   # Determine the member end moments and reactions for the frames shown in Figs. P15.17–P15.20 by using the slope-deflection method. FIG. P15.17, P15.21

#### Solutions

Chapter
Section
Chapter 15, Problem 17P
Textbook Problem
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## Determine the member end moments and reactions for the frames shown in Figs. P15.17–P15.20 by using the slope-deflection method. FIG. P15.17, P15.21

To determine

Find the member end moments and reaction for the frames.

### Explanation of Solution

Fixed end moment:

Formula to calculate the fixed moment for UDL is WL212.

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

Calculation:

Consider the flexural rigidity EI of the frame is constant.

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

Refer Figure 1,

Calculate the fixed end moment for AC.

FEMAC=75×68=56.25kNm

Calculate the fixed end moment for CA.

FEMCA=56.25kNm

Calculate the fixed end moment for CD.

FEMCD=25×(9)212=168.75kNm

Calculate the fixed end moment for EC.

FEMDC=168.75kNm

Calculate the slope deflection equation for the member AC.

MAC=2EIL(2θA+θC3ψ)+FEMAC

Substitute 0 for ψ, 0 for θA, 6 m for L and 56.25kNm for FEMAC.

MAC=2EI6(2(0)+θC3(0))+56.25=0.333EIθC+56.25        (1)

Calculate the slope deflection equation for the member CA.

MCA=2EIL(2θC+θA3ψ)+FEMCA

Substitute 0 for ψ, 0 for θA, 6 m for L and -56.25kNm for FEMCA.

MCA=2EI6((0)+2θC3(0))56.25=0.667EIθC56.25        (2)

Calculate the slope deflection equation for the member CD.

MCD=3EIL(θC+2θD3ψCD)+FEMCD+FEMCD2

Substitute 0 for ψ, 0 for θD, 9 m for L, and 168.75kNm for FEMCD.

MCD=3EI9(0+θC3(0))+168.75+168.752=0.333EIθC+253.13        (3)

Calculate the slope deflection equation for the member DC.

MDC=0

Write the equilibrium equation as below.

MCA+MCD=0

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

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