   # Solve Problem 15.18 for the loading shown in Fig. PlS.18 and a settlement of 6 mm at support A. 15.17 through 15.20 Determine the member end moments and reactions for the frames shown in Figs. P15.17–P15.20 by using the slope-deflection method. FIG. P15.18, P15.22

#### Solutions

Chapter
Section
Chapter 15, Problem 22P
Textbook Problem
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## Solve Problem 15.18 for the loading shown in Fig. PlS.18 and a settlement of 6 mm at support A.15.17 through 15.20 Determine the member end moments and reactions for the frames shown in Figs. P15.17–P15.20 by using the slope-deflection method. FIG. P15.18, P15.22

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.

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 BC.

FEMBC=18×(5)212=37.5kNm

Calculate the fixed end moment for CB.

FEMCB=37.5kNm

Calculate the fixed end moment for AC.

FEMAC=0kNm

Calculate the fixed end moment for CA.

FEMCA=0kNm

Calculate the fixed end moment for CD.

FEMCD=0kNm

Calculate the fixed end moment for DC.

FEMDC=0kNm

Chord rotations:

Show the free body diagram of the chord rotation of the frame as in Figure 2.

Calculate the chord rotation of the frame BC.

ψBC=6mm×1m1000mm5m=0.0012

Calculate the chord rotation of the frame CD.

ψCD=6mm×1m1000mm5m=0.0012

Calculate the slope deflection equation for the member AC.

MAC=2EIL(2θA+θC3ψ)+FEMAC

Substitute 200 GPa for E, 1360×106mm4 for I, 0 for ψ, 0 for θA, 5 m for L and 0kNm for FEMAC.

MAC=2×200×13605(2(0)+θC3(0))+0=108800θC        (1)

Calculate the slope deflection equation for the member CA.

MCA=2EIL(2θC+θA3ψ)+FEMCA

Substitute 200 GPa for E, 1360×106mm4 for I, 0 for ψ, 0 for θA, 5 m for L and 0kNm for FEMCA.

MCA=2×200×13605((0)+2θC3(0))0=217600θC        (2)

Calculate the slope deflection equation for the member BC.

MBC=2EIL(2θB+θC3ψBC)+FEMBC

Substitute 200 GPa for E, 1360×106mm4 for I, -0.0012 for ψBC, 0 for θB, 5 m for L and 37.5kNm for FEMBC.

MBC=2×200×13605(0+θC(3×0.0012))+37.5=108800θC+391.7+37.5=108800θC+429.2        (3)

Calculate the slope deflection equation for the member CB.

MCB=2EIL(2θC+θB3ψCB)+FEMCB

Substitute 200 GPa for E, 1360×106mm4 for I, -0.0012 for ψCB, 0 for θB, 5 m for L and -37.5kNm for FEMCB.

MCB=2×200×13605(0+2θC3(0.0012))37.5=217600θC+391.737.5=217600θC+354.2        (4)

Calculate the slope deflection equation for the member CD.

MCD=3EIL(θC+2θDψCD)+FEMCD

Substitute 200 GPa for E, 1360×106mm4 for I, 0.0012 for ψCD, 0 for θD, 5 m for L and 0kNm for FEMCD.

MCD=3×200×13605(0+θC(0.0012))+0=163200θC195

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