   # 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. P15.12, P15.16

#### Solutions

Chapter
Section
Chapter 15, Problem 12P
Textbook Problem
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## 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. P15.12, P15.16 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 unequal length are Pab2L2 and Pa2bL2.

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

Calculation:

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

Refer Figure 1,

Calculate the fixed end moment for AC.

FEMAC=120×6×42102=115.2kNm

Calculate the fixed end moment for CA.

FEMCA=120×4×62102=172.8kNm

Calculate the fixed end moment for CE.

FEMCE=120×6×42102=115.2kNm

Calculate the fixed end moment for EC.

FEMEC=120×4×62102=172.8kNm

Calculate the fixed end moment for EG.

FEMEG=150×88=150kNm

Calculate the fixed end moment for GE.

FEMGE=150×88=150kNm

Calculate the slope deflection equation for the member AC.

MAC=0

Calculate the slope deflection equation for the member CA.

MCA=3EIL(2θA+θC3ψ)+FEMCAFEMAC2

Substitute 0 for ψ, 0 for θA, 10 m for L, 172.8kNm for FEMCA, and 115.2kNm for FEMAC.

MCA=3EI10(θC+2(0)(0))172.8115.22=0.3EIθC230.4 (1)

Calculate the slope deflection equation for the member CE.

MCE=2E(2I)L(2θC+θE3ψ)+FEMCE

Substitute 0 for ψ, 10 m for L and 115.2kNm for FEMCE.

MCE=2E(2I)10(θE+2θC(0))+115.2=0.4EIθE+0.8EIθC+115.2 (2)

Calculate the slope deflection equation for the member EC.

MEC=2E(2I)L(2θE+θC3ψ)+FEMEC

Substitute 0 for ψ, 10 m for L and 172.8kNm for FEMEC.

MEC=2E(2I)10(θC+2θE(0))172.8=0.4EIθC+0.8EIθE172.8 (3)

Calculate the slope deflection equation for the member EG.

MEG=3EIL(2θG+θE3ψ)+FEMEGFEMGE2

Substitute 0 for ψ, 0 for θG, 8 m for L, 150kNm for FEMEG, and 150kNm for FEMGE.

MEG=3EI8(θE+2(0)(0))+150+1502=0.375EIθE+225 (4)

Calculate the slope deflection equation for the member GE.

MGE=0

Write the equilibrium equation as below.

MCA+MCE=0

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

0.3EIθC230.4+0.4EIθE+0.8EIθC+115.2=01.1EIθC+0.4EIθE115.2=01.1EIθC+0.4EIθE=115.2 (5)

Write the equilibrium equation as below.

MEC+MEG=0

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

0.4EIθC+0.8EIθE172.8+0.375EIθE+225=00.4EIθC+1.175EIθE+52.2=00.4EIθC+1.175EIθE=52.2 (6)

Solve the equation (5) and equation (6).

θC=137.96EIkNm2θE=91.39EIkNm2

Substitute 137.96EIkNm2 for θC in equation (1).

MCA=0.3EI(137.96EI)230.4=189kNm

Substitute 137.96EIkNm2 for θC and 91.39EIkNm2 for θE in equation (2).

MCE=0

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