   # 15.24 through 15.31 Determine the member end moments and reactions for the frames shown in Figs. P15.24–P15.31 by using the slope-deflection method. FIG. P15.26

#### Solutions

Chapter
Section
Chapter 15, Problem 26P
Textbook Problem
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## 15.24 through 15.31 Determine the member end moments and reactions for the frames shown in Figs. P15.24–P15.31 by using the slope-deflection method.FIG. P15.26 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 for equal length is 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=0kNm

Calculate the fixed end moment for CA.

FEMCA=0kNm

Calculate the fixed end moment for CD.

FEMCD=27×10212=225kNm

Calculate the fixed end moment for DC.

FEMDC=225kNm

Calculate the fixed end moment for DB.

FEMDB=0kNm

Calculate the fixed end moment for BD.

FEMBD=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 AC.

ψAC=Δ5

Calculate the chord rotation of the frame CD.

ψCD=0

Calculate the chord rotation of the frame BD.

ψBD=Δ5

Calculate the slope deflection equation for the member AC.

MAC=0

Calculate the slope deflection equation for the member CA.

MCA=3EIL(θC+2θAψCA)+FEMCA

Substitute Δ5 for ψCA, 0 for θA, 5 m for L, and 0kNm for FEMCA.

MCA=3EI5(θC+2(0)(Δ5))+0=0.6EIθC+0.12EIΔ        (1)

Calculate the slope deflection equation for the member CD.

MCD=2E(2I)L(2θC+θD3ψCD)+FEMCD

Substitute 0 for ψCD, 10 m for L, and 225kNm for FEMCD.

MCD=2EI10(2θC+θD(3×0))+225=0.4EIθC+0.2EIθD+225        (2)

Calculate the slope deflection equation for the member DC.

MDC=2EIL(2θD+θC3ψDC)+FEMDC

Substitute 0 for ψDC, 10 m for L, and 225kNm for FEMDC.

MDC=2EI10(2θD+θC(3×0))225=0.2EIθC+0.4EIθD225        (3)

Calculate the slope deflection equation for the member DB.

MDB=3EIL(θD+2θBψDB)+FEMDB

Substitute Δ5 for ψDB, 0 for θB, 30 m for L, and 0kNm for FEMDB.

MDB=3EI5(θD+0(Δ5))+0=0.6EIθD+0.12EIΔ        (4)

Calculate the slope deflection equation for the member BD.

MBD=0

Write the equilibrium equation as below.

MCA+MCD=0

Substitute equation (1) and equation (2) in above equation.

0.6EIθC+0.12EIΔ+0.4EIθC+0.2EIθD+225=0EIθC+0.2EIθD+0.12EIΔ=225        (5)

Write the equilibrium equation as below.

MDC+MDB=0

Substitute equation (3) and equation (4) in above equation.

0.2EIθC+0.4EIθD225+0.6EIθD+0.12EIΔ=00.2EIθC+EIθD+0.12EIΔ=225        (6)

Show the free body diagram of the frame due to sway force as in Figure 3.

Calculate the horizontal reaction at the member AC due to sway force by taking moment about point A.

MA=0(SAC×5)+MAC+MCA=0SAC=MAC+MCA5

Calculate the horizontal reaction at the member BD due to sway force by taking moment about point B.

MB=0(SBD×5)+MBD+MDB=0SBD=MBD+MDB5

Calculate the reaction of the support C and D due to sway force by considering the horizontal equilibrium

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