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Structural Analysis

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

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 13, Problem 29P
Textbook Problem
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13.26 through 13.29 Determine the reactions and the force in each member of the trusses shown in Figs. P13.26–P13.29 using the method of consistent deformations.

Chapter 13, Problem 29P, 13.26 through 13.29 Determine the reactions and the force in each member of the trusses shown in

To determine

Calculate the support reactions and the member forces of the truss using method of consistent deformation.

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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

  1. 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
  2. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the degree of indeterminacy of the structure:

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

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

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

Select the horizontal reaction Bx at the support B as redundant.

Sketch the FO forces of the primary truss subjected to external loads as shown in Figure 1.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

MA=0By(10)60(10)80(5)=0By=100kN

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

+Fy=0Ay+By80=0Ay+10080=0Ay=20kN

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

+Fx=0Ax+60=0Ax=60kN

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

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

+Fy=020+FACsin45°+FADsin63.435°=0FACsin45°+FADsin63.435°=20        (1)

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

+Fx=060+FACcos45°+FADcos63.435°=0FACcos45°+FADcos63.435°=60        (2)

Solve Equation (1) and Equation (2).

FAC=141.42kNFAD=89.44kN

Apply equilibrium equation to the joint B:

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

+Fy=0100+FBCsin45°+FBDsin63.435°=0FBCsin45°+FBDsin63.435°=100        (3)

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

+Fx=0FBCcos45°FBDcos63.435°=0        (4)

Solve Equation (3) and Equation (4).

FBC=141.42kNFBD=223.61kN

Apply equilibrium equation to the joint D:

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

+Fy=080FCDFADsin63.435°FBDsin63.435°=080FCD(89.44)sin63.435°(223.61)sin63.435°=0FCD=200kN

Sketch the uB forces of the primary truss subjected to unit value of redundant Bx as shown in Figure 2.

Find the reactions at the supports using equilibrium equations:

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

+Fx=0Ax1=0Ax=1kN

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

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

+Fy=0FACsin45°+FADsin63.435°=0        (5)

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

+Fx=01+FACcos45°+FADcos63.435°=0FACcos45°+FADcos63.435°=1        (6)

Solve Equation (5) and Equation (6).

FAC=2.828kNFAD=2.236kN

Apply equilibrium equation to the joint B:

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

+Fy=0FBCsin45°+FBDsin63

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Chapter 13 Solutions

Structural Analysis
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