   # 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. P15.25

#### Solutions

Chapter
Section
Chapter 15, Problem 25P
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. P15.25 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 elastic modulus E 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=0kft

Calculate the fixed end moment for CA.

FEMCA=0kft

Calculate the fixed end moment for CD.

FEMCD=1.5×40212=200kft

Calculate the fixed end moment for DC.

FEMDC=200kft

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 chord rotation of the frame AC.

ψAC=Δ30

Calculate the chord rotation of the frame CD.

ψCD=0

Calculate the chord rotation of the frame BD.

ψBD=Δ30

Calculate the slope deflection equation for the member AC.

MAC=2EIL(2θA+θC3ψAC)+FEMAC

Substitute Δ30 for ψAC, 0 for θA, 30 ft for L, and 0kft for FEMAC.

MAC=2EI30(2(0)+θC3(Δ30))+0=0.0667EIθC+0.00667EIΔ        (1)

Calculate the slope deflection equation for the member CA.

MCA=2EIL(2θC+θA3ψCA)+FEMCA

Substitute Δ30 for ψCA, 0 for θA, 30 ft for L and 0kft for FEMCA.

MCA=2EI30(2θC+03(Δ30))+0=0.1333EIθC+0.00667EIΔ        (2)

Calculate the slope deflection equation for the member CD.

MCD=2E(2I)L(2θC+θD3ψCD)+FEMCD

Substitute 0 for ψCD, 40 ft for L, and 200kft for FEMCD.

MCD=2E(2I)40(2θC+θD(3×0))+200=0.2EIθC+0.1EIθD+200        (3)

Calculate the slope deflection equation for the member DC.

MDC=2E(2I)L(2θD+θC3ψDC)+FEMDC

Substitute 0 for ψDC, 40 ft for L and 200kft for FEMDC.

MDC=2E(2I)40(2θD+θC(3×0))200=0.1EIθC+0.2EIθD200        (4)

Calculate the slope deflection equation for the member DB.

MDB=2EIL(2θD+θB3ψDB)+FEMDB

Substitute 0 for ψDB, 0 for θB, 30 ft for L, and 0kft for FEMDB.

MDB=2EI30(2θD+03(Δ30))+0=0.1333EIθD+0.00667EIΔ        (5)

Calculate the slope deflection equation for the member BD.

MBD=2EIL(2θB+θD3ψBD)+FEMBD

Substitute 0 for ψBD, 0 for θB, 30 ft for L, and 0kft for FEMBD.

MBD=2EI30(2(0)+θD3(Δ30))+0=0.0667EIθD+0.00667EIΔ        (6)

Write the equilibrium equation as below.

MCA+MCD=0

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

0.1333EIθC+0.00667EIΔ+0.2EIθC+0.1EIθD+200=00.3333EIθC+0.1EIθD+0.00667EIΔ=200        (7)

Write the equilibrium equation as below.

MDC+MDB=0

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

0.1EIθC+0.2EIθD200+0.1333EIθD+0.00667EIΔ=00.1EIθC+0.3333EIθD+0.00667EIΔ=200        (8)

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×30)+MAC+MCA=0SAC=MAC+MCA30

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

MB=0(SBD×30)+MBD+MDB=0SBD=MBD+MDB30

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

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