   # Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.1–P15.5 by using the slope-deflection method. FIG. P15.4, P15.7

#### Solutions

Chapter
Section
Chapter 15, Problem 4P
Textbook Problem
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## Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.1–P15.5 by using the slope-deflection method. 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 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.

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

Calculate the slope deflection equation for the member AB.

MAB=2EIL(2θA+θB3ψ)+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 for ψ, 0 for θA, 36ft for L and 216kft for FEMAB.

MAB=2×29,000×1,53036×144(2(0)+θB3(0))+216=17118.0556θB+216 (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψ)+FEMBA

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

MBA=2×29,000×1,53036×144(2θB+03(0))216=34,236.11θB216 (2)

Calculate the slope deflection equation for the member BC.

MBC=3EIL(θB+θCψ)+FEMBC+FEMBC2

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

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

34,236.11θB216+38,515.625θB+144=072,751.732θB72=0θB=7272,751.732θB=0.00099kft2

Substitute 0.00099kft2 for θB in equation (1).

MAB=17,118.0556(0.00099)+216=232.9kft

Substitute 0.000734kft2 for θB in equation (2).

MBA=34,236.11(0.00099)216=182.1kft

Substitute 0.000734kft2 for θB in equation (3).

MBC=38,515.625(0.00099)+144=182.1kft

Consider the member AB of the beam:

Show the section free body diagram of the member AB and BC as in Figure 2.

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