4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints. FIG. P4.14

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 14P
Textbook Problem
191 views

4.6 through 4.28 Determine the force in each member of the truss shown by the method of joints.FIG. P4.14

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 members AB, BC, CD, DE, EF, FG, AH, HI, IJ, JK, KL, LG, BH, CI, DJ, EK, FL, CH, DI, DK, and EL are FAB,FBC,FCD,FDE,FEF,FFG,FAH,FHI,FIJ,FJK,FKL,FLG,FBH,FCI,FDJ,FEK,FFL,FCH,FDI,FDK,FEL

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 G is Gy.

Calculate the value of the angle θ as follows:

tanθ=34θ=tan1(34)θ=36.86°

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

Fy=0Ay+Gy=60+60+60+30+30Ay+Gy=240        (1)

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

Fx=0Ax=0

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

MA=0(60×4)(60×8)(60×12)(30×16)(30×20)+Gy×24=0240480720480600+24Gy=0Gy=252024Gy=105kN

Substitute 105kN for Gy in Equation (1).

Ay+105=240Ay=135kN

Show the joint A as shown in Figure 2.

Refer Figure 2.

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

For Equilibrium of forces,

Fy=0AyFAHsin(36.86°)=01350.6FAH=0FAH=1350.6FAH=225kN(T)

Fx=0FAHcos(36.86°)+FAB=0FAB=FAHcos(36.86°)FAB=225cos(36.86°)FAB=180kN(C)

Show the joint B as shown in Figure 3.

Refer Figure 3.

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

For Equilibrium of forces,

Fy=0FBH60=0FBH=60kNFBH=60kN(C)

Fx=0FBCFAB=0FBC=FABFBC=180kNFBC=180kN(C)

Show the joint H as shown in Figure 4.

Refer Figure 4.

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

For Equilibrium of forces,

Fy=0FHCsin(36.86°)+FHAsin(36.86°)+FHB=0FHCsin(36.86°)+225sin(36.86°)60=00.6FHC=75FHC=125kN(C)

Fx=0FHCcos(36.86°)FHAcos(36.86°)+FHI=0125cos(36.86°)225cos(36.86°)+FHI=0280+FHI=0FHI=280kN(T)

Show the joint C as shown in Figure 5.

Refer Figure 5.

Find the forces in the member CD and CI as follows:

For Equilibrium of forces,

Fy=0FCHsin(36.86°)FCI60=0125sin(36.86°)FCI60=0FCI+15=0FCI=15kN(T)

Fx=0FCB+FCDFCHcos(36.86°)=0(180)+FCD(125)cos(36.86°)=0280+FCD=0FCD=280kNFCD=280kN(C)

Show the joint I as shown in Figure 6.

Refer Figure 6.

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

For Equilibrium of forces,

Fy=0FIDsin(36

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