# 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations. FIG. P13.24

### 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 13, Problem 24P
Textbook Problem
77 views

## 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations.FIG. P13.24

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

### Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Find the degree of indeterminacy of the structure:

Degree of determinacy of the frame is equal to the number of unknown reactions minus the number of equilibrium equations.

The frame is supported by 4 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the beam is i=1.

Select the vertical reaction By at the support B as redundant.

Sketch the support reactions and the member end forces of the frame as shown in Figure 1.

Find the reactions at the supports using equilibrium equations:

Summation of moments of all forces about A is equal to 0.

MA=0MA1.5(20)(202+15)=0MA=750k-ft

Summation of forces along y-direction is equal to 0.

+Fy=0Ay1.5(20)+18.37=0Ay=11.63k

The horizontal component of the reaction 30 k at A is 30cos53.13°=18k.

The vertical component of the reaction 30 k at A is 30sin53.13°=24k.

For unit value of the unknown redundant By:

Sketch the support reactions and the member end forces of the frame with unit value for the unknown redundant By as shown in Figure 2.

Summation of moments of all forces about A is equal to 0.

MA=0MA1(35)=0MA=35k-ft

Summation of forces along y-direction is equal to 0.

+Fy=0Ay+1=0Ay=1k

The horizontal component of the reaction 1 k at A is 1cos53.13°=0.6k.

The vertical component of the reaction 1 k at A is 1sin53.13°=0.8k.

Let the deflection at point B due to external loading is ΔBO and the flexibility coefficient representing the deflection at B due to unit value of redundant By is fBB.

Calculate the value of ΔBO as follows:

ΔBO=1EI[020(1.5x22)(1x)dx+025(18x750)(0.6x+35)dx]=1EI[020(1.5x22)dx+025(9x2+630x+375x26,250)dx]=1EI[[1.5x36]020+[9x33+1,005x2226,250x]025]=1EI(1.5(20)369(25)33+1,005(25)2226,250(25))

ΔBO=405,000k-ft3EI

Calculate the value of fBB as follows:

fBB=1EI[020(1x)2dx+025(0

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