analysis+of+structural+components

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Chapter : Civil Engineering For additional information on this subject, contact File Reference: CSE 106.02 PEDD Coordinator on 862-1026

Engineering Encyclopedia Saudi Aramco Desktop Standards

SAUDI ARAMCO ANALYSIS OF STRUCTURAL

COMPONENTS

Engineering Encyclopedia Civil Engineering: Basic Properties of Section

Analysis of Structural Components

Saudi Aramco Desktop Standards i

Section Page

OBJECTIVES ......................................................................................................... 1

Terminal Objective................................................................................................ 1

Enabling Objectives.............................................................................................. 1

INTRODUCTION ......................................................................................................... 3

ANALYZING BEAMS............................................................................................ 4

Definition And Function Of Beams............................................................. 4

Identifying Types Of Beams....................................................................... 6

Support Types................................................................................. 7

Pictures Showing Support Types .................................................... 8

Load Types ................................................................................................ 9

Static Determinacy .......................................................................... 9

Defining And Calculating Load-Shear-Moment Relationships ................. 10

Load .............................................................................................. 11

Shear ............................................................................................ 12

Bending Moment ........................................................................... 13

Sign Convention For Bending Moment ......................................... 14

Example 1: Reactions, Shear, And Bending Moment In A Simple Overhang Beam................................................................... 15

Drawing Shear And Moment Diagrams.................................................... 17

Procedures to Draw Diagrams ...................................................... 17

Example 2: Shear And Moment Diagram For Overhang Beam With Concentrated Load P ............................................................ 18

Example 3: Shear And Moment Diagram For Simple Beam With Uniform Load Distribution ................................................................... 21

Beam Diagrams - Key Points ................................................................... 24

Basic Shear And Moment Diagrams ........................................................ 24

Beam Diagrams - Superposition .............................................................. 26

Example 4: Beam Diagrams By Superposition........................................ 26



Saudi Aramco Desktop Standards ii

Calculating Beam Stresses ...................................................................... 28

Flexural (Bending) Stress ............................................................... 29

Shear Stresses............................................................................... 31

Example 5: Flexural And Shear Stresses In A Simple Overhang Beam . 34

Example 6: Flexural Stresses In A Simple T-Beam ................................ 36

Calculating Beam Deflection.................................................................... 38

Standard Tables For Beam Deflection............................................ 39

Conjugate Beam Method................................................................ 40

Example 7: Beam Deflection................................................................... 42

Finding And Using Beam Formulas ......................................................... 48

Beam Formula Work Aids............................................................... 48

Using Beam Formula Work Aids .................................................... 49

Example 8: Shear Moment And Deflection For Simple Beam With Uniform Load Distribution .................................................... 50

Superposition of Beam/Load Cases ............................................... 52

Example 9: Superposition Of Beam/Load Cases ..................................... 53

Example 10: Beam Diagrams And Deflection Using Superposition And Work Aids ..................................................................... 54

ANALYZING COLUMNS .................................................................................... 58

Introduction .............................................................................................. 58

Definition And Function............................................................................ 58

Identifying Column Materials And Sections.............................................. 59

Column Analysis And Considerations...................................................... 60

Identifying Column Types ........................................................................ 63

Column Type Based on Slenderness ............................................. 63

Column Type Based on Eccentricity............................................... 64

Column Type Based on Nature of Loading..................................... 65

Calculating Column Loads And Stresses................................................. 66

Ultimate Compressive Load ........................................................... 66

Compressive Yield Load................................................................. 67



Saudi Aramco Desktop Standards iii

Critical (Buckling) Load................................................................... 67

Failure Load ................................................................................... 68

Factor of Safety .............................................................................. 71

Allowable Load/Capacity ................................................................ 71

Example 11: Column Loads And Stresses.............................................. 72

Calculating Combined Bending And Axial Loading.................................. 76

Beam-Columns............................................................................... 76

Eccentrically Loaded Columns ....................................................... 77

Example 12: Combined Axial Load And Bending In Column .................. 79

ANALYZING FOOTINGS ................................................................................... 82

Identifying And Defining Types Of Footings............................................. 82

Analysis Procedure.................................................................................. 83

Identifying And Defining Applied Loads ................................................... 84

Identifying And Defining Eccentric Loads On Footings ............................ 84

Identifying And Defining Soil-Bearing Pressure On Footings................... 85

Centered/Concentric Load ............................................................ 85

Combined Vertical Load and Moment ........................................... 86

Critical Eccentricity........................................................................ 89

Partial Compression...................................................................... 89

Example 13: Soil-Bearing Pressure For A Square Footing ..................... 91

Identifying And Defining Stability Ratio On Footings................................ 93

Identifying And Defining Moment And Shear On Footings....................... 93

Example 14: Stability Ratio, Moment, And Shear For A Square Footing .............................................................................. 96

SUMMARY ....................................................................................................... 99

WORK AIDS ..................................................................................................... 101

PRACTICE PROBLEMS ............................................................................................. 149

GLOSSARY ..................................................................................................... 160



Saudi Aramco Desktop Standards iv

List of Figures

Figure 1. Beam (linear structural member having one or more supports) ...................... 4

Figure 2. Common Materials and Sections Used for Beams.......................................... 5

Figure 3a. Support Beams ............................................................................................. 7

Figure 3b. Support Beams ............................................................................................. 8

Figure 4. Beam Load Types ........................................................................................... 9

Figure 5. Load-Shear-Moment Relationships............................................................... 11

Figure 6. Determine Vertical Shear Force at Any Point Along Beam ........................... 12

Figure 7. Shear Force .................................................................................................. 13

Figure 8. Bending Moment ........................................................................................... 14

Figure 9. Overhang Beam ............................................................................................ 15

Figure 10. Shear and Moment at Support B................................................................. 17

Figure 11. Drawing Shear & Moment Diagrams........................................................... 18

Figure 12. Shear Diagram for the Overhang Beam in Algebraic Terms ....................... 19

Figure 13. Moment Diagram for the Overhang Beam in Algebraic Terms.................... 20

Figure 14. Uniform Load Distribution............................................................................ 21

Figure 15. Summing Vertical Forces ............................................................................ 22

Figure 16. Bending Moment ......................................................................................... 23

Figure 17. Cantilever Beam Subjected to Four Types of Load..................................... 25

Figure 18. Simple Support Beam Subjected to Three Types of Load .......................... 25

Figure 19. Cantilever Beam Loads............................................................................... 26

Figure 20a. Concentrated Load.................................................................................... 26

Figure 20b. Uniform Load............................................................................................. 27

Figure 21a. Cantilever Beam with Concentrated Load................................................. 27

Figure 21b. Cantilever Beam with Uniform Load.......................................................... 27

Figure 22. Combined Individual Cases......................................................................... 28

Figure 23. Stresses in a Beam ..................................................................................... 28

Figure 24. Flexural Stress ............................................................................................ 29



Saudi Aramco Desktop Standards v

Figure 25. Horizontal Shear in a Beam ........................................................................ 31

Figure 26. Plank Beams............................................................................................... 33

Figure 27. Simple Overhang Beam .............................................................................. 34

Figure 28. Maximum Shear at Center of Beam Section ............................................... 35

Figure 29. Cross Section of a Simple T-Beam ............................................................. 36

Figure 30. Elastic Curve of a Beam.............................................................................. 38

Figure 31. Deflections at Two Load Points (a & b) ....................................................... 42

Figure 32. Cover Plates Added on Flanges of Beam ................................................... 42

Figure 33a. Moment Diagram for Actual Beam ............................................................ 43

Figure 33b. Loading Diagram for Actual Beam ............................................................ 44

Figure 34a. Moment Diagram....................................................................................... 46

Figure 34b. EIM Diagram............................................................................................... 47

Figure 35. Superposition of Beam/Load Cases............................................................ 53

Figure 36. Beam Load Cases....................................................................................... 53

Figure 37. Example 10 Beam....................................................................................... 54

Figure 38. More Beam Load Cases.............................................................................. 54

Figure 39a. Beam Load Cases..................................................................................... 55

Figure 39b. More Beam Load Cases............................................................................ 56

Figure 40. Typical Structural Columns ......................................................................... 59

Figure 41. Typical Column Sections............................................................................. 60

Figure 42. Key Characteristic of a Single Column Loaded with P at Eccentricity e from the Centroidal Axis............................................................................... 62

Figure 43. Columns as Part of a Structural Frame ....................................................... 62

Figure 44a. Types of Columns Based on Slenderness ................................................ 64

Figure 44b. Types of Columns Based on Eccentricity .................................................. 65

Figure 44c. Typical Loadings for Beam-Columns......................................................... 66

Figure 45. Column Effective Length, Le = kL, Values of Column Coefficient, k ............ 69

Figure 46. Typical Plots of Pmax and σmax as a Function of the Column Slenderness Ratio, kL/r ................................................................................................... 70



Saudi Aramco Desktop Standards vi

Figure 47. Alternate Arrangements of Column Cross-Section...................................... 72

Figure 48. Biaxial Bending about x and y Axes of the Column Cross Section.............. 77

Figure 49. Eccentrically Loaded Column...................................................................... 78

Figure 50. Eccentrically Loaded Steel Column............................................................. 79

Figure 51. Types of Footings........................................................................................ 83

Figure 52. Loads Acting on a Support Footing ............................................................. 84

Figure 53. Footing Loads ............................................................................................. 85

Figure 54. Centered/Concentric Load .......................................................................... 86

Figure 55. Combined Vertical Load and Moment ......................................................... 87

Figure 56. Vertical Load and Moment .......................................................................... 88

Figure 57. Contact Area ............................................................................................... 88

Figure 58. Partial Compression.................................................................................... 90

Figure 59a. Soil-Bearing Pressure for Square Footing P = 100 k, M = 150 ft.k............ 91

Figure 59b. Maximum Soil-Bearing Pressure for Square Footing P = 100 k, M = 300 ft.k ............................................................................................................. 92

Figure 60. Stability Ratio .............................................................................................. 93

Figure 61. Moment and Shear on Footings .................................................................. 94

Figure 62. Critical Sections .......................................................................................... 95

Figure 63a. Wall and Footing Support Loads ............................................................... 96

Figure 63b. Wall and Footing Support Loads (Shear at Critical Section) ..................... 98

Figure 63c. Wall and Footing Support Loads (Moment at Critical Section) .................. 98

Figure 64. Overhang Beam ........................................................................................ 101

Figure 65. Work Aid 2, Beam with Loads ................................................................... 103

Figure 67. Work Aid 3, Beam with Uniform Load ....................................................... 105

Figure 68. ................................................................................................................. 105

Figure 69. ................................................................................................................. 106

Figure 70. Repeat of Figure 17, Work Aid 4, Basic Shear and Moment Diagrams..... 107

Figure 71. Repeat of Figure 18, Work Aid 4, Basic Shear and Moment Diagrams..... 108

Figure 72. Work Aid 6, Beam with Uniform and Concentrated Loads ........................ 110



Saudi Aramco Desktop Standards vii

Figure 73. Loads ......................................................................................................... 110

Figure 74. Load Cases ................................................................................................ 110

Figure 75. ................................................................................................................. 111

Figure 76. Repeat of Figure 23, Work Aid 7, Beam Stresses..................................... 112

Figure 77. Work Aid 8, Bending Moment and Shear Diagram.................................... 113

Figure 78. Work Aid 10, Calculating Beam Defelection.............................................. 116

Figure 79. Beam......................................................................................................... 117

Figure 80. Repeat of Figure 37. Work Aid 12 Beam................................................... 120

Figure 81. Repeat of Figure 38 More Beam Load Cases ........................................... 120

Figure 82a. Repeat of Figure 39a Beam Load Cases ................................................ 121

Figure 83b. Repeat of Figure 39b More Beam Load Cases ....................................... 122

Figure 84. Beam Diagrams and Formulas.................................................................. 127

Figure 85. Work Aid 15, Calculating Column Load and Stresses............................... 139

Figure 86. Work Aid 16, Calculating Combined Axial Load and Bending in Column.. 142



Saudi Aramco Desktop Standards 1

OBJECTIVES

Terminal Objective

After completing this module, the participant will be able to perform basic structural analysis for beams, columns, and footings.

Enabling Objectives

In order to meet the terminal objective, given the appropriate information, the participant will be able to:

• Identify types of beams.

• Define load-shear-moment relationships.

• Calculate load, shear, moment, and deflection in a beam.

• Draw shear and moment diagrams for simple beams.

• Calculate stresses and deflections in simple beams subjected to concentrated and distributed loads.

• Locate and use formulas to calculate moments, shears, and reactions in beams with selected end-support conditions.

• Identify column materials and sections.

• Calculate loads and stresses in columns subjected to axial, transverse and eccentric loads.

• Calculate combined bending and axial loading in columns.

• Identify footing types.

• Define applied and eccentric loads and soil-bearing pressure on footings.

• Calculate stresses and soil bearing pressures for eccentrically loaded footings.

• Define stability ratio, moment and shear on a footing.

• Calculate stability ratio, moment and shear in a footing

Note: Definitions of words in italics are contained in the Glossary




This Page Intentionally Blank




INTRODUCTION

This is the third and final module in the course, CSE-106, Basic Strength of Materials. The first module covered the various section properties needed to solve common civil/structural engineering problems. The second module identified and calculated the basic loads and stresses encountered in these problems. In this module, you will use this information to analyze various components of a structure such as beams, columns, and footings. The detailed analysis and design of these components and their connections are discussed in the applications courses CSE-104, CSE-108, and CSE-109 for timber, reinforced concrete, and steel structures, respectively.




ANALYZING BEAMS

Beams are one of the basic components of a civil engineering structure. To perform the structural analysis of beams you need to cover:

• Definition and function of beams.

• Types of beams based on the support conditions, load types, and whether or not they are statically determinate.

• Load-shear-moment relationship.

• Shear and bending moment diagrams.

• Beam stresses.

• Beam deflections.

Definition and Function of Beams

A beam is a linear structural member having one or more supports. It carries transverse or lateral loads, that is, loads perpendicular to its longitudinal axis. A beam is an efficient structural member used to support loads over an open span or clear area. Beams are commonly used for floors, roofs, and bridges.

Figure 1. Beam (linear structural member having one or more supports)

Span

Supports

Loads

x

y

Span

Supports

Loads

x

y

Span

Supports

Loads

x

y




Common materials and sections used for beams:

• Rolled shape structural steel.

• Reinforced or prestressed concrete.

• Structural grade timber.

• Composite sections comprising: - Structural steel and concrete (a) - Structural steel and timber (b)

Figure 2. Common Materials and Sections Used for Beams

Two main factors are involved in the analysis or design of a beam:

• Strength – the requirement to keep the stresses in a material below a specified level to ensure an adequate factor of safety against material failure.

• Serviceability – the requirement to keep the deflection less than a specified tolerable limit to ensure that a beam does not sag excessively. Excessive sag can impair its function or cause discomfort to people.

Therefore, the main focus in the analysis or design of a beam is to calculate and evaluate the stresses and deflection that will occur when it is loaded.

(a) Steel and Concrete

StructuralSteel Shape

Reinforced Concrete

(b) Steel and Timber

Steel Plate

Timber Sections



Reinforced Concrete



Reinforced Concrete


Steel Plate

Timber Sections


Steel Plate

Timber Sections




Identifying Types of Beams

The tables given in engineering handbooks for analyzing beams are often organized according to types or classes. For the purpose of structural analysis beams are usually classified according to:

• Support conditions.

• Type of loading.

• Static determinacy – whether they can be analyzed by the principle of statics alone.

The following are examples of the various types of beams usually encountered in civil engineering practice.




Support Types

Figure 3a. Support Beams

7

6

5

4

3

2

1

A cantilever beam with its free end restrained against rotation.

Guided Cantilever Beam

A cantilever beam fixed at one end and with a simple support at the other end.

Propped Cantilever Beam

A beam that is restrained from rotation and movement at its ends.

Fixed End Beam

A beam that rests on more then two supports.

Continuous Beam

A beam resting on two or more supports, which has one or both ends projecting beyond the support.

Overhanging Beam

A beam supported (fixed) at one end only

Cantilever Beam

A beam that nests on simple supports at each end.

Simple Beam

7

6

5

4

3

2

1

A cantilever beam with its free end restrained against rotation.

Guided Cantilever Beam

A cantilever beam fixed at one end and with a simple support at the other end.

Propped Cantilever Beam

A beam that is restrained from rotation and movement at its ends.

Fixed End Beam

A beam that rests on more then two supports.

Continuous Beam

A beam resting on two or more supports, which has one or both ends projecting beyond the support.

Overhanging Beam

A beam supported (fixed) at one end only

Cantilever Beam

A beam that nests on simple supports at each end.

Simple Beam

LSimple Support

LSimple Support

LSimple Support

FixedEnd

FreeEnd

FixedEnd

FreeEnd

OverhangOverhangOverhang




Pictures Showing Support Types

Hinge Supports

Roller Support

Figure 3b. Support Beams




Load Types

Four types of loads can act on a beam: the concentrated load, or point load; the uniformly distributed load; the nonuniformly distributed load; and the induced moment load. These beam loads are illustrated in Figure 4.

Figure 4. Beam Load Types

Static Determinacy

Statically determinate beams are those beams whose reactions can be found from the equations of equilibrium:

Σ Fx = 0

Σ Fy = 0

Σ Mz = 0

Examples include simple beams, cantilevers, and overhanging beams on two supports as shown in Figure 3, items 1, 2, and 3, respectively.

b) Uniform Loada) Concentrated Load

c) Nouniform Load d) Induced Moment Load

P w = lb/ft

M

b) Uniform Loada) Concentrated Load

c) Nouniform Load d) Induced Moment Load

P w = lb/ft

M




Statically indeterminate beams are beams whose reactions cannot be found from the equations of equilibrium only, but require additional equations to determine the reactions, as in continuous and fixed end beams and propped and guided cantilever beams (Figure 3, items 4 to 7).

The analysis of a statically indeterminate beam is performed with the static equilibrium equation and stiffness or flexibility relationships for the beam. Therefore, the analysis of a statically indeterminate beam depends on the material and section properties (E, I) of the beam.

Defining and Calculating Load-Shear-Moment Relationships

The lateral loads supported by a beam cause shear forces and bending moments to develop along the beam. A designer needs to calculate these forces and moments and the stresses that result in order to determine whether the beam can safely support a given load. These forces and moments are calculated by using the principles of equilibrium applied to a particular portion of the beam. The relationship between the loads, shears, and moments in a beam and the calculation procedures are discussed below.

Shear forces and bending moments in a beam can be calculated by the principles of equilibrium applied to any part of the beam as illustrated in Figure 5. For a specific loaded beam, the shear and moment can be represented graphically by a plot of the values of the shear force or bending moment along the x-axis of the beam to produce a Shear Diagram and a Moment Diagram. These diagrams and the associated formulas are provided in standard references for certain common beam types.




Figure 5. Load-Shear-Moment Relationships

Load (P or w)

The beam loading can be either concentrated, P, or distributed, w(x), along the longitudinal axis. These are the basic load types illustrated in Figure 4. The concentrated load has units of pounds or Newtons (lb, N), and the distributed load has units of lb/ft or N/m. The beam may also carry a combination of these basic loads.

Distributed loads and concentrated loads are positive when they act downward on the beam and negative when they act upward. A couple acting as a load on a beam is positive when it is counterclockwise and negative when it is clockwise.

Internal Forces at x- Shear Force, V- Bending Moment, M

Lateral Deflection, y

b) Beam Free-Body

y

VM

x

w

Reaction Support

LateralLoads Beam

x - axis

y - axis

a) Beam Loading

Pw

z

Internal Forces at x- Shear Force, V- Bending Moment, M

Lateral Deflection, y

b) Beam Free-Body

y

VM

x

w

Reaction Support

y

VM

x

w

Reaction Support

LateralLoads Beam

x - axis

y - axis

a) Beam Loading

Pw

z

LateralLoads Beam

x - axis

y - axis

a) Beam Loading

Pw

z




Shear (V)

Shear is the internal force required to maintain the equilibrium on any part of a beam. Figure 6 shows how to determine the vertical shear force at any point along the beam.

Figure 6. Determine Vertical Shear Force at Any Point Along Beam

The vertical shear force acting on a section at any point, x = xo, along the longitudinal axis of a beam is the algebraic sum of the forces on one side of the section. Although the forces on either side of the section may be used, the value of V will be the same. However, for convenience we generally deal with the forces to the left of the section. These forces include the applied loads and the reactions, with their proper signs. The shear in a beam has units of pounds or newtons.

The change in shear (V) along the longitudinal axis (x) of the section has the following relationship to the load (w):

wdxV ,dxdVw ∫−=−=

a) Loaded Beam

b) Free-Body:Shear found by summingvertical forces on either sideof the section:

xo

Pw

x

yPw

Rl RrV

V

0yF =∑

a) Loaded Beam

b) Free-Body:Shear found by summingvertical forces on either sideof the section:

xo

Pw

x

yPw

Rl RrV

V

0yF =∑




Shear is negative when it causes the right side of the beam to move down relative to the left side, and shear is positive when it causes the left side of the beam to move down relative to the right side.

Figure 7. Shear Force

Bending Moment

Bending moment (M) is the internal moment at a section required to maintain the equilibrium of any part of the beam.

The bending moment at any section of a beam is the algebraic sum of the moments on one side of the section. Moments include the moments of the applied external loads, the internal forces, and the reactions, with their proper sign. For equilibrium:

ΣMz = 0

Moment has units of pound-feet or newton-meters.

The relationships between bending moment and shear are:

VdxM ,dxdMV ∫==

Right LeftLeft Right

ShearForce

Positive ShearNegative Shear

vvRight Left

Left Right

ShearForce

Positive ShearNegative Shear

vv




That is, shear is the change in the bending moment and the bending moment is the area under the shear diagram. The relationships between bending moment and load are:

wdxM ,dx

Mdw 2

2

∫∫−=−=

Bending moment is positive when the upper part of the beam is in compression and the lower part is in tension. Conversely, bending moment is negative when the upper part of the beam is in tension and the lower part is in compression.

Sign Convention For Bending Moment

Figure 8. Bending Moment

-M

b) Negative Bending Moment

Lower fibersIn compression Upper fibers in tension

P

+M

Upper fibers in Compression

Lower fibers in tension

P

a) Positive Bending Moment

VCT

-M

b) Negative Bending Moment

Lower fibersIn compression Upper fibers in tension

P

+M

Upper fibers in Compression

Lower fibers in tension

P

a) Positive Bending Moment

VCT

CT




Example 1: Reactions, Shear, and Bending Moment in A Simple Overhang Beam

(Page 1 of 4)

For the overhang beam shown below, calculate:

a) Reactions R1 and R2.

b) The shear and bending moment at mid-span (C), which is 5 ft from left support (A) of the beam.

c) The shear and bending moment at support (B).

Figure 9. Overhang Beam

Example 1: Solution

a) Solve for reactions R1 and R2.

Taking moments about R1:

Σ M1 = 0 = (1200)(16) – R2 × 10 10R2 = 19,200

R2 = 1920 lb

P=1200 lb

A B

C

10 ft 6 ftR1

R2

P=1200 lb

A B

C

10 ft 6 ftR1

R2




Σ Fy = 0 = 1200 + R1 – R2 (Page 2 of 4)

R1 = 1920 – 1200

R1 = 720 lb

Note that R1 acts downward and R2 upward. The beam must be held down at R1 to be in equilibrium.

b) Solve for shear and moment at C.

(i) Draw free-body diagram of a beam to left of point C. A force (V) and moment (M) are required to keep it in equilibrium, where they are shown positive as per adopted sign convention.

(ii) Sum vertical forces to obtain shear.

ΣFy = 0, ∴V = −R1 = −720 lb

(iii) Sum moment about C to obtain bending moment.

ΣMC = 0, M + 5×720 = 0

M = −3600 lb-ft

c) Shear and moment at support B.

The shear is discontinuous at a support point. Again, sum the vertical forces and the moments on the free-body diagrams as shown:

R1 = 720 lbs

A V

C5 ft

R1 = 720 lbs

A V

C5 ft




(Page 3 of 4)

Figure 10. Shear and Moment at Support B

Σ F = 0, VB1 = −R1 = −720 lb VB2 = P = 1200 lb

R2 = − VB1 + VB2 = 1920 lb

Σ M = 0, MB1 = −10(R1) = − 10 × 720 = −7200 lb-ft

MB2 = −6(P) = − 6 × 1200 = −7200 lb-ft

Note MB2 = MB1

Drawing Shear and Moment Diagrams

A shear diagram is used to indicate the value of the vertical shear force at any point along the longitudinal axis of a beam. Similarly, a moment diagram shows the variation of the bending moment along the longitudinal axis of a beam.

Procedures to Draw Diagrams

The procedures to draw the shear and moment diagrams for beams are illustrated by the following two examples:

P = 1200 lbVB1 VB2

MB2MB1

VB1VB2

MB2MB1

R1

R2

P = 1200 lbVB1 VB2

MB2MB1

VB1VB2

MB2MB1

R1

R2




Example 2: Shear and Moment Diagram for Overhang Beam with Concentrated Load P

(Page 1 of 3)

Draw the shear and moment diagrams for the overhang beam in Figure 11.

Figure 11. Drawing Shear & Moment Diagrams

Again solving for R2 by Σ M1 = 0

( )L

aLPR2+

=

And summing vertical forces, Σ Fy = 0, for R1

( )L

PaL

aLPPLRPR 21 −=+−

=−=

Summing vertical forces on the free-body of the beam up to any point (x) between supports yields:

LPaRV 1 −=+= 0 ≤ x < L

ABC

L a

R1R2

x

P

ABC

L a

R1R2

x

ABC

L a

R1R2

x

P

R1

x

V M

R1

x

V M




(Page 2 of 3)

Similarly, summing forces on the free-body of beam to the right of any point (x) on the overhang results in:

V2 = +P L < x < L+a

Therefore, the shear diagram for the overhang beam in algebraic terms is as follows:

Figure 12. Shear Diagram for the Overhang Beam in Algebraic Terms

The bending moment at any point (x1) between supports is:

LPaxxRM 1 −== 0 ≤ x ≤ L

⎟⎠⎞

⎜⎝⎛ ≤≤=−= Lx0V

LPa

dxdMNote

At support 2, x = L and

PaM −=

+

--PaL

+P

+

--PaL

-PaL

+P

L+a-x

VPM

L+a-x

VPM




(Page 3 of 3)

On the overhang:

M = R1x + R2(x−L)

M = L)(xL

PaPL

Pax−⎟

⎠⎞

⎜⎝⎛ ++−

= −P(L+a) + Px L ≤ x ≤ L+a

At support 2, x = L ∴ M = −Pa

⎟⎠⎞

⎜⎝⎛ +≤≤=+= aLxLVP

dxdMNote

Therefore, the moment diagram is as follows:

Figure 13. Moment Diagram for the Overhang Beam in Algebraic Terms

Note that since the bending moment is negative (tension in the top fiber of the beam), it is plotted below the line.

R1x

V M

R2

L x-L

R1x

V M

R2

L x-L

- Pa- Pa




Example 3: Shear and Moment Diagram for Simple Beam with Uniform Load Distribution

(Page 1 of 3)

Draw the shear and moment diagrams for the beam shown.

Load:

Total Load = W = woL

LWwo =

Reactions:

Solve for Rb by summing moment about A

2Lw

2WR

W2LLR

0M

ob

b

A

==

=

=∑

Summing vertical forces,

2Lw

2WR-WR

0F

oba

Y

===

=∑

Figure 14. Uniform Load Distribution

Shear:

2WRVB At

2WRV AAt

bb

aa

−=−=

==

x

wo

V

M

Ra

x

wo

V

M

Ra

W = total loadx

L

A B

Ra Rb

wo

W = total loadx

L

A B

Ra Rb

wo




(Page 2 of 3)

At x, summing forces on the free body:

ΣF = Ra – wox – V = 0 V = Ra – wox

= ⎟⎠⎞

⎜⎝⎛ −=−

L2x1

2W

LWx

2W

For x = 2L , V = 0 .

Shear Diagram:

Figure 15. Summing Vertical Forces

-W/2 =L/2

L/2W/2 =

2Lwo

o-w L2

-W/2 =L/2

L/2W/2 =

2Lwo

o-w L2




(Page 3 of 3) Bending Moment:

At x, summing moments on the free-body to the left of x:

ΣMx = Rax − wo× x ×2x − M(x) = 0

M(x) = Rax – wo 2x2

= ⎟⎟⎠

⎞⎜⎜⎝

⎛−

2x

LW

2Wx 2

M(x) = ⎟⎠⎞

⎜⎝⎛ −

Lx1

2Wx

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛ −=−= V

L2x1

2W

LWx

2W

dxdMNote

Maximum Moment: Mmax at V = 0

V = 0 at x = 2L

∴Mmax = ( )( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛−

2LL1

22WL

Mmax = 8Lw

8WL 2

o=

Moment Diagram:

Figure 16. Bending Moment

L/2Mmax.=0.125 WL

= 0.125 woL22

ow L=8

L/2Mmax.=0.125 WL

= 0.125 woL2L/2Mmax.=0.125 WL

= 0.125 woL22

ow L=8




Beam Diagrams - Key Points

Note the following points from the foregoing examples:

• Positive values of shear and moment are plotted above the base line and negative values are plotted below the base line.

• The value of shear changes abruptly at concentrated loads and reactions and is indicated by a vertical line. From the relation V = dM/dx the slope of the moment diagram changes abruptly at concentrated loads.

• Where there is no change in load within a beam segment, the magnitude of shear is constant, and the shear diagram is horizontal. Also, since V = dM/dx, the slope of the moment diagram is constant where the shear is constant (positive slope where shear is positive, negative slope where shear is negative.)

• Since V = dM/dx, the maximum moment occurs where the shear is zero.

• The change of moment between any two points along a beam is equal to the area of the shear diagram between the two points.

Basic Shear and Moment Diagrams

Figure 17 shows the diagrams and expressions for the shear and bending moments for a cantilever beam subjected to four basic types of load. Figure 18 shows a simple-support beam for three load cases. You can use these diagrams and expressions for more complex, multiple-load cases with the aid of superposition, as illustrated in Example 4.




L

w

L

P

Lx

Moa) Loads

+

-Vx = 0 -P

+-

Vx = -P

+

-

Vx = -wx -wL

b) Shear Diagrams

-Mo

+-

Mx=-Mo

-PLMx=-Px2wxM

2

x−

=2

wL- 2

wo

L

+-

o

2

x w2LxV −

= 2Lwo−

6Lw- 2

o3ox x

6LwM −

=

c) Bending Moment Diagrams

L

w

L

w

L

P

L

P

Lx

Mo

Lx

Moa) Loads

+

-Vx = 0

+

-Vx = 0 -P

+-

Vx = -P

+-

Vx = -P

+

-

Vx = -wx -wL

+

-

Vx = -wx -wL

-

Vx = -wx -wL

b) Shear Diagrams

-Mo

+-

Mx=-Mo

+-

Mx=-Mo

-PLMx=-Px -PLMx=-Px2wxM

2

x−

=2

wL- 2

2wxM

2

x−

=2

wL- 2

wo

L

+-

o

2

x w2LxV −

= 2Lwo−

6Lw- 2

o3ox x

6LwM −

=

wo

L

+-

o

2

x w2LxV −

= 2Lwo−

+-

o

2

x w2LxV −

= 2Lwo−

6Lw- 2

o3ox x

6LwM −

=6

Lw- 2o3o

x x6LwM −

=

c) Bending Moment Diagrams

Figure 17. Cantilever Beam Subjected to Four Types of Load

Figure 18. Simple Support Beam Subjected to Three Types of Load

a bP

L

wwo

Total Load

LL

2LwW o=

-

+ +-

+-

LPb

LPa-

2wL

2wL-

3W

32W-

6Lw

2LxwV o

2o

x +−=

LPabMmax = 8

wLM2

max = Mmax=0.1283 WL( )xL

2wxMx −=

( )222x xL

3LWxM −=

a) Loads

b) Shear Diagrams

c) Moment Diagrams

a bP

L

wwo

Total Load

LL

2LwW o=

-

+

-

+ +-

+-

LPb

LPa-

2wL

2wL-

3W

32W-

6Lw

2LxwV o

2o

x +−=

LPabMmax = 8

wLM2

max = Mmax=0.1283 WL( )xL

2wxMx −=

( )222x xL

3LWxM −=

a) Loads

b) Shear Diagrams

c) Moment Diagrams




Beam Diagrams - Superposition

In certain situations in structural analysis or design you may encounter a beam with more than one load or with more than one type of load. The use of superposition is an effective method to obtain the shear and moment diagrams for such a beam. The steps involved in this approach are: 1. Separate problem into beams with individual loads. 2. Draw beam diagrams for each load case. 3. Construct beam diagrams for the overall beam loads by

combining diagrams for individual load cases.

Example 4: Beam Diagrams by Superposition

Draw the bending moment diagram for the cantilever beam with the loads shown.

Figure 19. Cantilever Beam Loads

Example 4: Solution

Step 1 Separate into two simpler individual cases:

a) Concentrated load, P = 5 kips L = 10 ft

Figure 20a. Concentrated Load

5 kips2 kips/ft

10 ftA

B

5 kips2 kips/ft

10 ftA

B

5 kips

10 ft

5 kips

10 ft




(Page 1 of 2)

b) Uniform load, wo = 2 kips/ft L = 10 ft

10 ft

2 kips/ft

10 ft

2 kips/ft

Figure 20b. Uniform Load

Step 2 Construct the moment diagrams for the individual load cases based on Figure 17.

a) Cantilever beam with concentrated load Mmax = PL = 5(10) = 50 kip-ft

50 kip-ft- 50 kip-ft-

Figure 21a. Cantilever Beam with Concentrated Load

b) Cantilever beam with uniform load

Mmax = ( ) ftkip1002

21022

2wL −==

100 kip-ft- 100 kip-ft-

Figure 21b. Cantilever Beam with Uniform Load




(Page 2 of 2)

Step 3 Construct the composite diagram by combining individual cases.

Mmax = 50 + 100 = 150 kip-ft

150 kip-ft

(a)

(b)

(c) = (a) + (b)

2 kips/ft

150 kip-ft

(a)

(b)

(c) = (a) + (b)

2 kips/ft

Figure 22. Combined Individual Cases

Calculating Beam Stresses

The analysis steps for determining the stresses in a beam are illustrated in Figure 23.

Beam

Lateral Loads

Bending Moment Shear Force

Bending orFlexural Stress

Vertical and Horizontal

Shear Stress

ConcentratedLoads

DistributedLoads

Figure 23. Stresses in a Beam




The bending moments and shear forces in a beam resulting from the transverse loads give rise to stresses in the beam as follows:

• Bending moment - normal bending or flexural stress.

• Shear force - vertical and horizontal shear stresses.

The procedure for calculating and evaluating these stresses is discussed below.

Flexural (Bending) Stress

When a beam is subjected to bending, the fibers on one side of the beam elongate and are in tension. The fibers on the other side shorten and are in compression. The plane that remains the same length, and that has neither compressive nor tensile stresses, is called the neutral axis. For a homogeneous beam, the neutral axis passes through the centroid of the beam cross section.

Figure 24. Flexural Stress

M M

Neutral Axis (N.A.)

N.A.

Compressioncc

ct

Tension

NeutralAxis

y

f(y)

a) Bending Moment b) Deformed Beam c) Stress Distribution

yM M

Neutral Axis (N.A.)

M M

Neutral Axis (N.A.)

N.A.

Compressioncc

ct

Tension

NeutralAxis

y

f(y)

NeutralAxis

y

f(y)

a) Bending Moment b) Deformed Beam c) Stress Distribution

y




The flexural stress in the beam is given by the formula:

IMy

=σ

where: σ = flexural stress (tension or compression) y = distance from neutral axis to fiber under consideration M = bending moment I = moment of inertia

Thus, the unit stress σ is directly proportional to the distance y from the neutral axis. The maximum stress in compression or tension is at the outermost fibers from the neutral axis, a distance cc or ct, respectively. That is, at y = cc and y = ct:

IMcc

c =σ compressive stress

IMc t

t =σ tensile stress

If the section is symmetrical, the flexural stress is expressed in terms of the section modulus, S

SM

=σ

where:

tc cI

cIS == the section modulus

The modulus of rupture is the flexural stress fr = Mmax/S, where the bending moment Mmax is the maximum value at rupture. At rupture, the variation of unit stress is no longer linear; therefore, the modulus of rupture is a fictitious value higher than the true stress in the beam. Modulus of rupture is useful in computing the ultimate beam strength of various species and grades of wood that are brittle under normal conditions.




Shear Stresses

In addition to the flexural stresses, a beam develops longitudinal (horizontal) and transverse (vertical) shear stresses due to the shear force, V.

Figure 25. Horizontal Shear in a Beam

Shear stresses occur only if the bending moment varies along the beam. Any beam, or portion of the beam’s length, that has a uniform bending moment has no vertical shear and therefore no horizontal shear.

Unlike flexural stress, the horizontal shear stress is zero at the outer fibers of the beam and maximum at the neutral axis of the beam. It tends to cause one part of the beam to slide past the other.

The horizontal and vertical shear stresses at any point in the beam are equal. The shear stress at any point in the cross-section of a beam is:

IbVQ

=τ

where:

V = external vertical shear on beam (lb, N)

Neutral AxisArea

yyd

b

ShearSection

FlexuralStress

ShearStress

ShearForce (V)

Element A

BendingMoment

M

VerticalShear

τ

Horizontal ShearElement A

(a) (b) (c)

Neutral AxisArea

yyyd

b

ShearSection

FlexuralStress

ShearStress

ShearForce (V)

Element A

BendingMoment

M

VerticalShear

τ

Horizontal ShearElement A

(a) (b) (c)




I = moment of inertia of whole section (in.4, mm4)

b = width of section at plane where stress is desired (in., mm)

A = area of section beyond plane where stress is desired (in.2, mm2)

y = distance of center of gravity of area to neutral axis (in., mm)

Q = yA , the static moment of the area beyond the plane being considered, taken about the neutral axis.

The average shear stress, τavg, on a section of the beam is:

AV

avg =τ

For a beam with a rectangular section; b x d:

A = bd, Qmax = 8

bd2

(for section at neutral axis) and

I = 12bd3

Therefore, τmax = bdV1.5 , τavg =

bdV

or

τmax = 1.5 τavg

A way to visualize the horizontal shear stress in a beam is to consider two smooth planks forming a simple beam as shown in Figure 26. Two cases are considered:




Lower fiber of upper plank stretchUpper fibers of lower plank shortenPlanks slide past each other

• Shear stress along this line

a) Planks Not Connected

b) Planks Connected

Lower fiber of upper plank stretchUpper fibers of lower plank shortenPlanks slide past each other

• Shear stress along this line• Shear stress along this line

a) Planks Not Connected

b) Planks Connected

Figure 26. Plank Beams

In the first case, the planks slide past each other because they are not connected. The bending resistance of the beam is the sum of the bending resistances of the two planks acting independently.

In the second case, if the planks are connected along their line of contact so as to develop adequate shear strength, the two planks can be made to act as one beam.




Example 5: Flexural and Shear Stresses in a Simple Overhang Beam

(Page 1 of 2)

Calculate the maximum flexural stress and the maximum shear stress in the beam given in Example 1. The beam has a rectangular cross-section, b = 6 in. and d = 10 in.

Figure 27. Simple Overhang Beam

Since the beam section is uniform, the maximum flexural stress occurs at the section with the maximum bending moment. The maximum shear stress is at the location of the maximum vertical shear.

From Example 1:

Mmax = 7200 lb-ft (at support B)

Vmax = 1200 lb

Flexural stress: SM

IMcσ ==

For rectangular section, ( ) 322

in. 1006106

6bdS ===

Max. flexural stress, 23max lb/in. 864

in. 100in. lb 12 x 7200

==σ

Maximum tensile stress is at the top of section over support B.

Maximum compression occurs at bottom of beam at support B.

A B10 ft 6 ft

720 1920

P = 1200 lb

N A

6 in.

5 in.

5 in.10 in.

A B10 ft 6 ft

720 1920

P = 1200 lb

A B10 ft 6 ft

720 1920

P = 1200 lb

N A

6 in.

5 in.

5 in.10 in.N A

6 in.

5 in.

5 in.10 in.




(Page 2 of 2)

Shear stress, IbVQ

=τ AV

avg =τ

For a rectangular section the maximum shear occurs at the center of the beam section (where the neutral axis is).

Qmax = 2bd d bd

2 4 8× =

I = 12bd3

t = b

2

max3

1bd 3 V8 V1 2 bdbd b12

τ = = ××

Figure 28. Maximum Shear at Center of Beam Section

avg1.5tAV1.5 ==

( ) psi 30106

1200 1.5 ==

A

b

d/4

d

Neutralaxes

τavg

τmax

A

b

d/4

d

Neutralaxes

τavg

τmax

τavg

τmax




Example 6: Flexural Stresses in a Simple T-Beam

(Page 1 of 2)

A standard rolled T-section (WT-6 in.-wide flange, 80.5 lb) is used as a beam. It is 100 in.-long, supported on each end, and bears a concentrated load of 10,000 lb at the mid-span. Find the maximum tensile and maximum compressive flexural stresses.

Figure 29 shows the cross-section of this beam, together with its load diagram.

Figure 29. Cross Section of a Simple T-Beam

Figure 18 shows that the formula for the bending moment for this type of beam is:

2Lba,

LPabMmax ===

Therefore,

( )( ) lb250,000in.4

10010,0004

PLMmax −===

P = 10,000 lb50 in.

100 in.5,000 lb5,000 lb

A

A

12.515 in.1.47 in.

N.A.

5.47 in.0.905 in.

1.485 in.

6.94 in.

Section AA

y = 1.47 in.I = 62.6 in.4

P = 10,000 lb50 in.

100 in.5,000 lb5,000 lb

A

A

12.515 in.1.47 in.

N.A.

5.47 in.0.905 in.

1.485 in.

6.94 in.

Section AA

y = 1.47 in.I = 62.6 in.4y = 1.47 in.I = 62.6 in.4




(Page 2 of 2)

Since the bottom portion of the beam is stressed in tension, substituting appropriate known values into the formula for flexural stress gives:

( )( )62.6

5.47250,000I

Mcσ t == = 21,845 psi (tension)

The top portion of the beam is in compression,

( )( )62.6

1.47250,000I

Mcσc == = 5870 psi (compression)




Calculating Beam Deflection

The module thus far has reviewed the types of beams commonly encountered in structural analysis and the relationships that exist between the loads, shear, and bending moment in a beam. You now know how to draw the shear and bending moment diagrams and to calculate the shear and flexural stresses at any section in the beam and at any point in that section. Therefore, you have completed the first requirement for the structural analysis of beams. The module now addresses the second requirement, calculating the beam deflections.

A loaded straight beam deforms due to shortening of the fibers in compression and elongation of the fibers in tension. As illustrated in Figure 30, the neutral axis retains its original length but curves, and the beam deflects downward under gravity loads. The longitudinal axis of the beam is called the elastic curve.

Figure 30. Elastic Curve of a Beam

Pw

x

y LateralLoads

L

Neutral Axis

x Beam Neutral Axis

Elastic Curve forDeformed Beam

¦ ymax¦-y(x)

θ

ρ

Internal Force:- Bending Moment, M- Shear Force, V

Beam Properties:- Elastic Modulus, E- Moment of Inertia, I

Pw

x

y LateralLoads

L

Neutral Axis

x Beam Neutral Axis

Elastic Curve forDeformed Beam

¦ ymax¦-y(x)

θ

ρ

Internal Force:- Bending Moment, M- Shear Force, V

Beam Properties:- Elastic Modulus, E- Moment of Inertia, I




The curvature (1/ρ), slope (θ), and deflection (y) of the beam are related to the bending moment (M) along the longitudinal axis (x) of the beam:

• Curvature, 1/ρ = 2

2

dxyd =

EIM , ρ = radius of curvature of beam

• Slope, θ = dxdy = ∫ dx

EIM

• Deflection, y(x) = ∫∫ dxEIM

The beam properties needed to calculate the deflection of a beam are the elastic modulus, E, and the moment of inertia, I.

Standard textbooks provide several methods for calculating beam deflection using the relationships in the above equations. The methods most applicable to the structural analysis and design of beams are standard tables and conjugate beams discussed in the following sections.

Standard Tables For Beam Deflection

The simplest and quickest way to determine the deflection of a beam is to use the formulas listed in standard beam tables provided in the Work Aids or from engineering handbooks such as the AISC Manual for Steel Construction. The use of these standard tables is covered in the section on Beam Formulas. These tables include only simple cases involving beams of uniform cross-sections. For cases not covered including beams having non-uniform cross-section, you must use a procedure such as the conjugate beam method to calculate deflection of the beam.




Conjugate Beam Method

The conjugate beam method changes a deflection problem into one of drawing moment diagrams. The method is able to handle beams of varying cross-sections and materials.

Step 1: Draw the moment diagram for the beam as it is actually loaded.

Step 2: Construct the M/EI diagram by dividing the value of M at every point along the beam by the product of EI at that point. If the beam is of constant cross-section, EI will be constant and the M/EI diagram will have the same shape as the moment diagram. However, if the beam cross-section varies with x, then I will change and the M/EI diagram will differ from the moment diagram.

Step 3: Draw a conjugate beam of the same length as the original beam. The material and cross-sectional area of this conjugate beam are not relevant.

• If the actual beam is simply supported at its end, the conjugate beam will be simply supported at its ends.

• If the actual beam is simply supported away from its ends, the conjugate beam has hinges at the support points

• If the actual beam has free ends, the conjugate beam has built-in ends.

• If the actual beam has built-in ends, the conjugate beam has free ends.

Step 4: Load the conjugate beam with the M/EI diagram. Find the conjugate reactions by methods of statics. Use the superscript (*) to indicate conjugate parameters.

Step 5: Find the conjugate moment at the point where the deflection is wanted. The deflection is numerically equal to the moment as calculated from the conjugate beam forces.




The relationships between the conjugate beam and the actual beam are as follows:

Conjugate Beam Actual Beam

Simple End* Simple End Hinge* Interior Support Free End* Built-in End Built-in End Free End

Load, w* EIM

Reaction, R* End Slope, θ Shear, V* Slope, θ Moment, M* Deflection, y

*Indicates parameters related to conjugate beam.




Example 7: Beam Deflection

(Page 1 of 6)

Use the conjugate beam method to determine the deflections at the two load points a and b for the beams shown below.

A. Uniform beam with constant value of EI = 2.356 × 106 lb-in.2.

Figure 31. Deflections at Two Load Points (a & b)

B. Nonuniform beam - Cover plates added on flanges of beam between the loads so that

EI of the portion of the beam, a-b, is doubled.

Figure 32. Cover Plates Added on Flanges of Beam

80 lb 120 lb

a b

30 in. 40 in. 20 in.

80 lb 120 lb

a b

30 in. 40 in. 20 in.

80 lb 120 lb

a b

30 in. 40 in. 20 in.

EI 2 EI EI

EI = 2.355 x 106 lb-in.2

Cover Plates80 lb 120 lb

a b

30 in. 40 in. 20 in.

EI 2 EI EI

EI = 2.355 x 106 lb-in.2

Cover Plates




Solution: Example 7A - Deflection of Uniform Beam

(Page 2 of 6)

80 lb 120 lb

a b

30 in. 40 in. 20 in.

80 lb 120 lb

a b

30 in. 40 in. 20 in.

Repeat of Figure 31

Step 1. Moment diagram for actual beam:

M

2400 lb-in.

M

2400 lb-in.

Figure 33a. Moment Diagram for Actual Beam

Steps 2 – 4: Since the cross-section is constant, the conjugate load has the same shape as the original moment diagram. The peak load on the conjugate beam is:

w* = EIM

= 26 in-lb 10 x 2.356in.-lb 2400

= 1.019 x 10-3 in.-1




(Page 3 of 6) The conjugate reaction R1* is found by the following method. The loading diagram is assumed to be made up of a rectangular load and two triangular loads.

Figure 33b. Loading Diagram for Actual Beam

Area x* M about Pt. 1*

(1) 40(1.019 x 10-3) = 4.076 x 10-2 x 50 = 203.8 x 10-2

(2) 2

30(1.019 x 10-3) = 1.529 x 10-2 x 20 = 30.58 x 10-2

(3) 2

20(1.019 x 10-3) = 1.019 x 10-2 x 76.67 = 78.13 x 10-2

Total 6.623 x 10-2 312.5 x 10-2 = 90R*2

R*2 =90

312.5x10-2 = 3.472 x 10-2

R*1 = 6.623 x 10-2 − R*2 = 3.151 x 10-2

90 in.

EIM

1.019 x 10-3

R*1 R*2x*

**

*

(1)(2) (3)

2050

76.67 13.33

90 in.

EIM

1.019 x 10-3

R*1 R*2x*

**

*

(1)(2) (3)

2050

76.67 13.33




(Page 4 of 6)

Step 5: Deflection - Conjugate Moment, M*

• Deflection at a:

ya = M*a = 3.151 x 10-2 (30) = 94.53 x 10-2

– (1.019 x 10-3) 30 302 3

× = −15.29 x 10-2

79.24 x 10-2 ya = 0.7924 in.

• Deflection at b:

yb = M*b = 3.472 x 10-2 (20) = 69.44 x 10-2

– (1.019 x 10-3) 20 202 3

× = −6.79 x 10-2

62.65 x 10-2 yb = 0.6265 in.




Solution: Example 7B - Deflection of Non-uniform Beam

(Page 5 of 6)

Using conjugate beam method to calculate the deflection of the beam in Figure 32 with cover plates added between the two loads so that EI in this section of the beam is double that of the remaining parts of the beam.

Repeat of Figure 32

Step 1: Moment diagram for actual beam is same as for Example 7A.

Figure 34a. Moment Diagram

MM

80 lb 120 lb

a b

30 in. 40 in. 20 in.

EI 2 EI EI

EI = 2.355 x 106 lb-in.2

Cover Plates80 lb 120 lb

a b

30 in. 40 in. 20 in.

EI 2 EI EI

EI = 2.355 x 106 lb-in.2

Cover Plates




(Page 6 of 6)

Steps 2 – 4: Conjugate beam load

Figure 34b. EIM Diagram

Area x* M about Pt. 1*

From Example 7A 6.623 x 10-2 312.5 x 10-2 For Items (1), (2), and (3)

-40(0.5095 x 10-3) = -2.038 x 10-2 X 50 = -101.9 x 10-2 For Item (4)

Total = 4.585 X 10-2 210.6 X 10-2 = 90R*2

R*2 =90 x10210.6 -2

= 2.34 X 10-2

R*1 = (4.585 – 2.34) x 10-2 = 2.245 x 10-2

Step 5: Deflection - Conjugate Moment, M*

• Deflection at a:

ya = M*a = 2.245 × 10-2 (30) = 67.35 × 10-2

– (1.019 ∗ 10-3) 30 302 3

× = −15.29 × 10-

2 52.06 × 10-2 ya = 0.5206 in.

1.019 x 10-3

* * *(1)(2) (3)

*(4)

R*1 R*2

1.019 x 10-3

0.5095 x 10-3

1.019 x 10-3

* * *(1)(2) (3)

*(4)

R*1 R*2

1.019 x 10-3

0.5095 x 10-3

* * *(1)(2) (3)

*(4)

R*1 R*2

1.019 x 10-3

0.5095 x 10-3




Finding and Using Beam Formulas

Previous sections have reviewed the procedures for calculating the reactions, shear, bending moment, and deflection of beams with various types of loads and supports. However, for certain standard type beams the standard beam formulas provided in engineering handbooks can be used. One useful source for such formulas is the AISC Manual of Steel Construction.

It is usually easier and quicker to use the listed diagrams and formulas from these sources for the analysis of a beam, if the loading and beam type are covered, than to do the calculation from the methods discussed earlier.

Beam Formula Work Aids

Beam diagrams and formulas from the AISC Manual are provided in Work Aids, where 33 cases are listed covering various beams and load types. Work Aid 13 lists the symbols used in the beam diagrams and formulas. Work Aid 14 is an index to the cases, and Work Aid 15 lists the beams and load types covered.

The types of beams covered in the table include: • Simple beam (Cases 1 to 11, 32, 33). • Beam fixed at one end and supported at the other; that is, a

propped cantilever (Cases 12 to 14). • Beam fixed at both ends (Cases 15 to 17). • Cantilever beam (Cases 18, 19, 21, 22). • Beam fixed at one end and free to deflect vertically but not to

rotate at the other end; that is, a guided cantilever (Cases 20, 23).

• Beam overhanging support (Cases 24 to 28). • Continuous beam with two equal spans (Cases 29 to 31).

The types of loads included in the beam diagrams of the Work Aids are as follows: • Concentrated loads. • Uniformly distributed loads.




• Linearly distributed (triangular) loads. • Variable end moments.

The results obtained from the beam diagrams and formulas listed in the Work Aids are:

• Beam/Load diagram.

• Shear and bending moment diagram.

• Expressions for:

- Beam reactions.

- Maximum shear and its location.

- Maximum bending moment and its location.

- Maximum deflection and its location. (Values for the shear, moment, and deflection at any point x along the beam.)

Using Beam Formula Work Aids

Use the beam formulas to perform beam analysis as follows:

1. Identify the beam/load type that corresponds to the problem being solved.

2. Obtain the appropriate diagrams and formulas from the Work Aids.

3. Use the superposition technique if your problem is not covered but can be composed from two or more beam/loading cases that are covered.

4. Calculate the desired values for the beam by substituting the appropriate known values into the formulas.

5. Draw the shear and moment diagram if needed.

These calculation steps are illustrated in the following examples and exercises.




Example 8: Shear Moment and Deflection for Simple Beam with Uniform Load Distribution

(Page 1 of 2)

Calculate the maximum values for shear, moment, and deflection for the beam and loading given in Example 3 (Figure 14).

Assume: W = 300 kips

L = 120 in.

E = 1.5 × 103 kips/in.2

I = 3000 in.4

Use the beam formulas in the Work Aids.

Repeat of Figure 14

W = total load

L

A B

x

wo

Ra Rb

W

A B

L/2 L/2RaRb

W = total load

L

A B

x

wo

Ra Rb

W

A B

L/2 L/2RaRb




Solution - Example 8

(Page 2 of 2)

1. Identify the applicable beam/load case for a simply supported beam. From Work Aid 14 - Beam/Load No. 1 applies.

2. From Work Aid 15, for Beam/Load No. 1, the applicable beam formulas are:

• Maximum shear, Vmax = 2W at the right and left

supports.

• Maximum moment, Mmax = 8

WL at x = L/2 from the left

support.

• Maximum deflection, ∆max = EI 384

5WL3

at x = L/2 from the

left support.

3. Superposition is not required.

4. Calculate required values based on the specified values for W, L, E, and I:

Vmax =2W =

2300 = 150 kips

Mmax = 8

WL = ( )( )8120300 = 4500 kip-in.

∆max = EI 384

5WL3

= ( )( )( )( )300010 x 1.5384

12030053

3

= 1.5 in.




Superposition of Beam/Load Cases

In some beam analysis problems, the beam loading may not be covered directly as a case in the Work Aids. However, in some situations, the principle of superposition can be used to combine two or more of the listed cases to obtain the results for the desired beam loading. For this approach to work correctly, the given beam must have the same support condition as the cases to be combined. The loads for the combined cases, when summed, should result in the loading being investigated. This approach to analyzing beams with complex loading is illustrated in Examples 9 and 10.




Example 9: Superposition of Beam/Load Cases

Given the beam in Figure 35, identify the beam/load cases from the Work Aids that can be used to perform the beam analysis.

Figure 35. Superposition of Beam/Load Cases


This beam/load case is not covered directly in the Work Aids. However, several listed cases can be combined to produce the desired result. These cases are as follows:

Figure 36. Beam Load Cases

P

w/2

w

a L b

P

w/2

w

a L b




Example 10: Beam Diagrams and Deflection Using Superposition and Work Aids

(Page 1 of 4)

Draw the shear and moment diagrams for the beam shown in Figure 37 and calculate the deflection at the midpoint between the supports.

Figure 37. Example 10 Beam


Use superposition to combine the following beam/load cases:

Figure 38. More Beam Load Cases

10 ft 10 ft 5 ft

R = 100 kipsP = 50 kipsw = 100 kips

C

BeamE = 29,000 kips/in.2 (steel)I = 1000 in.4

10 ft 10 ft 5 ft


C





(Page 2 of 4)


Figure 39a. Beam Load Cases




(Page 3 of 4)

Shear Diagrams Moment Diagrams

(Unit kips) (Unit: kip-ft)

d. Superposition: Combined Diagrams Case 7 + Case 24 + Case 26

Figure 39b. More Beam Load Cases

Deflection, ∆c, midpoint between support, x = 10

a. Case 7

∆c = 48EIRL3

= ( )( )( )100029,00048

240100 3

= 0.993 in.

b. Case 24

∆x = [ ]22223224o x2aL2aLxx2LL24EIL

xw+−+−

For x = ( ) ( ) ⎥⎦⎤

⎢⎣⎡ −= 22

2o

c 60324045

96EILw∆,/2L

∆c = ( )( )( ) ( ) ( ) ⎥⎦

⎤⎢⎣⎡ − 22

2

60324045

100029,00096/122404 = 0.422 in.

75

3570

50

65

105

75

3570

50

65

105

Mc = 550

M2 = 300

Mc = 550

M2 = 300




(Page 4 of 4)

c. Case 26

∆x = [ ]22 xL6EILPax

−

For x = 2L , ∆c =

16EIPaL2

(up)

∆c = ( )( )( )( )100029,00016

2406050 2

= 0.372 in. (up)

d. By superposition - Combine Case 7 + Case 24 + Case 26

Total ∆c = 0.993 + 0.422 – 0.372 = 1.043 in.




ANALYZING COLUMNS

Introduction

The first section of this module discussed the analysis of beams as basic components of civil engineering structures. Often the beams in a structure are framed into (or are supported by) vertical structural members, usually referred to as columns. Therefore, in evaluating or designing such structures you will often have to perform structural analysis of columns. This part of the module covers the analysis of columns, including:

• Definition and function of columns.

• Column materials and sections normally used.

• Column analysis considerations.

• Types of columns.

• Column loads and stresses - the load-carrying capacities of columns of various types.

• Combined actions of bending loading and axial loading in a structural member, called a beam-column.

Definition and Function

A column is a linear structural member loaded primarily along its longitudinal axis. A column usually has a uniform cross-section and is oriented vertically (or nearly vertically) in a structure. Generally, the loads are axial compression and result from the self weights and operating conditions of the structure. Columns are often connected to beams and other structural members to form structural frames to support the permanent and superimposed loads efficiently. Figure 40 shows examples of columns.




Figure 40. Typical Structural Columns

Identifying Column Materials and Sections

The materials commonly used for columns are:

• Structural steel.

• Reinforced concrete.

• Structural Timber.

• Composites. - Structural steel and concrete. - Structural steel and timber.

The section shapes used for columns include:

• Compact rolled shapes (AISC).

• Hollow sections - pipes and tubes.

• Simple solids - square, rectangular, circular, etc.

1st storycolumns

2nd storycolumns

columns

P

PP

a. IndividualColumns

b. Columns inBraced Steel Frame

c. Columns in RigidReinforced ConcreteSpace Frame

1st storycolumns

2nd storycolumns

1st storycolumns

2nd storycolumns

columns

P

PP

columns

P

PP

a. IndividualColumns

b. Columns inBraced Steel Frame

c. Columns in RigidReinforced ConcreteSpace Frame




• Compound or built-up sections.

Some of these column materials and sections are illustrated in Figure 41.

Figure 41. Typical Column Sections

Column Analysis and Considerations

The discussion of the analysis of beams showed that two key considerations in structural analysis are:

• Strength - the ability of the structure to safely support a specified load without experiencing excessive stresses, and

• Serviceability - the ability of the structure to support a specified load without undergoing unacceptable deflection, deformation, or movement.

These considerations apply to analysis and design of columns

a: Structural SteelRolled Shape

b: Reinforced Concretewith Round-Spiral Ties

c: Concrete-FilledTube

d: Reinforced ConcreteEncased Steel

e: Steel-TimberComposite

a: Structural SteelRolled Shape

b: Reinforced Concretewith Round-Spiral Ties

c: Concrete-FilledTube

d: Reinforced ConcreteEncased Steel

e: Steel-TimberComposite




as well. There is a third major consideration specific to columns, namely:

• Stability - the ability to support a given compressive load without experiencing a sudden change in geometry. This change in geometry is called buckling. The tendency to buckle under axial compression occurs even in a column that is initially straight. Deviation from straightness will further worsen this buckling tendency.

Based on these considerations, the parameters that can control or affect the behavior of a column, as shown in Figures 42 and 43, are as follows:

• Load magnitude, P.

• Load eccentricity, e - Distance from the center of the column cross-section to the load application point.

• Area of cross-section, A.

• End support conditions - Pinned, fixed, free, or partially restrained. See Figure 42.

• Initial out-of-straightness - imperfections.

• Effective length, - kL - The column length L is modified by a factor k to reflect the tendency to buckle due to the end conditions.

• Slenderness - usually expressed as a ratio of the effective kL of the column to a characteristic cross-section dimension (width b or radius of gyration r).

• Material yield stress, σy, and ultimate stress, σu.

• Material elastic modulus, E.




AAreaMoment of Inertia, I

AIr =

L

eP

d

b

x-axis

AAreaMoment of Inertia, I

AIr =

L

eP

d

b

x-axis

Figure 42. Key Characteristic of a Single Column Loaded with P at Eccentricity e from the Centroidal Axis

LoadsBeam

P

H

Columns

Footings

a) Structural Frame

Le = kL

L

HP PP

b) Structural Model

LoadsBeam

P

H

Columns

Footings

a) Structural Frame

Le = kL

L

HP PP

b) Structural Model

Figure 43. Columns as Part of a Structural Frame




Identifying Column Types

To analyze or design a column you need to know its type. This will help you to calculate the load capacity of the column. A column is classified to be one of the following three types as shown in Figure 44.

Column Type Based on Slenderness

Short column - Capacity is limited only by the compressive strength of the material; that is, buckling is not a consideration. A short column is one that has a slenderness ratio (kL/r for steel and reinforced concrete, kL/d for wood) below a specified limit:

• Structural Steel kL/r < 20.

• Reinforced Concrete kL/r < 22.

• Timber kL/d < 11.

Long column - Capacity is limited by elastic buckling. A long column will buckle sideways before the material crushes or yields under the compressive load. The limits on the slenderness ratio for a long column are:

• Structural steel kL/r > 120.

• Reinforced concrete kL/r > 100. • Timber - not permitted. (kL/d must be less than 50.)

Intermediate Length Column - Capacity is limited both by compressive crushing or yielding and by buckling. The slenderness ratios for this type of column fall in the range between those for a short column and long column. In most practical civil engineering applications, intermediate length columns are used.




Figure 44a. Types of Columns Based on Slenderness

Column Type Based on Eccentricity

Axially Loaded Column - The axial compressive load acts at or near the geometric center (centroid) of the column cross-section. The eccentricity e of the load (that is distance of load P to the centroid) must be less than 5% of the smaller cross-sectional dimension.

Eccentrically Loaded Column - The axial compressive load acts at a distance greater than 5% of the smaller cross-sectional dimension from the centroid of the column cross-section. The eccentric load can result in significant bending of the column. This bending increases the normal stresses in the column and the tendency of the column to buckle. Eccentricity in a column is often due to its construction details and to imperfections such as deviations from vertical straightness.




Figure 44b. Types of Columns Based on Eccentricity

Column Type Based on Nature of Loading

Beam-Column - Structural members do generally support both lateral loadings and axial loadings. The lateral loading can be caused by forces perpendicular to the column axis or by bending moments and shears due to the column framing into other structural members. The lateral load causes additional bending deflections and stresses as in a beam. These additional stresses combine with the stresses due to axial loading.

In most practical situations in civil engineering, structural members are subjected to both lateral and axial loads. However, if the lateral load effects predominate, these members are treated as beams. If the axial load effects are more significant, the members are treated as columns. In some cases both effects are important, and the members are treated as beam-columns.




MP

V w1

w2

w3

g) Beam-Columns

MP

V w1

w2

w3

g) Beam-Columns

Figure 44c. Typical Loadings for Beam-Columns

Calculating Column Loads and Stresses

It is often necessary to determine the load-carrying capacity of a column to safely support a prescribed level of load. To do this, you need to know how to calculate:

• Ultimate stress, σu and compressive load, Pu.

• Yield stress, σy and compressive yield load, Py.

• Critical stress, σcr and (buckling) load, Pcr.

• Failure load, Pmax, and stress, σmax.

• Factor of safety.

• Allowable axial load, Pa, and stress, σa.

Ultimate Compressive Load

The ultimate compressive load Pu or ultimate capacity of a column is the load that will produce crushing failure of the column material. That is, the compressive stress in the material will reach the ultimate value (σu). For a column with a cross-




sectional area A:

Pu = σuA or σu = APu

Compressive Yield Load

The compressive yield load or yield capacity of a column is the load that will cause compressive yielding of the column material. The stress in the material reaches the yield value, σy:

Py = σyA or σy = APy

Critical (Buckling) Load

The critical (buckling) load of a column is the load at which the column will become unstable (due to large geometric deformations needed for the column to move to a new equilibrium position) or begin to buckle. For an ideal, initially straight, pinned-end column, the buckling load is given by the Euler formula for elastic instability of a column:

Pcr = 2

2

(kL)EIπ with k = 1

where: E = the elastic modulus of the column material I = moment of inertia of the column section L = length of column between pinned supports k = effective length factor

Note that the buckling load of a column does not depend on the strength (ultimate stress or yield stress) of the material, but it depends on: i) elastic modulus, ii) shape of cross section in terms of moment of inertia; and iii) effective length factor k.

The corresponding critical stress in the Euler column due to the critical load is:

( )2

2cr

cr L/rEπ

APσ ==

where r = AI , the minimum radius of gyration of the column

cross section. The equation shows that the critical stress is proportional to the elastic modulus of the column material and inversely proportional to the square of the slenderness ratio, L/r,




for a column with pinned-end supports.

Generally, however, the critical (buckling) stress is given as

σcr = 2

2

(kL/r)Eπ

where: k = 1 for Euler column; k = 0.5 for a column with fixed ends and k = 2 for a cantilever column; and other intermediate values are determined depending on the end conditions.

For columns with other than pinned-end supports, the critical loads and critical stresses are based on the effective length, Le, of the column as defined in Figure 45. For such columns the length, L, is replaced by Le = kL in equations:

( )2

2

cr kLEIπP =

( )2

2

cr kL/rEπσ =

where values k are given for support conditions commonly encountered in civil engineering practice as provided in Figure 45.

Failure Load

The failure load, Pmax, for a column is the lower of two values, the ultimate load and the critical (buckling) load. The failure load is the largest load the column will support before it fails, or the minimum load that will cause failure of the column.

Pmax = Pu or Pcr, whichever is less.

The failure stress corresponding to the failure load is:

σmax = σu or σcr, whichever is less.

Figure 46 shows typical plots of Pmax and σmax as a function of the column slenderness ratio, kL/r.




Rotation fixed and translation fixed

Rotation free and translation fixed

Rotation fixed and translation free

Rotation free and translation free

End condition code

2.02.101.01.20.800.65

Recommended design value when ideal conditions are approximated

2.02.01.01.00.70.5Theoretical k value

(f)(e)(d)(c)(b)(a)

Buckled shape of column is shown by dashed line

Rotation fixed and translation fixed

Rotation free and translation fixed

Rotation fixed and translation free

Rotation free and translation free

End condition code

2.02.101.01.20.800.65

Recommended design value when ideal conditions are approximated

2.02.01.01.00.70.5Theoretical k value

(f)(e)(d)(c)(b)(a)

Buckled shape of column is shown by dashed line

Figure 45. Column Effective Length, Le = kL, Values of Column Coefficient, k




Figure 46. Typical Plots of Pmax and σmax as a Function of the Column Slenderness Ratio, kL/r

Note that in Figure 46 the transition point between the ultimate and critical values are based on setting σcr = σu such that

σU = 2

2

(kL/r)Eπ ; then:

For failure load: The critical length Lc = kL = Aσ

EIπu

For failure stress: The critical slenderness ratio Cc =

uσEπ

rkL

=

In practice, the ideal, theoretical curves for the column failure loads and stresses cannot be reached. Tests results show lower values for these loads and stresses as indicated by the lower curves in Figure 46, especially for columns of intermediate lengths. The actual column failure loads and stresses are lower than the theoretical values because of:

• Geometric imperfections in the column or deviation from




straightness.

• Actual or incidental eccentricity of the axial load.

• Effects of residual stresses (e.g. due to rolling and/or transportation) in rolled and fabricated column shapes.

Factor of Safety

A Factor of Safety (F.S.) is applied to failure load to obtain the load that the column can safely support. The factor of safety takes into account the uncertainties in the loads, in the material and geometric properties, and in the approximation and analysis. For column analysis and design, a value of 2.0 is usually selected for the factor of safety.

Allowable Load/Capacity

Allowable load or allowable capacity Pa of a column is the load that the column can safely support. It is obtained by applying a factor of safety to the column failure load, Pmax :

F.S.PP max

a =

Allowable stress σa corresponding to the allowable load for a column depends on the material compressive strength σu and the critical (buckling) stress σcr such that

σa = F.S.σ

AP maxa =

= F.S.σor

F.S.σ cru , whichever is less.

The stress due to axial compressive load for a column is limited to the allowable stress to ensure that the column functions safely; that is, so that the column has a low risk of failure due to crushing, yielding, or buckling.

Structural design handbooks such as the AISC Manuals [Ref. no. on ASD or LRFD methods] for Steel Construction and the National Forest Products Association (NFPA) National Design Specification for Lumber provide tables of allowable loads and stresses for columns of varying slenderness.




Example 11: Column Loads and Stresses

(Page 1 of 3)

A column, of 12 ft effective length, is to be constructed by securely nailing four wooden planks together. Each plank has an actual cross-section of 2 × 8 in. Two arrangements of the planks are to be considered, as shown in Figure 47 for an Euler column with k = 1.

Figure 47. Alternate Arrangements of Column Cross-Section

• Assume that:

E = 1.5 × 106 psi

σu = 4000 psi

F.S. = 3

• Calculate the following for each column arrangement:

a. Ultimate Load.

b. Critical buckling load and stress.

c. Failure load.

d. Allowable load and stress.

Note: It is assumed that the shear strength of each nail far exceeds the shear strength of the wooden planks and the




section will always act as if it were continuous (without nails).

Solution: Example 11 (Page 2 of 3)

Column 1.

a. Ultimate Load, Pu

Pu = Aσu

A = 8(8) = 64 in2

σu = 4000 psi

Pu = 64(4000) = 256 kips 8 in.

8 in.

b. Critical Euler Buckling Load, Pcr

Pcr = ( )2

2

kLEIπ (k = 1)

E = 1.5 × 106 psi

I = 12

8(8)12bd 33

= = 341.33 in.4

kL = 12 ft = 144 in.

Pcr = ( ) ( )( )( )2

62

144341.331.5x103.14 = 243.44 kips

Critical stress, σcr = 64

243.44A

Pcr = = 3.80 kips/in.2

c. Failure Load, Pmax

Pmax = min of [Pu or Pcr]

Pmax = Pcr = 243.44 kips

d. Allowable Load, Pa

Pa = F.S.Pmax




F.S. = 3.0

Pa =3.0

243.44 = 81.15 kips

(Page 3 of 3)

Allowable stress, σa = 64

81.15APa = = 1.27 kips/in.2

Column 2.


Pu = Aσu

A = (10)2 – (6)2 = 64 in.2

Pu = 64(4000) = 256 kips

b. Critical Buckling Load Pcr 6 in.

10 in.

6 in.10 in

Pcr = ( )2

2

kLEIπ

I = 1210 (10)3 –

126 (6)3 = 725.33 in.4

kL = 144 in.

Pcr = ( ) ( )( )( )2

62

144725.331.5x103.14 = 517.3 kips

Critical stress, σcr = 2cr

in. 64k 517.3

AP

= = 8.08 kips/in.2


Pmax = Pu or Pcr, whichever is less

Pmax = Pu = 256 kips


Pa = 3.0256

F.S.Pmax = = 85.33 kips




Allowable stress, σa = 2a

64in.85.33k

AP

= = 1.33 kips/in.2

Note: In the above problem, the box section is more efficient in carrying the applied axial loads.




Calculating Combined Bending and Axial Loading

This module has reviewed some of the principles and procedures required for analyzing beams subjected to bending from lateral loads and columns subjected primarily to axial compressive loads. However, in civil engineering structures, a structural member is usually subjected to both bending and axial loads, such as the column in a rigid frame (Figure 43) carrying both horizontal and vertical loads. A structural member that experiences a significant amount of bending as well as axial load is called a beam-column. The next section reviews the procedures and formulas for analyzing a beam-column.

Beam-Columns Recall that the basic stresses in a structural member due to the axial load, P, and bending moment, M, are:

• Axial stress: APfa ±=

• Bending (flexural) stress: SMfb ±=

• Combined stress: SM

APff f batot ±±=+=

The allowable combined stress formula (below) is valid only for a short member that has an allowable compressive stress governed by the yield or ultimate capacity of the material:

( )y

y

F.S.F

SM

APf ≤±=

and also u

utot (FS)

Ff ≤ .

For intermediate and long beam-columns, where buckling has to be considered, the allowable axial compressive stress is generally less than that for bending. Therefore, for such beam-columns, the interaction formulas are more meaningful and are applied as follows:

• Axial load and bending about one axis:




1F

M/SF

P/A,1Ff

Ff

bab

b

a

a ≤+≤+

where:

Fa = allowable stress when only axial load is present in the member (axially-loaded column).

Fb = allowable stress when only bending is present in the member (beam flexure).

• Axial load and bending about both axes:

1Ff

Ff

Ff

by

by

bx

bx

b

a ≤++

P y

xMy

Mx

P y

xMy

Mx

P y

xMy

Mx

Figure 48. Biaxial Bending about x and y Axes of the Column Cross Section

where the subscripts x and y denote values pertaining to the x and y axes of the section, about which the biaxial bending state is enforced.

Eccentrically Loaded Columns

Eccentrically loaded columns are a special case of combined axial load and bending in the column. Therefore, the combined




stress formula and the interaction formulas can also be used in the analysis of columns subjected to eccentric loads.

This is done by substituting the moment, the product of P multiplied by e, for the bending moment M in the formulas for combined stresses and in the beam-column interaction formulas. For example, using the short column formula, the combined stress due to an eccentric load is:

f = SM

AP

±

where:

M = Pe

Smin = cI

I = Moment of inertia.

c = Distance from the center of the column section to the extreme fiber.

Therefore:

⎟⎠⎞

⎜⎝⎛ ±=•±= 2r

ec1AP

IAc

APe

APf

where: the radius of gyration, r =

AI

Figure 49. Eccentrically Loaded Column

Load, PEccentricity, e

Area, A

c

Load, PEccentricity, e

Area, A

c




Example 12: Combined Axial Load and Bending in Column

(Page 1 of 2)

Determine the largest load P that can be safely carried by a W 310 x 74 steel column of 4.5-m effective length. Use E = 200 GPa and Fy = 250 MPa.

Figure 50. Eccentrically Loaded Steel Column

a) Based on allowable combined stress and Fa = 100 MPa.

b) Based on interaction formula with allowable axial stress Fa = 100 MPa, and allowable bending stress Fb = 0.6 Fy.

Solution: Example 12

a) Combined Stress: Eccentrically loaded column

e = 200 mm

fapp = ⎟⎠⎞

⎜⎝⎛ + 2r

ec1AP

≤ Fa

Maximum allowable P1 is obtained by setting fapp = Fa :

P1 = 2

a

rec1

AF

+

200 mm P

Cx

Cy

W 310 x 74A = 9480 mm2

rx = 131.6 mmry = 49.8 mmSx = 1058 x 103 mm3

200 mm P

Cx

Cy

W 310 x 74A = 9480 mm2

rx = 131.6 mmry = 49.8 mmSx = 1058 x 103 mm3




= ( )( )( )( )

( )2

22

131.6mm155mm200mm1

100N/mm9480mm

+ = 339.8 kN




(Page 2 of 2)

b) Interaction Formula:

( ) 2b

b

b

a

a 150N/mm2500.6F 1,Ff

Ff

==≤+

1SF

eAF

1P,1F

Pe/SF

P/Ababa

≤⎥⎦

⎤⎢⎣

⎡+≤+

Maximum allowable P2 is obtained by setting the sum of dimensionless stress ratio to 1:

P2 = ( )( ) ( )( )

1

3

1

ba 1501058x10200

10094801

SFe

AF1

−−

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+

= 431.9 kN

It is concluded: For the given conditions (e.g. cross section type, dimension, material properties, effective length, and value of e), the column can safely carry

Pmax = min [P1 ; P2] = 339.8 kN

Note: If one of the above conditions is changed, the load carrying capacity of the column will also change and one has to re-evaluate the column capacity.




ANALYZING FOOTINGS

In addition to beams and columns that have been reviewed in the two preceding sections, foundations are also important parts of civil engineering structures. This section discusses the analysis of footings, including the following topics:

• Definition and types of footings.

• Analysis procedure for footings.

• Applied loads.

• Effects of load eccentricity.

• Soil-bearing pressure.

• Stability ratio.

• Moment and shear in a footing.

Identifying and Defining Types of Footings

Footings are structural components used to transfer the loads on a structure to the ground. Footings are also called shallow foundations because they support the loads near to the ground surface.

The most common types of footings, as shown in Figure 51, are:

• Strip footing for walls.

• Spread (individual) footings for columns.

• Combined footings - supporting two or more columns.

• Octagonal or circular footings. Footings are generally made of reinforced concrete and are usually rectangular in shape. However, sometimes octagonal or circular footings are used, especially for large vessels and stacks.




db

d

d

d

L

a) Strip Footing (wall) b) Spread Footing (Column)

c) Combined Footing d) Octagonal Footing

db

db

dd

dd

d

L

d

L

a) Strip Footing (wall) b) Spread Footing (Column)

c) Combined Footing d) Octagonal Footing

Figure 51. Types of Footings

Analysis Procedure

The analysis of a footing involves the following:

• Determine the loads on the footing.

• Calculate the bearing pressure on the soil due to the loads and weight of the footing.

• Calculate the bending moment and shear force at the critical sections of the footing.

• Evaluate the soil-bearing pressure and footing stresses by comparing the calculated values with the specified allowable values.




Identifying and Defining Applied Loads

The loads acting on a support footing are the reactions from the structural member it supports. In general these loads are, as shown in Figure 52:

• Vertical (Axial) Load, P.

• Horizontal (Shear) Load, H.

• Moment M along one or both horizontal axes of the footing.

P

M

H

P

M

H

P

M

H

Figure 52. Loads Acting on a Support Footing

Identifying and Defining Eccentric Loads on Footings

Normally a wall or column is located at the center of the supporting footing, so that the vertical load is centered on the footing. However, sometimes the wall or column is off center. This results in an eccentric load on the footing as shown in Figure 53b.




P eP

a) Centered/Concentric Load b) Eccentric Load

P eP

a) Centered/Concentric Load b) Eccentric Load

Figure 53. Footing Loads

Load eccentricity can occur about one axis of the footing or about both axes. When a moment, M, is present on the footing, the centered or concentric load, P, acts as if it were eccentric. The apparent eccentricity, e, is the ratio M/P.

Identifying and Defining Soil-Bearing Pressure on Footings

The soil-bearing pressure, q, is the load per unit area produced by the footing on the underlying soil. A footing must support the applied load safely without soil failure or excessive settlement. Therefore, the soil-bearing pressure due to the loaded footing is limited to an allowable value specified by a geotechnical engineer or in a project specification.

To calculate the soil-bearing pressure, use the following formulas.

Centered/Concentric Load

The soil-bearing pressure for a centrally loaded footing is:

q = AP

where:

A = contact area of the footing.




Figure 54. Centered/Concentric Load

Combined Vertical Load and Moment

The standard combined stress formula is used to calculate the soil bearing pressure for a footing subject to both vertical load and overturning moment as follows:

q = I

McAP

±

qmax = I

McAP

+

qmin = I

McAP

−

where: I = moment of inertia of the contact area. c = distance from center to edge of footing.




Figure 55. Combined Vertical Load and Moment

• For a rectangular footing: A = bd

S = 6

bdcI 2

=

and the equation for soil bearing pressure, q, becomes:

q = ⎥⎦⎤

⎢⎣⎡ ±=±

0.167dM/P1

AP

bd6M

bdP

2

• For a circular footing, diameter = d:

A = 2d4π , S =

32πd3

Therefore:

q = ⎥⎦⎤

⎢⎣⎡ ±

0.125dM/P1

AP

• For an octagonal footing, d = distance across the parallel faces:

A = 0.8284 d2, S = 0.1094 d3

Therefore:

q = ⎥⎦⎤

⎢⎣⎡ ±

0.132dM/P1

AP




Eccentric Load

q = ( )I

cPeAP

±

= ⎥⎦⎤

⎢⎣⎡ ± 2r

ec1AP

where: e = Load eccentricity

r = Radius of gyration, AI

P

qminqmax

e P

qminqmax

e

Figure 56. Vertical Load and Moment

• For a rectangular footing contact area:

A = bd

r2 = 12d2

c = 2d

Therefore:

q = ⎥⎦⎤

⎢⎣⎡ ±

d6e1

bdP

Figure 57. Contact Area

*b

d

eP*b

d

eP




This is the same expression as that for a footing subjected to combined vertical load and moment. The same is true for footings of other shapes.

Critical Eccentricity

To prevent the uplift of any part of the footing, q > 0, the eccentricity of the load on the footing cannot exceed a limit value called the critical eccentricity, ecr. The expressions for ecr for specific footing shapes are as follows:

• For rectangular footing, ecr = 0.167 d.

• For circular footing, ecr = 0.125 d.

• For octagonal footing, ecr = 0.132 d.

Partial Compression

The standard combined stress equation for a footing with combined vertical load and moment or with eccentric load applies only for cases where the footing contact area is fully in compression. That is, the soil cannot take tensile stress, qmin < 0. This occurs if e > ecr. For a rectangular footing this means that:

e = PM > ecr = 0.167 d

That is, uplift occurs for a rectangular footing if the load falls outside the middle third or “kern” of the footing. For this case the rectangular footing is in only partial contact with the underlying soil. The following relationships apply:

From vertical equilibrium:

∑ Fy = 0 ⇒ P = 21 qmax × b

∴ qmax = bx2P




Also, resultant of soil pressure, “R”, should coincide with line of action of “P” for equilibrium, thus:

e2d

3x

−=

or:

x = ⎟⎠⎞

⎜⎝⎛ − e

2d3

Figure 58. Partial Compression




Example 13: Soil-Bearing Pressure for a Square Footing

(Page 1 of 2)

A column footing 10 ft square supports a concentric load, P = 100 k and an overturning moment, M = 150 k-ft. Calculate: a. Eccentricity, e.

b. Critical eccentricity, ecr.

c. Maximum and minimum soil bearing pressures. d. Maximum soil-bearing pressure, if the overturning moment is

doubled, M = 300 kip-ft.

Solution: Example 13

a. Eccentricity, e

e = kips 100

ftkip 150PM −

= = 1.5 ft

b. Critical eccentricity, ecr

ecr = 0.167d, d = 10ft

= 0.167(10) = 1.67 ft

Figure 59a. Soil-Bearing Pressure for Square Footing P = 100 k, M = 150 ft.k




(Page 2 of 2) c. Soil bearing pressure, q

e < ecr

q = ⎥⎦⎤

⎢⎣⎡ ±

0.167de1

AP

qmax = ⎥⎦⎤

⎢⎣⎡ +

0.167de1

AP

= ( ) ⎥⎦⎤

⎢⎣⎡ +

1.671.51

1010100 = 1.9 kips/ft2

qmin = ( ) =⎥⎦⎤

⎢⎣⎡ −=⎥

⎦

⎤⎢⎣

⎡−

1.671.51

1010100

ee1

AP

cr

0.1 kips/ft2

d. Maximum soil-bearing pressure

M = 300, e = 10030

PM

= = 3ft

e > ecr, footing uplift occurs

x = 3 ⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ − 3

210 3e

2d = 6ft

qmax = bx2P =

10(6)2(100) = 3.33 kips/ft2

qmax = 3.33qmax = 3.33

Figure 59b. Maximum Soil-Bearing Pressure for Square Footing P = 100 k, M = 300 ft.k




Identifying and Defining Stability Ratio on Footings

You know that the soil bearing pressure for a footing has to be limited to assure an adequate factor of safety against soil failure and to prevent excessive settlement of the footing. An adequate factor safety is also needed to prevent the overturning of a structure subjected to large overturning forces.

The factor of safety against the overturning of a footing or foundation is called the stability ratio. The stability ratio (SR) is the ratio of the sum of the moments preventing overturning and the sum of the moments causing overturning of the structure. The moments are taken about the point of rotation in the event of overturning. See Figure 60 for example:

w

PM

H

h

o

2d

( )HhM

d/2WP

ΣMΣMSR

o

r

++

=

=

Mr = Resisting Momentabout point O

Mo = Overturning Momentabout point O

w

PM

H

h

o

2d

( )HhM

d/2WP

ΣMΣMSR

o

r

++

=

=

Mr = Resisting Momentabout point O

Mo = Overturning Momentabout point O

Figure 60. Stability Ratio

Identifying and Defining Moment and Shear on Footings

The lesson so far has shown how to determine the loads on the footing and the resulting soil-bearing pressure. The next step in the analysis procedure for a footing is to determine the moments and shears at the critical section of the footing and to compute and evaluate the stresses that result from these internal forces. For this purpose, the footing is treated similarly to a beam, as shown in Figure 61. It is loaded vertically upward by the bearing pressure from the soil. With this structural model,




the moments and shears are calculated using the procedures covered earlier in the module.

PM

qmin qmax

Footing

w = qw = q

Cantilever Beams

PM

qmin qmax

Footing

w = qw = q

Cantilever Beams

Figure 61. Moment and Shear on Footings

The moment and shear forces and stresses are calculated at the critical sections defined as follows. See Figure 62.

• Critical section for bending moment — at face of support.

• Critical section for shear — 0.5t from support.




2t

t

Footing Section

d

d

b

Column Footing

Wall Footing

Critical Section:for bending momentfor shear force

2t

t

Footing Section

d

d

b

Column Footing

Wall Footing



Figure 62. Critical Sections




Example 14: Stability Ratio, Moment, and Shear for a Square Footing

(Page 1 of 3)

A retaining wall and footing support loads as shown in Figure 63. Calculate the following: a. Stability ratio. b. Eccentricity. c. Critical eccentricity. d. Soil-bearing pressure. e. Shear at critical section. f. Moment at critical section.

P e = 2CL

footing

x = 6 ft.qmax 5.33 kips/ft.2

P

c

H

4 ft.

12 in.

12 in.

5 ft. 3 ft.d = 8 ft., b = 1 ft.

wall

P = 16 kips/ft includingweight of wall andfooting.

H = 12 kips/ft

o

P e = 2CLCL

footing

x = 6 ft.qmax 5.33 kips/ft.2

P

c

H

4 ft.

12 in.

12 in.

5 ft. 3 ft.d = 8 ft., b = 1 ft.

wall

P = 16 kips/ft includingweight of wall andfooting.

H = 12 kips/ft

o

Figure 63a. Wall and Footing Support Loads




Solution: Example 14 (Page 2 of 3)

a. Stability ratio, SR

Taking moments about point O

SR = ( )( )412516

MM

o

r = = 1.67

b. Eccentricity, e

e = PM

M = Moment about center of footing

e = ( ) ( )16

116412 − = 2ft

c. Critical eccentricity, ecr

ecr = 0.167 d = 1.33ft

d. Soil-bearing pressure, q

e > ecr, footing uplifts

x = ⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ − 2

283e

2d3 = 6 ft

qmax = ( )( )61162

bx2P

= = 5.33 kips/ft2




(Page 3 of 3) e. Shear at critical section, Vc

Vc = ( )x412

1.785.33 + = 14.22 kips

0.5ft.2t

=

5.33

4 ft

t = 1 ft.

1.78

Critical Sectionfor Shear

0.5ft.2t

=

5.33

4 ft

t = 1 ft.

1.78

Critical Sectionfor Shear

Figure 63b. Wall and Footing Support Loads (Shear at Critical Section)

f. Moment at critical section, Mc

Mc = ( ) ( ) ( )( ) ( )4.5324.54.0

21

24.51.33

2

⎟⎠⎞

⎜⎝⎛+ = 40.5 kip-ft

4.0

4.5 ft

1.33

Critical Sectionfor Shear moment

1.33

4.0

4.5 ft

1.33

Critical Sectionfor Shear moment

1.33

Figure 63c. Wall and Footing Support Loads (Moment at Critical Section)




SUMMARY

This module has discussed information to be used in analyzing various components of structure such as beams, columns, and footings.




This Page Intentionally Blank




WORK AIDS

Work Aid 1: Calculating Reactions, Shear, and Bending Moment in a Simple Overhang Beam

(Page 1 of 2)

For the overhang beam shown below, calculate:

a) Reactions R1 and R2.

b) The shear and bending moment at (C), which is 5 ft from left support (A) of the beam.

A C B D

100 lb/ft300 lb

10' 6'

A C B D

100 lb/ft300 lb

10' 6'

R1 R2

A C B D

1000 lb 300 lb

5' 6'5'R1 R2

A C B D

1000 lb 300 lb

5' 6'5'

Figure 64. Overhang Beam




(Page 2 of 2) Procedure:

a) Solve for reactions R1 and R2.

+ Σ MB = 0 : −(10)(R1) + (1000)(5) − (300)(6) = 0

R1 = 320 lb ↑

+ ↑ Σ Fy = 0 : R2 + 320 − 1000 − 300 = 0

R2 = 980 lb ↑

b) Solve for shear and moment at C.

(i) Draw free-body diagram of a beam to left of point C.

A C

(100) (5) = 500 lb2.5'

5'

V

M

320 lb

A C

(100) (5) = 500 lb2.5'

5'

V

M

320 lb

(ii) Sum vertical forces to obtain shear.

+ ↑ Σ Fy = 0 : 320 − 500 − V = 0

V = −180 lb

(iii) Sum moment about C to obtain bending moment.

+ Σ MC = 0 : −(320)(5) + (500)(2.5) + M = 0

M = 350 lb.ft




Work Aid 2: Procedures to Draw Diagrams

(Page 1 of 2)

The procedures to draw the shear and moment diagrams for beams are illustrated by the following two examples:

Draw the shear and moment diagrams for the beam in Figure 65.

A C B

P

D

LLLR1 R2

5PL

A C B

P

D

LLLR1 R2

5PL

Figure 65. Work Aid 2, Beam with Loads

Procedure:

Solving for R1 by taking Σ MB = 0 :

R1 = −P ⇒ R1 = P ↓

Σ Fy = 0 ⇒ R2 = 2P ↑




(Page 2 of 2) 0 < x < L + ↑ Σ Fy = 0 ⇒ V = −P

+ Σ M = 0 : M = −Px L < x < 2L + ↑ Σ Fy = 0 : V = −2P

+ Σ M = 0 : Px + P(x−L) + M = 0 M = P(L−2x) 2L < x < 3L (Right to the section) + ↑ Σ Fy = 0 V = −2P + Σ M = 0 M = 2P(3L−x)

Fig. 74

Fig. 75

P5PL

P 2P

x

M

VP

x

M

VP

P

M

VP

P5PL

3L-x

M

V2PV

x-P

-2P

x

M

2PL

-3PL

-PL

L L L

P5PL

P 2P

P5PL

P 2P

x

M

VP x

M

VP

x

M

VP

P

x

M

V

M

VP

PP

M

VP

P5PL M

V

M

VP

PP5PL

3L-x

M

V2P

3L-x

M

V2PV

x-P

-2P

x-P

-2P

x

M

2PL

-3PL

-PLx

M

2PL

-3PL

-PL

L L L




Work Aid 3: Procedures to Draw Diagrams

(Page 1 of 2)

Drawing and Using Shear and Moment Diagram for Overhang Beam with Uniformly Distributed Load wo.

Draw the shear and moment diagrams for the overhang beam in Figure 66.

Figure 67. Work Aid 3, Beam with Uniform Load

Procedure:

Determine R1 by summing moments about B : + Σ MB = 0 :

− (R1)(2L) + (3woL) ⎟⎠⎞

⎜⎝⎛

2L = 0

R1 = 4

L3wo

Figure 68.

+ ↑ Σ Fy = 0 : 4

L3wo − 3woL + R2 = 0

R2 = 4

L9wo

A B

2L L

wo

A B

2L L

A B

2L L

wo

A B

LR1

3 woL

23L

2L

R2

A B

LR1

3 woL

23L

2L

R2




(Page 2 of 2) 0 < x < 2L + ↑ Σ Fy = 0 :

4

L3wo − wox − V = 0

V = wo ⎟⎠⎞

⎜⎝⎛ − x

43L

+ Σ M = 0 :

M2xw

4xL3w 2

oo ++− = 0

M = ⎟⎠⎞

⎜⎝⎛ − x

23L

2xwo

2L < x < 3L (Consider right part) + ↑ Σ Fy = 0 : V − wo(3L−x) = 0 V = wo(3L−x) + Σ M = 0 :

2

x)(3LwM2

o −−− = 0

M = −2

x)(3Lw 2o −

Maximum positive moment

= M at x = 4

3L

= 32

L9w4

3L2

3L4

3L2

w 2oo =⎟

⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛

Maximum negative moment = 2Lw 2

o−

Figure 69.

2L L

wo

4L3wo

4L9w o

wox

MV

4L3wo

2x

x

wo (3L-x)

MV

(3L-x)/2

3L-x

x

M

4L3w o woL

x

4L5wo−

43L

2Lw 2

o−

32L9w 2

o

V

2L L

wo

4L3wo

4L9w o

wox

MV

4L3wo

2x

x

wo (3L-x)

MV

(3L-x)/2

3L-x

x

M

4L3w o woL

x

4L5wo−

43L

2Lw 2

o−

32L9w 2

o

V




Work Aid 4: Basic Shear and Moment Diagrams

(Page 1 of 2)

You can use these diagrams and expressions for more complex, multiple-load cases with the aid of superposition.

Cantilever Beam subjected to four types of load

Figure 70. Repeat of Figure 17, Work Aid 4, Basic Shear and Moment Diagrams




Simple Support Beam subjected to three types of load

(Page 2 of 2)

Figure 71. Repeat of Figure 18, Work Aid 4, Basic Shear and Moment Diagrams




Work Aid 5: Beam Diagrams - Superposition

The steps involved in this approach are:

1. Separate problem into beams with individual loads.

2. Draw beam diagrams for each load case.

Construct beam diagrams for the overall beam loads by combining diagrams for individual load cases.




Work Aid 6: Using Beam Diagrams by Superposition

Draw the bending moment diagram for the beam with the loads shown.

(Page 1 of 2)

Figure 72. Work Aid 6, Beam with Uniform and Concentrated Loads

Step 1 Separate into two individual loads:

a) Concentrated load, P = 6 kips

b) Uniform load, wo = 3 kips/ft

Figure 73. Loads Step 2 Construct the moment diagrams for the

individual load cases based on Figure

a) Simply supported beam with a concentrated load

Mmax = 4

(6)(10)4

PL= = 15 kip-ft

b) Simply supported beam with uniform load

Mmax = 8

3(10)8

wL 22

= = 37.5 kip-ft

Figure 74. Load Cases

3 kips/ft

5'

6 kips

5'

3 kips/ft

5'

6 kips

5'

5' 5'

6 kips

3 kips/ft

10'

5' 5'

6 kips

3 kips/ft

10'

3 kips/ft

10'

37.5 kip-ft

15 kip-ft

+

+

37.5 kip-ft

15 kip-ft

+

+




(Page 2 of 2) Step 3 Construct the composite diagram by

combining individual cases

Mmax = 15 + 37.5

= 52.5 kip-ft

Figure 75.

52.5 kip-ft

+

52.5 kip-ft

+




Work Aid 7: Beam Stresses

The analysis steps for determining the stresses in a beam are illustrated below.

Beam

Lateral Loads

Bending Moment Shear Force

Bending orFlexural Stress

Vertical and Horizontal

Shear Stress

ConcentratedLoads

DistributedLoads

Figure 76. Repeat of Figure 23, Work Aid 7, Beam Stresses




Work Aid 8: Calculating Flexural and Shear Stresses in an Overhanging Steel Beam with Uniformly Distributed Load

(Page 1 of 2)

x-section

IN.A. = 211 in4

Figure 77. Work Aid 8, Bending Moment and Shear Diagram

From shear and moment diagrams:

Vmax = 62.5 kips

Mmax = 125.0 kip-ft (negative moment)

8"

9"

0.5"

0.5"

0.5"

8"

9"

0.5"

0.5"

0.5"

10 k ip/ft

10 ' 5 '

x

x

50.037.5

70.3

-62.5

-125.0

V(kip)

M(kip-ft)

10 k ip/ft

10 ' 5 '

x

x

50.037.5

70.3

-62.5

-125.0

V(kip)

M(kip-ft)




(Page 2 of 2)

Maximum flexural stress:

ksi35.55211

(5)12)(125I

Mcσmax =∗∗

==

Maximum shear stress:

bI

QVmaxmax =τ (at N.A.)

Q = (8)(0.5)(4.75) + (4.5)(0.5)(2.25)

= 24.06 in3

ksi14.26(211)(0.5)

06)(62.5)(24.max ==τ

8"

4.5"

0.5"

0.5"

4.75"2.25"

N.A.

8"

4.5"

0.5"

0.5"

4.75"2.25"

N.A.




Work Aid 9: Conjugate Beam Method

The conjugate beam method changes a deflection problem into one of drawing moment diagrams. The method is able to handle beams of varying cross-sections and materials.

Step 1: Draw the moment diagram for the beam as it is actually loaded.

Step 2: Construct the M/EI diagram by dividing the value of M at every point along the beam by the product of EI at that point. If the beam is of constant cross-section, EI will be constant and the M/EI diagram will have the same shape as the moment diagram. However, if the beam cross-section varies with x, then I will change and the M/EI diagram will differ from the moment diagram.

Step 3: Draw a conjugate beam of the same length as the original beam. The material and cross-sectional area of this conjugate beam are not relevant.

• If the actual beam is simply supported at its end, the conjugate beam will be simply supported at its ends.

• If the actual beam is simply supported away from its ends, the conjugate beam has hinges at the support points.

• If the actual beam has free ends, the conjugate beam has built-in ends.

• If the actual beam has built-in ends, the conjugate beam has free ends.

Step 4: Load the conjugate beam with the M/EI diagram. Find the conjugate reactions by methods of statics. Use the superscript (*) to indicate conjugate parameters.

Step 5: Find the conjugate moment at the point where the deflection is wanted. The deflection is numerically equal to the moment as calculated from the conjugate beam forces.

Relationships between conjugate beam and actual beam are shown below:

Conjugate Beam Actual Beam Simple End* Simple End Hinge* Interior Support Free End* Built-in End Built-in End Free End

Load, W* EIM

Reaction, R* End Slope, Shear, V* Slope, Moment, M* Deflection, y

*Indicates parameters related to conjugate beam




Work Aid 10: Calculating Beam Deflection

(Page 1 of 3)

Use the conjugate beam method to determine the deflections at the right end (point C).

Uniform beam with constant value of EI = 2.356 × 106 lb-in.2.

A CB

120 lb

60" 30"

A CB

120 lb

60" 30"

Figure 78. Work Aid 10, Calculating Beam Defelection




(Page 2 of 3)

Step 1: Moment diagram for actual

beam.

Step 2: Load the conjugate beam with M/EI

EIMw* =

w*max = 6102.3563600

∗

= 1.528∗10-3 in-1 Step 3: Consider 1st, the segment AB

and determine R*A and R*B.

Figure 79. Beam

+ Σ MB = 0 :

60− R*A + 21 (1.528∗10−3)(60) ⎟

⎠⎞

⎜⎝⎛

360 = 0

R*A = 1.528∗10−2

Σ Fy = 0 ⇒ R*B + 1.528∗10−2 − 21 (1.528∗10−3)(60) = 0

R*B = 3.06∗10−2

60" 30"x

-3600 lb-in

hinge

M/EI

1.528*10-3 in-1

(1)

*AR *

BR

(2) *CR

*CM

*BR

A CB

A CB

60" 30"xx

-3600 lb-in

hinge

M/EI

1.528*10-3 in-1

(1)

*AR *

BR

(2) *CR

*CM

*BR

A CB

A CB




(Page 3 of 3) Step 4: Consider segment BC.

Σ Fy = 0 : 2103.06 −∗− + R*C = 0

R*C = 3.06∗10−2

+ Σ MC = 0 :

21 (1.528∗10−3)(30)(20) + (R*B)(30) – M*C = 0

M*C = (3.06∗10−2)(30) + (½)(1.528∗10−3)(30)(20) = 1.38 in.

Therefore, the deflection at C = M*C = 1.38 in.




Work Aid 11: Using Beam Formula Work Aids

Use the beam formulas to perform beam analysis as follows:

1. Identify the beam/load type that corresponds to the problem being solved.

2. Obtain the appropriate diagrams and formulas from the Work Aids.

3. Use the superposition technique if your problem is not covered but can be composed from two or more beam/loading cases that are covered.

4. Calculate the desired values for the beam by substituting the appropriate known values into the formulas.

5. Draw the shear and moment diagram if needed.




Work Aid 12: Drawing Beam Diagrams and Deflection Using Superposition

(Page 1 of 4)

Draw the shear and moment diagrams for the beam shown in Figure 80 and calculate the deflection at the midpoint between the supports.

10 ft 10 ft 5 ft


C


10 ft 10 ft 5 ft


C


Figure 80. Repeat of Figure 37. Work Aid 12 Beam


Figure 81. Repeat of Figure 38 More Beam Load Cases




(Page 2 of 4)


Figure 82a. Repeat of Figure 39a Beam Load Cases




(Page 3 of 4)

Shear Diagrams Moment Diagrams

(Unit kips) (Unit: kip-ft)

d. Superposition: Combined Diagrams Case 7 + Case 24 + Case 26

Figure 83b. Repeat of Figure 39b More Beam Load Cases

Deflection, ∆c, midpoint between support, x = 10

a. Case 7

∆c = 48EIRL3

= ( )( )( )100029,00048

240100 3

= 0.993 in.

b. Case 24

∆x = [ ]22223224o x2aL2aLxx2LL24EIL

xw+−+−

For x = ( ) ( ) ⎥⎦⎤

⎢⎣⎡ −= 22

2o

c 60324045

96EILw∆,/2L

75

3570

50

65

105

75

3570

50

65

105

Mc = 550

M2 = 300

Mc = 550

M2 = 300




∆c = ( )( )( ) ( ) ( ) ⎥⎦

⎤⎢⎣⎡ − 22

2

60324045

100029,00096/122404 = 0.422 in.

(Page 4 of 4) c. Case 26

∆x = [ ]22 xL6EILPax

−

For x = 2L , ∆c =

16EIPaL2

(up)

∆c = ( )( )( )( )100029,00016

2406050 2

= 0.372 in. (up)

d. By superposition - Combine Case 7 + Case 24 + Case 26

Total ∆c = 0.993 + 0.422 – 0.372 = 1.043 in.




Work Aid 13: Beam Diagrams and Formulas

Nomenclature

E = modulus of elasticity of steel at 29,000 ksi I = moment of inertia of beam (in.4) L = total length of beam between reaction points (ft) Mmax = maximum moment (kip-in.) M1 = maximum moment in left section of beam (kip-in.) M2 = maximum moment in right section of beam (kip-in.) M3 = maximum positive moment in beam with combined end moment conditions (kip-in.) Mx = moment at distance x from end of beam (kip-in.) P = concentrated load (kips) P1 = concentrated load nearest left reaction (kips) P2 = concentrated load nearest right reaction, and of different magnitude than P1 (kips) R = end beam reaction for any condition of symmetrical loading (kips) R1 = left end beam reaction (kips) R2 = right end or intermediate beam reaction (kips) R3 = right end beam reaction (kips) V = maximum vertical shear for any condition of symmetrical loading (kips) V1 = maximum vertical shear in left section of beam (kips) V2 = vertical shear at right reaction point, or to left of intermediate reaction point of beam (kips) V3 = vertical shear at right reaction point, or to right of intermediate reaction point of beam (kips) Vx = vertical shear at distance x from end of beam (kips) W = total load on beam (kips) a = measured distance along beam (in.) b = measured distance along beam which may be greater or less than a (in.) l = total length of beam between reaction points (in.) w = uniformly distributed load per unit of length (kips/in.) w1 = uniformly distributed load per unit of length nearest left reaction (kips/in.) w2 = uniformly distributed load per unit of length nearest right reaction, and of different

magnitude than w1 (kips/in.) x = any distance measured along beam from left reaction (in.) x1 = any distance measured along overhang section of beam from nearest reaction point (in.) ∆max = maximum deflection (in.) ∆a = deflection at point of load (in.) ∆x = deflection at any point x distance from left reaction (in.) ∆x1 = deflection of overhang section of beam at any distance from nearest reaction point (in.)

AMERICAN INSTITUTE OF STEEL CONSTRUCTION




Work Aid 14: Beam Diagrams and Formulas

(Page 1 of 14)

Beam List

No. Beam Load Type 1. Simple - Uniformly Distributed Load

2. Simple - Load Increasing Uniformly to One End.

3. Simple - Load Increasing Uniformly to Center.

4. Simple - Uniform Load Partially Distributed.

5, Simple - Uniform Load Partially Distributed at One End

6. Simple - Uniform Load Partially Distributed at Each End.

7. Simple - Concentrated Load at Center.

8. Simple - Concentrated Load at Any Point.

9. Simple - Two Equal Concentrated Loads Symmetrically Placed.

10. Simple - Two Equal Concentrated Loads Unsymmetrically Placed.

11. Simple - Two Unequal Concentrated Loads Unsymmetrically Placed.

12. Fixed at One End, Supported at Other

- Uniformly Distributed Load.


- Concentrated Load at Center


- Concentrated Load at Any Point.

15. Fixed at Both Ends - Uniformly Distributed Load.

16. Fixed at Both Ends - Concentrated Load at Center.

17. Fixed at Both Ends - Concentrated Load at Any Point.

18. Cantilever - Load Increasing Uniformly to Fixed End.

19. Cantilever - Uniformly Distributed Load.




(Page 2 of 14)

20. Fixed at One End, Free to

Deflect Vertically But Not Rotate at Other


21. Cantilever - Concentrated Load at Any Point.

22. Cantilever - Concentrated Load at Free End.

23. Beam Fixed at One End, Free to Deflect Vertically but not Rotate at Other

- Concentrated Load at Deflected End.

24. Overhanging One Support



- Distributed Load on Overhang.


- Concentrated Load at End of Overhang.


- Uniformly Distributed Load Between Supports.


- Concentrated Load at Any Point Between Supports.

29. Continuous, Two Equal Spans

- Uniform Load on One Span.


- Concentrated Load at Center of One Span.


- Concentrated Load at Any Point.

32. Simple - Uniform Load and Variable End Moments.

33. Simple - Concentrated Load and Variable End Moments




(Page 3 of 14)

Figure 84. Beam Diagrams and Formulas; AMERICAN INSTITUTE OF STEEL CONSTRUCTION




(Page 4 of 14)

Figure 84 (cont’d.); AMERICAN INSTITUTE OF STEEL CONSTRUCTION




(Page 5 of 14)





(Page 6 of 14)





(Page 7 of 14)





(Page 8 of 14)





(Page 9 of 14)





(Page 10 of 14)





(Page 11 of 14)





(Page 12 of 14)





(Page 13 of 14)





(Page 14 of 14)





Work Aid 15: Calculating Column Load and Stress

(Page 1 of 3)

Consider a column of 15 ft effective length. Compare the load capacity for the following two solid sections (Figure 95). Assume that:

E = 1.0 × 106 psi

σu = 4000 psi

F.S. = 3

8 in.

8 in. 9 in.

(1) Square section (2) Circular section

8 in.

8 in. 9 in.

(1) Square section (2) Circular section

Figure 85. Work Aid 15, Calculating Column Load and Stresses

Procedure:

Column 1.


Pu = Aσu

A = 8(8) = 64 in.2

σu = 4000 psi

Pu = 64(4000) = 256 kips

8 in.

8 in.

8 in.

8 in.




(Page 2 of 3)

b. Critical Buckling Load, Pcr

Pcr = ( )2

2

kLEIπ

E = 1.0 × 106 psi

I = ( )1288 3

= 341.33 in.4

kL = 15 ft = 180 in.

Pcr = ( ) ( )( )

( )

2 6

2

3.14 1.0 x10 341.33

180 = 103.87 kips

c. Failure Load

Pmax = Pu or Pcr, whichever is less

Pmax = Pcr = 103.87 kips


Pa = F.S.Pmax

F.S. = 3.0

Pa = 3.0

103.87 = 34.62 kips

Column 2.


Pu = Aσu

A = 2rπ = (3.14) (4.5)2 = 63.59 in.2

σu = 4000 psi

Pu = 63.59 (4000) = 254.34 kips 9 in.9 in.




(Page 3 of 3)

b. Critical Buckling Load, Pcr

Pcr = ( )2

2

kLEIπ

I = 4r 4π = 321.90 in.4

kL = 180 in.

Pcr = ( ) ( )( )( )2

62

180321.9010 x 1.03.14 = 97.96 kips


Pu or Pcr, whichever is less

Pmax = Pu = 97.96 kips


Pa = F.S.Pmax =

3.097.96 = 32.65 kips




Work Aid 16: Calculating Combined Axial Load and Bending in Column

(Page 1 of 2)

Determine the largest load P that can be safely carried by a W 250 × 80 steel column of 4.5-m effective length. Use E = 200 GPa and fy = 250 MPa.

150 mm P

cx

cy

W 250 x 80A = 10200 mm2

rx = 111.2 mmSx = 985 x 103 mm3

255 mm

150 mm P

c

150 mm P

cx

cy

W 250 x 80A = 10200 mm2

rx = 111.2 mmSx = 985 x 103 mm3

255 mm

Figure 86. Work Aid 16, Calculating Combined Axial Load and Bending in Column

a) Based on allowable combined stress and Fa = 100 MPa.

b) Based on interaction formula with Fa = 100 MPa, Fb = 0.6 Fy.

Procedure:

a) Combined Stress: Eccentrically loaded column

e = 150 mm

f = ⎟⎠⎞

⎜⎝⎛ + 2r

ec1AP

≤ Fa

Maximum allowable P

P =

2

a

rec1

AF

+




= ( )( )( )( )

( )6

22

111.2mm127.5mmmm 1501

100N/mmmm 10200

+ = 400.5 kN




(Page 2 of 2)

b) Interaction Formula:

( ) 2b

b

b

a

a 150N/mm2500.6F1,Ff

Ff

==≤+

1SF

eAF

1P1,F

s / PeF / AP

baba

≤⎥⎦

⎤⎢⎣

⎡+≤+

Maximum allowable P

( )( ) ( )( )

1

3

1

ba 150985x10150

100102001

SFe

AF1

−−

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+ = 501.1 kN




Work Aid 17: Procedure for Analyzing Footings

The analysis of a footing involves the following:

• Determine the loads on the footing.

• Calculate the bearing pressure on the soil due to the loads and weight of the footing.

• Calculate the bending moment and shear force at the critical sections of the footing.

• Evaluate the soil-bearing pressure and footing stresses by comparing the calculated values with the specified allowable values.




Work Aid 18: Calculating Soil-Bearing Pressure for a Square Footing

A column footing 12 ft square supports a concentric load, P = 120 k and an overturning moment, M = 300 k-ft. Calculate:

a. Eccentricity, e. b. Critical eccentricity, ecr.

c. Distribution of soil bearing pressures. d. Minimum footing size to guarantee no uplifting

Procedure:

a. Eccentricity, e

e = kips 120

ftkip 300PM −

= = 2.5 ft

b. Critical eccentricity, ecr

ecr = 0.167d, d = 12 ft

= 0.167(12)

= 2.0 ft.

c. Soil bearing pressure, q

e > ecr, footing uplift occurs

x = ⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ − 5.2

2123e

2d3 = 10.5 ft

qmax = bx2P =

12(10.5)2(120) = 1.90 kips/ft2

d. Minimum footing size (no uplifting)

x = d

d2.52d3 =⎟

⎠⎞

⎜⎝⎛ −

d = 15 ft

Figure 95. Work Aid 18, Calculating Soil Bearing

Pressure for a Square Footing




Work Aid 19: Calculating Stability Ratio, Moment, and Shear for a Square Footing

(Page 1 of 2)

A retaining wall and footing support loads as shown in Figure 87. Calculate the following:

a. Stability ratio b. Eccentricity c. Critical eccentricity d. Soil-bearing pressure e. Shear at critical section f. Moment at critical section

Procedure:

a. Stability ratio, SR Taking moments about point “O”

SR = ( )( )51620

MM

o

r

0= = 2.4

b. Eccentricity, e

e = PM

M = Moment about center of footing

e = ( ) ( )0−10 5 2 1.520

= 1 ft

c. Critical eccentricity, ecr

ecr = 0.167d = 1.5 ft

e < ecr

q = ⎥⎦⎤

⎢⎣⎡ ±

0.167de1

AP

Figure 87a. Work Aid 19, Calculating Stability Ratio, Moment & Shear Pressure




(Page 2 of 2)

qmax = ⎥⎦⎤

⎢⎣⎡ +

0.167de1

AP

= ⎥⎦⎤

⎢⎣⎡ +

1.51.01

9(1)20 = 3.70 kips/ft2

qmin = ( ) ⎥⎦⎤

⎢⎣⎡ −=⎥

⎦

⎤⎢⎣

⎡−

1.51.01

1920

ee1

AP

cr

= 0.74 kips/ft2

e. Shear at critical section, Vc

Vc = ( )+3.70 2.06 1 x 52

= 14.4 kips

f. Moment at critical section, Mc

Mc = ( ) ( )2

5.51.892

( )( )1 21.81 5.5 5.52 3

⎛ ⎞+ ×⎜ ⎟⎝ ⎠

= 46.86 kips-ft

Figure 96b. Work Aid 19, Calculating Stability Ratio, Moment & Shear Pressure

Figure 96c. Work Aid 19, Calculating Stability Ratio, Moment & Shear Pressure




PRACTICE PROBLEMS

Practice Problems 1

(Page 1 of 2)

(a) Three 3x4 inches pieces of wood are glued together as shown. If the glue has a maximum shear strength of 500 psi, what load P will make the beam come apart? Assume that the joint connection governs the design load.

Sol INA = 3 3

23(4) 4 (3)2 (3 4)(3.5)12 12

⎡ ⎤ ×+ × +⎢ ⎥

⎣ ⎦

= 2(16+147) + 9 = 335 in4 max. V in beam = 0.6 P. Check τ at glue line:

τ = [ ]VQ 0.6P 3 4 (3.5)It 335 3

= × ××

= τall = 500

+

_

P

0.4PSF

-0.6

+

_

P

0.4PSF

-0.6




P = 500 335 30.6 12 3.5

× ×× ×

= 19,940 lbs = 19.94 kips

(Page 2 of 2)

(b) For P obtained, check maximum shear stress in the wood of the beam itself. Sol τmax in beam may be at neutral axis (depends on variation of “t”)

= VQ 0.6 19,940 1.512 3.5 4 1.5It 335 4 2

× ⎡ ⎤= × + × ×⎢ ⎥× ⎣ ⎦

= 415 psi (< 500 psi)

Check τmax in beam just above glue line = 0.6 19,940 (12 3.5)335 3×

××

= 500 psi

τmax of wood should be ≥ 500 psi (i.e. shear strength of wood)

Note

max. τ (through thickness) always where tQ is GREATEST.




Practice Problems 2

(Page 1 of 2) Shown below is a cantilever beam subjected to a force P (assume units lbs.) at the free end. The beam is made up of 3 planks of wood which are nailed as shown. Find the maximum permissible magnitude of P, given that (i) the allowable shear stress in the wood is 200 psi (ii) allowable bending stress (tension or compression) is 2,500 psi and (iii) shear rating of nail Rn is 400 lbs. The nails are spaced uniformly at S = 3 in. centers.

INA = 12

2(2.5)(4)12

6(6) 33

− = 81.33 in4

Shear in Wood Vmax = P

max τmax = maxV QmaxI t

⎛ ⎞× ⎜ ⎟⎝ ⎠

P




(Page 2 of 2)

200 = P 6 1 2.5 2 1 181.33 1

× × + × ×⎛ ⎞⎜ ⎟⎝ ⎠

⇒ P = 957 lbs.

Bending Stress in Wood

max σ = I

)(yM)(max max

2,500 = 60P 381.33

× ⇒ P = 1,130 lbs.

Nail Rating

q = I

VQ (to be calculated where flange is nailed to web)

q = nRP 400(6 1 2.5)81.33 S 3

× × = =

P = 723 lbs. Answer: Choose smallest P i.e. P = 723 lbs.




Practice Problems 3 A beam is to be fabricated from 2 sections of 2-in. standard steel pipe and a ¼ in. thick

steel plate by 4 fillet welds. If Vmax = 2000 lb, determine whether 81 -in. fillet welds with a

capacity of 100 lb/in. each will be sufficient.

Ixx = [ ]12

2.375)(1241

1.07(6)0.66623

2−

++

= 96.95 in4

2q = I

VQ ∴ q = [ ]2000 1.07 62 96.95

××

= 66.2 lb/in. (< 100 lb/in. OK)

4k

+2k

-2k

F

F

4k

+2k

-2k

F

F

qq qq

2.375xxload

2

4x-x

A = 1.07 inI = 0.666 in

2.375xxload

2.375xxload

2

4x-x

A = 1.07 inI = 0.666 in

x x

6"

12"

¼"

1 1

pipe

Steelplate

weld

x x

6"

12"

¼"

1 1

pipe

Steelplate

weld




Practice Problems 4

(Page 1 of 2)

Two 100 mm x 25 mm boards and two 150 mm x 25 mm boards are used in fabrication of a box beam. If Rn (shear resistance of each nail) = 1000 N and Vmax = 1 kN, determine maximum spacing s of nails for the two different schemes (a and b) shown.

(a)

INA = 3 31 10.15 (0.15) 0.1 (0.1)12 12

× × − × ×

= 33.85x10−6 m4

2q = I

VQ

q = [ ]3

6

VQ 1 10 0.15 0.025 0.06252I 2 33.85 10−

×= × ×

× ×

q = 3,462 N/m

& qs = Rn ⇒ s = 3,4621,000 = 0.288 m = 289 mm

q qq q

NA

25 mm

25 mm 25 mm

25 mm

100 mm

100 mm

NA

25 mm

25 mm 25 mm

25 mm

100 mm

100 mm




(Page 2 of 2) (b)

2q = I

VQ

where Q = 0.100×0.025×0.0625 = 1.5625×10−4 m3

q = 3

46

1 10 1.5625 102 33.85 10

−−

×× ×

× ×

= 2,308 N/m

& s = 2,3081,000 = 0.433 m

= 433 mm

(Scheme (b) is better Q Q(b) < Q(a)).

* Note: Shear stress in wood itself is not of interest in this problem.

qq

qq

NA

25 mm

25 mm 25 mm

25 mm

100 mm

100 mm

NA

25 mm

25 mm 25 mm

25 mm

100 mm

100 mm




Practice Problems 5

(Page 1 of 2)

The beam shown below is subjected to a uniform load w. Determine the maximum value of w if the allowable bending stress is 24 ksi and the allowable shear stress is 15 ksi. The moment of inertia about the neutral axis is 768 in4. (i) Bending

Note: 12824

4=⇒= T

T σσ

3"

4"

8"

12"

N.A.

3"

4"

8"

12"

N.A.+

82WLM =max

10 ft

+-

2WL

SFD

BMD

w(kips/ft)

2WL−

5'

+

82WLM =max

10 ft

+-

2WL

SFD

BMD

w(kips/ft)

2WL−

+

82WLM =max

10 ft

+-

2WL

SFD

BMD

w(kips/ft)

2WL−

5'




Sol Mmax = 2 2wL (10 12) w

8 8×

= (k-in.) (here, “w” in units of k/in.)

= 1800w k-in (Page 2 of 2)

σmax = maxMy Mc 1800w 8 24I I 768

×= = =

w = 1.28 k/in. = 1.28×12 = 15.36 k/ft.

(ii) Shear

Variation of τ with y

Vmax = wL w 10 122 2

× ×= (w ⇒ k/in.)

= 60w kips

Also, critical section through thickness for T-beam is NA (for shear).

max τmax = 15 = max

max

V Q 60w 12 4 2I t 768 3

× ×⎛ ⎞ ⎡ ⎤=⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

w = 6 k/in. = 72 k/ft.

Answer: w = 15.36 k/ft. (smaller of two)




Practice Problems 6

(Page 1 of 2)

Two 2-in.-by-6-in., full sized wooden planks are glued together to form a T-section as shown. If a moment M = +2,270 ft-lb. is applied around the z-axis, find

∗ bending stresses at top and bottom fibers (Iz = 136 in4)

∗ the total compressive force resultant C due to the bending stresses (above the neutral axis)

∗ the total tensile force resultant T due to the bending stresses (below the neutral axis) and compare it to the force C.

Cross-Section σ - Distribution

σt = 2270 12 3136

× × = 600.9 psi

σm = 600.9 ×1/3 = 200.3

σb = 2270 12 5136

× × = 1001.5 psi

tσ

bσ

1C2C

3C

T

z

y




T = 1 1001.5 5 22

× × × = 5007.5 lbs.

(Page 2 of 2)

C = 1

2 3

CC C

1 1200.3 6 2 400.6 2 6 200.3 1 22 2

× × + × × × + × × ×14424431442443 1442443

= 2403.6 + 2403.6 + 200.3 = 5007.5 lbs. = T OK



160

GLOSSARY

Allowable Load Load that the column or other structural member can safely support.

Applied Loads Loads that a structural member supports for design purpose or in actual service.

Beam Linear structural member having one or more supports and supports load perpendicular to its longitudinal axis.

Beam-Column A structural member that experiences a significant amount of bending as well as axial load.

Bending Moment Internal moment at a beam section required to maintain the equilibrium of any part of the beam.

Column Linear structural member loaded primarily along its longitudinal axis.

Compressive Yield Load Load that will cause yielding of the column material in compression.

Conjugate Beam Method A method to determine deflection of a beam. Critical Load Load at which the column will become unstable or

begin to buckle. Eccentric Load Load applied away from the geometric center of a

column or supporting footing. Load Eccentricity Distance from the center of the column cross-section or

of a footing to the load application point. Modulus of Rupture The flexural stress in a beam when the bending

moment of the beam is the maximum value at the point of rupture.

Moment Diagram Graphical representation which shows the variation of the bending moment along the longitudinal axis of a beam.

Neutral Axis Plane in a beam that remains the same length when it is subjected to bending.

Shear Diagram Graphical representation which shows the variation of the shear force along the longitudinal axis of a beam.

Shear Force The internal force at a beam section required to maintain the equilibrium on any part of a beam.

Serviceability The ability of the structure to support a specified load



161

without undergoing unacceptable deflection, deformation, or movement.

Slenderness Geometric property of a column that makes it tend to buckle under axial load.

Soil Bearing Pressure Load per unit area produced by the footing on the underlying soil.

Stability The ability to support a given compressive load without experiencing a sudden change in geometry, or buckling.

Stability Ratio The factor of safety against the overturning of a footing or foundation; the ratio of the resisting moment to the overturning moments.

Statically Determinate Beams

Beams whose reactions can be found from equations of equilibrium alone.

Statically Indeterminate Beams

Beams whose reactions cannot be found from the equations of equilibrium only, but require additional equations to determine the reactions.

Strength The ability of the structure to safely support a specified load without experiencing excessive stresses.

Superposition Method to determine the shear and moment diagrams for a beam by combining the results from known load cases.

Ultimate Compressive Load

Load that will produce crushing failure of the column material.

analysis+of+structural+components

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