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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 15, Problem 7P
Textbook Problem
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Solve Problem 15.4 for the loading shown in Fig. P15.4 and the support settlements of 1 in. at B and 1 4 in. at C.

Chapter 15, Problem 7P, Solve Problem 15.4 for the loading shown in Fig. P15.4 and the support settlements of 1 in. at B and

FIG. P15.4, P15.7

To determine

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

Explanation of Solution

Fixed end moment:

Formula to calculate the fixed moment for UDL is WL212.

Calculation:

Consider the flexural rigidity EI of the beam is constant.

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

Refer Figure 1,

Calculate the fixed end moment for AB.

FEMAB=2×36212=216kft

Calculate the fixed end moment for BA.

FEMBA=2×36212=216kft

Calculate the fixed end moment for BC.

FEMBC=2×24212=96kft

Calculate the fixed end moment for CB.

FEMCB=2×24212=96kft

Chord rotations:

Show the free body diagram of the chord rotation due to support reaction of the beam as in Figure 2.

Calculate the chord rotation of the beam AB.

ψAB=1in.36ft×12in.ft=0.002315

Calculate the chord rotation of the beam BC.

ψBC=(10.25)in.24ft×12in.ft=0.002604

Calculate the slope deflection equation for the member AB.

MAB=2EIL(2θA+θB3ψAB)+FEMAB

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

Substitute 29,000ksi for E, 1530in.4 for I, 0.002315 for ψAB, 0 for θA, 36ft for L and 216kft for FEMAB.

MAB=2×29,000×1,53036×144(2(0)+θB3(0.002315))+216=17,118.0556θB+118.89+216=17,118.0556θB+334.9 (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψBA)+FEMBA

Substitute 29,000ksi for E, 1530in.4 for I, 0.002315 for ψAB, 0 for θA, 36ft for L and 216kft for FEMBA.

MBA=2×29,000×1,53036×144(2θB+03(0.002315))216=34,236.11θB+118.89216=34,236.11θB97.11 (2)

Calculate the slope deflection equation for the member BC.

MBC=3EIL(θB+θCψBC)+FEMBC+FEMBC2

Substitute 29,000ksi for E, 1530in.4 for I, 0.002604 for ψBC, 0 for θC, 24ft for L and 96kft for FEMBC.

MBC=3×29,000×1,53024×144(θB+0(0.002604))+96+962=38,515.625θB100.3+144=38,515.625θB+43.7 (3)

Calculate the slope deflection equation for the member CB.

MCB=0

Write the equilibrium equation as below.

MBA+MBC=0

Substitute equation (1) and equation (2) in above equation.

34,236.11θB97.11+38,515.625θB+43.7=072,751.732θB53.41=0θB=53.4172,751.732θB=0.000734kft2

Calculate the moment about AB.

Substitute 0.000734kft2 for θB in equation (1).

MAB=17,118.0556(0.000734)+334.9=347.5kft

Calculate the moment about BA.

Substitute 0.000734kft2 for θB in equation (2).

MBA=34,236

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