structural systems in architecture

Professor Özgür ÖZÇELİK, Ph.D.

Structural Systems in ArchitectureAR 361

Fall Semester

Content

• Welcome to Introduction to Structural Systems in Architecture

• Course Syllabus

– What material will we cover?

• Administrative Issues

– Course Web Page, Text Book, Exams, Office Hours, Quizes, Grading, Cheating Policy, etc.

• Introduction

AR 361 – Structural Systems in Architecture

COURSE SYLLABUS



Course Hours

AR 361 – Structural Systems in

Architecture

Fall Semester 2014-2015

Class Room: ???

Two sections: Mondays 10:00 – 13:00

14:00 – 17:00

Office Hours

• Mondays 17:00-18:00

Email: [email protected]

Web page: kisi.deu.edu.tr/ozgur.ozcelik

Course Syllabus

General Course Description

• Classification of structural systems

• Loads, assumptions and Idealizations

• Statically Determinate vs. Indeterminate Structures

• Trusses – Analysis of Member Forces

• Shear and Moment Diagrams (Beams)

• Shear and Moment Diagrams (Frames)

• Deflection of Beams: Double Integration Method

• Deflection of Beams: Moment-Area Method


Course Syllabus

General Course Description

• Statically Indeterminate Structures

• Analysis of Indeterminate Beams and Frames:

Slope Deflection Method

• Analysis of Indeterminate Beams and Frames:

Moment Distribution Method


Course Syllabus

Textbook

TITLE: Fundamentals of Structural Analysis

AUTHOR: K.M. Leet, C.M. Uang, A.M. Gilbert

PUBLISHER:McGraw-Hill

Reference Book

TITLE: Fundamentals of Structural Analysis

AUTHOR: H.H. West, L.F. Geschwinder

PUBLISHER: John Wiley and Sons

Class notes and examples will be provided (check the website)


Course Text Book

• Chapter 1- Classification of structural system

• Chapter 2 – Design Load

• Chapter 3- Reactions, Determinate vs

Interdeterminate Structures

• Chapter 4 – Truss Analysis

• Chapter 5 – Beams and Frame Analysis – Shear and

Moment Diagrams

• Chapter 9 - Deflections of Beams and Frames: Double

Integration Method

• Chapter 9 - Deflections of Beams and Frames:

Moment-Area Method

• Chapter 12 - Analysis of Indeterminate Beams: Slope

Deflection Method

• Chapter 13 - Analysis of Indeterminate Beams:

Moment Distribution Method


Course Syllabus

Grading Policy

Your grade will be determined as follows

�Quizes 20% (3 quizes)

�Midterm 30 % (1 midterm exam)

�Final exam 50 %

Class participation will help your grade but

it is not mandatory!!!!


INTRODUCTION:

Introduction

analysis, design and construction of structural system

Structural Engineering:

Structural Systems:

Bridges, buildings, dams, transportation facilities, liquid or

gas storage facilities, industrial factories and plants, power

generation and transmission units

Introduction

Structural Analysis:

We will determine how a structure responds to specified

loads or actions: forces and deformations

We will learn the classical methods that have been used by

engineers for many years which are the foundation of

modern structural analysis

To be a good architects you need to be able to communicate

well with structural engineers!!!!

Basics Structural Elements

Axially Loaded Members in Tension

The material is used at optimum of

efficiency because the axially loaded

members are uniformly stressed in

tension.

We can use small cross sections but

they will be flexible and they will

vibrate under moving load

Building code try to limit the slenderness ratio (l/r)!!!

l = length of the member

r = sqrt(I/A) radius of gyration

Columns – Axially Loaded Members in Compression

The capacity to resist under compression

is a function of the slenderness,

If l/r is large the member is slender

and it will fail for buckling,

If l/r is small the member is stocky

and their capacity for axial load is

high,

The capacity of a slender column depends

also on the support conditions at its

ends.

Beams – Shear and Bending Moment

Beams are loaded perpendicular to

their longitudinal axis.

Except for short beams the shear stress

produced by V is much smaller than the

bending stress produced by M.

Planar Trusses – All Members Axially Loaded

A truss is a structural element composed of slender

bars that are connected by frictionless pin joints

They are very stiff longitudinally

but flexible when loaded

perpendicular to their

longitudinal axis.

Planar Trusses – All Members Axially Loaded

Arches – Direct Compression

Arches are in compression under their

dead load. To work properly the resultant

of the internal forces should passes

through the centroid.

On the abutment we will have a

horizontal as well as a vertical force.

For this reason we need massive

abutment to absorb the reaction.

Cables – Flexible Members in Tension

Cables are slender, flexible members

composed of high-strength steel wires

twisted mechanically.

They can only carry direct tensile stress and they have the

strength to support the large load of long-span structures.

Under distributed vertical load they will

deform as a parabola.


Cables undergoes large changes

in shape when concentrate load

are applied because of their

lack on bending stiffness

We want to minimize deformation

and vibrations produced by live loads.

To stiffen cable we can use

pretensioning, use tie-down

cables or adding extra loads

Rigid Frames

Rigid frames consists of beams and

columns and they carry axial load and

moment.

To have a rigid frame the joints

need to be rigid, meaning that

the right angle must not change

when the members are loaded.

It is necessary to make the joint

stiffer.

Plates or Slabs

Plates are planar elements,

their depth is smaller compared

to the length and width.

Their behaviour depends on

the position of supports

along the boundaries.

Thin Shells

One Story Building

One Story Building

Let’s consider a simple one-story

structure, consisting of structural steel

frame covered with light-gage metal

panels.

The metal roof is supported by

the beam CD that spans

between two pipe columns.

One story Building

The ends of the beam are

connected to the top of the

column by bolts that pass

through the bottom flange of

the beam and a cap plate that is

welded to the column.

This type of connection is not rigid enough. For this

reason we need additional diagonal bracing members to

strengthen the structure against lateral loads.

Vertical and Lateral Loads

For gravity load we

will neglect the

diagonal bracing,

these loads will be

carried by the roof

deck.

The lateral loads (for example the

wind load), that are applied to the top

of the roof, they will be carried by the

diagonal bracing.

Loads and Load Analysis

• Structures must be proportioned so that they will not

FAIL or DEFORM EXCESSIVELY under loads,

• The designer must decide which loads to apply based

on the type of structures (Shape and Function).

Dead Loads:• These are the loads associated with the WEIGHT of the

structure and its PERMANENT COMPONENTS (floors,

ceilings, ducts etc.)

• We assign to the members a dimension to evaluate the

dead loads, this value needs to be checked AFTER the

final dimensioning.

Loads and Load AnalysisDistribution of Dead Loads to Framed Floor Systems

• Many floor systems consist of reinforced

concrete slab supported on a grid of beams,

• We need to understand how to transfer the

load from the slab to the beams,

• In Case 1:• It is a square floor system,

• The edge beams support the same

triangular load

• The area of the slab portion that is

supported by a particular beam is called

the TRIBUTARY AREA.• Side Note:

• Total force of the slab is w*L2 where w is the load per

unit area (kN/m2),

• ¼ of that total load is w*L2/4,

• This total force has to be in triangular shape, therefore

the height of the load can be calculated as follows

w*L2/4 = ½ * height * L, height = wL/2

And from here reaction forces can be calculated…

Case 1:


• In Case 2:• It is a rectangular floor system,

• The rectangular slab is supported by two

parallel beams,

• The tributary area is (Ls/2) * Lb (shaded

grey area)

• Along the beam B1, the magnitude of the

distributed load will be

dist. load = w*Ls/2 [kN/m]

where w is the load per unit area

(kN/m2).

• What would be the reaction forces for the

beam B1?• Side Note:

• What would be the distributed load on a 1 m width of

slab beam Figure (c)?

• The total force on a 1 m width slab beam is

w*(1)*Ls = w*Ls [kN]

Therefore distributed force would be

w*Ls / Ls = w [kN/m] (Figure c)

• What would be the reaction forces?

Case 2:


• In Case 3:

• A slab carrying a uniformly distributed

load of w, and is supported on a

rectangular grid of beams (Figure d)

• Figure e shows the “assumed” and

“simplified” distributed forces acting on

the beam B2, respectively.

Case 3:

• Figures (f) and (g) show the

forces acting on the beam B,

• Notice that there are

distributed as well as

concentrated forces.


• Example:• The floor shown below is 12 cm thick reinforced concrete slab of unit weight 25 kN/m3

supported by STEEL BEAMS (see Figure b)

• The beams are connected by CLIP ANGLES (pin support)

• An acoustical board ceiling of unit weight 0.07 kN/m2 is suspended from the concrete slab,

• An additional dead load of 0.95 kN/m2 is considered to take into account duct, piping, conduit,

• The self-weight of beam B1 is 0.04 kN/m and for the beam B2 is 0.067 kN/m

• Find the magnitude of the dead load distribution on beam B1 and B2.

12 cm

3@2 m = 6 m

1m 1m

5 m


• Example:• For B1:

• Weight of the slab: (1 m + 1 m) * (0.12 m) * (25 kN/m3) = 6 kN/m

• Weight of the ceiling: (1 m + 1 m) * 0.07 kN/m2 = 0.14 kN/m

• Weight of the additional nonstructural elements: (1 m + 1 m) * 0.95 kN/m2 = 1.9 kN/m

• Weight of the beam itself: 0.04 kN/m

• Total weight per unit length: 6 + 0.14 + 1.9 + 0.04 = ~8.1 kN/m (rounded up)

3@2 m = 6 m

1m 1m

5 m

wD = 8.1 kN/m

5 m

20.25 kN 20.25 kN


• Example:• For B2:

• Notice that the slab load is carried by the B1 beam ONLY (due to the rectangular shape

of the slab),

• Therefore B2 only carries its own weight: 0.067 kN/m

• As well as the concentrated reaction forces due to B1 applied at the third points of girder

B2 (see figure f)

3@2 m = 6 m

1m 1m

5 m6 m

20.25 kN 20.25 kN

2 m 2 m 2 m

0.067 kN/m

20.45 kN 20.45 kN

Loads and Load AnalysisTributary Areas of Columns

• To determine the dead load

transmitted into a column from a

floor slab, the designer can

either:

• Determine the reactions of the

beams framing into the column, or• Multiply the tributary area of the

floor surrounding the column by the

magnitude of the dead load per unit

area acting on the floor,

• The tributary area of a column is

defined as the area surrounding the

column that is bounded by the panel

centerlines.

Notice that in the above case, an internal columns

support approximately 4 times more floor dead

load than corner columns.

CIV 301: Structural Analysis I


29/35

• Example:

• Using the tributary area method,

compute the floor dead loads

supported by columns A1 and B2

• Floor system is made of concrete and

weighing 75 lb/ft2

• Floor beams, utilities, suspended ceiling

weigh 15 lb/ft2

• Precast exterior wall supported by the

perimeter beams weighs 600 lb/ft


• Example:

• Total floor dead load:

• D = 75 + 15 = 90 lb/ft2 = 0.09 kip/ft2

• Dead Load for A1

• Tributary area = 9 ft * 10 ft = 90 ft2

• Floor dead load = 90 ft2 * 0.09 kip/ft2

= 8.1 kips

• Weight of exterior wall = 0.6 lb/ft *(10+9) ft = 11.4 kips

• Total dead load = 8.1 + 11.4 = 19.5 kips

• Dead Load for B2

• Tributary area = 18 ft *21 ft = 378 ft2

• Total dead load = 378 ft2 * 0.09 kip/ft2 = 34.02 kips

Loads and Load AnalysisLive Loads

• Live loads are the loads that can be moved on or off of a structure (weight of people, furniture,

machinery etc.)

• The live loads can change in function of the type of building and they change in time (can be

considered as dynamic loads)

• In building codes, specific tables are provided to evaluate live loads as a function of building type

(TS 498, Eurocode 1- Part 1, ASCE 7)

A portion of ASCE minimum

live loads are provided here.

Loads and Load AnalysisLive Loads

Eurocode 1991-1-1

Actions on Structures

Loads and Load AnalysisOther Loads

• Snow Loads:

• Considered for cold regions

• Values are regionalized and provided in building codes (EN 1991-

1-3 or TS-498)

• Wind Loads:

• The magnitude of wind pressure on a structure depends on the

wind velocity, shape and stiffness of the structure, roughness and

profile of the surrounding ground, influence of adjacent

structures,


• Wind Loads (continued):

Typical wind load distribution

on a multistory building.


• Earthquake Loads:

• Earthquakes occur in many regions of the world. In certain

locations where the intensity of the ground shaking is small, the

designer does not have to consider seismic effects.

• In other locations –particularly in regions near and active

geological fault, such as North Anatolian Fault Zone or San

Andreas Fault zone in western coast of CA, large ground motions

frequently occur that can damage or destroy buildings.


• Earthquake Loads (continued):

• The ground motion created by major earthquake forces cause

buildings to sway back and forth. Assuming the building is fixed at

its base, the displacement of floors will vary from zero at the base

to a maximum at the roof,


• Earthquake Loads (continued):

• Earthquake forces to be used in design of structures are defined

in building codes

• Turkish Earthquake Code (2007)

• EN-1998-1 (Eurocode 8) (2004)

• ASCE 7 (2005)

Loads and Load AnalysisLoad Combinations

• The forces produced by various ways discussed above need to be combined

in a proper manner,

• And need to be increased by a factor of safety (load factor) to produce the

desired level of safety,

• The combined load effect, sometimes called the required factored strength,

represents the minimum strength for which members need to be designed,

• Some examples of load combinations are given below (these combinations

are defined in building codes):

• 1.4D

• 1.2D + 1.0E + L + 0.2 S

where D: dead load, E: earthquake load, L: live load, S: snow load

Preparation for Computation

Preparation for Computations

Before starting an analysis it is important to prepare a set of

clear and complete computation to reduce the possibility of

errors and to be able to check your analysis in the future.

Each analysis needs to be clear and correct, without any

kind of numerical errors or missing units!!!!

Preparation for Computations

Suggestions:

• State the objective of the analysis

• Prepare a clear drawing of the structure, with loads and

dimensions

• Include all steps of your computations

• Check the results!!!!

• Verify that the direction of deflections is consistent with

the applied forces

Professor Özgür ÖZÇELİK, Ph.D.

Structural Systems in ArchitectureAR 361

Fall Semester

structural systems in architecture

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