   # 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. P15.13

#### Solutions

Chapter
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Chapter 15, Problem 13P
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. P15.13 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.

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

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

Refer Figure 1,

Calculate the fixed end moment for CD.

FEMCD=15×6212=45kNm

Calculate the fixed end moment for DC.

FEMDC=15×6212=45kNm

Calculate the fixed end moment for DE.

FEMDE=15×6212=45kNm

Calculate the fixed end moment for ED.

FEMED=15×6212=45kNm

Calculate the slope deflection equation for the member CD.

MCD=2EIL(2θC+θD3ψ)+FEMCD

Substitute 0 for ψ, 0 for θC, 6 m for L and 45kNm for FEMCD.

MCD=2EI6(2(0)+θD3(0))+45=0.333EIθD+45 (1)

Calculate the slope deflection equation for the member DC.

MDC=2EIL(2θD+θC3ψ)+FEMDC

Substitute 0 for ψ, 0 for θC, 6 m for L and 45kNm for FEMDC.

MDC=2EI6(2θD+(0)(0))45=0.667EIθD45 (2)

Calculate the slope deflection equation for the member DE.

MDE=3EIL(2θE+θD3ψ)+FEMDE+FEMDE2

Substitute 0 for ψ, 6 m for L and 45kNm for FEMDE.

MDE=3E6(θD+2(0)(0))+45+452=0.5EIθD+67.5 (3)

Calculate the slope deflection equation for the member ED.

MED=0

Write the equilibrium equation as below.

MDC+MDE=0

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

0.667EIθD45+0.5EIθD+67.5=01.167EIθD+22.5=01.167EIθD=22.5θD=19.28EIkNm2

Substitute 19.28EIkNm2 for θD in equation (1).

MCD=0.333EI(19.28EI)+45=38.6kNm

Substitute 19.28EIkNm2 for θD in equation (2).

MDC=0.667EI(19.28EI)45=57.9kNm

Substitute 19.28EIkNm2 for θD in equation (3).

MDE=0.5EI(19.28EI)+67.5=57.9kNm

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

Refer Figure 2,

Consider the member CD of the beam:

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

+MC=0Dy,L(6)15×(6)×(62)57.9+38.6=0Dy,L(6)=289.3Dy,L=289.36Dy,L=48.2kN

Calculate the vertical reaction at the right end of the joint C by resolving the vertical equilibrium.

+Fy=0Cy,R(15×6)+Dy,L=0Cy,R120+48

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