   # Solve Problem 15.9 for the loading shown in Fig. P15.9 and a settlement of 25 mm at support C . 15.8 through 15.14 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.8–P15.14 by using the slope-deflection method. FIG. P15.9, P15.15

#### Solutions

Chapter
Section
Chapter 15, Problem 15P
Textbook Problem
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## Solve Problem 15.9 for the loading shown in Fig. P15.9 and a settlement of 25 mm at support C.15.8 through 15.14 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P15.8–P15.14 by using the slope-deflection method.FIG. P15.9, P15.15 To determine

Find the reaction and plot the shear and bending moment diagram.

### Explanation of Solution

Fixed end moment:

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

Formula to calculate the fixed moment for point load with unequal length are Pab2L2 and Pa2bL2.

Formula to calculate the fixed moment for UDL is WL212.

Formula to calculate the fixed moment for deflection is 6EIΔL2

Calculation:

Consider the flexural rigidity EI of the beam is constant.

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

Refer Figure 1,

Calculate the fixed end moment for AB.

FEMAB=20×8212=106.7kNm

Calculate the fixed end moment for BA.

FEMBA=20×8212=106.7kNm

Calculate the fixed end moment for BC.

FEMBC=20×8212=106.7kNm

Calculate the fixed end moment for CB.

FEMCB=20×8212=106.7kNm

Calculate the fixed end moment for CE.

FEMCE=60×88=60kNm

Calculate the fixed end moment for EC.

FEMEC=60×88=60kNm

Chord rotations:

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

Calculate the chord rotation of the beam BC.

ψBC=25mm×1m1000mm8m=0.003125

Calculate the chord rotation of the beam CE.

ψCE=25mm×1m1000mm8m=0.003125

Calculate the slope deflection equation for the member AB.

MAB=2EIL(2θA+θB3ψ)+FEMAB

Here, ψ is the chord rotation, θA is the slope at the point A and θB is the slope at the point B.

Substitute 70GPa for E, 800×106mm4 for I, 0 for θA, 8 m for L and 106.7kNm for FEMAB.

MAB=2×70×8008(2(0)+θB3(0))+106.7=14000θB+106.7 (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψ)+FEMBA

Substitute 70GPa for E, 800×106mm4 for I, 0 for θA, 8 m for L and 106.7kNm for FEMBA.

MBA=2×70×8008(2θB+03(0))106.7=28000θB106.7 (2)

Calculate the slope deflection equation for the member BC.

MBC=2EIL(2θB+θC3ψBC)+FEMBC

Substitute 70GPa for E, 800×106mm4 for I, 0.003125 for ψBC, 8 m for L and 106.7kNm for FEMBC.

MBC=2×70×8008(2θB+θC(3×0.003125))+106.7=28000θB+14000θC+131.25+106.7=28000θB+14000θC+238 (3)

Calculate the slope deflection equation for the member CB.

MCB=2EIL(2θC+θB3ψCB)+FEMCB

Substitute 70GPa for E, 800×106mm4 for I, 0.003125 for ψCB, 8 m for L and 106.7kNm for FEMCB.

MCB=2×70×8008(θB+2θC(3×0.003125))106.7=14000θB+28000θC+131.25106.7=14000θB+28000θC+24.6 (4)

Calculate the slope deflection equation for the member CE.

MCE=2EIL(2θC+θE3ψCE)+FEMCE

Substitute 70GPa for E, 800×106mm4 for I, 0 for θE, 0.003125 for ψCE, 8 m for L and 60kNm for FEMCE.

MCE=2×70×8008(0+2θC(3×0.003125))+60=28000θC131.25+60=28000θC71.3 (5)

Calculate the slope deflection equation for the member EC.

MEC=2EIL(2θE+θC3ψEC)+FEMEC

Substitute 70GPa for E, 800×106mm4 for I, 0 for θE, 0.003125 for ψCE, 8 m for L and 60kNm for FEMEC.

MEC=2×70×8008(θC+2(0)(3×0.003125))60=14000θC131.2560=14000θC191.3 (6)

Write the equilibrium equation as below.

MBA+MBC=0

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

28000θB106.7+28000θB+14000θC+238=056000θB+14000θC=131.3 (7)

Write the equilibrium equation as below.

MCB+MCD=0

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

14000θB+28000θC+24.6+28000θC71.3=014000θB+56000θC46.7=014000θB+56000θC=46.7 (8)

Solve the equation (7) and equation (8).

θB=2.723×103kNm2θC=1.515×103kNm2

Substitute 2.723×103kNm2 for θB in equation (1).

MAB=14000(2.723×103)+106.7=68.6kNm

Substitute 2.723×103kNm2 for θB in equation (2).

MBA=28000(2

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