   # 4.46 through 4.50 Determine the force in each member of the truss shown. FIG. P4.46

#### Solutions

Chapter
Section
Chapter 4, Problem 46P
Textbook Problem
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## 4.46 through 4.50 Determine the force in each member of the truss shown. FIG. P4.46

To determine

Find the forces in the members of the truss.

### 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 and section.

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates Tension (T).

Calculation:

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 E is Ey.

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

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

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

Fy=0Ay+Ey=40+30Ay+Ey=70k        (1)

Take the sum of the moment about point A as zero.

MA=0[(10×20)(10×40)(40×30)(30×45)+(Ey×60)]=0Ey=3,15060Ey=52.5k

Substitute 52.5k for Ey in Equation (1).

Ay+52.5=70Ay=7052.5Ay=17.5k

Calculate the value of the angle θ as follows:

tanθ=(4015)θ=tan1(4015)θ=69.44°

Consider a section a-a passing through the members BC, CG, and JK.

Show the portion of the truss to the left of the section a-a as shown in Figure 2.

Refer Figure 2.

Take the sum of the moment about C as zero.

MF=0(10×20)(10×40)(17.5×30)FJK×40=0FJK=112540FJK=28.13k(C)

JOINT A.

Show the joint as shown in Figure 3.

Refer Figure 3.

For Equilibrium of forces,

Fy=0FAFsin(69.44°)+17.5=0FAF=17.5sin(69.44°)FAF=18.69k(C)

Fx=020+FAB+FAFcos(69.44°)=020+FAB+(18.69)cos(69.44°)=0FAB=18.69cos(69.44°)+20FAB=26.56k(T)

JOINT J.

Show the joint as shown in Figure 4.

Refer Figure 4.

For Equilibrium of forces,

Fy=0FJFsin(69.44°)FJGsin(69.44°)=0FJF=FJG        (2)

Fx=010FJFcos(69.44°)+FJK+FJGcos(69.44°)=010FJFcos(69.44°)+(28.13)+FJGcos(69.44°)=010(FJG)cos(69.44°)+(28.13)+FJGcos(69.44°)=0

10+0.3511FJG28.13+0.3511FJG=0FJG=18.130.7022FJG=25.81k(T)

Substitute 25.81k for FJG in Equation (2).

FJF=25.81kFJF=25.81k(C)

JOINT F.

Show the joint as shown in Figure 5.

Refer Figure 5.

For Equilibrium of forces,

Fy=0FFAsin(69.44°)FFBsin(69.44°)+FFJsin(69.44°)=0FFAFFB+FFJ=0(18.69)FFB+(25.81)=0FFB=7.12k(C)

Fx=010+FFJcos(69.44°)+FFBcos(69.44°)+FFGFFAcos(69.44°)=010+(25.81)cos(69.44°)+(7.12)cos(69.44°)+FFGFFAcos(69.44°)=0[10+(25.81)cos(69.44°)+(7.12)cos(69.44°)+FFG(18.69)cos(69.44°)]=0FFG+5=0FFG=5k(C)

JOINT G.

Show the joint as shown in Figure 6.

Refer Figure 6.

For Equilibrium of forces,

Fx=0FGFFGJcos(69.44°)FGBcos(69.44°)+FGCcos(69.44°)=0(5)25.81cos(69.44°)7.12cos(69.44°)+FGCcos(69.44°)=025.817.12+FGC=5cos(69.44°)25.817.12+FGC=14.237FGC=18.69k(T)

Fy=0FGBsin(69.44°)FGCsin(69.44°)+FGJsin(69.44°)=0FGBFGC+FGJ=0FGB18

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