   # Determine the forces in the members identified by “x” of the truss shown by the method of sections. FIG. P4.33

#### Solutions

Chapter
Section
Chapter 4, Problem 33P
Textbook Problem
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## Determine the forces in the members identified by “x” of the truss shown by the method of sections. FIG. P4.33

To determine

Find the forces in the members JK, DK, and DE of the truss by the method of sections.

### 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 sections:

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

The Figure of the truss is shown.

The members identified by “x” are JK, DK, and DE.

Consider the forces in the members JK, DK, and DE are denoted by FJK, FDK, and FDE.

Calculation:

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

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by Ax and Ay.

Consider the vertical reaction at G is denoted by Gy.

Ay=Gy=(20+20+20+20+202)=50k

Find the angle (θ) made by the member DK with the horizontal as follows:

tanθ=EKDEtanθ=2525θ=tan1(1)θ=45°

Consider a section a-a passing through JK, DK, and DE.

Show the portion of the truss to the right of the truss as shown in Figure 2.

Refer Figure 1 and Figure 2.

For Equilibrium of forces,

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

Fy=0FDK×sinθ2020+50=0FDK×sin45°+10=0FDK×sin45°=10

FDK=10sin45°FDK=14

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