Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 15, Problem 15P
To determine

Find the reaction and plot the shear and bending moment diagram.

Expert Solution & Answer
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Answer to Problem 15P

The end moments at the member A(MA), B(MB), C(MC), and E(ME) are 68.6kNm, 183kNm, 28.9kNm, and 170.1kNm respectively and the reaction at point A, B, C and E are 65.7kN_, 200.8kN_, 56.8kN_, and 54.9kN_ respectively.

Explanation of Solution

Fixed end moment:

Formula to calculate the fixed moment for point load with equal length are PL8.

Formula to calculate the fixed moment for point load with unequal length are Pab2L2 and Pa2bL2.

Formula to calculate the fixed moment for UDL is WL212.

Formula to calculate the fixed moment for deflection is 6EIΔL2

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the free body diagram of the entire beam as in Figure 1.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  1

Refer Figure 1,

Calculate the fixed end moment for AB.

FEMAB=20×8212=106.7kNm

Calculate the fixed end moment for BA.

FEMBA=20×8212=106.7kNm

Calculate the fixed end moment for BC.

FEMBC=20×8212=106.7kNm

Calculate the fixed end moment for CB.

FEMCB=20×8212=106.7kNm

Calculate the fixed end moment for CE.

FEMCE=60×88=60kNm

Calculate the fixed end moment for EC.

FEMEC=60×88=60kNm

Chord rotations:

Show the free body diagram of the chord rotation of the beam as in Figure 2.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  2

Calculate the chord rotation of the beam BC.

ψBC=25mm×1m1000mm8m=0.003125

Calculate the chord rotation of the beam CE.

ψCE=25mm×1m1000mm8m=0.003125

Calculate the slope deflection equation for the member AB.

MAB=2EIL(2θA+θB3ψ)+FEMAB

Here, ψ is the chord rotation, θA is the slope at the point A and θB is the slope at the point B.

Substitute 70GPa for E, 800×106mm4 for I, 0 for θA, 8 m for L and 106.7kNm for FEMAB.

MAB=2×70×8008(2(0)+θB3(0))+106.7=14000θB+106.7 (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψ)+FEMBA

Substitute 70GPa for E, 800×106mm4 for I, 0 for θA, 8 m for L and 106.7kNm for FEMBA.

MBA=2×70×8008(2θB+03(0))106.7=28000θB106.7 (2)

Calculate the slope deflection equation for the member BC.

MBC=2EIL(2θB+θC3ψBC)+FEMBC

Substitute 70GPa for E, 800×106mm4 for I, 0.003125 for ψBC, 8 m for L and 106.7kNm for FEMBC.

MBC=2×70×8008(2θB+θC(3×0.003125))+106.7=28000θB+14000θC+131.25+106.7=28000θB+14000θC+238 (3)

Calculate the slope deflection equation for the member CB.

MCB=2EIL(2θC+θB3ψCB)+FEMCB

Substitute 70GPa for E, 800×106mm4 for I, 0.003125 for ψCB, 8 m for L and 106.7kNm for FEMCB.

MCB=2×70×8008(θB+2θC(3×0.003125))106.7=14000θB+28000θC+131.25106.7=14000θB+28000θC+24.6 (4)

Calculate the slope deflection equation for the member CE.

MCE=2EIL(2θC+θE3ψCE)+FEMCE

Substitute 70GPa for E, 800×106mm4 for I, 0 for θE, 0.003125 for ψCE, 8 m for L and 60kNm for FEMCE.

MCE=2×70×8008(0+2θC(3×0.003125))+60=28000θC131.25+60=28000θC71.3 (5)

Calculate the slope deflection equation for the member EC.

MEC=2EIL(2θE+θC3ψEC)+FEMEC

Substitute 70GPa for E, 800×106mm4 for I, 0 for θE, 0.003125 for ψCE, 8 m for L and 60kNm for FEMEC.

MEC=2×70×8008(θC+2(0)(3×0.003125))60=14000θC131.2560=14000θC191.3 (6)

Write the equilibrium equation as below.

MBA+MBC=0

Substitute equation (2) and equation (3) in above equation.

28000θB106.7+28000θB+14000θC+238=056000θB+14000θC=131.3 (7)

Write the equilibrium equation as below.

MCB+MCD=0

Substitute equation (4) and equation (5) in above equation.

14000θB+28000θC+24.6+28000θC71.3=014000θB+56000θC46.7=014000θB+56000θC=46.7 (8)

Solve the equation (7) and equation (8).

θB=2.723×103kNm2θC=1.515×103kNm2

Calculate the moment about AB.

Substitute 2.723×103kNm2 for θB in equation (1).

MAB=14000(2.723×103)+106.7=68.6kNm

Calculate the moment about BA.

Substitute 2.723×103kNm2 for θB in equation (2).

MBA=28000(2.723×103)106.7=183kNm

Calculate the moment about BC.

Substitute 2.723×103kNm2 for θB and 1.515×103kNm2 for θC in equation (3).

MBC=28000(2.723×103)+14000(1.515×103)+238=183kNm

Calculate the moment about CB.

Substitute 2.723×103kNm2 for θB and 1.515×103kNm2 for θC in equation (4).

MCB=14000(2.723×103)+28000(1.515×103)+24.6=28.9kNm

Calculate the moment about CE.

Substitute 1.515×103kNm2 for θC in equation (5).

MCE=28000(1.515×103)71.3=28.9kNm

Calculate the moment about EC.

Substitute 1.515×103kNm2 for θC in equation (6).

MEC=14000(1.515×103)191.3=170.1kNm

Consider the member AB of the beam:

Show the section free body diagram of the member AB, BC and CE as in Figure 3.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  3

Calculate the vertical reaction at the left end of the joint B by taking moment about point A.

+MA=0By,L(8)20×(8)×(82)+68.6183=0By,L(8)=754.4By,L=754.48By,L=94.3kN

Calculate the horizontal reaction at point A by resolving the horizontal equilibrium.

+Fx=0Ax=0

Calculate the vertical reaction at point A by resolving the vertical equilibrium.

+Fy=0Ay(20×8)+By,L=0Ay160+94.3=0Ay=65.7kN

Consider the member BC of the beam:

Calculate the vertical reaction at the right end of the joint B by taking moment about point C.

+MC=0By,R(8)+20×(8)×(82)+183+28.9=0By,R(8)=851.9By,R=851.98By,R=106.5kN

Calculate the vertical reaction at the left end of joint C by resolving the vertical equilibrium.

+Fy=0Cy,L(20×8)+By,R=0Cy,L160+106.5=0Cy,L=53.5kN

Calculate the total reaction at point B.

By=By,R+By,L

Substitute 106.5kN for By,R and 94.3kN for By,L.

By=106.5kN+94.3kN=200.8kN

Calculate the vertical reaction at the right end of the joint C by taking moment about point E.

+ME=0Cy,R(8)+60×(4)28.9170.1=0Cy,R(8)=41Cy,R=418Cy,R=5.1kN

Calculate the horizontal reaction at point E by resolving the horizontal equilibrium.

+Fx=0Ex=0

Calculate the vertical reaction at point E by resolving the vertical equilibrium.

+Fy=0Ey60+Cy,R=0Ey60+5.1=0Ey=54.9kN

Calculate the total reaction at point C.

Cy=Cy,R+Cy,L

Substitute 5.1kN for Cy,R and 53.5kN for Cy,L.

Cy=5.1kN+53.5kN=58.6kN

Show the reactions of the beam in Figure 4.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  4

Refer Figure 4,

Shear diagram:

Point A:

SA,L=0SA,R=65.7kN

Point B:

SB,L=65.7(20×8)=94.3kNSB,R=94.3kN+200.8kN=106.5kN

Point C:

SC,L=106.5(20×8)=53.5kNSC,R=53.5+58.6=5.1kN

Point E:

SE,L=5.160=54.9kNSE,R=54.9+25.3=0kN

Plot the shear force diagram of the beam as in Figure 5.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  5

Show the shear diagram of the section AB as in Figure 6.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  6

Use the similar triangle concept, to find the location of the maximum bending moment.

65.7x=94.38x525.665.7x=94.3x160x=525.6x=3.3mfrompoint A

Show the shear diagram of the section BC as in Figure 7.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  7

Use the similar triangle concept, to find the location of the maximum bending moment.

106.5x=53.58x852106.5x=53.5x160x=852x=5.3mfrompoint B

Refer Figure 4,

Bending moment diagram:

Point A:

MA=68.6kNm

Point F:

MF=68.6+(65.7×3.3)(20×3.3×3.32)=39.3kNm

Point B:

MB=183kNm

Point G:

MG=68.6(65.7×13.3)(200.8×5.3)+(20×13.3×13.32)=100.6kNm

Point C:

MC=28.9kNm

Point D:

MD=(54.9×4)170.1=49.5kNm

Point E:

ME=170.1kNm

Plot the bending moment diagram of the beam as in Figure 8.

Structural Analysis, 5th Edition, Chapter 15, Problem 15P , additional homework tip  8

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