   # Determine the forces in the members identified by “x” of the truss shown by the method of sections. FIG. P4.44

#### Solutions

Chapter
Section
Chapter 4, Problem 44P
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. P4.44

To determine

Find the forces in the members AE, CD, and BF of the truss by the method of sections.

### Explanation of Solution

Given information:

The Figure of the truss is shown.

The members identified by “x” are AE, CD, and BF.

Consider the forces in the members AE, CD, and BF are denoted by FAE, FCD, and FBF.

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).

Calculation:

Show the truss as shown in Figure 1.

Refer Figure 1.

Find the slope of the member CD as follows:

Consider the slope of the member CD with the horizontal is (θ). Then,

tanθ=34θ=tan1(34)θ=36.869°

Consider a horizontal section passing through the members AE, CD, and BF denoted by a-a.

Show the portion of the truss just above the section as shown in Figure 2.

Refer Figure 2.

For Equilibrium of forces,

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

Fx=0100FCDcosθ=0100FCDcos(36.869°)=0FCD=100cos(36.869°)FCD=125kN(T)

Take sum of moments about F as zero.

MF=0FAE×16+50×6+FCDsinθ×6FCDcosθ×2=0FAE×16+50×6+125sin(36

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