   # 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. P15.14

#### Solutions

Chapter
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Chapter 15, Problem 14P
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. P15.14 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.

Formula to calculate the fixed moment for the point load of equal length is PL8.

Calculation:

Consider modulus of elasticity EI of the beam is constant.

Due to symmetric loading of the beam, consider the beam DG for the analysis.

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

Refer Figure 1,

Calculate the fixed end moment for DE.

FEMDE=1.5×20212=50kft

Calculate the fixed end moment for ED.

FEMED=50kft

Calculate the fixed end moment for EG.

FEMEG=40×208=100kft

Calculate the fixed end moment for GE.

FEMGE=100kft

Calculate the slope deflection equation for the member DE.

MDE=2E(2I)L(2θD+θE3ψ)+FEMDE

Substitute 0 for ψ, 0 for θD, 20 ft for L and 50kft for FEMDE.

MDE=2E(2I)20(2(0)+θE3(0))+50=0.2EIθE+50 (1)

Calculate the slope deflection equation for the member ED.

MED=2E(2I)L(θD+2θE3ψ)+FEMED

Substitute 0 for ψ, 0 for θD, 20 ft for L and 50kft for FEMED.

MED=2E(2I)20(2θE+(0)(0))50=0.4EIθE50 (2)

Calculate the slope deflection equation for the member EG.

MEG=2EIL(2θE+θG3ψ)+FEMEG

Substitute 0 for ψ, 0 for θG, 20 ft for L, and 100kft for FEMDE.

MEG=2EI20(θD+2(0)(0))+100=0.2EIθE+100 (3)

Calculate the slope deflection equation for the member GE.

MGE=2EIL(2θG+θE3ψ)+FEMGE

Substitute 0 for ψ, 0 for θG, 20 ft for L and 100kft for FEMGE.

MGE=2EI20(θE+2(0)(0))100=0.1EIθE100 (4)

Write the equilibrium equation as below.

MED+MEG=0

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

0.4EIθE50+0.2EIθE+100=00.6EIθE+50=00.6EIθE=50θE=83.33EIkft2

Substitute 83.33EIkft2 for θE in equation (1).

MDE=0.2EI(83.33EI)+50=33.33kft

Substitute 83.33EIkft2 for θE in equation (2).

MED=0.4EI(83.33EI)50=83.33kft

Substitute 83.33EIkft2 for θE in equation (3).

MEG=0.2EI(83.33EI)+100=83.33kft

Substitute 83.33EIkft2 for θE in equation (4).

MGE=0.1EI(83.33EI)100=108.3kft

Show the section free body diagram of the member DE and EG as in Figure 2.

Consider the member DE of the beam:

Calculate the vertical reaction at the left end of the joint E by taking moment about point D.

+MD=0Ey,L(20)1.5×(20)×(202)+33

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