# 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.17

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 4, Problem 17P
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. P4.17

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, DG, GF, and FE are FAB,FBC,FCD,FDG,FGF,FFE.

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

Refer Figure 1.

Calculate the length of the member CG, BF using the relation:

DAAE=CDCGCG=CDDA×AECG=515×5CG=53

DAAE=BDBFBF=BDDA×AEBF=1015×5BF=103

Calculate the value of the angle CDG as follows:

θ=tan1(515)=18.43°

Calculate the value of the angle ABF as follows:

α=tan1[(103)5]=33.69°

Consider the joint D as shown in Figure 2.

Refer Figure 2.

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

Fy=0FGDsinθ40=0FGDsin(18.43°)40=0FGD=40sin(18.43°)FGD=126.49kN(C)

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

Fx=0FGDcosθFCD=0FCD=FGDcos(18.43°)

Substitute 126.49kN for FGD.

FCD=(126.49)cos(18.43°)=120kN(T)

Consider the joint C as shown in Figure 3.

Refer Figure 3.

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

Fy=0FCG40=0FCG=40FCG=40kN(C)

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

Fx=0FCB=FCD

Substitute 120kN for FCD.

FCB=120kN(T)

Show the joint G as shown in Figure 4.

Refer Figure 4.

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

Fy=0FGC+FGBsin18.43°+FGDsin18.43°FGFsin18.43°=0(FGBFGF)sin18.43°=FGDsin18.43°FGCFGBFGF=FGDsin18.43°FGCsin18.43°FGBFGF=FGD3.163FGC

Substitute 40kN for GC and 126.49kN for FGD.

FGBFGF=(126.49)3.163×(40)FGBFGF=253        (1)

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

Fx=0FGBcos18.43°+FGDcos18

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