the upper part of the beam arein compression For this condition the bending moment is considered aspositive Another way to describe a positive bending moment is to saythat it is positive when the curve assumed by the bent beam is concaveupward When a beam projects beyond a support (Figure 412b) thisportion of the beam has tensile stresses in the upper part The bendingmoment for this condition is called negative the beam is bent concavedownward When constructing moment diagrams following the methodpreviously described the positive and negative moments are showngraphically

M x( ) ( ) ( )

ft-lb= = + times minus times minus times times

=4 5 10 600 4 5 7000 0 5 800 4 5

4 5

236 100

SENSE OF BENDING IN BEAMS 147

Figure 412 Sign of internal bending moment bending stress convention

3751 P-04 111301 1221 PM Page 147

Example 6 Draw the shear and bending moment diagrams for the over-hanging beam shown in Figure 413

Solution Computing the reactions

From ΣM about R1 R2 times 12 = 600 times 16 times 8 R2 = 6400 lbFrom ΣM about R2 R1 times 12 = 600 times 16 times 4 R1 = 3200 lb

With the reactions determined the construction of the shear diagram isquite evident For the location of the point of zero shear considering itsdistance from the left support as x

3200 ndash 600x = 0 x = 533 ft



3751 P-04 111301 1221 PM Page 148

For the critical values needed to plot the moment diagram

The form of the moment diagram for the distributed loading is a curve(parabolic) which may be verified by plotting some additional points onthe graph

For this case the shear diagram passes through zero twice both ofwhich points indicate peaks of the moment diagrammdashone positive andone negative As the peak in the positive portion of the moment diagramis actually the apex of the parabola the location of the zero momentvalue is simply twice the value previously determined as x This pointcorresponds to the change in the form of curvature on the elastic curve(deflected shape) of the beam this point is described as the inflectionpoint for the deflected shape The location of the point of zero momentcan also be determined by writing an equation for the sum of moments atthe unknown location In this case calling the new unknown point x

Solution of this quadratic equation should produce the value of x =1067 ft

Example 7 Compute the maximum bending moment for the overhang-ing beam shown in Figure 414

Solution Computing the reactions gives R1 = 3200 lb and R2 = 2800 lbAs usual the shear diagram can now be plotted as the graph of the loadsand reactions proceeding from left to right Note that the shear passesthrough zero at the location of the 4000-lb load and at both supports Asusual these are clues to the form of the moment diagram

With the usual moment summations values for the moment diagramcan now be found at the locations of the supports and all of the concen-trated loads From this plot it will be noted that there are two inflectionpoints (locations of zero moment) As the moment diagram is composedof straight-line segments in this case the locations of these points may be

M x xx= = + times minus times times

0 3200 600

2( )

M

M

x

x

( )

( )

( )

( ) ( )

ft-lb

ft-lb

=

=

= + times minus times times

=

= times minus times times = minus

5 33

12

3200 5 33 600 5 335 33

28533

3200 12 600 12 6 4800


3751 P-04 111301 1221 PM Page 149

found by writing simple linear equations for their locations Howeveruse can also be made of some relationships between the shear and mo-ment graphs One of these has already been used relating to the correla-tion of zero shear and maximum moment Another relationship is that thechange of the value of moment between any two points along the beamis equal to the total area of the shear diagram between the points If thevalue of moment is known at some point it is thus a simple matter to findvalues at other points For example starting from the left end the valueof moment is known to be zero at the left end of the beam then the valueof the moment at the support is the area of the rectangle on the shear



3751 P-04 111301 1221 PM Page 150

diagram with base of 4 ft and height of 800 lbmdashthe area being 4 times 800 =3200 ft-lb

Now proceeding along the beam to the point of zero moment (call itx distance from the support) the change is again 3200 which relates toan area of the shear diagram that is x times 2400 Thus

And now calling the distance from the right support to the point of zeromoment x

Problems 45AndashDDraw the shear and bending moment diagrams for the beams in Figure415 indicating all critical values for shear and moment and all signifi-cant dimensions

46 CANTILEVER BEAMS

In order to keep the signs for shear and moment consistent with those forother beams it is convenient to draw a cantilever beam with its fixed endto the right as shown in Figure 416 We then plot the values for the shearand moment on the diagrams as before proceeding from the left end

Example 8 The cantilever beam shown in Figure 416a projects 12 ftfrom the face of the wall and has a concentrated load of 800 lb at the un-supported end Draw the shear and moment diagrams What are the val-ues of the maximum shear and maximum bending moment

Solution The value of the shear is ndash800 lb throughout the entire lengthof the beam The bending moment is maximum at the wall its value is800 times 12 = ndash9600 ft-lb The shear and moment diagrams are as shown inFigure 416a Note that the moment is all negative for the cantilever beamcorresponding to its concave downward shape throughout its length

2600 400400

26000 154x x= = = ft

2400 32003200

24001 33x x= = = ft

CANTILEVER BEAMS 151

3751 P-04 111301 1221 PM Page 151



3751 P-04 111301 1221 PM Page 152

Although they are not shown in the figure the reactions in this caseare a combination of an upward force of 800 lb and a clockwise resistingmoment of 9600 ft-lb

Example 9 Draw the shear and bending moment diagrams for the can-tilever beam shown in Figure 416b which carries a uniformly distrib-uted load of 500 lbft over its full length

Solution The total load is 500 times 10 = 5000 lb The reactions are an up-ward force of 5000 lb and a moment determined as

whichmdashit may be notedmdashis also the total area of the shear diagram be-tween the outer end and the support

Example 10 The cantilever beam indicated in Figure 417 has a con-centrated load of 2000 lb and a uniformly distributed load of 600 lbft atthe positions shown Draw the shear and bending moment diagrams

M = minus times times = minus500 1010

225 000 ft-lb



3751 P-04 111301 1221 PM Page 153

What are the magnitudes of the maximum shear and maximum bendingmoment

Solution The reactions are actually equal to the maximum shear andbending moment Determined directly from the forces they are

The diagrams are quite easily determined The other moment valueneeded for the moment diagram can be obtained from the moment of theconcentrated load or from the simple rectangle of the shear diagram2000 times 8 = 16000 ft-lb

Note that the moment diagram has a straight-line shape from the outerend to the beginning of the distributed load and becomes a curve fromthis point to the support

It is suggested that Example 10 be reworked with Figure 417 re-versed left for right All numerical results will be the same but the sheardiagram will be positive over its full length

V

M

= + times =

= minus times minus times times

= minus

2000 600 6 5600

2000 14 600 66

238 800

( )

( )

lb

ft-lb



3751 P-04 111301 1221 PM Page 154


47 TABULATED VALUES FOR BEAM BEHAVIOR

Bending Moment Formulas

The methods of computing beam reactions shears and bending momentspresented thus far in this chapter make it possible to find critical valuesfor design under a wide variety of loading conditions However certainconditions occur so frequently that it is convenient to use formulas thatgive the maximum values directly Structural design handbooks containmany such formulas two of the most commonly used formulas are de-rived in the following examples

TABULATED VALUES FOR BEAM BEHAVIOR 155


3751 P-04 111301 1221 PM Page 155

Simple Beam Concentrated Load at Center of Span

A simple beam with a concentrated load at the center of the span occursvery frequently in practice Call the load P and the span length betweensupports L as indicated in the load diagram of Figure 419a For thissymmetrical loading each reaction is P2 and it is readily apparent thatthe shear will pass through zero at distance x = L2 from R1 Thereforethe maximum bending moment occurs at the center of the span under theload Computing the value of the bending moment at this section

Example 11 A simple beam 20 ft in length has a concentrated load of 8000 lb at the center of the span Compute the maximum bending moment

Solution As just derived the formula giving the value of the maximumbending moment for this condition is M = PL 4 Therefore

Simple Beam Uniformly Distributed Load

This is probably the most common beam loading it occurs time andagain For any beam its own dead weight as a load to be carried is usu-ally of this form Call the span L and the unit load w as indicated in Fig-ure 419b The total load on the beam is W = wL hence each reaction isW2 or wL 2 The maximum bending moment occurs at the center of thespan at distance L 2 from R1 Writing the value of M for this section

Note the alternative use of the unit load w or the total load W in this for-mula Both forms will be seen in various references It is important tocarefully identify the use of one or the other

MwL L

wL L wL WL= + times

minus times times

=

2 2 2 4 8 8

2

or

MPL= = times =4

8000 20

440 000 ft-lb

MP L PL= times =2 2 4


3751 P-04 111301 1221 PM Page 156

Example 12 A simple beam 14 ft long has a uniformly distributed loadof 800 lbft Compute the maximum bending moment

Solution As just derived the formula that gives the maximum bendingmoment for a simple beam with uniformly distributed load is M = wL28Substituting these values

or using the total load of 800 times 14 = 11200 lb

Use of Tabulated Values for Beams

Some of the most common beam loadings are shown in Figure 420 Inaddition to the formulas for the reactions R for maximum shear V andfor maximum bending moment M expressions for maximum deflection

MWL= = times =8

11 200 14

819 600

ft-lb

MwL= = times =

2 2

8

800 14

819 600 ft-lb


Figure 419 Loading and internal force diagrams for simple beams

3751 P-04 111301 1221 PM Page 157

D (or ∆) are given also Deflections formulas are not discussed in thischapter but are considered in Chapter 11

In Figure 420 if the loads P and W are in pounds or kips the verticalshear V will also be in units of pounds or kips When the loads are givenin pounds or kips and the span in feet the bending moment M will be inunits of foot-pounds or kip-feet


Figure 420 Values and diagrams for typical beam loadings

3751 P-04 111301 1221 PM Page 158

Problem 47AA simple-span beam has two concentrated loads of 4 kips [178 kN]each placed at the third points of the 24-ft [732-m] span Find the valuefor the maximum bending moment in the beam

Problem 47BA simple-span beam has a uniformly distributed load of 25 kipsft [365kNm] on a span of 18 ft [549 m] Find the value for the maximum bend-ing moment in the beam

Problem 47CA simple beam with a span of 32 ft [9745 m] has a concentrated load of12 kips [534 kN] at 12 ft [366 m] from one end Find the value for themaximum bending moment in the beam

Problem 47DA simple beam with a span of 36 ft [1097 m] has a distributed load thatvaries from a value of 0 at its ends to a maximum of 1000 lbft [1459kNm] at its center (Case 8 in Figure 420) Find the value for the maxi-mum bending moment in the beam


3751 P-04 111301 1221 PM Page 159

160

5CONTINUOUS AND

RESTRAINED BEAMS

Beams were used in combination with vertical posts in ancient culturesto produce early framed structures and this type of structure continues tobe used today In some forms of modern construction however a newfactor is the development of continuous members consisting of multiple-span beams and multistory columns (see Figure 51) In these forms ofconstruction beams are continuous through adjacent spans and some-times are restrained at their ends by rigid attachment to columns Thischapter presents some basic considerations for continuity and end re-straint for beams

51 BENDING MOMENTS FOR CONTINUOUS BEAMS

It is beyond the scope of this book to give a detailed discussion of bend-ing in members continuous over supports but the material presented inthis section will serve as an introduction to the subject A continuousbeam is a beam that rests on more than two supports Continuous beams

3751 P-05 111301 1222 PM Page 160

are characteristic of sitecast concrete construction but occur less often inwood and steel construction

The concepts underlying continuity and bending under restraint are il-lustrated in Figure 52 Figure 52a represents a single beam resting onthree supports and carrying equal loads at the centers of the two spans Ifthe beam is cut over the middle support as shown in Figure 52b the

BENDING MOMENTS FOR CONTINOUS BEAMS 161

Figure 51 Elaborate wood construction for forming of a sitecast concrete slaband beam structure supported by concrete columns Reproduced from Architectsand Builders Handbook by H Parker and F Kidder 1931 with permission of thepublisher John Wiley amp Sons New York The continuously cast concrete structureintroduced a degree of structural continuity not formerly experienced with ordinarywood and steel constructions necessitating more complex investigations for struc-tural behaviors to support design work

3751 P-05 111301 1222 PM Page 161

result will be two simple beams Each of these simple beams will deflectas shown However when the beam is made continuous over the middlesupport its deflected form is as indicated in Figure 52a

It is evident that there is no bending moment developed over the mid-dle support in Figure 52b while there must be a moment over the sup-port in Figure 52a In both cases there is positive moment at themidspan that is there is tension in the bottom and compression in the topof the beam at these locations In the continuous beam however there isa negative moment over the middle support that is there is tension in thetop and compression in the bottom of the beam The effect of the nega-tive moment over the support is to reduce the magnitudes of both maxi-mum bending moment and deflection at midspan which is a principaladvantage of continuity

Values for reaction forces and bending moments cannot be found forcontinuous beams by use of the equations for static equilibrium aloneFor example the beam in Figure 52a has three unknown reaction forceswhich constitute a parallel force system with the loads For this condi-tion there are only two conditions of equilibrium and thus only twoavailable equations for solving for the three unknowns This presents asituation in algebra that is qualified as indeterminate and the structure soqualified is said to be statically indeterminate

Solutions for investigation of indeterminate structures require addi-tional conditions to supplement those available from simple staticsThese additional conditions are derived from the deformation and thestress mechanisms of the structure Various methods for investigation of

162 CONTINUOUS AND RESTRAINED BEAMS

Figure 52 Deflected shape of the two-span beam (a) As a single-piece two-span member (b) With two separate pieces

3751 P-05 111301 1222 PM Page 162

indeterminate structures have been developed Of particular interest noware those that yield to application to computer-aided processes Just aboutany structure with any degree of indeterminacy can now be investigatedwith readily available programs

A procedural problem with highly indeterminate structures is thatsomething about the structure must be determined before an investigationcan be performed Useful for this purpose are shortcut methods that givereasonably approximate answers without an extensive investigation

Theorem of Three Moments

One method for determining reactions and constructing the shear andbending moment diagrams for continuous beams is based on the theoremof three moments This theorem deals with the relation among the bend-ing moments at any three consecutive supports of a continuous beam Ap-plication of the theorem produces an equation called the three-momentequation The three-moment equation for a continuous beam of two spanswith uniformly distributed loading and constant moment of inertia is

in which the various terms are as shown in Figure 53 The following ex-amples demonstrate the use of this equation

M L M L L M Lw L w L

1 1 2 1 2 3 21 1

32 2

3

24 4

+ + + = minus minus( )


Figure 53 Diagrams for the two-span beam with uniform load

3751 P-05 111301 1222 PM Page 163

Continuous Beam with Two Equal Spans

This is the simplest case with the formula reduced by the symmetry plusthe elimination of M1 and M2 due to the discontinuity of the beam at itsouter ends The equation is reduced to

With the loads and spans as given data a solution for this case is reducedto solving for M2 the negative moment at the center support Transform-ing the equation produces a form for direct solution of the unknown mo-ment thus

With this moment determined it is possible to now use the availableconditions of statics to solve the rest of the data for the beam The fol-lowing example demonstrates the process

Example 1 Compute the values for the reactions and construct the shearand moment diagrams for the beam shown in Figure 54a

Solution With only two conditions of statics for the parallel force sys-tem it is not possible to solve directly for the three unknown reactionsHowever use of the equation for the moment at the middle support yieldsa condition that can be used as shown in the following work

Next an equation for the bending moment at 10 ft to the right of the leftsupport is written in the usual manner and is equated to the now knownvalue of 1250 ft-lb

M(x = 10) = (R1 times 10) ndash (100 times 10 times 5) = ndash1250 ft-lb

from which

10R1 = 3750 R1 = 375 lb

MwL

2

2 2

8

100 10

81250= minus = minus = minus( )( )

ft-lb

MwL

2

2

8= minus

42

2

3

MwL= minus


3751 P-05 111301 1222 PM Page 164

By symmetry this is also the value for R3 The value for R2 can thenbe found by a summation of vertical forces thus

ΣFV = 0 = (375 + 375 + R2) ndash (100 times 20) R2 = 1250 lb

Sufficient data have now been determined to permit the complete con-struction of the shear diagram as shown in Figure 54b The location ofzero shear is determined by the equation for shear at the unknown dis-tance x from the left support

375 ndash (100 times x) = 0 x = 375 ft

The maximum value for positive moment at this location can be deter-mined with a moment summation or by finding the area of the shear dia-gram between the end and the zero shear location


Figure 54 Example 1

3751 P-05 111301 1222 PM Page 165

Because of symmetry the location of zero moment is determined as twicethe distance of the zero shear point from the left support Sufficient data arenow available to plot the moment diagram as shown in Figure 54c

Problems 51A BUsing the three-moment equation find the bending moments and reac-tions and draw the complete shear and moment diagrams for the follow-ing beams that are continuous over two equal spans and carry uniformlydistributed loadings

Beam Span Length ft Load lbft

A 16 200B 24 350

Continuous Beam with Unequal Spans

The following example shows the slightly more complex problem ofdealing with unequal spans

Example 2 Construct the shear and moment diagrams for the beam inFigure 55a

Solution In this case the moments at the outer supports are again zerowhich reduces the task to solving for only one unknown Applying thegiven values to the equation

Writing a moment summation about a point 14 ft to the right of the leftend support using the forces to the left of the point

14R1 ndash (1000 times 14 times 7) = ndash 19500 R1 = 5607 lb

Then writing an equation about a point 10 ft to the left of the right endusing the forces to the right of the point

2 14 101000 14

4

1000 10

419 500

2

3 3

2

M

M

( )

+ = minus times minus times

= minus ft-lb

M = times =375 3 75

2703 125

ft-lb


3751 P-05 111301 1222 PM Page 166


A vertical force summation will yield the value of R2 = 15343 lbWith the three reactions determined the shear values for completing theshear diagram are known Determination of the points of zero shear andzero moment and the values for positive moment in the two spans can bedone as demonstrated in Exercise 1 The completed diagrams are shownin Figures 55b and c

Problems 51C DFind the reactions and draw the complete shear and moment diagrams forthe following continuous beams with two unequal spans and uniformlydistributed loading


Figure 55 Example 2

3751 P-05 111301 1222 PM Page 167

Beam First Span ft Second Span ft Load lbft

C 12 16 2000D 16 20 1200

Continuous Beam with Concentrated Loads

In the previous examples the loads were uniformly distributed Figure56a shows a two-span beam with a single concentrated load in eachspan The shape for the moment diagram for this beam is shown in Fig-ure 56b For these conditions the form of the three-moment equation is

M1 L1 + 2M2 (L1 + L2) + M3 L2 = ndash P1 L12 [n1 (1 ndash n1)(1 + n1)] ndash

P2 L22 [n2 (1 ndash n2)(2 ndash n2)]

in which the various terms are as shown in Figure 56

Example 3 Compute the reactions and construct the shear and momentdiagrams for the beam in Figure 57a

Solution For this case note that L1 = L2 P1 = P2 M1 = M3 = 0 and bothn1 and n2 = 05 Substituting these conditions and given data into theequation

2M2 (20 + 20) = ndash4000(202)(05 times 05 times 15) ndash4000(202)(05 times 05 times 15)

from which M2 = 15000 ft-lb


Figure 56 Diagrams for the two-span beam with concentrated loads

3751 P-05 111301 1222 PM Page 168

The value of moment at the middle support can now be used as in theprevious examples to find the end reaction from which it is determined thatthe value is 1250 lb Then a summation of vertical forces will determine thevalue of R2 to be 5500 lb This is sufficient data for construction of the sheardiagram Note that points of zero shear are evident on the diagram

The values for maximum positive moment can be determined frommoment summations at the sections or simply from the areas of the rec-tangles in the shear diagrams The locations of points of zero moment canbe found by simple proportion since the moment diagram is composedof straight lines

Problems 51E FFind the reactions and draw the complete shear and moment diagrams forthe following continuous beams with two equal spans and a single con-centrated load at the center of each span

Beam Span Length ft Load kips

E 24 30F 32 24


Figure 57 Example 3

3751 P-05 111301 1222 PM Page 169

Continuous Beam with Three Spans

The preceding examples demonstrate that the key operation in investi-gation of continuous beams is the determination of negative momentvalues at the supports Use of the three-moment equation has beendemonstrated for a two-span beam but the method may be applied toany two adjacent spans of a beam with multiple spans For examplewhen applied to the three-span beam shown in Figure 58a it would firstbe applied to the left span and the middle span and next to the middlespan and right span This would produce two equations involving thetwo unknowns the negative moments at the two interior supports Inthis example case the process would be simplified by the symmetry ofthe beam but the application is a general one applicable to any arrange-ment of spans and loads

As with simple beams and cantilevers common situations of spansand loading may be investigated and formulas for beam behavior values derived for subsequent application in simpler investigationprocesses Thus the values of reactions shears and moments displayedfor the beam in Figure 58 may be used for any such support and loadingconditions Tabulations for many ordinary situations are available fromvarious references

Example 4 A continuous beam has three equal spans of 20 ft [6 m]each and a uniformly distributed load of 800 lbft [12 kNm] extendingover the entire length of the beam Compute the maximum bending mo-ment and the maximum shear

Solution Referring to Figure 58d the maximum positive moment(008wL2) occurs near the middle of each end span and the maximumnegative moment (010wL2) occurs over each of the interior supportsUsing the larger value the maximum bending moment on the beam is

M = ndash010wL2 = ndash(010 times 800 times 20 times 20)= ndash32000 ft-lb [432 kN-m]

Figure 58c shows that the maximum shear occurs at the face of the firstinterior support and is

V = 06wL = (06 times 800 times 20) = 9600 lb [432 kN]


3751 P-05 111301 1222 PM Page 170

Using this process it is possible to find the values of the reactions andthen to construct the complete shear and moment diagrams if the workat hand warrants it

Problem 51G HFor the following continuous beams with three equal spans and uni-formly distributed loading find the reactions and draw the completeshear and moment diagrams


G 24 1000H 32 1600


Figure 58 Diagrams and values for the three-span beam with uniform load

3751 P-05 111301 1222 PM Page 171

52 RESTRAINED BEAMS

A simple beam was previously defined as a beam that rests on a supportat each end there being no restraint against bending at the supports theends are simply supported The shape a simple beam tends to assumeunder load is shown in Figure 59a Figure 59b shows a beam whose leftend is restrained or fixed meaning that free rotation of the beam end isprevented Figure 59c shows a beam with both ends restrained End re-straint has an effect similar to that caused by the continuity of a beam atan interior support a negative bending moment is induced in the beamThe beam in Figure 59b therefore has a profile with an inflection pointindicating a change of sign of the moment within the span This span be-haves in a manner similar to one of the spans in the two-span beam

The beam with both ends restrained has two inflection points with aswitch of sign to negative bending moment near each end Although val-ues are slightly different for this beam the general form of the deflectedshape is similar to that for the middle span in the three-span beam (seeFigure 58)

Although they have only one span the beams in Figures 59b and c areboth indeterminate Investigation of the beam with one restrained end in-volves finding three unknowns the two reactions plus the restrainingmoment at the fixed end For the beam in Figure 59c there are four un-knowns There are however only a few ordinary cases that cover mostcommon situations and tabulations of formulas for these ordinary casesare readily available from references Figure 510 gives values for thebeams with one and two fixed ends under both uniformly distributedload and a single concentrated load at center span Values for other load-ings are also available from references


Figure 59 Deflected shape of the single-span beam (a) With simple supports(b) With one end fixed (c) With both ends fixed

3751 P-05 111301 1222 PM Page 172

Example 5 Figure 511a represents a 20-ft span beam with both endsfixed and a total uniformly distributed load of 8 kips Find the reactionsand construct the complete shear and moment diagrams

Solution Despite the fact that this beam is indeterminate to the seconddegree (four unknowns only two equations of static equilibrium) itssymmetry makes some investigation data self-evident Thus it can be ob-served that the two vertical reaction forces and thus the two end shearvalues are each equal to one half of the total load or 4000 lb Symmetryalso indicates that the location of the point of zero moment and thus the

RESTRAINED BEAMS 173

Figure 510 Values and diagrams for single-span beams with restrained supports

3751 P-05 111301 1222 PM Page 173

point of maximum positive bending moment is at the center of the spanAlso the end moments although indeterminate are equal to each otherleaving only a single value to be determined

From data in Figure 510a the negative end moment is 00833WL (ac-tually WL12) = (8000 times 20)12 = 13333 ft-lb The maximum positivemoment at midspan is 004167WL (actually WL24) = (8000 times 20)24 =6667 ft-lb And the point of zero moment is 0212L = (0212)(20) = 424ft from the beam end The complete shear and moment diagrams are asshown in Figures 511b and c

Example 6 A beam fixed at one end and simply supported at the otherend has a span of 20 ft and a total uniformly distributed load of 8000 lb(Figure 512a) Find the reactions and construct the shear and momentdiagrams

Solution This is the same span and loading as in the preceding exampleHere however one end is fixed and the other simply supported (the load-ing case shown in Figure 510c) The beam vertical reactions are equal tothe end shears thus from the data in Figure 510c



3751 P-05 111301 1222 PM Page 174

R1 = V1 = 0375(8000) = 3000 lbR2 = V2 = 0625(8000) = 5000 lb

and for the maximum moments

+M = 00703(8000 times 20) = 11248 ft-lbndashM = 0125(8000 times 20) = 20000 ft-lbk

The point of zero shear is at 0375(20) = 75 ft from the left end and thepoint of zero moment is at twice this distance 15 ft from the left endThe complete shear and moment diagrams are shown in Figures 512band c

Problem 52AA 22-ft [671-m] span beam is fixed at both ends and carries a single con-centrated load of 16 kips [712 kN] at midspan Find the reactions andconstruct the complete shear and moment diagrams



3751 P-05 111301 1222 PM Page 175

Problem 52BA 16-ft [488-m] span beam is fixed at one end and simply supported atthe other end A single concentrated load of 9600 lb [427 kN] is placedat the center of the span Find the vertical reactions and construct thecomplete shear and moment diagrams

53 BEAMS WITH INTERNAL PINS

In many structures conditions exist at supports or within the structurethat modify the behavior of the structure often eliminating some poten-tial components of force actions Qualification of supports as fixed orpinned (not rotation-restrained) has been a situation in most of the struc-tures presented in this work We now consider some qualification of con-ditions within the structure that modify its behavior

Internal Pins

Within a structure members may be connected in a variety of ways If astructural joint is qualified as pinned it is considered to be capable onlyof transfer of direct forces of shear tension or compression Such jointsare commonly used for wood and steel framed structures In some casesa pinned joint may deliberately be used to eliminate the possibility fortransfer of bending moment through the joint such is the case in the fol-lowing examples

Continuous Beams with Internal Pins

The typical continuous beam such as that shown in Figure 513a is sta-tically indeterminate in this case having a number of reaction compo-nents (three) in excess of the conditions of equilibrium for the parallelforce system (two) The continuity of such a beam results in the deflectedshape and variation of moment as shown beneath the beam in Figure513a If the beam is made discontinuous at the middle support as shownin Figure 513b the two spans each behave independently as simplebeams with the deflected shapes and moment as shown

If a multiple-span beam is made internally discontinuous at somepoint off of the supports its behavior may emulate that of a truly contin-


3751 P-05 111301 1222 PM Page 176

BEAMS WITH INTERNAL PINS 177

uous beam For the beam shown in Figure 513c the internal pin is lo-cated at the point where the continuous beam inflects Inflection of thedeflected shape is an indication of zero moment and thus the pin doesnot actually change the continuous nature of the structure The deflectedshape and moment variation for the beam in Figure 513c is therefore thesame as for the beam in Figure 513a This is true of course only for

Figure 513 Behavior of two-spanbeams (a) As a continuous single-piece beam (b) As separate piecesin each span (c) With internal pin inone span

3751 P-05 111301 1222 PM Page 177

the single loading pattern that results in the inflection point at the samelocation as the internal pin

In the first of the following examples the internal pin is deliberatelyplaced at the point where the beam would inflect if it were continuous Inthe second example the pins are placed slightly closer to the supportrather than in the location of the natural inflection points The modifica-tion in the second example results in slightly increasing the positive mo-ment in the outer spans while reducing the negative moments at thesupports thus the values of maximum moment are made closer If it isdesired to use a single-size beam for the entire length the modification inExample 8 permits design selection of a slightly smaller size member

Example 7 Investigate the beam shown in Figure 514a Find the reac-tions draw the shear and moment diagrams and sketch the deflected shape

Solution Because of the internal pin the first 12 ft of the left-hand spanacts as a simple beam Its two reactions are therefore equal being one-half the total load and its shear moment and deflected shape diagramsare those for a simple beam with a uniformly distributed load (See Case2 Figure 420) As shown in Figures 514b and c the simple beam reac-tion at the right end of the 12-ft portion of the left span becomes a 6-kipconcentrated load at the left end of the remainder of the beam This beam(Figure 514c) is then investigated as a beam with one overhanging endcarrying a single concentrated load at the cantilevered end and the totaldistributed load of 20 kips (Note that on the diagram the total uniformlydistributed load is indicated in the form of a single force representing itsresultant) The second portion of the beam is statically determinate andits reactions can now be determined by statics equations

With the reactions known the shear diagram can be completed Notethe relation between the point of zero shear in the span and the locationof maximum positive moment For this loading the positive momentcurve is symmetrical and thus the location of the zero moment (andbeam inflection) is at twice the distance from the end as the point of zeroshear As noted previously the pin in this example is located exactly atthe inflection point of the continuous beam (For comparison see Section51 Example 1)

Example 8 Investigate the beam shown in Figure 515


3751 P-05 111301 1222 PM Page 178

Solution The procedure is essentially the same as for Example 7 Notethat this beam with four supports requires two internal pins to becomestatically determinate As before the investigation begins with theconsideration of the two end portions acting as simple beams The secondstep is to consider the center portion as a beam with two overhanging ends



3751 P-05 111301 1222 PM Page 179

Problems 53AndashCInvestigate the beams shown in Figures 516andashc Find the reactions anddraw the shear and moment diagrams indicating all critical valuesSketch the deflected shapes and determine the locations of any inflectionpoints not related to the internal pins (Note Problem 53B has the samespans and loading as Example 2 in Section 51)



3751 P-05 111301 1222 PM Page 180

54 APPROXIMATE ANALYSIS OF CONTINUOUS BEAMS

In some situations it may be acceptable to perform an approximateanalysis of a continuous beam for the purpose of its design This processmay be adequate for actual construction or may be simply a first approx-imation in a multistage design process in which some aspects of the beammust be defined before an exact analysis can proceed

The ACI Code (Ref 4) permits analysis of some continuous rein-forced concrete beams by approximate methods Use of these methods islimited by several conditions including those of only uniformly distrib-uted loads a relatively high dead load in proportion to live load and ap-proximately equal values for the beam spans Figure 517 shows asummary of the approximation factors used to establish design momentsand design shears with this method Values displayed may be comparedwith those indicated for various load span and support conditions in Figures 420 58 and 510

APPROXIMATE ANALYSIS OF CONTINUOUS BEAMS 181


3751 P-05 111301 1222 PM Page 181


Figure 517 Approximate design factors for continuous beams

3751 P-05 111301 1222 PM Page 182

183

6RETAINING WALLS

Strictly speaking any wall that sustains significant lateral soil pressure isa retaining wall That definition includes basement walls but the term isusually applied to site structures outside of buildings (see Figure 61) Forthe site retaining wall a critical concern is for the dimension of the dif-ference in the ground surface elevation on the two sides of the wall Thegreater this dimension the more the lateral force that will be exerted onthe wall attempting to topple the wall onto the lower side This chaptertreats some aspects of the structural behavior of the cantilever retainingwall an example of which is shown in the upper figure in Figure 61 The three major concerns for such a structure are its stability againstsliding against overturning (toppling) and the maximum soil pressuredeveloped on the bottom of the footing The latter two effects will beconsidered here

3751 P-06 111301 1223 PM Page 183

61 HORIZONTAL EARTH PRESSURE

Horizontal earth pressures are classified as either active or passive Pas-sive pressure is the resistance offered by a soil mass to something beingpushed against it For example passive pressure against the sides of abuildingrsquos below-grade construction is generally what resists the overallpush of the wind against the building

184 RETAINING WALLS

Figure 61 Achieving abrupt changes in the elevation of the ground surface hasbeen accomplished by various means over the years Shown here are two formsof construction in current use depending on various requirements The semi-openinterlocking units shown in the lower drawing permit easy drainage of the soil massbehind the wall and let air get to roots of plant growth behind the wall But a com-mon solution for abrupt changes of significant height is the cantilever structure ofreinforced concrete or masonry as shown in the upper drawing

3751 P-06 111301 1223 PM Page 184

Active pressure is that exerted by a soil mass against some restrainingstructure such as a basement wall or a retaining wall This is the form ofpressure that will be treated here The nature of active horizontal soilpressure can be visualized by considering the situation of an unrestrainedvertical cut in a soil mass as shown in Figure 62a In most soils such acut will not stand for long Under the action of various influencesmdashpri-marily gravitymdashthe soil mass will tend to move to the profile shown inFigure 62b

There are two force effects that tend to move the soil mass at the ver-tical cut First is the simple downward push of the soil at the top of thecut The second effect is the outward horizontal push by the soil at thebottom of the cut responding to the downward push of the soil above Acommon form of the actual soil movement consists of the rotational slipof the soil mass along a curved slip plane as shown in Figure 62c withthe slip plane indicated by the dashed line

HORIZONTAL EARTH PRESSURE 185

Figure 62 Aspects of the development of lateral soil pressure (a) Unrestrainedvertical cut (b) General form of failure at the face of a vertical cut (c) Common formof failure by rotational slip (d ) Net force effect by the soil on a bracing structure atthe cut soil face and indication of the form of horizontal pressure assumed in theequivalent fluid method

3751 P-06 111301 1223 PM Page 185

If a restraining structure is placed at the cut face the force effects de-scribed for the unrestrained soil will be exerted against the restrainingstructure as shown in Figure 62d The most critical part of this effect onthe restraining structure is the horizontal push consequently a commonpractice for design is to consider the soil mass to behave in the manner ofan equivalent fluid with pressure varying directly with the height as itdoes on the side of a water tank This pressure variation is shown in Figure 62d with the maximum pressure at the base of the wall indicatedas some constant times the wall height For a pure fluid this constantwould be the unit density (weight) of the fluid For soil it is some partialfraction of the soil weight typically about one-third

62 STABILITY OF RETAINING WALLS

The two basic concerns for stability of a retaining wall are with regard toits toppling (rotation) and its sliding in a horizontal direction away fromthe cut face of soil A typical investigation for toppling (more oftencalled overturning) is to do a summation of the rotational moments of allthe forces on the wall about a point at the low side toe of its foundationThis analysis is demonstrated in the following example

Example 1 Investigate the safety of the concrete retaining wall shownin Figure 63a with regard to rotation about the toe of its footing Use thefollowing data

Lateral soil pressure = 30 psfft of height

Soil weight = 100 pcf

Concrete weight = 150 pcf

Solution The loading condition for this analysis is shown in Figure 63bRotation about the lower left corner of the footing (toe) is induced by thesingle horizontal force acting as a resultant at one-third the height of the triangular pressure variation Resistance to this rotation is offered by theweight of the wall itself and by the weight of the soil above the footing Ata minimum the effect of the soil behind the wall is taken as the componentW3 which is the soil mass directly above the footing The computation ofthe component forces and their moments is summarized in Table 61

186 RETAINING WALLS

3751 P-06 111301 1223 PM Page 186

STABILITY OF RETAINING WALLS 187

Figure 63 Example 1

3751 P-06 111301 1223 PM Page 187

Safety is indicated by the ratio of the resisting moment to the over-turning moment a computation usually described as the safety factor Inthis example the safety factor SF against overturn is thus

Whether this is adequate safety or not is a judgement for the designerIn most cases building codes require a minimum safety factor of 15 forthis situation in which case the wall seems quite adequate

Problems 62A BInvestigate the stability of the concrete retaining walls shown in Figure64 with regard to overturning Use the data given in Example 1

63 VERTICAL SOIL PRESSURE

Stability of a cantilever retaining wall depends partly on the resistance ofthe supporting soil beneath the wall footing If this is a highly compress-ible soil the footing may settle considerably While a direct vertical set-tlement of some minor dimension is to be expected of greater concern isthe effect of a nonuniformly distributed pressure on the bottom of thefooting With a major horizontal force exerted on the retaining wall thismay well be the case thus an investigation is often made for the actualvertical pressure

SF = = =(resisting moment)

(overturning moment)

21 700

99882 17

188 RETAINING WALLS

TABLE 61 Analysis for Overturning Effect

Force lb Moment Arm in Moment lb-in

OverturnH = 1frasl2 times 55 times 165 = 454 22 M1 = ndash9988

Restoring Momentw1 = 0667 times 4667 times 150 = 467 18 8406w2 = (1012) times 25 times 150 = 312 15 4680w3 = 0667 times 4667 times 100 = 311 26 8086w4 = (1412) times 0667 times 100 = 78 7 546____ ____ ______Totals ΣW = 1168 M2 = +21718

3751 P-06 111301 1223 PM Page 188

Unless the vertical loads are exactly centered on the footing and theresisting moment exactly equals the overturning moments there is likelyto be some net moment at the bottom of the footing The usual practiceis to investigate for the combination of vertical compression due to thevertical forces and add to it any vertical stress due to a bending momentwith respect to the center of the footing The general form of such ananalysis is demonstrated in Section 132 Example 1 The method pre-sented there is used in the following example

Example 2 Investigate the retaining wall in Example 1 (Figure 63) forthe maximum vertical soil pressure at the bottom of the footing

Solution The vertical soil pressure at the bottom of the footing is pro-duced by the combination of the vertical load and the net moment withregard to the center of the footing The true loading condition as a resultof the vertical and horizontal loads shown in Figure 63 is indicated by the resultant shown in Figure 65a At the base of the footing theeccentricity of this resultant from the toe of the footing can be computedfrom the sum of the vertical load and the net moment about the toe Thedata for this computation are provided in Table 61 Thus the eccentric-ity e1 is found as

VERTICAL SOIL PRESSURE 189


3751 P-06 111301 1223 PM Page 189

Referring to Figure 65a with the value for e1 determined the distanceindicated as e2 may be found by subtraction from the dimension of one-half the footing width Thus e2 = 15 ndash 1004 = 496 in This is the ec-centricity that relates to the combined stress analysis for the footingvertical soil pressure

A first determination at this point is that made with regard to the sig-nificance of the eccentricity with respect to the kern of the footing (seediscussion in Section 132) For this case the kern limit is one-sixth ofthe footing width or 5 in The eccentricity as computed is thus seen to bejust inside the limit allowing for an investigation for Case 1 as shown inFigure 135 The analysis for this is illustrated in Figure 65b and thecomputation of the stress is shown in Figure 65c The two componentsfor this computation are as follows

1 For the normal compression stress

2 For the bending stress

Then

The limiting values of the combined stress as shown in Figure 65care thus 930 psf and 4 psf

Problems 63A BCompute the values for the vertical soil pressure for the retaining wallsin Figures 64a and b

pM

S

W e

S= = Σ times = times =2 1168 4 96 12

1 042463

( )

psf

Sbd= = times =

2 2

6

1 2 5

61 042

( ) ft3

pN

A

W

A= = Σ =

times=1168

1 2 5467

psf

eM M

W1

2 1 21 718 9988

116810 04= minus

Σ= minus =

in

190 RETAINING WALLS

3751 P-06 111301 1223 PM Page 190


Figure 65 Example 2

3751 P-06 111301 1223 PM Page 191

192

7RIGID FRAMES

Frames in which two or more of the members are attached to each otherwith connections that are capable of transmitting bending between theends of the members are called rigid frames The connections used toachieve such a frame are called moment connections or moment-resistingconnections Most rigid frame structures are statically indeterminate anddo not yield to investigation by consideration of static equilibrium aloneThe rigid-frame structure occurs quite frequently as a multiple-levelmultiple-span bent constituting part of the structure for a multistorybuilding (see Figure 71) In most cases such a bent is used as a lateralbracing element although once it is formed as a moment-resistive frame-work it will respond as such for all types of loads The computational ex-amples presented in this section are all rigid frames that have conditionsthat make them statically determinate and thus capable of being fully in-vestigated by methods developed in this book

3751 P-07 111301 1223 PM Page 192

71 CANTILEVER FRAMES

Consider the frame shown in Figure 72a consisting of two membersrigidly joined at their intersection The vertical member is fixed at itsbase providing the necessary support condition for stability of the frameThe horizontal member is loaded with a uniformly distributed loadingand functions as a simple cantilever beam The frame is described as acantilever frame because of the single fixed support The five sets of fig-ures shown in Figures 72b through f are useful elements for the investi-gation of the behavior of the frame They consist of the following

CANTILEVER FRAMES 193

Figure 71 The rigid frame derives its name from the nature of the joint betweenthe frame membersmdashbeing one that rigidly resists the rotation of member endswith respect to each other at the joint Sitecast concrete frames develop this qual-ity naturally and steel frames may be formed with special connections to developthe rigid joints Individual rows of beams and columns may be visualized as planarrigid frames in such constructionmdashas shown here

3751 P-07 111301 1223 PM Page 193

1 The free-body diagram of the entire frame showing the loadsand the components of the reactions (Figure 72b) Study of thisfigure will help in establishing the nature of the reactions and inthe determination of the conditions necessary for stability of theframe as a whole

2 The free-body diagrams of the individual elements (Figure 72c)These are of great value in visualizing the interaction of the partsof the frame They are also useful in the computations for the in-ternal forces in the frame

3 The shear diagrams of the individual elements (Figure 72d )These are sometimes useful for visualizing or for actually com-puting the variations of moment in the individual elements Noparticular sign convention is necessary unless in conformity withthe sign used for moment

4 The moment diagrams for the individual elements (Figure 72e)These are very useful especially in determination of the defor-mation of the frame The sign convention used is that of plottingthe moment on the compression (concave) side of the flexedelement

5 The deformed shape of the loaded frame (Figure 72f ) This is theexaggerated profile of the bent frame usually superimposed onan outline of the unloaded frame for reference This is very use-ful for the general visualization of the frame behavior It is par-ticularly useful for determination of the character of the externalreactions and the form of interaction between the parts of theframe Correlation between the deformed shape and the form ofthe moment diagram is a useful check

When performing investigations these elements are not usually pro-duced in the sequence just described In fact it is generally recommendedthat the deformed shape be sketched first so that its correlation with otherfactors in the investigation may be used as a check on the work The fol-lowing examples illustrate the process of investigation for simple can-tilever frames

Example 1 Find the components of the reactions and draw the free-body diagrams shear and moment diagrams and the deformed shape ofthe frame shown in Figure 73a

194 RIGID FRAMES

3751 P-07 111301 1223 PM Page 194

Solution The first step is the determination of the reactions Consider-ing the free-body diagram of the whole frame (Figure 73b)

ΣF = 0 = +8 ndash Rv Rv = 8 kips (up)

and with respect to the support

ΣM = 0 = MR ndash (8 times 4) MR = 32 kip-ft (clockwise)


Figure 72 Diagrams for investigation of the rigid frame

3751 P-07 111301 1223 PM Page 195

196 RIGID FRAMES

Figure 73 Example 1

3751 P-07 111301 1223 PM Page 196

Note that the sense or sign of the reaction components is visualizedfrom the logical development of the free-body diagram

Consideration of the free-body diagrams of the individual memberswill yield the actions required to be transmitted by the moment connec-tion These may be computed by application of the conditions for equi-librium for either of the members of the frame Note that the sense of theforce and moment is opposite for the two members simply indicatingthat what one does to the other is the opposite of what is done to it

In this example there is no shear in the vertical member As a resultthere is no variation in the moment from the top to the bottom of themember The free-body diagram of the member the shear and momentdiagrams and the deformed shape should all corroborate this fact Theshear and moment diagrams for the horizontal member are simply thosefor a cantilever beam

It is possible with this example as with many simple frames to visual-ize the nature of the deformed shape without recourse to any mathematicalcomputations It is advisable to attempt to do so as a first step in investiga-tion and to check continually during the work that individual computationsare logical with regard to the nature of the deformed structure

Example 2 Find the components of the reactions and draw the shear andmoment diagrams and the deformed shape of the frame in Figure 74a

Solution In this frame there are three reaction components required forstability since the loads and reactions constitute a general coplanar forcesystem Using the free-body diagram of the whole frame (Figure 74b)the three conditions for equilibrium for a coplanar system are used to findthe horizontal and vertical reaction components and the moment compo-nent If necessary the reaction force components could be combined into a single-force vector although this is seldom required for designpurposes

Note that the inflection occurs in the larger vertical member becausethe moment of the horizontal load about the support is greater than thatof the vertical load In this case this computation must be done before thedeformed shape can be accurately drawn

The reader should verify that the free-body diagrams of the individualmembers are truly in equilibrium and that there is the required correlationbetween all the diagrams


3751 P-07 111301 1223 PM Page 197

198 RIGID FRAMES

Figure 74 Example 2

Problems 71AndashCFor the frames shown in Figure 75andashc find the components of the reac-tions draw the free-body diagrams of the whole frame and the individualmembers draw the shear and moment diagrams for the individual mem-bers and sketch the deformed shape of the loaded structure

3751 P-07 111301 1223 PM Page 198

72 SINGLE-SPAN FRAMES

Single-span rigid frames with two supports are ordinarily statically inde-terminate The following example illustrates the case of a statically de-terminate single-span frame made so by the particular conditions of itssupport and internal construction In fact these conditions are technicallyachievable but a little weird for practical use The example is offeredhere as an exercise for readers an exercise that is within the scope of thework in this section

Example 3 Investigate the frame shown in Figure 76 for the reactionsand internal conditions Note that the right-hand support allows for anupward vertical reaction only whereas the left-hand support allows forboth vertical and horizontal components Neither support provides mo-ment resistance

Solution The typical elements of investigation as illustrated for the pre-ceding examples are shown in Figure 76 The suggested procedure forthe work is as follows

SINGLE-SPAN FRAMES 199


3751 P-07 111301 1223 PM Page 199

200 RIGID FRAMES

Figure 76 Example 3

3751 P-07 111301 1223 PM Page 200

1 Sketch the deflected shape (a little tricky in this case but a good

exercise)

2 Consider the equilibrium of the free-body diagram for the wholeframe to find the reactions

3 Consider the equilibrium of the left-hand vertical member to findthe internal actions at its top

4 Proceed to the equilibrium of the horizontal member

5 Finally consider the equilibrium of the right-hand verticalmember

6 Draw the shear and moment diagrams and check for correlationof all work

Before attempting the exercise problems the reader is advised to at-tempt to produce the results shown in Figure 76 independently

Problems 72A BInvestigate the frames shown in Figures 77a and b for reactions and in-ternal conditions using the procedure shown for the preceding examples



3751 P-07 111301 1223 PM Page 201

202

8NONCOPLANAR

FORCE SYSTEMS

Forces and structures exist in reality in a three-dimensional world (seeFigure 81) The work in preceding chapters has been limited mostly tosystems of forces operating in two-dimensional planes This is com-monly done in design practice primarily for the same reasons that wehave done it here it makes both visualization and computations easierAs long as the full three-dimensional character of the forces and thestructures is eventually dealt with this approach is usually quite ade-quate For visualization as well as for some computations however it issometimes necessary to work directly with forces in noncoplanar sys-tems This chapter presents some exercises that will help in the develop-ment of an awareness of the problems of working with such forcesystems

Graphical representation visualization and any mathematical com-putation all become more complex with noncoplanar systems The fol-lowing discussions rely heavily on the examples to illustrate basicconcepts and procedures The orthogonal axis system x-y-z is used forease of both visualization and computation

3751 P-08 111301 1224 PM Page 202

Units of measurement for both forces and dimensions are of small sig-nificance in this work Because of this and because of the complexity ofboth the graphical representations and the mathematical computationsthe conversions for metric units have been omitted except for the dataand answers for the exercise problems

81 CONCURRENT SYSTEMS

Figure 82 shows a single force acting in such a manner that it has com-ponent actions in three dimensions That is it has x y and z componentsIf this force represents the resultant of a system of forces it may be iden-tified as follows

For its magnitude

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2

CONCURRENT SYSTEMS 203

Figure 81 All building structures are three-dimensional in their general formNevertheless most can be broken down into component linear and planar (two-dimensional) elements for investigation of behavior However some systems arefundamentally three-dimensional and must be treated as such for investigationThe two-way spanning trussmdashalso called a space framemdashis one such structure

3751 P-08 111301 1224 PM Page 203

and for its direction

Equilibrium for this system can be established by fulfilling the followingconditions

ΣFx = 0 ΣFy = 0 ΣFz = 0

Example 1 Find the resultant of the three forces shown in Figure 83a

Solution Various methods may be used employing trigonometry polarcoordinates and so on The method used here is to first find the geome-try of the force lines for the three forces Then the vectors for the forcesand their x y and z components can be expressed using the proportion-ate values from the force line geometry The construction for this com-putation is shown in Figure 83a

Referring to the line lengths shown in Figure 83a

L

L

12 2

22

5 3 34 5 83

12 34 178 13 34

= + = =

= + = =

( ) ( )

( )

cos cos cos θ θ θxx

yy

zzF

R

F

R

F

R= Σ =

Σ= Σ

204 NONCOPLANNAR FORCE SYSTEMS

Figure 82 Components of a noncoplanar force

3751 P-08 111301 1224 PM Page 204

Note To reinforce the point that the unit of measurement for dimensionsis not relevant for these computations it is omitted

The other line lengths can be determined in the same manner Theirvalues are shown on the figure The determination of the force compo-nents and their summation is presented in Table 81 Note that the senseof the components is established with reference to the positive directionsindicated for the three axes as shown in Figure 83a To aid in visual-ization the sense of the forces in Table 81 is indicated with arrowsrather than with plus and minus signs

Using the summations from the table the value of the resultant is de-termined as

R = + + = =( ) ( ) ( ) 2 4 466 1 22 4 217 757 466 72 2 2


Figure 83 Example 1

TABLE 81 Summation of Forces Example 1

Force x Component y Component z Component

F1 200(51334) = 75 200(121334) = 180 darr 200(31334) = 45

F2 160(21356) = 236 160(121356) = 1417 darr 160(61356) = 708

F3 180(81497) = 962 180(121497) = 1444 darr 180(41497) = 482____ _____ ____

ΣFx = 24 lb ΣFy = 4661 lb darr ΣFz = 224 lb larrlarr

larrlarr

larrlarrlarr

larr

3751 P-08 111301 1224 PM Page 205

The direction of R may be established by expressing the three cosineequations as described earlier or by establishing its points of intersec-tion with the x-z plane as shown in Figure 83b Using the latter methodand calling the x distance from the z-axis L3

Then

And similarly calling the z distance from the x-axis L4

Example 2 For the structure shown in Figure 84a find the tension inthe guy wires and the compression in the mast for the loading indicated

Solution The basic problem here is the resolution of the concentric forcesystem at the top of the mast As in Example 1 the geometry of the wiresis established first Thus

Consider the concentric forces at the top of the mast For equilibrium inthe x direction

ΣFx = 0 = +1000 ndash 2(Tx) Tx = 500 lb

Then from the geometry of the wire

T

T

T T

x

x

=

= = =

25

12

25

12

25

12500 1041 67( ) ( ) lb

L = + + = =( ) ( ) ( )9 12 20 626 252 2 2

L422 4

466 112 0 578= =

( )

L322 4

466 112 0 062= =

( )

ΣΣ

= =F

F

Lx

y

3

12

2 4

466 1


3751 P-08 111301 1224 PM Page 206

For the compression in the mast consider the equilibrium of the forces inthe y direction Thus

ΣFy = 0 = +C ndash 2(Ty) C = 2(Ty)

where

Thus

C Ty= = =2 220

12500 1666 67( ) ( ) lb

T

T

T T

y

x

y x

=

= =

20

12

20

12

20

12500( ) ( )



3751 P-08 111301 1224 PM Page 207

Example 3 Find the tension in each of the three wires in Figure 84c dueto the force indicated

Solution As before the first step is to find the lengths of the three wiresThus

The three static equilibrium equations for the concentric forces are thus

Solution of these three simultaneous equations with three unknownsyields the following

T1 = 525 lb T2 = 271 lb T3 = 290 lb

Problem 81AFind the resultant of the three forces shown in Figure 85a Establish thedirection of the resultant by finding the coordinates of its intersectionwith the x-z plane

Problem 81BFind the compression force in the struts and the tension force in the wirefor the structure in Figure 85b

Problem 81CFind the tension force in each of the wires for the system shown in Figure 85c

Σ = = + minus

Σ = = + + + minus

Σ = = + + minus

F T T

F T T T

F T T T

x

y

z

04

21

8

21 63

020

21

20

21 63

20

23 321000

05

21

2

21 63

12

23 32

1 2

1 2 3

1 2 3

( )

( )

( )

( )

( )

( )

( )

( )

L

L

L

12 2 2

22 2 2

32 2

5 4 20 441 21

2 8 20 468 21 63

12 20 544 23 32

= + + = =

= + + = =

= + = =

( ) ( ) ( )

( ) ( ) ( )

( ) ( )


3751 P-08 111301 1224 PM Page 208

82 PARALLEL SYSTEMS

Consider the force system shown in Figure 86 Assuming the directionof the forces to be parallel to the y-axis the resultant force can be statedas

R = ΣFy

and its location in the x-z plane can be established by two moment equa-tions taken with respect to the x-axis and the z-axis thus

LM

RL

M

Rx

zz

x= Σ = Σand

PARALLEL SYSTEMS 209


3751 P-08 111301 1224 PM Page 209

The static equilibrium for the system can be established by fulfillingthe following conditions

ΣFy = 0 ΣMx = 0 ΣMz = 0

As with the coplanar parallel systems the resultant may be a coupleThat is the summation of forces may be zero but there may be a net ro-tational effect about the x-axis andor the z-axis When this is the casethe resultant couple may be visualized in terms of two component cou-ples one in the x-y plane (for ΣMz) and one in the z-y plane (for ΣMx) SeeExample 5 in the following work

Example 4 Find the resultant of the system shown in Figure 87a

Solution The magnitude of the resultant is found as the simple alge-braic sum of the forces Thus

R = ΣF = 50 + 60 + 160 + 80 = 350 lb

Then for its location in the x-z plane

ΣMx = +(160 times 8) ndash(60 times 6) = 920 ft-lbΣMz = +(50 times 8) ndash(80 times 15) = 800 ft-lb


Figure 86 Resultant of a parallel noncoplanar force system

3751 P-08 111301 1224 PM Page 210

and the distances from the axes are

Example 5 Find the resultant of the system shown in Figure 87b

Solution As in the previous example three summations are made

ΣF = R = +40 +20 ndash10 ndash50 = 0ΣMx = +( 40 times 8) ndash(20 times 8) = 160 ft-lbΣMz = +(10 times 6) ndash(50 times 10) = 440 ft-lb

The resultant is seen to be a couple with the two moment componentsdescribed by the moment summations If necessary these two compo-

L Lx z= = = =800

3502 29

920

3502 63 ft ft


Figure 87 Examples 4 5 and 6

3751 P-08 111301 1224 PM Page 211



nents can be combined into a single couple about an axis at some angleto the x-axis or the z-axis although it may be sufficient to use the com-ponents for some problems

Example 6 Find the tension in the three wires in the system shown inFigure 87c

Solution Using the three static equilibrium equations

ΣF = 0 = T1 + T2 + T3 ndash 1000ΣMx = 0 = 4T1 ndash 6T2

ΣMz = 0 = 6T1 ndash 8T3

Solution of these three simultaneous equations yields

T1 = 414 lb T2 = 276 lb T3 = 310 lb

Problem 82AFind the resultant and its location with respect to the x- and z-axes for thesystem shown in Figure 88a

Problem 82BFind the tension in the three wires of the system shown in Figure 88b

3751 P-08 111301 1224 PM Page 212

83 GENERAL NONCOPLANAR SYSTEMS

This is the general spatial force system with no simplifying conditions re-garding geometry The resultant for such a system may be any of fourpossibilities as follows

1 Zero if the system is in equilibrium

2 A force if the sum of forces is not zero

3 A couple if the sum of moments is not zero

4 A force plus a couple which is the general case when equilibriumdoes not exist

If the resultant is a force its magnitude is determined as

and its direction by

If the resultant is a couple it may be determined in terms of its com-ponent moments about the three axes in a procedure similar to that shownfor the parallel systems in Section 82

Solution of general spatial force systems is often quite complex andlaborious However in some situations the existence of symmetry orother qualifications may simplify the work In structural design practicesuch systems are usually broken down into simpler component systemsfor investigation and design

cos cos cos Θ Θ Θxx

yy

zzF

R

F

R

F

R= Σ =

Σ= Σ

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2

GENERAL NONCOPLANAR SYSTEMS 213

3751 P-08 111301 1224 PM Page 213

214

9PROPERTIES

OF SECTIONS

This chapter deals with various geometric properties of plane (two-dimensional) areas The areas referred to are the cross-sectional areas ofstructural members The geometric properties are used in the analysis ofstresses and deformations and in the design of the structural membersMost structural members used for building structures have cross sectionsthat are standardized for the industrial production of products In the toprow in Figure 91 are shown four such common shapes produced fromsteel and frequently used for building columns the round pipe the squareor oblong tube and the I- or H-shape (actually called a W-shape) How-ever these and other elements are sometimes combined to produce built-up sections such as those shown in the middle and bottom rows in Figure91 Geometric properties for standard cross sections are tabulated in in-dustry publications but properties for special sections that are cut fromor built up from standard shapes must be computed This chapter presentssome of the basic structural geometric properties and the processes fortheir computation

3751 P-09 111301 1224 PM Page 214

91 CENTROIDS

The center of gravity of a solid is the imaginary point at which all itsweight may be considered to be concentrated or the point through whichthe resultant weight passes Since a two-dimensional planar area has noweight it has no center of gravity The point in a plane area that corre-sponds to the center of gravity of a very thin plate of the same area andshape is called the centroid of the area The centroid is a useful referencefor various geometric properties of a planar area

For example when a beam is subjected to forces that cause bendingthe fibers above a certain plane in the beam are in compression and the

CENTROIDS 215

Figure 91 Cross sections for steel compression members Top row shows com-mon single-piece sections pipe tubes and I-shape (called W-shape) Other sec-tions are combinations of various individual elements Geometric properties forthese planar sections must be obtained for use in the investigation of stresses andstrains induced by loading of the structural member

3751 P-09 111301 1224 PM Page 215

fibers below the plane are in tension This plane is the neutral stressplane also called simply the neutral surface (see Section 111) For across section of the beam the intersection of the neutral surface with theplane of the cross section is a line this line passes through the centroidof the section and is called the neutral axis of the beam The neutral axisis important for investigation of flexural stresses in a beam

The location of the centroid for symmetrical shapes is usually quitereadily apparent If an area possesses a line (axis) of symmetry the cen-troid will be on that line If there are two distinct lines of symmetry thecentroid will lie at their intersection point Consider the rectangular areashown in Figure 92a obviously its centroid is at its geometric centerwhich is readily determined This point may be located by measured dis-tances (half the width and half the height) or may be obtained by geo-metric construction as the intersection of the two diagonals of therectangle

(Note Tables 93 through 97 and Figure 913 referred to in the dis-cussion that follows are located at the end of this chapter)

For more complex forms such as those of rolled steel members(called shapes) the centroid will also lie on any axis of symmetry Thusfor a W-shape (actually I- or H-shaped) the two bisecting major axeswill define the centroid by their intersection (See reference figure forTable 93) For a channel shape (actually U-shaped) there is only oneaxis of symmetry (the axis labeled X-X in the reference figure for Table94) and it is therefore necessary to determine the location of the centroidalong this line by computation Given the dimensions of a channel shapethis determination is possible it is listed as dimension x in the propertiesin Table 94

216 PROPERTIES OF SECTIONS

Figure 92 Centroids of various planar shapes

3751 P-09 111301 1224 PM Page 216

For many structural members their cross sections are symmetricalabout two axes squares rectangles circles hollow circular cylinders(pipe) and so on Or their properties are defined in a reference sourcesuch as the Manual of Steel Construction (Ref 3) from which propertiesof steel shapes are obtained However it is sometimes necessary to de-termine some geometric properties such as the centroid for compositeshapes produced by combinations of multiple parts The process for de-termining centroids involves the use of the statical moment which is defined as the product of an area times the perpendicular distance of thecentroid of the area from a reference axis in the plane of the area If the area can be reduced to simple components then its total statical mo-ment can be obtained by summation of the moments of the componentsSince this sum is equal to the total area times its centroidal distance fromthe reference axis the centroidal distance may be determined by dividingthe summation of moments by the total area As with many geometricpostulations the saying is more difficult than the doing as the followingsimple demonstrations will show

Example 1 Figure 93 is a beam cross section unsymmetrical with re-spect to the horizontal axis (X-X in Figure 93c) Find the location of thehorizontal centroidal axis for this shape

Solution The usual process for this problem is to first divide the shapeinto units for which both the area and centroid of the unit are easily

CENTROIDS 217

Figure 93 Example 1

3751 P-09 111301 1224 PM Page 217

determined The division chosen here is shown in Figure 93b with twoparts labeled 1 and 2

The second step is to chose an arbitrary reference axis about which tosum statical moments and from which the centroid of the shape is read-ily measured A convenient reference axis for this shape is one at eitherthe top or bottom of the shape With the bottom chosen the distancesfrom the centroids of the parts to this reference axis are as shown in Figure 93b

The computation next proceeds to the determination of the unit areasand the unit statical moments This work is summarized in Table 91which shows the total area to be 80 in2 and the total statical moment tobe 520 in3 Dividing the moment by the area produces the value of 65in which is the distance from the reference axis to the centroid of thewhole shape as shown in Figure 93c

Problems 91AndashFFind the location of the centroid for the cross-sectional areas shown inFigures 94andashf Use the reference axes and indicate the distances from thereference axes to the centroid as cx and cy as shown in Figure 94b

92 MOMENT OF INERTIA

Consider the area enclosed by the irregular line in Figure 95a In thisarea designated A a small unit area a is indicated at z distance from theaxis marked X-X If this unit area is multiplied by the square of its dis-tance from the reference axis the quantity a times z2 is defined If all of the


TABLE 91 Summary of Computations forCentroid Example 1

Area y A times yPart (in2) (in) (in3)

1 2 times 10 = 20 11 2202 6 times 10 = 60 5 300Σ 6 times 10 = 80 520

yx = 52080 = 65 in

3751 P-09 111301 1224 PM Page 218

units of the total area are thus identified and the summation of theseproducts is made the result is defined as the moment of inertia or the sec-ond moment of the area indicated as I thus

Σ az2 = I or specifically IX-X

MOMENT OF INERTIA 219


3751 P-09 111301 1224 PM Page 219

which is identified as the moment of inertia of the area about the X-Xaxis

The moment of inertia is a somewhat abstract item somewhat harderto grasp than the concepts of area weight or center of gravity It is nev-ertheless a real geometric property that becomes an essential factor in in-vestigations for stresses and deformations in structural members Ofparticular interest is the moment of inertia about a centroidal axis andmdashmost significantlymdashabout a principal axis for a shape Figures 95b c eand f indicate such axes for various shapes

Inspection of Tables 93 through 97 will reveal the properties of mo-ment of inertia about the principal axes of the shapes in the tables Use ofthese values is demonstrated in various computations in this book

Moment of Inertia of Geometric Figures

Values for moments of inertia can often be obtained from tabulations ofstructural properties Occasionally it is necessary to compute values fora given shape This may be a simple shape such as a square rectangularcircular or triangular area For such shapes simple formulas are derivedto express the value for the moment of inertia (as they are for area cir-cumference etc)

Squares and Other Rectangles Consider the rectangle shown inFigure 95c Its width is b and its depth is d The two principal axes areX-X and Y-Y both passing through the centroid (in this case the simplecenter) of the area For this case the moment of inertia with respect to thecentroidal axis X-X is computed as

and the moment of inertia with respect to the axis Y-Y is

Example 2 Find the value of the moment of inertia for a 6 times 12-in woodbeam about an axis through its centroid and parallel to the narrow base ofthe section

Idb

Y-Y =3

12

Ibd

X-X =3

12


3751 P-09 111301 1224 PM Page 220

Solution Referring to Table 97 the actual dimensions of the section are55 times 115 in Then

which is in agreement with the value of IX-X in the table

Circles Figure 95e shows a circular area with diameter d and axis X-Xpassing through its center For the circular area the moment of inertia is

Example 3 Compute the moment of inertia of a circular cross section10 in in diameter about an axis through its centroid

Solution The moment of inertia about any axis through the center of thecircle is

Triangles The triangle in Figure 95f has a height d and base b Withrespect to the base of the triangle the moment of inertia about the cen-troidal axis parallel to the base is

Ibd=

3

36

Id= = times =π 4 4

4

64

3 1416 10

64490 9

in

Id= π 4

64

Ibd= = ( )( ) =

3 34

12

5 5 11 5

12697 1

in


Figure 95 Consideration of reference axes for the moment of inertia of variousshapes of cross sections

3751 P-09 111301 1224 PM Page 221

Example 4 Assuming that the base of the triangle in Figure 95f is 12 inand that the height is 10 in find the value for the centroidal moment ofinertia parallel to the base

Solution Using the given values in the formula

Open and Hollow Shapes Values of moment of inertia for shapesthat are open or hollow may sometimes be computed by a method of sub-traction This consists of finding the moment of inertia of a solid areamdashthe outer boundary of the areamdashand subtracting the voided parts Thefollowing examples demonstrate the process Note that this is possibleonly for symmetrical shapes

Example 5 Compute the moment of inertia for the hollow box sectionshown in Figure 96a about a horizontal axis through the centroid paral-lel to the narrow side

Solution Find first the moment of inertia of the shape defined by theouter limits of the box

Then find the moment of inertia for the area defined by the void space

The value for the hollow section is the difference thus

I = 500 ndash 1707 = 3293 in4

Example 6 Compute the moment of inertia about an axis through thecentroid of the pipe cross section shown in Figure 96b The thickness ofthe shell is 1 in

Solution As in the preceding example the two values may be found andsubtracted Alternatively a single computation may be made as follows

I = times =4 8

12170 7

34 in

Ibd= = times =

3 34

12

6 10

12500 in

Ibd= = times =

3 34

36

12 10

36333 3 in


3751 P-09 111301 1224 PM Page 222

Example 7 Referring to Figure 96c compute the moment of inertia ofthe I-section about a horizontal axis through the centroid and parallel tothe flanges

Solution This is essentially similar to the computation for Example 5The two voids may be combined into a single one that is 7-in-wide thus

Note that this method can only be used when the centroid of the outershape and the voids coincide For example it cannot be used to find themoment of inertia for the I-shaped section in Figure 96c about its verti-cal centroidal axis For this computation the method discussed in the fol-lowing section may be used

93 TRANSFERRING MOMENTS OF INERTIA

Determination of the moment of inertia of unsymmetrical and complexshapes cannot be done by the simple processes illustrated in the preced-ing examples An additional step that must be used is that involving the

I = times minus times = minus =8 10

12

7 8

12667 299 368

3 34 in

I d do i=

( ) minus ( )[ ]

=

minus( ) = minus =

π64

3 1416

6410 8 491 201 290

4 4

4 4 4 in

TRANSFERRING MOMENTS OF INERTIA 223


3751 P-09 111301 1224 PM Page 223

transfer of moment of inertia about a remote axis The formula forachieving this transfer is as follows

I = Io + Az2

In this formula

I = moment of inertia of the cross section about the requiredreference axis

Io = moment of inertia of the cross section about its own centroidalaxis parallel to the reference axis

A = area of the cross section

z = distance between the two parallel axes

These relationships are illustrated in Figure 97 where X-X is the cen-troidal axis of the area and Y-Y is the reference axis for the transferredmoment of inertia

Application of this principle is illustrated in the following examples

Example 8 Find the moment of inertia of the T-shaped area in Figure98 about its horizontal (X-X) centroidal axis (Note the location of thecentroid for this section was solved as Example 1 in Section 91)

Solution A necessary first step in these problems is to locate the posi-tion of the centroidal axis if the shape is not symmetrical In this case theT-shape is symmetrical about its vertical axis but not about the horizon-tal axis Locating the position of the horizontal axis was the problemsolved in Example 1 in Section 91


Figure 97 Transfer of moment ofinertia to a parallel axis

3751 P-09 111301 1224 PM Page 224

The next step is to break the complex shape down into parts for whichcentroids areas and centroidal moments of inertia are readily found Aswas done in Example 1 the shape here is divided between the rectangu-lar flange part and the rectangular web part

The reference axis to be used here is the horizontal centroidal axisTable 92 summarizes the process of determining the factors for the par-allel axis transfer process The required value for I about the horizontalcentroidal axis is determined to be 10467 in4

A common situation in which this problem must be solved is in thecase of structural members that are built up from distinct parts One suchsection is that shown in Figure 99 where a box-shaped cross section iscomposed by attaching two plates and two rolled channel sections Whilethis composite section is actually symmetrical about both its principalaxes and the locations of these axes are apparent the values for momentof inertia about both axes must be determined by the parallel axis trans-fer process The following example demonstrates the process


Figure 98 Example 8

TABLE 92 Summary of Computations for Moment of Inertia Example 9

Area y Io A times y2 Ix

Part (in2) (in) (in4) (in4) (in4)

1 20 45 10(2)312 = 67 20(45)2 = 405 41172 60 15 6(10)312 = 500 60(15)2 = 135 6357

ndashmdashmdashΣ 10467

3751 P-09 111301 1224 PM Page 225

Example 9 Compute the moment of inertia about the centroidal X-Xaxis of the built-up section shown in Figure 99

Solution For this situation the two channels are positioned so that theircentroids coincide with the reference axis Thus the value of Io for thechannels is also their actual moment of inertia about the required refer-ence axis and their contribution to the required value here is simplytwice their listed value for moment of inertia about their X-X axis asgiven in Table 94 2(162) = 324 in4

The plates have simple rectangular cross sections and the centroidalmoment of inertia of one plate is thus determined as

The distance between the centroid of the plate and the reference X-Xaxis is 625 in and the area of one plate is 8 in2 The moment of inertiafor one plate about the reference axis is thus

Io + Az2 = 01667 + (8)(625)2 = 3127 in4

and the value for the two plates is twice this or 6254 in4Adding the contributions of the parts the answer is 324 + 6254 =

9494 in4

Ibd

o = = times =3 3

4

12

16 0 5

120 1667

in


Figure 99 Example 9

3751 P-09 111301 1224 PM Page 226

Problems 93AndashFCompute the moments of inertia about the indicated centroidal axes forthe cross-sectional shapes in Figure 910

Problems 93GndashICompute the moments of inertia with respect to the centroidal X-X axesfor the built-up sections in Figure 911 Make use of any appropriate datafrom the tables of properties for steel shapes



3751 P-09 111301 1224 PM Page 227

94 MISCELLANEOUS PROPERTIES

Section Modulus

As noted in Section 112 the term Ic in the formula for flexural stress iscalled the section modulus (or S) Use of the section modulus permits aminor shortcut in the computations for flexural stress or the determina-tion of the bending moment capacity of members However the realvalue of this property is in its measure of relative bending strength ofmembers As a geometric property it is a direct index of bending strengthfor a given member cross section Members of various cross sectionsmay thus be rank-ordered in terms of their bending strength strictly onthe basis of their S values Because of its usefulness the value of S islisted together with other significant properties in the tabulations for steeland wood members

For members of standard form (structural lumber and rolled steelshapes) the value of S may be obtained from tables similar to those pre-sented at the end of this chapter For complex forms not of standard formthe value of S must be computed which is readily done once the cen-troidal axes are located and moments of inertia about the centroidal axesare determined

Example 10 Verify the tabulated value for the section modulus of a 6 times 12 wood beam about the centroidal axis parallel to its narrow side



3751 P-09 111301 1224 PM Page 228

Solution From Table 97 the actual dimensions of this member are 55 times 115 in and the value for the moment of inertia is 697068 in4Then

which agrees with the value in Table 97

Radius of Gyration

For design of slender compression members an important geometricproperty is the radius of gyration defined as

Just as with moment of inertia and section modulus values the radiusof gyration has an orientation to a specific axis in the planar cross sectionof a member Thus if the I used in the formula for r is that with respect tothe X-X centroidal axis then that is the reference for the specific value of r

A value of r with particular significance is that designated as the leastradius of gyration Since this value will be related to the least value of Ifor the cross section and since I is an index of the bending stiffness of themember then the least value for r will indicate the weakest response ofthe member to bending This relates specifically to the resistance of slen-der compression members to buckling Buckling is essentially a sidewaysbending response and its most likely occurrence will be on the axis iden-tified by the least value of I or r Use of these relationships for columnsis discussed in Chapter 12

95 TABLES OF PROPERTIES OF SECTIONS

Figure 912 presents formulas for obtaining geometric properties of var-ious simple plane sections Some of these may be used for single-piecestructural members or for the building up of complex members

rI

A=

SI

c= = =697 068

5 75121 229

TABLES OF PROPERTIES OF SECTIONS 229

3751 P-09 111301 1224 PM Page 229


Figure 912 Properties of various geometric shapes of cross sections

3751 P-09 111301 1224 PM Page 230

Tables 93 through 97 present the properties of various plane sec-tions These are sections identified as those of standard industry-produced sections of wood and steel Standardization means that theshapes and dimensions of the sections are fixed and each specific sectionis identified in some way

Structural members may be employed for various purposes and thusthey may be oriented differently for some structural uses Of note for anyplane section are the principal axes of the section These are the two mu-tually perpendicular centroidal axes for which the values will be great-est and least respectively for the section thus the axes are identified asthe major and minor axes If sections have an axis of symmetry it will al-ways be a principal axismdasheither major or minor

For sections with two perpendicular axes of symmetry (rectangle HI etc) one axis will be the major axis and the other the minor axis In thetables of properties the listed values for I S and r are all identified as toa specific axis and the reference axes are identified in a figure for thetable

Other values given in the tables are for significant dimensions totalcross-sectional area and the weight of a 1-ft-long piece of the memberThe weight of wood members is given in the table assuming an averagedensity for structural softwood of 35 lbft3 The weight of steel membersis given for W and channel shapes as part of their designation thus a W 8 times 67 member weighs 67 lbft For steel angles and pipes the weightis given in the table as determined from the density of steel at 490 lbft3

The designation of some members indicates their true dimensionsThus a 10-in channel and a 6-in angle have true dimensions of 10 and 6in For W-shapes pipe and structural lumber the designated dimensionsare nominal and the true dimensions must be obtained from the tables


3751 P-09 111301 1224 PM Page 231


TABLE 93 Properties of W-Shapes

Flange Elastic PropertiesWeb

Width Thickness Axis X-X Axis Y-YPlastic

Area Depth Thickness ModulusA d tw bf tf k I S r I S r Zx

Shape (in2) (in) (in) (in) (in) (in) (in4) (in3) (in) (in4) (in3) (in) (in3)

W 30 times 116 342 3001 0565 10495 0850 1625 49300 3290 1200 1640 3130 219 3780times 108 317 2983 0545 10475 0760 1562 4470 2990 1190 1460 2790 215 3460times 099 291 2965 0520 10450 0670 1437 3990 2690 1170 1280 2450 210 3120

W 27 times 094 277 2692 0490 09990 0745 1437 3270 2430 1090 1240 2480 212 2780times 084 248 2671 0460 09960 0640 1375 2850 2130 1070 1060 2120 207 2440


W 21 times 083 243 2143 0515 08355 0835 1562 1830 1710 0867 0814 1950 183 1960times 073 215 2124 0455 08295 0740 1500 1600 1510 0864 0706 1700 181 1720times 057 167 2106 0405 06555 0650 1375 1170 1110 0836 0306 0935 135 1290times 050 147 2083 0380 06530 0535 1312 0984 0945 0818 0249 0764 130 1100

W 18 times 086 253 1839 0480 11090 0770 1437 1530 1660 0777 1750 3160 263 1860times 076 223 1821 0425 11035 0680 1375 1330 1460 0773 1520 2760 261 1630times 060 176 1824 0415 07555 0695 1375 0984 1080 0747 0501 1330 169 1230times 055 162 1811 0390 07530 0630 1312 0890 0983 0741 0449 1190 167 1120times 050 147 1799 0355 07495 0570 1250 0800 0889 0738 0401 1070 165 1010times 046 135 1806 0360 06060 0605 1250 0712 0788 0725 0225 0743 129 0907times 040 118 1790 0315 06015 0525 1187 0612 0684 0721 0191 0635 127 0784


W 14 times 216 620 1572 0980 15800 1560 2250 2660 338 655 1030 130 407 3900times 176 518 1522 0830 15650 1310 2000 2140 281 643 838 107 402 3200times 132 388 1466 0645 14725 1030 1687 1530 209 628 548 745 376 2340times 120 353 1448 0590 14670 0940 1625 1380 190 624 495 675 374 2120times 74 218 1417 0450 10070 0785 1562 796 112 604 134 266 248 1260times 68 200 1404 0415 10035 0720 1500 723 103 601 121 242 246 1150times 48 141 1379 0340 8030 0595 1375 485 703 585 514 128 191 784times 43 126 1366 0305 7995 0530 1312 428 627 582 452 113 189 696times 34 100 1398 0285 6745 0455 1000 340 486 583 233 691 153 546times 30 885 1384 0270 6730 0385 0937 291 420 573 196 582 149 473

3751 P-09 111301 1224 PM Page 232


TABLE 93 (continued )





W 12 times 136 399 1341 0790 12400 1250 1937 12400 1860 558 3980 642 316 214times 120 353 1312 0710 12320 1105 1812 10700 1630 551 345 560 313 186times 72 211 1225 0430 12040 0670 1375 5970 974 531 195 324 304 108times 65 191 1212 0390 12000 0605 1312 5330 879 528 174 291 302 968times 53 156 1206 0345 9995 0575 1250 4250 706 523 958 192 248 779times 45 132 1206 0335 8045 0575 1250 3500 581 515 500 124 194 647times 40 118 1194 0295 8005 0515 1250 3100 519 513 441 110 193 575times 30 879 1234 0260 6520 0440 0937 2380 386 521 203 624 152 431times 26 765 1222 0230 6490 0380 0875 2040 334 517 173 534 151 372


Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction Chicago IL This table is a sample from anextensive set of tables in the reference document

3751 P-09 111301 1224 PM Page 233



Width Thickness Axis X-X Axis Y-YArea Depth ThicknessA d tw bf tf k I S r I S r xa eo

b

Shape (in2) (in) (in) (in) (in) (in) (in4) (in3) (in) (in4) (in3) (in) (in) (in)

TABLE 94 Properties of American Standard Channels

C 15 times 50 147 150 0716 3716 0650 144 404 538 524 110 378 0867 0798 0583times 40 118 150 0520 3520 0650 144 349 465 544 923 337 0886 0777 0767times 339 996 150 0400 3400 0650 144 315 420 562 813 311 0904 0787 0896


C 10 times 30 882 100 0673 3033 0436 100 103 207 342 394 165 0669 0649 0369times 25 735 100 0526 2886 0436 100 912 182 352 336 148 0676 0617 0494times 20 588 100 0379 2739 0436 100 789 158 366 281 132 0692 0606 0637times 153 449 100 0240 2600 0436 100 674 135 387 228 116 0713 0634 0796





Source Adapted from data in the Manual of Steel Construction 8th edition with permission of the publishersAmerican Institute of Steel Construction Chicago IL This table is a sample from an extensive set of tables in thereference documentaDistance to centroid of sectionbDistance to shear center of section

3751 P-09 111301 1224 PM Page 234


TABLE 95 Properties of Single-Angle Shapes

8 times 8 times 11frasl8 175 569 167 980 175 242 241 980 175 242 241 156 1000times 1 162 510 150 890 158 244 237 890 158 244 237 156 1000

8 times 6 times 3frasl4 125 338 994 634 117 253 256 307 692 176 156 129 0551times 1frasl2 100 230 675 443 802 256 247 217 479 179 147 130 0558


6 times 4 times 5frasl8 112 200 586 211 531 190 203 752 254 113 103 0864 0435times 1frasl2 100 162 475 174 433 191 199 627 208 115 0987 0870 0440times 3frasl8 087 123 361 135 332 193 194 490 160 117 0941 0877 0446

5 times 31frasl2 times 1frasl2 100 136 400 999 299 158 166 405 156 101 0906 0755 0479times 3frasl8 087 104 305 778 229 160 161 318 121 102 0861 0762 0486




31frasl2 times 31frasl2 times 3frasl8 075 85 248 287 115 107 101 287 115 107 101 0687 1000times 5frasl16 069 72 209 245 0976 108 0990 245 0976 108 0990 0690 1000






21frasl2 times 2 times 3frasl8 069 53 155 0912 0547 0768 0831 0514 0363 0577 0581 0420 0614times 5frasl16 062 45 131 0788 0466 0776 0809 0446 0310 0584 0559 0422 0620

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of the publishersAmerican Institute of Steel Construction Chicago IL This table is a sample from an extensive set of tables in thereference document

Axis X-X Axis Y-Y Axis Z-ZSize Weightand per Area

Thickness k ft A I S r y I S r x r tan a(in) (in) (lb) (in2) (in4) (in3) (in) (in) (in4) (in3) (in) (in) (in)

3751 P-09 111301 1224 PM Page 235


TABLE 96 Properties of Standard Weight Steel Pipe

Dimensions Properties

Nominal Outside Inside Wall WeightDiameter Diameter Diameter Thickness per ft A I S r(in) (in) (in) (in) (lb) (in2) (in4) (in3) (in)

1frasl2 0840 0622 0109 085 0250 0017 0041 02613frasl4 1050 0824 0113 113 0333 0037 0071 0334

1 1315 1049 0133 168 0494 0087 0133 042111frasl4 1660 1380 0140 227 0669 0195 0235 054011frasl2 1900 1610 0145 272 0799 0310 0326 06232 2375 2067 0154 365 1070 0666 0561 078721frasl2 2875 2469 0203 579 1700 153 1060 09473 3500 3068 0216 758 2230 302 1720 116031frasl2 4000 3548 0226 911 2680 479 2390 13404 4500 4026 0237 1079 3170 723 3210 15105 5563 5047 0258 1462 4300 152 5450 18806 6625 6065 0280 1897 5580 281 8500 22508 8625 7981 0322 2855 8400 725 16800 2940

10 10750 10020 0365 4048 1190 161 29900 367012 12750 12000 0375 4956 1460 279 43800 4380

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of the publishers

American Institute of Steel Construction Chicago IL This table is a sample from an extensive set of tables in the

reference document

3751 P-09 111301 1224 PM Page 236


TABLE 97 Properties of Structural Lumber

Dimensions (in)Section Moment

Area Modulus of InertiaNominal Actual A S I Weighta

b times h b times h (in2) (in3) (in4) (lbft)

2 times 3 15 times 25 03750 001563 0001953 0092 times 4 15 times 35 05250 003063 0005359 0132 times 6 15 times 55 08250 007563 0020797 0202 times 8 15 times 725 10875 013141 0047635 0262 times 10 15 times 925 13875 021391 0098932 0342 times 12 15 times 1125 16875 031641 0177979 0412 times 14 15 times 1325 19875 043891 0290775 0483 times 2 25 times 15 03750 000938 0000703 0093 times 4 25 times 35 08750 005104 0008932 0213 times 6 25 times 55 13750 012604 0034661 0333 times 8 25 times 725 18125 021901 0079391 0443 times 10 25 times 925 23125 035651 0164886 0563 times 12 25 times 1125 28125 052734 0296631 0683 times 14 25 times 1325 33125 073151 0484625 0813 times 16 25 times 1525 38125 096901 0738870 0934 times 2 35 times 15 05250 001313 0000984 0134 times 3 35 times 25 08750 003646 0004557 0214 times 4 35 times 35 12250 007146 0012505 0304 times 6 35 times 55 19250 017646 0048526 0474 times 8 35 times 725 25375 030661 0111148 0624 times 10 35 times 925 32375 049911 0230840 0794 times 12 35 times 1125 39375 073828 0415283 0964 times 14 35 times 1325 46375 102411 0678475 1134 times 16 35 times 1525 53375 135661 1034418 13006 times 2 55 times 15 00825 0002063 00001547 02006 times 3 55 times 25 01375 0005729 00007161 03306 times 4 55 times 35 01925 0011229 00019651 04706 times 6 55 times 55 03025 0027729 00076255 07416 times 10 55 times 95 05225 0082729 00392963 12706 times 12 055 times 115 06325 0121229 00697068 154

3751 P-09 111301 1224 PM Page 237


06 times 14 055 times 135 07425 0167063 01127672 18006 times 16 055 times 155 08525 0220229 01706776 20708 times 2 725 times 150 010875 0002719 00002039 02608 times 3 725 times 250 018125 0007552 00009440 04408 times 4 725 times 350 025375 0014802 00025904 06208 times 6 75 times 55 04125 0037813 00103984 10008 times 8 75 times 75 05625 0070313 00263672 13708 times 10 75 times 95 07125 0112813 00535859 17308 times 12 075 times 115 08625 0165313 00950547 21008 times 14 075 times 135 10125 0227813 01537734 24608 times 16 075 times 155 11625 0300313 02327422 28308 times 18 075 times 175 13125 0382813 03349609 31908 times 20 075 times 195 14625 0475313 04634297 35510 times 10 95 times 95 09025 0142896 00678755 21910 times 12 95 times 115 10925 0209396 01204026 26610 times 14 95 times 135 12825 0288563 01947797 31210 times 16 95 times 155 14725 0380396 02948068 35810 times 18 95 times 175 16625 0484896 04242836 40410 times 20 95 times 195 18525 0602063 05870109 45012 times 12 115 times 115 13225 0253479 01457505 32112 times 14 115 times 135 15525 0349313 02357859 37712 times 16 115 times 155 17825 0460479 03568713 43312 times 18 115 times 175 20125 0586979 05136066 48912 times 20 115 times 195 22425 0728813 07105922 54512 times 22 115 times 215 24725 0885979 09524273 60112 times 24 115 times 235 27025 1058479 12437129 65714 times 14 135 times 135 18225 0410063 02767922 44316 times 16 155 times 155 24025 0620646 04810004 584

Source Compiled from data in the National Design Specification for Wood Construction 1982 edwith permission of the publishers National Forest Products Association Washington DCaBased on an assumed average density of 35 psf

TABLE 97 (Continued )




3751 P-09 111301 1224 PM Page 238

239

10STRESS AND

DEFORMATION

Structural actions develop stresses in the material of a structure and ac-companying shape changes or deformations (see Figure 101) Simpleforces of compression and tension produce corresponding direct stressesof compression or tension in the material and accompanying shorteningor lengthening as shape changes Shear produces a slipping type of stressand an angular change as deformation All other force actions and com-binations of actions produce some combination of these three basic typesof stress compression tension and shear For example bending pro-duces a combination of opposed compression and tension in the affectedstructural member the accumulation of which over the memberrsquos lengthresults in curvature of the member

This chapter presents some basic considerations for the structural be-havior of materials

3751 P-10 111301 1225 PM Page 239

240 STRESS AND DEFORMATION

Figure 101 Force actions produce stresses in the materials of a structure Theseincremental stresses accumulate to achieve overall deformations of structuressuch as the deflection of beams

3751 P-10 111301 1225 PM Page 240

101 MECHANICAL PROPERTIES OF MATERIALS

Stresses are visualized as unit stresses and are measured in terms of forceper unit area The unit area is usually an increment of the area of a crosssection of a structural member and the force is that required to be devel-oped at the cross section Thus in Figure 102a the force of 6400 lb pro-duces a unit stress of 100 psi on the 64 in2 of the cross section of the postIn a similar manner it can be determined that the tension force of the1500-lb block in Figure 102c produces a tension stress of 7653 psi in the1frasl2-in diameter rod

Direct shear actions can also be treated in this manner Thus if thebolt in Figure 102d is 3frasl4-in in diameter and is subjected to a force of5000 lb as shown the direct slicing shear stress will be 11317 psi

In these situations the relationship between the force the area of thecross section and the unit stress may be stated in general terms as

in which

P = axial direct force in pounds newtons and so on

f = unit stress in pounds per square inch (psi) and so on

A = area of the stressed cross section in units of in2 and so on

The first form of the stress equation is used to determine the capacityof a member with a given cross section and a specific limiting stress Thesecond form is used to investigate a stress condition for a given memberunder a specified load The third form is used directly in design work todetermine the required cross-sectional area for a member with a limitingstress and a required load

Deformation

Whenever a force acts on a body there is an accompanying change inshape or size of the body In structural mechanics this is called defor-mation Regardless of the magnitude of the force some deformation is al-ways present although often it is so small that it is difficult to measureeven with the most sensitive instruments In the design of structures it is

P f A fP

AA

P

f= times = =or or

MECHANICAL PROPERTIES OF MATERIALS 241

3751 P-10 111301 1225 PM Page 241

often necessary to know what the deformation in certain members willbe A floor joist for instance may be large enough to support a givenload safely but may deflect (the term for deformation that occurs withbending) to such an extent that the plaster ceiling below it will crack orthe floor may feel excessively springy to persons walking on it For theusual cases we can readily determine what the deformation will be Thisis considered in more detail later

Strength

The strength of a material or a structural member is the measure of its ca-pacity to resist force Strength of a material may be expressed in terms of



3751 P-10 111301 1225 PM Page 242

its resistance to the three basic stresses compression tension and shearStrength of a structural member may be expressed in terms of its resis-tance to a particular structural action such as direct compression directtension bending and so on

102 DESIGN USE OF DIRECT STRESS

In the examples and problems dealing with the direct stress equation dif-ferentiation was made between the unit stress developed in a membersustaining a given load ( f = PA) and the allowable unit stress used whendetermining the size of a member required to carry a given load (A =Pf ) The latter form of the equation is of course the one used in designThe procedures for establishing allowable unit stresses in tension com-pression shear and bending are different for different materials and areprescribed in industry-prepared specifications A sample of data fromsuch references is presented in Table 101

In actual design work the building code governing the construction ofbuildings in the particular locality must be consulted for specific re-quirements Many municipal codes are revised infrequently and conse-quently may not be in agreement with current editions of the industryrecommended allowable stresses

DESIGN USE OF DIRECT STRESS 243

TABLE 101 Selected Values for Common Structural Materials

Common Values

Material and Property psi kPa

Structural SteelYield strength 36000 248220Allowable tension 22000 151690Modulus of elasticity E 29000000 200000000

Concretef centc (specified compressive strength) 3000 20685Usable compression in bearing 900 6206Modulus of elasticity E 3100000 21374500

Structural Lumber (Douglas FirndashLarchSelect Structural Grade Posts and Timbers)Compression parallel to grain 1150 7929Modulus of elasticity E 1600000 11032000

3751 P-10 111301 1225 PM Page 243

Except for shear the stresses discussed so far have been direct or axialstresses This means they are assumed to be uniformly distributed overthe cross section The examples and problems presented fall under threegeneral types first the design of structural members (A = Pf ) secondthe determination of safe loads (P = fA) third the investigation of mem-bers for safety ( f = PA) The following examples will serve to fix inmind each of these types

Example 1 Design (determine the size of) a short square post of Dou-glas fir select structural grade to carry a compressive load of 30000 lb[133440 N]

Solution Referring to Table 101 the allowable unit compressive stressfor this wood parallel to the grain is 1150 psi [7929 kPa] The requiredarea of the post is

From Table 97 an area of 3025 in2 [19517 mm2] is provided by a 6 times 6 in post with a dressed size of 51frasl2 times 51frasl2 in [1397 mm]

Example 2 Determine the safe axial compressive load for a shortsquare concrete pier with a side dimension of 2 ft [06096 m]

Solution The area of the pier is 4 ft2 or 576 in2 [03716 m2] Table 101gives the allowable unit compressive stress for concrete as 900 psi [6206kPa] Therefore the safe load on the pier is

P = ( f )(A) = (900)(576) = 528400 lb [206 kN]

Example 3 A running track in a gymnasium is hung from the rooftrusses by steel rods each of which supports a tensile load of 11200 lb[49818 N] The round rods have a diameter of 7frasl8 in [2223 mm] with theends upset that is made larger by forging This upset allows the fullcross-sectional area of the rod (0601 in2) [388 mm2] to be utilized oth-erwise the cutting of the threads will reduce the cross section of the rodInvestigate this design to determine whether it is safe

AP

f= = = [ ]30 000

115026 09 16 8292 2

in mm


3751 P-10 111301 1225 PM Page 244

Solution Since the gross area of the hanger rod is effective the unitstress developed is

Table 101 gives the allowable unit tensile stress for steel as 22000psi [151690 kPa] which is greater than that developed by the loadingTherefore the design is safe

Shearing Stress Formula

The foregoing manipulations of the direct stress formula can of coursebe carried out also with the shearing stress formula fv = PA However itmust be borne in mind that the shearing stress acts transversely to thecross sectionmdashnot at right angles to it Furthermore while the shearingstress equation applies directly to the situation illustrated by Figures102d and e it requires modification for application to beams The lattersituation is considered in more detail in Section 115

Problem 102AWhat should be the minimum cross-sectional area of a steel rod to sup-port a tensile load of 26 kips [115648 kN]

Problem 102BA short square post of Douglas fir select structural grade is to support anaxial load of 61 kips [2713 kN] What should its nominal dimensions be

Problem 102CA steel rod has a diameter of 125 in [3175 mm] What safe tensile loadwill it support if its ends are upset

Problem 102DWhat safe load will a short 12 times 12 in [actually 2921 mm] Douglas firpost support if the grade of the wood is select structural grade

Problem 102EA short post of Douglas fir select structural grade with nominal dimen-sions of 6 times 8 in [actually 1397 times 1905 mm] supports an axial load of 50kips [2224 kN] Investigate this design to determine whether it is safe

fP

A= = = [ ]11 200

0 60118 636 128 397

psi kPa


3751 P-10 111301 1225 PM Page 245

Problem 102FA short concrete pier 1 ft 6 in [4572 mm] square supports an axial loadof 150 kips [6672 kN] Is the construction safe

103 DEFORMATION AND STRESS RELATIONS AND ISSUES

Stress is a major issue primarily for determination of the strength of struc-tures However deformation due to stress is often of concern and the re-lation of stress to strain is one that must be quantitatively establishedThese relations and the issues they raise are discussed in this section

Hookersquos Law

As a result of experiments with clock springs Robert Hooke a mathe-matician and physicist working in the seventeenth century developed thetheory that ldquodeformations are directly proportional to stressesrdquo In otherwords if a force produces a certain deformation twice the force will pro-duce twice the amount of deformation This law of physics is of utmostimportance in structural engineering although as we shall find Hookersquoslaw holds true only up to a certain limit

Elastic Limit and Yield Point

Suppose that a bar of structural steel with a cross-sectional area of 1 in2

is placed into a machine for making tension tests Its length is accuratelymeasured and then a tensile force of 5000 lb is applied which of courseproduces a unit tensile stress of 5000 psi in the bar Measuring the lengthagain it is found that the bar has lengthened a definite amount call it xinches On applying 5000 lb more the amount of lengthening is now2(x) or twice the amount noted after the first 5000 lb If the test is con-tinued it will be found that for each 5000 lb increment of additionalload the length of the bar will increase the same amount as noted whenthe initial 5000 lb was applied that is the deformations (length changes)are directly proportional to the stresses So far Hookersquos law has heldtrue but when a unit stress of about 36000 psi is reached the length in-creases more than x for each additional 5000 lb of load This unit stressis called the elastic limit or the yield stress Beyond this stress limitHookersquos law will no longer apply


3751 P-10 111301 1225 PM Page 246

Another phenomenon may be noted in this connection In the test justdescribed it will be observed that when any applied load that producesa unit stress less than the elastic limit is removed the bar returns to itsoriginal length If the load producing a unit stress greater than the elas-tic limit is removed it will be found that the bar has permanently in-creased its length This permanent deformation is called the permanentset This fact permits another way of defining the elastic limit it is thatunit stress beyond which the material does not return to its original lengthwhen the load is removed

If this test is continued beyond the elastic limit a point is reachedwhere the deformation increases without any increase in the load Theunit stress at which this deformation occurs is called the yield point it hasa value only slightly higher than the elastic limit Since the yield point oryield stress as it is sometimes called can be determined more accuratelyby test than the elastic limit it is a particularly important unit stressNonductile materials such as wood and cast iron have poorly definedelastic limits and no yield point

Ultimate Strength

After passing the yield point the steel bar of the test described in the pre-ceding discussion again develops resistance to the increasing load Whenthe load reaches a sufficient magnitude rupture occurs The unit stress inthe bar just before it breaks is called the ultimate strength For the gradeof steel assumed in the test the ultimate strength may occur at a stress ashigh as about 80000 psi

Structural members are designed so that stresses under normal serviceconditions will not exceed the elastic limit even though there is consid-erable reserve strength between this value and the ultimate strength Thisprocedure is followed because deformations produced by stresses abovethe elastic limit are permanent and hence change the shape of the struc-ture in a permanent manner

Factor of Safety

The degree of uncertainty that exists with respect to both actual loadingof a structure and uniformity in the quality of materials requires thatsome reserve strength be built into the design This degree of reservestrength is the factor of safety Although there is no general agreement on

DEFORMATION AND STRESS RELATIONS AND ISSUES 247

3751 P-10 111301 1225 PM Page 247

the definition of this term the following discussion will serve to fix theconcept in mind

Consider a structural steel that has an ultimate tensile unit stress of58000 psi a yield-point stress of 36000 psi and an allowable stress of22000 psi If the factor of safety is defined as the ratio of the ultimatestress to the allowable stress its value is 58000 divide 22000 or 264 On theother hand if it is defined as the ratio of the yield-point stress to the al-lowable stress its value is 36000 divide 22000 or 164 This is a consider-able variation and since deformation failure of a structural memberbegins when it is stressed beyond the elastic limit the higher value maybe misleading Consequently the term factor of safety is not employedextensively today Building codes generally specify the allowable unitstresses that are to be used in design for the grades of structural steel tobe employed

If one should be required to pass judgment on the safety of a structurethe problem resolves itself into considering each structural element find-ing its actual unit stress under the existing loading conditions and com-paring this stress with the allowable stress prescribed by the localbuilding regulations This procedure is called structural investigation

Modulus of Elasticity

Within the elastic limit of a material deformations are directly propor-tional to the stresses The magnitude of these deformations can be com-puted by use of a number (ratio) called the modulus of elasticity thatindicates the degree of stiffness of a material

A material is said to be stiff if its deformation is relatively small whenthe unit stress is high As an example a steel rod 1 in2 in cross-sectionalarea and 10 ft long will elongate about 0008 in under a tensile load of2000 lb But a piece of wood of the same dimensions will stretch about024 in with the same tensile load The steel is said to be stiffer than thewood because for the same unit stress the deformation is not as great

Modulus of elasticity is defined as the unit stress divided by the unitdeformation Unit deformation refers to the percent of deformation and isusually called strain It is dimensionless since it is expressed as a ratio asfollows

strain = =se

L


3751 P-10 111301 1225 PM Page 248

in which

s = the strain or the unit deformation

e = the actual dimensional change

L = the original length of the member

The modulus of elasticity for direct stress is represented by the letterE expressed in pounds per square inch and has the same value in com-pression and tension for most structural materials Letting f represent theunit stress and s the strain then by definition

From Section 101 f = PA It is obvious that if L represents thelength of the member and e the total deformation then s the deformationper unit of length must equal the total deformation divided by the lengthor s = eL Now by substituting these values in the equation determinedby definition

This can also be written in the form

in which

e = total deformation in inches

P = force in pounds

L = length in inches

A = cross-sectional area in square inchesE = modulus of elasticity in pounds per square inch

Note that E is expressed in the same units as f (pounds per square inch[kilopascals]) because in the equation E = fs s is a dimensionlessnumber For steel E = 29000000 psi [200000000 kPa] and for wooddepending on the species and grade it varies from something less than1000000 psi [6895000 kPa] to about 1900000 psi [13100000 kPa]

ePL

AE=

Ef

s

P A

e L

PL

Ae= = =

Ef

s=


3751 P-10 111301 1225 PM Page 249

For concrete E ranges from about 2000000 psi [13790000 kPa] toabout 5000000 psi [34475000 kPa] for common structural grades

Example 4 A 2-in [508-mm] diameter round steel rod 10 ft [305 m]long is subjected to a tensile force of 60 kips [26688 kN] How muchwill it elongate under the load

Solution The area of the 2-in rod is 31416 in2 [2027 mm2] Checkingto determine whether the stress in the bar is within the elastic limit wefind that

which is within the elastic limit of ordinary structural steel (36 ksi) so theformula for finding the deformation is applicable From data P = 60kips L = 120 (length in inches) A = 31416 and E = 29000000 Sub-stituting these values we calculate the total lengthening of the rod as

Problem 103AWhat force must be applied to a steel bar 1 in [254 mm] square and 2ft [610 mm] long to produce an elongation of 0016 in [04064 mm]

Problem 103BHow much will a nominal 8 times 8 in [actually 1905 mm] Douglas fir post12 ft [3658 m] long shorten under an axial load of 45 kips [200 kN]

Problem 103CA routine quality control test is made on a structural steel bar that is 1 in[254 mm] square and 16 in [406 mm] long The data developed duringthe test show that the bar elongated 00111 in [0282 mm] when sub-jected to a tensile force of 205 kips [91184 kN] Compute the modulusof elasticity of the steel

Problem 103DA 1frasl2 in [127-mm] diameter round steel rod 40 ft [1219 m] long supportsa load of 4 kips [1779 kN] How much will it elongate

ePL

AE= = times

times= [ ]60 000 120

3 1416 29 000 0000 079 2 0

in mm

fP

A= = = [ ]60

3 141619 1 131 663

ksi kPa


3751 P-10 111301 1225 PM Page 250

104 INELASTIC AND NONLINEAR BEHAVIOR

Most of the discussion of stress and strain behavior presented so far inthis book relates to the idealized theories of classic structural analysisbased on elastic and linear conditions of stressstrain interaction Whilethis assumption is useful for simple definitions and for derivations of fun-damental relationships from which the basic equations for stress andstrain computations are obtained actual behavior of common structuralmaterials often varies considerably from this ideal condition

Figure 103 is a repeat of Figure 137 where it was initially used to de-fine some fundamental terms and relationships Linear stressstrain be-havior is that represented by curves 1 and 2 in the figure A nonlinearstressstrain relationship is demonstrated by curve 3 in the figure While asingle value for the modulus of elasticity (E ) may be obtained for thematerials represented by curves 1 and 2 such is not the case for the mate-rial represented by curve 3 Metals and ceramics generally exhibit thebehavior shown by curves 1 and 2 and thus a single value for E maybe used for these materials throughout a considerable range of stressmagnitude Wood and concrete have responses more of the form of curve

INELASTIC AND NONLINEAR BEHAVIOR 251

Figure 103 Stress-strain relationships through a range from zero stress to failure

3751 P-10 111301 1225 PM Page 251

3 and thus some adjustment must be made when stressstrain involve-ments extend over some range of magnitude

A second consideration concerns the relative elasticity of materialsThis generally refers to how much of the strain is recoverable when thestress is removed from the material A good rubber band for examplecan be stretched considerably and be expected to return entirely to itsoriginal length when let go The same goes for structural materialsmdashupto some limit Consider curve 4 in Figure 103 which represents the gen-eral nature of stressstrain response of a ductile material such as ordinarystructural steel Initially this material displays a linear stressstrain re-sponse However when the yield point of the stress magnitude isreached considerable strain occurs without an increase in stress Up tothe yield point the strain is recoverable (material remains elastic) but de-formations beyond this limit will produce some permanent change Thisphenomenon is illustrated in Figure 104 in which the portion of the linewith downward-pointing arrows indicates what the stressstrain responsewill be when the stress magnitude is reduced to zero after strain beyondthe yield point occurs

These issues relate to the general behavior of real structural materialsThey become increasingly of concern when behaviors are projected tothe ultimate response limits of materials It is possible that they may beof less concern for behaviors within the general usage limits that is upto the maximum anticipated service conditions for a structure Thereforethey are not so much a concern for the expected actual use of the struc-


Figure 104 Stress-strain behaviorfor a ductile material

3751 P-10 111301 1225 PM Page 252

ture However they maymdashand indeed domdashrelate quite significantly tothe character of responses at the ultimate capacity of the structure interms of material behavior

The work in this book being of an introductory nature deals primarilywith simple idealized material responses This is both a logical startingpoint and a necessary reference point for more complex investigationsMost current structural design work uses methods that are based on eval-uation of ultimate load conditions called strength behavior or strength de-sign The use of the term strength here refers to ultimate strength of thematerials or of the whole structure It is not possible here to fully presentthe background for these methods which are unavoidably based consid-erably on inelastic and nonlinear behaviors However some discussion ofnonlinear behavior of steel is provided in Section 1110 and the ultimatestress limit for concrete is discussed in Chapter 15


3751 P-10 111301 1225 PM Page 253

254

11STRESS AND

STRAIN IN BEAMS

The behavior of beams with regard to their resolution of the externalforces of the loads and support reactions is discussed in Chapter 4 Alsodiscussed in Chapter 4 is the development of the internal force effects ofshear and bending moment as generated by the external forces In thischapter the discussion relates to how the beam produces the necessaryinternal resistance to shear and bending through stresses in the materialof the beam Since stress is unavoidably accompanied by strain it is alsonecessary to consider the deformation of the beam the major effect con-sists of deflection manifested as a curving of the beam away from itsform prior to loading

Primary considerations for beam stresses and strains have influencedthe development of widely used structural products such as the I-shapedsteel shapes shown in Figure 111 For the I-shaped beam the verticalweb is ideally oriented for resistance to vertical shear forces and thewidely separated flanges are ideally oriented for opposed tensioncom-pression forces to resist bending moment

3751 P-11 111301 1225 PM Page 254

111 DEVELOPMENT OF BENDING RESISTANCE

As developed in the preceding sections bending moment is a measure ofthe tendency of the external forces on a beam to deform it by bendingThe purpose of this section is to consider the action within the beam thatresists bending called the resisting moment

Figure 112a shows a simple beam rectangular in cross section sup-porting a single concentrated load P Figure 112b is an enlarged sketchof the left-hand portion of the beam between the reaction and section X-X It is observed that the reaction R1 tends to cause a clockwise rotationabout point A in the section under consideration this is defined as the

DEVELOPMENT OF BENDING RESISTANCE 255

Figure 111 The standard cross section of steel I-shaped beams has been es-tablished with consideration of the properties of the material the basic productionprocess of hot rolling and the means of attachment of steel members in buildingframeworks However the primary consideration is the usage of the member as abeam with its web oriented in the plane of vertical gravity force The form and spe-cific dimensions of each of the several hundred standard shapes thus respondsessentially to the stress and strain functions for beam action Reproduced fromFundamentals of Building Construction 2nd edition by E Allen 1990 with per-mission of the publisher John Wiley amp Sons New York

3751 P-11 111301 1225 PM Page 255

bending moment in the section In this type of beam the fibers in theupper part are in compression and those in the lower part are in tensionThere is a horizontal plane separating the compressive and tensilestresses it is called the neutral surface and at this plane there are neithercompressive nor tensile stresses with respect to bending The line inwhich the neutral surface intersects the beam cross section (Figure 112c)is called the neutral axis NA

Call C the sum of all the compressive stresses acting on the upper partof the cross section and call T the sum of all the tensile stresses acting onthe lower part It is the sum of the moments of those stresses at the sec-tion that holds the beam in equilibrium this is called the resisting mo-ment and is equal to the bending moment in magnitude The bendingmoment about A is R1 times x and the resisting moment about the same pointis (C times y) + (T times y) The bending moment tends to cause a clockwise ro-tation and the resisting moment tends to cause a counterclockwise rota-tion If the beam is in equilibrium these moments are equal or

R1 times x = (C times y) + (T times y)

256 STRESS AND STRAIN IN BEAMS

Figure 112 Development of bending stress in a beam

3751 P-11 111301 1225 PM Page 256

that is the bending moment equals the resisting moment This is the the-ory of flexure (bending) in beams For any type of beam it is possible tocompute the bending moment and to design a beam to withstand this ten-dency to bend this requires the selection of a member with a cross sec-tion of such shape area and material that it is capable of developing aresisting moment equal to the bending moment

The Flexure Formula

The flexure formula M = fS is an expression for resisting moment (rep-resented by M) that involves the size and shape of the beam cross section(represented by S in the formula) and the material of which the beam ismade (represented by f ) It is used in the design of all homogeneousbeams that is beams made of one material only such as steel or woodThe following brief derivation is presented to show the principles onwhich the formula is based

Figure 113 represents a partial side elevation and the cross section ofa homogeneous beam subjected to bending stresses The cross sectionshown is unsymmetrical about the neutral axis but this discussion ap-plies to a cross section of any shape In Figure 113a let c be the distanceof the fiber farthest from the neutral axis and let f be the unit stress on thefiber at distance c If f the extreme fiber stress does not exceed the elas-tic limit of the material the stresses in the other fibers are directly pro-portional to their distances from the neutral axis That is to say if onefiber is twice as far from the neutral axis as another fiber the fiber at thegreater distance will have twice the stress The stresses are indicated in


Figure 113 Distribution of bending stress on a beam cross section

3751 P-11 111301 1225 PM Page 257

the figure by the small lines with arrows which represent the compres-sive and tensile stresses acting toward and away from the section re-spectively If c is in inches the unit stress on a fiber at 1-in distance isfc Now imagine an infinitely small area a at z distance from the neutralaxis The unit stress on this fiber is ( f c) times z and because this small areacontains a square inches the total stress on fiber a is ( fc) times z times a Themoment of the stress on fiber a at z distance is

There is an extremely large number of these minute areas Using thesymbol Σ to represent the sum of this very large number

means the sum of the moments of all the stresses in the cross section withrespect to the neutral axis This is the resisting moment and it is equal tothe bending moment

Therefore

The quantity Σ(a times z2) may be read ldquothe sum of the products of all the el-ementary areas times the square of their distances from the neutral axisrdquoThis is called the moment of inertia and is represented by the letter I (seeSection 92) Therefore substituting in the above

This is know as the flexure formula or beam formula and by its use it ispossible to design any beam that is composed of a single material Theexpression may be simplified further by substituting S for Ic called thesection modulus a term that is described more fully in Section 94 Mak-ing this substitution the formula becomes

M = fS

Mf

cI M

fI

cR R= times =or

Mf

ca zR = times sum times( )2

sum times times

f

ca z2

f

cz a z

f

ca ztimes times times times timesor 2


3751 P-11 111301 1225 PM Page 258

112 INVESTIGATION OF BEAMS

One use of the flexure formula is for the investigation of beams A pri-mary investigation is that performed to determine whether a beam isstrong enough for a certain loading In regard to bending the flexure for-mula may be used to determine the maximum bending stress caused bythe loading This stress is then compared to the maximum permitted forthe material of the beam

Another method for achieving the same investigation is to determinethe section modulus required based on the loading and the limiting bend-ing stress This value for S is then compared to that for the given beam

Finally a third method for achieving this investigation is to computethe maximum bending moment produced by the loading and then com-pare it to the maximum resisting moment for the beam as determined byits section modulus and the limiting bending stress

These three methods for investigating the same problem simply usethree variations of the form of the basic flexure formula The followingexample demonstrates these methods

Example 1 A W 10 times 33 steel beam is proposed to carry a total uni-formly distributed load of 30 kips on a span of 13 ft (see Figure 114)The maximum allowable bending stress is 24 ksi Determine whether thebeam is safe by (a) finding the maximum bending stress caused by theloading (b) comparing the required section modulus to that of the givenbeam (c) comparing the maximum bending moment due to the loadingto the maximum resisting moment of the beam

Solution From Case 2 in Figure 420 the equation for maximum bend-ing moment for the loading is found and computed as

MWL= = times = times =8

30 13

848 8 48 8 12 585 6 kip-ft or kip-in

INVESTIGATION OF BEAMS 259


3751 P-11 111301 1225 PM Page 259

From Table 93 the section modulus for the beam is 366 in3 Then

(a) The maximum bending stress due to the maximum moment is

As this is less than the allowable stress the beam is safe

(b) The section modulus required for the maximum moment of 5856kip-ft with the allowable stress of 24 ksi is

As this is less than that of the beam the beam is safe

(c) With the beamrsquos given section modulus and the limiting stressthe maximum resisting moment for the beam is

MR = fS = 24 times 366 = 8784 kip-in

As this is greater than the required maximum moment the beamis safe

Obviously it is not necessary to perform all three of these computa-tions as they all use the same basic equation and produce the same an-swer We use all three here to gain familiarity with the use of the flexureformula for different situations

Problem 112AA W 12 times 30 has a span of 10 ft with a uniformly distributed load of 36kips The allowable bending stress is 24 ksi Is the beam safe with respectto bending stress

Problem 112BA W 16 times 45 has a loading consisting of 10 kips at each of the quarterpoints of a 24-ft span (Figure 420 Case 5) and a uniformly distributedload of 52 kips The allowable bending stress is 24 ksi Is the beam safewith regard to bending stress

SM

f= = =585 6

2424 4 3

in

fM

S= = =585 6

36 616 0

ksi


3751 P-11 111301 1225 PM Page 260

113 COMPUTATION OF SAFE LOADS

The flexure formula can also be used to determine the allowable load thata given beam may carry In this case the given data include the beam spanthe beam section modulus and the allowable bending stress This basicproblem is used to establish data for tabulation of safe loads for beams forvarious spans The following examples demonstrate the process

Example 2 A W 12 times 30 has a span of 14 ft Find the maximum con-centrated load it will support at midspan if the allowable bending stressis 22 ksi

Solution From Table 93 the section modulus for the beam is 386 in3The maximum resisting moment for the beam is thus

This is the maximum resisting moment but part of it will be used up bythe beam in supporting its own weight As a uniformly distributed loadthe 30 lbft on the 14-ft span will produce a moment of

The resisting moment available for carrying the applied load is thus

M = 70767 ndash 735 = 70032 ft-lb

From Case 1 of Figure 420 the maximum moment for the concentratedloading is PL4 To solve for P find

Example 3 A W 12 times 40 is used as a simple beam on a span of 14 ftWhat is the maximum uniformly distributed load that this beam willcarry if the allowable stress is 24 ksi

MPL

PM

L= = = times =

4

4 4 70 032

1420 009

lb

MwL= = times =

2 2

8

30 14

8735 ft-lb

M fSR = = times =

=

22 000 38 6 849 200

849 200

1270 767

in-lb

or ft-lb

COMPUTATION OF SAFE LOADS 261

3751 P-11 111301 1225 PM Page 261

Solution From Table 93 the section modulus for this beam is 519 in3For this loading from Case 2 of Figure 420 the maximum moment isWL8 Then

The maximum resisting moment for the beam is

MR = fS = 24 times 519 = 12456 kip-in

Equating these two moments

The beam weighs a total of 14 times 40 = 560 lb or approximately 06 kipsDeducting this the total load the beam can support is 593 ndash 06 = 587kips

For the steel beams in the following problems ignore the beam weightand use an allowable bending stress of 24 ksi

Problem 113ACompute the maximum allowable uniformly distributed load for a sim-ple beam with a span of 16 ft if the section used is a W 12 times 30

Problem 113BAn 8 times 12 wood beam for which the allowable bending stress is 1400psi has a span of 15 ft with equal concentrated loads at the third pointsof the span (Case 3 in Figure 420) Compute the maximum permittedvalue for the individual load

Problem 113CA W 14 times 30 having a span of 14 ft supports a uniformly distributed loadof 7 kips and also a concentrated load at the center of the span Computethe maximum allowable value for the concentrated load

Problem 113DWhat is the maximum concentrated load that may be placed at the freeend of a cantilever beam 9 ft long if the section used is a W 12 times 26

21 1245 61245 6

2159 3W W= = =

kips

MWL W

W= = times times =8

14 12

821 kip-in


3751 P-11 111301 1225 PM Page 262

Problem 113EA simple beam has a span of 20 ft with a concentrated load placed 4 ftfrom one of the supports If the section is a W 16 times 36 compute the al-lowable value for the concentrated load

114 DESIGN OF BEAMS FOR FLEXURE

The flexure formula is used primarily to determine the size of a beamwith respect to strength in bending Shear and deflection must also beconsidered but it is common to first pick a size required for bending andthen to investigate its adequacy for shear and deflection The flexureformula may be used directly for this taskmdashas demonstrated in the fol-lowing examplesmdashbut the frequency of occurrence of the problem en-courages the use of various aids to shorten the process Professionaldesigners will commonly avail themselves of these aids

Example 4 A simple beam spans 22 ft and supports a uniformly dis-tributed load of 36 kips including the beam weight If allowable bendingstress is 24 ksi design a steel beam for strength in bending

Solution From Figure 420 Case 2

Using the flexure formula the required section modulus is found as

From Table 93 a W 16 times 36 has an S of 565 in3 and is therefore ac-ceptable Other sections having a section modulus of at least 495 in3 arealso acceptable If there is no other criteria the lightest-weight section isusually the most economical (The last number in the designation for theW-shape indicates its weight in pounds per foot of length)

Example 5 A simple beam of wood has a span of 16 ft and supports auniformly distributed load of 6500 lb including its own weight If the

SM

f= = =1188

2449 5 3 in


36 22

899 99 12 1188 kip-ft or kip-in

DESIGN OF BEAMS FOR FLEXURE 263

3751 P-11 111301 1225 PM Page 263

wood is to be Douglas fir Select Structural grade with allowable bend-ing stress of 1600 psi determine the required size of the beam with theleast cross-sectional area on the basis of limiting bending stress

Solution The maximum bending moment is

The allowable bending stress is 1600 psi Then the required section mod-ulus is determined as

From Table 97 the wood timber section with the least area to satisfy thisrequirement is a 6 times 14 with S = 167 in3

Ignore the beam weight in the following problems Use allowablebending stresses of 24 ksi for steel and 1600 psi for wood

Problem 114AA simple beam has a span of 17 ft and supports a uniformly distributedload of 23 kips Determine the size required for a steel W-shape with theleast weight to carry this load

Problem 114BTwo loads of 11 kips each occur at the third points of the span of a sim-ple beam with a span of 18 ft Find the least-weight W-shape that is acceptable

Problem 114CA simple beam with a 20-ft span has a concentrated load of 20 kips at itscenter and also a uniformly distributed load of 200 lbft over its entirelength Find the least-weight W-shape that is acceptable

Problem 114DA wood beam of Douglas fir Select Structural grade has a span of 15 ftand carries a concentrated load of 96 kips at 5 ft from one end Find theleast-weight (least cross-sectional area) member that is acceptable

SM

f= = =156 000

160097 5 3

in

MWL= = times =8

6500 16

813 000 156 000 lb-ft or lb-in


3751 P-11 111301 1225 PM Page 264

115 SHEAR STRESS IN BEAMS

Shear is developed in beams in direct resistance to the vertical force at abeam cross section Because of the interaction of shear and bending inthe beam the exact nature of stress resistance within the beam dependson the form and materials of the beam For an example in wood beamsthe wood grain is normally oriented in the direction of the span and thewood material has a very low resistance to horizontal splitting along thegrain An analogy to this is represented in Figure 115 which shows astack of loose boards subjected to a beam loading With nothing butminor friction between the boards the individual boards will slide overeach other to produce the loaded form indicated in the bottom figureThis is the failure tendency in the wood beam and the shear phenomenonfor wood beams is usually described as one of horizontal shear

Shear stresses in beams are not distributed evenly over the cross sec-tion of the beam as was assumed for the case of simple direct shear (seeSection 23) From observations of tested beams and derivations consid-ering the equilibrium of beam segments under combined actions of shearand bending the following expression has been obtained for shear stressin a beam

fVQ

Ibv =

SHEAR STRESS IN BEAMS 265

Figure 115 Nature of horizontal shear in beams

3751 P-11 111301 1225 PM Page 265

in which

V = shear force at the beam section

Q = moment about the neutral axis of the portion of the cross-sectional area between the edge of the section and the pointwhere stress is being computed

I = moment of inertia of the section with respect to the neutral(centroidal) axis

b = width of the section at the point where stress is beingcomputed

It may be observed that the highest value for Q and thus for shearstress will occur at the neutral axis and that shear stress will be zero atthe top and bottom edges of the section This is essentially opposite to theform of distribution of bending stress on a section The form of shear dis-tribution for various geometric shapes of beam sections is shown inFigure 116

The following examples illustrate the use of the general shear stressformula

Example 6 A rectangular beam section with depth of 8 in and width of4 in sustains a shear force of 4 kips Find the maximum shear stress (seeFigure 117a)


Figure 116 Distribution of shear stress in beams with various cross sections

3751 P-11 111301 1225 PM Page 266

Solution For the rectangular section the moment of inertia about thecentroidal axis is (see Figure 912)

The static moment (Q) is the product of the area acent and its centroidal dis-tance from the neutral axis of the section (y as shown in Figure 117b)This is the greatest value that can be obtained for Q and will produce thehighest shear stress for the section Thus

Q = acenty = (4 times 4)(2) = 32 in3

and

Example 7 A beam with the T-section shown in Figure 118a is sub-jected to a shear force of 8 kips Find the maximum shear stress and thevalue of shear stress at the location of the juncture of the web and theflange of the T

Solution Since this section is not symmetrical with respect to its hori-zontal centroidal axis the first steps for this problem consist of locatingthe neutral axis and determining the moment of inertia for the sectionwith respect to the neutral axis To save space this work is not shownhere although it is performed as Examples 1 and 8 in Chapter 9 From

fVQ

Ibv = = times

times=4000 32

170 7 4187 5

psi

Ibd= = times =

3 34

12

4 8

12170 7 in



3751 P-11 111301 1225 PM Page 267

that work it is found that the centroidal neutral axis is located at 65 infrom the bottom of the T and the moment of inertia about the neutral axisis 10467 in4

For computation of the maximum shear stress at the neutral axis thevalue of Q is found by using the portion of the web below the neutralaxis as shown in Figure 118c Thus

and the maximum stress at the neutral axis is thus

For the stress at the juncture of the web and flange Q is determinedusing the area shown in Figure 118d Thus

Q = (2 times 10)(45) = 90 in3

fVQ

Ibv = = times

times=8000 126 75

1046 7 6161 5

psi

Q a y= prime = times( ) times

=6 5 6

6 5

2126 75 3

in



3751 P-11 111301 1225 PM Page 268

And the two values for shear stress at this location as displayed in Fig-ure 118b are

In many situations it is not necessary to use the complex form of thegeneral expression for shear stress in a beam For wood beams the sec-tions are mostly simple rectangles for which the following simplificationcan be made

For the simple rectangle from Figure 912 I = bd312 Also

Thus

This is the formula specified by design codes for investigation of shear inwood beams

Problem 115AA beam has an I-shaped cross section with an overall depth of 16 in [400 mm] web thickness of 2 in [50 mm] and flanges that are 8 in wide [200 mm] and 3 in [75 mm] thick Compute the critical shear stressesand plot the distribution of shear stress on the cross section if the beamsustains a shear force of 20 kips [89 kN]

Problem 115BA T-shaped beam cross section has an overall depth of 18 in [450 mm]web thickness of 4 in [100 mm] flange width of 8 in [200 mm] andflange thickness of 3 in [75 mm] Compute the critical shear stresses andplot the distribution of shear stress on the cross section if the beam sus-tains a shear force of 12 kips [534 kN]

fVQ

Ib

V bd

bd b

V

bdv = =

times ( )( ) times

=

2

3

8

121 5

Q bd d bd= times

=

2 4 8

2

f

f

v

v

= timestimes

=

= timestimes

=

8000 90

1046 7 6114 6

8000 90

1046 7 1068 8

psi (in the web)

psi (in the flange)


3751 P-11 111301 1225 PM Page 269

116 SHEAR IN STEEL BEAMS

Shear in beams consists of the vertical slicing effect produced by the op-position of the vertical loads on the beams (downward) and the reactiveforces at the beam supports (upward) The internal shear force mecha-nism is visualized in the form of the shear diagram for the beam With auniformly distributed load on a simply supported beam this diagramtakes the form shown in Figure 119a

As the shear diagram for the uniformly loaded beam shows this loadcondition results in an internal shear force that peaks to a maximum valueat the beam supports and steadily decreases in magnitude to zero at thecenter of the beam span With a beam having a constant cross sectionthroughout the span the critical location for shear is thus at the supportsandmdashif conditions there are adequatemdashthere is no concern for shear atother locations along the beam Since this is the common condition ofloading for many beams it is therefore necessary only to investigate thesupport conditions for such beams


Figure 119 Development of shear in beams (a) Shear force in a uniformlyloaded beam (b) Shear force in a beam with a large concentrated load (c) Rec-tangular section (d) I-shaped section (e) Assumed stress in W-shape

3751 P-11 111301 1225 PM Page 270

Figure 119b shows another loading condition that of a major con-centrated load within the beam span Framing arrangements for roof andfloor systems frequently employ beams that carry the end reactions ofother beams so this is also a common condition In this case a major internal shear force is generated over some length of the beam If the con-centrated load is close to one support a critical internal shear force is cre-ated in the shorter portion of the beam length between the load and thecloser support

For a simple rectangular cross section such as that of a wood beamthe distribution of beam shear stress is as shown in Figure 119c takingthe form of a parabola with a maximum shear stress value at the beamneutral axis and a decrease to zero stress at the extreme fiber distances(top and bottom edges)

For the I-shaped cross section of the typical W-shape rolled steel beamthe beam shear stress distribution takes the form shown in Figure 119d(referred to as the ldquoderby hatrdquo form) Again the shear stress is a maximumat the beam neutral axis but the falloff is less rapid between the neutralaxis and the inside of the beam flanges Although the flanges indeed takesome shear force the sudden increase in beam width results in an abruptdrop in the beam unit shear stress A traditional shear stress investigationfor the W-shape therefore is based on ignoring the flanges and assumingthe shear-resisting portion of the beam to be an equivalent vertical plate(Figure 119e) with a width equal to the beam web thickness and a heightequal to the full beam depth An allowable value is established for a unitshear stress on this basis and the computation is performed as

in which

fv = the average unit shear stress based on an assumed distributionas shown in Figure 119e

V = the value for the internal shear force at the cross section

tw = the beam web thickness

db = the overall beam depth

For ordinary situations the allowable shear stress for W-shapes is 040Fywhere Fy is the elastic yield value This is rounded off to 145 ksi for A36steel

fV

t dv

w b

=

SHEAR IN STEEL BEAMS 271

3751 P-11 111301 1225 PM Page 271

Example 8 A simple beam of A36 steel is 6 ft [183 m] long and has aconcentrated load of 36 kips [160 kN] applied 1 ft [03 m] from one endIt is found that a W 10 times 33 is adequate for the bending moment Inves-tigate the beam for shear

Solution The two reactions for this loading are 30 kips [133 kN] and 6kips [27 kN] The maximum shear in the beam is equal to the larger re-action force

From Table 93 for the given shape d = 973 in and tw = 0435 in Then

As this is less than the allowable value of 145 ksi the shape is acceptable

Problems 116AndashCCompute the maximum permissible shears for the following beams ofA36 steel

(A) W 24 times 84 (B) W 12 times 40 (C) W 10 times 19

117 FLITCHED BEAMS

The discussion of bending stresses presented thus far in this chapter per-tains to beams consisting of a single material that is to homogeneousbeams Reinforced concrete construction utilizes beams of two mate-rialsmdashsteel and concretemdashacting together (Chapter 15) Another exam-ple of this condition of mixed materials is a flitched beam in which steeland wood elements are fastened together so as to act as a single unit Twomeans of achieving such a built-up beam section are shown in Figure1110 The stress behavior in a two-material beam will be illustrated byinvestigating the flitched beam

A basic premise for an elastic stressstrain investigation is that the twomaterials deform equally when the beam is bent Then let

s1 and s2 = the deformations per unit length (strain) of the outermostfibers of the two materials respectively

f1 and f2 = the unit bending stresses in the outermost fibers of the twomaterials respectively

E1 and E2 = the modulus of elasticity of the two materials respectively

fV

t dv

w b

= =times

=30

0 435 9 737 09

ksi


3751 P-11 111301 1225 PM Page 272

Since by definition the modulus of elasticity of a material is equal to

the unit stress divided by the unit deformation (strain) then

and transposing for an expression of the unit deformations

Since the two deformations must be equal

from which a basic expression for the relation between the two stresses is

This basic relationship may be used for the investigation or design of thetwo-material beam as demonstrated in the following example

Example 9 A flitched beam is formed as shown in Figure 1110a con-sisting of two 2 times 12 planks of Douglas fir Select Structural grade anda 05 times 1125 steel plate Compute the allowable uniformly distributedload this beam will carry on a span of 14 ft

f fE

E1 2

1

2

= times

s sf

E

f

E1 2

1

1

2

2

= =or

sf

Es

f

E1

1

12

2

2

= =and

Ef

sE

f

s1

1

12

2

2

= =and

FLITCHED BEAMS 273

Figure 1110 Forms of fliched beams

3751 P-11 111301 1225 PM Page 273

Solution From other sources the following data are obtained for the twomaterials

For the steel E = 29000000 psi and the maximum allowable bendingstress is 22 ksi

For the wood E = 1900000 psi and the maximum allowable bendingstress is 1500 psi

For a trial assume the stress in the steel to be the limiting condition Thenfind the stress in the wood that corresponds to this limit in the steel

As this produces a stress lower than that of the limit for the wood the as-sumption is correct That is if a stress of 1500 psi is permitted in thewood the stress in the steel will exceed 22000 psi

Now find the load that can be carried individually by each unit of thebeam using the limiting stresses just established A procedure for this isas follows

For the wood the maximum bending resistance is 1441 psi and thecombined section modulus for the two members is 2 times 316 = 632 in3 (Sfor the 2 times 12 from Table 97) Then the limiting moment for the wood is

Mw = fw times Sw = 1441 times 632 = 91071 in-lb or 7589 ft-lb

For the plate the value of S must be computed From Figure 912 thesection modulus for the rectangle is bd26 thus for the plate with b = 05in and d = 1125 in

Then

Ms = fs times Ss = 22000 times 1055 = 232100 in-lb or 19342 ft-lb

The total capacity for the combined wood and steel section is thus

M = Mw + Ms = 7589 + 19342 = 26931 ft-lb

Sbd

s = = times ( ) =2 2

3

6

0 5 11 25

610 55

in

f fE

Ew s

w

s

= times

= ( ) times

=22 0001 900 000

29 000 0001441

psi


3751 P-11 111301 1225 PM Page 274

Equating this to the maximum moment for a uniformly loaded simplebeam (Figure 420 Case 2) and solving for W

This value for W includes the beam weight which must be deducted todetermine the allowable superimposed load

Although the load-carrying capacity of the wood elements is slightlyreduced in this beam the total capacity is substantially greater than thatof the wood members alone This significant increase in strengthachieved with a small increase in size is a principal reason for popularityof the flitched beam However often of greater interest is the substantialreduction of deflection and the virtual elimination of sag over timemdashanatural phenomenon in the ordinary wood beam

For the following problems use the same allowable stress and modu-lus of elasticity values for the materials as those given in the example andneglect the beam weight

Problem 117AA flitched beam consists of two 2 times 10 pieces of Douglas fir SelectStructural grade and a single 0375 times 925 in steel plate of A36 steel(Figure 1110a) Find the magnitude of the total uniformly distributedload this beam will carry on a span of 18 ft

Problem 117BA flitched beam consists of a single 10 times 14 of Douglas fir Select Struc-tural grade and two A36 steel plates each 05 times 135 in (Figure 1110b)Find the magnitude of the single concentrated load that this beam willcarry at the center of a 16 ft span

118 DEFLECTION OF BEAMS

Deformations of structures must often be controlled for various reasonsThese reasons sometimes relate to the proper functioning of the structureitself but more often relate to effects on the supported construction or theoverall purposes of the structure

MWL W

W

= = = ( )

= times =

26 9318

14

88 26 931

1415 389

lb

DEFLECTION OF BEAMS 275

3751 P-11 111301 1225 PM Page 275

To steelrsquos advantage is the relative stiffness of the material itself Witha modulus of elasticity of 29000 ksi it is 8 to 10 times as stiff as averagestructural concrete and 15 to 20 times as stiff as structural lumber How-ever it is usually the overall deformation of whole structural elements orassemblages that must be controlled in this regard steel structures are fre-quently quite deformable and flexible Because of its cost steel is usuallyformed into elements with thin parts (beam flanges and webs for exam-ple) and because of its high strength it is frequently formed into rela-tively slender elements (beams and columns for example)

For a beam in a horizontal position the critical deformation is usuallythe maximum sag called the beamrsquos deflection For most beams this de-flection will be too small in magnitude to be detected by eye However anyload on a beam such as that in Figure 1111 will cause some amount of de-flection beginning with the beamrsquos own weight In the case of the simplysupported symmetrical single-span beam in Figure 1111 the maximumdeflection will occur at midspan this is usually the only deformation valueof concern for design However as the beam deflects its ends rotate unlessrestrained and this deformation may also be of concern in some situations

If deflection is determined to be excessive the usual remedy is to se-lect a deeper beam Actually the critical property of the beam cross sec-tion is its moment of inertia (I ) about its major axis (Ix for a W-shape)which is typically affected significantly by increases in depth of thebeam Formulas for deflection of beams take a typical form that involvesvariables as follows

(Note the Greek uppercase letter delta (∆) is also used as the symbol for deflection)

in which

D = the deflection measured vertically in units of inches ormillimeters

C = a constant related to the form of the load and support conditionsfor the beam

W = the load on the beamL = the span of the beamE = the modulus of elasticity of the material of the beam

D CWL

EI=

3


3751 P-11 111301 1225 PM Page 276

I = the moment of inertia of the beam cross section for the axis

about which bending occurs

Note that the magnitude of the deflection is directly proportional tothe magnitude of the load double the load double the deflection How-ever the deflection is proportional to the third power of the span doublethe span and you get 23 or eight times the deflection For resistance to de-flection increases in either the materialrsquos stiffness or the beams geomet-ric form (I ) will cause direct proportional reduction of the deflection

Allowable Deflections

What is permissible for beam deflection is mostly a matter of judgementby experienced designers It is difficult to provide specific limitations toavoid various deflection problems Each situation must be investigatedindividually and some cooperative decisions made about the necessarydesign controls by the designers of the structure and those who developthe rest of the building construction

For spanning beams in ordinary situations some rules of thumb havebeen derived over many years of experience These usually consist of es-tablishing some maximum degree of beam curvature described in the formof a limiting ratio of the deflection to the beam span L expressed as a frac-tion of the span for example L 100 These are sometimes although not al-ways specified in general design codes or legally enacted building codesSome typical limitations recognized by designers are the following

For a minimum limit to avoid visible sag on short to medium spans a total load deflection of 1150

For total load deflection of a roof structure 1180

For deflection under live load only for a roof structure 1240

For total load deflection of a floor structure 1240

For deflection under live load only for a floor structure 1360


Figure 1111 Deflection of a simple beam under symmetrical loading

3751 P-11 111301 1225 PM Page 277

Deflection of Uniformly Loaded Simple Beams

The most frequently used beam in flat roof and floor systems is the uni-formly loaded beam with a single simple span (no end restraint) Thissituation is shown in Figure 420 as Case 2 For this case the followingvalues may be obtained for the beam behavior

Maximum bending moment

Maximum stress on the beam cross section

Maximum midspan deflection

Using these relationships together with the typical case of a knownmodulus of elasticity (E = 29000 ksi for steel) and a common limit forbending stress for W-shapes of 24 ksi a convenient formula can be de-rived for deflection of steel beams Noting that the dimension c in thebending stress formula is d2 for symmetrical shapes and substituting the expression for M we can say

Then

DWL

EI

WLd

I

L

Ed

fL

Ed

fL

Ed

=

times

=

=

=

5

384

16

5

24

5

24

5

24

3

2

2

2

( )

fMc

I

WL d

I

WLd

I= = ( )( ) = 8 2

16

DWL

EI=

5

384

3

fMc

I=

MWL=8


3751 P-11 111301 1225 PM Page 278

This is a basic formula for any beam symmetrical about its bending axisFor a shorter version use values of 24 ksi for f and 29000 ksi for E Alsofor convenience spans are usually measured in feet not inches so a fac-tor of 12 is figured in Thus

In metric units with f = 165 MPa E = 200 GPa and the span in meters

119 DEFLECTION COMPUTATIONS

The following examples illustrate the investigation for deflection of theuniformly loaded simple beam

Example 10 A simple beam has a span of 20 ft [610 m] and a total uni-formly distributed load of 39 kips [1735 kN] The beam is a steel W 14times 34 Find the maximum deflection

Solution First determine the maximum bending moment as

Then from Table 93 S = 486 in3 and the maximum bending stress is

which is sufficiently close to the value of the limiting stress of 24 ksi toconsider the beam stressed exactly to its limit Thus the derived formulamay be used without modification From Table 93 the true depth of thebeam is 1398 in Then

fM

S= = times =97 5 12

48 624 07

ksi

MWL= = times =8

39 20

897 5 kip-ft

DL

d= 0 00017179 2

DfL

Ed

L

d

L d

=

=

times

times ( )

=

5

24

5

24

24

29 000

12

0 02483

2

2

2

DEFLECTION COMPUTATIONS 279

3751 P-11 111301 1225 PM Page 279

For a check the general formula for deflection of the simple beam withuniformly distributed load can be used For this it is found that the valueof I for the beam from Table 93 is 340 in4 Then

which is close enough for a verificationIn a more typical situation the chosen beam is not precisely stressed at

24 ksi The following example illustrates the procedure for this situation

Example 11 A simple beam consisting of a W 12 times 26 carries a totaluniformly distributed load of 24 kips [107 kN] on a span of 19 ft [579 m] Find the maximum deflection

Solution As in Example 1 find the maximum bending moment and themaximum bending stress

From Table 93 S for the beam is 334 in3 thus

With the deflection formula that is based only on span and beam depththe basis for bending stress is a value of 24 ksi Therefore an adjustmentmust be made consisting of the ratio of true bending stress to 24 ksi thus

DL

d=

times

= ( ) timestimes( )

= [ ]

20 48

24

0 02483

0 85330 02483 19

12 22

0 626 16

2

2

in mm

fM

S= = times =57 12

33 420 48

ksi

MWL= = times =8

24 19

857 kip-ft

DWL

EI= = ( ) times( )

times times=5

384

5 39 20 12

384 29 000 3400 712

3 3

in

DL

d= = times = [ ]0 2483 0 02483 20

13 980 7104 18 05

2 2

in mm


3751 P-11 111301 1225 PM Page 280

The derived deflection formula involving only span and beam depthcan be used to plot a graph that displays the deflection of a beam of aconstant depth for a variety of spans Figure 1112 consists of a series ofsuch graphs for beams from 6 to 36 in in depth Use of these graphs pre-sents yet another means for determining beam deflections The readermay verify that deflections may be found from the graphs for the beamsin Examples 1 and 2 with reasonable agreement with the computed re-sults An answer within about 5 should be considered reasonable fromthe graphs

A second deflection graph is shown in Figure 1113 in this case forwood beams While the value of E is constant for steel it varies over aconsiderable range for various species and grades of wood The graphs inFigure 1113 are based on an assumed value of 1500000 psi for themodulus of elasticity The value of allowable bending stress also varieswith a value assumed here of 1500 psi These values are average for themembers typically used for timber beams

The real value of the graphs in Figures 1112 and 1113 is in the de-sign process Once the necessary span is known the designer can deter-mine from the graphs what beam depth is required for a given deflectionThe limiting deflection may be given in an actual dimension or morecommonly as a limiting percentage of the span (1240 1360 etc) aspreviously discussed To aid in the latter situation lines are drawn on thegraph representing the usual percentage limits of 1360 1240 and 1180(see previous discussion in this section for deflection limits) Thus if asteel beam is to be used for a span of 36 ft and the total load deflectionlimit is L 240 it may be observed in Figure 1112 that the lines for a spanof 36 ft and a ratio of 1240 intersect almost precisely on the curve for an18-in deep beam This means that an 18-in deep beam will deflect al-most precisely 1240th of the span if stressed in bending to 24 ksi Thusany beam chosen with less depth will be inadequate for deflection andany beam greater in depth will be conservative in regard to deflection

Determination of deflections for other than uniformly loaded simplebeams is considerably more complicated However many handbooksprovide formulas for computation of deflections for a variety of beamloading and support situations

Problems 119AndashCFind the maximum deflection in inches for the following simple beamsof A36 steel with uniformly distributed load Find the values using


3751 P-11 111301 1225 PM Page 281

(a) the equation for Case 2 in Figure 420 (b) the formula involving onlyspan and beam depth (c) the curves in Figure 1112

(A) W 10 times 33 span = 18 ft total load = 30 kips [55 m 133 kN]

(B) W 16 times 36 span = 20 ft total load = 50 kips [6 m 222 kN]

(C) W 18 times 46 span = 24 ft total load = 55 kips [73 m 245 kN]


Figure 1112 Deflection of uniformly loaded simple-span steel beams with amaximum bending stress of 24 ksi [165 Mpa]

3751 P-11 111301 1225 PM Page 282

1110 PLASTIC BEHAVIOR IN STEEL BEAMS

The maximum resisting moment by elastic theory is predicted to occurwhen the stress at the extreme fiber reaches the elastic yield value Fy andit may be expressed as

PLASTIC BEHAVIOR IN STEEL BEAMS 283

Figure 1113 Deflection of uniformly loaded simple-span wood beans with max-imum bending stress of 1500 psi [10 Mpa] and modulus of elasticity of 1500000psi [10 Gpa]

3751 P-11 111301 1225 PM Page 283

My = Fy times S

Beyond this condition the resisting moment can no longer be expressedby elastic theory equations since an inelastic or plastic stress conditionwill start to develop on the beam cross section

Figure 1114 represents an idealized form of a load-test response fora specimen of ductile steel The graph shows that up to the yield pointthe deformations are proportional to the applied stress and that beyondthe yield point there is a deformation without an increase in stress ForA36 steel this additional deformation called the plastic range is ap-proximately 15 times that produced just before yield occurs This relativemagnitude of the plastic range is the basis for qualification of the mate-rial as significantly ductile

Note that beyond the plastic range the material once again stiffenscalled the strain-hardening effect which indicates a loss of the ductilityand the onset of a second range in which additional deformation is pro-duced only by additional increase in stress The end of this range estab-lishes the ultimate stress limit for the material

For plastic failure to be significant the extent of the plastic range ofdeformation must be several times that of the elastic range as it is indeedfor A36 steel As the yield limit of steel is increased in higher grades the


Figure 1114 Idealized form of the stress-strain behavior of ductile steel

3751 P-11 111301 1225 PM Page 284

plastic range decreases so that the plastic theory of behavior is at presentgenerally limited in application to steels with a yield point not exceeding65 ksi [450 MPa]

The following example illustrates the application of the elastic theoryand will be used for comparison with an analysis of plastic behavior

Example 12 A simple beam has a span of 16 ft [488 m] and supports asingle concentrated load of 18 kips [80 kN] at its center If the beam is aW 12 times 30 compute the maximum flexural stress

Solution See Figure 1115 For the maximum value of the bendingmoment

In Table 93 find the value of S for the shape as 386 in3 [632 times103 mm3] Thus the maximum stress is

and it occurs as shown in Figure 1115d Note that this stress conditionoccurs only at the beam section at midspan Figure 1115e shows theform of the deformations that accompany the stress condition This stresslevel is well below the elastic stress limit (yield point) and in this exam-ple below the allowable stress of 24 ksi

fM

S= = times = [ ]72 12

38 622 4 154

ksi MPa

MPL= = times = [ ]4

18 16

472 kip-ft 98 kN-m


Figure 1115 Example 12 elastic behavior of the beam

3751 P-11 111301 1225 PM Page 285

The limiting moment that may be expressed in allowable stress termsis that which occurs when the maximum flexural stress reaches the yieldstress limit as stated before in the expression for My This condition is il-lustrated by the stress diagram in Figure 1116a

If the loading (and the bending moment) that causes the yield limitflexural stress is increased a stress condition like that illustrated in Fig-ure 1116b begins to develop as the ductile material deforms plasticallyThis spread of the higher stress level over the beam cross section indi-cates the development of a resisting moment in excess of My With a highlevel of ductility a limit for this condition takes a form as shown in Fig-ure 1116c and the limiting resisting moment is described as the plasticmoment designated Mp Although a small percentage of the cross sectionnear the beamrsquos neutral axis remains in an elastic stress condition its ef-fect on the development of the resisting moment is quite negligible Thusit is assumed that the full plastic limit is developed by the conditionshown in Figure 1116d

Attempts to increase the bending moment beyond the value of Mp willresult in large rotational deformation with the beam acting as though itwere hinged (pinned) at this location For practical purposes thereforethe resisting moment capacity of the ductile beam is considered to be ex-hausted with the attaining of the plastic moment additional loading willmerely cause a free rotation at the location of the plastic moment Thislocation is thus described as a plastic hinge (see Figure 1117) and its ef-fect on beams and frames is discussed further in what follows


Figure 1116 Progression of development of bending stress from the elastic tothe plastic range of stress magnitude

3751 P-11 111301 1225 PM Page 286

In a manner similar to that for elastic stress conditions the value of theresisting plastic moment is expressed as

M = Fy times Z

The term Z is called the plastic section modulus and its value is deter-mined as follows

Referring to Figure 1118 which shows a W-shape subjected to a levelof flexural stress corresponding to the fully plastic section (Figure 1116d)

Au = the upper area of the cross section above the neutral axis

yu = distance of the centroid of Au from the neutral axis

Al = the lower area of the cross section below the neutral axis

yl = distance of the centroid of Al from the neutral axis

For equilibrium of the internal forces on the cross section (the result-ing forces C and T developed by the flexural stresses) the condition canbe expressed as

ΣFh = 0

or

[Au times (+fy)] + [Al times (ndashfy)] = 0


Figure 1117 Development of the plastic hinge

3751 P-11 111301 1225 PM Page 287

and thus

Au = Al

This shows that the plastic stress neutral axis divides the cross sectioninto equal areas which is apparent for symmetrical sections but it ap-plies to unsymmetrical sections as well The resisting moment equals the sum of the moments of the stresses thus the value for Mp may be expressed as

Mp = (Au times fy times yu) + (Al times fy times yl)

or

Mp = fy[(Au times yu) + (Al times yl)]

or

Mp = fy times Z

and the quantity [(Au times yu) + (Al times yl)] is the property of the cross sectiondefined as the plastic section modulus designated Z

Using the expression for Z just derived its value for any cross sectioncan be computed However values of Z are tabulated in the AISC Man-ual (Ref 3) for all rolled sections used as beams

Comparison of the values for Sx and Zx for the same W shape willshow that the values for Z are larger This presents an opportunity tocompare the fully plastic resisting moment to the yield stress limitingmoment by elastic stress


Figure 1118 Development of the plastic resisting moment

3751 P-11 111301 1225 PM Page 288

Example 13 A simple beam consisting of a W 21 times 57 is subjected tobending Find the limiting moments (a) based on elastic stress conditionsand a limiting stress of Fy = 36 ksi and (b) based on full development ofthe plastic moment

Solution For (a) the limiting moment is expressed as

My = Fy times Sx

From Table 93 for the W 21 times 57 Sx is 111 in3 so the limiting moment is

For (b) the limiting plastic moment using the value of Zx = 129 in3 fromTable 93 is

The increase in moment resistance represented by the plastic moment in-dicates an increase of 387 ndash 333 = 54 kip-ft or a percentage gain of(54333)(100) = 162

Advantages of use of the plastic moment for design are not so simplydemonstrated A different process must be used regarding safety fac-torsmdashand if the load and resistance factor design (LRFD) method isused a whole different approach In general little difference will befound for the design of simple beams Significant differences occur withcontinuous beams restrained beams and rigid columnbeam frames asdemonstrated in the following discussion

Problem 1110AA simple-span uniformly loaded beam consists of a W 18 times 50 with Fy

= 36 ksi Find the percentage of gain in the limiting bending moment if afully plastic condition is assumed instead of a condition limited by elas-tic stress

Problem 1110BA simple-span uniformly loaded beam consists of a W 16 times 45 with Fy = 36 ksi Find the percentage of gain in the limiting bending moment

M F Zp y= times = times = =36 129 46444644

12387 kip-in or kip-ft

My = times = =36 111 39963996



3751 P-11 111301 1225 PM Page 289

if a fully plastic condition is assumed instead of a condition limited byelastic stress

Plastic Hinging in Continuous and Restrained Beams

The general behavior of restrained and continuous beams is presented inChapter 5 Figure 1119 shows a uniformly distributed load of w lbft ona beam that is fixed (restrained from rotation) at both ends The momentinduced by this condition is distributed along the beam length in a man-ner represented by the moment diagram for a simple-span beam (seeFigure 420 Case 2) consisting of a symmetrical parabola with maxi-mum height (maximum moment) of wL28 For other conditions of sup-port or continuity this distribution of moment will be altered howeverthe total moment remains the same

In Figure 1119a the fixed ends result in the distribution shown be-neath the beam with maximum end moments of wL212 and a moment atthe center of wL28 ndash wL212 = wL224 This distribution will continue aslong as stress does not exceed the yield limit Thus the limiting condition


Figure 1119 Development of the fully plastic restrained beam

3751 P-11 111301 1225 PM Page 290

for elastic conditions is shown in Figure 1119b with a load limit of wy

corresponding to the yield stress limitOnce the flexural stress at the point of maximum moment reaches the

fully plastic state further loading will result in the development of a plas-tic hinge and the resisting moment at that location will not exceed theplastic moment for any additional loadings However additional loadingof the beam may be possible with the moment at the plastic hinge re-maining constant this may proceed until an additional fully plastic con-dition occurs at some other location

For the beam in Figure 1119 the plastic limit for the beam is shownin Figure 1119c this condition is arrived at when both maximum mo-ments are equal to the beamrsquos plastic limit Thus if 2(Mp) = wpL28 thenthe plastic limit (Mp) is equal to wpL 16 as shown in the figure The fol-lowing is a simple example of the form of investigation that is carried outin the LRFD method

Example 14 A beam with fixed ends carries a uniformly distributed loadThe beam consists of a W 21 times 57 of A36 steel with Fy = 36 ksi Find thevalue for the expression of the uniform load if (a) the limit for flexure isthe limit for elastic behavior of the beam and (b) the beam is permitted todevelop the fully plastic moment at critical moment locations

Solution This is the same shape for which limiting yield stress momentand limiting fully plastic moment were found in Example 13 As foundthere these are

My = 333 kip-ft (the elastic stress limit at yield)

Mp = 387 kip-ft (the fully plastic moment)

(a) Referring to Figure 1119b maximum moment for elastic stressis wL212 and equating this to the limiting value for moment

from which

wL L

y = times =333 12 39962 2

(in kip-ft units)

Mw L

yy= =33312

2


3751 P-11 111301 1225 PM Page 291

(b) Referring to Figure 1119c the maximum value for plastic mo-ments with hinging at the fixed ends is wL216 and equating thisto the limiting value for moment

from which

Combining the increase due to the plastic moment with the effectof the redistribution of moments due to plastic hinging the totalincrease is 6192 ndash 3996 = 2196L2 and the percentage gain is

This is a substantially greater gain than that indicated in Example 13(only 162) where difference in moments alone was considered It isthis combined effect that is significant for applications of plastic analy-sis and the LRFD method for continuous and rigid frame structures

Problem 1110CIf the beam in Problem 1110A has fixed ends instead of simple sup-ports find the percentage gain in load-carrying capacity if a fully plasticcondition is assumed rather than a condition limited by elastic stress

Problem 1110DIf the beam in Problem 1110B has fixed ends instead of simple sup-ports find the percentage gain in load-carrying capacity if a fully plasticcondition is assumed rather than a condition limited by elastic stress

2196

3996100 55times =

wL L

p = times =387 16 61922 2

(in kip-ft units)

Mw L

pp= =

times387

16

2


3751 P-11 111301 1225 PM Page 292

293

12COMPRESSION MEMBERS

Compression is developed in a number of ways in structures includingthe compression component that accompanies the development of inter-nal bending In this section consideration is given to elements whoseprimary purpose is resistance of compression In general this includestruss members piers bearing walls and bearing footings although majortreatment here is given to columns which are linear compression mem-bers Building columns may be free-standing architectural elements withthe structural column itself exposed to view However for fire or weatherprotection the structural column must often be incorporated into otherconstruction (see Figure 121) and may in some cases be fully concealedfrom view

121 SLENDERNESS EFFECTS

Structural columns are for the most part quite slender although the spe-cific aspect of slenderness (called relative slenderness) must be consid-ered (see Figure 122) At the extremes the limiting situations are those

3751 P-12 111301 1226 PM Page 293

of the very stout or short column that fails by crushing and the very slen-der or tall column that fails by lateral buckling

The two basic limiting response mechanismsmdashcrushing and buck-lingmdashare entirely different in nature Crushing is a stress resistance phe-nomenon and its limit is represented on the graph in Figure 122 as ahorizontal line basically established by the compression resistance of thematerial and the amount of material (area of the cross section) in thecompression member This behavior is limited to the range labeled zone1 in Figure 122

294 COMPRESSION MEMBERS

Figure 121 Steel column incor-porated in the construction of amultistory building Primarily carry-ing a vertical compression loadthe column also serves a majorfunction as part of the general steelframework for the building typi-cally supporting steel beams asshown in the illustration here

3751 P-12 111301 1226 PM Page 294

Buckling actually consists of lateral deflection in bending and its ex-treme limit is affected by the bending stiffness of the member as relatedto the stiffness of the material (modulus of elasticity) and to the geomet-ric property of the cross section directly related to deflectionmdashthe mo-ment of inertia of the cross-sectional area The classic expression forelastic buckling is stated in the form of the equation developed by Euler

The curve produced by this equation is of the form shown in Figure 122It closely predicts the failure of quite slender compression members inthe range labeled zone 3 in Figure 122

PEI

L= π 2

2

SLENDERNESS EFFECTS 295

Figure 122 Effect of column slenderness on axial compression capacity

3751 P-12 111301 1226 PM Page 295

In fact most building columns fall somewhere between very stoutand very slender in other words in the range labeled zone 2 in Figure122 Their behavior therefore is one of an intermediate form some-where between pure stress response and pure elastic buckling Predic-tions of structural response in this range must be established by empiricalequations that somehow make the transition from the horizontal line tothe Euler curve Equations currently used are explained in Section 122for wood columns and in Section 123 for steel columns

Buckling may be affected by constraints such as lateral bracing thatprevents sideways movement or support conditions that restrain the ro-tation of the memberrsquos ends Figure 123a shows the case for the mem-ber that is the general basis for response as indicated by the Eulerformula This form of response can be altered by lateral constraints asshown in Figure 123b that result in a multimode deflected shape Themember in Figure 123c has its ends restrained against rotation (de-scribed as a fixed end) This also modifies the deflected shape and thusthe value produced from the buckling formula One method used for ad-justment is to modify the column length used in the buckling formula tothat occurring between inflection points thus the effective bucklinglength for the columns in both Figures 123b and c would be one half thatof the true column total length Inspection of the Euler formula will indi-cate the impact of this modified length on buckling resistance


Figure 123 Form of buckling of a column as affected by various end conditionsand lateral constraint

3751 P-12 111301 1226 PM Page 296

122 WOOD COLUMNS

The wood column that is used most frequently is the solid-sawn sectionconsisting of a single piece of wood square or oblong in cross sectionSingle piece round columns are also used as building columns or founda-tion piles This section deals with these common elements and some otherspecial forms used as compression members in building construction

For all columns a fundamental consideration is the column slender-ness For the solid-sawn wood column slenderness is established as theratio of the laterally unbraced length to the least side dimension or Ld(Figure 124a) The unbraced length (height) is typically the overall ver-tical length of the column However it takes very little force to brace acolumn from moving sideways (buckling under compression) so thatwhere construction constrains a column there may be a shorter unbracedlength on one or both axes (Figure 124b)

An important point to make here is that the short compression mem-ber is limited by stress resistance while the very slender member is lim-ited essentially by its stiffnessmdashthat is by the resistance of the member

WOOD COLUMNS 297

Figure 124 Determination of relative slenderness for investigation of buckling(a) The relative slenderness for this column is determined as L d (b) For bucklingin the direction of the broader dimension (d1) slenderness is determined as L1d1while for buckling in the direction of the narrow dimension (d2) slenderness is de-termined as L2 d2

3751 P-12 111301 1226 PM Page 297

to lateral deflection Deflection resistance is measured in terms of thestiffness (modulus of elasticity) of the material of the column and thegeometric property of its cross section (moment of inertia) It is there-fore stress that establishes the limit at the low range of relative stiffnessand stiffness (modulus of elasticity slenderness ratio) that establishesthe limit at extreme values of relative stiffness

Most building columns however fall in a range of stiffness that istransitional between these extremes (Zone 2 as described in Section121) It becomes necessary therefore to establish some means for de-termination of the axial capacity of columns that treats the completerangemdashfrom very short to very tall and all points between Current col-umn design standards establish complex formulas for description of asingle curve that makes the full transition of column behavior related toslenderness It is important to understand the effect of the variables inthese formulas although for practical design work use is generally madeof one or more design aids that permit shortcuts to pragmatic answers

Excessively slender building columns are neither safe nor practical Infact the point of separation between Zones 2 and 3 in Figure 122 gener-ally represents a practical limit for maximum slenderness for columnsSome codes specify a limit but this degree of slenderness is a generalguide for designers For wood columns a limit used in the past was aslenderness ratio of 150

Column Load Capacity

The following discussion presents materials from the NDS (National De-sign SpecificationmdashRef 2) for design of axially loaded columns Thebasic formula for determination of the capacity of a wood column basedon the working stress method is

P = (Fc)(Cp)(A)

in which

A = area of the column cross section

Fc = the allowable design value for compression parallel to the

grain as modified by applicable factors except Cp

Cp = the column stability factor

P = the allowable column axial compression load


3751 P-12 111301 1226 PM Page 298

The column stability factor is determined as follows

in which

FcE = the Euler buckling stress as determined by the formula below

c = 08 for sawn lumber 085 for round poles 09 for glued-laminated timbers

For the buckling stress

in which

KcE = 03 for visually graded lumber and machine evaluated lumber0418 for machine stress rated lumber and glued-laminatedtimber

E = modulus of elasticity for the wood species and grade

Le = the effective length (unbraced height as modified by anyfactors for support conditions) of the column

d = the column cross-sectional dimension (column width)measured in the direction that buckling occurs

The values to be used for the effective column length and the corre-sponding column width should be considered as discussed for the condi-tions displayed in Figure 124 For a basic reference the bucklingphenomenon typically uses a member that is pinned at both ends and pre-vented from lateral movement only at the ends for which no modifica-tion for support conditions is made this is a common condition for woodcolumns The NDS presents methods for modified buckling lengths thatare essentially similar to those used for steel design (see Section 123)These are illustrated for steel columns in Section 123 but not here

The following examples illustrate the use of the NDS formulas forcolumns

FK E

L dcE

cE

e

= ( )( )( ) 2

CF F

c

F F

c

F F

cp

cE c cE cE c=+ ( )

minus+( )

minus1

2

1

2

2

WOOD COLUMNS 299

3751 P-12 111301 1226 PM Page 299

Example 1 A wood column consists of a 6 times 6 of Douglas fir larch No1 grade Find the safe axial compression load for unbraced lengths of (a)2 ft (b) 8 ft (c) 16 ft

Solution From the NDS (Ref 2) find values of Fc = 1000 psi and E =1600000 psi With no basis for adjustment given the Fc value is used di-rectly as the Fc

value in the column formulasFor (a) Ld = 2(12)55 = 436 Then

And the allowable compression load is

P = (Fc)(Cp)(A) = (1000)(0993)(55)2 = 30038 lb

For (b) Ld = 8(12)55 = 1745 for which FcE = 1576 psi FcEFc =

1576 Cp = 0821 and thus

P = (1000)(0821)(55)2 = 24835 lb

For (c) Ld = 16(12)55 = 349 for which FcE = 394 psi FcEFc = 0394

Cp = 0355 and thus

P = (1000)(0355)(55)2 = 10736 lb

Example 2 Wood 2 times 4 elements are to be used as vertical compressionmembers to form a wall (ordinary stud construction) If the wood isDouglas fir larch stud grade and the wall is 85 ft high what is the col-umn load capacity of a single stud

Solution It is assumed that the wall has a covering attached to the studsor blocking between the studs to brace them on their weak (15-in

FK E

L d

F

F

C

cEcE

e

cE

c

p

= ( )( )( )

= ( )( )( )

=

= =

= + minus +

minus =

2 2

2

0 3 1 600 000

4 3625 250

25 250

100025 25

1 25 25

1 6

1 25 25

1 6

25 25

0 80 993

psi


3751 P-12 111301 1226 PM Page 300

dimension) axis Otherwise the practical limit for the height of the wallis 50 times 15 = 75 in Therefore using the larger dimension

From the NDS (Ref 2) Fc = 850 psi E = 1400000 psi with the valuefor Fc adjusted to 105(850) = 8925 psi Then

Problems 122AndashCFind the allowable axial compression load for the following woodcolumns Use Fc = 700 psi and E = 1300000 psi

Nominal SizeUnbraced Length

Column (in) (ft) (mm)

A 6 times 6 10 305B 8 times 8 18 549C 10 times 10 14 427

123 STEEL COLUMNS

Steel compression members range from small single-piece columns andtruss members to huge built-up sections for high-rise buildings and largetower structures The basic column function is one of simple compressiveforce resistance but is often complicated by the effects of buckling andthe possible presence of bending actions

FK E

L d

F

F

C

P F C A

cEcE

e

cE

c

p

c p

= ( )( )( )

= ( )( )( )

=

= =

= minus

minus =

= ( )( )( ) = ( )( )

2 2

2

0 3 1 400 000

29 14495

495

892 50 555

1 555

1 6

1 555

1 6

0 555

0 80 471

892 5 0 471 1 5

psi

timestimes( ) =3 5 2207 lb

L

d= ( ) =8 5 12

3 529 14

STEEL COLUMNS 301

3751 P-12 111301 1226 PM Page 301

Column Shapes

For modest load conditions the most frequently used shapes are theround pipe the rectangular tube and the H-shaped rolled sectionmdashmostoften the W shapes that approach a square form (see Figure 125) Accommodation of beams for framing is most easily achieved with W-shapes of 10-in or larger nominal depth

For various reasons it is sometimes necessary to make up a columnsection by assembling two or more individual steel elements The cus-tomized assemblage of built-up sections is usually costly so a singlepiece is typically favored if one is available

Slenderness and End Conditions

The general effect of slenderness on the axial compression load capacityof columns is discussed in Section 121 For steel columns the value ofthe allowable stress in compression is determined from formulas in theAISC Specification (found in Ref 3) it includes variables of the steelyield stress and modulus of elasticity the relative slenderness of the col-umn and special considerations for any bracing or rotational restraint atthe column ends

Column slenderness is determined as the ratio of the column unbracedlength to the radius of gyration of the column section Lr Effects of endrestraint are considered by use of a modifying factor (K ) resulting insome reduced or magnified value for L (see Figure 126) The modifiedslenderness is thus expressed as KL r

Figure 127 is a graph of the allowable axial compressive stress for acolumn for two grades of steel with Fy of 36 ksi and 50 ksi Values forfull number increments of KL r are also given in Table 121 Values in-dicated on the graph curve for 36 ksi may be compared with those ob-tained for the corresponding Lr values in Table 121


Figure 125 Common shapes of cross sections for steel columns

3751 P-12 111301 1226 PM Page 302

STEEL COLUMNS 303

Figure 126 Determination of modified effective column length for buckling of steelcolumns Reproduced from the Manual of Steel Construction 8th edition with per-mission of the publisher the American Institute of Steel Construction Chicago IL

Figure 127 Allowable axial compressive stress for steel columns as a functionof yield limit and column slenderness Range 1 involves essentially a yield stressfailure condition Range 3 involves essentially an elastic buckling limit based onsteel stiffness which is independent of stress magnitude Range 2 is the inelasticbuckling condition which is transitional between the other two ranges

3751 P-12 111301 1226 PM Page 303

It may be noted in Figure 127 that the two curves converge at an Lrvalue of approximately 125 This is a manifestation of the fact that elas-tic buckling takes over beyond this point making the materialrsquos stiffness(modulus of elasticity) the only significant property for stiffness valueshigher than about 125 Thus the usefullness of the higher grade steel be-comes moot for very slender members

For practical reasons most building columns tend to have relativestiffnesses between about 50 and 100 with only very heavily loadedcolumns falling below this and most designers avoid using extremelyslender columns


TABLE 121 Allowable Unit Stress Fa for Columns of A36 Steel (ksi)a

KLr Fa KLr Fa KLr Fa KLr Fa KLr Fa KLr Fa KLr Fa KLr Fa

1 2156 26 2022 51 1826 76 1579 101 1285 126 941 151 655 176 482

2 2152 27 2015 52 1817 77 1569 102 1272 127 926 152 646 177 477

3 2148 28 2008 53 1808 78 1558 103 1259 128 911 153 638 178 471

4 2144 29 2001 54 1799 79 1547 104 1247 129 897 154 630 179 466

5 2139 30 1994 55 1790 80 1536 105 1233 130 884 155 622 180 461

6 2135 31 1987 56 1781 81 1524 106 1220 131 870 156 614 181 456

7 2130 32 1980 57 1771 82 1513 107 1207 132 857 157 606 182 451

8 2125 33 1973 58 1762 83 1502 108 1194 133 844 158 598 183 446

9 2121 34 1965 59 1753 84 1490 109 1181 134 832 159 591 184 441

10 2116 35 1958 60 1743 85 1479 110 1167 135 819 160 583 185 436

11 2110 36 1950 61 1733 86 1467 111 1154 136 807 161 576 186 432

12 2105 37 1942 62 1724 87 1456 112 1140 137 796 162 569 187 427

13 2100 38 1935 63 1714 88 1444 113 1126 138 784 163 562 188 423

14 2095 39 1927 64 1704 89 1432 114 1113 139 773 164 555 189 418

15 2089 40 1919 65 1694 90 1420 115 1099 140 762 165 549 190 414

16 2083 41 1911 66 1684 91 1409 116 1085 141 751 166 542 191 409

17 2078 42 1903 67 1674 92 1397 117 1071 142 741 167 535 192 405

18 2072 43 1895 68 1664 93 1384 118 1057 143 730 168 529 193 401

19 2066 44 1886 69 1653 94 1372 119 1043 144 720 169 523 194 397

20 2060 45 1878 70 1643 95 1360 120 1028 145 710 170 517 195 393

21 2054 46 1870 71 1633 96 1348 121 1014 146 701 171 511 196 389

22 2048 47 1861 72 1622 97 1335 122 999 147 691 172 505 197 385

23 2041 48 1853 73 1612 98 1323 123 985 148 682 173 499 198 381

24 2035 49 1844 74 1601 99 1310 124 970 149 673 174 493 199 377

25 2028 50 1835 75 1590 100 1298 125 955 150 664 175 488 200 373

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction Chicago ILaValue of K is taken as 10 Fy = 36 ksi

3751 P-12 111301 1226 PM Page 304

Safe Axial Loads for Steel Columns

The allowable axial load for a column is computed by multiplying the al-lowable stress (Fa) by the cross-sectional area of the column The follow-ing examples demonstrate the process For single-piece columns a moredirect process consists of using column load tables For built-up sectionshowever it is necessary to compute the properties of the section

Example 3 A W 12 times 53 is used as a column with an unbraced lengthof 16 ft [488 m] Compute the allowable load

Solution Referring to Table 93 A = 156 in2 rx = 523 in and ry = 248in If the column is unbraced on both axes it is limited by the lower rvalue for the weak axis With no stated end conditions Case (d) in Figure126 is assumed for which K = 10 that is no modification is made (Thisis the unmodified condition) Thus the relative stiffness is computed as

In design work it is usually considered acceptable to round the slender-ness ratio off to the nearest whole number Thus with a KLr value of 77Table 121 yields a value for Fa of 1569 ksi The allowable load for thecolumn is then

P = (Fa)(A) = (1569)(156) = 2448 kips [1089 kN]

Example 4 Compute the allowable load for the column in Example 3 ifthe top is pinned but prevented from lateral movement and the bottom istotally fixed

Solution Referring to Figure 126 this is Case (b) and the modifyingfactor is 08 Then

From Table 121 Fa = 1724 ksi and thus

P = (1724)(156) = 2689 kips [1196 kN]

KL

r= times times =0 8 16 12

2 4862

KL

r= times times =1 16 12

2 4877 4

STEEL COLUMNS 305

3751 P-12 111301 1226 PM Page 305

The following example illustrates the situation in which a W-shape isbraced differently on its two axes

Example 5 Figure 128a shows an elevation of the steel framing at thelocation of an exterior wall The column is laterally restrained but rota-tionally free at the top and bottom in both directions (end condition as forCase (d) in Figure 126) With respect to the x-axis of the section the col-umn is laterally unbraced for its full height However the existence of thehorizontal framing in the wall plane provides lateral bracing with respectto the y-axis of the section thus the buckling of the column in this di-rection takes the form shown in Figure 128b If the column is a W 12 times53 of A36 steel L1 is 30 ft and L2 is 18 ft what is the allowable com-pression load

Solution The basic procedure here is to investigate both axes separatelyand to use the highest value for relative stiffness obtained to find the al-lowable stress (Note This is the same section used in Example 1 forwhich properties were previously obtained from Table 93) For the x-axis the situation is Case (d) from Figure 126 Thus

For the y-axis the situation is also assumed to be Case (d) from Figure126 except that the deformation occurs in two parts (see Figure 128b)The lower part is used as it has the greater unbraced length Thus

Despite the bracing the column is still critical on its weak axis FromTable 121 the value for Fa is 1456 ksi and the allowable load is thus

P = Fa A = (1456)(156) = 2271 kips [1010 kN]

For the following problems use A36 steel with Fy = 36 ksi

Problem 123ADetermine the allowable axial compression load for a W 10 times 49 columnwith an unbraced height of 15 ft [457 m] Assume K = 10

yKL

r-axis say = times times =1 18 12

2 4887 1 87

xKL

r-axis = times times =1 30 12

5 2368 8


3751 P-12 111301 1226 PM Page 306

Problem 123BDetermine the allowable axial compression load for a W 12 times 120 col-umn with an unbraced height of 22 ft [671 m] if both ends are fixedagainst rotation and horizontal movement

STEEL COLUMNS 307

Figure 128 Example 5 biaxial bracing conditions for the column

3751 P-12 111301 1226 PM Page 307

Problem 123CDetermine the allowable axial compression load in Problem 123A if theconditions are as shown in Figure 128 with L1 = 15 ft [46 m] and L2 = 8ft [244 m]

Problem 123DDetermine the allowable axial compression load in Problem 103B if theconditions are as shown in Figure 128 with L1 = 40 ft [12 m] and L2 = 22ft [67 m]


3751 P-12 111301 1226 PM Page 308

309

13COMBINED FORCES

AND STRESSES

Many structural elements perform singular tasks as simple tension mem-bers compression members beams and so on As such their stress con-ditions may be quite simply visualized and investigated However it isnot uncommon for structural tasks to be multiple for a given structuralmember as shown in Figure 131 In this case it is necessary to considerboth the individual tasks and the effects of their combination Design ofsuch multitask members may well produce different results from theforms taken by simpler elements Demonstrated versatility in performingmultiple structural tasks establishes some popularity for particular struc-tural elements such as the cylindrical steel shape (pipe) shown in Figure131 This chapter presents some considerations for combinations offorce effects and stresses

131 COMBINED ACTION TENSION PLUS BENDING

Various situations occur in which both an axial force of tension and abending moment occur at the same cross section in a structural member

3751 P-13 111301 1226 PM Page 309

Consider the hanger shown in Figure 132 in which a 2-in square steelbar is welded to a plate and the plate is bolted to the bottom of a wood beamA short piece of steel plate with a hole is welded to the face of the barand a load is hung from the hole In this situation the steel bar is sub-jected to combined actions of tension and bending both of which are pro-duced by the hung load The bending moment is the product of the loadtimes its eccentricity from the centroid of the bar cross section thus

M = 5000 times 2 = 10000 in-lb[22 times 50 = 1100 kN-m]

310 COMBINED FORCES AND STRESSES

Figure 131 The multifunction structure In some circumstances structural ele-ments must perform several different structural functions simultaneously Such isthe case for the vertical post in the structure shown heremdashpossibly the support fora cantilevered sign Under a combined loading of wind and gravity the post mustresist compression twisting (torsion) bending in two directions and lateral shearThe single most effective element for this situation is the steel cylinder producedfor use as piping No other single element has the versatility and efficiency of thepipe for multiple functions

3751 P-13 111301 1226 PM Page 310

For this simple case the stresses due to the two phenomena are foundseparately and added as follows For the direct tension effect (Figure 133a)

For the bending stress the section modulus of the bar is found as

Then for the bending stress (Figure 133b)

and the stress combinations are (Figure 133c)

maximum f = 1250 + 7502 = 8752 psi [616 MPa] (tension)minimum f = 1250 ndash 7502 = ndash6252 psi [440 MPa] (compression)

Although the reversal compression stress is less than the maximumtension stress there are situations in which it may be critical The 2-insquare bar in this example is probably capable of developing the com-pression but other member cross sections may not be so versatile A thinbar for example may become critical in buckling due to the compres-sion even though the tension stress is higher

fM

Sb = = = [ ]10 000

1 3337500 52 8

psi MPa

Sbd= = times = times[ ]

2 23 3 3

6

2 2

61 333 20 82 10 in mm

fN

Aa = = = [ ]5

41250 8 8 psi MPa

COMBINED ACTION TENSION PLUS BENDING 311

Figure 132 An example of combined tension and bending

3751 P-13 111301 1226 PM Page 311

Problem 131AFor the hanger rod shown in Figure 134 find the maximum and mini-mum values of the tension stress

Problem 131BA hanger rod similar to that shown in Figure 134 consists of a 1-insquare steel rod Find the maximum and minimum values of the tensionstress if the load is 120 lb and the eccentricity is 25 in

132 COMBINED ACTION COMPRESSION PLUS BENDING

Combined actions of compression plus bending produce various effectson structures In some situations the actual stress combinations may ofthemselves be critical one such case being the development of bearing


Figure 133 Consideration for the combined stress shown in Figure 132 (a) Di-rect tension (b) Bending (c) Combined stress

Figure 134 Problem 131

3751 P-13 111301 1226 PM Page 312

stress on soils At the contact face of a bearing footing and its supportingsoil the ldquosectionrdquo for stress investigation is the contact face that is thebottom of the footing The following discussion deals with an approachto this investigation

Figure 135 illustrates a classical approach to the combined directforce and bending moment at a cross section In this case the ldquocross sec-tionrdquo is the contact face of the footing bottom with the soil However thecombined force and moment originate a common analytical technique isto make a transformation into an equivalent eccentric force that produces

COMBINED ACTION COMPRESSION PLUS BENDING 313

Figure 135 Investigation for combined stress due to compression and bending

3751 P-13 111301 1226 PM Page 313

the same combined effect The value for the hypothetical eccentricity e isestablished by dividing the moment by the force as shown in the figureThe net or combined stress distribution at the section is visualized as the sum of separate stresses created by the force and the bending For thelimiting stresses at the edges of the section the general equation for the combined stress is

p = (direct stress) plusmn (bending stress)

or

Four cases for this combined stress are shown in the figure The firstcase occurs when e is small resulting in very little bending stress Thesection is thus subjected to all compressive stress varying from a maxi-mum value at one edge to a minimum on the opposite edge

The second case occurs when the two stress components are equal sothat the minimum stress becomes zero This is the boundary conditionbetween the first and third cases since any increase in e will tend to pro-duce some reversal stress (in this situation tension) on the section

The second stress case is a significant one for the footing since ten-sion stress is not possible for the soil-to-footing interface Case 3 is onlypossible for a beam or column or some other continuously solid elementThe value for e that produces Case 2 can be derived by equating the twostress components as follows

This value for e establishes what is known as the kern limit of the sec-tion The kern is defined as a zone around the centroid of the sectionwithin which an eccentric force will not cause reversal stress on the sec-tion The form and dimensions of this zone may be established for anygeometric shape by application of the derived formula for e The kernlimit zones for three common geometric shapes are shown in Figure 136

When tension stress is not possible larger eccentricities of the normalforce will produce a so-called cracked section which is shown as Case 4in Figure 135 In this situation some portion of the cross section be-comes unstressed or cracked and the compressive stress on the remain-

N

A

Nec

Ie

I

Ac= = thus

pN

A

Nec

I= plusmn


3751 P-13 111301 1226 PM Page 314

der of the section must develop the entire resistance to the loading effectsof the combined force and moment

Figure 137 shows a technique for the analysis of a cracked sectioncalled the pressure wedge method The ldquowedgerdquo is a volume that repre-sents the total compressive force as developed by the soil pressure (stresstimes stressed area) Analysis of the static equilibrium of this wedge pro-duces two relationships that may be used to establish the dimensions ofthe stress wedge These relationships are

1 The volume of the wedge is equal to the vertical force (Sum ofvertical forces equals zero)

2 The centroid (center of gravity) of the wedge is located on a ver-tical line that coincides with the location of the hypothetical ec-centric force (Sum of moments equals zero)

Referring to Figure 137 the three dimensions of the wedge are w(width of the footing) p (maximum soil pressure) and x (limiting di-mension of the stressed portion of the cracked section) In this situationthe footing width is known so the definition of the wedge requires onlythe determination of p and x

For the rectangular section the centroid of the wedge is at the thirdpoint of the triangle Defining this distance from the edge as a as shownin the figure then x is equal to three times a And it may be observed thata is equal to half the footing width minus e Thus once the eccentricityis computed the values of a and x can be determined

The volume of the stress wedge may be expressed in terms of its threedimensions as

V = 1frasl2 (wpx)


Figure 136 Kern limits for common shapes

3751 P-13 111301 1226 PM Page 315

With w and x established the remaining dimension of the wedge maythen be established by transforming the equation for the volume to

All four cases of combined stress shown in Figure 135 will cause ro-tation (tilt) of the footing due to deformation of the compressible soilThe extent of this rotation and the concern for its effect on the supportedstructure must be carefully considered in the design of the footing It is

pN

wx= 2


Figure 137 Investigation of combined stress on a cracked section by the pres-sure wedge method

3751 P-13 111301 1226 PM Page 316

generally desirable that long-term loads (such as dead load) not developuneven stress on the footing Thus the extreme situations of stress shownin Cases 2 and 4 in Figure 135 should be allowed only for short durationloads See discussion of the cantilever retaining wall in Chapter 6

Example 1 Find the maximum value of soil pressure for a square foot-ing The axial compression force at the bottom of the footing is 100 kipsand the moment is 100 kip-ft Find the pressure for footing widths of (a)8 ft (b) 6 ft and (c) 5 ft

Solution The first step is to determine the equivalent eccentricity andcompare it to the kern limit for the footing to establish which of the casesshown in Figure 135 applies

(a) For all parts the eccentricity is

For the 8-ft-wide footing the kern limit is 86 = 133 ft thus Case1 applies

For the computation of soil pressure the properties of the sec-tion (the 8 ft by 8 ft square) must be determined Thus

and the maximum soil pressure is determined as

(b) For the 6-ft-wide footing the kern limit is 1 ft the same as the ec-centricity Thus the situation is stress Case 2 in Figure 135 withNA = McI Thus

pN

A=

=

=2 2

100

365 56 ksf

pN

A

Mc

I= + = + times

= + =

100

64

100 4

341 31 56 1 17 2 73

ksf

A

Ibd

= times =

= = times =

8 8 64

12

8 8

12341 3

2

3 34

ft

ft

eM

N= = =100

1001 ft

DEVELOPMENT OF SHEAR STRESS 317

3751 P-13 111301 1226 PM Page 317

(c) The eccentricity exceeds the kern limit and the investigationmust be done as illustrated in Figure 137

Problem 132AThe compression force at the bottom of a square footing is 40 kips [178kN] and the bending moment is 30 kip-ft [407 kN-m] Find the maxi-mum soil pressure for widths of (a) 5 ft [15 m] (b) 4 ft [12 m]

Problem 132BThe compression force at the bottom of a square footing is 60 kips [267kN] and the bending moment is 60 kip-ft [814 kN-m] Find the maxi-mum soil pressure for widths of (a) 7 ft [213 m] (b) 5 ft [15 m]

133 DEVELOPMENT OF SHEAR STRESS

Shear force generates a lateral slicing effect in materials Visualized intwo dimensions this direct effect is as shown in Figure 138a For stabil-ity within the material there will be a counteracting or reactive shearstress developed at right angles to the active stress as shown in Figure138b The interaction of the active and reactive shears produces bothdiagonal tension and diagonal compression stresses as shown in Figures138c and d

a e

x a

pN

wx

= minus = minus =

= = =

= timestimes

=

5

22 5 1 1 5

3 3 1 5 4 5

2 2 100

5 4 58 89

( )

ft

ft

= ksf


Figure 138 Development of shear and resulting diagonal stresses

3751 P-13 111301 1226 PM Page 318

Referring to Figure 138 it may be observed that

1 The unit reactive shear stress is equal in magnitude to the unit ac-tive shear stress

2 The diagonal effect (tension or compression) is the vector com-bination of the active and reactive shear and thus has a magni-tude of 1414 times the unit shear

3 The diagonal stress is developed on a diagonal plane which hasan area 1414 times that of the area on which the unit shear is de-veloped thus the unit diagonal stresses are the same magnitudeas the shear stresses

Accepting the observations just made it is possible to determine thecritical diagonal tension or diagonal compression by simply computingthe unit shear stress However the direction of these stresses must also benoted

134 STRESS ON AN OBLIQUE SECTION

In the preceding section it has been shown that shear produces directstresses as well as shear stresses We now demonstrate that direct forceproduces shear stresses as well as direct stresses Consider the objectshown in Figure 139a subjected to a tension force If a section is cut thatis not at a right angle to the force (an oblique section) there may be seento exist two components of the internal resistance One component is ata right angle to the cut section and the other is in the plane of the cut sec-tion These two components produce respectively direct tension stress( f ) and shear stress (v) at the cut section

With the angle of the cut section defined as shown in the figure andthe right-angle cross section area defined as A these stresses may be ex-pressed as follows

fP

A

P

A

vP

A

P

A

= =

( )

= =

( )

cos

coscos

sin

cossin cos

ΘΘ

Θ

ΘΘ

Θ Θ

2

STRESS ON AN OBLIQUE SECTION 319

3751 P-13 111301 1226 PM Page 319

The following may be noted for two special values of the angle Θ

1 For a right-angle section Θ = 0 cos Θ = 1 and sin Θ = 0 then

2 When Θ = 45deg and cos Θ = sin Θ = 0707 then

It may be demonstrated that the value for the diagonal shear stress onthe 45deg cut section is the highest value generated by the direct forceAlso the value for the direct stress on an oblique section will always beless than that on a right-angle section since any value for the cosine of Θwill be less than one if Θ is greater than zero

In some situations the specific value for these stresses on a particularoblique plane may be of concern The following example demonstratesthe use of the derived stress formulas for this situation

Example 2 The wood block shown in Figure 1310a has its grain at anangle of 30deg to the direction of a compression force of 1200 lb on theblock Find the compression and shear stresses on a section that is paral-lel to the wood grain

fP A

vP A= =

2 2and

fP

Av= =and 0


Figure 139 Stresses on an oblique section

3751 P-13 111301 1226 PM Page 320

Solution Note that as used in Figure 139 Θ = 60deg Then for the free-body diagram shown in Figure 1310b

N = P cos 60deg V = P sin 60deg A = 3 times 4 = 120 in2

Then applying the data to the stress formulas

Problems 134AndashCA structural member such as that in Figure 139 has a right-angle crosssection of 10 in2 and is loaded in compression with a force of 10000 lbFind the direct and shear stresses on an oblique section with Θ as shownin Figure 139 equal to (A) 15deg (B) 20deg (C) 30deg

135 COMBINED DIRECT AND SHEAR STRESSES

The stress actions shown in Figure 138 represent the conditions that occurwhen an internal force of shear alone is considered When internal shearoccurs simultaneously with other effects the various resulting stress

fP

A

vP

A

=

=

( ) =

=

=

( )( ) =

cos

sin cos

2 21200

120 5 25

1200

120 5 0 866 43 3

Θ

Θ Θ

psi

psi

COMBINED DIRECT AND SHEAR STRESSES 321


3751 P-13 111301 1226 PM Page 321

conditions must be combined to produce the net stress effect Figure 1311shows the result of combining a shear stress effect with a direct tensionstress effect For shear alone the critical tension stress plane is at 45deg as shown in Figure 1311a For tension alone the critical stress plane is at 90deg as shown in Figure 1311b For the combined shear plus tension(Figure 1311c) the net unit tension stress will be some magnitude higherthan either the shear or direct tension stress and the plane on which thiscritical tension stress acts will be somewhere between 45deg and 90deg

A common example of the stress condition shown in Figure 1311 oc-curs in a beam in which some combination of internal vertical shear andinternal bending moment exists at all points in the beam span Considerthe beam shown in profile in Figure 1312 At all cross sections the formof distribution of shear and bending stresses considered alone are asshown in Figures 1312b and c Various combinations of shear and directstress may be visualized in terms of the conditions at the cross section la-beled S-S in the figure With reference to the points on the section la-beled 1 through 5 the following may be observed

1 At point 1 the vertical shear stress is zero and the dominant stressis compressive stress due to bending oriented in a horizontal di-rection Tension stress here approaches zero in a vertical direction

2 At point 5 the vertical shear stress is zero and the dominant stressis tension stress due to bending oriented in a horizontal direction

3 At point 3 the vertical shear stress is maximum bending stress iszero and the maximum tension stress is the diagonal stress due toshear oriented in a 45deg direction


Figure 1311 Development of principal net stress due to combined shear and di-rect stresses

3751 P-13 111301 1226 PM Page 322

4 At point 2 the net tension stress acts in a direction between 45degand 90deg


The direction of the net tension stress is indicated for various points inthe beam by the short dark bars on the beam elevation in Figure 1312dThe light dashed lines indicate the direction of flow of internal tensionstress If Figure 1212d were inverted it would show the flow of internalcompression This is a highly informative device for visualization of thebasic nature of beams


Figure 1312 Direction of development of net tension stresses in a beam

3751 P-13 111301 1227 PM Page 323

324

14CONNECTIONS FOR

STEEL STRUCTURES

Making a steel structure for a building typically involves the connectingof many parts (see Figure 141) The technology available for achievingconnections is subject to considerable variety depending on the form andsize of the connected parts the structural forces transmitted betweenparts and the nature of the connecting materials At the scale of buildingstructures the primary connecting methods utilized presently are thoseusing electric arc welding and high strength steel bolts these are themethods treated in this chapter

141 BOLTED CONNECTIONS

Elements of steel are often connected by mating flat parts with commonholes and inserting a pin-type device to hold them together In times pastthe device was a rivet today it is usually a bolt Many types and sizes ofbolt are available as are many connections in which they are used

3751 P-14 111301 1227 PM Page 324

Structural Actions of Bolted Connections

Figures 142a and b show plan and section of a simple connection be-tween two steel bars that functions to transfer a tension force from onebar to another Although this is a tension-transfer connection it is also re-ferred to as a shear connection because of the manner in which the con-necting device (the bolt) works in the connection (see Figure 142c) For

BOLTED CONNECTIONS 325

Figure 141 At the scale ofbuilding structures the con-necting of individual membersin a frame system usually in-volves matching of the flatparts of members Slipping atthe contact face is then pre-vented by welding or by in-serting bolts in matching holesin the members Contact be-tween members may be direct(a) or through an auxiliary ele-ment such as the gusset plateshown in (b)

3751 P-14 111301 1227 PM Page 325

structural connections this type of joint is now achieved mostly with so-called high strength bolts which are special bolts that are tightened in acontrolled manner that induces development of yield stress in the boltshaft For a connection using such bolts there are many possible formsof failure that must be considered including the following

Bolt Shear In the connection shown in Figures 142a and b the fail-ure of the bolt involves a slicing (shear) failure that is developed as ashear stress on the bolt cross section The resistance of the bolt can be ex-pressed as an allowable shear stress Fv times the area of the bolt crosssection or

R = Fv times A

With the size of the bolt and the grade of steel known it is a simplematter to establish this limit In some types of connections it may be nec-essary to slice the same bolt more than once to separate the connectedparts This is the case in the connection shown in Figure 142f in which

326 CONNECTIONS FOR STEEL STRUCTURES

Figure 142 Actions of bolted joints

3751 P-14 111301 1227 PM Page 326

it may be observed that the bolt must be sliced twice to make the jointfail When the bolt develops shear on only one section (Figure 142c) itis said to be in single shear when it develops shear on two sections (Figure 142f) it is said to be in double shear

Bearing If the bolt tension (due to tightening of the nut) is relativelylow the bolt serves primarily as a pin in the matched holes bearingagainst the sides of the holes as shown in Figure 142d When the bolt di-ameter is larger or the bolt is made of very strong steel the connectedparts must be sufficiently thick if they are to develop the full capacity ofthe bolt The maximum bearing stress permitted for this situation by theAISC Specification (see Ref 3) is Fp = 15Fu where Fu is the ultimatetensile strength of the steel in the connected part in which the hole occurs

Tension on Net Section of Connected Parts For the connectedbars in Figure 142b the tension stress in the bars will be a maximum ata section across the bar at the location of the hole This reduced sectionis called the net section for tension resistance Although this is indeed alocation of critical stress it is possible to achieve yield here without se-rious deformation of the connected parts for this reason allowable stressat the net section is based on the ultimatemdashrather than the yieldmdashstrength of the bars The value normally used is 050Fu

Bolt Tension While the shear (slip-resisting) connection shown inFigures 142a and b is common some joints employ bolts for their resis-tance in tension as shown in Figure 142g For the threaded bolt themaximum tension stress is developed at the net section through the cutthreads However it is also possible for the bolt to have extensive elon-gation if yield stress develops in the bolt shaft (at an unreduced section)However stress is computed bolt tension resistance is established on thebasis of data from destructive tests

Bending in the Connection Whenever possible bolted connec-tions are designed to have a bolt layout that is symmetrical with regard to the directly applied forces This is not always possible so that in ad-dition to the direct force actions the connection may be subjected totwisting due to a bending moment or torsion induced by the loads Figure 143 shows some examples of this situation


3751 P-14 111301 1227 PM Page 327

In Figure 143a two bars are connected by bolts but the bars are notaligned in a way to transmit tension directly between the bars This mayinduce a rotational effect on the bolts with a torsional twist equal to theproduct of the tension force and the eccentricity due to misalignment ofthe bars Shearing forces on individual bolts will be increased by thistwisting action And of course the ends of the bars will also be twisted

Figure 143b shows the single-shear joint as shown in Figure 142aand b When viewed from the top such a joint may appear to have thebars aligned however the side view shows that the basic nature of thesingle-shear joint is such that a twisting action is inherent in the jointThis twisting increases with thicker bars It is usually not highly criticalfor steel structures where connected elements are usually relatively thinfor connecting of wood elements however this is not a favored form of joint

Figure 143c shows a side view of a beam end with a typical form ofconnection that employs a pair of angles As shown the angles grasp thebeam web between their legs and turn the other legs out to fit flat against


Figure 143 Development of bending in bolted joints

3751 P-14 111301 1227 PM Page 328

a column or the web of another beam Vertical load from the beamvested in the shear in the beam web is transferred to the angles by theconnection of the angles to the beam webmdashwith bolts as shown hereThis load is then transferred from the angles at their outward-turned faceresulting in a separated set of forces due to the eccentricity shown Thisaction must be considered with others in design of these connections

Slipping of Connected Parts Highly tensioned high-strengthbolts develop a very strong clamping action on the mated flat parts beingconnected analogous to the situation shown in Figure 144 As a resultthere is a strong development of friction at the slip face which is the ini-tial form of resistance in the shear-type joint Development of bolt shearbearing and even tension on the net section will not occur until this slip-ping is allowed For service level loads therefore this is the usual formof resistance and the bolted joint with high-strength bolts is consideredto be a very rigid form of joint

Block Shear One possible form of failure in a bolted connection isthat of tearing out the edge of one of the attached members This is calleda block shear failure The diagrams in Figure 145a show this potential-ity in a connection between two plates The failure in this case involvesa combination of shear and tension to produce the torn-out form shownThe total tearing force is computed as the sum required to cause both


Figure 144 Clamping action of highly tightened bolts

3751 P-14 111301 1227 PM Page 329

forms of failure The allowable stress on the net tension area is specifiedat 050Fu where Fu is the maximum tensile strength of the steel The al-lowable stress on the shear areas is specified as 030Fu With the edgedistance hole spacing and diameter of the holes known the net widthsfor tension and shear are determined and multiplied by the thickness of the part in which the tearing occurs These areas are then multiplied bythe appropriate stress to find the total tearing force that can be resisted Ifthis force is greater than the connection design load the tearing problemis not critical

Another case of potential tearing is shown in Figure 145b This is thecommon situation for the end framing of a beam in which support is pro-vided by another beam whose top is aligned with that of the supportedbeam The end portion of the top flange of the supported beam must be


Figure 145 Tearing (block shear) failure in bolted joints

3751 P-14 111301 1227 PM Page 330

cut back to allow the beam web to extend to the side of the supportingbeam With the use of a bolted connection the tearing condition shownmay develop

Types of Steel Bolts

Bolts used for the connection of structural steel members come in twobasic types Bolts designated A307 and called unfinished have the low-est load capacity of the structural bolts The nuts for these bolts are tight-ened just enough to secure a snug fit of the attached parts because of thislow resistance to slipping plus the over-sizing of the holes to achievepractical assemblage there is some movement in the development of fullresistance These bolts are generally not used for major connections es-pecially when joint movement or loosening under vibration or repeatedloading may be a problem They are however used extensively for tem-porary connections during erection of frames

Bolts designated A325 or A490 are called high-strength bolts Thenuts of these bolts are tightened to produce a considerable tension forcewhich results in a high degree of friction resistance between the attachedparts Different specifications for installation of these bolts results in dif-ferent classifications of their strength relating generally to the criticalmode of failure

When loaded in shear-type connections bolt capacities are based onthe development of shearing action in the connection The shear capac-ity of a single bolt is further designated as S for single shear (Figure142c) or D for double shear (Figure 142 f ) The capacities of structuralbolts in both tension and shear are given in Table 141 These bolts rangein size from 5frasl8 to 11frasl2 in in diameter and capacities for these sizes aregiven in tables in the AISC Manual (Ref 3) However the most com-monly used sizes for light structural steel framing are 3frasl4 and 7frasl8 in How-ever for larger connections and large frameworks sizes of 1 to 11frasl4 arealso used This is the size range for which data are given in Table 1413frasl4 to 11frasl4

Bolts are ordinarily installed with a washer under both head and nutSome manufactured high-strength bolts have specially formed heads ornuts that in effect have self-forming washers eliminating the need for aseparate loose washer When a washer is used it is sometimes the limit-ing dimensional factor in detailing for bolt placement in tight locationssuch as close to the fillet (inside radius) of angles or other rolled shapes


3751 P-14 111301 1227 PM Page 331

For a given diameter of bolt there is a minimum thickness requiredfor the bolted parts in order to develop the full shear capacity of the boltThis thickness is based on the bearing stress between the bolt and the sideof the hole which is limited to a maximum of Fp = 15Fu The stress limitmay be established by either the bolt steel or the steel of the bolted parts

Steel rods are sometimes threaded for use as anchor bolts or tie rodsWhen they are loaded in tension their capacities are usually limited bythe stress on the reduced section at the threads Tie rods are sometimesmade with upset ends which consist of larger diameter portions at theends When these enlarged ends are threaded the net section at the threadis the same as the gross section in the remainder of the rods the result isno loss of capacity for the rod

Layout of Bolted Connections

Design of bolted connections generally involves a number of considera-tions in the dimensional layout of the bolt-hole patterns for the attached


TABLE 141 Capacity of Structural Bolts (kips)a

Nominal Diameter of Bolt (in)

3frasl4 7frasl8 1 11frasl8 11frasl4Area Based on Nominal Diameter (in2)

ASTM LoadingDesignation Conditionb 04418 06013 07854 09940 1227

A307 S 44 60 79 99 123D 88 120 157 199 245T 88 120 157 199 245

A325 S 75 102 134 169 209D 150 204 267 338 417T 194 265 346 437 540

A490 S 93 126 165 209 258D 186 253 330 417 515T 239 325 424 537 663

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction Chicago ILaSlip-critical connections assuming there is no bending in the connection and that bearing onconnected materials is not criticalbS = single shear D = double shear T = tension

3751 P-14 111301 1227 PM Page 332

structural members The material in this section presents some basic fac-tors that often must be included in the design of bolted connections Insome situations the ease or difficulty of achieving a connection may af-fect the choice for the form of the connected members

Figure 146a shows the layout of a bolt pattern with bolts placed intwo parallel rows Two basic dimensions for this layout are limited by thesize (nominal diameter) of the bolt The first is the center-to-center spac-ing of the bolts usually called the pitch The AISC Specification (seeRef 3) limits this dimension to an absolute minimum of 22frasl3 times thebolt diameter The preferred minimum however which is used in thisbook is 3 times the diameter

The second critical layout dimension is the edge distance which is thedistance from the center line of the bolt to the nearest edge of the mem-ber containing the bolt hole There is also a specified limit for this as a


Figure 146 Layout considerations for bolted joints

3751 P-14 111301 1227 PM Page 333

function of bolt size and the nature of the edge the latter referring towhether the edge is formed by rolling or is cut Edge distance may alsobe limited by edge tearing in block shear which is discussed later

Table 142 gives the recommended limits for pitch and edge distancefor the bolt sizes used in ordinary steel construction

In some cases bolts are staggered in parallel rows (Figure 146b) Inthis case the diagonal distance labeled m in the illustration must also beconsidered For staggered bolts the spacing in the direction of the rowsis usually referred to as the pitch the spacing of the rows is called thegage The usual reason for staggering the bolts is that sometimes therows must be spaced closer (gage spacing) than the minimum spacing re-quired for the bolts selected However staggering the bolt holes alsohelps to create a slightly less critical net section for tension stress in thesteel member with the holes

Location of bolt lines is often related to the size and type of structuralmembers being attached This is especially true of bolts placed in the legsof angles or in the flanges of W- M- S- C- and structural tee shapesFigure 146c shows the placement of bolts in the legs of angles When asingle row is placed in a leg its recommended location is at the distancelabeled g from the back of the angle When two rows are used the firstrow is placed at the distance g1 and the second row is spaced a distance


TABLE 142 Pitch and Edge Distances for Bolts

Minimum Edge Distance for PunchedReamed or Drilled Holes (in)

Rivet or Bolt At Rolled EdgesDiameter of Plates Shapes Minimum Recommended

d or Bars or Gas- Pitch Center-to-Center (in)

(in) At Sheared Edges Cut Edgesa 2667d 3d

0625 1125 0875 167 18750750 125 10 20 2250875 15b 1125 233 26251000 175b 125 267 30

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction Chicago ILaMay be reduced to 0125 in when the hole is at a point where stress does not exceed 25 of themaximum allowed in the connected elementbMay be 125 in at the ends of beam connection angles

3751 P-14 111301 1227 PM Page 334

g2 from the first Table 143 gives the recommended values for these distances

When placed at the recommended locations in rolled shapes bolts willend up a certain distance from the edge of the part Based on the recom-mended edge distance for rolled edges given in Table 142 it is thus pos-sible to determine the maximum size of bolt that can be accommodatedFor angles the maximum fastener may be limited by the edge distanceespecially when two rows are used however other factors may in somecases be more critical The distance from the center of the bolts to the in-side fillet of the angle may limit the use of a large washer where one is re-quired Another consideration may be the stress on the net section of theangle especially if the member load is taken entirely by the attached leg

Tension Connections

When tension members have reduced cross sections two stress investi-gations must be considered This is the case for members with holes forbolts or for bolts or rods with cut threads For the member with a hole theallowable tension stress at the reduced cross section through the hole is050Fu where Fu is the ultimate tensile strength of the steel The total re-sistance at this reduced section (also called the net section) must be com-pared with the resistance at other unreduced sections at which theallowable stress is 060Fy

For threaded steel rods the maximum allowable tension stress at thethreads is 033Fu For steel bolts the allowable stress is specified as avalue based on the type of bolt The tension load capacities of three typesof bolt for various sizes are given in Table 141


TABLE 143 Usual Gage Dimensions for Angles (in)

Width of Angle Leg (in)

GageDimension 8 7 6 5 4 35 3 25 2

g 45 40 350 300 25 20 175 1375 1125g1 30 25 225 200g2 30 30 250 175

Source Adapted from data in the Manual of Steel Construction 8th edition with permission of thepublishers American Institute of Steel Construction Chicago IL

3751 P-14 111301 1227 PM Page 335

For W- M- S- C- and tee shapes the tension connection is usuallynot made in a manner that results in the attachment of all the parts of thesection (eg both flanges plus the web for a W) In such cases the AISCSpecification (see Ref 3) requires the determination of a reduced effec-tive net area Ae that consists of

Ae = C1An

in which

An = actual net area of the member

C1 = reduction coefficient

Unless a larger coefficient can be justified by tests the following val-ues are specified

1 For W- M- or S-shapes with flange widths not less than two-thirds the depth and structural tees cut from such shapes whenthe connection is to the flanges and has at least three fasteners perline in the direction of stress C1 = 075

2 For W- M- or S-shapes not meeting the above conditions and fortees cut from such shapes provided the connection has not fewerthan three fasteners per line in the direction of stress C1 = 085

3 For all members with connections that have only two fastenersper line in the direction of stress C1 = 075

Angles used as tension members are often connected by only one legIn a conservative design the effective net area is only that of the con-nected leg less the reduction caused by bolt holes

Rivet and bolt holes are punched larger in diameter than the nominaldiameter of the fastener The punching damages a small amount of thesteel around the perimeter of the hole consequently the diameter of the hole to be deducted in determining the net section is 1frasl8 in greaterthan the nominal diameter of the fastener

When only one hole is involved as in Figure 142 or in a similar con-nection with a single row of fasteners along the line of stress the net areaof the cross section of one of the plates is found by multiplying the platethickness by its net width (width of member minus diameter of hole)


3751 P-14 111301 1227 PM Page 336

When holes are staggered in two rows along the line of stress (Figure

147) the net section is determined somewhat differently The AISCSpecification (see Ref 3) reads

In the case of a chain of holes extending across a part in any diagonal orzigzag line the net width of the part shall be obtained by deducting from thegross width the sum of the diameters of all the holes in the chain and addingfor each gage space in the chain the quantity s24g where

s = longitudinal spacing (pitch) in inches or any two successive holesg = transverse spacing (gage) in inches for the same two holes

The critical net section of the part is obtained from that chain that gives theleast net width

The AISC Specification also provides that in no case shall the net sec-tion through a hole be considered as more than 85 of the correspond-ing gross section

142 DESIGN OF A BOLTED CONNECTION

The issues raised in the preceding sections are illustrated in the follow-ing design example

Example 1 The connection shown in Figure 148 consists of a pair ofnarrow plates that transfer a tension force of 100 kips [445 kN] to a sin-gle 10-in-wide [250-mm] plate All plates are of A36 steel with Fy = 36ksi [250 MPa] and Fu = 58 ksi [400 MPa] and are attached with 3frasl4-in

DESIGN OF A BOLTED CONNECTION 337

Figure 147 Determination of net cross-sectional area for a member in a boltedjoint

3751 P-14 111301 1227 PM Page 337

A325 bolts placed in two rows Using data from Table 141 determinethe number of bolts required the width and thickness of the narrowplates the thickness of the wide plate and the layout for the connection

Solution From Table 141 the capacity of a single bolt in double shear isfound as 155 kips [69 kN] The required number for the connection is thus

Although placement of seven bolts in the connection is possible most de-signers would choose to have a symmetrical arrangement with eightbolts four to a row The average bolt load is thus

From Table 142 for the 3frasl4-in bolts minimum edge distance for a cutedge is 125 in and minimum recommended spacing is 225 in The min-imum required width for the plates is thus (see Figure 146)

w = b + 2(a) = 225 + 2(125) = 475 in [121 mm]

If space is tightly constrained this actual width could be specified for thenarrow plates For this example a width of 6 in is used Checking for

P = = [ ]100

812 5 55 6 kips kN

n say= =100

15 56 45 7


Figure 148 Example 1 General considerations

3751 P-14 111301 1227 PM Page 338

the requirement of stress on the gross area of the plate cross section wherethe allowable stress is 060Fy = 060(36) = 216 ksi the required area is

and with the 6-in width the required thickness is

This permits the use of a minimum thickness of 7frasl16 in (04375 in) [11 mm]The next step is to check the stress on the net section where the al-

lowable stress is 050Fu = 050(58) = 29 ksi [200 MPa] For the compu-tations it is recommended to use a bolt-hole size at least 1frasl8-in larger thanthe bolt diameter This allows for the true over-size (usually 1frasl16-in) andsome loss due to the roughness of the hole edges Thus the hole is as-sumed to be 7frasl8-in (0875) in diameter and the net width is

w = 6 ndash 2(0875) = 425 in [108 mm]

and the stress on the net section is

As this is lower than the allowable stress the narrow plates are adequatefor tension stress

The bolt capacities in Table 141 are based on a slip-critical conditionwhich assumes a design failure limit to be that of the friction resistance(slip resistance) of the bolts However the back-up failure mode is theone in which the plates slip to permit development of the pin-action ofthe bolts against the sides of the holes this then involves the shear ca-pacity of the bolts and the bearing resistance of the plates Bolt shear ca-pacities are higher than the slip failures so the only concern for this is thebearing on the plates For this the AISC Specification (see Ref 3) allowsa value of Fp = 12Fu = 12(58) = 696 ksi [480 MPa]

Bearing stress is computed by dividing the load for a single bolt by theproduct of the bolt diameter and the plate thickness Thus for the narrowplates

ft =times( )

= [ ]100

2 0 4375 4 2526 9 185

ksi MPa

t =times

= [ ]4 63

2 60 386 9 8

in mm

A = = [ ]100

21 64 63 29872 2

in mm


3751 P-14 111301 1227 PM Page 339

which is clearly not a critical concernFor the middle plate the procedure is essentially the same except

that the width is given and there is a single plate As before the stress onthe unreduced cross section requires an area of 463 in2 so the requiredthickness of the 10-in-wide plate is

which indicates the use of a 1frasl2-in thicknessFor the middle plate the width at the net section is

w = 10 ndash (2 times 0875) = 825 in [210 mm]


which compares favorably with the allowable of 29 ksi as determinedpreviously

The computed bearing stress on the sides of the holes in the middleplate is

which is less than the allowable value of 696 ksi as determined previouslyIn addition to the layout restrictions described in Section 141 the

AISC Specification (see Ref 3) requires that the minimum spacing in thedirection of the load be

and that the minimum edge distance in the direction of the load be

2P

F tu

2

2

P

F t

D

u

+

fp =times

= [ ]12 5

0 75 0 5033 3 230

ksi MPa

ft =times

= [ ]100

8 25 0 524 24 167

ksi MPa

t = = [ ]4 63

100 463 11 8

in mm

fp =times times

= [ ]12 5

2 0 75 0 437519 05 131

ksi MPa


3751 P-14 111301 1227 PM Page 340

in which

D = the diameter of the bolt

P = the force transmitted by one bolt to the connected part

t = the thickness of the connected part

For this example for the middle plate the minimum edge distance is thus

which is considerably less than the distance listed in Table 142 for the 3frasl4-in bolt at a cut edge 125 in

For the minimum spacing

which is also not criticalA final problem that must be considered is the possibility for tearing

out of the two bolts at the end of a plate in a block shear failure (Figure149) Because the combined thicknesses of the outer plates is greaterthan that of the middle plate the critical case for this connection is thatof the middle plate Figure 149 shows the condition for tearing which

2

20 862 0 375 1 237

P

F t

D

u

+ = + = in

2 2 12 5

58 0 50 862

P

F tu

= timestimes

=

in


Figure 149 Example 1 Tearing in themiddle plate

3751 P-14 111301 1227 PM Page 341

involves a combination of tension on the section labeled 1 and shear onthe two sections labeled 2 For the tension section

net w = 3 ndash 0875 = 2125 in [54 mm]

and the allowable stress for tension is

Ft = 050Fu = 29 ksi [200 MPa]

For the two shear sections

and the allowable stress for shear is

Fv = 030Fu = 174 ksi [120 MPa]

The total resistance to tearing is thus

T = (2125 times 05 times 29) + (1625 times 05 times 174) = 4495 kips [205 kN]

Because this is greater than the combined load on the two end bolts (25kips) the plate is not critical for tearing in block shear

The solution for the connection is displayed in the top and side viewsin Figure 1410

Connections that transfer compression between the joined parts are es-sentially the same with regard to the bolt stresses and bearing on theparts Stress on the net section in the joined parts is not likely to be criti-cal since the compression members are likely to be designed for a rela-tively low stress due to column action

Problem 142AA bolted connection of the general form shown in Fig 148 is to be usedto transmit a tension force of 175 kips [780 kN] by using 7frasl8-in A325bolts and plates of A36 steel The outer plates are to be 8 in wide [200mm] and the center plate is to be 12 in wide [300 mm] Find the requiredthicknesses of the plates and the number of bolts needed if the bolts areplaced in two rows Sketch the final layout of the connection

net in mmw = minus

= [ ]2 1 25

0 875

21 625 41 3


3751 P-14 111301 1227 PM Page 342

Problem 142BDesign the connection for the data in Problem 142A except that theouter plates are 9 in wide and the bolts are placed in three rows

143 WELDED CONNECTIONS

Welding is in some instances an alternative means of making connec-tions in a structural joint the other principal option being structural boltsA common situation is that of a connecting device (bearing plate fram-ing angles etc) that is welded to one member in the shop and fastened bybolting to a connecting member in the field However there are alsomany instances of joints that are fully welded whether done in the shop

WELDED CONNECTIONS 343

Figure 1410 Example 1 Layout for the final solution

3751 P-14 111301 1227 PM Page 343

or at the site of the building construction For some situations the use ofwelding may be the only reasonable means of making an attachment fora joint As in many other situations the design of welded joints requiresconsiderable awareness of the problems encountered by the welder andthe fabricator of the welded parts

One advantage of welding is that it offers the possibility for directconnection of members often eliminating the need for intermediate de-vices such as gusset plates or framing angles Another advantage is thelack of need for holes (required for bolts) which permits development of the capacity of the unreduced cross section of tension members Weld-ing also offers the possibility of developing exceptionally rigid joints an advantage in moment-resistive connections or generally nondeformingconnections

Electric Arc Welding

Although there are many welding processes electric arc welding is theone generally used in steel building construction In this type of weldingan electric arc is formed between an electrode and the pieces of metal thatare to be joined The term penetration is used to indicate the depth fromthe original surface of the base metal to the point at which fusion ceasesThe melted metal from the electrode flows into the molten seat and whencool unites with the members that are to be welded together Partial pen-etration is the failure of the weld metal and base metal to fuse at the rootof a weld It may result from a number of items and such incomplete fu-sion produces welds that are inferior to those of full penetration (calledcomplete penetration welds)

Types of Welded Connections

There are three common forms of joints butt joints lap joints and teejoints Several variations of these joints are shown in Figure 1411 Whentwo members are to be joined the ends or edges may or may not beshaped in preparation for welding The scope of this book prevents a de-tailed discussion of the many joints and their uses and limitations

A weld commonly used for structural steel in building construction isthe fillet weld It is approximately triangular in cross section and isformed between the two intersecting surfaces of the joined members (seeFigures 1411e f and g) As shown in Figure 1412a the size of a fillet


3751 P-14 111301 1227 PM Page 344

weld is determined by the leg length AB or BC of the largest isoscelesright triangle that can be inscribed within the weld cross section Thethroat of a fillet weld is the distance from the root to the hypotenuse ofthis same right triangle distance BD in Figure 1412a The exposed sur-face of a weld is not the plane surface indicated in Figure 1412a but is


Figure 1411 Common forms for welded joints

3751 P-14 111301 1227 PM Page 345

usually somewhat convex as shown in Figure 1412b Therefore the ac-tual throat may be greater than that shown in Figure 1412a This addi-tional material is called reinforcement It is not included in thedetermination of the strength of a weld

Stresses in Fillet Welds

If the weld size (dimension AB in Figure 1412a) is one unit in length thethroat dimension of the weld (BD in Figure 1412a) is

BD = 1frasl2(12 + 12)12 = 1frasl2(2)12 = 0707

Therefore the throat of a fillet weld is equal to the size of the weld mul-tiplied by 0707 As an example consider a 1frasl2-in fillet weld This wouldbe a weld with dimensions AB or BC equal to 1frasl2 in In accordance withthe above the throat would be 05 times 0707 or 03535 in Then if the al-lowable unit shearing stress on the throat is 21 ksi the allowable work-ing strength of a 1frasl2-in fillet weld is 03535 times 21 = 742 kips per lin inof weld If the allowable unit stress is 18 ksi the allowable workingstrength is 03535 times 18 = 636 kips per lin in of weld

The permissible unit stresses used in the preceding paragraph are forwelds made with E 70 XX- and E 60 XX-type electrodes on A36 steelParticular attention is called to the fact that the stress in a fillet weld isconsidered as shear on the throat regardless of the direction of the ap-plied load The allowable working strengths of fillet welds of varioussizes are given in Table 144 with values rounded to 010 kip

The stresses allowed for the metal of the connected parts (known asthe base metal) apply to complete penetration groove welds that arestressed in tension or compression parallel to the axis of the weld or are


Figure 1412 Dimensional considerations for welds

3751 P-14 111301 1227 PM Page 346

stressed in tension perpendicular to the effective throat They apply alsoto complete or partial penetration groove welds stressed in compressionnormal to the effective throat and in shear on the effective throat Conse-quently allowable stresses for butt welds are the same as for the basemetal

The relation between the weld size and the maximum thickness ofmaterial in joints connected only by fillet welds is shown in Table 145The maximum size of a fillet weld applied to the square edge of a plateor section that is 1frasl4 in or more in thickness should be 1frasl16 in less than thenominal thickness of the edge Along edges of material less than 1frasl4 inthick the maximum size may be equal to the thickness of the material

The effective area of butt and fillet welds is considered to be the ef-fective length of the weld multiplied by the effective throat thicknessThe minimum effective length of a fillet weld should not be less than four


TABLE 144 Safe Service Loads for Fillet Welds

Allowable Load Allowable Load(kipsin) (kNm)

Size of Size ofWeld E 60 XX E 70 XX E 60 XX E 70 XX Weld(in) Electrodes Electrodes Electrodes Electrodes (mm)

3frasl16 24 28 042 049 4761frasl4 32 37 056 065 6355frasl16 40 46 070 081 7943frasl8 48 56 084 098 9521frasl2 64 74 112 130 12705frasl8 80 93 140 163 15903frasl4 95 111 166 194 1910

TABLE 145 Relation Between Material Thickness and Size of Fillet Welds

Material Thickness of the Minimum SizeThicker Part Joined of Fillet Weld

in mm in mm

To 1frasl4 inclusive To 635 inclusive 1frasl8 318Over 1frasl4 to 1frasl2 Over 635 to 127 3frasl16 476Over 1frasl2 to 3frasl4 Over 127 to 191 1frasl4 635Over 3frasl4 Over 191 5frasl16 794

3751 P-14 111301 1227 PM Page 347

times the weld size For starting and stopping the arc a distance approx-imately equal to the weld size should be added to the design length of fil-let welds for specification to the welder

Figure 1413a represents two plates connected by fillet welds Thewelds marked A are longitudinal B indicates a transverse weld If a loadis applied in the direction shown by the arrow the stress distribution inthe longitudinal weld is not uniform and the stress in the transverse weldis approximately 30 higher per unit of length

Added strength is given to a transverse fillet weld that terminates atthe end of a member as shown in Figure 1413b if the weld is returnedaround the corner for a distance not less than twice the weld size Theseend returns sometimes called boxing afford considerable resistance tothe tendency of tearing action on the weld

The 1frasl4-in fillet weld is considered to be the minimum practical sizeand a 5frasl16-in weld is probably the most economical size that can be ob-tained by one pass of the electrode A small continuous weld is generallymore economical than a larger discontinuous weld if both are made inone pass Some specifications limit the single-pass fillet weld to 5frasl16 inLarge fillet welds require two or more passes (multipass welds) of theelectrode as shown in Figure 1413c

Design of Welded Connections

Welding in the shop (factory) is now often achieved by automatedprocesses However in the field welding is almost always achieved byldquohandrdquo and details must be developed on this basis The following ex-amples demonstrate the design for simple fillet welds for some ordinaryconnections


Figure 1413 Welding of lapped steel elements with fillet welds

3751 P-14 111301 1227 PM Page 348

Example 2 A bar of A36 steel 3 times 7frasl16 in [762 times 11 mm] in cross sec-tion is to be welded with E 70 XX electrodes to the back of a channel asshow in the two views in Figures 1414a and b Determine the size of thefillet weld required to develop the full tensile strength of the bar

Solution The usual allowable tension stress for this situation is 06Fy thus

Fa = 06(Fy) = 06(36) = 216 ksi

and the tension capacity of the bar is thus

T = Fa A = 216(3 times 04375) = 2835 kips

The weld must be of ample size to resist this forceA practical weld size is 3frasl8 in for which Table 144 yields a strength

of 56 kipsin The required length to develop the bar strength is thus

Adding a minimum distance equal to the weld size to each end for startand stop of the weld a practical length for specification would be 6 in

Figure 1414 shows three possibilities for the arrangement of the weldFor Figure 1414a the total weld is divided into two equal parts As thereare now two starts and stops some additional length should be usedPlacing 4 in of weld on each side of the bar should be adequate

For the weld in Figure 1414c there are three parts the first being a 3-in-long weld across the end of the bar That leaves another 3 in of re-quired weld which can be split between the two sides of the barmdasheachbeing a 2-in weld to assure a total of 3 in of effective weld

L = =28 35

5 65 06

in


Figure 1414 Example 2 Variations of form of the welded joint

3751 P-14 111301 1227 PM Page 349

Neither of the welds shown in Figures 1414a or c provides good re-sistance to the twisting action on the unsymmetrical joint To accommo-date this action most designers would provide some additional weld ifeither of these options is selected The better weld is that shown in Fig-ure 1414d where a weld is provided on the back of the bar between thebar and the corner of the channel This weld could be developed as an ad-dition to either of the welds in Figures 1414a or c The weld on the backis primarily only a stabilizing weld and would not be counted for directresistance of the required tension force

As may be seen there is more than computation involved in develop-ing a welded jointmdashand some judgements are those of individual designers

Example 3 A 31frasl2 times 31frasl2 times 5frasl16-in [89 times 89 times 8 mm] angle of A36 steelsubjected to a tensile load is to be connected to a plate by fillet weldsusing E 70 XX electrodes (see Figure 1415) What should the dimen-sions of the welds be to develop the full tensile strength of the angle

Solution From Table 95 the cross-sectional area of the angle is 209in2 [1348 mm2] The maximum allowable tension stress is 060Fy =060(36) = 216 ksi [150 MPa] thus the tensile capacity of the angle is

T = Ft A = (216)(209) = 451 kips [200 kN]

For the 5frasl16-in angle leg thickness the maximum recommended weldis 1frasl4 in From Table 144 the weld capacity is 37 kipsin The total lengthof weld required is thus

This total length could be divided between the two sides of the angleHowever assuming the tension load in the angle to coincide with its cen-troid the distribution of the load to the two sides is not in equal sharesThus some designers prefer to proportion the lengths of the two welds sothat they correspond to their positions on the angle If this is desired thefollowing procedure may be used

From Table 95 the centroid of the angle is at 099 in from the backof the angle Referring to the two weld lengths as shown in Figure 1415

L = = [ ]45 1

3 712 2 310

in mm


3751 P-14 111301 1227 PM Page 350

their lengths should be in inverse proportion to their distances from thecentroid Thus

and

These are the design lengths required and as noted earlier each shouldbe made at least 1frasl4 in longer at each end Reasonable specified lengthsare thus L1 = 925 in L2 = 40 in

When angle shapes are used as tension members and are connected attheir ends by fastening only one leg it is questionable to assume a stressdistribution of equal magnitude on the entire angle cross section Somedesigners therefore prefer to ignore the development of stress in the un-connected leg and to limit the member capacity to the force obtained byconsidering only the connected leg If this is done in this example themaximum tension is thus reduced to

T = Ft A = (216)(35 times 03125) = 23625 kips [105 kN]

and the required total weld length is

L = = [ ]23 625

3 76 39 162

in mm

L20 99

3 512 2 3 45 88=

( ) = [ ]

in mm

L12 51

3 512 2 8 75 222=

( ) = [ ]

in mm


Figure 1415 Example 3 Form of the welded joint

3751 P-14 111301 1227 PM Page 351

This length would then be divided evenly between the two sides Addingan extra length of twice the weld size a specified length would be for375 in on each side

Problem 143AA 4 times 4 times 1frasl2-in angle of A36 steel is to be welded to a plate with E 70XX electrodes to develop the full tensile strength of the angle Using 3frasl8-in fillet welds compute the design lengths for the welds on the twosides of the angle assuming development of tension on the full cross sec-tion of the angle

Problem 143BSame as Problem 143A except the angle is a 3 times 3 times 3frasl8 welds are madewith E 60 XX electrodes and are 5frasl16-in fillet welds

Problem 143CRedesign the welded connection in Problem 143A assuming that thetension force is developed only in the connected leg of the angle

Problem 143DRedesign the welded connection in Problem 143B assuming that thetension force is developed only in the connected leg of the angle


3751 P-14 111301 1227 PM Page 352

353

15REINFORCED

CONCRETE BEAMS

This chapter deals primarily with concrete formed with the common bind-ing agent of Portland cement and a loose mass consisting of sand andgravel With minor variations this is the material most used for structuralconcretemdashto produce building structures pavements and foundations

151 GENERAL CONSIDERATIONS

Concrete made from natural materials was used by ancient builders thou-sands of years ago Modern concrete made with industrially producedcement was developed in the early part of the nineteenth century whenthe process for producing Portland cement was developed Because of itslack of tensile strength however concrete was used principally forcrude massive structuresmdashfoundations bridge piers and heavy walls

In the mid to late nineteenth century several builders experimentedwith the technique of inserting iron or steel rods into relatively thin struc-tures of concrete to enhance their ability to resist tensile forces This was

3751 P-15 111301 1229 PM Page 353

the beginning of what we now know as reinforced concrete (see Figure151)

For building structures concrete is mostly used with a method calledsitecast concrete in which the wet concrete mix is deposited in someforming at the location where it is to be used This method is also de-scribed as cast-in-place or in situ construction

Design Methods

Traditional structural design was developed primarily with a method nowreferred to as stress design This method utilizes basic relationships de-rived from classic theories of elastic behavior of materials The adequacy

354 REINFORCED CONCRETE BEAMS

Figure 151 Elements of a concreteframe structure for a building Most rein-forced concrete beams occur in this situa-tion interacting with simultaneously castcolumns and spanning slabs to form awhole concrete system

3751 P-15 111301 1229 PM Page 354

or safety of designs is measured by comparison with two primary limitsan acceptable level for maximum stress and a tolerable limit for the ex-tent of deformation (deflection stretch etc) These limits are calculatedas they occur in response to the service loads that is the loads caused bythe normal usage conditions visualized for the structure This method isalso called the working stress method The stress limits are called allow-able working stresses and the tolerable movements are called allowabledeflection allowable elongation and so on

The Stress Method

The stress method generally consists of the following

1 The service (working) load conditions are visualized and quanti-fied as intelligently as possible Adjustments may be made hereby the determination of various statistically likely load combina-tions (dead load plus live load plus wind load etc) by consid-eration of load duration and so on

2 Stress stability and deformation limits are set by standards forthe various responses of the structure to the loads in tensionbending shear buckling deflection and so on

3 The structure is then evaluated (investigated) for its adequacy oris proposed (designed) for an adequate response

An advantage obtained in working with the stress method is that thereal usage condition (or at least an intelligent guess about it) is kept con-tinuously in mind The principal disadvantage comes from its detachednature regarding real failure conditions since most structures developmuch different forms of stress and strain as they approach their failurelimits

The Strength Method

In essence the working stress method consists of designing a structureto work at some established appropriate percentage of its total capacityThe strength method consists of designing a structure to fail but at aload condition well beyond what it should have to experience in use Amajor reason for favoring of strength methods is that the failure of astructure is relatively easily demonstrated by physical testing What is

GENERAL CONSIDERATIONS 355

3751 P-15 111301 1229 PM Page 355

truly appropriate as a working condition however is pretty much a theo-retical speculation The strength method is now largely preferred in pro-fessional design work It was first largely developed for design of concretestructures but is now generally taking over all areas of structural design

Nevertheless it is considered necessary to study the classic theories ofelastic behavior as a basis for visualization of the general ways that struc-tures work Ultimate responses are usually some form of variant from theclassic responses (because of inelastic materials secondary effects mul-timode responses etc) In other words the usual study procedure is tofirst consider a classic elastic response and then to observe (or speculateabout) what happens as failure limits are approached

For the strength method the process is as follows

1 The service loads are quantified as in step 1 of the stress methodand then are multiplied by an adjustment factor (essentially asafety factor) to produce the factored load

2 The form of response of the structure is visualized and its ulti-mate (maximum failure) resistance is quantified in appropriateterms (resistance to compression to buckling to bending etc)Sometimes this quantified resistance is also subject to an adjust-ment factor called the resistance factor

3 The usable resistance of the structure is then compared to the ul-timate resistance required (an investigation procedure) or a struc-ture with an appropriate resistance is proposed (a designprocedure)

When the design process using the strength method employs both loadand resistance factors it is now sometimes called load and resistancefactor design (abbreviated LRFD)

Strength of Concrete

The property of concrete of greatest significance for structural purposesis its resistance to compressive stress As such the common practice is tospecify a desired limiting capacity of compressive stress to design a con-crete mix to achieve that limit and to test samples of cast and hardenedconcrete to verify its true capacity for compression This stress is giventhe symbol fccent


3751 P-15 111301 1229 PM Page 356

For design work the capacity of concrete for all purposes is estab-lished as some percentage off fccent Attainment of a quality of concrete toachieve a particular level of compressive resistance generally also servesto certify various other properties such as hardness density and dura-bility Choice for the desired strength is typically based on the form ofconstruction For most purposes a strength of 3000 to 5000 psi for fccent isusually adequate However strengths of 20000 psi and higher have re-cently been achieved for lower columns in very tall structures

Stiffness of Concrete

As with other materials the stiffness of concrete is measured by the mod-ulus of elasticity designated E This modulus is established by tests andis the ratio of stress to strain Since strain has no unit designation (mea-sured as inchinch etc) the unit for E thus becomes the unit for stressusually psi or ksi [MPa]

The magnitude of elasticity for concrete Ec depends on the weight ofthe concrete and its strength For values of unit weight between 90 and155 lbft3 or pcf the value of Ec is determined as

The unit weight for ordinary stone-aggregate concrete is usually as-sumed to be an average of 145 pcf Substituting this value for w in theequation results in an average concrete modulus of

For metric units with stress measured in megapascals the expression be-comes

Distribution of stresses and strains in reinforced concrete is dependenton the concrete modulus the steel modulus being a constant This is dis-cussed in Section 152 In the design of reinforced concrete members theterm n is employed This is the ratio of the modulus of elasticity of steel

E fc c= prime4730

E fc c= prime57 000

E w fc c= prime1 533


3751 P-15 111301 1229 PM Page 357

to that of concrete or n = EsEc Es is taken as 29000 ksi [200000 MPa]a constant Values for n are usually given in tables of properties althoughthey are typically rounded off

In truth the modulus of elasticity of concrete is a variable not a con-stant value The general form of the stressstrain graph for concrete isshown in Figure 152 When plotted all the way to the failure of the mate-rial it describes a considerably curved form Thus the material has itsgreatest stiffness at low stress and loses stiffness continuously as it ap-proaches its stress limit It therefore becomes necessary to decide on thestress range at which a value for stiffness is to be considered For deter-mination of structural deformations (deflection of beams etc) at usageloading well below the ultimate limit an average value for E may betaken for the somewhat less curved lower portion of the graph This isgenerally the procedure for computation of E values used for designwork

Cement

The cement used most extensively in building construction is Portlandcement Of the five types of standard Portland cement generally availablein the United States and for which the American Society for Testing and


Figure 152 Consideration of the value for modulus of elasticity of concrete

3751 P-15 111301 1229 PM Page 358

Materials has established specifications two types account for most ofthe cement used in buildings These are a general-purpose cement for usein concrete designed to reach its required strength in about 28 days anda high-early-strength cement for use in concrete that attains its designstrength in a period of a week or less All Portland cements set andharden by reacting with water and this hydration process is accompaniedby generation of heat

Reinforcement

The steel used in reinforced concrete consists of round bars mostly of thedeformed type with lugs or projections on their surfaces The surface de-formations help to develop a greater bond between the steel rods and theenclosing concrete mass

Purpose of Reinforcement The essential purpose of steel rein-forcing is to reduce the failure of the concrete due to tensile stresses (seeFigure 153) Structural actions are investigated for the development oftension in the structural members and steel reinforcement in the properamount is placed within the concrete mass to resist the tension In somesituations steel reinforcement may also be used to increase compressiveresistance since the ratio of magnitudes of strength of the two materialsis quite high thus the steel displaces a much weaker material and themember gains significant strength

Tension stress can be induced by shrinkage of the concrete during itsdrying out from the initial wet mix Temperature variations may also in-duce tension in many situations To provide for these latter actions aminimum amount of reinforcing is used in surface-type members such aswalls and paving slabs even when no structural action is visualized

Stress-Strain Considerations The most common types of steelused for ordinary reinforcing bars are Grade 40 and Grade 60 havingyield strengths of 40 ksi [276 MPa] and 60 ksi [414 MPa] respectivelyThe yield strength of the steel is of primary interest for two reasons Plas-tic yielding of the steel generally represents the limit of its practical uti-lization for reinforcing of the concrete since the extensive deformation ofthe steel in its plastic range results in major cracking of the concreteThus for service load conditions it is desirable to keep the stress in thesteel within its elastic range of behavior where deformation is minimal


3751 P-15 111301 1229 PM Page 359

The second reason for the importance of the yield character of the re-inforcing is its ability to impart a generally yielding nature (plastic de-formation character) to the otherwise typically very brittle concretestructure This is of particular importance for dynamic loading and is amajor consideration in design for earthquake forces Also of importanceis the residual strength of the steel beyond its yield stress limit The steelcontinues to resist stress in its plastic range and then gains a secondhigher strength before failure Thus the failure induced by yielding isonly a first stage response and a second level of resistance is reserved


Figure 153 Utilization of reinforcement in concrete beams (a) Simple beam (b)Form of the moment diagram for uniformly distributed loading on a simple beam(c) Use of flexural reinforcement for a simple beam (d ) Continuous beam typicalof concrete construction (e) Form of the moment diagram for uniformly distributedloading on a continuous beam (f ) Use of flexural reinforcement for a continuousbeam

3751 P-15 111301 1229 PM Page 360

Cover Ample concrete protection called cover must be provided forthe steel reinforcement This is important to protect the steel from rustingand to be sure that it is well engaged by the mass of concrete Cover ismeasured as the distance from the outside face of the concrete to the edgeof the reinforcing bar

Code minimum requirements for cover are 3frasl4 in for walls and slabsand 11frasl2 in for beams and columns Additional distance of cover is re-quired for extra fire protection or for special conditions of exposure ofthe concrete surface to weather or by contact with the ground

Spacing of Bars Where multiple bars are used in concrete members(which is the common situation) there are both upper and lower limitsfor the spacing of the bars Lower limits are intended to facilitate the flowof wet concrete during casting and to permit adequate development of theconcrete-to-steel stress transfers for individual bars

Maximum spacing is generally intended to assure that there is somesteel that relates to a concrete mass of limited size that is there is not tooextensive a mass of concrete with no reinforcement For relatively thinwalls and slabs there is also a concern of scale of spacing related to thethickness of the concrete

Amount of Reinforcement For structural members the amount ofreinforcement is determined from structural computations as that re-quired for the tension force in the member This amount (in total cross-sectional area of the steel) is provided by some combination of bars Invarious situations however there is a minimum amount of reinforcementthat is desirable which may on occasion exceed the amount determinedby computation Minimum reinforcement may be specified as a mini-mum number of bars or as a minimum amount of steel cross-sectionalarea the latter usually based on the amount of the cross-sectional area ofthe concrete member

Standard Reinforcing Bars In early concrete work reinforcingbars took various shapes An early problem that emerged was the properbonding of the steel bars within the concrete mass due to the tendency ofthe bars to slip or pull out of the concrete

In order to anchor the bars in the concrete various methods were usedto produce something other than the usual smooth surfaces on bars (seeFigure 154) After much experimentation and testing a single set of bars


3751 P-15 111301 1229 PM Page 361

was developed with a form similar to that for the top left bar in Figure154 Deformed bars were produced in graduated sizes with bars identi-fied by a single number (see Table 151)

For bars numbered 2 through 8 the cross-sectional area is equivalentto a round bar having a diameter of as many eighths of an inch as the barnumber Thus a No 4 bar is equivalent to a round bar of 4frasl8 or 05 in di-ameter Bars numbered from 9 up lose this identity and are essentiallyidentified by the tabulated properties in a reference document

The bars in Table 151 are developed in US units but can of coursebe used with their properties converted to metric units However a new setof bars has recently been developed deriving their properties more logi-cally from metric units The general range of sizes is similar for both setsof bars and design work can readily be performed with either set Metric-based bars are obviously more popular outside the United States but fordomestic use (nongovernment) in the United States the old bars are still inwide use This is part of a wider conflict over units that is still going on

The work in this book uses the old inch-based bars simply becausethe computational examples are done in US units In addition manywidely used references still use US units and the old bar sizes


Figure 154 Forms of early steel reinforcing bars Reproduced from ConcretemdashPlain and Reinforced by Frederick W Taylor and Sanford E Thompson 1916with permission of the publisher John Wiley amp Sons New York

3751 P-15 111301 1229 PM Page 362

152 FLEXURE STRESS METHOD

For wood or steel beams the usual concern is only for the singular max-imum values of bending and shear in a given beam For concrete beamson the other hand it is necessary to provide for the values of bending andshear as they vary along the entire length of a beam even through multi-ple spans in the case of continuous beams which are a common occur-rence in concrete structures For simplification of the work it is necessaryto consider the actions of a beam at a specific location but it should beborne in mind that this action must be integrated with all the other effectson the beam throughout its length

When a member is subjected to bending such as the beam shown inFigure 155a internal resistances of two basic kinds are generally re-quired Internal actions are ldquoseenrdquo by visualizing a cut section such asthat taken at X-X in Figure 155a Removing the portion of the beam tothe left of the cut section its free-body actions are as shown in Figure155b At the cut section consideration of static equilibrium requires thedevelopment of the internal shear force (V in the figure) and the internalresisting moment (represented by the force couple C and T in the figure)

If a beam consists of a simple rectangular concrete section with ten-sion reinforcement only as shown in Figure 155c the force C is consid-ered to be developed by compressive stresses in the concretemdashindicated

FLEXURE STRESS METHOD 363

TABLE 151 Properties of Deformed Reinforcing Bars

Nominal Dimensions

Cross-Sectional

Bar Size Nominal Weight Diameter Area

Designation lbft kgm in mm in2 mm2

No 3 0376 0560 0375 95 011 71No 4 0668 0994 0500 127 020 129No 5 1043 1552 0625 159 031 200No 6 1502 2235 0750 191 044 284No 7 2044 3042 0875 222 060 387No 8 2670 3974 1000 254 079 510No 9 3400 5060 1128 287 100 645No 10 4303 6404 1270 323 127 819No 11 5313 7907 1410 358 156 1006No 14 7650 11390 1693 430 225 1452No 18 13600 20240 2257 573 400 2581

3751 P-15 111301 1229 PM Page 363

by the shaded area above the neutral axis The tension force however isconsidered to be developed by the steel alone ignoring the tensile resis-tance of the concrete For low-stress conditions the latter is not true butat a serious level of stress the tension-weak concrete will indeed crackvirtually leaving the steel unassisted as assumed

At moderate levels of stress the resisting moment is visualized asshown in Figure 156a with a linear variation of compressive stress fromzero at the neutral axis to a maximum value of fc at the edge of the sec-tion As stress levels increase however the nonlinear stress-strain char-acter of the concrete becomes more significant and it becomes necessaryto acknowledge a more realistic form for the compressive stress varia-tion such as that shown in Figure 156b As stress levels approach thelimit of the concrete the compression becomes vested in an almost con-stant magnitude of unit stress concentrated near the top of the sectionFor strength design in which the moment capacity is expressed at the ul-timate limit it is common to assume the form of stress distribution shownin Figure 156c with the limit for the concrete stress set at 085 times fccentExpressions for the moment capacity derived from this assumed distrib-ution have been shown to compare reasonably with the response ofbeams tested to failure in laboratory experiments

Response of the steel reinforcement is more simply visualized and ex-pressed Since the steel area in tension is concentrated at a small location


Figure 155 Development of bending in a reinforced concrete beam

3751 P-15 111301 1229 PM Page 364

with respect to the size of the beam the stress in the bars is considered tobe a constant Thus at any level of stress the total value of the internaltension force may be expressed as

T = As fs

and for the practical limit of T

T = As fy

In stress design a maximum allowable (working) value for the ex-treme fiber stress is established and the formulas are predicated on elas-tic behavior of the reinforced concrete member under service load Thestraight-line distribution of compressive stress is valid at working stresslevels because the stresses developed vary approximately with the dis-tance from the neutral axis in accordance with elastic theory

The following is a presentation of the formulas and procedures used inthe stress method The discussion is limited to a rectangular beam sectionwith tension reinforcement only

Referring to Figure 157 the following are defined

b = width of the concrete compression zone

d = effective depth of the section for stress analysis from thecentroid of the steel to the edge of the compressive zone


Figure 156 Distribution of bending stress in a reinforced concrete beam (a) Atlow levels of stress (b) At the point where the maximum stress in the concrete ap-proaches the limit (c) As assumed for investigation by the strength method

3751 P-15 111301 1229 PM Page 365

As = cross-sectional area of reinforcing bars

p = percentage of reinforcement defined as p = Asbd

n = elastic ratio defined as n = (E of the steel)(E of the concrete)

kd = height of the compression stress zone used to locate theneutral axis of the stressed section expressed as a decimalfraction (k) of d

jd = internal moment arm between the net tension force and thenet compression force expressed as a decimal fraction ( j) of d

fc = maximum compressive stress in the concrete

fs = tensile stress in the reinforcement

The compression force C may be expressed as the volume of the com-pression stress ldquowedgerdquo as shown in the figure

C = 1frasl2(kd )(b)( fc) = 1frasl2kfcbd

Using this force we may express the moment resistance of the section as

M = Cjd = (1frasl2kfcbd )( jd ) = 1frasl2kjfcbd 2 (1521)

This may be used to derive an expression for the concrete stress

(1522)f

M

kjbdc = 2

2


Figure 157 Development of bending resistance stress method

3751 P-15 111301 1229 PM Page 366

The resisting moment may also be expressed in terms of the steel andthe steel stress as

M = Tjd = As fs jd

This may be used for determination of the steel stress as

(1523)

or for finding the required area of steel as

(1524)

A useful reference is the so-called balanced section which occurswhen use of the exact amount of reinforcement results in the simultane-ous development of the limiting stresses in the concrete and steel Theproperties that establish this relationship may be expressed as follows

(1525)

(1526)

(1527)

(1528)

in which

R = 1frasl2kjfc (1529)

derived from equation (1521)If the limiting compression stress in the concrete ( fc = 045fccent) and the

limiting stress in the steel are entered in equation (1525) the balanced

M Rbd= 2

pf k

fc

s

=2

jk= minus

1

3

kf nfs c

=+ ( )

1

1

AM

f jds

s

=

fM

A jds

s

=


3751 P-15 111301 1229 PM Page 367

section value for k may be found Then the corresponding values for j pand R may be found The balanced p may be used to determine the max-imum amount of tensile reinforcement that may be used in a sectionwithout the addition of compressive reinforcing If less tensile reinforce-ment is used the moment will be limited by the steel stress the maxi-mum stress in the concrete will be below the limit of 045fccent the value ofk will be slightly lower than the balanced value and the value of j will beslightly higher than the balanced value These relationships are useful indesign for the determination of approximate requirements for cross sections

Table 152 gives the balanced section properties for various combina-tions of concrete strength and limiting steel stress The values of n k jand p are all without units However R must be expressed in particularunits the units in the table are kips per square inch (ksi) and kilopascals(kPa)

When the area of steel used is less than the balanced p the true valueof k may be determined by the following formula

Figure 158 may be used to find approximate k values for various com-binations of p and n

Beams with reinforcement less than that required for the balancedmoment are called under-balanced sections or under-reinforced sectionsIf a beam must carry bending moment in excess of the balanced moment

k np np np= minus ( ) minus2 2


TABLE 152 Balanced Section Properties for Rectangular Sections withTension Reinforcement Only

fs fccent R

ksi MPa ksi MPa n k j p ksi kPa

20 138 2 1379 113 0337 0888 00076 0135 9283 2068 92 0383 0872 00129 0226 15544 2758 80 0419 0860 00188 0324 22285 3448 71 0444 0852 00250 0426 2937

24 165 2 1379 113 0298 0901 00056 0121 8323 2068 92 0341 0886 00096 0204 14034 2758 80 0375 0875 00141 0295 20285 3448 71 0400 0867 00188 0390 2690

3751 P-15 111301 1229 PM Page 368

for the section it is necessary to provide some compressive reinforce-ment The balanced section is not necessarily a design ideal but is use-ful in establishing the limits for the section

In the design of concrete beams there are two situations that com-monly occur The first occurs when the beam is entirely undeterminedthat is when both the concrete dimensions and the reinforcement neededare unknown The second occurs when the concrete dimensions aregiven and the required reinforcement for a specific bending momentmust be determined The following examples illustrate the use of the for-mulas just developed for each of these problems


Figure 158 Flexural k factors for rectangular beams with tensile reinforcing onlyas a function of p and n

3751 P-15 111301 1229 PM Page 369

Example 1 A rectangular concrete beam with concrete having fccent of3000 psi [207 MPa] and steel reinforcement with fs = 20 ksi [138 MPa]must sustain a bending moment of 200 kip-ft [271 kN-m] Select thebeam dimensions and the reinforcement for a section with tension rein-forcement only

Solution With tension reinforcement only the minimum size beam is abalanced section since a smaller beam would have to be stressed beyondthe capacity of the concrete to develop the required moment Using equa-tion (1528)

M = Rbd 2 = 200 kip-ft [271 kN-m]

Then from Table 152 for fccentof 3000 psi and fs of 20 ksi

R = 0226 (in units of kip-in) [1554 in units of kN-m]

Therefore

M = 200 times 12 = 0226(bd 2) and bd 2 = 10619

Various combinations of b and d may be found for example

Although they are not given in this example there are often consider-ations other than flexural behavior alone that influence the choice of spe-cific dimensions for a beam These may include

Design for shear

Coordination of the depths of a set of beams in a framing system

Coordination of the beam dimensions and placement of reinforcementin adjacent beam spans

b d

d

b d

d

= = =

=[ ]

= = =

=[ ]

1010 619

1032 6

0 829

1510 619

1526 6

0 677

in in

b = 0254 m m

in in

b = 0381 m m


3751 P-15 111301 1229 PM Page 370

Coordination of beam dimensions with supporting columns

Limiting beam depth to provide overhead clearance beneath the structure

If the beam is of the ordinary form shown in Figure 159 the specifieddimension is usually that given as h Assuming the use of a No 3 U-stirrup a cover of 15 in [38 mm] and an average-size reinforcing bar of1-in [25-mm] diameter (No 8 bar) the design dimension d will be lessthan h by 2375 in [60 mm] Lacking other considerations assume a b of15 in [380 mm] and an h of 29 in [740 mm] with the resulting d of29 ndash 2375 = 26625 in [680 mm]

Next use the specific value for d with equation (1524) to find the re-quired area of steel As Since the selection is very close to the balancedsection use the value of j from Table 152 Thus

Or using the formula for the definition of p and the balanced p valuefrom Table 152

As = pbd = 00129(15 times 26625) = 515 in2 [3312 mm2]

Next select a set of reinforcing bars to obtain this area For the pur-pose of the example select bars all of a single size (see Table 152) thenumber required will be

AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 872 26 6255 17 33122 2

in mm


Figure 159 Common form of reinforcement for a rectangular concrete beamsection

3751 P-15 111301 1229 PM Page 371

No 6 bars 517044 = 1175 or 12 [3312284 = 1166]

No 7 bars 517060 = 862 or 9 [3312387 = 856]

No 8 bars 517079 = 654 or 7 [3312510 = 649]

No 9 bars 517100 = 517 or 6 [3312645 = 513]

No 10 bars 517127 = 407 or 5 [3312819 = 404]

No 11 bars 517156 = 331 or 4 [33121006 = 329]

In real design situations there are always various additional consider-ations that influence the choice of the reinforcing bars One general de-sire is that of having the bars in a single layer as this keeps the centroidof the steel as close as possible to the edge (bottom in this case) of themember giving the greatest value for d with a given height (h) of a con-crete section With the section as shown in Figure 159 a beam width of15 in will yield a net width of 1125 in inside the No 3 stirrups (outsidewidth of 15 less 2 times 15 cover and 2 times 0375 stirrup diameter) Applyingthe code criteria for minimum spacing for this situation the requiredwidth for the various bar combinations can be determined Minimumspace required between bars is one bar diameter or a lower limit of oneinch Two examples for this are shown in Figure 1510 It will be foundthat the four No 11 bars are the only choice that will fit this beam width


Figure 1510 Consideration of beam width for proper spacing of reinforcement ina single layer

3751 P-15 111301 1229 PM Page 372

Example 2 A rectangular beam of concrete with fccent of 3000 psi [207MPa] and steel with fs of 20 ksi [138 MPa] has dimensions of b = 15 in[380 mm] and h = 36 in [910 mm] Find the area required for the steelreinforcement for a moment of 200 kip-ft [271 kN-m]

Solution The first step in this case is to determine the balanced momentcapacity of the beam with the given dimensions If we assume the sectionto be as shown in Figure 159 we may assume an approximate value ford to be h minus 25 in [64 mm] or 335 in [851 mm] Then with thevalue for R from Table 152

Since this value is considerably larger than the required moment it isthus established that the given section is larger than that required for abalanced stress condition As a result the concrete flexural stress will belower than the limit of 045fccent and the section is qualified as being under-reinforced which is to say that the reinforcement required will be lessthan that required to produce a balanced section (with moment capacityof 317 kip-ft) In order to find the required area of steel we use equation(1524) just as we did in the preceding example However the truevalue for j in the equation will be something greater than that for the bal-anced section (0872 from Table 152)

As the amount of reinforcement in the section decreases below the fullamount required for a balanced section the value of k decreases and thevalue of j increases However the range for j is small from 0872 up tosomething less than 10 A reasonable procedure is to assume a value forj find the corresponding required area and then perform an investigationto verify the assumed value for j as follows Assume j = 090 Then

and

pA

bds= =

( )( )=3 98

15 33 50 00792

AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 90 33 53 98 25672 2

in mm

M Rbd

M M

= = ( )( )( ) =

= = = ( )( )( ) =[ ]

2 2

2

0 226 15 33 5 3804

3804

12317 1554 0 380 0 850 427

kip-in

or

kip-ft kN-M


3751 P-15 111301 1229 PM Page 373

Using this value for p in Figure 158 find k = 0313 Using equation(1526) j is found to be

which is reasonably close to the assumption so the computed area is ad-equate for design

For beams that are classified as under-reinforced (section dimensionslarger than the limit for a balanced section) a check should be made forthe minimum required reinforcement For the rectangular section theACI Code (Ref 4) specifies that a minimum area be

but not less than

On the basis of these requirements values for the minimum reinforce-ment for rectangular sections with tension reinforcement only are givenin Table 153 for the two common grades of steel and a range of concretestrengths

For the example with fccent of 3000 psi and fy of 40 ksi the minimumarea of steel is thus

As = 0005(bd) = 0005(15 times 335) = 251 in2

which is not critical in this case

AF

bdsy

=

( )200

Af

fbds

c

y

= ( )prime3

jk= minus

= minus

=1

31

0 313

30 896


TABLE 153 Minimum Required TensionReinforcement for Rectangular Sectionsa

fccent (psi) fy = 40 ksi fy = 60 ksi

3000 00050 0003334000 00050 0003335000 00053 000354

aRequired As equals table value times bd of the beam

3751 P-15 111301 1229 PM Page 374

Problem 152AA rectangular concrete beam has concrete with fccent= 3000 psi [207 MPa]and steel reinforcement with fs = 20 ksi [138 MPa] Select the beam di-mensions and reinforcement for a balanced section if the beam sustainsa bending moment of 240 kip-ft [326 kN-m]

Problem 152BSame as Problem 152A except fccent= 4000 psi fs = 24 ksi M = 160 kip-ft

Problem 152CFind the area of steel required and select the bars for the beam in Prob-lem 152A if the section dimensions are b = 16 in and d = 32 in

Problem 152DFind the area of steel required and select the bars for the beam in Prob-lem 152B if the section dimensions are b = 14 in and d = 25 in

153 GENERAL APPLICATION OF STRENGTH METHODS

Application of the working stress method consists of designing membersto work in an adequate manner (without exceeding established stress lim-its) under actual service load conditions Strength design in effect con-sists of designing members to fail thus the ultimate strength of themember at failure (called its design strength) is the only type of resis-tance considered The basic procedure of the strength method consists ofdetermining a factored (increased) design load and comparing it to thefactored (usually reduced) ultimate resistance of the structural member

The ACI Code (Ref 4) provides various combinations of loads thatmust be considered for design Each type of load (live dead wind earth-quake snow etc) is given an individual factor in these load equationsFor an example with only live and dead load considered the equation forthe factored design load U is

U = 14D + 17L

in which

D = the effect of dead load

L = the effect of live load

GENERAL APPLICATION OF STRENGTH METHODS 375

3751 P-15 111301 1229 PM Page 375

The design strength of individual members (ie their usable ultimatestrength) is determined by the application of assumptions and require-ments given in the code and is further modified by the use of a strengthreduction factor f as follows

f = 090 for flexure axial tension and combinations of flexureand tension

= 075 for columns with spirals

= 070 for columns with ties

= 085 for shear and torsion

= 070 for compressive bearing

= 065 for flexure in plain (not reinforced) concrete

Thus while the formulas for U may imply a somewhat low safety factoran additional margin of safety is provided by the strength reduction factors

154 FLEXURE STRENGTH METHOD

Figure 1511 shows the rectangular ldquostress blockrdquo that is used for analy-sis of the rectangular section with tension reinforcing only by the strengthmethod This is the basis for investigation and design as provided for inthe ACI Code (Ref 4)

The rectangular stress block is based on the assumption that a concretestress of 085fccent is uniformly distributed over the compression zone


Figure 1511 Development of bending resistance strength method

3751 P-15 111301 1229 PM Page 376

which has dimensions equal to the beam width b and the distance a thatlocates a line parallel to and above the neutral axis The value of a is de-termined from the expression a = b1 times c where b1 (beta one) is a factorthat varies with the compressive strength of the concrete and c is the dis-tance from the extreme fiber to the neutral axis For concrete having fccentequal to or less than 4000 psi [276 MPa] the Code gives a maximumvalue for a = 085c

With the rectangular stress block the magnitude of the compressiveforce in the concrete is expressed as

C = (085fccent )(b)(a)

and it acts at a distance of a2 from the top of the beam The arm of theresisting force couple then becomes d ndash (a2) and the developed resist-ing moment as governed by the concrete is

(1541)

With T expressed as As times fy the developed moment as governed by the re-inforcement is

(1542)

A formula for the dimension a of the stress block can be derived byequating the compression and tension forces thus

(1543)

By expressing the area of steel in terms of a percentage p the formulafor a may be modified as follows

(1544)

pA

bdA pbd

apbd f

f b

pdf

f

ss

y

c

y

c

= =

=( )( )

prime=

prime

0 85 0 85

0 850 85

f ba A f aA f

f bc s y

s y

c

prime = =prime

M T da

A f da

t s y= minus

= minus

2 2

M C da

f ba da

c c= minus

= prime minus

2

0 852

FLEXURE STRENGTH METHOD 377

3751 P-15 111301 1229 PM Page 377

The balanced section for strength design is visualized in terms ofstrain rather than stress The limit for a balanced section is expressed inthe form of the percentage of steel required to produce balanced condi-tions The formula for this percentage is

(1545)

in which fcprime and fy are in units of ksi Although this is a precise formulait is advisable to limit the percentage of steel to 75 of this balancedvalue in beams with tension reinforcing only

Returning to the formula for the developed resisting moment as ex-pressed in terms of the steel a useful formula may be derived as follows

Thus

Mt = Rbd 2 (1546)

where

(1547)

With the reduction factor applied the design moment for a section islimited to nine-tenths of the theoretical resisting moment

Values for the balanced section factors ( p R and ad ) are given inTable 154 for various combinations of fccentand fy The balanced section asdiscussed in the preceding section is not necessarily a practical one fordesign In most cases economy will be achieved by using less than thebalanced reinforcing for a given concrete section In special circum-

R pfa

dy= minus

12

M A f da

pbd f da

pbd f da

d

bd pfa

d

t s y

y

y

y

= minus

= ( )( ) minus

= ( )( )( ) minus

= ( ) minus

2

2

12

12

2

p f ff

b c yy

= prime

times+( )

0 8587

87


3751 P-15 111301 1229 PM Page 378

stances it may also be possible or even desirable to use compressive re-inforcing in addition to tension reinforcing Nevertheless just as in theworking stress method the balanced section is often a useful referencewhen design is performed

The following example illustrates a procedure for the design of a sim-ple rectangular beam section with tension reinforcing only

Example 3 The service load bending moments on a beam are 58 kip-ft[786 kN-m] for dead load and 38 kip-ft [515 kN-m] for live load Thebeam is 10 in [254 mm] wide fccent is 4000 psi [276 MPa] and fy is 60 ksi[414 MPa] Determine the depth of the beam and the tensile reinforcingrequired

Solution The first step is to determine the required moment using theload factors Thus

U = 14D + 17LMu = 14(MDL) + 17(MLL)

= 14(58) + 17(38) = 1458 kip-ft [1977 kN-m]

With the capacity reduction of 090 applied the desired moment ca-pacity of the section is determined as

MM

tu= = =

times [ ]0 90

145 8

0 90162

220

kip-ft

= 162 12 = 1944 kip-in kN-m


TABLE 154 Balanced Section Properties for Rectangular Sections withTension Reinforcement Only Strength Method

fy fccent Usable RBalanced Usable ad Usable

ksi MPa ksi MPa ad (75 Balanced) p ksi kPa

40 276 2 1379 05823 04367 00186 0580 40003 2068 05823 04367 00278 0870 60004 2758 05823 04367 00371 1161 80005 3448 05480 04110 00437 1388 9600

60 414 2 1379 05031 03773 00107 0520 36003 2068 05031 03773 00160 0781 54004 2758 05031 03773 00214 1041 72005 3448 04735 03551 00252 1241 8600

3751 P-15 111301 1229 PM Page 379

The maximum usable reinforcement ratio as given in Table 154 isp = 00214 If a balanced section is used the required area of reinforce-ment may thus be determined from the relationship

As = pbd

While there is nothing especially desirable about a balanced section itdoes represent the beam section with least depth if tension reinforcingonly is used Therefore proceed to find the required balanced section forthis example

To determine the required effective depth d use equation (1546)thus

M1 = Rbd 2

With the value of R = 1041 from Table 154

M1 = 1944 = 1041(10)(d )2

and

If this value is used for d the required steel area may be found as

As = pbd = 00214(10)(1366) = 292 in2 [1880 mm2]

From Table 154 the minimum ratio of reinforcing is 000333 which isclearly not critical for this example

Selection of the actual beam dimensions and the actual number andsize of reinforcing bars involves various considerations as discussed inSection 152

If there are reasons as there often are for not selecting the least deepsection with the greatest amount of reinforcing a slightly different pro-cedure must be used as illustrated in the following example

Example 4 Using the same data as in Example 3 find the reinforcementrequired if the desired beam section has b = 10 in [254 mm] and d = 18in [457 mm]

d =( )

= = [ ]1944

1 041 10186 7 13 66 347

in mm


3751 P-15 111301 1229 PM Page 380

Solution The first two steps in this situation would be the same as inExample 1mdashto determine Mu and Mt The next step would be to deter-mine whether the given section is larger than smaller than or equal to abalanced section Since this investigation has already been done in Ex-ample 1 observe that the 10 times 18 in section is larger than a balanced sec-tion Thus the actual value of a d will be less than the balanced sectionvalue of 03773 The next step would then be as follows

Estimate a value for a dmdashsomething smaller than the balanced valueFor example try a d = 025 Then

a = 025d = 025(18) = 45 in [114 mm]

With this value for a use equation (1542) to find a required value for AsReferring to Figure 1511

Next test to see if the estimate for a d was close by finding ad usingequation (1544) Thus

and

Thus

a da= ( ) = minus

= [ ]0 202 18 3 63

216 2 400 in in mm

apdf

f

a

d

pf

f

y

c

y

c

=prime

=prime

= ( )( )

=

0 85

0 85

0 0114 60

0 85 40 202

pA

bds= =

( )=2 057

10 180 0114

M T jd A F da

AM

f d a

t s y

st

y

= ( ) = minus

=minus ( )[ ] =

( )= [ ]

2

2

1944

60 15 752 057 13272 2

in mm


3751 P-15 111301 1229 PM Page 381

If this value for d ndash (a 2) is used to replace that used earlier the re-quired value of As will be slightly reduced In this example the correctionwill be only a few percent If the first guess of a d had been way off itmight have justified another run through the analysis to get closer to anexact answer

Problems 154AndashCUsing fccent= 3 ksi [207 MPa] and fy = 60 ksi [414 MPa] find the minimumdepth required for a balanced section for the given data Also find thearea of reinforcement required if the depth chosen is 15 times that re-quired for the balanced section Use strength design methods

Moment Due to

Dead Load Live Load Beam Width

kip-ft kN-m kip-ft kN-m (in) (mm)

A 40 542 20 271 12 305B 80 1085 40 542 15 381C 100 1356 50 678 18 457

155 T-BEAMS

When a floor slab and its supporting beams are cast at the same time theresult is monolithic construction in which a portion of the slab on eachside of the beam serves as the flange of a T-beam The part of the sectionthat projects below the slab is called the web or stem of the T-beam Thistype of beam is shown in Figure 1512a For positive moment the flangeis in compression and there is ample concrete to resist compressivestresses as shown in Figures 1512b or c However in a continuousbeam there are negative bending moments over the supports and theflange here is in the tension stress zone with compression in the web

It is important to remember that only the area formed by the width ofthe web bw and the effective depth d is to be considered in computing re-sistance to shear and to bending moment over the supports This is thehatched area shown in Figure 1512d


3751 P-15 111301 1229 PM Page 382

The effective flange width (bf) to be used in the design of symmetri-cal T-beams is limited to one-fourth the span length of the beam In ad-dition the overhanging width of the flange on either side of the web islimited to eight times the thickness of the slab or one-half the clear dis-tance to the next beam

In monolithic construction with beams and one-way solid slabs the ef-fective flange area of the T-beams is usually quite capable of resisting thecompressive stresses caused by positive bending moments With a largeflange area as shown in Figure 1512a the neutral axis of the section usu-ally occurs quite high in the beam web If the compression developed inthe web is ignored the net compression force may be considered to be lo-cated at the centroid of the trapezoidal stress zone that represents the stressdistribution in the flange On this basis the compression force is locatedat something less than t2 from the top of the beam

It is possible to conduct an approximate analysis of the T-section bythe working stress method while avoiding the need to find the location ofthe neutral axis and the centroid of the trapezoidal stress zone the pro-cedure consists of the following steps

T-BEAMS 383

Figure 1512 Considerations for T-beams

3751 P-15 111301 1229 PM Page 383

1 Determine the effective flange width for the T as previouslydescribed

2 Ignore compression in the web and assume a constant value forcompressive stress in the flange (see Figure 1513) Thus

Then find the required steel area as

3 Check the compressive stress in the concrete as

where

The actual value of maximum compressive stress will be slightlyhigher but will not be critical if this computed value is signifi-cantly less than the limit of 045fccent

4 T-beams ordinarily function for positive moments in continuousbeams Since these moments are typically less than those at thebeam supports and the required section is typically derived forthe more critical bending at the supports the T-beam is typicallyconsiderably under-reinforced This makes it necessary to con-sider the problem of minimum reinforcement as discussed for therectangular section The ACI Code (Ref 4) provides special re-quirements for this for the T-beam for which the minimum arearequired is defined as the greater value of

Af

fb ds

c

yw=

prime( )6

CM

jd

M

d t= =

minus ( )2

fC

b tc

f

=

AM

f jd

M

f d ts

s s

= =minus ( )[ ]2

jd dt= minus

2


3751 P-15 111301 1229 PM Page 384

or

in which

bw = the width of the beam web

bf = the effective width of the T-flange

The following example illustrates the use of this procedure It assumesa typical design situation in which the dimensions of the section (bf bwd and t) are all predetermined by other design considerations and the de-sign of the T-section is reduced to the work of determining the requiredarea of tension reinforcement

Example 5 A T-section is to be used for a beam to resist positive mo-ment The following data are given beam span = 18 ft [549 m] beamsare 9 ft [274 m] center to center slab thickness is 4 in [0102 m] beamstem dimensions are bw = 15 in [0381 m] and d = 22 in [0559 m] fccent=4 ksi [276 MPa] fy = 60 ksi [414 MPa] fs = 24 ksi [165 MPa] Find therequired area of steel and select reinforcing bars for a moment of 200 kip-ft [272 kN-m]

Solution Determine the effective flange width (necessary only for acheck on the concrete stress) The maximum value for the flange width is

Af

fb ds

c

yf= ( )

prime3

T-BEAMS 385

Figure 1513 Basis for simplified analysis of a T-beam

3751 P-15 111301 1229 PM Page 385

or

bf = center-to-center beam spacing = 9(12) = 108 in [274 m]

or

bf = beam stem width plus 16 times the slab thickness = 15 + 16(4) = 79 in [201 m]

The limiting value is therefore 54 in [137 m]Next find the required steel area as

Select bars using Table 155 which incorporates consideration for theadequacy of the stem width From the table choose five No 9 bars ac-tual As = 500 in2 Consideration for the beam width and adequate spac-ing of the bars is discussed in Example 1 in Section 152

Check the concrete stress

CM

jd

fC

b tc

f

= = times = [ ]

= =times

= [ ]

200 12

20120 535

120

54 40 556 3 83

kips kN

ksi MPa

AM

f d ts

s

=minus ( )[ ] = times

minus ( )[ ] = [ ]

2

200 12

24 22 4 25 00 33642 2 in mm

bf = = times = [ ]beam span

4 in m

18 12

454 1 37


TABLE 155 Options for the T-Beam Reinforcement

Actual Area Provided Width RequiredBar Size Number of Bars (in2) (in)

7 9 540 228 7 553 179 5 500 14

10 4 508 1311 4 628 14

3751 P-15 111301 1229 PM Page 386

Compare this to the limiting stress of

045fccent= 045(4) = 18 ksi [124 MPa]

Thus compressive stress in the flange is clearly not criticalUsing the beam stem width of 15 in and the effective flange width of

54 in the minimum area of reinforcement is determined as the greater of

or

As both of these are less than the computed area minimum area is notcritical in this case

The example in this section illustrates a procedure that is reasonably ad-equate for beams that occur in ordinary beam and slab construction Whenspecial T-sections occur with thin flanges (t less than d 8 or so) thesemethods may not be valid In such cases more accurate investigationshould be performed using the requirements of the ACI Code (Ref 4)

Problem 155AFind the area of steel reinforcement required for a concrete T-beam forthe following data fccent= 3 ksi allowable fs = 20 ksi [138 MPa] d = 28 in[711 mm] t = 6 in [152 mm] bw = 16 in [406 mm] and the section sus-tains a bending moment of 240 kip-ft [326 kN-m]

Problem 155BSame as Problem 155A except fccent = 4 ksi fs = 24 ksi d = 32 in t = 5in bw = 18 in M = 320 kip-ft

156 SHEAR IN CONCRETE BEAMS

From general consideration of shear effects as developed in the scienceof mechanics of materials the following observations can be made

Af

fb ds

c

yf= ( ) = ( )( ) = [ ]

prime3 3 4000

60 00054 22 2 56 16502 2

in mm

Af

fb ds

c

yw= ( ) = ( )( ) = [ ]

prime6 6 4000

60 00015 22 2 09 13502 2

in mm

SHEAR IN CONCRETE BEAMS 387

3751 P-15 111301 1229 PM Page 387

1 Shear is an ever-present phenomenon produced directly by slic-ing actions by transverse loading in beams and on oblique sec-tions in tension and compression members

2 Shear forces produce shear stress in the plane of the force andequal unit shear stresses in planes that are perpendicular to theshear force

3 Diagonal stresses of tension and compression having magnitudesequal to that of the shear stress are produced in directions of 45degfrom the plane of the shear force

4 Direct slicing shear force produces a constant magnitude shearstress on affected sections but beam shear action produces shearstress that varies on the affected sections having magnitude ofzero at the edges of the section and a maximum value at the cen-troidal neutral axis of the section

In the discussions that follow it is assumed that the reader has a gen-eral familiarity with these relationships

Consider the case of a simple beam with uniformly distributed loadand end supports that provides only vertical resistance (no moment re-straint) The distribution of internal shear and bending moment are asshown in Figure 1514a For flexural resistance it is necessary to providelongitudinal reinforcing bars near the bottom of the beam These bars areoriented for primary effectiveness in resistance to tension stresses thatdevelop on a vertical (90deg) plane (which is the case at the center of thespan where the bending moment is maximum and the shear approacheszero)

Under the combined effects of shear and bending the beam tends todevelop tension cracks as shown in Figure 1514b Near the center of thespan where the bending is predominant and the shear approaches zerothese cracks approach 90deg Near the support however where the shearpredominates and bending approaches zero the critical tension stressplane approaches 45deg and the horizontal bars are only partly effective inresisting the cracking

Shear Reinforcement for Beams

For beams the most common form of added shear reinforcement consistsof a series of U-shaped bent bars (Figure 1514d) placed vertically and


3751 P-15 111301 1229 PM Page 388

spaced along the beam span as shown in Figure 1514c These barscalled stirrups are intended to provide a vertical component of resis-tance working in conjunction with the horizontal resistance provided bythe flexural reinforcement In order to develop flexural tension near thesupport face the horizontal bars must be anchored in the concrete beyondthe point where the stress is developed Where the ends of simple beamsextend only a short distance over the support (a common situation) it isoften necessary to bend or hook the bars to achieve adequate anchorageas shown in Figure 1514c

The simple span beam and the rectangular section shown in Figure1514d occur only infrequently in building structures The most commoncase is that of the beam section shown in Figure 1515a which occurswhen a beam is cast continuously with a supported concrete slab In ad-dition these beams normally occur in continuous spans with negative


Figure 1514 Considerationsfor shear in concrete beams

3751 P-15 111301 1229 PM Page 389

moments at the supports Thus the stress in the beam near the support isas shown in Figure 1515a with the negative moment producing com-pressive flexural stress in the bottom of the beam stem This is substan-tially different from the case of the simple beam where the momentapproaches zero near the support

For the purpose of shear resistance the continuous T-shaped beam isconsidered to consist of the section indicated in Figure 1515b The effectof the slab is ignored and the section is considered to be a simple rec-tangular one Thus for shear design there is little difference between thesimple span beam and the continuous beam except for the effect of thecontinuity on the distribution of internal shear forces along the beamspan It is important however to understand the relationships betweenshear and moment in the continuous beam

Figure 1516 illustrates the typical condition for an interior span of acontinuous beam with uniformly distributed load Referring to the por-tions of the beam span numbered 1 2 and 3 on the moment diagram

1 In zone 1 the high negative moment requires major flexural re-inforcing consisting of horizontal bars near the top of the beam

2 In zone 2 the moment reverses sign moment magnitudes arelow and if shear stress is high the design for shear is a predom-inant concern

3 In zone 3 shear consideration is minor and the predominant con-cern is for positive moment requiring major flexural reinforcingin the bottom of the beam

(Note See Figure 153f for a typical layout of flexural reinforcement ina continuous beam)


Figure 1515 Development of negative bending moment and shear in concreteT-beams

3751 P-15 111301 1229 PM Page 390

Vertical U-shaped stirrups similar to those shown in Figure 1517a maybe used in the T-shaped beam An alternate detail for the U-shaped stirrupis shown in Figure 1517b in which the top hooks are turned outward thismakes it possible to spread the negative moment reinforcing bars to makeplacing of the concrete somewhat easier Figures 1517c and d show possi-bilities for stirrups in L-shaped beams that occur at the edges of large open-ings or at the outside edge of the structure This form of stirrup is used toenhance the torsional resistance of the section and also assists in develop-ing the negative moment resistance in the slab at the edge of the beam

So-called closed stirrups similar to ties in columns are sometimesused for T- and L-shaped beams as shown in Figures 1517c through fThese are generally used to improve the torsional resistance of the beamsection

Stirrup forms are often modified by designers or by the reinforcingfabricatorrsquos detailers to simplify the fabrication andor the field installa-tion The stirrups shown in Figures 1517d and f are two such modifica-tions of the basic details in Figures 1517c and e respectively

Design Considerations

The following are some of the general considerations and code require-ments that apply to current practices of design for beam shear


Figure 1516 Shear and bending in continuous concrete beams

3751 P-15 111301 1229 PM Page 391

Concrete Capacity Whereas the tensile strength of the concrete isignored in design for flexure the concrete is assumed to take some por-tion of the shear in beams If the capacity of the concrete is not ex-ceededmdashas is sometimes the case for lightly loaded beamsmdashthere maybe no need for reinforcement The typical case however is as shown inFigure 1518 where the maximum shear V exceeds the capacity of theconcrete alone (Vc) and the steel reinforcement is required to absorb theexcess indicated as the shaded portion in the shear diagram

Minimum Shear Reinforcement Even when the maximum com-puted shear stress falls below the capacity of the concrete the presentcode requires the use of some minimum amount of shear reinforcementExceptions are made in some situations such as for slabs and very shal-low beams The objective is essentially to toughen the structure with asmall investment in additional reinforcement

Type of Stirrup The most common stirrups are the simple U-shapedor closed forms shown in Figure 1517 placed in a vertical position at in-tervals along the beam It is also possible to place stirrups at an incline(usually 45deg) which makes them somewhat more effective in direct re-sistance to the potential shear cracking near the beam ends (see Figure


Figure 1517 Forms for vertical stirrups

3751 P-15 111301 1229 PM Page 392

1514b) In large beams with excessively high unit shear stress both ver-tical and inclined stirrups are sometimes used at the location of the great-est shear

Size of Stirrups For beams of moderate size the most common sizefor U-stirrups is a No 3 bar These bars can be bent relatively tightly atthe corners (small radius of bend) in order to fit within the beam sectionFor larger beams a No 4 bar is sometimes used its strength (as a func-tion of its cross-sectional area) being almost twice that of a No 3 bar

Spacing of Stirrups Stirrup spacings are computed (as discussed inthe following sections) on the basis of the amount of reinforcing requiredfor the unit shear stress at the location of the stirrups A maximum spac-ing of d 2 (ie one-half the effective beam depth d ) is specified in orderto ensure that at least one stirrup occurs at the location of any potentialdiagonal crack (see Figure 1514b) When shear stress is excessive themaximum spacing is limited to d4

Critical Maximum Design Shear Although the actual maximumshear value occurs at the end of the beam the ACI Code (Ref 4) permitsthe use of the shear stress at a distance of d (effective beam depth) fromthe beam end as the critical maximum for stirrup design Thus as shownin Figure 1519 the shear requiring reinforcing is slightly different fromthat shown in Figure 1518

Total Length for Shear Reinforcement On the basis of com-puted shear stresses reinforcement must be provided along the beamlength for the distance defined by the shaded portion of the shear stressdiagram shown in Figure 1519 For the center portion of the span theconcrete is theoretically capable of the necessary shear resistance withoutthe assistance of reinforcement However the code requires that some


Figure 1518 Sharing of shearresistance in reinforced concretebeams

3751 P-15 111301 1229 PM Page 393

shear reinforcement be provided for a distance beyond this computed cut-off point Earlier codes required that stirrups be provided for a distanceequal to the effective depth of the beam beyond the computed cutoff pointCurrently codes require that minimum shear reinforcement be providedas long as the computed shear stress exceeds one-half of the capacity ofthe concrete However it is established the total extended range overwhich reinforcement must be provided is indicated as R on Figure 1519

157 DESIGN FOR SHEAR IN CONCRETE BEAMS

The following is a description of the procedure for design of shear rein-forcement for beams that is in compliance with Appendix A of the 1995ACI Code (Ref 4)

Shear stress is computed as

vV

bd=


Figure 1519 Layout for shear stress analysis ACI Code requirements

3751 P-15 111301 1229 PM Page 394

in which

V = total shear force at the section

b = beam width (of the stem for T-shapes)

d = effective depth of the section

For beams of normal weight concrete subjected only to flexure andshear shear stress in the concrete is limited to

When v exceeds the limit for vc reinforcement must be providedcomplying with the general requirements discussed previously Althoughthe code does not use the term the notation of vcent is used here for the ex-cess unit shear for which reinforcement is required Thus

vcent = v ndash vc

Required spacing of shear reinforcement is determined as followsReferring to Figure 1520 note that the capacity in tensile resistance of asingle two-legged stirrup is equal to the product of the total steel cross-sectional area Av times the allowable steel stress Thus

T = Av fs

v fc c= prime1 1

DESIGN FOR SHEAR IN CONCRETE BEAMS 395

Figure 1520 Consideration forspacing of a single stirrup

3751 P-15 111301 1229 PM Page 395

This resisting force opposes the development of shear stress on thearea s times b in which b is the width of the beam and s is the spacing(half the distance to the next stirrup on each side) Equating the stirruptension to this force an equilibrium equation is obtained

Av fs = bsvcent

From this equation an expression for the required spacing can be de-rived thus

The following examples illustrate the design procedure for a simplebeam

Example 6 Design the required shear reinforcement for the simplebeam shown in Figure 1521a Use fccent = 3 ksi [207 MPa] and fs = 20 ksi[138 MPa] and single U-shaped stirrups

Solution The maximum value for the shear is 40 kips [178 kN] and themaximum value for shear stress is computed as

Now construct the shear stress diagram for one-half of the beam asshown in Figure 1521c For the shear design the critical shear stress isat 24 in (the effective depth of the beam) from the support Using pro-portionate triangles this value is

The capacity of the concrete without reinforcing is

At the point of critical stress therefore there is an excess shear stressof 104 ndash 60 = 44 psi [718 ndash 414 = 304 kPa] that must be carried by

v fc c= prime = = [ ]1 1 1 1 3000 60 414 psi KPa

72

96139 104 718

( ) = [ ] psi kPa

vV

bd= =

times= [ ]40 000

12 24139 957

psi KPa

sA f

v bv s=prime


3751 P-15 111301 1229 PM Page 396

reinforcement Next complete the construction of the diagram in Figure1521c to define the shaded portion which indicates the extent of the re-quired reinforcement Observe that the excess shear condition extends to544 in[1382 m] from the support

In order to satisfy the requirements of the ACI Code shear reinforce-ment must be used wherever the computed unit stress exceeds one-half ofvc As shown in Figure 1521c this is a distance of 753 in from the sup-port The code further stipulates that the minimum cross-sectional area ofthis reinforcing be

Abs

fv

y

=

50



3751 P-15 111301 1229 PM Page 397

Assuming an fy value of 40 ksi [276 MPa] and the maximum allow-able spacing of one-half the effective depth the required area is

which is less than the area of 2 times 011 = 022 in2 provided by the two legsof the No 3 stirrup

For the maximum vcent value of 44 ksi the maximum spacing permittedis determined as

Since this is less than the maximum allowable of one-half the depth or 12in it is best to calculate at least one more spacing at a short distance be-yond the critical point For example at 36 in from the support the stressis

and the value of vcent at this point is 87 ndash 60 = 27 psi The spacing requiredat this point is thus

which indicates that the required spacing drops to the maximum allowedat less than 12 in from the critical point A possible choice for the stir-rup spacings is shown in Figure 1521d with a total of eight stirrups thatextend over a range of 74 in from the support There are thus a total of16 stirrups in the beam 8 at each end Note that the first stirrup is placedat 4 in from the support which is one-half the computed required spac-ing this is a common practice with designers

Example 7 Determine the required number and spacings for No 3 U-stirrups for the beam shown in Figure 1522 Use fccent = 3 ksi [207 MPa]and fs = 20 ksi [138 MPa]

sA f

v bv s=prime

= timestimes

=0 22 20 000

27 1013 6

in

v =

( ) =60

96139 87 psi

sA f

v bv s=prime

= timestimes

=0 22 20 000

44 128 3

in

Av = times

=5012 12

40 0000 18 2

in


3751 P-15 111301 1229 PM Page 398

Solution As in Example 1 the shear values and corresponding stressesare determined and the diagram in Figure 1522c is constructed In thiscase the maximum critical shear stress of 89 psi results in a maximum vcentvalue to 29 psi for which the required spacing is

Since this value exceeds the maximum limit of d2 = 10 in the stir-rups may all be placed at the limited spacing and a possible arrangementis as shown in Figure 1522d As in Example 6 note that the first stirrupis placed at one-half the required distance from the support

sA f

v bv s=prime

= timestimes

=0 22 20 000

29 1015 2

in



3751 P-15 111301 1229 PM Page 399


Solution In this case the maximum critical design shear stress is foundto be less than vc which in theory indicates that reinforcement is not re-quired To comply with the code requirement for minimum reinforce-ment however provide stirrups at the maximum permitted spacing out to



3751 P-15 111301 1229 PM Page 400

the point where the shear stress drops to 30 psi (one-half of vc) To ver-ify that the No 3 stirrup is adequate compute

which is less than the area provided so the No 3 stirrup at 10-in isadequate

Problem 157AA concrete beam similar to that shown in Figure 1521 sustains a totalload of 60 kips [267 kN] on a span of 24 ft [732 m] Determine the lay-out for a set of No 3 U-stirrups with fs = 20 ksi [138 MPa] and fccent = 3000psi [207 MPa] The beam section dimensions are b = 12 in [305 mm]and d = 26 in [660 mm]

Problem 157BSame as Problem 157A except load is 50 kips [222 kN] span is 20 ft[61 m] b = 10 in [254 mm] d = 23 in [584 mm]

Problem 157CDetermine the layout for a set of No 3 U-stirrups for a beam with thesame data as Problem 157A except the total load on the beam is 30 kips[133 kN]

Problem 157DDetermine the layout for a set of No 3 U-stirrups for a beam with thesame data as Problem 157B except the total load on the beam is 25 kips[111 kN]

Abs

fv

y

=

= times

=50 5010 10

40 0000 125 2 in (See Example 6)


3751 P-15 111301 1229 PM Page 401

402

REFERENCES

1 Uniform Building Code Volume 2 Structural Engineering DesignProvisions 1997 ed International Conference of Building OfficialsWhittier CA (Called simply the UBC)

2 National Design Specification for Wood Construction (NDS) 1997ed American Forest and Paper Association Washington DC(Called simply the NDS)

3 Manual of Steel Construction 8th ed American Institute of SteelConstruction Chicago IL 1981 (Called simply the AISC Manual)

4 Building Code Requirements for Reinforced Concrete ACI 318-95American Concrete Institute Detroit MI 1995 (Called simply theACI Code)

5 Timber Construction Manual 3rd ed American Institute of TimberConstruction Wiley New York 1985

6 James Ambrose Design of Building Trusses Wiley New York 19947 James Ambrose Simplified Design of Building Foundations 2nd ed

Wiley New York 1988

3751 P-16 (refs) 111301 1230 PM Page 402

403

ANSWERS TO SELECTED EXERCISE PROBLEMS

Chapter 2

27A R = 8062 lb upward to the right 2974deg from the horizontal

27C R = 9487 lb downward to the right 1843deg from the horizontal

27E R = 100 lb downward to the left 5313deg from the horizontal

27G R = 5807 lb downward to the right 749deg from the horizontal

27I R = 9113 lb upward to the right 9495deg from the horizontal

28A 1414 lb T

28C 300 lb C

210A 193deg

210C 07925 lb

211A 400 lb

211C 1250 lb

212A Sample M about R1 = + (500 times 4) + (400 times 6) + (600 times 10) ndash(650 times 16)

212B R1 = 359375 lb [1598 kN] R2 = 440625 lb [1960 kN]

3751 P-17 (answers) 111301 1234 PM Page 403

404 ANSWERS TO SELECTED EXERCISE PROBLEMS

212D R1 = 7667 lb [3411 kN] R2 = 9333 lb [4153 kN]

212F R1 = 7143 lb [3179 kN] R2 = 11857 lb [5276 kN]

Chapter 3

31A Sample values CI = 2000 C IJ = 8125 T JG = 1250 T

32A Same as 31A

33A Sample values in kips DN 5333 C KL 1500 T OI 6000 TLM 2500 C

Chapter 4

43A Maximum shear = 10 kips [445 kN]

43C Maximum shear = 1114 lb [4956 kN]

43E Maximum shear = 9375 kips [41623 kN]

44A Maximum M = 60 kip-ft [801 kN-m]

44C Maximum M = 4286 ft-lb [5716 kN-m]

44E Maximum M = 1835 kip-ft [2445 kN-m]

45A R1 = 1860 lb [827 kN] maximum V = 1360 lb [605 kN] max-imum ndash M = 2000 ft-lb [266 kN-m] maximum + M = 3200 ft-lb [427 kN-m]

45C R1 = 2760 lb [1228 kN] maximum V = 2040 lb [907 kN]maximum ndash M = 2000 ft-lb [267 kN-m] maximum + M = 5520ft-lb [737 kN-m]

46A Maximum V = 1500 lb [667 kN] maximum M = 12800 ft-lb[171 kN-m]

46C Maximum V = 1200 lb [527 kN] maximum M = 8600 ft-lb[1133 kN-m]

47A M = 32 kip-ft [434 kN-m]

47C M = 90 kip-ft [122 kN-m]

Chapter 5

51A R1 = R3 = 1200 lb [534 kN] R2 = 4000 lb [1779 kN] + M =3600 ft-lb [499 kN-m] ndash M = 6400 ft-lb [868 kN-m]

51C R1 = 767 kips [3335 kN] R2 = 3558 kips [15479 kN] R3 =1275 kips [5546 kN]

3751 P-17 (answers) 111301 1234 PM Page 404

ANSWERS TO SELECTED EXERCISE PROBLEMS 405

51E R1 = R3 = 9375 lb [417 kN] R2 = 4125 lb [1835 kN] + M =7031 ft-lb [953 kN-m] ndash M = 13500 lb-ft [1831 kN-m]

51G R1 = R4 = 9600 lb [427 kN] R2 = R3 = 26400 lb [1174 kN] +M1 = 46080 ft-lb [6248 kN-m] + M2 = 14400 lb-ft [1953 kN-m] ndash M = 57600 ft-lb [7811 kN-m]

52A Maximum V = 8 kips maximum + M = maximum ndash M = 44 kip-ft inflection at 55 ft from both ends

53A R1 = 16 kips [72 kN] R2 = 48 kips [216 kN] maximum + M =64 kip-ft [864 kN-m] maximum ndash M = 80 kip-ft [108 kN-m]inflection at pin location in both spans

53C R1 = 64 kips [288 kN] R2 = 196 kips [882 kN] + M = 2048kip-ft [277 kN-m] in end span and 244 kip-ft [331 kN-m] incenter span ndash M = 256 kip-ft [344 kN-m] inflection at 32 ftfrom R2 in end span

Chapter 6

62A SF = 253

63A Maximum pressure = 1098 psf minimum pressure = 133 psf

Chapter 7

71A R = 10 kips up and 110 kip-ft counterclockwise

71C R = 6 kips to the left and 72 kip-ft counterclockwise

72A R1 = 45 kips down R2 = 45 kips up and 12 kips to the right

Chapter 8

81A R = 21605 lb x = 0769 ft z = 1181 ft

81C T1 = 508 lb T2 = 197 lb T3 = 450 lb

82A R = 4 lb down x = 1075 ft right z = 155 ft left

Chapter 9

91A cy = 26 in [70 mm]

91C cy = 42895 in [10724 mm]

3751 P-17 (answers) 111301 1234 PM Page 405


91E cy = 44375 in [1109 mm] cx = 10625 in [266 mm]

93A I = 53586 in4 [211 times 108 mm4]

93C I = 44733 in4 [17474 times 106 mm4]

93E I = 20533 in4 [ 8021 times 106 mm4]

93G I = 438 in4

93I I = 167245 in4

Chapter 10

102A 1182 in2 [762 mm2]

102C 270 kips [120 kN]

102E Not acceptable actual stress exceeds allowable

103A 19333 lb [86 kN]

103C 29550000 psi [203 GPa]

Chapter 11

112A Okay actual stress = 1399 ksi less than allowable of 24 ksi

113A 386 kips

113C 205 kips

113E 226 kips

114A W 12 times 22 or W 14 times 22 (lightest) also W 10 times 26 W 8 times 31

114C W 18 times 35

115A At neutral axis fv = 8114 psi at junction of web and flange fv

= 175 and 700 psi

116A 1683 kips

116C 371 kips

117A 6735 kips

119A 080 in [20 mm]

1110A 136

1110C 515

Chapter 12

122A 15720 lb

3751 P-17 (answers) 111301 1234 PM Page 406


123A 235 kips [1046 kN]

123C 274 kips [1219 kN]

Chapter 13

131A 3183 psi tension 2929 psi compression

132A (a) 304 ksf [151 kPa] (b) 533 ksf [266 kPa]

134A f = 933 psi [643 Mpa] v = 250 psi [172 MPa]

134C f = 750 psi [517 Mpa] v = 433 psi [299 MPa]

Chapter 14

142A 6 bolts outer plates 1frasl2 in middle plate 5frasl8 in

143A Rounding up to the next full inch L1 = 11 in L2 = 5 in

143C Minimum of 425 in weld on each side

Chapter 15

152A Width required to get bars into one layer is critical least widthis 16 in with h = 31 in and five No 10 bars

152C From work for Problem 152A this section is under-reinforcedfind actual k = 0347 j = 0884 required area of steel = 509 in2use four No 10 bars

154A With d = 11 in As = 367 in2 with d = 165 in As = 197 in2

155A 576 in2 [371 times 103 mm2]

157A Possible choice for spacing 1 at 6 in 8 at 13 in

157C 1 at 6 in 4 at 13 in

3751 P-17 (answers) 111301 1234 PM Page 407

3751 P-17 (answers) 111301 1234 PM Page 408

409

INDEX

Accuracy of computations 3Active soil pressure 184Algebraic analysis of

forces 89truss 120

Allowable deflection 277Allowable loads for

fillet welds 347steel

bolts 322columns 301

welded joints 347wood columns 297

Allowable stress design 354Allowable stress 243 355Angles structural steel 235

gage for bolts 334properties of 235in tension 336

Approximate analysis of structures 181Architectural elements 23Areas of steel reinforcing bars 363

Balanced reinforcementstrength design 378working stress design 367

Balanced section 367 378Bars reinforcing 363Beams

analysis of 132 259bending in 140cantilever 133 151concentrated load 134concrete 363connections steel 324continuous 134 160deflection 275diagrams typical loadings 155

3751 P-18 (index) 111301 1236 PM Page 409

distributed load 134effective depth 365fixed-end 172flexure formula 257flitched 272forces on 102indeterminate 162inflection 147with internal pins 176internal resisting moment 255investigation of 132 259loading 13moment diagram 142moment in 255neutral axis 216 256overhanging 107 133reactions 105resisting moment in 255restrained 172 290safe load tables for steelsense (sign) of bending 147shear diagram 139shear in 135 265 270 387simple 105 133statically indeterminate 162steel 270strength design of 355stresses in 254T-beams 382tabular data for 155theorem of three moments 163types of 133typical loadings 155under-reinforced 373uniformly distributed load 134vertical shear 136width concrete 372

Bending 41in bolted connection 327in concrete beams 383resistance 255stress 257

Bending momentin beam 140diagrams 142negative 147positive 147

Blast load 12Block shear failure 329 341Bolted connections 324

bearing in 327bending in 328effective net area 327 336 339gage for angles 334layout of 332tearing in 329 341tension stress in 327 335

Boltscapacity of in steel 332edge distance for in steel 333high-strength 331spacing in steel 334unfinished 331

Bowrsquos notation 85Buckling 30 294Built-up sections in steel 226

Cantileverbeam 133 151frame 193retaining wall 183

Cement 353 358Center of gravity 2Centroid 215Channels steel 234Classification of force systems 75Coefficient of friction 92Columns

buckling 294end conditions 302relative slenderness of 293slenderness of 293steel 301wood 297

410 INDEX

3751 P-18 (index) 111301 1236 PM Page 410

Combinedaxial force and moment 309load 20stress 49 321

Component of a force 81Composite construction 58Compression 39

in columns 293Computations 3Concentrated load 134Concurrent force systems 76 203Concrete

balanced section properties 367378

beam 363cast-in-place 354cover 361design methods 253 375k-factors for beams 369modulus of elasticity 357reinforced 354reinforcement for 358 363shear in 387sitecast 354stirrup 389stiffness 357strength of 356T-beam 382

Connectionbolted steel 324tension 327 335welded 343

Continuous action of beams 134 160Conversion factors for units 3 6Couple 99Cover of reinforcement 361Cracked section 314Cut section 66

Damping effect on harmonic motion63

Dead load of building construction 14

Deflectionallowable 277of beams general 275computations 279formulas typical loadings 155of steel beams 280of wood beams 281

Deformation 2 241and stress 239 246

Design methods 354Development of resisting moment

255in concrete beam 363

Dispersion of load 16Direct stress 46Double shear 48 252 327Ductility 48Dynamic behavior 61Dynamic effects 61

harmonic motion 62motion 61

Dynamics 2

Earth pressure 184Earthquake 12 19Eccentric load 309Edge distance of bolts in steel 353Effective

column length 296depth of concrete beam 365net area in tension 336 339width of concrete T-beam flange

382Elastic limit 48 246Equilibrant 82Equilibrium 36 77 203

static 2Euler buckling formula 33

Factored load 356Factor of safety 247Fillet weld 344

INDEX 411

3751 P-18 (index) 111301 1236 PM Page 411

Fixed end beam 172Flexure formula 257Flitched beam 272Floors 23Force

actions 69algebraic solution 89on beam 102classification of systems 75composition 78 81equilibrant 82equilibrium 36 77 203graphical analysis 83 88internal 28 39investigation 87line of action 74notation for truss analysis 85parallelogram 79point of application 74polygon 83properties 74resolution 78resultant 78systems 75 202types 72

Free body diagram 65 194Friction 91Functional requirements of structures

30Fundamental period 5

Gage in angles 334Generation of structures 21Graphical analysis of

forces 83 88truss 11

Handling load 12Harmonic motion 62 68High strength bolts steel 331Hookersquos Law 246

Horizontalearth pressure 184shear 136

Hydraulic load 12

Indeterminate structures 162Inelastic behavior 251Inflection in beams 147Internal forces 12 28 39

in beams 43 255combined 45in rigid framesin trusses 111

Internal pins in continuous beams 176Internal resisting moment 255Investigation of

beams 132 259columns 297 301forces 87frames 192trusses 111 120

Joints method of 111 120

Kern 314k factors for concrete beams 369K factors for steel columns 302

Lateral unsupported length ofcolumns 30 297 301

Line of action of a force 74Live load 14Load 11 69

blast 12combined 20concentrated 16dead (DL) 14dispersion 16distributed 16earthquake 12eccentric 309

412 INDEX

3751 P-18 (index) 111301 1236 PM Page 412

factored 356gravity 11handling 12hydraulic 12internal action 12live (LL) 14service 355shrinkage 12uniformly distributed 16vibration 12wind 11 17

LRFD (load and resistance factordesign) 356

Lumber properties of standard sizes237

Maxwell diagram 113Measurement units of 2Mechanical couple 99Mechanics 2Method of joints 111 120Minimum reinforcement 373Modulus

of elasticity for direct stress 48248 357

section 228Moment 97

arm 97beams 140 255diagram for beam 142of a force 97general definition 97of inertia 218internal bending moment 43 255of a mechanical couple 99negative 147overturning 186positive 147resistance 255restoring 188sense of 147

stabilizing 188statical 217

Moment of inertia 218Motion 76

Net sectionin shear329 341in tension 327 336 339

Neutral axis 216 256Nomenclature 7Noncoplanar forces 202Nonlinear behavior 251

Oblique section 49 319Overturning moment on retaining

wall 186

Passive soil pressure 184Parallel forces 76 102 209Parallelogram of forces 79Permanent set 247Pin internal 176Pitch of bolts 334Plastic

behavior of steel 283hinge 286moment 286section modulus 287

Point of inflection 147Polygon of forces 83Portland cement 353Pressure

soil 184wedge method 315

Properties offorces 74geometric shapes 230reinforcing bars 363sections 214

Properties of sectionsbuilt-up 217 226

INDEX 413

3751 P-18 (index) 111301 1236 PM Page 413

Properties of sections (cont)centroid 215channels 234geometric shapes 230kern 314lumber 237moment of inertia 218plastic section modulus 287radius of gyration 30 229section modulus 228single angle shapes 235statical moment 217steel angles 235steel channels 234steel pipe 236structural lumber 237transfer axis theorem 223W shapes 232

Radius of gyration 30 229Reactions 24 105Rectangular

beam in concrete 363stress block strength method 364

376Reinforced concrete 354Reinforcement for concrete 358Relation of shear and moment 144Relative

slenderness 30 293stiffness of columns 295

Resistance factor 356Restrained beam 172 290Restoring moment 188Resultant of forces 78Retaining wall 183Rigid frame 192Roofs 23

S elastic section modulus 228Safe load tables for

fillet welds 347steel bolts 332

Safety 247Section

balanced 367cut 66cracked 314net 323 336 339properties 214

Section moduluselastic 228plastic 287

Sense offorce 74bending 147

Separated joint diagram 113Service load 355Shapes steel 302Shear 40

in beams 135 265in bolts 326in concrete beam 387diagram for beam 139double 327horizontal 136 265reinforcement 389single 327in steel beam 70stress 245 317

Shrinkage load 12Sitecast concrete 354Simple beam 105Simple support 105Single shear 327Slenderness 30

of columns 293Soil pressure 184Solid-sawn wood element 297Space diagram 112 372Spacing of

bars in concrete 361steel bolts 334stirrups 393

Spanning structure 22

414 INDEX

3751 P-18 (index) 111301 1236 PM Page 414

Stability 30of retaining wall 186

Standard notation 7Static equilibrium 2 77Static versus dynamic force 14Statical moment 217Statically indeterminate

beam 162frame 199

Statics 2Steel

allowable stresses 243angle 235bolts 324columns 301connections 324pipe 236reinforcement 358

Stiffness 36 248relative 30

Stirrups 389spacing of 393

Straingeneral definition 2 46 243hardening 247

Strength 34 242of concrete 356design method 253 375of materials 2 242yield 246ultimate 253

Stress 2 46 71 243allowable 243 355in beams 254bending 140 255combined 49compression 46design 354and deformation 239 246development of internal force 46direct 46in ductile material 48

flexural 43general definition 2inelastic 257kinds of 46method 354 363on oblique section 49 319in plastic range 251shear 245 31in soils 184strain behavior 46strength method 355tensile 46thermally-induced 12 53types of 46ultimate 247unit 243working method 354 363yield 246

Stress-strain 46diagram 48 248 357ductility 48modulus of elasticity 48 248 357proportional limit 243relations 46time-related 59yield stress 243

Stirrup 389Structural

analysis 1investigation 1 64mechanics 2response 64

Structurescomposite 58functional requirements 30generation of 21spanning 21

Symbols 7

T-beam concrete 382Tearing in bolted connection 329 341Tension 39

INDEX 415

3751 P-18 (index) 111301 1236 PM Page 415

and combined bending 309in bolted connection 327 335

Tension elementseffective area in 336 339net section in 336 339upset end threaded rod 244

Three Moment Theorem 163Thermally-induced stress 12 53Throat of weld 345Time-dependent behavior 61Torsion 41 45Trusses

algebraic analysis 120Bowrsquos notation of forces for 85graphical analysis 111internal forces in 111investigation methods 111 127joints method of 111 120Maxwell diagram for 113notation of forces for 85sections method of 127separated joint diagram 113space diagram 112

Ultimate stress 247Under-reinforced concrete beam 373Unfinished bolt 331Uniform Building Code UBC 7Uniformly distributed load 134US units 2

Units of measurement 2conversion of 3 6

Unit stress 243Upset end 244

Vector 73Velocity of wind 18Vertical

shear 136soil pressure 188

Vibration 12

W shapes steel properties of 232Walls 23Welded connections 343Welds fillet 344Width of concrete beam 372Wind 11 17Wood

allowable stresses for 243columns 297deflection of beams 281properties of structural lumber 237

Working stress method 355

Yieldpoint 246stress 246

Z plastic section modulus 287

416 INDEX

3751 P-18 (index) 111301 1236 PM Page 416


CONTENTS

Preface to the Sixth Edition

Preface to the First Edition

Introduction

Structural Mechanics

Units of Measurement

Accuracy of Computations

Symbols

Nomenclature

1 Structures Purpose and Function

11 Loads

12 Special Considerations for Loads

13 Generation of Structures

14 Reactions

15 Internal Forces

16 Functional Requirements of Structures

17 Types of Internal Force

18 Stress and Strain

19 Dynamic Effects

110 Design for Structural Response

2 Forces and Force Actions

21 Loads and Resistance

22 Forces and Stresses

23 Types of Forces

24 Vectors

25 Properties of Forces

26 Motion

27 Force Components and Combinations

28 Graphical Analysis of Forces

29 Investigation of Force Actions

210 Friction

211 Moments

212 Forces on a Beam

3 Analysis of Trusses

31 Graphical Analysis of Trusses

32 Algebraic Analysis of Trusses

33 The Method of Sections

4 Analysis of Beams

41 Types of Beams

42 Loads and Reactions

43 Shear in Beams

44 Bending Moments in Beams

45 Sense of Bending in Beams

46 Cantilever Beams

47 Tabulated Values for Beam Behavior

5 Continuous and Restrained Beams

51 Bending Moments for Continuous Beams

52 Restrained Beams

53 Beams with Internal Pins

54 Approximate Analysis of Continuous Beams

6 Retaining Walls

61 Horizontal Earth Pressure

62 Stability of Retaining Walls

63 Vertical Soil Pressure

7 Rigid Frames

71 Cantilever Frames

72 Single-Span Frames

8 Noncoplanar Force Systems

81 Concurrent Systems

82 Parallel Systems

83 General Noncoplanar Systems

9 Properties of Sections

91 Centroids

92 Moment of Inertia

93 Transferring Moments of Inertia

94 Miscellaneous Properties

95 Tables of Properties of Sections

10 Stress and Deformation

101 Mechanical Properties of Materials

102 Design Use of Direct Stress

103 Deformation and Stress Relations and Issues

104 Inelastic and Nonlinear Behavior

11 Stress and Strain in Beams

111 Development of Bending Resistance

112 Investigation of Beams

113 Computation of Safe Loads

114 Design of Beams for Flexure

115 Shear Stress in Beams

116 Shear in Steel Beams

117 Flitched Beams

118 Deflection of Beams

119 Deflection Computations

1110 Plastic Behavior in Steel Beams

12 Compression Members

121 Slenderness Effects

122 Wood Columns

123 Steel Columns

13 Combined Forces and Stresses

131 Combined Action Tension Plus Bending

132 Combined Action Compression Plus Bending

133 Development of Shear Stress

134 Stress on an Oblique Section

135 Combined Direct and Shear Stresses

14 Connections for Structural Steel

141 Bolted Connections

142 Design of a Bolted Connection

143 Welded Connections

15 Reinforced Concrete Beams

151 General Considerations

152 Flexure Stress Method

153 General Application of Strength Methods

154 Flexure Strength Method

155 T-Beams

156 Shear in Concrete Beams

157 Design for Shear in Concrete Beams

References

Answers to Selected Exercise Problems

Index

3751 P- FM 111301 1214 PM Page xii



3751 P- FM 111301 1214 PM Page i

















3751 P- FM 111301 1214 PM Page ii


Sixth Edition

JAMES AMBROSE







3751 P- FM 111301 1214 PM Page iii






fcopyebkqxd 11702 943 AM Page 1

v

CONTENTS






Symbols 7

Nomenclature 7




14 Reactions 24



3751 P- FM 111301 1214 PM Page v








24 Vectors 73


26 Motion 76




210 Friction 91

211 Moments 97














vi CONTENTS

3751 P- FM 111301 1214 PM Page vi








CONTENTS vii

3751 P- FM 111301 1214 PM Page vii




















155 T-Beams 382



References 402


Index 409

viii CONTENTS

3751 P- FM 111301 1214 PM Page viii

ix




3751 P- FM 111301 1214 PM Page ix








3751 P- FM 111301 1214 PM Page x




JAMES AMBROSE

2002


3751 P- FM 111301 1214 PM Page xi

3751 P- FM 111301 1214 PM Page xii

xiii





3751 P- FM 111301 1214 PM Page xiii




HARRY PARKER



3751 P- FM 111301 1214 PM Page xiv

1

INTRODUCTION



3751 P-00 (intro) 111301 1217 PM Page 1








2 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM








3751 P-00 (intro) 111301 1217 PM Page 3

4 INTRODUCTION





cross sections


sections









3751 P-00 (intro) 111301 1217 PM Page 4






cross sections


cross sections









3751 P-00 (intro) 111301 1217 PM Page 5

6 INTRODUCTION




254 in mm 00393703048 ft m 3281

6452 in2 mm2 1550 times 10-3

1639 times 103 in3 mm3 6102 times 10-6

4162 times 103 in4 mm4 2403 times 10-6









3751 P-00 (intro) 111301 1217 PM Page 6


SYMBOLS


Symbol Reading


NOMENCLATURE




An = net area


NOMENCLATURE 7

3751 P-00 (intro) 111301 1217 PM Page 7





M = bending moment


S = section modulus

T = tension force


a = unit area







8 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 8

9

1STRUCTURES PURPOSE

AND FUNCTION



3751 P-01 111301 1217 PM Page 9



3751 P-01 111301 1217 PM Page 10







11 LOADS


Gravity




Wind

Source Moving air


LOADS 11

3751 P-01 111301 1217 PM Page 11






Blast




Hydraulic Pressure




Thermal Change




3751 P-01 111301 1217 PM Page 12


Shrinkage


Vibration


Internal Actions


Handling





3751 P-01 111301 1217 PM Page 13

Live and Dead Loads







3751 P-01 111301 1217 PM Page 14





3751 P-01 111301 1217 PM Page 15

Load Dispersion





3751 P-01 111301 1217 PM Page 16

Wind






3751 P-01 111301 1217 PM Page 17


p = 0003V 2

in which






3751 P-01 111301 1217 PM Page 18


Earthquakes




3751 P-01 111301 1217 PM Page 19










Load Combinations



3751 P-01 111301 1217 PM Page 20






3751 P-01 111301 1217 PM Page 21





3751 P-01 111301 1218 PM Page 22







3751 P-01 111301 1218 PM Page 23

14 REACTIONS




3751 P-01 111301 1218 PM Page 24





REACTIONS 25


3751 P-01 111301 1218 PM Page 25




3751 P-01 111301 1218 PM Page 26



REACTIONS 27


3751 P-01 111301 1218 PM Page 27




15 INTERNAL FORCES






3751 P-01 111301 1218 PM Page 28




INTERNAL FORCES 29


3751 P-01 111301 1218 PM Page 29



Stability




Lr

in which




3751 P-01 111301 1218 PM Page 30


In this formula



rI

A=

1 2



3751 P-01 111301 1218 PM Page 31






3751 P-01 111301 1218 PM Page 32





3751 P-01 111301 1218 PM Page 33


Strength





3751 P-01 111301 1218 PM Page 34








3751 P-01 111301 1218 PM Page 35

Stiffness








3751 P-01 111301 1218 PM Page 36






3751 P-01 111301 1218 PM Page 37




3751 P-01 111301 1218 PM Page 38




Tension


Compression




(a)

(b)

3751 P-01 111301 1218 PM Page 39


Shear





3751 P-01 111301 1218 PM Page 40

Bending





3751 P-01 111301 1218 PM Page 41






3751 P-01 111301 1218 PM Page 42





3751 P-01 111301 1218 PM Page 43




3751 P-01 111301 1218 PM Page 44

Torsion








3751 P-01 111301 1218 PM Page 45









3751 P-01 111301 1218 PM Page 46






3751 P-01 111301 1218 PM Page 47






3751 P-01 111301 1218 PM Page 48

Stress Combinations






3751 P-01 111301 1218 PM Page 49





3751 P-01 111301 1218 PM Page 50







3751 P-01 111301 1218 PM Page 51




3751 P-01 111301 1218 PM Page 52

Thermal Stress





3751 P-01 111301 1218 PM Page 53




3751 P-01 111301 1218 PM Page 54






3751 P-01 111301 1218 PM Page 55







3751 P-01 111301 1218 PM Page 56




3751 P-01 111301 1218 PM Page 57







3751 P-01 111301 1218 PM Page 58








3751 P-01 111301 1218 PM Page 59







3751 P-01 111301 1218 PM Page 60

19 DYNAMIC EFFECTS




DYNAMIC EFFECTS 61


3751 P-01 111301 1218 PM Page 61





3751 P-01 111301 1218 PM Page 62





DYNAMIC EFFECTS 63


3751 P-01 111301 1218 PM Page 63













3751 P-01 111301 1218 PM Page 64







3751 P-01 111301 1218 PM Page 65




3751 P-01 111301 1218 PM Page 66





3751 P-01 111301 1218 PM Page 67



se

P Qt Rt

=

+[ ]1

sin( )



3751 P-01 111301 1218 PM Page 68

69

2FORCES AND

FORCE ACTIONS




3751 P-02 111301 1219 PM Page 69





3751 P-02 111301 1219 PM Page 70








3751 P-02 111301 1219 PM Page 71





3751 P-02 111301 1219 PM Page 72

23 TYPES OF FORCES




24 VECTORS


VECTORS 73

3751 P-02 111301 1219 PM Page 73











3751 P-02 111301 1219 PM Page 74




3751 P-02 111301 1219 PM Page 75






26 MOTION




Qualifications




3751 P-02 111301 1219 PM Page 76



Static Equilibrium



MOTION 77


3751 P-02 111301 1219 PM Page 77





Resultant of Forces




3751 P-02 111301 1219 PM Page 78







3751 P-02 111301 1219 PM Page 79








3751 P-02 111301 1219 PM Page 80



Combined Resultants







3751 P-02 111301 1219 PM Page 81


Equilibrant





3751 P-02 111301 1219 PM Page 82




Force Polygon




3751 P-02 111301 1219 PM Page 83







3751 P-02 111301 1219 PM Page 84

Bowrsquos Notation





3751 P-02 111301 1219 PM Page 85







3751 P-02 111301 1219 PM Page 86







3751 P-02 111301 1219 PM Page 87







3751 P-02 111301 1219 PM Page 88


Algebraic Solution





ΣFH = 0 = CA + BCH

ΣFV = 0 = AB + BCV





3751 P-02 111301 1219 PM Page 89



from which

BCV = +200 or 200 lb up





Two-Force Members



BCBCV=

deg= =

sin lb

30

200

0 5400


3751 P-02 111301 1219 PM Page 90

210 FRICTION





FRICTION 91


3751 P-02 111301 1219 PM Page 91




F cent = F cos Q




F cent = mN






3751 P-02 111301 1219 PM Page 92






micro φφ

φ= prime = =F

N

W

W

sin

cos tan

FRICTION 93





3751 P-02 111301 1219 PM Page 93








P = Fcent = mN = 030(100) = 30 lb




3751 P-02 111301 1219 PM Page 94




FRICTION 95



3751 P-02 111301 1219 PM Page 95


Solution For (a)





F cent = 1732(040) = 693 lb


For (c)


or

040(20 cos f) = (20 sin f) ndash 15






3751 P-02 111301 1219 PM Page 96

211 MOMENTS




MOMENTS 97



3751 P-02 111301 1219 PM Page 97



Increasing Moments





3751 P-02 111301 1219 PM Page 98






MOMENTS 99


3751 P-02 111301 1219 PM Page 99








3751 P-02 111301 1219 PM Page 100





MOMENTS 101



3751 P-02 111301 1219 PM Page 101










3751 P-02 111301 1219 PM Page 102










3751 P-02 111301 1219 PM Page 103














3751 P-02 111301 1219 PM Page 104











3751 P-02 111301 1219 PM Page 105






ΣF = 0 = +450 ndash1800 +1350




1213502 2( ) ( )

thus lb


124501 1( ) ( ) thus lb



3751 P-02 111301 1219 PM Page 106



Thus






R2400 3 1000 5 600 11

15

12 800


lb

R14800 10 000 2400

15

17 200

151146 7= + + = =

lb



3751 P-02 111301 1219 PM Page 107



from which



Thus



R230 400

181688 9= =

lb

R116 400

18911 1= =

lb



3751 P-02 111301 1219 PM Page 108






2018601 1( ) ( ) ( )

lb



3751 P-02 111301 1219 PM Page 109



3751 P-02 111301 1219 PM Page 110

111





3751 P-03 111301 1221 PM Page 111





3751 P-03 111301 1221 PM Page 112






3751 P-03 111301 1221 PM Page 113





3751 P-03 111301 1221 PM Page 114




3751 P-03 111301 1221 PM Page 115







3751 P-03 111301 1221 PM Page 116






3751 P-03 111301 1221 PM Page 117



3751 P-03 111301 1221 PM Page 118






3751 P-03 111301 1221 PM Page 119






3751 P-03 111301 1221 PM Page 120





3751 P-03 111301 1221 PM Page 121






from which

and

BI = =1 000

0 5552000 3606

( ) lb

BIh = =0 832

0 5552000 3000

( ) lb

BI BI BIv h

1 000 0 555 0 832 = =


3751 P-03 111301 1221 PM Page 122












3751 P-03 111301 1221 PM Page 123





or

0 = + 1000 ndash 0555CK ndash 0555KJ (321)



or

0 = + 3000 ndash 0832CK + 0832KJ (322)



00 832

0 5551000

0 832

0 5550 555

0 832

0 5550 555= + + minus + minus

( )

( )

( )CK KJ


3751 P-03 111301 1221 PM Page 124



3751 P-03 111301 1221 PM Page 125

or

0 = + 1500 ndash 0832CK ndash 0832KJ



0 = + 1000 ndash 0555(2704) ndash 0555(KJ)

and





KJ = = minus500

0 555901

lb

0 4500 1 6644500

1 6642704= + minus = =

CK CK lb


3751 P-03 111301 1221 PM Page 126






3751 P-03 111301 1221 PM Page 127





3751 P-03 111301 1221 PM Page 128





3751 P-03 111301 1221 PM Page 129




or




Thus


DO = =54 000

105400

lb

10 72 000 12 000 12 000 48 000

48 000

104800

( )

NI

NI

= + minus minus = +

= = lb


3751 P-03 111301 1221 PM Page 130







3751 P-03 111301 1221 PM Page 131

132

4ANALYSIS OF BEAMS


3751 P-04 111301 1221 PM Page 132

41 TYPES OF BEAMS





TYPES OF BEAMS 133


3751 P-04 111301 1221 PM Page 133







3751 P-04 111301 1221 PM Page 134








43 SHEAR IN BEAMS


w = =16 000

20800


SHEAR IN BEAMS 135

3751 P-04 111301 1221 PM Page 135





3751 P-04 111301 1221 PM Page 136

Vertical Shear






SHEAR IN BEAMS 137


3751 P-04 111301 1221 PM Page 137


V(x = 1) = R1 ndash 0 or V(x = 1) = 1000 lb



V(x = 2+) = 1000 ndash 600 = 400 lb


V(x = 6+) = 1000 ndash (600 + 1000) = ndash600 lb




3751 P-04 111301 1221 PM Page 138




Shear Diagrams



SHEAR IN BEAMS 139

3751 P-04 111301 1221 PM Page 139








3751 P-04 111301 1221 PM Page 140

141

Fig

ure

47

Pro

blem

s 4

3A

ndashF

3751 P-04 111301 1221 PM Page 141







3751 P-04 111301 1221 PM Page 142









3751 P-04 111301 1221 PM Page 143






Figure 49 Example 3

3751 P-04 111301 1221 PM Page 144






3751 P-04 111301 1221 PM Page 145





V x xx( ) at ft= minus + = = =7600 800 07600

8009 5



3751 P-04 111301 1221 PM Page 146






M x( ) ( ) ( )


=4 5 10 600 4 5 7000 0 5 800 4 5

4 5

236 100



3751 P-04 111301 1221 PM Page 147





3200 ndash 600x = 0 x = 533 ft



3751 P-04 111301 1221 PM Page 148









0 3200 600

2( )

M

M

x

x

( )

( )

( )

( ) ( )

ft-lb

ft-lb

=

=


=


5 33

12

3200 5 33 600 5 335 33

28533

3200 12 600 12 6 4800


3751 P-04 111301 1221 PM Page 149




3751 P-04 111301 1221 PM Page 150





46 CANTILEVER BEAMS




2600 400400

26000 154x x= = = ft

2400 32003200

24001 33x x= = = ft


3751 P-04 111301 1221 PM Page 151



3751 P-04 111301 1221 PM Page 152







225 000 ft-lb



3751 P-04 111301 1221 PM Page 153






V

M

= + times =


= minus

2000 600 6 5600

2000 14 600 66

238 800

( )

( )

lb

ft-lb



3751 P-04 111301 1221 PM Page 154







3751 P-04 111301 1221 PM Page 155








MwL L

wL L wL WL= + times

minus times times

=

2 2 2 4 8 8

2

or

MPL= = times =4

8000 20

440 000 ft-lb



3751 P-04 111301 1221 PM Page 156






MWL= = times =8

11 200 14

819 600

ft-lb

MwL= = times =

2 2

8

800 14

819 600 ft-lb



3751 P-04 111301 1221 PM Page 157





3751 P-04 111301 1221 PM Page 158






3751 P-04 111301 1221 PM Page 159

160

5CONTINUOUS AND

RESTRAINED BEAMS




3751 P-05 111301 1222 PM Page 160





3751 P-05 111301 1222 PM Page 161







3751 P-05 111301 1222 PM Page 162







1 1 2 1 2 3 21 1

32 2

3

24 4




3751 P-05 111301 1222 PM Page 163









from which

10R1 = 3750 R1 = 375 lb

MwL

2

2 2

8

100 10


ft-lb

MwL

2

2

8= minus

42

2

3

MwL= minus


3751 P-05 111301 1222 PM Page 164




375 ndash (100 times x) = 0 x = 375 ft



Figure 54 Example 1

3751 P-05 111301 1222 PM Page 165




A 16 200B 24 350








2 14 101000 14

4

1000 10

419 500

2

3 3

2

M

M

( )


= minus ft-lb

M = times =375 3 75

2703 125

ft-lb


3751 P-05 111301 1222 PM Page 166





Figure 55 Example 2

3751 P-05 111301 1222 PM Page 167


C 12 16 2000D 16 20 1200












3751 P-05 111301 1222 PM Page 168





E 24 30F 32 24


Figure 57 Example 3

3751 P-05 111301 1222 PM Page 169










3751 P-05 111301 1222 PM Page 170




G 24 1000H 32 1600



3751 P-05 111301 1222 PM Page 171

52 RESTRAINED BEAMS






3751 P-05 111301 1222 PM Page 172





3751 P-05 111301 1222 PM Page 173







3751 P-05 111301 1222 PM Page 174

R1 = V1 = 0375(8000) = 3000 lbR2 = V2 = 0625(8000) = 5000 lb







3751 P-05 111301 1222 PM Page 175




Internal Pins






3751 P-05 111301 1222 PM Page 176




3751 P-05 111301 1222 PM Page 177








3751 P-05 111301 1222 PM Page 178




3751 P-05 111301 1222 PM Page 179




3751 P-05 111301 1222 PM Page 180






3751 P-05 111301 1222 PM Page 181



3751 P-05 111301 1222 PM Page 182

183

6RETAINING WALLS


3751 P-06 111301 1223 PM Page 183



184 RETAINING WALLS


3751 P-06 111301 1223 PM Page 184





3751 P-06 111301 1223 PM Page 185









186 RETAINING WALLS

3751 P-06 111301 1223 PM Page 186


Figure 63 Example 1

3751 P-06 111301 1223 PM Page 187








21 700

99882 17

188 RETAINING WALLS





3751 P-06 111301 1223 PM Page 188






3751 P-06 111301 1223 PM Page 189





Then



pM

S

W e

S= = Σ times = times =2 1168 4 96 12

1 042463

( )

psf

Sbd= = times =

2 2

6

1 2 5

61 042

( ) ft3

pN

A

W

A= = Σ =

times=1168

1 2 5467

psf

eM M

W1

2 1 21 718 9988

116810 04= minus

Σ= minus =

in

190 RETAINING WALLS

3751 P-06 111301 1223 PM Page 190


Figure 65 Example 2

3751 P-06 111301 1223 PM Page 191

192

7RIGID FRAMES


3751 P-07 111301 1223 PM Page 192





3751 P-07 111301 1223 PM Page 193








194 RIGID FRAMES

3751 P-07 111301 1223 PM Page 194







3751 P-07 111301 1223 PM Page 195

196 RIGID FRAMES

Figure 73 Example 1

3751 P-07 111301 1223 PM Page 196










3751 P-07 111301 1223 PM Page 197

198 RIGID FRAMES

Figure 74 Example 2


3751 P-07 111301 1223 PM Page 198







3751 P-07 111301 1223 PM Page 199

200 RIGID FRAMES

Figure 76 Example 3

3751 P-07 111301 1223 PM Page 200


exercise)










3751 P-07 111301 1223 PM Page 201

202

8NONCOPLANAR

FORCE SYSTEMS



3751 P-08 111301 1224 PM Page 202




For its magnitude

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2



3751 P-08 111301 1224 PM Page 203



ΣFx = 0 ΣFy = 0 ΣFz = 0




L

L

12 2

22

5 3 34 5 83

12 34 178 13 34

= + = =

= + = =

( ) ( )

( )


yy

zzF

R

F

R

F

R= Σ =

Σ= Σ



3751 P-08 111301 1224 PM Page 204




R = + + = =( ) ( ) ( ) 2 4 466 1 22 4 217 757 466 72 2 2


Figure 83 Example 1



F1 200(51334) = 75 200(121334) = 180 darr 200(31334) = 45

F2 160(21356) = 236 160(121356) = 1417 darr 160(61356) = 708

F3 180(81497) = 962 180(121497) = 1444 darr 180(41497) = 482____ _____ ____


larrlarr

larrlarrlarr

larr

3751 P-08 111301 1224 PM Page 205


Then





ΣFx = 0 = +1000 ndash 2(Tx) Tx = 500 lb


T

T

T T

x

x

=

= = =

25

12

25

12

25

12500 1041 67( ) ( ) lb

L = + + = =( ) ( ) ( )9 12 20 626 252 2 2

L422 4

466 112 0 578= =

( )

L322 4

466 112 0 062= =

( )

ΣΣ

= =F

F

Lx

y

3

12

2 4

466 1


3751 P-08 111301 1224 PM Page 206



where

Thus

C Ty= = =2 220

12500 1666 67( ) ( ) lb

T

T

T T

y

x

y x

=

= =

20

12

20

12

20

12500( ) ( )



3751 P-08 111301 1224 PM Page 207





T1 = 525 lb T2 = 271 lb T3 = 290 lb




Σ = = + minus

Σ = = + + + minus

Σ = = + + minus

F T T

F T T T

F T T T

x

y

z

04

21

8

21 63

020

21

20

21 63

20

23 321000

05

21

2

21 63

12

23 32

1 2

1 2 3

1 2 3

( )

( )

( )

( )

( )

( )

( )

( )

L

L

L

12 2 2

22 2 2

32 2

5 4 20 441 21

2 8 20 468 21 63

12 20 544 23 32

= + + = =

= + + = =

= + = =

( ) ( ) ( )

( ) ( ) ( )

( ) ( )


3751 P-08 111301 1224 PM Page 208

82 PARALLEL SYSTEMS


R = ΣFy


LM

RL

M

Rx

zz

x= Σ = Σand



3751 P-08 111301 1224 PM Page 209


ΣFy = 0 ΣMx = 0 ΣMz = 0




R = ΣF = 50 + 60 + 160 + 80 = 350 lb





3751 P-08 111301 1224 PM Page 210






L Lx z= = = =800

3502 29

920

3502 63 ft ft



3751 P-08 111301 1224 PM Page 211









T1 = 414 lb T2 = 276 lb T3 = 310 lb



3751 P-08 111301 1224 PM Page 212












yy

zzF

R

F

R

F

R= Σ =

Σ= Σ

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2


3751 P-08 111301 1224 PM Page 213

214

9PROPERTIES

OF SECTIONS


3751 P-09 111301 1224 PM Page 214

91 CENTROIDS



CENTROIDS 215


3751 P-09 111301 1224 PM Page 215







3751 P-09 111301 1224 PM Page 216




CENTROIDS 217

Figure 93 Example 1

3751 P-09 111301 1224 PM Page 217











yx = 52080 = 65 in

3751 P-09 111301 1224 PM Page 218





3751 P-09 111301 1224 PM Page 219









Idb

Y-Y =3

12

Ibd

X-X =3

12


3751 P-09 111301 1224 PM Page 220







Ibd=

3

36

Id= = times =π 4 4

4

64

3 1416 10

64490 9

in

Id= π 4

64

Ibd= = ( )( ) =

3 34

12

5 5 11 5

12697 1

in



3751 P-09 111301 1224 PM Page 221








I = 500 ndash 1707 = 3293 in4



I = times =4 8

12170 7

34 in

Ibd= = times =

3 34

12

6 10

12500 in

Ibd= = times =

3 34

36

12 10

36333 3 in


3751 P-09 111301 1224 PM Page 222







12

7 8

12667 299 368

3 34 in

I d do i=

( ) minus ( )[ ]

=

minus( ) = minus =

π64

3 1416

6410 8 491 201 290

4 4

4 4 4 in



3751 P-09 111301 1224 PM Page 223


I = Io + Az2

In this formula











3751 P-09 111301 1224 PM Page 224





Figure 98 Example 8




1 20 45 10(2)312 = 67 20(45)2 = 405 41172 60 15 6(10)312 = 500 60(15)2 = 135 6357


3751 P-09 111301 1224 PM Page 225





Io + Az2 = 01667 + (8)(625)2 = 3127 in4


9494 in4

Ibd

o = = times =3 3

4

12

16 0 5

120 1667

in


Figure 99 Example 9

3751 P-09 111301 1224 PM Page 226





3751 P-09 111301 1224 PM Page 227


Section Modulus






3751 P-09 111301 1224 PM Page 228



Radius of Gyration






rI

A=

SI

c= = =697 068

5 75121 229


3751 P-09 111301 1224 PM Page 229



3751 P-09 111301 1224 PM Page 230







3751 P-09 111301 1224 PM Page 231








W 27 times 094 277 2692 0490 09990 0745 1437 3270 2430 1090 1240 2480 212 2780times 084 248 2671 0460 09960 0640 1375 2850 2130 1070 1060 2120 207 2440






3751 P-09 111301 1224 PM Page 232










3751 P-09 111301 1224 PM Page 233




b











3751 P-09 111301 1224 PM Page 234





















3751 P-09 111301 1224 PM Page 235





1frasl2 0840 0622 0109 085 0250 0017 0041 02613frasl4 1050 0824 0113 113 0333 0037 0071 0334


10 10750 10020 0365 4048 1190 161 29900 367012 12750 12000 0375 4956 1460 279 43800 4380



reference document

3751 P-09 111301 1224 PM Page 236







3751 P-09 111301 1224 PM Page 237








3751 P-09 111301 1224 PM Page 238

239

10STRESS AND

DEFORMATION



3751 P-10 111301 1225 PM Page 239



3751 P-10 111301 1225 PM Page 240





in which





Deformation


P f A fP

AA

P

f= times = =or or


3751 P-10 111301 1225 PM Page 241


Strength




3751 P-10 111301 1225 PM Page 242







Common Values





3751 P-10 111301 1225 PM Page 243







P = ( f )(A) = (900)(576) = 528400 lb [206 kN]


AP

f= = = [ ]30 000

115026 09 16 8292 2

in mm


3751 P-10 111301 1225 PM Page 244










fP

A= = = [ ]11 200

0 60118 636 128 397

psi kPa


3751 P-10 111301 1225 PM Page 245




Hookersquos Law






3751 P-10 111301 1225 PM Page 246



Ultimate Strength



Factor of Safety



3751 P-10 111301 1225 PM Page 247








strain = =se

L


3751 P-10 111301 1225 PM Page 248

in which







in which


P = force in pounds




ePL

AE=

Ef

s

P A

e L

PL

Ae= = =

Ef

s=


3751 P-10 111301 1225 PM Page 249









ePL

AE= = times

times= [ ]60 000 120

3 1416 29 000 0000 079 2 0

in mm

fP

A= = = [ ]60

3 141619 1 131 663

ksi kPa


3751 P-10 111301 1225 PM Page 250






3751 P-10 111301 1225 PM Page 251






3751 P-10 111301 1225 PM Page 252




3751 P-10 111301 1225 PM Page 253

254

11STRESS AND

STRAIN IN BEAMS



3751 P-11 111301 1225 PM Page 254






3751 P-11 111301 1225 PM Page 255






3751 P-11 111301 1225 PM Page 256


The Flexure Formula





3751 P-11 111301 1225 PM Page 257




Therefore



M = fS

Mf

cI M

fI

cR R= times =or

Mf


sum times times

f

ca z2

f

cz a z

f



3751 P-11 111301 1225 PM Page 258









30 13

848 8 48 8 12 585 6 kip-ft or kip-in



3751 P-11 111301 1225 PM Page 259







MR = fS = 24 times 366 = 8784 kip-in





SM

f= = =585 6

2424 4 3

in

fM

S= = =585 6

36 616 0

ksi


3751 P-11 111301 1225 PM Page 260







M = 70767 ndash 735 = 70032 ft-lb



MPL

PM

L= = = times =

4

4 4 70 032

1420 009

lb

MwL= = times =

2 2

8

30 14

8735 ft-lb

M fSR = = times =

=

22 000 38 6 849 200

849 200

1270 767

in-lb

or ft-lb


3751 P-11 111301 1225 PM Page 261



MR = fS = 24 times 519 = 12456 kip-in








21 1245 61245 6

2159 3W W= = =

kips

MWL W

W= = times times =8

14 12

821 kip-in


3751 P-11 111301 1225 PM Page 262









SM

f= = =1188

2449 5 3 in


36 22

899 99 12 1188 kip-ft or kip-in


3751 P-11 111301 1225 PM Page 263










SM

f= = =156 000

160097 5 3

in

MWL= = times =8

6500 16

813 000 156 000 lb-ft or lb-in


3751 P-11 111301 1225 PM Page 264




fVQ

Ibv =



3751 P-11 111301 1225 PM Page 265

in which










3751 P-11 111301 1225 PM Page 266



Q = acenty = (4 times 4)(2) = 32 in3

and



fVQ

Ibv = = times

times=4000 32

170 7 4187 5

psi

Ibd= = times =

3 34

12

4 8

12170 7 in



3751 P-11 111301 1225 PM Page 267





Q = (2 times 10)(45) = 90 in3

fVQ

Ibv = = times

times=8000 126 75

1046 7 6161 5

psi


=6 5 6

6 5

2126 75 3

in



3751 P-11 111301 1225 PM Page 268




Thus




fVQ

Ib

V bd

bd b

V

bdv = =

times ( )( ) times

=

2

3

8

121 5

Q bd d bd= times

=

2 4 8

2

f

f

v

v

= timestimes

=

= timestimes

=

8000 90

1046 7 6114 6

8000 90

1046 7 1068 8

psi (in the web)

psi (in the flange)


3751 P-11 111301 1225 PM Page 269






3751 P-11 111301 1225 PM Page 270




in which






fV

t dv

w b

=


3751 P-11 111301 1225 PM Page 271







117 FLITCHED BEAMS






fV

t dv

w b

= =times

=30

0 435 9 737 09

ksi


3751 P-11 111301 1225 PM Page 272








f fE

E1 2

1

2

= times

s sf

E

f

E1 2

1

1

2

2

= =or

sf

Es

f

E1

1

12

2

2

= =and

Ef

sE

f

s1

1

12

2

2

= =and

FLITCHED BEAMS 273


3751 P-11 111301 1225 PM Page 273










Then



M = Mw + Ms = 7589 + 19342 = 26931 ft-lb

Sbd

s = = times ( ) =2 2

3

6

0 5 11 25

610 55

in

f fE

Ew s

w

s

= times

= ( ) times

=22 0001 900 000

29 000 0001441

psi


3751 P-11 111301 1225 PM Page 274









MWL W

W

= = = ( )

= times =

26 9318

14

88 26 931

1415 389

lb


3751 P-11 111301 1225 PM Page 275





in which




D CWL

EI=

3


3751 P-11 111301 1225 PM Page 276














3751 P-11 111301 1225 PM Page 277







Then

DWL

EI

WLd

I

L

Ed

fL

Ed

fL

Ed

=

times

=

=

=

5

384

16

5

24

5

24

5

24

3

2

2

2

( )

fMc

I

WL d

I

WLd

I= = ( )( ) = 8 2

16

DWL

EI=

5

384

3

fMc

I=

MWL=8


3751 P-11 111301 1225 PM Page 278









fM

S= = times =97 5 12

48 624 07

ksi

MWL= = times =8

39 20

897 5 kip-ft

DL

d= 0 00017179 2

DfL

Ed

L

d

L d

=

=

times

times ( )

=

5

24

5

24

24

29 000

12

0 02483

2

2

2


3751 P-11 111301 1225 PM Page 279








DL

d=

times

= ( ) timestimes( )

= [ ]

20 48

24

0 02483

0 85330 02483 19

12 22

0 626 16

2

2

in mm

fM

S= = times =57 12

33 420 48

ksi

MWL= = times =8

24 19

857 kip-ft

DWL

EI= = ( ) times( )

times times=5

384

5 39 20 12

384 29 000 3400 712

3 3

in

DL

d= = times = [ ]0 2483 0 02483 20

13 980 7104 18 05

2 2

in mm


3751 P-11 111301 1225 PM Page 280







3751 P-11 111301 1225 PM Page 281







3751 P-11 111301 1225 PM Page 282





3751 P-11 111301 1225 PM Page 283

My = Fy times S







3751 P-11 111301 1225 PM Page 284







fM

S= = times = [ ]72 12

38 622 4 154

ksi MPa

MPL= = times = [ ]4

18 16

472 kip-ft 98 kN-m



3751 P-11 111301 1225 PM Page 285






3751 P-11 111301 1225 PM Page 286


M = Fy times Z








ΣFh = 0

or




3751 P-11 111301 1225 PM Page 287

and thus

Au = Al



or


or

Mp = fy times Z






3751 P-11 111301 1225 PM Page 288



My = Fy times Sx










My = times = =36 111 39963996



3751 P-11 111301 1225 PM Page 289







3751 P-11 111301 1225 PM Page 290










from which

wL L

y = times =333 12 39962 2

(in kip-ft units)

Mw L

yy= =33312

2


3751 P-11 111301 1225 PM Page 291


from which





2196

3996100 55times =

wL L

p = times =387 16 61922 2

(in kip-ft units)

Mw L

pp= =

times387

16

2


3751 P-11 111301 1225 PM Page 292

293





3751 P-12 111301 1226 PM Page 293





3751 P-12 111301 1226 PM Page 294



PEI

L= π 2

2



3751 P-12 111301 1226 PM Page 295





3751 P-12 111301 1226 PM Page 296

122 WOOD COLUMNS




WOOD COLUMNS 297


3751 P-12 111301 1226 PM Page 297






P = (Fc)(Cp)(A)

in which







3751 P-12 111301 1226 PM Page 298


in which




in which







FK E

L dcE

cE

e

= ( )( )( ) 2

CF F

c

F F

c

F F

cp

cE c cE cE c=+ ( )

minus+( )

minus1

2

1

2

2

WOOD COLUMNS 299

3751 P-12 111301 1226 PM Page 299





P = (Fc)(Cp)(A) = (1000)(0993)(55)2 = 30038 lb


1576 Cp = 0821 and thus

P = (1000)(0821)(55)2 = 24835 lb


Cp = 0355 and thus

P = (1000)(0355)(55)2 = 10736 lb



FK E

L d

F

F

C

cEcE

e

cE

c

p

= ( )( )( )

= ( )( )( )

=

= =

= + minus +

minus =

2 2

2

0 3 1 600 000

4 3625 250

25 250

100025 25

1 25 25

1 6

1 25 25

1 6

25 25

0 80 993

psi


3751 P-12 111301 1226 PM Page 300







123 STEEL COLUMNS


FK E

L d

F

F

C

P F C A

cEcE

e

cE

c

p

c p

= ( )( )( )

= ( )( )( )

=

= =

= minus

minus =

= ( )( )( ) = ( )( )

2 2

2

0 3 1 400 000

29 14495

495

892 50 555

1 555

1 6

1 555

1 6

0 555

0 80 471

892 5 0 471 1 5

psi


L

d= ( ) =8 5 12

3 529 14

STEEL COLUMNS 301

3751 P-12 111301 1226 PM Page 301

Column Shapes









3751 P-12 111301 1226 PM Page 302

STEEL COLUMNS 303



3751 P-12 111301 1226 PM Page 303






1 2156 26 2022 51 1826 76 1579 101 1285 126 941 151 655 176 482

2 2152 27 2015 52 1817 77 1569 102 1272 127 926 152 646 177 477

3 2148 28 2008 53 1808 78 1558 103 1259 128 911 153 638 178 471

4 2144 29 2001 54 1799 79 1547 104 1247 129 897 154 630 179 466

5 2139 30 1994 55 1790 80 1536 105 1233 130 884 155 622 180 461

6 2135 31 1987 56 1781 81 1524 106 1220 131 870 156 614 181 456

7 2130 32 1980 57 1771 82 1513 107 1207 132 857 157 606 182 451

8 2125 33 1973 58 1762 83 1502 108 1194 133 844 158 598 183 446

9 2121 34 1965 59 1753 84 1490 109 1181 134 832 159 591 184 441

10 2116 35 1958 60 1743 85 1479 110 1167 135 819 160 583 185 436

11 2110 36 1950 61 1733 86 1467 111 1154 136 807 161 576 186 432

12 2105 37 1942 62 1724 87 1456 112 1140 137 796 162 569 187 427

13 2100 38 1935 63 1714 88 1444 113 1126 138 784 163 562 188 423

14 2095 39 1927 64 1704 89 1432 114 1113 139 773 164 555 189 418

15 2089 40 1919 65 1694 90 1420 115 1099 140 762 165 549 190 414

16 2083 41 1911 66 1684 91 1409 116 1085 141 751 166 542 191 409

17 2078 42 1903 67 1674 92 1397 117 1071 142 741 167 535 192 405

18 2072 43 1895 68 1664 93 1384 118 1057 143 730 168 529 193 401

19 2066 44 1886 69 1653 94 1372 119 1043 144 720 169 523 194 397

20 2060 45 1878 70 1643 95 1360 120 1028 145 710 170 517 195 393

21 2054 46 1870 71 1633 96 1348 121 1014 146 701 171 511 196 389

22 2048 47 1861 72 1622 97 1335 122 999 147 691 172 505 197 385

23 2041 48 1853 73 1612 98 1323 123 985 148 682 173 499 198 381

24 2035 49 1844 74 1601 99 1310 124 970 149 673 174 493 199 377

25 2028 50 1835 75 1590 100 1298 125 955 150 664 175 488 200 373


3751 P-12 111301 1226 PM Page 304






P = (Fa)(A) = (1569)(156) = 2448 kips [1089 kN]




P = (1724)(156) = 2689 kips [1196 kN]

KL


2 4862

KL


2 4877 4

STEEL COLUMNS 305

3751 P-12 111301 1226 PM Page 305






P = Fa A = (1456)(156) = 2271 kips [1010 kN]



yKL


2 4887 1 87

xKL


5 2368 8


3751 P-12 111301 1226 PM Page 306


STEEL COLUMNS 307


3751 P-12 111301 1226 PM Page 307




3751 P-12 111301 1226 PM Page 308

309

13COMBINED FORCES

AND STRESSES




3751 P-13 111301 1226 PM Page 309





3751 P-13 111301 1226 PM Page 310







fM

Sb = = = [ ]10 000

1 3337500 52 8

psi MPa


2 23 3 3

6

2 2

61 333 20 82 10 in mm

fN

Aa = = = [ ]5

41250 8 8 psi MPa



3751 P-13 111301 1226 PM Page 311








3751 P-13 111301 1226 PM Page 312





3751 P-13 111301 1226 PM Page 313



or






N

A

Nec

Ie

I

Ac= = thus

pN

A

Nec

I= plusmn


3751 P-13 111301 1226 PM Page 314








V = 1frasl2 (wpx)



3751 P-13 111301 1226 PM Page 315



pN

wx= 2



3751 P-13 111301 1226 PM Page 316









pN

A=

=

=2 2

100

365 56 ksf

pN

A

Mc

I= + = + times

= + =

100

64

100 4

341 31 56 1 17 2 73

ksf

A

Ibd

= times =

= = times =

8 8 64

12

8 8

12341 3

2

3 34

ft

ft

eM

N= = =100

1001 ft


3751 P-13 111301 1226 PM Page 317






a e

x a

pN

wx

= minus = minus =

= = =

= timestimes

=

5

22 5 1 1 5

3 3 1 5 4 5

2 2 100

5 4 58 89

( )

ft

ft

= ksf



3751 P-13 111301 1226 PM Page 318









fP

A

P

A

vP

A

P

A

= =

( )

= =

( )

cos

coscos

sin

cossin cos

ΘΘ

Θ

ΘΘ

Θ Θ

2


3751 P-13 111301 1226 PM Page 319







fP A

vP A= =

2 2and

fP

Av= =and 0



3751 P-13 111301 1226 PM Page 320







fP

A

vP

A

=

=

( ) =

=

=

( )( ) =

cos

sin cos

2 21200

120 5 25

1200

120 5 0 866 43 3

Θ

Θ Θ

psi

psi



3751 P-13 111301 1226 PM Page 321








3751 P-13 111301 1226 PM Page 322






3751 P-13 111301 1227 PM Page 323

324

14CONNECTIONS FOR

STEEL STRUCTURES




3751 P-14 111301 1227 PM Page 324





3751 P-14 111301 1227 PM Page 325



R = Fv times A




3751 P-14 111301 1227 PM Page 326







3751 P-14 111301 1227 PM Page 327






3751 P-14 111301 1227 PM Page 328






3751 P-14 111301 1227 PM Page 329





3751 P-14 111301 1227 PM Page 330








3751 P-14 111301 1227 PM Page 331










A307 S 44 60 79 99 123D 88 120 157 199 245T 88 120 157 199 245

A325 S 75 102 134 169 209D 150 204 267 338 417T 194 265 346 437 540

A490 S 93 126 165 209 258D 186 253 330 417 515T 239 325 424 537 663


3751 P-14 111301 1227 PM Page 332






3751 P-14 111301 1227 PM Page 333











0625 1125 0875 167 18750750 125 10 20 2250875 15b 1125 233 26251000 175b 125 267 30


3751 P-14 111301 1227 PM Page 334



Tension Connections






GageDimension 8 7 6 5 4 35 3 25 2

g 45 40 350 300 25 20 175 1375 1125g1 30 25 225 200g2 30 30 250 175


3751 P-14 111301 1227 PM Page 335


Ae = C1An

in which











3751 P-14 111301 1227 PM Page 336












3751 P-14 111301 1227 PM Page 337





w = b + 2(a) = 225 + 2(125) = 475 in [121 mm]


P = = [ ]100

812 5 55 6 kips kN

n say= =100

15 56 45 7



3751 P-14 111301 1227 PM Page 338





w = 6 ndash 2(0875) = 425 in [108 mm]





ft =times( )

= [ ]100

2 0 4375 4 2526 9 185

ksi MPa

t =times

= [ ]4 63

2 60 386 9 8

in mm

A = = [ ]100

21 64 63 29872 2

in mm


3751 P-14 111301 1227 PM Page 339











2P

F tu

2

2

P

F t

D

u

+

fp =times

= [ ]12 5

0 75 0 5033 3 230

ksi MPa

ft =times

= [ ]100

8 25 0 524 24 167

ksi MPa

t = = [ ]4 63

100 463 11 8

in mm

fp =times times

= [ ]12 5

2 0 75 0 437519 05 131

ksi MPa


3751 P-14 111301 1227 PM Page 340

in which









2

20 862 0 375 1 237

P

F t

D

u

+ = + = in

2 2 12 5

58 0 50 862

P

F tu

= timestimes

=

in



3751 P-14 111301 1227 PM Page 341


net w = 3 ndash 0875 = 2125 in [54 mm]


Ft = 050Fu = 29 ksi [200 MPa]



Fv = 030Fu = 174 ksi [120 MPa]







net in mmw = minus

= [ ]2 1 25

0 875

21 625 41 3


3751 P-14 111301 1227 PM Page 342






3751 P-14 111301 1227 PM Page 343









3751 P-14 111301 1227 PM Page 344




3751 P-14 111301 1227 PM Page 345




BD = 1frasl2(12 + 12)12 = 1frasl2(2)12 = 0707






3751 P-14 111301 1227 PM Page 346











in mm in mm


3751 P-14 111301 1227 PM Page 347









3751 P-14 111301 1227 PM Page 348



Fa = 06(Fy) = 06(36) = 216 ksi


T = Fa A = 216(3 times 04375) = 2835 kips






L = =28 35

5 65 06

in



3751 P-14 111301 1227 PM Page 349





T = Ft A = (216)(209) = 451 kips [200 kN]




L = = [ ]45 1

3 712 2 310

in mm


3751 P-14 111301 1227 PM Page 350


and





L = = [ ]23 625

3 76 39 162

in mm

L20 99

3 512 2 3 45 88=

( ) = [ ]

in mm

L12 51

3 512 2 8 75 222=

( ) = [ ]

in mm



3751 P-14 111301 1227 PM Page 351







3751 P-14 111301 1227 PM Page 352

353

15REINFORCED

CONCRETE BEAMS





3751 P-15 111301 1229 PM Page 353



Design Methods




3751 P-15 111301 1229 PM Page 354


The Stress Method






The Strength Method



3751 P-15 111301 1229 PM Page 355











3751 P-15 111301 1229 PM Page 356








E fc c= prime4730

E fc c= prime57 000



3751 P-15 111301 1229 PM Page 357



Cement




3751 P-15 111301 1229 PM Page 358


Reinforcement






3751 P-15 111301 1229 PM Page 359




3751 P-15 111301 1229 PM Page 360









3751 P-15 111301 1229 PM Page 361







3751 P-15 111301 1229 PM Page 362







Nominal Dimensions

Cross-Sectional




3751 P-15 111301 1229 PM Page 363






3751 P-15 111301 1229 PM Page 364


T = As fs


T = As fy








3751 P-15 111301 1229 PM Page 365













(1522)f

M

kjbdc = 2

2



3751 P-15 111301 1229 PM Page 366


M = Tjd = As fs jd


(1523)


(1524)


(1525)

(1526)

(1527)

(1528)

in which




M Rbd= 2

pf k

fc

s

=2

jk= minus

1

3

kf nfs c

=+ ( )

1

1

AM

f jds

s

=

fM

A jds

s

=


3751 P-15 111301 1229 PM Page 367









fs fccent R


20 138 2 1379 113 0337 0888 00076 0135 9283 2068 92 0383 0872 00129 0226 15544 2758 80 0419 0860 00188 0324 22285 3448 71 0444 0852 00250 0426 2937

24 165 2 1379 113 0298 0901 00056 0121 8323 2068 92 0341 0886 00096 0204 14034 2758 80 0375 0875 00141 0295 20285 3448 71 0400 0867 00188 0390 2690

3751 P-15 111301 1229 PM Page 368





3751 P-15 111301 1229 PM Page 369



M = Rbd 2 = 200 kip-ft [271 kN-m]



Therefore

M = 200 times 12 = 0226(bd 2) and bd 2 = 10619



Design for shear



b d

d

b d

d

= = =

=[ ]

= = =

=[ ]

1010 619

1032 6

0 829

1510 619

1526 6

0 677

in in

b = 0254 m m

in in

b = 0381 m m


3751 P-15 111301 1229 PM Page 370






As = pbd = 00129(15 times 26625) = 515 in2 [3312 mm2]


AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 872 26 6255 17 33122 2

in mm



3751 P-15 111301 1229 PM Page 371

No 6 bars 517044 = 1175 or 12 [3312284 = 1166]

No 7 bars 517060 = 862 or 9 [3312387 = 856]

No 8 bars 517079 = 654 or 7 [3312510 = 649]

No 9 bars 517100 = 517 or 6 [3312645 = 513]

No 10 bars 517127 = 407 or 5 [3312819 = 404]

No 11 bars 517156 = 331 or 4 [33121006 = 329]




3751 P-15 111301 1229 PM Page 372





and

pA

bds= =

( )( )=3 98

15 33 50 00792

AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 90 33 53 98 25672 2

in mm

M Rbd

M M

= = ( )( )( ) =

= = = ( )( )( ) =[ ]

2 2

2

0 226 15 33 5 3804

3804

12317 1554 0 380 0 850 427

kip-in

or

kip-ft kN-M


3751 P-15 111301 1229 PM Page 373




but not less than



As = 0005(bd) = 0005(15 times 335) = 251 in2


AF

bdsy

=

( )200

Af

fbds

c

y

= ( )prime3

jk= minus

= minus

=1

31

0 313

30 896




3000 00050 0003334000 00050 0003335000 00053 000354


3751 P-15 111301 1229 PM Page 374








U = 14D + 17L

in which




3751 P-15 111301 1229 PM Page 375














3751 P-15 111301 1229 PM Page 376





(1541)


(1542)


(1543)


(1544)

pA

bdA pbd

apbd f

f b

pdf

f

ss

y

c

y

c

= =

=( )( )

prime=

prime

0 85 0 85

0 850 85

f ba A f aA f

f bc s y

s y

c

prime = =prime

M T da

A f da

t s y= minus

= minus

2 2

M C da

f ba da

c c= minus

= prime minus

2

0 852


3751 P-15 111301 1229 PM Page 377


(1545)



Thus

Mt = Rbd 2 (1546)

where

(1547)



R pfa

dy= minus

12

M A f da

pbd f da

pbd f da

d

bd pfa

d

t s y

y

y

y

= minus

= ( )( ) minus

= ( )( )( ) minus

= ( ) minus

2

2

12

12

2

p f ff

b c yy

= prime

times+( )

0 8587

87


3751 P-15 111301 1229 PM Page 378





U = 14D + 17LMu = 14(MDL) + 17(MLL)

= 14(58) + 17(38) = 1458 kip-ft [1977 kN-m]


MM

tu= = =

times [ ]0 90

145 8

0 90162

220

kip-ft

= 162 12 = 1944 kip-in kN-m





40 276 2 1379 05823 04367 00186 0580 40003 2068 05823 04367 00278 0870 60004 2758 05823 04367 00371 1161 80005 3448 05480 04110 00437 1388 9600

60 414 2 1379 05031 03773 00107 0520 36003 2068 05031 03773 00160 0781 54004 2758 05031 03773 00214 1041 72005 3448 04735 03551 00252 1241 8600

3751 P-15 111301 1229 PM Page 379


As = pbd



M1 = Rbd 2


M1 = 1944 = 1041(10)(d )2

and


As = pbd = 00214(10)(1366) = 292 in2 [1880 mm2]





d =( )

= = [ ]1944

1 041 10186 7 13 66 347

in mm


3751 P-15 111301 1229 PM Page 380



a = 025d = 025(18) = 45 in [114 mm]



and

Thus

a da= ( ) = minus

= [ ]0 202 18 3 63

216 2 400 in in mm

apdf

f

a

d

pf

f

y

c

y

c

=prime

=prime

= ( )( )

=

0 85

0 85

0 0114 60

0 85 40 202

pA

bds= =

( )=2 057

10 180 0114

M T jd A F da

AM

f d a

t s y

st

y

= ( ) = minus

=minus ( )[ ] =

( )= [ ]

2

2

1944

60 15 752 057 13272 2

in mm


3751 P-15 111301 1229 PM Page 381



Moment Due to



A 40 542 20 271 12 305B 80 1085 40 542 15 381C 100 1356 50 678 18 457

155 T-BEAMS




3751 P-15 111301 1229 PM Page 382




T-BEAMS 383


3751 P-15 111301 1229 PM Page 383





where



Af

fb ds

c

yw=

prime( )6

CM

jd

M

d t= =

minus ( )2

fC

b tc

f

=

AM

f jd

M

f d ts

s s

= =minus ( )[ ]2

jd dt= minus

2


3751 P-15 111301 1229 PM Page 384

or

in which






Af

fb ds

c

yf= ( )

prime3

T-BEAMS 385


3751 P-15 111301 1229 PM Page 385

or


or





CM

jd

fC

b tc

f

= = times = [ ]

= =times

= [ ]

200 12

20120 535

120

54 40 556 3 83

kips kN

ksi MPa

AM

f d ts

s


minus ( )[ ] = [ ]

2

200 12

24 22 4 25 00 33642 2 in mm


4 in m

18 12

454 1 37




7 9 540 228 7 553 179 5 500 14

10 4 508 1311 4 628 14

3751 P-15 111301 1229 PM Page 386


045fccent= 045(4) = 18 ksi [124 MPa]



or







Af

fb ds

c

yf= ( ) = ( )( ) = [ ]

prime3 3 4000

60 00054 22 2 56 16502 2

in mm

Af

fb ds

c

yw= ( ) = ( )( ) = [ ]

prime6 6 4000

60 00015 22 2 09 13502 2

in mm


3751 P-15 111301 1229 PM Page 387











3751 P-15 111301 1229 PM Page 388





3751 P-15 111301 1229 PM Page 389










3751 P-15 111301 1229 PM Page 390








3751 P-15 111301 1229 PM Page 391






3751 P-15 111301 1229 PM Page 392








3751 P-15 111301 1229 PM Page 393





vV

bd=



3751 P-15 111301 1229 PM Page 394

in which






vcent = v ndash vc


T = Av fs

v fc c= prime1 1



3751 P-15 111301 1229 PM Page 395


Av fs = bsvcent









72

96139 104 718

( ) = [ ] psi kPa

vV

bd= =

times= [ ]40 000

12 24139 957

psi KPa

sA f

v bv s=prime


3751 P-15 111301 1229 PM Page 396



Abs

fv

y

=

50



3751 P-15 111301 1229 PM Page 397








sA f

v bv s=prime

= timestimes

=0 22 20 000

27 1013 6

in

v =

( ) =60

96139 87 psi

sA f

v bv s=prime

= timestimes

=0 22 20 000

44 128 3

in

Av = times

=5012 12

40 0000 18 2

in


3751 P-15 111301 1229 PM Page 398



sA f

v bv s=prime

= timestimes

=0 22 20 000

29 1015 2

in



3751 P-15 111301 1229 PM Page 399





3751 P-15 111301 1229 PM Page 400







Abs

fv

y

=

= times

=50 5010 10

40 0000 125 2 in (See Example 6)


3751 P-15 111301 1229 PM Page 401

402

REFERENCES







Wiley New York 1988

3751 P-16 (refs) 111301 1230 PM Page 402

403


Chapter 2






28A 1414 lb T

28C 300 lb C

210A 193deg

210C 07925 lb

211A 400 lb

211C 1250 lb


212B R1 = 359375 lb [1598 kN] R2 = 440625 lb [1960 kN]

3751 P-17 (answers) 111301 1234 PM Page 403


212D R1 = 7667 lb [3411 kN] R2 = 9333 lb [4153 kN]

212F R1 = 7143 lb [3179 kN] R2 = 11857 lb [5276 kN]

Chapter 3


32A Same as 31A


Chapter 4











47A M = 32 kip-ft [434 kN-m]

47C M = 90 kip-ft [122 kN-m]

Chapter 5



3751 P-17 (answers) 111301 1234 PM Page 404







Chapter 6

62A SF = 253


Chapter 7




Chapter 8

81A R = 21605 lb x = 0769 ft z = 1181 ft

81C T1 = 508 lb T2 = 197 lb T3 = 450 lb


Chapter 9

91A cy = 26 in [70 mm]

91C cy = 42895 in [10724 mm]

3751 P-17 (answers) 111301 1234 PM Page 405


91E cy = 44375 in [1109 mm] cx = 10625 in [266 mm]

93A I = 53586 in4 [211 times 108 mm4]

93C I = 44733 in4 [17474 times 106 mm4]

93E I = 20533 in4 [ 8021 times 106 mm4]

93G I = 438 in4

93I I = 167245 in4

Chapter 10

102A 1182 in2 [762 mm2]

102C 270 kips [120 kN]


103A 19333 lb [86 kN]

103C 29550000 psi [203 GPa]

Chapter 11


113A 386 kips

113C 205 kips

113E 226 kips


114C W 18 times 35


= 175 and 700 psi

116A 1683 kips

116C 371 kips

117A 6735 kips

119A 080 in [20 mm]

1110A 136

1110C 515

Chapter 12

122A 15720 lb

3751 P-17 (answers) 111301 1234 PM Page 406


123A 235 kips [1046 kN]

123C 274 kips [1219 kN]

Chapter 13





Chapter 14




Chapter 15




155A 576 in2 [371 times 103 mm2]


157C 1 at 6 in 4 at 13 in

3751 P-17 (answers) 111301 1234 PM Page 407

3751 P-17 (answers) 111301 1234 PM Page 408

409

INDEX


forces 89truss 120











3751 P-18 (index) 111301 1236 PM Page 409











410 INDEX

3751 P-18 (index) 111301 1236 PM Page 410
















Dynamics 2






INDEX 411

3751 P-18 (index) 111301 1236 PM Page 411









Hydraulic load 12










412 INDEX

3751 P-18 (index) 111301 1236 PM Page 412













wall 186







INDEX 413

3751 P-18 (index) 111301 1236 PM Page 413









Safety 247Section










414 INDEX

3751 P-18 (index) 111301 1236 PM Page 414



beam 162frame 199

Statics 2Steel












Symbols 7


INDEX 415

3751 P-18 (index) 111301 1236 PM Page 415










Vibration 12






416 INDEX

3751 P-18 (index) 111301 1236 PM Page 416


CONTENTS



Introduction




Symbols

Nomenclature


11 Loads



14 Reactions

15 Internal Forces




19 Dynamic Effects





23 Types of Forces

24 Vectors


26 Motion




210 Friction

211 Moments






4 Analysis of Beams

41 Types of Beams


43 Shear in Beams



46 Cantilever Beams




52 Restrained Beams



6 Retaining Walls




7 Rigid Frames





82 Parallel Systems



91 Centroids

















117 Flitched Beams






122 Wood Columns

123 Steel Columns
















155 T-Beams



References


Index



3751 P- FM 111301 1214 PM Page i

















3751 P- FM 111301 1214 PM Page ii


Sixth Edition

JAMES AMBROSE







3751 P- FM 111301 1214 PM Page iii







v

CONTENTS






Symbols 7

Nomenclature 7




14 Reactions 24



3751 P- FM 111301 1214 PM Page v








24 Vectors 73


26 Motion 76




210 Friction 91

211 Moments 97














vi CONTENTS

3751 P- FM 111301 1214 PM Page vi








CONTENTS vii

3751 P- FM 111301 1214 PM Page vii




















155 T-Beams 382



References 402


Index 409

viii CONTENTS

3751 P- FM 111301 1214 PM Page viii

ix




3751 P- FM 111301 1214 PM Page ix








3751 P- FM 111301 1214 PM Page x




JAMES AMBROSE

2002


3751 P- FM 111301 1214 PM Page xi

3751 P- FM 111301 1214 PM Page xii

xiii





3751 P- FM 111301 1214 PM Page xiii




HARRY PARKER



3751 P- FM 111301 1214 PM Page xiv

1

INTRODUCTION



3751 P-00 (intro) 111301 1217 PM Page 1








2 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 2








3751 P-00 (intro) 111301 1217 PM

4 INTRODUCTION





cross sections


sections









3751 P-00 (intro) 111301 1217 PM Page 4






cross sections


cross sections









3751 P-00 (intro) 111301 1217 PM Page 5

6 INTRODUCTION




254 in mm 00393703048 ft m 3281

6452 in2 mm2 1550 times 10-3

1639 times 103 in3 mm3 6102 times 10-6

4162 times 103 in4 mm4 2403 times 10-6









3751 P-00 (intro) 111301 1217 PM Page 6


SYMBOLS


Symbol Reading


NOMENCLATURE




An = net area


NOMENCLATURE 7

3751 P-00 (intro) 111301 1217 PM Page 7





M = bending moment


S = section modulus

T = tension force


a = unit area







8 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 8

9

1STRUCTURES PURPOSE

AND FUNCTION



3751 P-01 111301 1217 PM Page 9



3751 P-01 111301 1217 PM Page 10







11 LOADS


Gravity




Wind

Source Moving air


LOADS 11

3751 P-01 111301 1217 PM Page 11






Blast




Hydraulic Pressure




Thermal Change




3751 P-01 111301 1217 PM Page 12


Shrinkage


Vibration


Internal Actions


Handling





3751 P-01 111301 1217 PM Page 13

Live and Dead Loads







3751 P-01 111301 1217 PM Page 14





3751 P-01 111301 1217 PM Page 15

Load Dispersion





3751 P-01 111301 1217 PM Page 16

Wind






3751 P-01 111301 1217 PM Page 17


p = 0003V 2

in which






3751 P-01 111301 1217 PM Page 18


Earthquakes




3751 P-01 111301 1217 PM Page 19










Load Combinations



3751 P-01 111301 1217 PM Page 20






3751 P-01 111301 1217 PM Page 21





3751 P-01 111301 1218 PM Page 22







3751 P-01 111301 1218 PM Page 23

14 REACTIONS




3751 P-01 111301 1218 PM Page 24





REACTIONS 25


3751 P-01 111301 1218 PM Page 25




3751 P-01 111301 1218 PM Page 26



REACTIONS 27


3751 P-01 111301 1218 PM Page 27




15 INTERNAL FORCES






3751 P-01 111301 1218 PM Page 28




INTERNAL FORCES 29


3751 P-01 111301 1218 PM Page 29



Stability




Lr

in which




3751 P-01 111301 1218 PM Page 30


In this formula



rI

A=

1 2



3751 P-01 111301 1218 PM Page 31






3751 P-01 111301 1218 PM Page 32





3751 P-01 111301 1218 PM Page 33


Strength





3751 P-01 111301 1218 PM Page 34








3751 P-01 111301 1218 PM Page 35

Stiffness








3751 P-01 111301 1218 PM Page 36






3751 P-01 111301 1218 PM Page 37




3751 P-01 111301 1218 PM Page 38




Tension


Compression




(a)

(b)

3751 P-01 111301 1218 PM Page 39


Shear





3751 P-01 111301 1218 PM Page 40

Bending





3751 P-01 111301 1218 PM Page 41






3751 P-01 111301 1218 PM Page 42





3751 P-01 111301 1218 PM Page 43




3751 P-01 111301 1218 PM Page 44

Torsion








3751 P-01 111301 1218 PM Page 45









3751 P-01 111301 1218 PM Page 46






3751 P-01 111301 1218 PM Page 47






3751 P-01 111301 1218 PM Page 48

Stress Combinations






3751 P-01 111301 1218 PM Page 49





3751 P-01 111301 1218 PM Page 50







3751 P-01 111301 1218 PM Page 51




3751 P-01 111301 1218 PM Page 52

Thermal Stress





3751 P-01 111301 1218 PM Page 53




3751 P-01 111301 1218 PM Page 54






3751 P-01 111301 1218 PM Page 55







3751 P-01 111301 1218 PM Page 56




3751 P-01 111301 1218 PM Page 57







3751 P-01 111301 1218 PM Page 58








3751 P-01 111301 1218 PM Page 59







3751 P-01 111301 1218 PM Page 60

19 DYNAMIC EFFECTS




DYNAMIC EFFECTS 61


3751 P-01 111301 1218 PM Page 61





3751 P-01 111301 1218 PM Page 62





DYNAMIC EFFECTS 63


3751 P-01 111301 1218 PM Page 63













3751 P-01 111301 1218 PM Page 64







3751 P-01 111301 1218 PM Page 65




3751 P-01 111301 1218 PM Page 66





3751 P-01 111301 1218 PM Page 67



se

P Qt Rt

=

+[ ]1

sin( )



3751 P-01 111301 1218 PM Page 68

69

2FORCES AND

FORCE ACTIONS




3751 P-02 111301 1219 PM Page 69





3751 P-02 111301 1219 PM Page 70








3751 P-02 111301 1219 PM Page 71





3751 P-02 111301 1219 PM Page 72

23 TYPES OF FORCES




24 VECTORS


VECTORS 73

3751 P-02 111301 1219 PM Page 73











3751 P-02 111301 1219 PM Page 74




3751 P-02 111301 1219 PM Page 75






26 MOTION




Qualifications




3751 P-02 111301 1219 PM Page 76



Static Equilibrium



MOTION 77


3751 P-02 111301 1219 PM Page 77





Resultant of Forces




3751 P-02 111301 1219 PM Page 78







3751 P-02 111301 1219 PM Page 79








3751 P-02 111301 1219 PM Page 80



Combined Resultants







3751 P-02 111301 1219 PM Page 81


Equilibrant





3751 P-02 111301 1219 PM Page 82




Force Polygon




3751 P-02 111301 1219 PM Page 83







3751 P-02 111301 1219 PM Page 84

Bowrsquos Notation





3751 P-02 111301 1219 PM Page 85







3751 P-02 111301 1219 PM Page 86







3751 P-02 111301 1219 PM Page 87







3751 P-02 111301 1219 PM Page 88


Algebraic Solution





ΣFH = 0 = CA + BCH

ΣFV = 0 = AB + BCV





3751 P-02 111301 1219 PM Page 89



from which

BCV = +200 or 200 lb up





Two-Force Members



BCBCV=

deg= =

sin lb

30

200

0 5400


3751 P-02 111301 1219 PM Page 90

210 FRICTION





FRICTION 91


3751 P-02 111301 1219 PM Page 91




F cent = F cos Q




F cent = mN






3751 P-02 111301 1219 PM Page 92






micro φφ

φ= prime = =F

N

W

W

sin

cos tan

FRICTION 93





3751 P-02 111301 1219 PM Page 93








P = Fcent = mN = 030(100) = 30 lb




3751 P-02 111301 1219 PM Page 94




FRICTION 95



3751 P-02 111301 1219 PM Page 95


Solution For (a)





F cent = 1732(040) = 693 lb


For (c)


or

040(20 cos f) = (20 sin f) ndash 15






3751 P-02 111301 1219 PM Page 96

211 MOMENTS




MOMENTS 97



3751 P-02 111301 1219 PM Page 97



Increasing Moments





3751 P-02 111301 1219 PM Page 98






MOMENTS 99


3751 P-02 111301 1219 PM Page 99








3751 P-02 111301 1219 PM Page 100





MOMENTS 101



3751 P-02 111301 1219 PM Page 101










3751 P-02 111301 1219 PM Page 102










3751 P-02 111301 1219 PM Page 103














3751 P-02 111301 1219 PM Page 104











3751 P-02 111301 1219 PM Page 105






ΣF = 0 = +450 ndash1800 +1350




1213502 2( ) ( )

thus lb


124501 1( ) ( ) thus lb



3751 P-02 111301 1219 PM Page 106



Thus






R2400 3 1000 5 600 11

15

12 800


lb

R14800 10 000 2400

15

17 200

151146 7= + + = =

lb



3751 P-02 111301 1219 PM Page 107



from which



Thus



R230 400

181688 9= =

lb

R116 400

18911 1= =

lb



3751 P-02 111301 1219 PM Page 108






2018601 1( ) ( ) ( )

lb



3751 P-02 111301 1219 PM Page 109



3751 P-02 111301 1219 PM Page 110

111





3751 P-03 111301 1221 PM Page 111





3751 P-03 111301 1221 PM Page 112






3751 P-03 111301 1221 PM Page 113





3751 P-03 111301 1221 PM Page 114




3751 P-03 111301 1221 PM Page 115







3751 P-03 111301 1221 PM Page 116






3751 P-03 111301 1221 PM Page 117



3751 P-03 111301 1221 PM Page 118






3751 P-03 111301 1221 PM Page 119






3751 P-03 111301 1221 PM Page 120





3751 P-03 111301 1221 PM Page 121






from which

and

BI = =1 000

0 5552000 3606

( ) lb

BIh = =0 832

0 5552000 3000

( ) lb

BI BI BIv h

1 000 0 555 0 832 = =


3751 P-03 111301 1221 PM Page 122












3751 P-03 111301 1221 PM Page 123





or

0 = + 1000 ndash 0555CK ndash 0555KJ (321)



or

0 = + 3000 ndash 0832CK + 0832KJ (322)



00 832

0 5551000

0 832

0 5550 555

0 832

0 5550 555= + + minus + minus

( )

( )

( )CK KJ


3751 P-03 111301 1221 PM Page 124



3751 P-03 111301 1221 PM Page 125

or

0 = + 1500 ndash 0832CK ndash 0832KJ



0 = + 1000 ndash 0555(2704) ndash 0555(KJ)

and





KJ = = minus500

0 555901

lb

0 4500 1 6644500

1 6642704= + minus = =

CK CK lb


3751 P-03 111301 1221 PM Page 126






3751 P-03 111301 1221 PM Page 127





3751 P-03 111301 1221 PM Page 128





3751 P-03 111301 1221 PM Page 129




or




Thus


DO = =54 000

105400

lb

10 72 000 12 000 12 000 48 000

48 000

104800

( )

NI

NI

= + minus minus = +

= = lb


3751 P-03 111301 1221 PM Page 130







3751 P-03 111301 1221 PM Page 131

132

4ANALYSIS OF BEAMS


3751 P-04 111301 1221 PM Page 132

41 TYPES OF BEAMS





TYPES OF BEAMS 133


3751 P-04 111301 1221 PM Page 133







3751 P-04 111301 1221 PM Page 134








43 SHEAR IN BEAMS


w = =16 000

20800


SHEAR IN BEAMS 135

3751 P-04 111301 1221 PM Page 135





3751 P-04 111301 1221 PM Page 136

Vertical Shear






SHEAR IN BEAMS 137


3751 P-04 111301 1221 PM Page 137


V(x = 1) = R1 ndash 0 or V(x = 1) = 1000 lb



V(x = 2+) = 1000 ndash 600 = 400 lb


V(x = 6+) = 1000 ndash (600 + 1000) = ndash600 lb




3751 P-04 111301 1221 PM Page 138




Shear Diagrams



SHEAR IN BEAMS 139

3751 P-04 111301 1221 PM Page 139








3751 P-04 111301 1221 PM Page 140

141

Fig

ure

47

Pro

blem

s 4

3A

ndashF

3751 P-04 111301 1221 PM Page 141







3751 P-04 111301 1221 PM Page 142









3751 P-04 111301 1221 PM Page 143






Figure 49 Example 3

3751 P-04 111301 1221 PM Page 144






3751 P-04 111301 1221 PM Page 145





V x xx( ) at ft= minus + = = =7600 800 07600

8009 5



3751 P-04 111301 1221 PM Page 146






M x( ) ( ) ( )


=4 5 10 600 4 5 7000 0 5 800 4 5

4 5

236 100



3751 P-04 111301 1221 PM Page 147





3200 ndash 600x = 0 x = 533 ft



3751 P-04 111301 1221 PM Page 148









0 3200 600

2( )

M

M

x

x

( )

( )

( )

( ) ( )

ft-lb

ft-lb

=

=


=


5 33

12

3200 5 33 600 5 335 33

28533

3200 12 600 12 6 4800


3751 P-04 111301 1221 PM Page 149




3751 P-04 111301 1221 PM Page 150





46 CANTILEVER BEAMS




2600 400400

26000 154x x= = = ft

2400 32003200

24001 33x x= = = ft


3751 P-04 111301 1221 PM Page 151



3751 P-04 111301 1221 PM Page 152







225 000 ft-lb



3751 P-04 111301 1221 PM Page 153






V

M

= + times =


= minus

2000 600 6 5600

2000 14 600 66

238 800

( )

( )

lb

ft-lb



3751 P-04 111301 1221 PM Page 154







3751 P-04 111301 1221 PM Page 155








MwL L

wL L wL WL= + times

minus times times

=

2 2 2 4 8 8

2

or

MPL= = times =4

8000 20

440 000 ft-lb



3751 P-04 111301 1221 PM Page 156






MWL= = times =8

11 200 14

819 600

ft-lb

MwL= = times =

2 2

8

800 14

819 600 ft-lb



3751 P-04 111301 1221 PM Page 157





3751 P-04 111301 1221 PM Page 158






3751 P-04 111301 1221 PM Page 159

160

5CONTINUOUS AND

RESTRAINED BEAMS




3751 P-05 111301 1222 PM Page 160





3751 P-05 111301 1222 PM Page 161







3751 P-05 111301 1222 PM Page 162







1 1 2 1 2 3 21 1

32 2

3

24 4




3751 P-05 111301 1222 PM Page 163









from which

10R1 = 3750 R1 = 375 lb

MwL

2

2 2

8

100 10


ft-lb

MwL

2

2

8= minus

42

2

3

MwL= minus


3751 P-05 111301 1222 PM Page 164




375 ndash (100 times x) = 0 x = 375 ft



Figure 54 Example 1

3751 P-05 111301 1222 PM Page 165




A 16 200B 24 350








2 14 101000 14

4

1000 10

419 500

2

3 3

2

M

M

( )


= minus ft-lb

M = times =375 3 75

2703 125

ft-lb


3751 P-05 111301 1222 PM Page 166





Figure 55 Example 2

3751 P-05 111301 1222 PM Page 167


C 12 16 2000D 16 20 1200












3751 P-05 111301 1222 PM Page 168





E 24 30F 32 24


Figure 57 Example 3

3751 P-05 111301 1222 PM Page 169










3751 P-05 111301 1222 PM Page 170




G 24 1000H 32 1600



3751 P-05 111301 1222 PM Page 171

52 RESTRAINED BEAMS






3751 P-05 111301 1222 PM Page 172





3751 P-05 111301 1222 PM Page 173







3751 P-05 111301 1222 PM Page 174

R1 = V1 = 0375(8000) = 3000 lbR2 = V2 = 0625(8000) = 5000 lb







3751 P-05 111301 1222 PM Page 175




Internal Pins






3751 P-05 111301 1222 PM Page 176




3751 P-05 111301 1222 PM Page 177








3751 P-05 111301 1222 PM Page 178




3751 P-05 111301 1222 PM Page 179




3751 P-05 111301 1222 PM Page 180






3751 P-05 111301 1222 PM Page 181



3751 P-05 111301 1222 PM Page 182

183

6RETAINING WALLS


3751 P-06 111301 1223 PM Page 183



184 RETAINING WALLS


3751 P-06 111301 1223 PM Page 184





3751 P-06 111301 1223 PM Page 185









186 RETAINING WALLS

3751 P-06 111301 1223 PM Page 186


Figure 63 Example 1

3751 P-06 111301 1223 PM Page 187








21 700

99882 17

188 RETAINING WALLS





3751 P-06 111301 1223 PM Page 188






3751 P-06 111301 1223 PM Page 189





Then



pM

S

W e

S= = Σ times = times =2 1168 4 96 12

1 042463

( )

psf

Sbd= = times =

2 2

6

1 2 5

61 042

( ) ft3

pN

A

W

A= = Σ =

times=1168

1 2 5467

psf

eM M

W1

2 1 21 718 9988

116810 04= minus

Σ= minus =

in

190 RETAINING WALLS

3751 P-06 111301 1223 PM Page 190


Figure 65 Example 2

3751 P-06 111301 1223 PM Page 191

192

7RIGID FRAMES


3751 P-07 111301 1223 PM Page 192





3751 P-07 111301 1223 PM Page 193








194 RIGID FRAMES

3751 P-07 111301 1223 PM Page 194







3751 P-07 111301 1223 PM Page 195

196 RIGID FRAMES

Figure 73 Example 1

3751 P-07 111301 1223 PM Page 196










3751 P-07 111301 1223 PM Page 197

198 RIGID FRAMES

Figure 74 Example 2


3751 P-07 111301 1223 PM Page 198







3751 P-07 111301 1223 PM Page 199

200 RIGID FRAMES

Figure 76 Example 3

3751 P-07 111301 1223 PM Page 200


exercise)










3751 P-07 111301 1223 PM Page 201

202

8NONCOPLANAR

FORCE SYSTEMS



3751 P-08 111301 1224 PM Page 202




For its magnitude

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2



3751 P-08 111301 1224 PM Page 203



ΣFx = 0 ΣFy = 0 ΣFz = 0




L

L

12 2

22

5 3 34 5 83

12 34 178 13 34

= + = =

= + = =

( ) ( )

( )


yy

zzF

R

F

R

F

R= Σ =

Σ= Σ



3751 P-08 111301 1224 PM Page 204




R = + + = =( ) ( ) ( ) 2 4 466 1 22 4 217 757 466 72 2 2


Figure 83 Example 1



F1 200(51334) = 75 200(121334) = 180 darr 200(31334) = 45

F2 160(21356) = 236 160(121356) = 1417 darr 160(61356) = 708

F3 180(81497) = 962 180(121497) = 1444 darr 180(41497) = 482____ _____ ____


larrlarr

larrlarrlarr

larr

3751 P-08 111301 1224 PM Page 205


Then





ΣFx = 0 = +1000 ndash 2(Tx) Tx = 500 lb


T

T

T T

x

x

=

= = =

25

12

25

12

25

12500 1041 67( ) ( ) lb

L = + + = =( ) ( ) ( )9 12 20 626 252 2 2

L422 4

466 112 0 578= =

( )

L322 4

466 112 0 062= =

( )

ΣΣ

= =F

F

Lx

y

3

12

2 4

466 1


3751 P-08 111301 1224 PM Page 206



where

Thus

C Ty= = =2 220

12500 1666 67( ) ( ) lb

T

T

T T

y

x

y x

=

= =

20

12

20

12

20

12500( ) ( )



3751 P-08 111301 1224 PM Page 207





T1 = 525 lb T2 = 271 lb T3 = 290 lb




Σ = = + minus

Σ = = + + + minus

Σ = = + + minus

F T T

F T T T

F T T T

x

y

z

04

21

8

21 63

020

21

20

21 63

20

23 321000

05

21

2

21 63

12

23 32

1 2

1 2 3

1 2 3

( )

( )

( )

( )

( )

( )

( )

( )

L

L

L

12 2 2

22 2 2

32 2

5 4 20 441 21

2 8 20 468 21 63

12 20 544 23 32

= + + = =

= + + = =

= + = =

( ) ( ) ( )

( ) ( ) ( )

( ) ( )


3751 P-08 111301 1224 PM Page 208

82 PARALLEL SYSTEMS


R = ΣFy


LM

RL

M

Rx

zz

x= Σ = Σand



3751 P-08 111301 1224 PM Page 209


ΣFy = 0 ΣMx = 0 ΣMz = 0




R = ΣF = 50 + 60 + 160 + 80 = 350 lb





3751 P-08 111301 1224 PM Page 210






L Lx z= = = =800

3502 29

920

3502 63 ft ft



3751 P-08 111301 1224 PM Page 211









T1 = 414 lb T2 = 276 lb T3 = 310 lb



3751 P-08 111301 1224 PM Page 212












yy

zzF

R

F

R

F

R= Σ =

Σ= Σ

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2


3751 P-08 111301 1224 PM Page 213

214

9PROPERTIES

OF SECTIONS


3751 P-09 111301 1224 PM Page 214

91 CENTROIDS



CENTROIDS 215


3751 P-09 111301 1224 PM Page 215







3751 P-09 111301 1224 PM Page 216




CENTROIDS 217

Figure 93 Example 1

3751 P-09 111301 1224 PM Page 217











yx = 52080 = 65 in

3751 P-09 111301 1224 PM Page 218





3751 P-09 111301 1224 PM Page 219









Idb

Y-Y =3

12

Ibd

X-X =3

12


3751 P-09 111301 1224 PM Page 220







Ibd=

3

36

Id= = times =π 4 4

4

64

3 1416 10

64490 9

in

Id= π 4

64

Ibd= = ( )( ) =

3 34

12

5 5 11 5

12697 1

in



3751 P-09 111301 1224 PM Page 221








I = 500 ndash 1707 = 3293 in4



I = times =4 8

12170 7

34 in

Ibd= = times =

3 34

12

6 10

12500 in

Ibd= = times =

3 34

36

12 10

36333 3 in


3751 P-09 111301 1224 PM Page 222







12

7 8

12667 299 368

3 34 in

I d do i=

( ) minus ( )[ ]

=

minus( ) = minus =

π64

3 1416

6410 8 491 201 290

4 4

4 4 4 in



3751 P-09 111301 1224 PM Page 223


I = Io + Az2

In this formula











3751 P-09 111301 1224 PM Page 224





Figure 98 Example 8




1 20 45 10(2)312 = 67 20(45)2 = 405 41172 60 15 6(10)312 = 500 60(15)2 = 135 6357


3751 P-09 111301 1224 PM Page 225





Io + Az2 = 01667 + (8)(625)2 = 3127 in4


9494 in4

Ibd

o = = times =3 3

4

12

16 0 5

120 1667

in


Figure 99 Example 9

3751 P-09 111301 1224 PM Page 226





3751 P-09 111301 1224 PM Page 227


Section Modulus






3751 P-09 111301 1224 PM Page 228



Radius of Gyration






rI

A=

SI

c= = =697 068

5 75121 229


3751 P-09 111301 1224 PM Page 229



3751 P-09 111301 1224 PM Page 230







3751 P-09 111301 1224 PM Page 231








W 27 times 094 277 2692 0490 09990 0745 1437 3270 2430 1090 1240 2480 212 2780times 084 248 2671 0460 09960 0640 1375 2850 2130 1070 1060 2120 207 2440






3751 P-09 111301 1224 PM Page 232










3751 P-09 111301 1224 PM Page 233




b











3751 P-09 111301 1224 PM Page 234





















3751 P-09 111301 1224 PM Page 235





1frasl2 0840 0622 0109 085 0250 0017 0041 02613frasl4 1050 0824 0113 113 0333 0037 0071 0334


10 10750 10020 0365 4048 1190 161 29900 367012 12750 12000 0375 4956 1460 279 43800 4380



reference document

3751 P-09 111301 1224 PM Page 236







3751 P-09 111301 1224 PM Page 237








3751 P-09 111301 1224 PM Page 238

239

10STRESS AND

DEFORMATION



3751 P-10 111301 1225 PM Page 239



3751 P-10 111301 1225 PM Page 240





in which





Deformation


P f A fP

AA

P

f= times = =or or


3751 P-10 111301 1225 PM Page 241


Strength




3751 P-10 111301 1225 PM Page 242







Common Values





3751 P-10 111301 1225 PM Page 243







P = ( f )(A) = (900)(576) = 528400 lb [206 kN]


AP

f= = = [ ]30 000

115026 09 16 8292 2

in mm


3751 P-10 111301 1225 PM Page 244










fP

A= = = [ ]11 200

0 60118 636 128 397

psi kPa


3751 P-10 111301 1225 PM Page 245




Hookersquos Law






3751 P-10 111301 1225 PM Page 246



Ultimate Strength



Factor of Safety



3751 P-10 111301 1225 PM Page 247








strain = =se

L


3751 P-10 111301 1225 PM Page 248

in which







in which


P = force in pounds




ePL

AE=

Ef

s

P A

e L

PL

Ae= = =

Ef

s=


3751 P-10 111301 1225 PM Page 249









ePL

AE= = times

times= [ ]60 000 120

3 1416 29 000 0000 079 2 0

in mm

fP

A= = = [ ]60

3 141619 1 131 663

ksi kPa


3751 P-10 111301 1225 PM Page 250






3751 P-10 111301 1225 PM Page 251






3751 P-10 111301 1225 PM Page 252




3751 P-10 111301 1225 PM Page 253

254

11STRESS AND

STRAIN IN BEAMS



3751 P-11 111301 1225 PM Page 254






3751 P-11 111301 1225 PM Page 255






3751 P-11 111301 1225 PM Page 256


The Flexure Formula





3751 P-11 111301 1225 PM Page 257




Therefore



M = fS

Mf

cI M

fI

cR R= times =or

Mf


sum times times

f

ca z2

f

cz a z

f



3751 P-11 111301 1225 PM Page 258









30 13

848 8 48 8 12 585 6 kip-ft or kip-in



3751 P-11 111301 1225 PM Page 259







MR = fS = 24 times 366 = 8784 kip-in





SM

f= = =585 6

2424 4 3

in

fM

S= = =585 6

36 616 0

ksi


3751 P-11 111301 1225 PM Page 260







M = 70767 ndash 735 = 70032 ft-lb



MPL

PM

L= = = times =

4

4 4 70 032

1420 009

lb

MwL= = times =

2 2

8

30 14

8735 ft-lb

M fSR = = times =

=

22 000 38 6 849 200

849 200

1270 767

in-lb

or ft-lb


3751 P-11 111301 1225 PM Page 261



MR = fS = 24 times 519 = 12456 kip-in








21 1245 61245 6

2159 3W W= = =

kips

MWL W

W= = times times =8

14 12

821 kip-in


3751 P-11 111301 1225 PM Page 262









SM

f= = =1188

2449 5 3 in


36 22

899 99 12 1188 kip-ft or kip-in


3751 P-11 111301 1225 PM Page 263










SM

f= = =156 000

160097 5 3

in

MWL= = times =8

6500 16

813 000 156 000 lb-ft or lb-in


3751 P-11 111301 1225 PM Page 264




fVQ

Ibv =



3751 P-11 111301 1225 PM Page 265

in which










3751 P-11 111301 1225 PM Page 266



Q = acenty = (4 times 4)(2) = 32 in3

and



fVQ

Ibv = = times

times=4000 32

170 7 4187 5

psi

Ibd= = times =

3 34

12

4 8

12170 7 in



3751 P-11 111301 1225 PM Page 267





Q = (2 times 10)(45) = 90 in3

fVQ

Ibv = = times

times=8000 126 75

1046 7 6161 5

psi


=6 5 6

6 5

2126 75 3

in



3751 P-11 111301 1225 PM Page 268




Thus




fVQ

Ib

V bd

bd b

V

bdv = =

times ( )( ) times

=

2

3

8

121 5

Q bd d bd= times

=

2 4 8

2

f

f

v

v

= timestimes

=

= timestimes

=

8000 90

1046 7 6114 6

8000 90

1046 7 1068 8

psi (in the web)

psi (in the flange)


3751 P-11 111301 1225 PM Page 269






3751 P-11 111301 1225 PM Page 270




in which






fV

t dv

w b

=


3751 P-11 111301 1225 PM Page 271







117 FLITCHED BEAMS






fV

t dv

w b

= =times

=30

0 435 9 737 09

ksi


3751 P-11 111301 1225 PM Page 272








f fE

E1 2

1

2

= times

s sf

E

f

E1 2

1

1

2

2

= =or

sf

Es

f

E1

1

12

2

2

= =and

Ef

sE

f

s1

1

12

2

2

= =and

FLITCHED BEAMS 273


3751 P-11 111301 1225 PM Page 273










Then



M = Mw + Ms = 7589 + 19342 = 26931 ft-lb

Sbd

s = = times ( ) =2 2

3

6

0 5 11 25

610 55

in

f fE

Ew s

w

s

= times

= ( ) times

=22 0001 900 000

29 000 0001441

psi


3751 P-11 111301 1225 PM Page 274









MWL W

W

= = = ( )

= times =

26 9318

14

88 26 931

1415 389

lb


3751 P-11 111301 1225 PM Page 275





in which




D CWL

EI=

3


3751 P-11 111301 1225 PM Page 276














3751 P-11 111301 1225 PM Page 277







Then

DWL

EI

WLd

I

L

Ed

fL

Ed

fL

Ed

=

times

=

=

=

5

384

16

5

24

5

24

5

24

3

2

2

2

( )

fMc

I

WL d

I

WLd

I= = ( )( ) = 8 2

16

DWL

EI=

5

384

3

fMc

I=

MWL=8


3751 P-11 111301 1225 PM Page 278









fM

S= = times =97 5 12

48 624 07

ksi

MWL= = times =8

39 20

897 5 kip-ft

DL

d= 0 00017179 2

DfL

Ed

L

d

L d

=

=

times

times ( )

=

5

24

5

24

24

29 000

12

0 02483

2

2

2


3751 P-11 111301 1225 PM Page 279








DL

d=

times

= ( ) timestimes( )

= [ ]

20 48

24

0 02483

0 85330 02483 19

12 22

0 626 16

2

2

in mm

fM

S= = times =57 12

33 420 48

ksi

MWL= = times =8

24 19

857 kip-ft

DWL

EI= = ( ) times( )

times times=5

384

5 39 20 12

384 29 000 3400 712

3 3

in

DL

d= = times = [ ]0 2483 0 02483 20

13 980 7104 18 05

2 2

in mm


3751 P-11 111301 1225 PM Page 280







3751 P-11 111301 1225 PM Page 281







3751 P-11 111301 1225 PM Page 282





3751 P-11 111301 1225 PM Page 283

My = Fy times S







3751 P-11 111301 1225 PM Page 284







fM

S= = times = [ ]72 12

38 622 4 154

ksi MPa

MPL= = times = [ ]4

18 16

472 kip-ft 98 kN-m



3751 P-11 111301 1225 PM Page 285






3751 P-11 111301 1225 PM Page 286


M = Fy times Z








ΣFh = 0

or




3751 P-11 111301 1225 PM Page 287

and thus

Au = Al



or


or

Mp = fy times Z






3751 P-11 111301 1225 PM Page 288



My = Fy times Sx










My = times = =36 111 39963996



3751 P-11 111301 1225 PM Page 289







3751 P-11 111301 1225 PM Page 290










from which

wL L

y = times =333 12 39962 2

(in kip-ft units)

Mw L

yy= =33312

2


3751 P-11 111301 1225 PM Page 291


from which





2196

3996100 55times =

wL L

p = times =387 16 61922 2

(in kip-ft units)

Mw L

pp= =

times387

16

2


3751 P-11 111301 1225 PM Page 292

293





3751 P-12 111301 1226 PM Page 293





3751 P-12 111301 1226 PM Page 294



PEI

L= π 2

2



3751 P-12 111301 1226 PM Page 295





3751 P-12 111301 1226 PM Page 296

122 WOOD COLUMNS




WOOD COLUMNS 297


3751 P-12 111301 1226 PM Page 297






P = (Fc)(Cp)(A)

in which







3751 P-12 111301 1226 PM Page 298


in which




in which







FK E

L dcE

cE

e

= ( )( )( ) 2

CF F

c

F F

c

F F

cp

cE c cE cE c=+ ( )

minus+( )

minus1

2

1

2

2

WOOD COLUMNS 299

3751 P-12 111301 1226 PM Page 299





P = (Fc)(Cp)(A) = (1000)(0993)(55)2 = 30038 lb


1576 Cp = 0821 and thus

P = (1000)(0821)(55)2 = 24835 lb


Cp = 0355 and thus

P = (1000)(0355)(55)2 = 10736 lb



FK E

L d

F

F

C

cEcE

e

cE

c

p

= ( )( )( )

= ( )( )( )

=

= =

= + minus +

minus =

2 2

2

0 3 1 600 000

4 3625 250

25 250

100025 25

1 25 25

1 6

1 25 25

1 6

25 25

0 80 993

psi


3751 P-12 111301 1226 PM Page 300







123 STEEL COLUMNS


FK E

L d

F

F

C

P F C A

cEcE

e

cE

c

p

c p

= ( )( )( )

= ( )( )( )

=

= =

= minus

minus =

= ( )( )( ) = ( )( )

2 2

2

0 3 1 400 000

29 14495

495

892 50 555

1 555

1 6

1 555

1 6

0 555

0 80 471

892 5 0 471 1 5

psi


L

d= ( ) =8 5 12

3 529 14

STEEL COLUMNS 301

3751 P-12 111301 1226 PM Page 301

Column Shapes









3751 P-12 111301 1226 PM Page 302

STEEL COLUMNS 303



3751 P-12 111301 1226 PM Page 303






1 2156 26 2022 51 1826 76 1579 101 1285 126 941 151 655 176 482

2 2152 27 2015 52 1817 77 1569 102 1272 127 926 152 646 177 477

3 2148 28 2008 53 1808 78 1558 103 1259 128 911 153 638 178 471

4 2144 29 2001 54 1799 79 1547 104 1247 129 897 154 630 179 466

5 2139 30 1994 55 1790 80 1536 105 1233 130 884 155 622 180 461

6 2135 31 1987 56 1781 81 1524 106 1220 131 870 156 614 181 456

7 2130 32 1980 57 1771 82 1513 107 1207 132 857 157 606 182 451

8 2125 33 1973 58 1762 83 1502 108 1194 133 844 158 598 183 446

9 2121 34 1965 59 1753 84 1490 109 1181 134 832 159 591 184 441

10 2116 35 1958 60 1743 85 1479 110 1167 135 819 160 583 185 436

11 2110 36 1950 61 1733 86 1467 111 1154 136 807 161 576 186 432

12 2105 37 1942 62 1724 87 1456 112 1140 137 796 162 569 187 427

13 2100 38 1935 63 1714 88 1444 113 1126 138 784 163 562 188 423

14 2095 39 1927 64 1704 89 1432 114 1113 139 773 164 555 189 418

15 2089 40 1919 65 1694 90 1420 115 1099 140 762 165 549 190 414

16 2083 41 1911 66 1684 91 1409 116 1085 141 751 166 542 191 409

17 2078 42 1903 67 1674 92 1397 117 1071 142 741 167 535 192 405

18 2072 43 1895 68 1664 93 1384 118 1057 143 730 168 529 193 401

19 2066 44 1886 69 1653 94 1372 119 1043 144 720 169 523 194 397

20 2060 45 1878 70 1643 95 1360 120 1028 145 710 170 517 195 393

21 2054 46 1870 71 1633 96 1348 121 1014 146 701 171 511 196 389

22 2048 47 1861 72 1622 97 1335 122 999 147 691 172 505 197 385

23 2041 48 1853 73 1612 98 1323 123 985 148 682 173 499 198 381

24 2035 49 1844 74 1601 99 1310 124 970 149 673 174 493 199 377

25 2028 50 1835 75 1590 100 1298 125 955 150 664 175 488 200 373


3751 P-12 111301 1226 PM Page 304






P = (Fa)(A) = (1569)(156) = 2448 kips [1089 kN]




P = (1724)(156) = 2689 kips [1196 kN]

KL


2 4862

KL


2 4877 4

STEEL COLUMNS 305

3751 P-12 111301 1226 PM Page 305






P = Fa A = (1456)(156) = 2271 kips [1010 kN]



yKL


2 4887 1 87

xKL


5 2368 8


3751 P-12 111301 1226 PM Page 306


STEEL COLUMNS 307


3751 P-12 111301 1226 PM Page 307




3751 P-12 111301 1226 PM Page 308

309

13COMBINED FORCES

AND STRESSES




3751 P-13 111301 1226 PM Page 309





3751 P-13 111301 1226 PM Page 310







fM

Sb = = = [ ]10 000

1 3337500 52 8

psi MPa


2 23 3 3

6

2 2

61 333 20 82 10 in mm

fN

Aa = = = [ ]5

41250 8 8 psi MPa



3751 P-13 111301 1226 PM Page 311








3751 P-13 111301 1226 PM Page 312





3751 P-13 111301 1226 PM Page 313



or






N

A

Nec

Ie

I

Ac= = thus

pN

A

Nec

I= plusmn


3751 P-13 111301 1226 PM Page 314








V = 1frasl2 (wpx)



3751 P-13 111301 1226 PM Page 315



pN

wx= 2



3751 P-13 111301 1226 PM Page 316









pN

A=

=

=2 2

100

365 56 ksf

pN

A

Mc

I= + = + times

= + =

100

64

100 4

341 31 56 1 17 2 73

ksf

A

Ibd

= times =

= = times =

8 8 64

12

8 8

12341 3

2

3 34

ft

ft

eM

N= = =100

1001 ft


3751 P-13 111301 1226 PM Page 317






a e

x a

pN

wx

= minus = minus =

= = =

= timestimes

=

5

22 5 1 1 5

3 3 1 5 4 5

2 2 100

5 4 58 89

( )

ft

ft

= ksf



3751 P-13 111301 1226 PM Page 318









fP

A

P

A

vP

A

P

A

= =

( )

= =

( )

cos

coscos

sin

cossin cos

ΘΘ

Θ

ΘΘ

Θ Θ

2


3751 P-13 111301 1226 PM Page 319







fP A

vP A= =

2 2and

fP

Av= =and 0



3751 P-13 111301 1226 PM Page 320







fP

A

vP

A

=

=

( ) =

=

=

( )( ) =

cos

sin cos

2 21200

120 5 25

1200

120 5 0 866 43 3

Θ

Θ Θ

psi

psi



3751 P-13 111301 1226 PM Page 321








3751 P-13 111301 1226 PM Page 322






3751 P-13 111301 1227 PM Page 323

324

14CONNECTIONS FOR

STEEL STRUCTURES




3751 P-14 111301 1227 PM Page 324





3751 P-14 111301 1227 PM Page 325



R = Fv times A




3751 P-14 111301 1227 PM Page 326







3751 P-14 111301 1227 PM Page 327






3751 P-14 111301 1227 PM Page 328






3751 P-14 111301 1227 PM Page 329





3751 P-14 111301 1227 PM Page 330








3751 P-14 111301 1227 PM Page 331










A307 S 44 60 79 99 123D 88 120 157 199 245T 88 120 157 199 245

A325 S 75 102 134 169 209D 150 204 267 338 417T 194 265 346 437 540

A490 S 93 126 165 209 258D 186 253 330 417 515T 239 325 424 537 663


3751 P-14 111301 1227 PM Page 332






3751 P-14 111301 1227 PM Page 333











0625 1125 0875 167 18750750 125 10 20 2250875 15b 1125 233 26251000 175b 125 267 30


3751 P-14 111301 1227 PM Page 334



Tension Connections






GageDimension 8 7 6 5 4 35 3 25 2

g 45 40 350 300 25 20 175 1375 1125g1 30 25 225 200g2 30 30 250 175


3751 P-14 111301 1227 PM Page 335


Ae = C1An

in which











3751 P-14 111301 1227 PM Page 336












3751 P-14 111301 1227 PM Page 337





w = b + 2(a) = 225 + 2(125) = 475 in [121 mm]


P = = [ ]100

812 5 55 6 kips kN

n say= =100

15 56 45 7



3751 P-14 111301 1227 PM Page 338





w = 6 ndash 2(0875) = 425 in [108 mm]





ft =times( )

= [ ]100

2 0 4375 4 2526 9 185

ksi MPa

t =times

= [ ]4 63

2 60 386 9 8

in mm

A = = [ ]100

21 64 63 29872 2

in mm


3751 P-14 111301 1227 PM Page 339











2P

F tu

2

2

P

F t

D

u

+

fp =times

= [ ]12 5

0 75 0 5033 3 230

ksi MPa

ft =times

= [ ]100

8 25 0 524 24 167

ksi MPa

t = = [ ]4 63

100 463 11 8

in mm

fp =times times

= [ ]12 5

2 0 75 0 437519 05 131

ksi MPa


3751 P-14 111301 1227 PM Page 340

in which









2

20 862 0 375 1 237

P

F t

D

u

+ = + = in

2 2 12 5

58 0 50 862

P

F tu

= timestimes

=

in



3751 P-14 111301 1227 PM Page 341


net w = 3 ndash 0875 = 2125 in [54 mm]


Ft = 050Fu = 29 ksi [200 MPa]



Fv = 030Fu = 174 ksi [120 MPa]







net in mmw = minus

= [ ]2 1 25

0 875

21 625 41 3


3751 P-14 111301 1227 PM Page 342






3751 P-14 111301 1227 PM Page 343









3751 P-14 111301 1227 PM Page 344




3751 P-14 111301 1227 PM Page 345




BD = 1frasl2(12 + 12)12 = 1frasl2(2)12 = 0707






3751 P-14 111301 1227 PM Page 346











in mm in mm


3751 P-14 111301 1227 PM Page 347









3751 P-14 111301 1227 PM Page 348



Fa = 06(Fy) = 06(36) = 216 ksi


T = Fa A = 216(3 times 04375) = 2835 kips






L = =28 35

5 65 06

in



3751 P-14 111301 1227 PM Page 349





T = Ft A = (216)(209) = 451 kips [200 kN]




L = = [ ]45 1

3 712 2 310

in mm


3751 P-14 111301 1227 PM Page 350


and





L = = [ ]23 625

3 76 39 162

in mm

L20 99

3 512 2 3 45 88=

( ) = [ ]

in mm

L12 51

3 512 2 8 75 222=

( ) = [ ]

in mm



3751 P-14 111301 1227 PM Page 351







3751 P-14 111301 1227 PM Page 352

353

15REINFORCED

CONCRETE BEAMS





3751 P-15 111301 1229 PM Page 353



Design Methods




3751 P-15 111301 1229 PM Page 354


The Stress Method






The Strength Method



3751 P-15 111301 1229 PM Page 355











3751 P-15 111301 1229 PM Page 356








E fc c= prime4730

E fc c= prime57 000



3751 P-15 111301 1229 PM Page 357



Cement




3751 P-15 111301 1229 PM Page 358


Reinforcement






3751 P-15 111301 1229 PM Page 359




3751 P-15 111301 1229 PM Page 360









3751 P-15 111301 1229 PM Page 361







3751 P-15 111301 1229 PM Page 362







Nominal Dimensions

Cross-Sectional




3751 P-15 111301 1229 PM Page 363






3751 P-15 111301 1229 PM Page 364


T = As fs


T = As fy








3751 P-15 111301 1229 PM Page 365













(1522)f

M

kjbdc = 2

2



3751 P-15 111301 1229 PM Page 366


M = Tjd = As fs jd


(1523)


(1524)


(1525)

(1526)

(1527)

(1528)

in which




M Rbd= 2

pf k

fc

s

=2

jk= minus

1

3

kf nfs c

=+ ( )

1

1

AM

f jds

s

=

fM

A jds

s

=


3751 P-15 111301 1229 PM Page 367









fs fccent R


20 138 2 1379 113 0337 0888 00076 0135 9283 2068 92 0383 0872 00129 0226 15544 2758 80 0419 0860 00188 0324 22285 3448 71 0444 0852 00250 0426 2937

24 165 2 1379 113 0298 0901 00056 0121 8323 2068 92 0341 0886 00096 0204 14034 2758 80 0375 0875 00141 0295 20285 3448 71 0400 0867 00188 0390 2690

3751 P-15 111301 1229 PM Page 368





3751 P-15 111301 1229 PM Page 369



M = Rbd 2 = 200 kip-ft [271 kN-m]



Therefore

M = 200 times 12 = 0226(bd 2) and bd 2 = 10619



Design for shear



b d

d

b d

d

= = =

=[ ]

= = =

=[ ]

1010 619

1032 6

0 829

1510 619

1526 6

0 677

in in

b = 0254 m m

in in

b = 0381 m m


3751 P-15 111301 1229 PM Page 370






As = pbd = 00129(15 times 26625) = 515 in2 [3312 mm2]


AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 872 26 6255 17 33122 2

in mm



3751 P-15 111301 1229 PM Page 371

No 6 bars 517044 = 1175 or 12 [3312284 = 1166]

No 7 bars 517060 = 862 or 9 [3312387 = 856]

No 8 bars 517079 = 654 or 7 [3312510 = 649]

No 9 bars 517100 = 517 or 6 [3312645 = 513]

No 10 bars 517127 = 407 or 5 [3312819 = 404]

No 11 bars 517156 = 331 or 4 [33121006 = 329]




3751 P-15 111301 1229 PM Page 372





and

pA

bds= =

( )( )=3 98

15 33 50 00792

AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 90 33 53 98 25672 2

in mm

M Rbd

M M

= = ( )( )( ) =

= = = ( )( )( ) =[ ]

2 2

2

0 226 15 33 5 3804

3804

12317 1554 0 380 0 850 427

kip-in

or

kip-ft kN-M


3751 P-15 111301 1229 PM Page 373




but not less than



As = 0005(bd) = 0005(15 times 335) = 251 in2


AF

bdsy

=

( )200

Af

fbds

c

y

= ( )prime3

jk= minus

= minus

=1

31

0 313

30 896




3000 00050 0003334000 00050 0003335000 00053 000354


3751 P-15 111301 1229 PM Page 374








U = 14D + 17L

in which




3751 P-15 111301 1229 PM Page 375














3751 P-15 111301 1229 PM Page 376





(1541)


(1542)


(1543)


(1544)

pA

bdA pbd

apbd f

f b

pdf

f

ss

y

c

y

c

= =

=( )( )

prime=

prime

0 85 0 85

0 850 85

f ba A f aA f

f bc s y

s y

c

prime = =prime

M T da

A f da

t s y= minus

= minus

2 2

M C da

f ba da

c c= minus

= prime minus

2

0 852


3751 P-15 111301 1229 PM Page 377


(1545)



Thus

Mt = Rbd 2 (1546)

where

(1547)



R pfa

dy= minus

12

M A f da

pbd f da

pbd f da

d

bd pfa

d

t s y

y

y

y

= minus

= ( )( ) minus

= ( )( )( ) minus

= ( ) minus

2

2

12

12

2

p f ff

b c yy

= prime

times+( )

0 8587

87


3751 P-15 111301 1229 PM Page 378





U = 14D + 17LMu = 14(MDL) + 17(MLL)

= 14(58) + 17(38) = 1458 kip-ft [1977 kN-m]


MM

tu= = =

times [ ]0 90

145 8

0 90162

220

kip-ft

= 162 12 = 1944 kip-in kN-m





40 276 2 1379 05823 04367 00186 0580 40003 2068 05823 04367 00278 0870 60004 2758 05823 04367 00371 1161 80005 3448 05480 04110 00437 1388 9600

60 414 2 1379 05031 03773 00107 0520 36003 2068 05031 03773 00160 0781 54004 2758 05031 03773 00214 1041 72005 3448 04735 03551 00252 1241 8600

3751 P-15 111301 1229 PM Page 379


As = pbd



M1 = Rbd 2


M1 = 1944 = 1041(10)(d )2

and


As = pbd = 00214(10)(1366) = 292 in2 [1880 mm2]





d =( )

= = [ ]1944

1 041 10186 7 13 66 347

in mm


3751 P-15 111301 1229 PM Page 380



a = 025d = 025(18) = 45 in [114 mm]



and

Thus

a da= ( ) = minus

= [ ]0 202 18 3 63

216 2 400 in in mm

apdf

f

a

d

pf

f

y

c

y

c

=prime

=prime

= ( )( )

=

0 85

0 85

0 0114 60

0 85 40 202

pA

bds= =

( )=2 057

10 180 0114

M T jd A F da

AM

f d a

t s y

st

y

= ( ) = minus

=minus ( )[ ] =

( )= [ ]

2

2

1944

60 15 752 057 13272 2

in mm


3751 P-15 111301 1229 PM Page 381



Moment Due to



A 40 542 20 271 12 305B 80 1085 40 542 15 381C 100 1356 50 678 18 457

155 T-BEAMS




3751 P-15 111301 1229 PM Page 382




T-BEAMS 383


3751 P-15 111301 1229 PM Page 383





where



Af

fb ds

c

yw=

prime( )6

CM

jd

M

d t= =

minus ( )2

fC

b tc

f

=

AM

f jd

M

f d ts

s s

= =minus ( )[ ]2

jd dt= minus

2


3751 P-15 111301 1229 PM Page 384

or

in which






Af

fb ds

c

yf= ( )

prime3

T-BEAMS 385


3751 P-15 111301 1229 PM Page 385

or


or





CM

jd

fC

b tc

f

= = times = [ ]

= =times

= [ ]

200 12

20120 535

120

54 40 556 3 83

kips kN

ksi MPa

AM

f d ts

s


minus ( )[ ] = [ ]

2

200 12

24 22 4 25 00 33642 2 in mm


4 in m

18 12

454 1 37




7 9 540 228 7 553 179 5 500 14

10 4 508 1311 4 628 14

3751 P-15 111301 1229 PM Page 386


045fccent= 045(4) = 18 ksi [124 MPa]



or







Af

fb ds

c

yf= ( ) = ( )( ) = [ ]

prime3 3 4000

60 00054 22 2 56 16502 2

in mm

Af

fb ds

c

yw= ( ) = ( )( ) = [ ]

prime6 6 4000

60 00015 22 2 09 13502 2

in mm


3751 P-15 111301 1229 PM Page 387











3751 P-15 111301 1229 PM Page 388





3751 P-15 111301 1229 PM Page 389










3751 P-15 111301 1229 PM Page 390








3751 P-15 111301 1229 PM Page 391






3751 P-15 111301 1229 PM Page 392








3751 P-15 111301 1229 PM Page 393





vV

bd=



3751 P-15 111301 1229 PM Page 394

in which






vcent = v ndash vc


T = Av fs

v fc c= prime1 1



3751 P-15 111301 1229 PM Page 395


Av fs = bsvcent









72

96139 104 718

( ) = [ ] psi kPa

vV

bd= =

times= [ ]40 000

12 24139 957

psi KPa

sA f

v bv s=prime


3751 P-15 111301 1229 PM Page 396



Abs

fv

y

=

50



3751 P-15 111301 1229 PM Page 397








sA f

v bv s=prime

= timestimes

=0 22 20 000

27 1013 6

in

v =

( ) =60

96139 87 psi

sA f

v bv s=prime

= timestimes

=0 22 20 000

44 128 3

in

Av = times

=5012 12

40 0000 18 2

in


3751 P-15 111301 1229 PM Page 398



sA f

v bv s=prime

= timestimes

=0 22 20 000

29 1015 2

in



3751 P-15 111301 1229 PM Page 399





3751 P-15 111301 1229 PM Page 400







Abs

fv

y

=

= times

=50 5010 10

40 0000 125 2 in (See Example 6)


3751 P-15 111301 1229 PM Page 401

402

REFERENCES







Wiley New York 1988

3751 P-16 (refs) 111301 1230 PM Page 402

403


Chapter 2






28A 1414 lb T

28C 300 lb C

210A 193deg

210C 07925 lb

211A 400 lb

211C 1250 lb


212B R1 = 359375 lb [1598 kN] R2 = 440625 lb [1960 kN]

3751 P-17 (answers) 111301 1234 PM Page 403


212D R1 = 7667 lb [3411 kN] R2 = 9333 lb [4153 kN]

212F R1 = 7143 lb [3179 kN] R2 = 11857 lb [5276 kN]

Chapter 3


32A Same as 31A


Chapter 4











47A M = 32 kip-ft [434 kN-m]

47C M = 90 kip-ft [122 kN-m]

Chapter 5



3751 P-17 (answers) 111301 1234 PM Page 404







Chapter 6

62A SF = 253


Chapter 7




Chapter 8

81A R = 21605 lb x = 0769 ft z = 1181 ft

81C T1 = 508 lb T2 = 197 lb T3 = 450 lb


Chapter 9

91A cy = 26 in [70 mm]

91C cy = 42895 in [10724 mm]

3751 P-17 (answers) 111301 1234 PM Page 405


91E cy = 44375 in [1109 mm] cx = 10625 in [266 mm]

93A I = 53586 in4 [211 times 108 mm4]

93C I = 44733 in4 [17474 times 106 mm4]

93E I = 20533 in4 [ 8021 times 106 mm4]

93G I = 438 in4

93I I = 167245 in4

Chapter 10

102A 1182 in2 [762 mm2]

102C 270 kips [120 kN]


103A 19333 lb [86 kN]

103C 29550000 psi [203 GPa]

Chapter 11


113A 386 kips

113C 205 kips

113E 226 kips


114C W 18 times 35


= 175 and 700 psi

116A 1683 kips

116C 371 kips

117A 6735 kips

119A 080 in [20 mm]

1110A 136

1110C 515

Chapter 12

122A 15720 lb

3751 P-17 (answers) 111301 1234 PM Page 406


123A 235 kips [1046 kN]

123C 274 kips [1219 kN]

Chapter 13





Chapter 14




Chapter 15




155A 576 in2 [371 times 103 mm2]


157C 1 at 6 in 4 at 13 in

3751 P-17 (answers) 111301 1234 PM Page 407

3751 P-17 (answers) 111301 1234 PM Page 408

409

INDEX


forces 89truss 120











3751 P-18 (index) 111301 1236 PM Page 409











410 INDEX

3751 P-18 (index) 111301 1236 PM Page 410
















Dynamics 2






INDEX 411

3751 P-18 (index) 111301 1236 PM Page 411









Hydraulic load 12










412 INDEX

3751 P-18 (index) 111301 1236 PM Page 412













wall 186







INDEX 413

3751 P-18 (index) 111301 1236 PM Page 413









Safety 247Section










414 INDEX

3751 P-18 (index) 111301 1236 PM Page 414



beam 162frame 199

Statics 2Steel












Symbols 7


INDEX 415

3751 P-18 (index) 111301 1236 PM Page 415










Vibration 12






416 INDEX

3751 P-18 (index) 111301 1236 PM Page 416


CONTENTS



Introduction




Symbols

Nomenclature


11 Loads



14 Reactions

15 Internal Forces




19 Dynamic Effects





23 Types of Forces

24 Vectors


26 Motion




210 Friction

211 Moments






4 Analysis of Beams

41 Types of Beams


43 Shear in Beams



46 Cantilever Beams




52 Restrained Beams



6 Retaining Walls




7 Rigid Frames





82 Parallel Systems



91 Centroids

















117 Flitched Beams






122 Wood Columns

123 Steel Columns
















155 T-Beams



References


Index

















3751 P- FM 111301 1214 PM Page ii


Sixth Edition

JAMES AMBROSE







3751 P- FM 111301 1214 PM Page iii







v

CONTENTS






Symbols 7

Nomenclature 7




14 Reactions 24



3751 P- FM 111301 1214 PM Page v








24 Vectors 73


26 Motion 76




210 Friction 91

211 Moments 97














vi CONTENTS

3751 P- FM 111301 1214 PM Page vi








CONTENTS vii

3751 P- FM 111301 1214 PM Page vii




















155 T-Beams 382



References 402


Index 409

viii CONTENTS

3751 P- FM 111301 1214 PM Page viii

ix




3751 P- FM 111301 1214 PM Page ix








3751 P- FM 111301 1214 PM Page x




JAMES AMBROSE

2002


3751 P- FM 111301 1214 PM Page xi

3751 P- FM 111301 1214 PM Page xii

xiii





3751 P- FM 111301 1214 PM Page xiii




HARRY PARKER



3751 P- FM 111301 1214 PM Page xiv

1

INTRODUCTION



3751 P-00 (intro) 111301 1217 PM Page 1








2 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 2








3751 P-00 (intro) 111301 1217 PM Page 3

4 INTRODUCTION





cross sections


sections









3751 P-00 (intro) 111301 1217 PM






cross sections


cross sections









3751 P-00 (intro) 111301 1217 PM Page 5

6 INTRODUCTION




254 in mm 00393703048 ft m 3281

6452 in2 mm2 1550 times 10-3

1639 times 103 in3 mm3 6102 times 10-6

4162 times 103 in4 mm4 2403 times 10-6









3751 P-00 (intro) 111301 1217 PM Page 6


SYMBOLS


Symbol Reading


NOMENCLATURE




An = net area


NOMENCLATURE 7

3751 P-00 (intro) 111301 1217 PM Page 7





M = bending moment


S = section modulus

T = tension force


a = unit area







8 INTRODUCTION

3751 P-00 (intro) 111301 1217 PM Page 8

9

1STRUCTURES PURPOSE

AND FUNCTION



3751 P-01 111301 1217 PM Page 9



3751 P-01 111301 1217 PM Page 10







11 LOADS


Gravity




Wind

Source Moving air


LOADS 11

3751 P-01 111301 1217 PM Page 11






Blast




Hydraulic Pressure




Thermal Change




3751 P-01 111301 1217 PM Page 12


Shrinkage


Vibration


Internal Actions


Handling





3751 P-01 111301 1217 PM Page 13

Live and Dead Loads







3751 P-01 111301 1217 PM Page 14





3751 P-01 111301 1217 PM Page 15

Load Dispersion





3751 P-01 111301 1217 PM Page 16

Wind






3751 P-01 111301 1217 PM Page 17


p = 0003V 2

in which






3751 P-01 111301 1217 PM Page 18


Earthquakes




3751 P-01 111301 1217 PM Page 19










Load Combinations



3751 P-01 111301 1217 PM Page 20






3751 P-01 111301 1217 PM Page 21





3751 P-01 111301 1218 PM Page 22







3751 P-01 111301 1218 PM Page 23

14 REACTIONS




3751 P-01 111301 1218 PM Page 24





REACTIONS 25


3751 P-01 111301 1218 PM Page 25




3751 P-01 111301 1218 PM Page 26



REACTIONS 27


3751 P-01 111301 1218 PM Page 27




15 INTERNAL FORCES






3751 P-01 111301 1218 PM Page 28




INTERNAL FORCES 29


3751 P-01 111301 1218 PM Page 29



Stability




Lr

in which




3751 P-01 111301 1218 PM Page 30


In this formula



rI

A=

1 2



3751 P-01 111301 1218 PM Page 31






3751 P-01 111301 1218 PM Page 32





3751 P-01 111301 1218 PM Page 33


Strength





3751 P-01 111301 1218 PM Page 34








3751 P-01 111301 1218 PM Page 35

Stiffness








3751 P-01 111301 1218 PM Page 36






3751 P-01 111301 1218 PM Page 37




3751 P-01 111301 1218 PM Page 38




Tension


Compression




(a)

(b)

3751 P-01 111301 1218 PM Page 39


Shear





3751 P-01 111301 1218 PM Page 40

Bending





3751 P-01 111301 1218 PM Page 41






3751 P-01 111301 1218 PM Page 42





3751 P-01 111301 1218 PM Page 43




3751 P-01 111301 1218 PM Page 44

Torsion








3751 P-01 111301 1218 PM Page 45









3751 P-01 111301 1218 PM Page 46






3751 P-01 111301 1218 PM Page 47






3751 P-01 111301 1218 PM Page 48

Stress Combinations






3751 P-01 111301 1218 PM Page 49





3751 P-01 111301 1218 PM Page 50







3751 P-01 111301 1218 PM Page 51




3751 P-01 111301 1218 PM Page 52

Thermal Stress





3751 P-01 111301 1218 PM Page 53




3751 P-01 111301 1218 PM Page 54






3751 P-01 111301 1218 PM Page 55







3751 P-01 111301 1218 PM Page 56




3751 P-01 111301 1218 PM Page 57







3751 P-01 111301 1218 PM Page 58








3751 P-01 111301 1218 PM Page 59







3751 P-01 111301 1218 PM Page 60

19 DYNAMIC EFFECTS




DYNAMIC EFFECTS 61


3751 P-01 111301 1218 PM Page 61





3751 P-01 111301 1218 PM Page 62





DYNAMIC EFFECTS 63


3751 P-01 111301 1218 PM Page 63













3751 P-01 111301 1218 PM Page 64







3751 P-01 111301 1218 PM Page 65




3751 P-01 111301 1218 PM Page 66





3751 P-01 111301 1218 PM Page 67



se

P Qt Rt

=

+[ ]1

sin( )



3751 P-01 111301 1218 PM Page 68

69

2FORCES AND

FORCE ACTIONS




3751 P-02 111301 1219 PM Page 69





3751 P-02 111301 1219 PM Page 70








3751 P-02 111301 1219 PM Page 71





3751 P-02 111301 1219 PM Page 72

23 TYPES OF FORCES




24 VECTORS


VECTORS 73

3751 P-02 111301 1219 PM Page 73











3751 P-02 111301 1219 PM Page 74




3751 P-02 111301 1219 PM Page 75






26 MOTION




Qualifications




3751 P-02 111301 1219 PM Page 76



Static Equilibrium



MOTION 77


3751 P-02 111301 1219 PM Page 77





Resultant of Forces




3751 P-02 111301 1219 PM Page 78







3751 P-02 111301 1219 PM Page 79








3751 P-02 111301 1219 PM Page 80



Combined Resultants







3751 P-02 111301 1219 PM Page 81


Equilibrant





3751 P-02 111301 1219 PM Page 82




Force Polygon




3751 P-02 111301 1219 PM Page 83







3751 P-02 111301 1219 PM Page 84

Bowrsquos Notation





3751 P-02 111301 1219 PM Page 85







3751 P-02 111301 1219 PM Page 86







3751 P-02 111301 1219 PM Page 87







3751 P-02 111301 1219 PM Page 88


Algebraic Solution





ΣFH = 0 = CA + BCH

ΣFV = 0 = AB + BCV





3751 P-02 111301 1219 PM Page 89



from which

BCV = +200 or 200 lb up





Two-Force Members



BCBCV=

deg= =

sin lb

30

200

0 5400


3751 P-02 111301 1219 PM Page 90

210 FRICTION





FRICTION 91


3751 P-02 111301 1219 PM Page 91




F cent = F cos Q




F cent = mN






3751 P-02 111301 1219 PM Page 92






micro φφ

φ= prime = =F

N

W

W

sin

cos tan

FRICTION 93





3751 P-02 111301 1219 PM Page 93








P = Fcent = mN = 030(100) = 30 lb




3751 P-02 111301 1219 PM Page 94




FRICTION 95



3751 P-02 111301 1219 PM Page 95


Solution For (a)





F cent = 1732(040) = 693 lb


For (c)


or

040(20 cos f) = (20 sin f) ndash 15






3751 P-02 111301 1219 PM Page 96

211 MOMENTS




MOMENTS 97



3751 P-02 111301 1219 PM Page 97



Increasing Moments





3751 P-02 111301 1219 PM Page 98






MOMENTS 99


3751 P-02 111301 1219 PM Page 99








3751 P-02 111301 1219 PM Page 100





MOMENTS 101



3751 P-02 111301 1219 PM Page 101










3751 P-02 111301 1219 PM Page 102










3751 P-02 111301 1219 PM Page 103














3751 P-02 111301 1219 PM Page 104











3751 P-02 111301 1219 PM Page 105






ΣF = 0 = +450 ndash1800 +1350




1213502 2( ) ( )

thus lb


124501 1( ) ( ) thus lb



3751 P-02 111301 1219 PM Page 106



Thus






R2400 3 1000 5 600 11

15

12 800


lb

R14800 10 000 2400

15

17 200

151146 7= + + = =

lb



3751 P-02 111301 1219 PM Page 107



from which



Thus



R230 400

181688 9= =

lb

R116 400

18911 1= =

lb



3751 P-02 111301 1219 PM Page 108






2018601 1( ) ( ) ( )

lb



3751 P-02 111301 1219 PM Page 109



3751 P-02 111301 1219 PM Page 110

111





3751 P-03 111301 1221 PM Page 111





3751 P-03 111301 1221 PM Page 112






3751 P-03 111301 1221 PM Page 113





3751 P-03 111301 1221 PM Page 114




3751 P-03 111301 1221 PM Page 115







3751 P-03 111301 1221 PM Page 116






3751 P-03 111301 1221 PM Page 117



3751 P-03 111301 1221 PM Page 118






3751 P-03 111301 1221 PM Page 119






3751 P-03 111301 1221 PM Page 120





3751 P-03 111301 1221 PM Page 121






from which

and

BI = =1 000

0 5552000 3606

( ) lb

BIh = =0 832

0 5552000 3000

( ) lb

BI BI BIv h

1 000 0 555 0 832 = =


3751 P-03 111301 1221 PM Page 122












3751 P-03 111301 1221 PM Page 123





or

0 = + 1000 ndash 0555CK ndash 0555KJ (321)



or

0 = + 3000 ndash 0832CK + 0832KJ (322)



00 832

0 5551000

0 832

0 5550 555

0 832

0 5550 555= + + minus + minus

( )

( )

( )CK KJ


3751 P-03 111301 1221 PM Page 124



3751 P-03 111301 1221 PM Page 125

or

0 = + 1500 ndash 0832CK ndash 0832KJ



0 = + 1000 ndash 0555(2704) ndash 0555(KJ)

and





KJ = = minus500

0 555901

lb

0 4500 1 6644500

1 6642704= + minus = =

CK CK lb


3751 P-03 111301 1221 PM Page 126






3751 P-03 111301 1221 PM Page 127





3751 P-03 111301 1221 PM Page 128





3751 P-03 111301 1221 PM Page 129




or




Thus


DO = =54 000

105400

lb

10 72 000 12 000 12 000 48 000

48 000

104800

( )

NI

NI

= + minus minus = +

= = lb


3751 P-03 111301 1221 PM Page 130







3751 P-03 111301 1221 PM Page 131

132

4ANALYSIS OF BEAMS


3751 P-04 111301 1221 PM Page 132

41 TYPES OF BEAMS





TYPES OF BEAMS 133


3751 P-04 111301 1221 PM Page 133







3751 P-04 111301 1221 PM Page 134








43 SHEAR IN BEAMS


w = =16 000

20800


SHEAR IN BEAMS 135

3751 P-04 111301 1221 PM Page 135





3751 P-04 111301 1221 PM Page 136

Vertical Shear






SHEAR IN BEAMS 137


3751 P-04 111301 1221 PM Page 137


V(x = 1) = R1 ndash 0 or V(x = 1) = 1000 lb



V(x = 2+) = 1000 ndash 600 = 400 lb


V(x = 6+) = 1000 ndash (600 + 1000) = ndash600 lb




3751 P-04 111301 1221 PM Page 138




Shear Diagrams



SHEAR IN BEAMS 139

3751 P-04 111301 1221 PM Page 139








3751 P-04 111301 1221 PM Page 140

141

Fig

ure

47

Pro

blem

s 4

3A

ndashF

3751 P-04 111301 1221 PM Page 141







3751 P-04 111301 1221 PM Page 142









3751 P-04 111301 1221 PM Page 143






Figure 49 Example 3

3751 P-04 111301 1221 PM Page 144






3751 P-04 111301 1221 PM Page 145





V x xx( ) at ft= minus + = = =7600 800 07600

8009 5



3751 P-04 111301 1221 PM Page 146






M x( ) ( ) ( )


=4 5 10 600 4 5 7000 0 5 800 4 5

4 5

236 100



3751 P-04 111301 1221 PM Page 147





3200 ndash 600x = 0 x = 533 ft



3751 P-04 111301 1221 PM Page 148









0 3200 600

2( )

M

M

x

x

( )

( )

( )

( ) ( )

ft-lb

ft-lb

=

=


=


5 33

12

3200 5 33 600 5 335 33

28533

3200 12 600 12 6 4800


3751 P-04 111301 1221 PM Page 149




3751 P-04 111301 1221 PM Page 150





46 CANTILEVER BEAMS




2600 400400

26000 154x x= = = ft

2400 32003200

24001 33x x= = = ft


3751 P-04 111301 1221 PM Page 151



3751 P-04 111301 1221 PM Page 152







225 000 ft-lb



3751 P-04 111301 1221 PM Page 153






V

M

= + times =


= minus

2000 600 6 5600

2000 14 600 66

238 800

( )

( )

lb

ft-lb



3751 P-04 111301 1221 PM Page 154







3751 P-04 111301 1221 PM Page 155








MwL L

wL L wL WL= + times

minus times times

=

2 2 2 4 8 8

2

or

MPL= = times =4

8000 20

440 000 ft-lb



3751 P-04 111301 1221 PM Page 156






MWL= = times =8

11 200 14

819 600

ft-lb

MwL= = times =

2 2

8

800 14

819 600 ft-lb



3751 P-04 111301 1221 PM Page 157





3751 P-04 111301 1221 PM Page 158






3751 P-04 111301 1221 PM Page 159

160

5CONTINUOUS AND

RESTRAINED BEAMS




3751 P-05 111301 1222 PM Page 160





3751 P-05 111301 1222 PM Page 161







3751 P-05 111301 1222 PM Page 162







1 1 2 1 2 3 21 1

32 2

3

24 4




3751 P-05 111301 1222 PM Page 163









from which

10R1 = 3750 R1 = 375 lb

MwL

2

2 2

8

100 10


ft-lb

MwL

2

2

8= minus

42

2

3

MwL= minus


3751 P-05 111301 1222 PM Page 164




375 ndash (100 times x) = 0 x = 375 ft



Figure 54 Example 1

3751 P-05 111301 1222 PM Page 165




A 16 200B 24 350








2 14 101000 14

4

1000 10

419 500

2

3 3

2

M

M

( )


= minus ft-lb

M = times =375 3 75

2703 125

ft-lb


3751 P-05 111301 1222 PM Page 166





Figure 55 Example 2

3751 P-05 111301 1222 PM Page 167


C 12 16 2000D 16 20 1200












3751 P-05 111301 1222 PM Page 168





E 24 30F 32 24


Figure 57 Example 3

3751 P-05 111301 1222 PM Page 169










3751 P-05 111301 1222 PM Page 170




G 24 1000H 32 1600



3751 P-05 111301 1222 PM Page 171

52 RESTRAINED BEAMS






3751 P-05 111301 1222 PM Page 172





3751 P-05 111301 1222 PM Page 173







3751 P-05 111301 1222 PM Page 174

R1 = V1 = 0375(8000) = 3000 lbR2 = V2 = 0625(8000) = 5000 lb







3751 P-05 111301 1222 PM Page 175




Internal Pins






3751 P-05 111301 1222 PM Page 176




3751 P-05 111301 1222 PM Page 177








3751 P-05 111301 1222 PM Page 178




3751 P-05 111301 1222 PM Page 179




3751 P-05 111301 1222 PM Page 180






3751 P-05 111301 1222 PM Page 181



3751 P-05 111301 1222 PM Page 182

183

6RETAINING WALLS


3751 P-06 111301 1223 PM Page 183



184 RETAINING WALLS


3751 P-06 111301 1223 PM Page 184





3751 P-06 111301 1223 PM Page 185









186 RETAINING WALLS

3751 P-06 111301 1223 PM Page 186


Figure 63 Example 1

3751 P-06 111301 1223 PM Page 187








21 700

99882 17

188 RETAINING WALLS





3751 P-06 111301 1223 PM Page 188






3751 P-06 111301 1223 PM Page 189





Then



pM

S

W e

S= = Σ times = times =2 1168 4 96 12

1 042463

( )

psf

Sbd= = times =

2 2

6

1 2 5

61 042

( ) ft3

pN

A

W

A= = Σ =

times=1168

1 2 5467

psf

eM M

W1

2 1 21 718 9988

116810 04= minus

Σ= minus =

in

190 RETAINING WALLS

3751 P-06 111301 1223 PM Page 190


Figure 65 Example 2

3751 P-06 111301 1223 PM Page 191

192

7RIGID FRAMES


3751 P-07 111301 1223 PM Page 192





3751 P-07 111301 1223 PM Page 193








194 RIGID FRAMES

3751 P-07 111301 1223 PM Page 194







3751 P-07 111301 1223 PM Page 195

196 RIGID FRAMES

Figure 73 Example 1

3751 P-07 111301 1223 PM Page 196










3751 P-07 111301 1223 PM Page 197

198 RIGID FRAMES

Figure 74 Example 2


3751 P-07 111301 1223 PM Page 198







3751 P-07 111301 1223 PM Page 199

200 RIGID FRAMES

Figure 76 Example 3

3751 P-07 111301 1223 PM Page 200


exercise)










3751 P-07 111301 1223 PM Page 201

202

8NONCOPLANAR

FORCE SYSTEMS



3751 P-08 111301 1224 PM Page 202




For its magnitude

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2



3751 P-08 111301 1224 PM Page 203



ΣFx = 0 ΣFy = 0 ΣFz = 0




L

L

12 2

22

5 3 34 5 83

12 34 178 13 34

= + = =

= + = =

( ) ( )

( )


yy

zzF

R

F

R

F

R= Σ =

Σ= Σ



3751 P-08 111301 1224 PM Page 204




R = + + = =( ) ( ) ( ) 2 4 466 1 22 4 217 757 466 72 2 2


Figure 83 Example 1



F1 200(51334) = 75 200(121334) = 180 darr 200(31334) = 45

F2 160(21356) = 236 160(121356) = 1417 darr 160(61356) = 708

F3 180(81497) = 962 180(121497) = 1444 darr 180(41497) = 482____ _____ ____


larrlarr

larrlarrlarr

larr

3751 P-08 111301 1224 PM Page 205


Then





ΣFx = 0 = +1000 ndash 2(Tx) Tx = 500 lb


T

T

T T

x

x

=

= = =

25

12

25

12

25

12500 1041 67( ) ( ) lb

L = + + = =( ) ( ) ( )9 12 20 626 252 2 2

L422 4

466 112 0 578= =

( )

L322 4

466 112 0 062= =

( )

ΣΣ

= =F

F

Lx

y

3

12

2 4

466 1


3751 P-08 111301 1224 PM Page 206



where

Thus

C Ty= = =2 220

12500 1666 67( ) ( ) lb

T

T

T T

y

x

y x

=

= =

20

12

20

12

20

12500( ) ( )



3751 P-08 111301 1224 PM Page 207





T1 = 525 lb T2 = 271 lb T3 = 290 lb




Σ = = + minus

Σ = = + + + minus

Σ = = + + minus

F T T

F T T T

F T T T

x

y

z

04

21

8

21 63

020

21

20

21 63

20

23 321000

05

21

2

21 63

12

23 32

1 2

1 2 3

1 2 3

( )

( )

( )

( )

( )

( )

( )

( )

L

L

L

12 2 2

22 2 2

32 2

5 4 20 441 21

2 8 20 468 21 63

12 20 544 23 32

= + + = =

= + + = =

= + = =

( ) ( ) ( )

( ) ( ) ( )

( ) ( )


3751 P-08 111301 1224 PM Page 208

82 PARALLEL SYSTEMS


R = ΣFy


LM

RL

M

Rx

zz

x= Σ = Σand



3751 P-08 111301 1224 PM Page 209


ΣFy = 0 ΣMx = 0 ΣMz = 0




R = ΣF = 50 + 60 + 160 + 80 = 350 lb





3751 P-08 111301 1224 PM Page 210






L Lx z= = = =800

3502 29

920

3502 63 ft ft



3751 P-08 111301 1224 PM Page 211









T1 = 414 lb T2 = 276 lb T3 = 310 lb



3751 P-08 111301 1224 PM Page 212












yy

zzF

R

F

R

F

R= Σ =

Σ= Σ

R F F Fx y z= Σ( ) + Σ( ) + Σ( )2 2 2


3751 P-08 111301 1224 PM Page 213

214

9PROPERTIES

OF SECTIONS


3751 P-09 111301 1224 PM Page 214

91 CENTROIDS



CENTROIDS 215


3751 P-09 111301 1224 PM Page 215







3751 P-09 111301 1224 PM Page 216




CENTROIDS 217

Figure 93 Example 1

3751 P-09 111301 1224 PM Page 217











yx = 52080 = 65 in

3751 P-09 111301 1224 PM Page 218





3751 P-09 111301 1224 PM Page 219









Idb

Y-Y =3

12

Ibd

X-X =3

12


3751 P-09 111301 1224 PM Page 220







Ibd=

3

36

Id= = times =π 4 4

4

64

3 1416 10

64490 9

in

Id= π 4

64

Ibd= = ( )( ) =

3 34

12

5 5 11 5

12697 1

in



3751 P-09 111301 1224 PM Page 221








I = 500 ndash 1707 = 3293 in4



I = times =4 8

12170 7

34 in

Ibd= = times =

3 34

12

6 10

12500 in

Ibd= = times =

3 34

36

12 10

36333 3 in


3751 P-09 111301 1224 PM Page 222







12

7 8

12667 299 368

3 34 in

I d do i=

( ) minus ( )[ ]

=

minus( ) = minus =

π64

3 1416

6410 8 491 201 290

4 4

4 4 4 in



3751 P-09 111301 1224 PM Page 223


I = Io + Az2

In this formula











3751 P-09 111301 1224 PM Page 224





Figure 98 Example 8




1 20 45 10(2)312 = 67 20(45)2 = 405 41172 60 15 6(10)312 = 500 60(15)2 = 135 6357


3751 P-09 111301 1224 PM Page 225





Io + Az2 = 01667 + (8)(625)2 = 3127 in4


9494 in4

Ibd

o = = times =3 3

4

12

16 0 5

120 1667

in


Figure 99 Example 9

3751 P-09 111301 1224 PM Page 226





3751 P-09 111301 1224 PM Page 227


Section Modulus






3751 P-09 111301 1224 PM Page 228



Radius of Gyration






rI

A=

SI

c= = =697 068

5 75121 229


3751 P-09 111301 1224 PM Page 229



3751 P-09 111301 1224 PM Page 230







3751 P-09 111301 1224 PM Page 231








W 27 times 094 277 2692 0490 09990 0745 1437 3270 2430 1090 1240 2480 212 2780times 084 248 2671 0460 09960 0640 1375 2850 2130 1070 1060 2120 207 2440






3751 P-09 111301 1224 PM Page 232










3751 P-09 111301 1224 PM Page 233




b











3751 P-09 111301 1224 PM Page 234





















3751 P-09 111301 1224 PM Page 235





1frasl2 0840 0622 0109 085 0250 0017 0041 02613frasl4 1050 0824 0113 113 0333 0037 0071 0334


10 10750 10020 0365 4048 1190 161 29900 367012 12750 12000 0375 4956 1460 279 43800 4380



reference document

3751 P-09 111301 1224 PM Page 236







3751 P-09 111301 1224 PM Page 237








3751 P-09 111301 1224 PM Page 238

239

10STRESS AND

DEFORMATION



3751 P-10 111301 1225 PM Page 239



3751 P-10 111301 1225 PM Page 240





in which





Deformation


P f A fP

AA

P

f= times = =or or


3751 P-10 111301 1225 PM Page 241


Strength




3751 P-10 111301 1225 PM Page 242







Common Values





3751 P-10 111301 1225 PM Page 243







P = ( f )(A) = (900)(576) = 528400 lb [206 kN]


AP

f= = = [ ]30 000

115026 09 16 8292 2

in mm


3751 P-10 111301 1225 PM Page 244










fP

A= = = [ ]11 200

0 60118 636 128 397

psi kPa


3751 P-10 111301 1225 PM Page 245




Hookersquos Law






3751 P-10 111301 1225 PM Page 246



Ultimate Strength



Factor of Safety



3751 P-10 111301 1225 PM Page 247








strain = =se

L


3751 P-10 111301 1225 PM Page 248

in which







in which


P = force in pounds




ePL

AE=

Ef

s

P A

e L

PL

Ae= = =

Ef

s=


3751 P-10 111301 1225 PM Page 249









ePL

AE= = times

times= [ ]60 000 120

3 1416 29 000 0000 079 2 0

in mm

fP

A= = = [ ]60

3 141619 1 131 663

ksi kPa


3751 P-10 111301 1225 PM Page 250






3751 P-10 111301 1225 PM Page 251






3751 P-10 111301 1225 PM Page 252




3751 P-10 111301 1225 PM Page 253

254

11STRESS AND

STRAIN IN BEAMS



3751 P-11 111301 1225 PM Page 254






3751 P-11 111301 1225 PM Page 255






3751 P-11 111301 1225 PM Page 256


The Flexure Formula





3751 P-11 111301 1225 PM Page 257




Therefore



M = fS

Mf

cI M

fI

cR R= times =or

Mf


sum times times

f

ca z2

f

cz a z

f



3751 P-11 111301 1225 PM Page 258









30 13

848 8 48 8 12 585 6 kip-ft or kip-in



3751 P-11 111301 1225 PM Page 259







MR = fS = 24 times 366 = 8784 kip-in





SM

f= = =585 6

2424 4 3

in

fM

S= = =585 6

36 616 0

ksi


3751 P-11 111301 1225 PM Page 260







M = 70767 ndash 735 = 70032 ft-lb



MPL

PM

L= = = times =

4

4 4 70 032

1420 009

lb

MwL= = times =

2 2

8

30 14

8735 ft-lb

M fSR = = times =

=

22 000 38 6 849 200

849 200

1270 767

in-lb

or ft-lb


3751 P-11 111301 1225 PM Page 261



MR = fS = 24 times 519 = 12456 kip-in








21 1245 61245 6

2159 3W W= = =

kips

MWL W

W= = times times =8

14 12

821 kip-in


3751 P-11 111301 1225 PM Page 262









SM

f= = =1188

2449 5 3 in


36 22

899 99 12 1188 kip-ft or kip-in


3751 P-11 111301 1225 PM Page 263










SM

f= = =156 000

160097 5 3

in

MWL= = times =8

6500 16

813 000 156 000 lb-ft or lb-in


3751 P-11 111301 1225 PM Page 264




fVQ

Ibv =



3751 P-11 111301 1225 PM Page 265

in which










3751 P-11 111301 1225 PM Page 266



Q = acenty = (4 times 4)(2) = 32 in3

and



fVQ

Ibv = = times

times=4000 32

170 7 4187 5

psi

Ibd= = times =

3 34

12

4 8

12170 7 in



3751 P-11 111301 1225 PM Page 267





Q = (2 times 10)(45) = 90 in3

fVQ

Ibv = = times

times=8000 126 75

1046 7 6161 5

psi


=6 5 6

6 5

2126 75 3

in



3751 P-11 111301 1225 PM Page 268




Thus




fVQ

Ib

V bd

bd b

V

bdv = =

times ( )( ) times

=

2

3

8

121 5

Q bd d bd= times

=

2 4 8

2

f

f

v

v

= timestimes

=

= timestimes

=

8000 90

1046 7 6114 6

8000 90

1046 7 1068 8

psi (in the web)

psi (in the flange)


3751 P-11 111301 1225 PM Page 269






3751 P-11 111301 1225 PM Page 270




in which






fV

t dv

w b

=


3751 P-11 111301 1225 PM Page 271







117 FLITCHED BEAMS






fV

t dv

w b

= =times

=30

0 435 9 737 09

ksi


3751 P-11 111301 1225 PM Page 272








f fE

E1 2

1

2

= times

s sf

E

f

E1 2

1

1

2

2

= =or

sf

Es

f

E1

1

12

2

2

= =and

Ef

sE

f

s1

1

12

2

2

= =and

FLITCHED BEAMS 273


3751 P-11 111301 1225 PM Page 273










Then



M = Mw + Ms = 7589 + 19342 = 26931 ft-lb

Sbd

s = = times ( ) =2 2

3

6

0 5 11 25

610 55

in

f fE

Ew s

w

s

= times

= ( ) times

=22 0001 900 000

29 000 0001441

psi


3751 P-11 111301 1225 PM Page 274









MWL W

W

= = = ( )

= times =

26 9318

14

88 26 931

1415 389

lb


3751 P-11 111301 1225 PM Page 275





in which




D CWL

EI=

3


3751 P-11 111301 1225 PM Page 276














3751 P-11 111301 1225 PM Page 277







Then

DWL

EI

WLd

I

L

Ed

fL

Ed

fL

Ed

=

times

=

=

=

5

384

16

5

24

5

24

5

24

3

2

2

2

( )

fMc

I

WL d

I

WLd

I= = ( )( ) = 8 2

16

DWL

EI=

5

384

3

fMc

I=

MWL=8


3751 P-11 111301 1225 PM Page 278









fM

S= = times =97 5 12

48 624 07

ksi

MWL= = times =8

39 20

897 5 kip-ft

DL

d= 0 00017179 2

DfL

Ed

L

d

L d

=

=

times

times ( )

=

5

24

5

24

24

29 000

12

0 02483

2

2

2


3751 P-11 111301 1225 PM Page 279








DL

d=

times

= ( ) timestimes( )

= [ ]

20 48

24

0 02483

0 85330 02483 19

12 22

0 626 16

2

2

in mm

fM

S= = times =57 12

33 420 48

ksi

MWL= = times =8

24 19

857 kip-ft

DWL

EI= = ( ) times( )

times times=5

384

5 39 20 12

384 29 000 3400 712

3 3

in

DL

d= = times = [ ]0 2483 0 02483 20

13 980 7104 18 05

2 2

in mm


3751 P-11 111301 1225 PM Page 280







3751 P-11 111301 1225 PM Page 281







3751 P-11 111301 1225 PM Page 282





3751 P-11 111301 1225 PM Page 283

My = Fy times S







3751 P-11 111301 1225 PM Page 284







fM

S= = times = [ ]72 12

38 622 4 154

ksi MPa

MPL= = times = [ ]4

18 16

472 kip-ft 98 kN-m



3751 P-11 111301 1225 PM Page 285






3751 P-11 111301 1225 PM Page 286


M = Fy times Z








ΣFh = 0

or




3751 P-11 111301 1225 PM Page 287

and thus

Au = Al



or


or

Mp = fy times Z






3751 P-11 111301 1225 PM Page 288



My = Fy times Sx










My = times = =36 111 39963996



3751 P-11 111301 1225 PM Page 289







3751 P-11 111301 1225 PM Page 290










from which

wL L

y = times =333 12 39962 2

(in kip-ft units)

Mw L

yy= =33312

2


3751 P-11 111301 1225 PM Page 291


from which





2196

3996100 55times =

wL L

p = times =387 16 61922 2

(in kip-ft units)

Mw L

pp= =

times387

16

2


3751 P-11 111301 1225 PM Page 292

293





3751 P-12 111301 1226 PM Page 293





3751 P-12 111301 1226 PM Page 294



PEI

L= π 2

2



3751 P-12 111301 1226 PM Page 295





3751 P-12 111301 1226 PM Page 296

122 WOOD COLUMNS




WOOD COLUMNS 297


3751 P-12 111301 1226 PM Page 297






P = (Fc)(Cp)(A)

in which







3751 P-12 111301 1226 PM Page 298


in which




in which







FK E

L dcE

cE

e

= ( )( )( ) 2

CF F

c

F F

c

F F

cp

cE c cE cE c=+ ( )

minus+( )

minus1

2

1

2

2

WOOD COLUMNS 299

3751 P-12 111301 1226 PM Page 299





P = (Fc)(Cp)(A) = (1000)(0993)(55)2 = 30038 lb


1576 Cp = 0821 and thus

P = (1000)(0821)(55)2 = 24835 lb


Cp = 0355 and thus

P = (1000)(0355)(55)2 = 10736 lb



FK E

L d

F

F

C

cEcE

e

cE

c

p

= ( )( )( )

= ( )( )( )

=

= =

= + minus +

minus =

2 2

2

0 3 1 600 000

4 3625 250

25 250

100025 25

1 25 25

1 6

1 25 25

1 6

25 25

0 80 993

psi


3751 P-12 111301 1226 PM Page 300







123 STEEL COLUMNS


FK E

L d

F

F

C

P F C A

cEcE

e

cE

c

p

c p

= ( )( )( )

= ( )( )( )

=

= =

= minus

minus =

= ( )( )( ) = ( )( )

2 2

2

0 3 1 400 000

29 14495

495

892 50 555

1 555

1 6

1 555

1 6

0 555

0 80 471

892 5 0 471 1 5

psi


L

d= ( ) =8 5 12

3 529 14

STEEL COLUMNS 301

3751 P-12 111301 1226 PM Page 301

Column Shapes









3751 P-12 111301 1226 PM Page 302

STEEL COLUMNS 303



3751 P-12 111301 1226 PM Page 303






1 2156 26 2022 51 1826 76 1579 101 1285 126 941 151 655 176 482

2 2152 27 2015 52 1817 77 1569 102 1272 127 926 152 646 177 477

3 2148 28 2008 53 1808 78 1558 103 1259 128 911 153 638 178 471

4 2144 29 2001 54 1799 79 1547 104 1247 129 897 154 630 179 466

5 2139 30 1994 55 1790 80 1536 105 1233 130 884 155 622 180 461

6 2135 31 1987 56 1781 81 1524 106 1220 131 870 156 614 181 456

7 2130 32 1980 57 1771 82 1513 107 1207 132 857 157 606 182 451

8 2125 33 1973 58 1762 83 1502 108 1194 133 844 158 598 183 446

9 2121 34 1965 59 1753 84 1490 109 1181 134 832 159 591 184 441

10 2116 35 1958 60 1743 85 1479 110 1167 135 819 160 583 185 436

11 2110 36 1950 61 1733 86 1467 111 1154 136 807 161 576 186 432

12 2105 37 1942 62 1724 87 1456 112 1140 137 796 162 569 187 427

13 2100 38 1935 63 1714 88 1444 113 1126 138 784 163 562 188 423

14 2095 39 1927 64 1704 89 1432 114 1113 139 773 164 555 189 418

15 2089 40 1919 65 1694 90 1420 115 1099 140 762 165 549 190 414

16 2083 41 1911 66 1684 91 1409 116 1085 141 751 166 542 191 409

17 2078 42 1903 67 1674 92 1397 117 1071 142 741 167 535 192 405

18 2072 43 1895 68 1664 93 1384 118 1057 143 730 168 529 193 401

19 2066 44 1886 69 1653 94 1372 119 1043 144 720 169 523 194 397

20 2060 45 1878 70 1643 95 1360 120 1028 145 710 170 517 195 393

21 2054 46 1870 71 1633 96 1348 121 1014 146 701 171 511 196 389

22 2048 47 1861 72 1622 97 1335 122 999 147 691 172 505 197 385

23 2041 48 1853 73 1612 98 1323 123 985 148 682 173 499 198 381

24 2035 49 1844 74 1601 99 1310 124 970 149 673 174 493 199 377

25 2028 50 1835 75 1590 100 1298 125 955 150 664 175 488 200 373


3751 P-12 111301 1226 PM Page 304






P = (Fa)(A) = (1569)(156) = 2448 kips [1089 kN]




P = (1724)(156) = 2689 kips [1196 kN]

KL


2 4862

KL


2 4877 4

STEEL COLUMNS 305

3751 P-12 111301 1226 PM Page 305






P = Fa A = (1456)(156) = 2271 kips [1010 kN]



yKL


2 4887 1 87

xKL


5 2368 8


3751 P-12 111301 1226 PM Page 306


STEEL COLUMNS 307


3751 P-12 111301 1226 PM Page 307




3751 P-12 111301 1226 PM Page 308

309

13COMBINED FORCES

AND STRESSES




3751 P-13 111301 1226 PM Page 309





3751 P-13 111301 1226 PM Page 310







fM

Sb = = = [ ]10 000

1 3337500 52 8

psi MPa


2 23 3 3

6

2 2

61 333 20 82 10 in mm

fN

Aa = = = [ ]5

41250 8 8 psi MPa



3751 P-13 111301 1226 PM Page 311








3751 P-13 111301 1226 PM Page 312





3751 P-13 111301 1226 PM Page 313



or






N

A

Nec

Ie

I

Ac= = thus

pN

A

Nec

I= plusmn


3751 P-13 111301 1226 PM Page 314








V = 1frasl2 (wpx)



3751 P-13 111301 1226 PM Page 315



pN

wx= 2



3751 P-13 111301 1226 PM Page 316









pN

A=

=

=2 2

100

365 56 ksf

pN

A

Mc

I= + = + times

= + =

100

64

100 4

341 31 56 1 17 2 73

ksf

A

Ibd

= times =

= = times =

8 8 64

12

8 8

12341 3

2

3 34

ft

ft

eM

N= = =100

1001 ft


3751 P-13 111301 1226 PM Page 317






a e

x a

pN

wx

= minus = minus =

= = =

= timestimes

=

5

22 5 1 1 5

3 3 1 5 4 5

2 2 100

5 4 58 89

( )

ft

ft

= ksf



3751 P-13 111301 1226 PM Page 318









fP

A

P

A

vP

A

P

A

= =

( )

= =

( )

cos

coscos

sin

cossin cos

ΘΘ

Θ

ΘΘ

Θ Θ

2


3751 P-13 111301 1226 PM Page 319







fP A

vP A= =

2 2and

fP

Av= =and 0



3751 P-13 111301 1226 PM Page 320







fP

A

vP

A

=

=

( ) =

=

=

( )( ) =

cos

sin cos

2 21200

120 5 25

1200

120 5 0 866 43 3

Θ

Θ Θ

psi

psi



3751 P-13 111301 1226 PM Page 321








3751 P-13 111301 1226 PM Page 322






3751 P-13 111301 1227 PM Page 323

324

14CONNECTIONS FOR

STEEL STRUCTURES




3751 P-14 111301 1227 PM Page 324





3751 P-14 111301 1227 PM Page 325



R = Fv times A




3751 P-14 111301 1227 PM Page 326







3751 P-14 111301 1227 PM Page 327






3751 P-14 111301 1227 PM Page 328






3751 P-14 111301 1227 PM Page 329





3751 P-14 111301 1227 PM Page 330








3751 P-14 111301 1227 PM Page 331










A307 S 44 60 79 99 123D 88 120 157 199 245T 88 120 157 199 245

A325 S 75 102 134 169 209D 150 204 267 338 417T 194 265 346 437 540

A490 S 93 126 165 209 258D 186 253 330 417 515T 239 325 424 537 663


3751 P-14 111301 1227 PM Page 332






3751 P-14 111301 1227 PM Page 333











0625 1125 0875 167 18750750 125 10 20 2250875 15b 1125 233 26251000 175b 125 267 30


3751 P-14 111301 1227 PM Page 334



Tension Connections






GageDimension 8 7 6 5 4 35 3 25 2

g 45 40 350 300 25 20 175 1375 1125g1 30 25 225 200g2 30 30 250 175


3751 P-14 111301 1227 PM Page 335


Ae = C1An

in which











3751 P-14 111301 1227 PM Page 336












3751 P-14 111301 1227 PM Page 337





w = b + 2(a) = 225 + 2(125) = 475 in [121 mm]


P = = [ ]100

812 5 55 6 kips kN

n say= =100

15 56 45 7



3751 P-14 111301 1227 PM Page 338





w = 6 ndash 2(0875) = 425 in [108 mm]





ft =times( )

= [ ]100

2 0 4375 4 2526 9 185

ksi MPa

t =times

= [ ]4 63

2 60 386 9 8

in mm

A = = [ ]100

21 64 63 29872 2

in mm


3751 P-14 111301 1227 PM Page 339











2P

F tu

2

2

P

F t

D

u

+

fp =times

= [ ]12 5

0 75 0 5033 3 230

ksi MPa

ft =times

= [ ]100

8 25 0 524 24 167

ksi MPa

t = = [ ]4 63

100 463 11 8

in mm

fp =times times

= [ ]12 5

2 0 75 0 437519 05 131

ksi MPa


3751 P-14 111301 1227 PM Page 340

in which









2

20 862 0 375 1 237

P

F t

D

u

+ = + = in

2 2 12 5

58 0 50 862

P

F tu

= timestimes

=

in



3751 P-14 111301 1227 PM Page 341


net w = 3 ndash 0875 = 2125 in [54 mm]


Ft = 050Fu = 29 ksi [200 MPa]



Fv = 030Fu = 174 ksi [120 MPa]







net in mmw = minus

= [ ]2 1 25

0 875

21 625 41 3


3751 P-14 111301 1227 PM Page 342






3751 P-14 111301 1227 PM Page 343









3751 P-14 111301 1227 PM Page 344




3751 P-14 111301 1227 PM Page 345




BD = 1frasl2(12 + 12)12 = 1frasl2(2)12 = 0707






3751 P-14 111301 1227 PM Page 346











in mm in mm


3751 P-14 111301 1227 PM Page 347









3751 P-14 111301 1227 PM Page 348



Fa = 06(Fy) = 06(36) = 216 ksi


T = Fa A = 216(3 times 04375) = 2835 kips






L = =28 35

5 65 06

in



3751 P-14 111301 1227 PM Page 349





T = Ft A = (216)(209) = 451 kips [200 kN]




L = = [ ]45 1

3 712 2 310

in mm


3751 P-14 111301 1227 PM Page 350


and





L = = [ ]23 625

3 76 39 162

in mm

L20 99

3 512 2 3 45 88=

( ) = [ ]

in mm

L12 51

3 512 2 8 75 222=

( ) = [ ]

in mm



3751 P-14 111301 1227 PM Page 351







3751 P-14 111301 1227 PM Page 352

353

15REINFORCED

CONCRETE BEAMS





3751 P-15 111301 1229 PM Page 353



Design Methods




3751 P-15 111301 1229 PM Page 354


The Stress Method






The Strength Method



3751 P-15 111301 1229 PM Page 355











3751 P-15 111301 1229 PM Page 356








E fc c= prime4730

E fc c= prime57 000



3751 P-15 111301 1229 PM Page 357



Cement




3751 P-15 111301 1229 PM Page 358


Reinforcement






3751 P-15 111301 1229 PM Page 359




3751 P-15 111301 1229 PM Page 360









3751 P-15 111301 1229 PM Page 361







3751 P-15 111301 1229 PM Page 362







Nominal Dimensions

Cross-Sectional




3751 P-15 111301 1229 PM Page 363






3751 P-15 111301 1229 PM Page 364


T = As fs


T = As fy








3751 P-15 111301 1229 PM Page 365













(1522)f

M

kjbdc = 2

2



3751 P-15 111301 1229 PM Page 366


M = Tjd = As fs jd


(1523)


(1524)


(1525)

(1526)

(1527)

(1528)

in which




M Rbd= 2

pf k

fc

s

=2

jk= minus

1

3

kf nfs c

=+ ( )

1

1

AM

f jds

s

=

fM

A jds

s

=


3751 P-15 111301 1229 PM Page 367









fs fccent R


20 138 2 1379 113 0337 0888 00076 0135 9283 2068 92 0383 0872 00129 0226 15544 2758 80 0419 0860 00188 0324 22285 3448 71 0444 0852 00250 0426 2937

24 165 2 1379 113 0298 0901 00056 0121 8323 2068 92 0341 0886 00096 0204 14034 2758 80 0375 0875 00141 0295 20285 3448 71 0400 0867 00188 0390 2690

3751 P-15 111301 1229 PM Page 368





3751 P-15 111301 1229 PM Page 369



M = Rbd 2 = 200 kip-ft [271 kN-m]



Therefore

M = 200 times 12 = 0226(bd 2) and bd 2 = 10619



Design for shear



b d

d

b d

d

= = =

=[ ]

= = =

=[ ]

1010 619

1032 6

0 829

1510 619

1526 6

0 677

in in

b = 0254 m m

in in

b = 0381 m m


3751 P-15 111301 1229 PM Page 370






As = pbd = 00129(15 times 26625) = 515 in2 [3312 mm2]


AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 872 26 6255 17 33122 2

in mm



3751 P-15 111301 1229 PM Page 371

No 6 bars 517044 = 1175 or 12 [3312284 = 1166]

No 7 bars 517060 = 862 or 9 [3312387 = 856]

No 8 bars 517079 = 654 or 7 [3312510 = 649]

No 9 bars 517100 = 517 or 6 [3312645 = 513]

No 10 bars 517127 = 407 or 5 [3312819 = 404]

No 11 bars 517156 = 331 or 4 [33121006 = 329]




3751 P-15 111301 1229 PM Page 372





and

pA

bds= =

( )( )=3 98

15 33 50 00792

AM

f jds

s

= = times( )( )( )

= [ ]200 12

20 0 90 33 53 98 25672 2

in mm

M Rbd

M M

= = ( )( )( ) =

= = = ( )( )( ) =[ ]

2 2

2

0 226 15 33 5 3804

3804

12317 1554 0 380 0 850 427

kip-in

or

kip-ft kN-M


3751 P-15 111301 1229 PM Page 373




but not less than



As = 0005(bd) = 0005(15 times 335) = 251 in2


AF

bdsy

=

( )200

Af

fbds

c

y

= ( )prime3

jk= minus

= minus

=1

31

0 313

30 896




3000 00050 0003334000 00050 0003335000 00053 000354


3751 P-15 111301 1229 PM Page 374








U = 14D + 17L

in which




3751 P-15 111301 1229 PM Page 375














3751 P-15 111301 1229 PM Page 376





(1541)


(1542)


(1543)


(1544)

pA

bdA pbd

apbd f

f b

pdf

f

ss

y

c

y

c

= =

=( )( )

prime=

prime

0 85 0 85

0 850 85

f ba A f aA f

f bc s y

s y

c

prime = =prime

M T da

A f da

t s y= minus

= minus

2 2

M C da

f ba da

c c= minus

= prime minus

2

0 852


3751 P-15 111301 1229 PM Page 377


(1545)



Thus

Mt = Rbd 2 (1546)

where

(1547)



R pfa

dy= minus

12

M A f da

pbd f da

pbd f da

d

bd pfa

d

t s y

y

y

y

= minus

= ( )( ) minus

= ( )( )( ) minus

= ( ) minus

2

2

12

12

2

p f ff

b c yy

= prime

times+( )

0 8587

87


3751 P-15 111301 1229 PM Page 378





U = 14D + 17LMu = 14(MDL) + 17(MLL)

= 14(58) + 17(38) = 1458 kip-ft [1977 kN-m]


MM

tu= = =

times [ ]0 90

145 8

0 90162

220

kip-ft

= 162 12 = 1944 kip-in kN-m





40 276 2 1379 05823 04367 00186 0580 40003 2068 05823 04367 00278 0870 60004 2758 05823 04367 00371 1161 80005 3448 05480 04110 00437 1388 9600

60 414 2 1379 05031 03773 00107 0520 36003 2068 05031 03773 00160 0781 54004 2758 05031 03773 00214 1041 72005 3448 04735 03551 00252 1241 8600

3751 P-15 111301 1229 PM Page 379


As = pbd



M1 = Rbd 2


M1 = 1944 = 1041(10)(d )2

and


As = pbd = 00214(10)(1366) = 292 in2 [1880 mm2]





d =( )

= = [ ]1944

1 041 10186 7 13 66 347

in mm


3751 P-15 111301 1229 PM Page 380



a = 025d = 025(18) = 45 in [114 mm]



and

Thus

a da= ( ) = minus

= [ ]0 202 18 3 63

216 2 400 in in mm

apdf

f

a

d

pf

f

y

c

y

c

=prime

=prime

= ( )( )

=

0 85

0 85

0 0114 60

0 85 40 202

pA

bds= =

( )=2 057

10 180 0114

M T jd A F da

AM

f d a

t s y

st

y

= ( ) = minus

=minus ( )[ ] =

( )= [ ]

2

2

1944

60 15 752 057 13272 2

in mm


3751 P-15 111301 1229 PM Page 381



Moment Due to



A 40 542 20 271 12 305B 80 1085 40 542 15 381C 100 1356 50 678 18 457

155 T-BEAMS




3751 P-15 111301 1229 PM Page 382




T-BEAMS 383


3751 P-15 111301 1229 PM Page 383





where



Af

fb ds

c

yw=

prime( )6

CM

jd

M

d t= =

minus ( )2

fC

b tc

f

=

AM

f jd

M

f d ts

s s

= =minus ( )[ ]2

jd dt= minus

2


3751 P-15 111301 1229 PM Page 384

or

in which






Af

fb ds

c

yf= ( )

prime3

T-BEAMS 385


3751 P-15 111301 1229 PM Page 385

or


or





CM

jd

fC

b tc

f

= = times = [ ]

= =times

= [ ]

200 12

20120 535

120

54 40 556 3 83

kips kN

ksi MPa

AM

f d ts

s


minus ( )[ ] = [ ]

2

200 12

24 22 4 25 00 33642 2 in mm


4 in m

18 12

454 1 37




7 9 540 228 7 553 179 5 500 14

10 4 508 1311 4 628 14

3751 P-15 111301 1229 PM Page 386


045fccent= 045(4) = 18 ksi [124 MPa]



or







Af

fb ds

c

yf= ( ) = ( )( ) = [ ]

prime3 3 4000

60 00054 22 2 56 16502 2

in mm

Af

fb ds

c

yw= ( ) = ( )( ) = [ ]

prime6 6 4000

60 00015 22 2 09 13502 2

in mm


3751 P-15 111301 1229 PM Page 387











3751 P-15 111301 1229 PM Page 388





3751 P-15 111301 1229 PM Page 389










3751 P-15 111301 1229 PM Page 390








3751 P-15 111301 1229 PM Page 391






3751 P-15 111301 1229 PM Page 392








3751 P-15 111301 1229 PM Page 393





vV

bd=



3751 P-15 111301 1229 PM Page 394

in which






vcent = v ndash vc


T = Av fs

v fc c= prime1 1



3751 P-15 111301 1229 PM Page 395


Av fs = bsvcent









72

96139 104 718

( ) = [ ] psi kPa

vV

bd= =

times= [ ]40 000

12 24139 957

psi KPa

sA f

v bv s=prime


3751 P-15 111301 1229 PM Page 396



Abs

fv

y

=

50



3751 P-15 111301 1229 PM Page 397








sA f

v bv s=prime

= timestimes

=0 22 20 000

27 1013 6

in

v =

( ) =60

96139 87 psi

sA f

v bv s=prime

= timestimes

=0 22 20 000

44 128 3

in

Av = times

=5012 12

40 0000 18 2

in


3751 P-15 111301 1229 PM Page 398



sA f

v bv s=prime

= timestimes

=0 22 20 000

29 1015 2

in



3751 P-15 111301 1229 PM Page 399





3751 P-15 111301 1229 PM Page 400







Abs

fv

y

=

= times

=50 5010 10

40 0000 125 2 in (See Example 6)


3751 P-15 111301 1229 PM Page 401

402

REFERENCES







Wiley New York 1988

3751 P-16 (refs) 111301 1230 PM Page 402

403


Chapter 2






28A 1414 lb T

28C 300 lb C

210A 193deg

210C 07925 lb

211A 400 lb

211C 1250 lb


212B R1 = 359375 lb [1598 kN] R2 = 440625 lb [1960 kN]

3751 P-17 (answers) 111301 1234 PM Page 403


212D R1 = 7667 lb [3411 kN] R2 = 9333 lb [4153 kN]

212F R1 = 7143 lb [3179 kN] R2 = 11857 lb [5276 kN]

Chapter 3


32A Same as 31A


Chapter 4











47A M = 32 kip-ft [434 kN-m]

47C M = 90 kip-ft [122 kN-m]

Chapter 5



3751 P-17 (answers) 111301 1234 PM Page 404







Chapter 6

62A SF = 253


Chapter 7




Chapter 8

81A R = 21605 lb x = 0769 ft z = 1181 ft

81C T1 = 508 lb T2 = 197 lb T3 = 450 lb


Chapter 9

91A cy = 26 in [70 mm]

91C cy = 42895 in [10724 mm]

3751 P-17 (answers) 111301 1234 PM Page 405


91E cy = 44375 in [1109 mm] cx = 10625 in [266 mm]

93A I = 53586 in4 [211 times 108 mm4]

93C I = 44733 in4 [17474 times 106 mm4]

93E I = 20533 in4 [ 8021 times 106 mm4]

93G I = 438 in4

93I I = 167245 in4

Chapter 10

102A 1182 in2 [762 mm2]

102C 270 kips [120 kN]


103A 19333 lb [86 kN]

103C 29550000 psi [203 GPa]

Chapter 11


113A 386 kips

113C 205 kips

113E 226 kips


114C W 18 times 35


= 175 and 700 psi

116A 1683 kips

116C 371 kips

117A 6735 kips

119A 080 in [20 mm]

1110A 136

1110C 515

Chapter 12

122A 15720 lb

3751 P-17 (answers) 111301 1234 PM Page 406


123A 235 kips [1046 kN]

123C 274 kips [1219 kN]

Chapter 13





Chapter 14




Chapter 15




155A 576 in2 [371 times 103 mm2]


157C 1 at 6 in 4 at 13 in

3751 P-17 (answers) 111301 1234 PM Page 407

3751 P-17 (answers) 111301 1234 PM Page 408

409

INDEX


forces 89truss 120











3751 P-18 (index) 111301 1236 PM Page 409











410 INDEX

3751 P-18 (index) 111301 1236 PM Page 410
















Dynamics 2






INDEX 411

3751 P-18 (index) 111301 1236 PM Page 411









Hydraulic load 12










412 INDEX

3751 P-18 (index) 111301 1236 PM Page 412













wall 186







INDEX 413

3751 P-18 (index) 111301 1236 PM Page 413









Safety 247Section










414 INDEX

3751 P-18 (index) 111301 1236 PM Page 414



beam 162frame 199

Statics 2Steel












Symbols 7


INDEX 415

3751 P-18 (index) 111301 1236 PM Page 415










Vibration 12






416 INDEX

3751 P-18 (index) 111301 1236 PM Page 416


CONTENTS



Introduction




Symbols

Nomenclature


11 Loads



14 Reactions

15 Internal Forces




19 Dynamic Effects





23 Types of Forces

24 Vectors


26 Motion




210 Friction

211 Moments






4 Analysis of Beams

41 Types of Beams


43 Shear in Beams



46 Cantilever Beams




52 Restrained Beams



6 Retaining Walls




7 Rigid Frames





82 Parallel Systems



91 Centroids

















117 Flitched Beams






122 Wood Columns

123 Steel Columns
















155 T-Beams



References


Index

simplified mechanics & strength of materials for architects and builders

Documents