   # 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.22

#### Solutions

Chapter
Section
Chapter 4, Problem 22P
Textbook Problem
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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. P4.22

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 AE, EG, AC, CG, BF, FG, BD, DG, CD, CE, and DF are FAE,FEG,FAC,FCG,FBF,FFG,FBD,FDG, FCD,FCE,FDF.

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 vertical reaction at B is By.

Calculate the value of the angle θ as follows:

tanθ=2525θ=tan1(1)θ=45°

Calculate the value of the angle α as follows:

tanβ=1010β=tan1(1)β=45°

Calculate the value of the angle α as follows:

tanα=3515α=tan1(3515)α=66.80°

Calculate the value of the angle γ as follows:

tanγ=1535γ=tan1(1535)γ=23.198°

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

Fy=0Ay+By=40        (1)

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

Fx=0Ax+10+20+10=0Ax=40k

Take the sum of the moments at A is zero.

MA=0(By×100)(10×35)(20×25)(20×50)(20×75)(10×35)=0By×100=3700By=37k

Substitute 37k for By in Equation (1).

Ay+37=40Ay=3k

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the forces in the member AC and AE as follows:

For Equilibrium of forces,

Fy=0Ay+FAEsinα+FACsinθ=03+FAEsin(66.80°)+FACsin(45°)=03+0.9191FAE+0.7071FAC=00.9191FAE+0.7071FAC=3        (2)

Fx=0Ax+FAEcosα+FACcosθ=040+FAEcos(66.80°)+FACcos(45°)=040+0.3939FAE+0.7071FAC=00.3939FAE+0.7071FAC=40        (3)

Subtract Equation (2) from Equation (3).

0.5252FAE=43FAE=430.5252FAE=81.87k(C)

Substitute 81.87k for FAE in Equation (3).

0.3939×(81.87)+0.7071FAC=400.7071FAC=40+0.3939×(81.87)FAC=40+0.3939×(81.87)0.7071FAC=102.18k(T)

Show the joint E as shown in Figure 3.

Refer Figure 3.

Find the forces in the member EC and EG as follows:

For Equilibrium of forces,

Fy=0FEAsinαFECsinβ+FEGsinγ=0(81.87)sin66.80°FECsin45°+FEGsin23.198°=075.2490.7071FEC+0.3939FEG=00.3939FEG0.7071FEC=75.249        (4)

Fx=010FAEcosα+FECcosβ+FEGcosγ=010(81.87)cos66.80°+FECcos45°+FEGcos23.198°=010+32.252+0.7071FEC+0.9191FEG=00.9191FEG+0.7071FEC=42.252        (5)

Solve Equation (4) and (5).

FEG=89.49kFEG=89.49k(C)

FEC=56.57kFEC=56.57k(T)

Show the joint C as shown in Figure 4

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