   # 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. P15.2, P15.6

#### Solutions

Chapter
Section
Chapter 15, Problem 2P
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. P15.2, P15.6

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:

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

Refer Figure 1,

Calculate the fixed end moment for AB.

FEMAB=1.5×30212+20×308=112.5+75=187.5kft

Calculate the fixed end moment for BA.

FEMBA=1.5×3021220×308=112.575=187.5kft

Calculate the fixed end moment for BC.

FEMBC=3×20212=100kft

Calculate the fixed end moment for CB.

FEMCB=3×20212=100kft

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 0 for ψ, 0 for θA, 30ft for L, and 187.5kft for FEMAB.

MAB=2EI30(2(0)+θB3ψ)+187.5=0.0667EIθB+187.5 (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψ)+FEMBA

Substitute 0 for ψ, 0 for θA, 30ft for L, and 187.5kft for FEMBA.

MBA=2EI30(2θB+03ψ)187.5=0.1333EIθB187.5 (2)

Calculate the slope deflection equation for the member BC.

MBC=2EIL(2θB+θC3ψ)+FEMBC

Substitute 0 for ψ, 0 for θC, 20ft for L, and 100kft for FEMBC.

MBC=2EI20(2θB+03(0))+100=0.2EIθB+100 (3)

Calculate the slope deflection equation for the member CB.

MCB=2EIL(2θC+θB3ψ)+FEMCB

Substitute 0 for ψ, 0 for θC, 20ft for L and 100kft for FEMCB.

MCB=2EI20(2(0)+θB3(0))100=0.1EIθB100 (4)

Write the equilibrium equation as below.

MBA+MBC=0

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

0.1333EIθB187.5+0.2EIθB+100=00.3333EIθB87.5=0θB=87.50.3333EIθB=262.5EIkft2

Substitute 262.5EIkft2 for θB in equation (1).

MAB=0.0667EI(262.5EI)+187.5=205kft

Substitute 262.5EIkft2 for θB in equation (2).

MBA=0.1333EI(262.5EI)187.5=152.5kft

Substitute 262.5EIkft2 for θB in equation (3).

MBC=0.2EI(262.5EI)+100=152.5kft

Substitute 262.5EIkft2 for θB in equation (4)

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