   # 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.28

#### Solutions

Chapter
Section
Chapter 4, Problem 28P
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. P4.28

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 AB, BC, CD, DE, BF, DG, AF, CF, CG, and EG are FAB,FBC,FCD,FDE,FBF,FDG,FAF,FCF,FCG,FEG.

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 D and E are Dy and Ey.

Find the value of the angle θ as follows:

tanθ=34θ=tan1(34)θ=36.869°

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

Fx=0Ax+50=0Ax=50kN

Consider the member AC.

Take the sum of the moments at C is zero. Then,

MCAC=0(Ay×8)+(120×4)=08Ay=480Ay=(4808)Ay=60kN

Take the sum of the moments at E is zero. Then,

ME=0(Ay×16)+(120×12)+(120×8)(50×3)Dy×4=016Ay+1440+9604Dy150=016Ay+24004Dy150=0

16Ay+24004Dy150=0Dy=16×60+24001504Dy=12904Dy=322.5kN

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

Fy=0Ay+Dy+Ey120120=0(60)+(322.5)+Ey240=0Ey=142.5kN

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the forces in the members AF and AB.

For the Equilibrium of forces,

Fy=0FAFsin(36.869°)+Ay=0FAFsin(36.869°)=AyFAF=60sin(36.869°)FAF=100kNFAF=100kN(C)

Fx=0FAFcos(36.869°)+FAB+Ax=0100cos(36.869°)+FAB+(50)=0FAB=100cos(36.869°)+50FAB=130kN(T)

Show the joint B as shown in Figure 3.

Refer Figure 3.

Find the force in the member BC and BF as follows:

For the Equilibrium of forces,

Fy=0FBF120=0FBF=120kN(T)

Fx=0FAB+FBC=0FBC=FABFBC=130kN(T)

Show the joint E as shown in Figure 4

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