   # 13.34 through 13.36 Determine the reactions and the force in each member of the trusses shown in Figs. P13.34–P13.36 using the method of consistent deformations. FIG. P13.34

#### Solutions

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Chapter 13, Problem 34P
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## 13.34 through 13.36 Determine the reactions and the force in each member of the trusses shown in Figs. P13.34–P13.36 using the method of consistent deformations.FIG. P13.34 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:

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(16)20(16)15(12)=0By=31.25k

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

+Fy=0Ay+By2020=0Ay+31.252020=0Ay=8.75k

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

+Fx=0Ax+15=0Ax=15k

Find the member forces using method of joints:

The force in the member CD is zero, as it satisfies the zero force member condition.

Apply equilibrium equation to the joint A:

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

+Fy=08.75+FAC=0FAC=8.75k

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

+Fx=015+FAB=0FAB=15k

Apply equilibrium equation to the joint B:

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

+Fx=0FABFBCcos36.87°=015FBCcos36.87°=0FBC=18.75k

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

+Fy=031.25+FBCsin36.87°+FBD=031.25+(18.75)sin36.87°+FBD=0FBD=20k

Sketch the uAD forces of the primary truss subjected to unit value of FAD as shown in Figure 2.

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.

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

+Fx=0(1)cos36.87°+FAB=0FAB=0.8k

Apply equilibrium equation to the joint B:

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

+Fx=0FABFBCcos36.87°=0(0.8)FBCcos36.87°=0FBC=1k

Summation of forces along y-direction is equal to 0

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