# 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.15

### 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 15P
Textbook Problem
60 views

## 4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints.FIG. P4.15

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, EF, FG, AH, HI, IJ, JK, KL, LG, BH, BI, CI, CJ, FL, FK, EK, EJ, DJ are FAB,FBC,FCD,FDE,FEF,FFG,FAH,FHI,FIJ,FJK,FKL,FLG,FBH, FBI, FCI, FCJ, FFL, FFK, FEK, FEJ, FDJ.

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

Refer Figure 1.

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

Consider the vertical reaction at F is Fy.

By=Fy=10+20+20+20+102By=Fy=40k

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

Fx=0Bx=0

Consider the angle made by each of the inclined member with the horizontal is denoted by θ.

Calculate the value of the angle θ as follows:

tanθ=2525θ=tan1(1)θ=45°

Show the joint A as shown in Figure 2.

Refer Figure 2.

Find the forces in the member AB and AH as follows:

For Equilibrium of forces,

Fy=0FAHsin(45°)=10FAH=14.14k(T)

Fx=0FAHcos(45°)+FAB=0FAB=FAHcos(45°)FAB=14.14cos(45°)FAB=10k(C)

Show the joint H as shown in Figure 3.

Refer Figure 3.

Find the forces in the member HI and HB as follows:

For Equilibrium of forces,

Fx=0FHI=FHAsin(45°)FHI=14.14sin(45°)FHI10k(T)

Fy=0FHB=FAHcos(45°)FHB=14.14cos(45°)FHB10k(C)

Show the joint B as shown in Figure 4.

Refer Figure 4.

Find the forces in the member BI and BC as follows:

For Equilibrium of forces,

Fy=040+FBH+FBIsin(45°)=0FBI=(40FBHsin(45°))FBI=(40(10)sin(45°))FBI=42.43k(C)

Fx=0FBA+FBC+FBIcos45°=0FBC=FBAFBIcos45°FBC=(10)(42.43)cos45°FBC=20k(T)

Show the joint I as shown in Figure 5.

Refer Figure 5.

Find the forces in the member IC and IJ as follows:

For Equilibrium of forces,

Fy=0FICFIBcos45°=0FIC=FIBcos45°FIC=(42

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