# 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

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 15, Problem 9P
Textbook Problem
29 views

## 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=106.7kNm

Calculate the fixed end moment for BC.

FEMBC=106.7kNm

Calculate the fixed end moment for CB.

FEMCB=106.7kNm

Calculate the fixed end moment for CE.

FEMCE=60×88=60kNm

Calculate the fixed end moment for EC.

FEMEC=60kNm

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 0 for θA, 8 m for L and 106.7kNm for FEMAB.

MAB=2EI8(2(0)+θB3(0))+106.7=0.25EIθB+106.7        (1)

Calculate the slope deflection equation for the member BA.

MBA=2EIL(2θB+θA3ψ)+FEMBA

Substitute 0 for θA, 8 m for L and 106.7kNm for FEMBA.

MBA=2EI8(2θB+03(0))106.7=0.5EIθB106.7        (2)

Calculate the slope deflection equation for the member BC.

MBC=2EIL(2θB+θCψ)+FEMBC

Substitute 8 m for L and 106.7kNm for FEMBC.

MBC=2EI8(2θB+θC(0))+106.7=0.5EIθB+0.25EIθC+106.7        (3)

Calculate the slope deflection equation for the member CB.

MCB=2EIL(2θC+θBψ)+FEMCB

Substitute 8 m for L and 106.7kNm for FEMCB.

MCB=2EI8(θB+2θC(0))106.7=0.25EIθB+0.5EIθC106.7        (4)

Calculate the slope deflection equation for the member CE.

MCE=2EIL(2θC+θEψ)+FEMCE

Substitute 0 for θE, 8 m for L, and 60kNm for FEMCE.

MCE=2EI8(0+2θC(0))+60=0.5EIθC+60        (5)

Calculate the slope deflection equation for the member EC.

MEC=2EIL(2θE+θCψ)+FEMEC

Substitute 0 for θE, 8 m for L, and 60kNm for FEMEC.

MEC=2EI8(θC+2(0)(0))60=0.25EIθC60        (6)

Write the equilibrium equation as below.

MBA+MBC=0

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

0.5EIθB106.7+0.5EIθB+0.25EIθC+106.7=0EIθB+0.25EIθC=0        (7)

Write the equilibrium equation as below.

MCB+MCE=0

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

0.25EIθB+0.5EIθC106.7+0.5EIθC+60=00.25EIθB+EIθC46.7=00.25EIθB+EIθC=46.7        (8)

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

θB=12.453EIkNm2θC=49.81EIkNm2

Substitute 12.453EIkNm2 for θB in equation (1).

MAB=0.25EI(12.453EI)+106.7=103.6kNm

Substitute 12.453EIkNm2 for θB in equation (2).

MBA=0.5EI(12.453EI)106.7=112.9kNm

Substitute 12.453EIkNm2 for θB and 49.81EIkNm2 for θC in equation (3).

MBC=0.5EI(12.453EI)+0.25EI(49.81EI)+106.7=112.9kNm

Substitute 12.453EIkNm2 for θB and 49.81EIkNm2 for θC in equation (4)

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