   # 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. P15.30

#### Solutions

Chapter
Section
Chapter 15, Problem 30P
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. P15.30 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 AC.

FEMAC=0

Calculate the fixed end moment for CA.

FEMCA=0

Calculate the fixed end moment for CD.

FEMCD=3×16212=64kft

Calculate the fixed end moment for DC.

FEMDC=64kft

Calculate the fixed end moment for DB.

FEMDB=0kft

Calculate the fixed end moment for BD.

FEMBD=0kft

Chord rotations:

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

Calculate the length of AC by using Pythagoras theorem.

LAC=(162)+(42)=16.49ft

Calculate the length of BD by using Pythagoras theorem.

LBD=(162)+(42)=16.49ft

Calculate the chord rotation of the frame AC.

ψAC=CCLAC=174Δ16.49=0.0625Δ

Calculate the chord rotation of the frame BD.

ψBD=DDLBD=174Δ5=0.0625Δ

Calculate the chord rotation of the frame CD.

ψCD=D1DLCD=2(14)Δ16=0.03125Δ

Calculate the slope deflection equation for the member AC.

MAC=2EIL(2θA+θC3ψAC)+FEMAC

Substitute 16.49 ft for L, 0 for θA, 0.0625Δ for ψAC, and 0 for FEMAC.

MAC=2EI16.49(2(0)+θC3(0.0625Δ))+0=0.121EIθC+0.0227EIΔ        (1)

Calculate the slope deflection equation for the member CA.

MCA=2EIL(2θC+θA3ψCA)+FEMCA

Substitute 16.49 ft for L, 0 for θA, 0.0625Δ for ψCA, and 0 for FEMCA.

MCA=2EI16.49(2θC+03(0.0625Δ))0=0.2426EIθC+0.0227EIΔ        (2)

Calculate the slope deflection equation for the member CD.

MCD=2EIL(2θC+θDψCD)+FEMCD

Substitute 16 ft for L, 0.03125Δ for ψCD, and 64kft for FEMCD.

MCD=2EI16(2θC+θD3(0.03125Δ))+64=0.25EIθC+0.125EIθD0.01172EIΔ+64        (3)

Calculate the slope deflection equation for the member DC.

MDC=2EIL(2θD+θCψDC)+FEMDC

Substitute 16 ft for L, 0.03125Δ for ψDC, and 64kft for FEMDC.

MDC=2EI16(2θD+θC3(0.15Δ))64=0.125EIθC+0.25EIθD0.01172EIΔ64        (4)

Calculate the slope deflection equation for the member DB.

MDB=2EIL(2θD+θB3ψDB)+FEMDB

Substitute 16.49 ft for L, 0 for θB, 0.0625Δ for ψDB, and 0kft for FEMDB.

MDB=2EI16.49(2θD+(0)3(0.0625Δ))+0=0.2426EIθD+0.0227EIΔ        (5)

Calculate the slope deflection equation for the member BD.

MBD=2EIL(2θB+θD3ψBD)+FEMBD

Substitute 16.49 ft for L, 0 for θB, 0.0625Δ for ψBD, and 0kft for FEMBD.

MBD=2EI16.49(θD+2(0)3(0.0625Δ))+0=0.121EIθD+0.0227EIΔ        (6)

Write the equilibrium equation as below.

MCA+MCD=0

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

0.2426EIθC+0.0227EIΔ+0.25EIθC+0.125EIθD0.01172EIΔ+64=00.4926EIθC+0.125EIθD+0.011EIΔ=64        (7)

Write the equilibrium equation as below.

MDC+MDB=0

Substitute equation (4) and equation (5) in above equation.

0.125EIθC+0.25EIθD0.01172EIΔ64+0.2426EIθD+0.0227EIΔ=00.125EIθC+0.4926EIθD+0.011EIΔ=64        (8)

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

Show the free body diagram of the frame due to sway force as in Figure 4.

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

MA=0(SAC×16.49)+MAC+MCA=0SAC=MAC+MCA16.49

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

MA=0(SBD×16.49)+MBD+MDB=0SBD=MBD+MDB16

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