   # 4.33 through 4.45 Determine the forces in the members identified by “×” of the truss shown by the method of sections. FIG. P4.45

#### Solutions

Chapter
Section
Chapter 4, Problem 45P
Textbook Problem
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## 4.33 through 4.45 Determine the forces in the members identified by “×” of the truss shown by the method of sections. FIG. P4.45

To determine

Find the forces in the members IN, ID, and CD 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).

Calculation:

Show the section a-a and b-b on the truss as shown in Figure 1.

Refer Figure 1.

Consider the section a-a passing through the members CD, CI, IM, and MN.

Calculate the value of the angle θ as follows:

tanθ=1530θ=tan1(1530)θ=26.565°

Consider the vertical reaction at D as Dy.

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

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

Fy=0Ay+Dy=40+40+40+40+20Ay+Dy=180        (1)

Determine the reaction at support A using the relation;

MD=0[(Ay×90)+(40×60)+(40×30)(40×30)(40×60)(20×90)]=0(Ay×90)800=0Ay=80090Ay=20k

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

Fx=0Ax=0

Show the portion of the truss left to the section a-a as shown in Figure 2.

Refer Figure 2.

Consider the forces in the members CD, CI, IM and MN are denoted by FCD, FCI, FIM, and FMN.

Take the sum of the moments acting at M as zero.

MM=0(20×60)+(40×30)+FCD×30=0FCD=240030FCD=80k(C)

Consider the section b-b passing through members CD, DI, IN, and MN

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