Analysis of Structural Components

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SAUDI ARAMCO ANALYSIS OF STRUCTURAL COMPONENTS

Note: The source of the technical material in this volume is the Professional Engineering Development Program (PEDP) of Engineering Services. Warning: The material contained in this document was developed for Saudi Aramco and is intended for the exclusive use of Saudi Aramco’s employees. Any material contained in this document which is not already in the public domain may not be copied, reproduced, sold, given, or disclosed to third parties, or otherwise used in whole, or in part, without the written permission of the Vice President, Engineering Services, Saudi Aramco.

Chapter : Civil Engineering File Reference: CSE 106.02

For additional information on this subject, contact PEDD Coordinator on 862-1026

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Civil Engineering: Basic Properties of Section Analysis of Structural Components

Section OBJECTIVES

Page ......................................................................................................... 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

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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

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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 GLOSSARY ..................................................................................................... 101 ..................................................................................................... 160 PRACTICE PROBLEMS ............................................................................................. 149

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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
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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.

M Diagram............................................................................................... 47 EI

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

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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

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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

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

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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.

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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.

y

Loads

x

Supports Span

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

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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)

Timber Sections
Reinforced Concrete

Structural Steel Shape Steel Plate (a) Steel and Concrete (b) Steel and Timber

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.

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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.

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Support Types

1

Simple Beam

A beam that nests on simple supports at each end.

Simple Support L

2

Cantilever Beam

A beam supported (fixed) at one end only A beam resting on two or more supports, which has one or both ends projecting beyond the support. A beam that rests on more then two supports. A beam that is restrained from rotation and movement at its ends. A cantilever beam fixed at one end and with a simple support at the other end. A cantilever beam with its free end restrained against rotation.

Fixed End

Free End

3

Overhanging Beam

Overhang

4

Continuous Beam

5

Fixed End Beam

6

Propped Cantilever Beam Guided Cantilever Beam

7

Figure 3a. Support Beams

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Pictures Showing Support Types

Hinge Supports

Roller Support

Figure 3b. Support Beams

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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.
P w = lb/ft

a) Concentrated Load

b) Uniform Load

M

c) Nouniform Load

d) Induced Moment Load

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.

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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 xaxis 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.

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y - axis w

Lateral Loads

P

Beam x - axis

z a) Beam Loading

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

Lateral Deflection, y
V M

x Reaction Support

b) Beam Free-Body

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.
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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.
w P
a) Loaded Beam

xo y w
V

P x

Rl

V

Rr

b) Free-Body: Shear found by summing vertical forces on either side of the section:

∑F

y

=0

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):

w=−

dV , dx

V = − ∫ wdx

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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.
Shear Force Left v Right Left v Right

Negative Shear

Positive Shear

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: V= dM , dx M = ∫ Vdx

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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: d 2M w=− , dx 2 M = − ∫ ∫ wdx

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

P

Upper fibers in Compression +M V

C T

Lower fibers in tension

a) Positive Bending Moment P

-M

Lower fibers In compression

Upper fibers in tension

b) Negative Bending Moment

Figure 8. Bending Moment

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Example 1:

Reactions, Shear, and Bending Moment in A Simple Overhang Beam
(Page 1 of 4) For the overhang beam shown below, calculate: a) b) c) Reactions R1 and R2. The shear and bending moment at mid-span (C), which is 5 ft from left support (A) of the beam. The shear and bending moment at support (B).

P=1200 lb
A C B

10 ft R1 R2

6 ft

Figure 9. Overhang Beam

Example 1:

Solution
a) Solve for reactions R1 and R2. Taking moments about R1:

Σ M1
10R2 R2

= 0 = (1200)(16) – R2 × 10 = 19,200 = 1920 lb

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Σ Fy
R1 R1

= 0 = 1200 + R1 – R2 (Page 2 of 4) = 1920 – 1200 = 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.
A V C

5 ft R1 = 720 lbs

(ii)

Sum vertical forces to obtain shear.

ΣFy = 0, ∴V = −R1 = −720 lb
(iii) Sum moment about C to obtain bending moment.

ΣM C
M

= 0,

M + 5 × 720

=0

= −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:

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(Page 3 of 4)
P = 1200 lb VB1 MB1 R1 VB1 MB1 MB2 VB2 MB2 VB2

R2

Figure 10. Shear and Moment at Support B

Σ F = 0,

VB1 VB2

= −R1 =P

= −720 lb = 1200 lb

R2 = − VB1 + VB2 = 1920 lb

Σ M = 0,

MB1 = −10(R1) MB2 = −6(P)

= − 10 × 720 = −7200 lb-ft = − 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:
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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.
P
A C x R1 L R2

B

a

Figure 11. Drawing Shear & Moment Diagrams Again solving for R2 by Σ M1 = 0 R2 = P(L + a ) L

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

M

R1 = P − R 2 =

PL − P(L + a ) Pa =− L L

Summing vertical forces on the free-body of the beam up to any point (x) between supports yields: V = +R1 = − Pa L 0≤x<L

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(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:

M

V L+a-x

P

V2 = +P

L < x < L+a

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

+P + -Pa L -

Figure 12. Shear Diagram for the Overhang Beam in Algebraic Terms The bending moment at any point (x1) between supports is: M = R 1x = − Pax L 0≤x≤L
⎞ 0 ≤ x ≤ L⎟ ⎠

dM Pa ⎛ =− =V ⎜ Note dx L ⎝ At support 2, x = L and
M = −Pa
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(Page 3 of 3) On the overhang:

V L R1 x x-L R2

M

M = R1x + R2(x−L) M= − Pax ⎛ Pa ⎞ + ⎜P + ⎟(x − L) L L ⎠ ⎝ L ≤ x ≤ L+a

= −P(L+a) + Px At support 2, x = L ∴ M = −Pa

dM ⎛ =+P = V ⎜ Note dx ⎝

⎞ L ≤ x ≤ L + a⎟ ⎠

Therefore, the moment diagram is as follows:

- Pa

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.

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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 wo = W L
A L Ra x

W = total load
wo B

Reactions: Solve for Rb by summing moment about A

Rb

∑M

A

=0

R bL =

L W 2 W w oL Rb = = 2 2
=0 W w oL = 2 2
x

Summing vertical forces,

∑F
Shear: At A At B

Figure 14. Uniform Load Distribution

Y

Ra = W - Rb =

Va = R a =

W 2 −W 2

wo

M

Vb = − R b =

V

Ra

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(Page 2 of 3) At x, summing forces on the free body:

ΣF = Ra – wox – V = 0
V = Ra – wox = For x = W Wx W ⎛ 2x ⎞ − = ⎜1 − ⎟ 2 L 2 ⎝ L ⎠

L , V=0. 2

Shear Diagram:

W/2 =
w oL 2

L/2

L/2

-W/2 =
-w o L 2

Figure 15. Summing Vertical Forces

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(Page 3 of 3) Bending Moment: At x, summing moments on the free-body to the left of x: x ΣMx = Rax − wo × x × − M(x) = 0 2 2 x M(x) = Rax – wo 2

= M(x) =

Wx W ⎛ x 2 ⎞ ⎜ ⎟ − 2 L ⎜ 2 ⎟ ⎝ ⎠

Wx ⎛ x ⎞ ⎜1− ⎟ 2 ⎝ L⎠

⎞ ⎛ dM W Wx W ⎛ 2x ⎞ = − = ⎜ Note ⎜1− ⎟ = V ⎟ ⎟ ⎜ dx 2 L 2 ⎝ L ⎠ ⎠ ⎝

Maximum Moment: Mmax at V = 0 V = 0 at x =
∴Mmax =

L 2

WL ⎛ L ⎞ ⎜ 1− ⎜ L(2) ⎟ ⎟ (2)(2) ⎝ ⎠ WL w oL2 = 8 8

Mmax =

Moment Diagram:
Mmax.=0.125 WL = 0.125 woL2

L/2

w o L2 = 8

Figure 16. Bending Moment

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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.

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a) Loads

Mo x L

P L

w L

wo

L

+ b) Shear Diagrams Vx = 0

+ -P Vx = -P

+ Vx = -wx -wL
Vx =

+ − x2 wo 2L − w oL 2

+ c) Bending Moment Diagrams Mx=-Mo

-Mo

Mx=-Px

-PL

Mx =

− wx 2 2

- wL2 2

Mx =

− wo 3 x 6L

- w oL2 6

Figure 17. Cantilever Beam Subjected to Four Types of Load

a a) Loads L

Pb

w

Total Load W =

w oL 2

wo L L

b) Shear Diagrams

Pb L

+ - Pa L

wL 2

Vx = −

+

-

W 3 - wL 2

w o x 2 w oL + 2L 6

+

-

- 2W 3

Mx = Mmax = Pab L Mmax = wL2 8

wx (L − x ) 2

Mx =

Wx 2 (L − x 2 ) 3L2

Mmax=0.1283 WL

c) Moment Diagrams

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

2 kips/ft

A

10 ft

B

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

5 kips

10 ft

Figure 20a. Concentrated Load
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(Page 1 of 2) b) Uniform load, wo = 2 kips/ft L = 10 ft

2 kips/ft

10 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

Figure 21a. Cantilever Beam with Concentrated Load

b)

Cantilever beam with uniform load 2 wL2 = 2(10 ) = 100 kip − ft Mmax = 2 2

-

100 kip-ft

Figure 21b. Cantilever Beam with Uniform Load

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(Page 2 of 2) Step 3 Construct the composite diagram by combining individual cases. Mmax = 50 + 100 = 150 kip-ft
2 kips/ft
(a) (b)

150 kip-ft

(c) = (a) + (b)

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

Concentrated Loads

Distributed Loads

Bending Moment

Shear Force

Bending or Flexural Stress

Vertical and Horizontal Shear Stress

Figure 23. Stresses in a Beam

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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.

Compression cc M M

y

N.A.

ct

y

Neutral Axis f(y)

Neutral Axis (N.A.) a) Bending Moment

Tension b) Deformed Beam c) Stress Distribution

Figure 24. Flexural Stress

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

σ=

My I

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:

σc = σt =

Mc c I Mc t I

compressive stress

tensile stress

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

σ=
where: S=

M S

I I = cc c t

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.

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Shear Stresses

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

Neutral Axis Area d y b Shear Section (a) Flexural Stress y

Element A

Bending Moment M

Vertical Shear

τ

Shear Stress (b)

Shear Force (V)

Horizontal Shear Element A (c)

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 crosssection of a beam is:

τ=
where: V

VQ Ib

= external vertical shear on beam (lb, N)

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I b A

= moment of inertia of whole section (in.4, mm4) = width of section at plane where stress is desired (in., mm) = area of section beyond plane where stress is desired (in.2, mm2) = distance of center of gravity of area to neutral axis (in., mm) =
Ay , the static moment of the area beyond the plane being considered, taken about the neutral axis.

y
Q

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

τ avg =

V A

For a beam with a rectangular section; b x d: A = bd, Qmax = bd3 I= 12 Therefore, or bd2 (for section at neutral axis) and 8

τmax = 1.5

V V , τavg = bd bd

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:

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Lower fiber of upper plank stretch Upper fibers of lower plank shorten Planks slide past each other

a) Planks Not Connected • Shear stress along this line

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.

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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.

P = 1200 lb A 10 ft 720
5 in.

6 in.

B

6 ft

N 5 in.

A 10 in.

1920

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: σ= Mc M = I S
2

bd2 6(10 ) For rectangular section, S = = = 100 in.3 6 6 Max. flexural stress,

σ max =

7200 x 12 lb in. = 864 lb/in.2 3 100 in.

Maximum tensile stress is at the top of section over support B. Maximum compression occurs at bottom of beam at support B.
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(Page 2 of 2) Shear stress,

τ=

VQ Ib

τ avg =

V A

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

Qmax = I= bd3 12

bd d bd2 × = 2 4 8
d/4

b

τavg

A d

t=b 1 2 bd 8 3 V × 2 bd
Neutral axes

τmax

τ max =

1 3 bd × b 12

V=

Figure 28. Maximum Shear at Center of Beam Section

= 1.5 = 1.5

V = 1.5t avg A 1200 = 30 psi 6(10 )

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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.

50 in.

P = 10,000 lb A 100 in. 12.515 in. A 5,000 lb y = 1.47 in. I = 62.6 in.4

5,000 lb

1.485 in.

1.47 in. N.A.

6.94 in.

0.905 in.

5.47 in.

Section AA

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: Mmax = Therefore, Mmax = PL (10,000 )(100 ) = = 250,000in. − lb 4 4 Pab L ,a = b = L 2

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(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: σt = Mc (250,000 )(5.47 ) = 21,845 psi (tension) = I 62.6

The top portion of the beam is in compression, σc = Mc (250,000 )(1.47 ) = = 5870 psi (compression) I 62.6

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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.

L y

w

Lateral Loads

P
Neutral Axis x

Internal Force: - Bending Moment, M - Shear Force, V x -y(x)

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

ρ
¦ ymax¦

θ

Elastic Curve for Deformed Beam

Figure 30. Elastic Curve of a Beam

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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 = Slope, Deflection,

d2 y M = , ρ = radius of curvature of beam 2 dx EI dy = dx y(x) = M ∫ EI dx

θ =

∫∫ EI dx

M

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.

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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.

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The relationships between the conjugate beam and the actual beam are as follows: Conjugate Beam Simple End* Hinge* Free End* Built-in End Load, w* Reaction, R* Shear, V* Moment, M* Actual Beam Simple End Interior Support Built-in End Free End M EI End Slope, θ Slope, θ Deflection, y

*Indicates parameters related to conjugate beam.

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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.

80 lb

120 lb

a 30 in. 40 in.

b 20 in.

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.

80 lb

Cover Plates

120 lb

EI 30 in.

a

2 EI 40 in.

b

EI 20 in.

EI = 2.355 x 106 lb-in.2

Figure 32. Cover Plates Added on Flanges of Beam

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Solution:

Example 7A - Deflection of Uniform Beam
(Page 2 of 6)

80 lb

120 lb

a 30 in. 40 in.

b 20 in.

Repeat of Figure 31

Step 1.

Moment diagram for actual beam:

2400 lb-in.

M

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* = = M EI 2400 lb - in. 2.356 x 10 6 lb - in 2

= 1.019 x 10-3 in.-1

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(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.
1.019 x 10-3

M EI

R*1

x*
90 in.

R*2

*
20

(2)

*

(1)

*

(3)

50 76.67 13.33

Figure 33b. Loading Diagram for Actual Beam
Area (1) (2) 40(1.019 x 10-3) = 4.076 x 10-2 x x* 50 M about Pt. 1* = 203.8 x 10-2

30 2

(1.019 x 10-3) = 1.529 x 10-2

x

20

= 30.58 x 10-2

(3)

20 2

(1.019 x 10-3) = 1.019 x 10-2 6.623 x 10-2

x

76.67

= 78.13 x 10-2 312.5 x 10-2 = 90R*2
= 3.472 x 10-2

Total

R*2
R*1

=
=

312.5x10 -2 90

6.623 x 10-2 − R*2 = 3.151 x 10-2

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(Page 4 of 6) Step 5: Deflection - Conjugate Moment, M*

•

Deflection at a:
ya = M*a = 3.151 x 10 (30) – (1.019 x 10 )
-3 -2

= 94.53 x 10-2 = −15.29 x 10-2 79.24 x 10 = 0.7924 in.
-2

30 30 × 2 3
ya

•

Deflection at b:
yb = M*b = 3.472 x 10 (20) 20 -3 20 × – (1.019 x 10 ) 2 3 yb
-2

= 69.44 x 10-2 = −6.79 x 10-2 62.65 x 10 = 0.6265 in.
-2

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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.

80 lb

Cover Plates

120 lb

EI 30 in.

a

2 EI 40 in.

b

EI 20 in.

EI = 2.355 x 106 lb-in.2

Repeat of Figure 32

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

M

Figure 34a. Moment Diagram

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(Page 6 of 6) Steps 2 – 4: Conjugate beam load
1.019 x 10-3

1.019 x 10-3
(2)

* 0.5095 x 10-3 * (1) *

(4)

* R*1

(3)

R*2

Figure 34b.
Area From Example 7A -40(0.5095 x 10-3) Total 6.623 x 10-2 = -2.038 x 10-2 X 50 = 4.585 X 10-2 R*2
R*1

M Diagram EI
M about Pt. 1* 312.5 x 10-2 = -101.9 x 10-2 210.6 X 10-2 =
210.6 x10-2 90

x*

For Items (1), (2), and (3) For Item (4) = 90R*2
= 2.34 X 10-2

= (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 (30) – (1.019 ∗ 10 )
-3 -2

= 67.35 × 10-2 =
2

30 30 × 2 3
ya

−15.29 × 10

-

52.06 × 10 = 0.5206 in.

-2

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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.

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• •

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. 2. 3. Identify the beam/load type that corresponds to the problem being solved. Obtain the appropriate diagrams and formulas from the Work Aids. Use the superposition technique if your problem is not covered but can be composed from two or more beam/loading cases that are covered. Calculate the desired values for the beam by substituting the appropriate known values into the formulas. Draw the shear and moment diagram if needed.

4. 5.

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

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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.

x

W = total load wo

A L W

B

Ra

Rb

A Ra L/2 L/2

B Rb

Repeat of Figure 14

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

2.

•

Maximum shear, Vmax = supports.

W at the right and left 2 WL at x = L/2 from the left 8 5WL3 at x = L/2 from the 384 EI

•

Maximum moment, Mmax = support.

•

Maximum deflection, ∆max = left support.

3. 4.

Superposition is not required. Calculate required values based on the specified values for W, L, E, and I: Vmax = Mmax = W 300 = = 150 kips 2 2 WL = 8

(300 )(120 )
8

= 4500 kip-in.
3

∆max =

5WL3 5(300 )(120 ) = = 1.5 in. 384 EI 384(1.5 x 10 3 )(3000 )

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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.

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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.
P w/2 w

a

L

b

Figure 35. Superposition of Beam/Load Cases

Solution - Example 9
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
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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.
R = 100 kips w = 100 kips

P = 50 kips

C
10 ft 10 ft Beam E = 29,000 kips/in.2 (steel) I = 1000 in.4 5 ft

Figure 37. Example 10 Beam Solution - Example 10

Use superposition to combine the following beam/load cases:

Figure 38. More Beam Load Cases

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(Page 2 of 4) Use superposition to combine the following beam/load cases:

Figure 39a. Beam Load Cases

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(Page 3 of 4)

Shear Diagrams (Unit kips) d. Superposition: Combined Diagrams Case 7 + Case 24 + Case 26
75 35 70 50

Moment Diagrams (Unit: kip-ft)

Mc = 550

65

M2 = 300
105

Figure 39b. More Beam Load Cases

Deflection, ∆c, midpoint between support, x = 10 a. Case 7 RL3 ∆c = 48EI 100(240 ) = = 0.993 in. 48(29,000 )(1000 )
3

b.

Case 24

∆x

=

wox 4 L − 2L2 x 2 + Lx 3 − 2a 2L2 + 2a 2 x 2 24EIL
w oL2 ⎡ 5 2⎤ 2 ⎢ 4 240 − 3(60 ) ⎥ 96EI ⎣ ⎦

[

]

For x = L /2, ∆ c =

(

)

∆c

=

4(240 ) /12 ⎡5 2⎤ 2 ⎢ 4 (240 ) − 3(60 ) ⎥ = 0.422 in. 96(29,000 )(1000 ) ⎣ ⎦
2

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(Page 4 of 4) c. Case 26

∆x

=

Pax 2 L − x2 6EIL

[

]

For x =

L PaL2 (up) , ∆c = 16EI 2
2

∆c
d.

50(60 )(240 ) = = 0.372 in. (up) 16(29,000 )(1000 )

By superposition - Combine Case 7 + Case 24 + Case 26 Total ∆c = 0.993 + 0.422 – 0.372 = 1.043 in.

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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.

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P P P 2nd story columns

columns

1st story columns

a. Individual Columns

b. Columns in Braced Steel Frame

c. Columns in Rigid Reinforced Concrete Space Frame

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.

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•

Compound or built-up sections.

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

a: Structural Steel Rolled Shape

b: Reinforced Concrete with Round-Spiral Ties

c: Concrete-Filled Tube

d: Reinforced Concrete Encased Steel

e: Steel-Timber Composite

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

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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.

•

• •

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P

x-axis e b d

Area Moment of Inertia, I

A L

r=

I A

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

Loads H

P

Beam

P Le = kL H

P

P

Columns Footings

L

a) Structural Frame

b) Structural Model

Figure 43. Columns as Part of a Structural Frame

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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 Reinforced Concrete Timber

kL/r < 20. kL/r < 22. 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 Reinforced concrete

kL/r > 120. 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.

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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 crosssection. The eccentricity e of the load (that is distance of load P to the centroid) must be less than 5% of the smaller crosssectional 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.

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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.

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M V

P 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-

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sectional area A: Pu = σuA or σu =
Compressive Yield Load

Pu A

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: P Py = σyA or σy = y A
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: π 2EI with k = 1 Pcr = (kL) 2 where: I E = the elastic modulus of the column material = 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: P π 2E σ cr = cr = A (L/r )2
I , the minimum radius of gyration of the column A 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,

where r =

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for a column with pinned-end supports. Generally, however, the critical (buckling) stress is given as

σcr =

π 2E (kL/r) 2

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: Pcr =

(kL )2

π 2EI

σ cr =

(kL/r )2

π 2E

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.

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(a)

(b)

(c)

(d)

(e)

(f)

Buckled shape of column is shown by dashed line

Theoretical k value Recommended design value when ideal conditions are approximated

0.5

0.7

1.0

1.0

2.0

2.0

0.65

0.80

1.2

1.0

2.10

2.0

Rotation fixed and translation fixed End condition code Rotation free and translation fixed Rotation fixed and translation free Rotation free and translation free

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

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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 =

π 2E (kL/r) 2

;

then:
EI σuA

For failure load: For failure stress:

The critical length Lc = kL = π

The critical slenderness ratio Cc =
E kL =π σu r

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

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straightness.

• •
Factor of Safety

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.

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 :

Pa =

Pmax F.S.

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 =
=

Pa σ max = A F.S. σ σu or cr , whichever is less. F.S. F.S.

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.
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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 = 4000 psi =3

σu
F.S.

•

Calculate the following for each column arrangement: a. b. c. d. Ultimate Load. Critical buckling load and stress. Failure load. 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

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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 A = Aσu = 8(8) = 4000 psi = 64(4000) = 256 kips
8 in.
2

8 in.

= 64 in

σu
Pu b.

Critical Euler Buckling Load, Pcr Pcr = E I kL

(kL )2

π 2EI

(k = 1)

= 1.5 × 106 psi bd3 8(8)3 = = = 341.33 in.4 12 12 = 12 ft = 144 in.

Pcr =

(3.14 )2 (1.5x106 )(341.33 ) (144 )2

= 243.44 kips

Critical stress, σcr = c. Failure Load, Pmax

Pcr 243.44 2 = = 3.80 kips/in. A 64

Pmax = min of [Pu or Pcr] Pmax = Pcr = 243.44 kips d. Allowable Load, Pa Pa = Pmax F.S.

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F.S. = 3.0 Pa = 243.44 = 81.15 kips 3.0 (Page 3 of 3) Allowable stress, σa = Column 2. a. Ultimate Load, Pu Pu A Pu b. = Aσu = (10)2 – (6)2 = 64 in.2 = 64(4000) = 256 kips
6 in. 10 in. 6 in. 10 in

Pa 81.15 2 = 1.27 kips/in. = A 64

Critical Buckling Load Pcr Pcr = I kL =

(kL )2

π 2EI

10 6 (10)3 – (6)3 = 725.33 in.4 12 12

= 144 in.

Pcr =

(3.14 )2 (1.5x10 6 )(725.33 ) (144 )2

= 517.3 kips

Critical stress, σcr = c. Failure Load, Pmax

Pcr 517.3 k 2 = 8.08 kips/in. = 2 A 64 in.

Pmax = Pu or Pcr, whichever is less Pmax = Pu = 256 kips d. Allowable Load, Pa Pa = Pmax 256 = = 85.33 kips F.S. 3.0

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Allowable stress, σa =

Pa 85.33k 2 = 1.33 kips/in. = 2 A 64in.

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

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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: Bending (flexural) stress: Combined stress:

fa = ± fb = ±

P A M S P M ± A S

f tot = fa + fb = ±

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: f = Fy P M ± ≤ A S (F.S.)y

and also f tot ≤

Fu . (FS)u 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 beamcolumns, the interaction formulas are more meaningful and are applied as follows:

•

Axial load and bending about one axis:

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fa f P/A M/S + b ≤1 , + ≤1 Fa Fb Fa Fb

where: Fa Fb • = allowable stress when only axial load is present in the member (axially-loaded column). = allowable stress when only bending is present in the member (beam flexure).

Axial load and bending about both axes: fby fa f + bx + ≤1 Fb Fbx Fby

P My Mx

y

x

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

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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 = P M ± A S

where: M = Pe Smin = I I c

= Moment of inertia.

c = Distance from the center of the column section to the extreme fiber. Therefore: f = P Pe Ac P ⎛ ec ⎞ ± • = ⎜1 ± 2 ⎟ A A I A⎝ r ⎠
Load, P
Eccentricity, e

where: the radius of gyration, r = I A
Area, A

c

Figure 49. Eccentrically Loaded Column

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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.

200 mm x W 310 x 74 A = 9480 mm2 rx = 131.6 mm ry = 49.8 mm Sx = 1058 x 103 mm3

P

C

y

C

Figure 50. Eccentrically Loaded Steel Column
a) b) Based on allowable combined stress and Fa = 100 MPa. 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 P ⎛ ec ⎞ ⎜1+ 2 ⎟ ≤ Fa A⎝ r ⎠

fapp =

Maximum allowable P1 is obtained by setting fapp = Fa : P1 =

AFa ec 1+ 2 r

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=

(9480mm )(100N/mm )
2 2

(200mm )(155mm ) 1+ (131.6mm )2

= 339.8 kN

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(Page 2 of 2) b) Interaction Formula: fa fb + ≤ 1, Fb = 0.6(250 ) = 150N/mm 2 Fa Fb
⎡ 1 e ⎤ P/A Pe/S + ≤ 1 , P⎢ + ⎥ ≤1 Fa Fb ⎣ AFa SFb ⎦

Maximum allowable P2 is obtained by setting the sum of dimensionless stress ratio to 1: ⎡ 1 e ⎤ P2 = ⎢ + ⎥ ⎣ AFa SFb ⎦
−1

⎡ ⎤ 1 200 =⎢ + ⎥ 3 ⎣ (9480 )(100 ) 1058x10 (150 ) ⎦

−1

(

)

= 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 reevaluate the column capacity.

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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.

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b d d

a) Strip Footing (wall)

b) Spread Footing (Column)
d

d L

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.

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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

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.

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P

P e

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 = where: A = contact area of the footing. P A

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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 = P Mc ± A I P Mc + A I P Mc − A I

qmax = qmin where: I c =

= moment of inertia of the contact area. = distance from center to edge of footing.

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Figure 55. Combined Vertical Load and Moment

•

For a rectangular footing: A = bd I bd2 S = = c 6 and the equation for soil bearing pressure, q, becomes: q = P 6M P ⎡ M/P ⎤ ± 2 = ⎢1 ± bd bd A ⎣ 0.167d ⎥ ⎦

•

For a circular footing, diameter = d: A = Therefore: q = P⎡ M/P ⎤ ⎢1 ± 0.125d ⎥ A⎣ ⎦ S = 0.1094 d3 π 2 d , 4 S = πd3 32

•

For an octagonal footing, d = distance across the parallel faces: A = 0.8284 d2, Therefore: q = P⎡ M/P ⎤ ⎢1± 0.132d ⎥ A⎣ ⎦

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Eccentric Load q = = where: e = Load eccentricity r = Radius of gyration,
e

P (Pe )c ± A I P ⎡ ec ⎤ 1± A ⎢ r2 ⎥ ⎣ ⎦

I A

P

qmin

qmax

Figure 56. Vertical Load and Moment

•

For a rectangular footing contact area: A = bd r
2

d2 = 12

e b

d c = 2 Therefore: q = P ⎡ 6e ⎤ 1± ⎥ bd ⎢ d⎦ ⎣

*P
d
Figure 57. Contact Area

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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:
• • •
Partial Compression

For rectangular footing, For circular footing, For octagonal footing,

ecr = 0.167 d. ecr = 0.125 d. ecr = 0.132 d.

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 = M > ecr = 0.167 d P

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 ∴ qmax = ⇒

P=

1 qmax × b 2

2P bx

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Also, resultant of soil pressure, “R”, should coincide with line of action of “P” for equilibrium, thus: x d = −e 3 2 or:
⎛d ⎞ x = 3⎜ − e ⎟ ⎝2 ⎠

Figure 58. Partial Compression

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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 = b. M 150 kip − ft = = 1.5 ft P 100 kips

Critical eccentricity, ecr ecr = 0.167d, = 0.167(10) = 1.67 ft d = 10ft

Figure 59a. Soil-Bearing Pressure for Square Footing P = 100 k, M = 150 ft.k
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(Page 2 of 2) c. Soil bearing pressure, q e < ecr q = P⎡ e ⎤ ⎢1± 0.167d ⎥ A⎣ ⎦ P⎡ e ⎤ ⎢1+ 0.167d ⎥ A⎣ ⎦ 100 ⎡ 1.5 ⎤ 2 ⎢1 + 1.67 ⎥ = 1.9 kips/ft 10(10 ) ⎣ ⎦
1.5 ⎤ e ⎤ 100 ⎡ P⎡ 2 ⎥= ⎢1 − ⎢1 − 1.67 ⎥ = 0.1 kips/ft A ⎣ e cr ⎦ 10(10 ) ⎣ ⎦

qmax = = qmin d. =

Maximum soil-bearing pressure M = 300, e = M 30 = = 3ft P 100

e > ecr, footing uplift occurs x qmax
⎛d ⎞ ⎛ 10 ⎞ = 3 ⎜ − e ⎟ = 3 ⎜ − 3 ⎟ = 6ft ⎝2 ⎠ ⎝ 2 ⎠

=

2P 2(100) 2 = = 3.33 kips/ft bx 10(6)

qmax = 3.33

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

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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:

P H M SR = = ΣMr ΣMo

(P + W )d/2
M + Hh

w d 2 o

h

Mr = Resisting Moment about point O Mo = Overturning Moment about 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,

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

P

M

qmin Footing

qmax

w=q

w=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.

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d

t

2

b t

Footing Section

Column Footing d

Critical Section: for bending moment for shear force

Wall Footing

Figure 62. Critical Sections

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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 = 16 kips/ft including weight of wall and footing. 12 in. H = 12 kips/ft

H wall

P
12 in.

4 ft.

o

3 ft. 5 ft. d = 8 ft., b = 1 ft. P C L e = 2 footing

c

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

Figure 63a. Wall and Footing Support Loads

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Solution:

Example 14
(Page 2 of 3) a. Stability ratio, SR Taking moments about point O SR = b. Mr 16(5 ) = = 1.67 Mo 12(4 )

Eccentricity, e e M e = M P 12(4 ) − 16(1) = 2ft 16

= Moment about center of footing =

c.

Critical eccentricity, ecr ecr = 0.167 d = 1.33ft

d.

Soil-bearing pressure, q e > ecr, footing uplifts x
⎛d ⎞ ⎛8 ⎞ = 3⎜ − e ⎟ = 3⎜ − 2 ⎟ = 6 ft ⎝2 ⎠ ⎝2 ⎠

qmax =

2P 2(16 ) 2 = = 5.33 kips/ft bx 1(6 )

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(Page 3 of 3) e. Shear at critical section, Vc Vc = 5.33 + 1.78 (1)x4 = 14.22 kips 2
t = 0.5ft. 2
t = 1 ft.

Critical Section for Shear

5.33

1.78

4 ft

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

f.

Moment at critical section, Mc

Mc = (1.33 )

(4.5 )2
2

+

1 (4.0 )(4.5 )⎛ 2 ⎞(4.5 ) = 40.5 kip-ft ⎜ ⎟ 2 ⎝3⎠

Critical Section for Shear moment

1.33 4.0 1.33

4.5 ft

Figure 63c. Wall and Footing Support Loads (Moment at Critical Section)
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SUMMARY
This module has discussed information to be used in analyzing various components of structure such as beams, columns, and footings.

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This Page Intentionally Blank

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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) b) Reactions R1 and R2. The shear and bending moment at (C), which is 5 ft from left support (A) of the beam.
100 lb/ft 300 lb

A

C 10'

B 6'

D

1000 lb

300 lb

A R1 5'

C 5' R2

B 6'

D

Figure 64. Overhang Beam

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(Page 2 of 2)
Procedure:

a)

Solve for reactions R1 and R2. +

Σ MB = 0 ↑ Σ Fy = 0

:

−(10)(R1) + (1000)(5) − (300)(6) = 0

R1 = 320 lb ↑ + : 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.
2.5' (100) (5) = 500 lb

M A 320 lb 5' C

V

(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

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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.

P 5PL A R1 L C L D L B R2

Figure 65. Work Aid 2, Beam with Loads

Procedure:

Solving for R1 by taking Σ MB = 0 : R1 = −P
⇒ ⇒

R1 = P ↓ R2 = 2P ↑

Σ Fy = 0

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(Page 2 of 2) 0<x<L + +
P 5PL

↑ Σ Fy = 0 Σ M= 0 :
M = −Px

⇒

V = −P
P L M V P x P M L L 2P

L < x < 2L + +

↑ Σ Fy = 0

:

V = −2P

V P x P 5PL V M

Σ M= 0 :
Px + P(x−L) + M = 0 M = P(L−2x)
P

2L < x < 3L (Right to the section) + +

M V V 3L-x 2P

↑ Σ Fy = 0 Σ M= 0

V = −2P M = 2P(3L−x)

x -P -2P

Fig. 74
M 2PL x -PL -3PL

Fig. 75
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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.

wo

A 2L

B L

Figure 67. Work Aid 3, Beam with Uniform Load

Procedure:

Determine R1 by summing moments about B : +

3 woL

Σ MB = 0 :
⎛L⎞ − (R1)(2L) + (3woL) ⎜ ⎟ = 0 ⎝2⎠
A R1 B R2 L

3w oL R1 = 4 +

3L 2

L 2

Figure 68.

↑ Σ Fy = 0 :

3w oL − 3woL + R2 = 0 4 9w oL R2 = 4

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(Page 2 of 2) 0 < x < 2L +
wo

↑ Σ Fy = 0 :
3w oL − wox − V = 0 4 ⎛ 3L ⎞ V = wo ⎜ − x⎟ ⎝ 4 ⎠
3w oL 4
w ox M V 2L

9w oL 4

L

+

Σ M= 0 :
3w oL x w o x 2 − + +M = 0 4 2 w x ⎛ 3L ⎞ M = o ⎜ −x⎟ 2 ⎝ 2 ⎠
3w oL 4 x 2
x

wo (3L-x) M V
(3L-x)/2 3L-x

2L < x < 3L (Consider right part) +

↑ Σ Fy = 0 :
V − wo(3L−x) = 0 V = wo(3L−x)
V

+

Σ M= 0 :
w o (3L − x)2 = 0 2 w (3L − x)2 M=− o 2
−M −

3w oL 4

w oL

x

3L 4 −

5w oL 4

Maximum positive moment 3L = M at x = 4 w o ⎛ 3L ⎞ ⎛ 3L 3L ⎞ 9w oL2 − = ⎜ ⎟⎜ ⎟= 2 ⎝ 4 ⎠⎝ 2 4 ⎠ 32 Maximum negative moment = − w oL 2
2

M

9w oL2 32
x

−

w oL2 2

Figure 69.

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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

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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

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Work Aid 5: Beam Diagrams - Superposition
The steps involved in this approach are: 1. 2. Separate problem into beams with individual loads. Draw beam diagrams for each load case.

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

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Work Aid 6: Using Beam Diagrams by Superposition
Draw the bending moment diagram for the beam with the loads shown. (Page 1 of 2)

6 kips

3 kips/ft

5'

5'

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

Step 1

Separate into two individual loads: a) b) Concentrated load, P = 6 kips Uniform load, wo = 3 kips/ft
5'
6 kips

5'
3 kips/ft

10'

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 = b) PL (6)(10) = = 15 kip-ft 4 4
15 kip-ft

+
37.5 kip-ft

Simply supported beam with uniform load wL2 3(10) 2 = = 37.5 kip-ft Mmax = 8 8

+

Figure 74. Load Cases
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(Page 2 of 2) Step 3 Construct the composite diagram by combining individual cases Mmax = 15 + 37.5 = 52.5 kip-ft
+

52.5 kip-ft

Figure 75.

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Work Aid 7: Beam Stresses
The analysis steps for determining the stresses in a beam are illustrated below.

Beam

Lateral Loads

Concentrated Loads

Distributed Loads

Bending Moment

Shear Force

Bending or Flexural Stress

Vertical and Horizontal Shear Stress

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

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Work Aid 8: Calculating Flexural and Shear Stresses in an Overhanging Steel Beam with Uniformly Distributed Load
(Page 1 of 2)
10 k ip/ft

8" 0.5"
10' 5'

0.5"

9" 0.5"
V (kip) 50.0

37.5

x-section IN.A. = 211 in4
-62.5

x

M (k ip-ft)

70.3

x
-125.0

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)

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(Page 2 of 2) Maximum flexural stress: σ max = Mc (125 ∗ 12) ∗ (5) = = 35.55 ksi I 211
8" 0.5"

Maximum shear stress:
τ max =

Vmax Q Ib

(at N.A.)

4.75" 2.25" N.A. 0.5"

4.5"

Q = (8)(0.5)(4.75) + (4.5)(0.5)(2.25) = 24.06 in
τmax =
3

(62.5)(24.06) = 14.26 ksi (211)(0.5)

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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* Hinge* Free End* Built-in End Load, W* Reaction, R* Shear, V* Moment, M*

Simple End Interior Support Built-in End Free End M EI End Slope, Slope, Deflection, y

*Indicates parameters related to conjugate beam

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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.

120 lb

A 60"

B 30"

C

Figure 78. Work Aid 10, Calculating Beam Defelection

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(Page 2 of 3) Step 1: Moment diagram for actual beam. Step 2: Load the conjugate beam with M/EI M w* = EI w*max = 3600 2.356 ∗10 6
-3 -1
* MC

60"

A

B

30"

C

-3600 lb-in

A

B

hinge

C

M/EI 1.528*10-3 in-1

x x x
* RB

R* A

(1)

= 1.528∗10 in

Step 3: Consider 1st, the segment AB and determine R*A and R*B.

* RB

(2)

* RC

Figure 79. Beam

+

Σ MB = 0 :
− 60 R*A +

1 ⎛ 60 ⎞ (1.528∗10−3)(60) ⎜ ⎟ = 0 2 ⎝ 3 ⎠

R*A = 1.528∗10−2

Σ Fy = 0

⇒

R*B + 1.528∗10−2 − R*B = 3.06∗10−2

1 (1.528∗10−3)(60) = 0 2

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

Σ Fy = 0 :

− 3.06 ∗ 10 −2 + R*C = 0

R*C = 3.06∗10−2 +

Σ MC = 0 :
1 (1.528∗10−3)(30)(20) + (R*B)(30) – M*C = 0 2 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.

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Work Aid 11: Using Beam Formula Work Aids
Use the beam formulas to perform beam analysis as follows: 1. 2. 3. 4. 5. Identify the beam/load type that corresponds to the problem being solved. Obtain the appropriate diagrams and formulas from the Work Aids. Use the superposition technique if your problem is not covered but can be composed from two or more beam/loading cases that are covered. Calculate the desired values for the beam by substituting the appropriate known values into the formulas. Draw the shear and moment diagram if needed.

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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.
R = 100 kips w = 100 kips

P = 50 kips

C
10 ft 10 ft Beam E = 29,000 kips/in.2 (steel) I = 1000 in.4 5 ft

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

Use superposition to combine the following beam/load cases:

Figure 81. Repeat of Figure 38 More Beam Load Cases
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(Page 2 of 4) Use superposition to combine the following beam/load cases:

Figure 82a. Repeat of Figure 39a Beam Load Cases
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(Page 3 of 4)

Shear Diagrams (Unit kips) d. Superposition: Combined Diagrams Case 7 + Case 24 + Case 26
75 35 70 50

Moment Diagrams (Unit: kip-ft)

Mc = 550

65

M2 = 300
105

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

Deflection, ∆c, midpoint between support, x = 10 a. Case 7
∆c =

RL3 48EI 100(240 ) = 0.993 in. 48(29,000 )(1000 )
3

= b.

Case 24
∆x

=

wox 4 L − 2L2 x 2 + Lx 3 − 2a 2L2 + 2a 2 x 2 24EIL
w oL2 ⎡ 5 2⎤ 2 ⎢ 4 (240 ) − 3(60 ) ⎥ 96EI ⎣ ⎦

[

]

For x = L /2, ∆ c =

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∆c

4(240 ) /12 ⎡5 2⎤ 2 = ⎢ 4 240 − 3(60 ) ⎥ = 0.422 in. 96(29,000 )(1000 ) ⎣ ⎦ (Page 4 of 4)
2

(

)

c.

Case 26
∆x

=

Pax 2 L − x2 6EIL

[

]

For x =

L PaL2 , ∆c = (up) 16EI 2 50(60 )(240 ) = 0.372 in. (up) 16(29,000 )(1000 )
2

∆c

=

d.

By superposition - Combine Case 7 + Case 24 + Case 26 Total ∆c = 0.993 + 0.422 – 0.372 = 1.043 in.

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Work Aid 13: Beam Diagrams and Formulas
Nomenclature
E I L Mmax M1 M2 M3 Mx P P1 P2 R R1 R2 R3 V V1 V2 V3 Vx W a b = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = modulus of elasticity of steel at 29,000 ksi moment of inertia of beam (in.4) total length of beam between reaction points (ft) maximum moment (kip-in.) maximum moment in left section of beam (kip-in.) maximum moment in right section of beam (kip-in.) maximum positive moment in beam with combined end moment conditions (kip-in.) moment at distance x from end of beam (kip-in.) concentrated load (kips) concentrated load nearest left reaction (kips) concentrated load nearest right reaction, and of different magnitude than P1 (kips) end beam reaction for any condition of symmetrical loading (kips) left end beam reaction (kips) right end or intermediate beam reaction (kips) right end beam reaction (kips) maximum vertical shear for any condition of symmetrical loading (kips) maximum vertical shear in left section of beam (kips) vertical shear at right reaction point, or to left of intermediate reaction point of beam (kips) vertical shear at right reaction point, or to right of intermediate reaction point of beam (kips) vertical shear at distance x from end of beam (kips) total load on beam (kips) measured distance along beam (in.) measured distance along beam which may be greater or less than a (in.) total length of beam between reaction points (in.) uniformly distributed load per unit of length (kips/in.) uniformly distributed load per unit of length nearest left reaction (kips/in.) uniformly distributed load per unit of length nearest right reaction, and of different magnitude than w1 (kips/in.) any distance measured along beam from left reaction (in.) any distance measured along overhang section of beam from nearest reaction point (in.) maximum deflection (in.) deflection at point of load (in.) deflection at any point x distance from left reaction (in.) deflection of overhang section of beam at any distance from nearest reaction point (in.)

l
w w1 w2

x x1
∆max ∆a ∆x ∆x1

AMERICAN INSTITUTE OF STEEL CONSTRUCTION

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Work Aid 14: Beam Diagrams and Formulas
(Page 1 of 14)
Beam List No. Beam Load Type

1. 2. 3. 4. 5, 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

Simple Simple Simple Simple Simple Simple Simple Simple Simple Simple Simple Fixed at One End, Supported at Other Fixed at One End, Supported at Other Fixed at One End, Supported at Other Fixed at Both Ends Fixed at Both Ends Fixed at Both Ends Cantilever Cantilever

- Uniformly Distributed Load - Load Increasing Uniformly to One End. - Load Increasing Uniformly to Center. - Uniform Load Partially Distributed. - Uniform Load Partially Distributed at One End - Uniform Load Partially Distributed at Each End. - Concentrated Load at Center. - Concentrated Load at Any Point. - Two Equal Concentrated Loads Symmetrically Placed. - Two Equal Concentrated Loads Unsymmetrically Placed. - Two Unequal Concentrated Loads Unsymmetrically Placed. - Uniformly Distributed Load. - Concentrated Load at Center - Concentrated Load at Any Point. - Uniformly Distributed Load. - Concentrated Load at Center. - Concentrated Load at Any Point. - Load Increasing Uniformly to Fixed End. - Uniformly Distributed Load.

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(Page 2 of 14) 20. Fixed at One End, Free to - Uniformly Distributed Load. Deflect Vertically But Not Rotate at Other Cantilever Cantilever Beam Fixed at One End, Free to Deflect Vertically but not Rotate at Other Overhanging One Support Overhanging One Support Overhanging One Support Overhanging One Support Overhanging One Support Continuous, Two Equal Spans Continuous, Two Equal Spans Continuous, Two Equal Spans Simple Simple - Concentrated Load at Any Point. - Concentrated Load at Free End. - Concentrated Load at Deflected End.

21. 22. 23.

24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

- Uniformly Distributed Load. - Distributed Load on Overhang. - Concentrated Load at End of Overhang. - Uniformly Distributed Load Between Supports. - Concentrated Load at Any Point Between Supports. - Uniform Load on One Span. - Concentrated Load at Center of One Span. - Concentrated Load at Any Point. - Uniform Load and Variable End Moments. - Concentrated Load and Variable End Moments

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

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

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

= 1.0 × 106 psi = 4000 psi =3

F.S.

8 in.

8 in. (1) Square section

9 in. (2) Circular section

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

Procedure: Column 1.

a.

Ultimate Load, Pu Pu A
σu

8 in.

= Aσu = 8(8) = 4000 psi = 64(4000) = 256 kips = 64 in.2
8 in.

Pu

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(Page 2 of 3) b. Critical Buckling Load, Pcr Pcr = E I kL

(kL )2

π 2EI

= 1.0 × 106 psi = 8(8 ) = 341.33 in.4 12
3

= 15 ft = 180 in.

Pcr = c.

( 3.14 )

2

(1.0 x10 ) ( 341.33 )
6

(180 )

2

= 103.87 kips

Failure Load Pmax Pmax = Pu or Pcr, whichever is less = Pcr = 103.87 kips

d.

Allowable Load, Pa Pa =
Pmax F.S.

F.S. = 3.0 Pa = 103.87 = 3.0 34.62 kips

Column 2.

a.

Ultimate Load, Pu Pu A
σu

= Aσu = π r 2 = (3.14) (4.5)2 = 4000 psi = 63.59 (4000) = 254.34 kips
9 in.

= 63.59 in.2

Pu

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(Page 3 of 3) b. Critical Buckling Load, Pcr Pcr =

(kL )2
π r4
4

π 2EI

I kL

=

= 321.90 in.4

= 180 in.

Pcr =

(3.14 )2 (1.0 x 10 6 )(321.90 ) (180 )2

=

97.96 kips

c.

Failure Load, Pmax Pu or Pcr, whichever is less Pmax = Pu = 97.96 kips

d.

Allowable Load, Pa Pa =
Pmax 97.96 = F.S. 3.0

=

32.65 kips

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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 x W 250 x 80 A = 10200 mm2 rx = 111.2 mm Sx = 985 x 103 mm3

P

c

y

c 255 mm

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

a) b)

Based on allowable combined stress and Fa = 100 MPa. Based on interaction formula with Fa = 100 MPa, Fb = 0.6 Fy.

Procedure:

a)

Combined Stress: Eccentrically loaded column e f = 150 mm = P ⎛ ec ⎞ ⎜1 + 2 ⎟ ≤ Fa A⎝ r ⎠

Maximum allowable P P = AFa ec 1+ 2 r

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=

(10200 mm )(100N/mm )
2 2

(150 mm )(127.5mm ) 1+ (111.2mm )6

= 400.5 kN

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(Page 2 of 2) b) Interaction Formula:
f a fb + ≤ 1, Fb = 0.6(250 ) = 150N/mm 2 Fa Fb

⎡ 1 e ⎤ P / A Pe / s + ≤ 1, P ⎢ + ⎥ ≤1 Fa Fb ⎣ AFa SFb ⎦

Maximum allowable P
⎡ 1 e ⎤ + ⎢ ⎥ ⎣ AFa SFb ⎦
−1

⎡ ⎤ 1 150 =⎢ + ⎥ = 501.1 kN 3 ⎣ (10200 )(100 ) 985x10 (150 )⎦

−1

(

)

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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.

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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. b. c. d. Eccentricity, e. Critical eccentricity, ecr. Distribution of soil bearing pressures. Minimum footing size to guarantee no uplifting

Procedure:
a. Eccentricity, e e b. = M 300 kip − ft = = 2.5 ft P 120 kips

Critical eccentricity, ecr ecr = 0.167d, = 0.167(12) = 2.0 ft. d = 12 ft

c.

Soil bearing pressure, q e > ecr, footing uplift occurs x qmax = ⎛d ⎞ ⎛ 12 ⎞ = 3⎜ − e ⎟ = 3⎜ − 2.5 ⎟ = 10.5 ft ⎝2 ⎠ ⎝ 2 ⎠ 2P 2(120) 2 = = 1.90 kips/ft bx 12(10.5)

Figure 95. Work Aid 18, Calculating Soil Bearing Pressure for a Square Footing

d.

Minimum footing size (no uplifting)
x=d

⎛d ⎞ 3⎜ − 2.5 ⎟ = d ⎝2 ⎠ d = 15 ft
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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. b. c. d. e. f. Stability ratio Eccentricity Critical eccentricity Soil-bearing pressure Shear at critical section Moment at critical section

Procedure:
a. Stability ratio, SR Taking moments about point “O” SR = b.
Mr 20(6 ) = = 2.4 Mo 10(5 )

Eccentricity, e e M e M P = Moment about center of footing = = 10 ( 5 ) − 20 (1.5 ) 20 = 1 ft

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

c.

Critical eccentricity, ecr ecr = 0.167d e < ecr q = P⎡ e ⎤ ⎢1 ± 0.167d ⎥ A⎣ ⎦ = 1.5 ft

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(Page 2 of 2) P⎡ e ⎤ ⎢1 + 0.167d ⎥ A⎣ ⎦ 20 ⎡ 1.0 ⎤ 1+ = 9(1) ⎢ 1.5 ⎥ ⎣ ⎦ 3.70 kips/ft
2

qmax

=

=

qmin

=

e ⎤ 20 ⎡ 1.0 ⎤ P⎡ = 1− ⎢1 − ⎥= A ⎣ e cr ⎦ 9(1) ⎢ 1.5 ⎥ ⎣ ⎦

0.74 kips/ft

2

e.

Shear at critical section, Vc Vc = 3.70 + 2.06 (1) x 5 2

= 14.4 kips

f.

Moment at critical section, Mc Mc = (1.89 ) +

(5.5 )2
2

1 2 (1.81)( 5.5 ) ⎛ × 5.5 ⎞ ⎜3 ⎟ 2 ⎝ ⎠

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

= 46.86 kips-ft

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

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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

⎡ 3(4)3 ⎤ 4 × (3)3 INA = 2 ⎢ + (3 × 4)(3.5)2 ⎥ + 12 ⎣ 12 ⎦

= 2(16+147) + 9 = 335 in4 max. V in beam = 0.6 P.
0.4P +

P

_

SF

Check τ at glue line: τ= VQ 0.6P = [3 × 4 × (3.5)] = τall = 500 It 335 × 3

-0.6

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P=

500 × 335 × 3 = 19,940 lbs = 19.94 kips 0.6 × 12 × 3.5 (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 ⎡ 1.5 ⎤ = ⎢12 × 3.5 + 4 × 1.5 × 2 ⎥ It 335 × 4 ⎣ ⎦

= 415 psi (< 500 psi) Check τmax in beam just above glue line = 0.6 × 19,940 (12 × 3.5) = 500 psi 335 × 3

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

Note
max. τ (through thickness) always where Q is GREATEST. t

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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.

P INA = 6(6) 3 2(2.5)(4) 3 − = 81.33 in4 12 12

Shear in Wood
Vmax = P max τmax = Vmax ⎛Q⎞ × max ⎜ ⎟ I ⎝ t ⎠

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(Page 2 of 2) 200 = P ⎛ 6 × 1× 2.5 + 2 × 1× 1 ⎞ ⎟ 81.33 ⎜ 1 ⎝ ⎠ ⇒ P = 957 lbs.

Bending Stress in Wood
max σ =
(max M) (y max ) I

2,500 =

60P × 3 81.33

⇒

P = 1,130 lbs.

Nail Rating
q= VQ I (to be calculated where flange is nailed to web)

q=

R P 400 (6 × 1× 2.5) = n = 81.33 S 3

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

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Practice Problems 3
A beam is to be fabricated from 2 sections of 2-in. standard steel pipe and a ¼ in. thick 1 steel plate by 4 fillet welds. If Vmax = 2000 lb, determine whether -in. fillet welds with a 8 capacity of 100 lb/in. each will be sufficient.

pipe

1
4k

1 ¼"

6"

F F

load x
12" weld

x 2.375

x Steel plate
-2k

x

+2k

A = 1.07 in2 Ix-x = 0.666 in4

Ixx = 2 0.666 + 1.07(6) 2 = 96.95 in4

[

]

1 (12 − 2.375) 3 + 4 12

2q =

VQ I

∴ q=

2000 [1.07 × 6] 2 × 96.95

q

q

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

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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)
25 mm 100 mm 25 mm

25 mm

NA

100 mm

25 mm

INA =

1 1 × 0.15 × (0.15)3 − × 0.1× (0.1)3 12 12

= 33.85x10−6 m4 2q = VQ I
q q

q=

VQ 1× 103 = [0.15 × 0.025 × 0.0625] 2I 2 × 33.85 × 10−6

q = 3,462 N/m & qs = Rn ⇒ s= 1,000 = 0.288 m = 289 mm 3,462

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(Page 2 of 2) (b)
25 mm 100 mm 25 mm

25 mm

NA

100 mm

25 mm

q

VQ 2q = I

q

where Q = 0.100× 0.025× 0.0625 = 1.5625× 10−4 m3 q= 1× 103 × 1.5625 × 10−4 2 × 33.85 × 10−6

= 2,308 N/m & s= 1,000 = 0.433 m 2,308

= 433 mm (Scheme (b) is better

Q Q(b) < Q(a)).

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

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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.

w (kips/ft)
12"

10 ft

Mmax = WL 8
BMD 5' +

2

4" N.A.

8"
WL 2

SFD

+

− WL 2

3"

(i) Bending

Note:

σT
4

=

24 ⇒ σ T = 12 8

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Sol

Mmax =

wL2 (10 × 12)2 = w 8 8 k-in

(k-in.)

(here, “w” in units of k/in.) (Page 2 of 2)

= 1800w

σmax =

Mymax Mc 1800w × 8 = = = 24 I I 768

w = 1.28 k/in. = 1.28× 12 = 15.36 k/ft. (ii) Shear

Variation of τ with y wL w × 10 × 12 = 2 2 kips

Vmax =

(w ⇒ k/in.)

= 60w

Also, critical section through thickness for T-beam is NA (for shear). max τmax = 15 =
Vmax ⎛ Q ⎞ 60w ⎡ 12 × 4 × 2 ⎤ = ⎜ t ⎟ ⎢ ⎥ I ⎝ ⎠max 768 ⎣ 3 ⎦

w = 6 k/in. = 72 k/ft. Answer: w = 15.36 k/ft. (smaller of two)

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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.

σt
C2 C1 C3

y

z

T

σb
Cross-Section 2270 × 12 × 3 = 600.9 psi 136 σ - Distribution

σt

=

σm = 600.9 × 1/3 = 200.3 σb = 2270 × 12 × 5 = 1001.5 psi 136

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T

=

1 × 1001.5 × 5 × 2 = 5007.5 lbs. 2 (Page 2 of 2)

C

1 1 = 200.3 × 6 × 2 + × 400.6 × 2 × 6 + × 200.3 × 1× 2 14 244 2 4 3 2 C1 1442443 1442443
C2 C3

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

OK

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GLOSSARY
Allowable Load Applied Loads Beam Beam-Column Bending Moment Column Compressive Yield Load Conjugate Beam Method Critical Load Eccentric Load Load Eccentricity Modulus of Rupture
Load that the column or other structural member can safely support. Loads that a structural member supports for design purpose or in actual service. Linear structural member having one or more supports and supports load perpendicular to its longitudinal axis. A structural member that experiences a significant amount of bending as well as axial load. Internal moment at a beam section required to maintain the equilibrium of any part of the beam. Linear structural member loaded primarily along its longitudinal axis. Load that will cause yielding of the column material in compression. A method to determine deflection of a beam. Load at which the column will become unstable or begin to buckle. Load applied away from the geometric center of a column or supporting footing. Distance from the center of the column cross-section or of a footing to the load application point. The flexural stress in a beam when the bending moment of the beam is the maximum value at the point of rupture. Graphical representation which shows the variation of the bending moment along the longitudinal axis of a beam. Plane in a beam that remains the same length when it is subjected to bending. Graphical representation which shows the variation of the shear force along the longitudinal axis of a beam. The internal force at a beam section required to maintain the equilibrium on any part of a beam. The ability of the structure to support a specified load

Moment Diagram

Neutral Axis Shear Diagram Shear Force Serviceability

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without undergoing unacceptable deflection, deformation, or movement.

Slenderness Soil Bearing Pressure Stability

Geometric property of a column that makes it tend to buckle under axial load. Load per unit area produced by the footing on the underlying soil. The ability to support a given compressive load without experiencing a sudden change in geometry, or buckling. The factor of safety against the overturning of a footing or foundation; the ratio of the resisting moment to the overturning moments. Beams whose reactions can be found from equations of equilibrium alone. Beams whose reactions cannot be found from the equations of equilibrium only, but require additional equations to determine the reactions. The ability of the structure to safely support a specified load without experiencing excessive stresses. Method to determine the shear and moment diagrams for a beam by combining the results from known load cases. Load that will produce crushing failure of the column material.

Stability Ratio

Statically Determinate Beams Statically Indeterminate Beams Strength Superposition

Ultimate Compressive Load

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Description: Analysis of Structural Components