   # 4.6 through 4.28 Determine the force in each member of the truss shown by the mehod of joints. FIG. P4.27

#### Solutions

Chapter
Section
Chapter 4, Problem 27P
Textbook Problem
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## 4.6 through 4.28 Determine the force in each member of the truss shown by the mehod of joints. FIG. P4.27

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 AC, CG, BF, FG, AD, DG, BE, EG, CD, FE are FAC,FCG,FBF,FFG,FAD,FDG,FBE,FEG,FCD,FFE.

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

Refer Figure 1.

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

Consider the horizontal and vertical reactions at B are Bx and By.

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

Fy=0Ay+By=20+20+20Ay+Dy=60        (1)

Take the sum of the moments at A is zero.

MA=0(20×3)(10×6)(20×6)(20×9)+By×12=06060120180+By×12=0By×12=420By=35kN

Substitute 35kN for By in Equation (1).

Ay+35=60Ay=25kN

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

Fx=0Ax+10Bx=0        (2)

Consider the portion of the truss to the left of the section passing through G.

Take sum of the moments about G is zero.

MG=0Ay×6+Ax×6+20×3=025×6+Ax×6+20×3=0Ax=906Ax=15kN

Substitute 15kN for Ax in Equation (1).

15+10Bx=10Bx=25kN

Calculate the value of the angle θ as follows:

tanθ=33θ=tan1(33)θ=45°

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the forces in the members AC and AD.

For Equilibrium of forces,

Solve Equation (3) and (4).

Show the joint C as shown in Figure 3.

Refer Figure 3.

Find the forces in the members CD and CG.

For Equilibrium of forces,

Fx=0FCA+FCG20cos45°=0(47.71)+FCG20cos45°=0(47

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