   # 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.13

#### Solutions

Chapter
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Chapter 4, Problem 13P
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## 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints.FIG. P4.13 To determine

Find the forces in the members of the truss by the method of joints.

### Explanation of Solution

Given information:

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).

Calculation:

Consider the forces in the member AB, BC, BD, BE, AD, DE, and EC are FAB,FBC,FBD,FBE,FAD,FDE,FEC.

Show the free body diagram of the truss as shown in Figure 1.

Refer Figure 1.

Consider the horizontal and vertical reactions at D are Dx and Dy.

Consider the horizontal reaction at A is Ax.

Consider the angle made by member BD, CE and BE with the horizontal is θ.

Calculate the value of the angle θ as follows:

tanθ=512θ=tan1(512)θ=22.61°

Take the sum of the forces in the vertical direction as zero.

Fy=0Dy+6.5sin(22.61°)1020=0Dy=6.5sin(22.61°)1020Dy=27.5k        (1)

Take the sum of the forces in the horizontal direction as zero.

Fx=0Ax+Dx6.5cosθ=0Ax+Dx6.5cos(22.61°)=0Ax+Dx=6        (2)

Take the sum of the moments at D is zero.

MD=0(Ax×5)(20×12)+(6.5sinθ×24)10×36=0(Ax×5)(20×12)+(6.5sin(22.61°))×2410×36=05Ax240+60360=0

5Ax=540Ax=(5405)Ax=108k

Substitute 108k for Ax in Equation (2).

108+Dx=6Dx=102k

Show the joint C as shown in Figure 2.

Refer Figure 2.

Take the sum of the forces in the vertical direction as zero.

Fy=010+FCEsinθ=010+FCEsin(22.61°)=0FCE=[10sin(22.61°)]FCE=26k(T)

Take the sum of the forces in the horizontal direction as zero.

Fx=0FCEcosθFCB=0FCB=FCEcos(22.61°)

Substitute 26k for FEC.

FCB=26cos(22.61°)FCB=24k(C)

Show the joint E as shown in Figure 3.

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