   # 15.8 through 15.14 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.8–P15.14 by using the slope-deflection method. FIG. P15.11

#### Solutions

Chapter
Section
Chapter 15, Problem 11P
Textbook Problem
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## 15.8 through 15.14 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.8–P15.14 by using the slope-deflection method.FIG. P15.11 To determine

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

### Explanation of Solution

Fixed end moment:

Formula to calculate the overhang beam for UDL is WL22.

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

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.

Refer Figure 1,

Calculate the fixed end moment for AB.

FEMAB=1×1022=50kft

Calculate the fixed end moment for BA.

FEMBA=50kft

Calculate the fixed end moment for BD.

FEMBD=35×208=87.5kft

Calculate the fixed end moment for DB.

FEMDB=87.5kft

Calculate the fixed end moment for DE.

FEMDE=2×20212=66.67kft

Calculate the fixed end moment for ED.

FEMED=66.67kft

Calculate the slope deflection equation for the member BD.

MBD=2EIL(2θB+θD3ψ)+FEMBD

Substitute 20 ft for L and 87.5kft for FEMBD.

MBD=2EI20(2θB+θD(0))+87.5=0.2EIθB+0.1EIθD+87.5 (1)

Calculate the slope deflection equation for the member DB.

MDB=2EIL(2θD+θB6ΔL)+FEMDB

Substitute 20 ft for L and 87.5kft for FEMDB.

MDB=2EI10(θB+2θD(0))87.5=0.1EIθB+0.2EIθD87.5 (2)

Calculate the slope deflection equation for the member DE.

MDE=2EIL(2θD+θE3ψ)+FEMDE

Substitute 0 for θE, 20 ft for L, and 66.67kft for FEMDE.

MDE=2EI20(0+2θD(0))+66.67=0.2EIθD+66.67 (3)

Calculate the slope deflection equation for the member ED.

MED=2EIL(2θE+θD3ψ)+FEMED

Substitute 0 for θE, 20 ft for L, and 66.67kft for FEMED.

MED=2EI20(θD+2(0)(0))66.67=0.1EIθD66.67 (4)

Write the equilibrium equation as below.

MBD50=0

Substitute equation (1) in above equation.

0.2EIθB+0.1EIθD+87.550=00.2EIθB+0.1EIθD+37.5=00.2EIθB+0.1EIθD=37.5 (5)

Write the equilibrium equation as below.

MDB+MDE=0

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

0.1EIθB+0.2EIθD87.5+0.2EIθD+66.67=00.1EIθB+0.4EIθD20.83=00.1EIθB+0.4EIθD=20.83 (6)

Solve the equation (5) and equation (6).

θB=244.05EIkft2θD=113.1EIkft2

Substitute 244.05EIkft2 for θB and 113.1EIkft2 for θD in equation (1).

MBD=0.2EI(244.5EI)+0.1EI(113.1EI)+87.5=50kft

Substitute 244.05EIkft2 for θB and 113.1EIkft2 for θD in equation (2).

MDB=0.1EI(244.05EI)+0.2EI(113.1EI)87

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