   # For the truss of Problem 8.47, determine the maximum compressive axial force in member GH due to a concentrated live load of30 k, a uniformly distributed live load of 2 k/ ft, and a uniformly distributed dead load of 1 k/ft. FIG. P8.47

#### Solutions

Chapter
Section
Chapter 9, Problem 9P
Textbook Problem
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## For the truss of Problem 8.47, determine the maximum compressive axial force in member GH due to a concentrated live load of30 k, a uniformly distributed live load of 2 k/ ft, and a uniformly distributed dead load of 1 k/ft. FIG. P8.47

To determine

Find the maximum compressive axial force in member GH.

### Explanation of Solution

Given Information:

The concentrated live load (P) is 30 k.

The uniformly distributed live load (wL) is 2 k/ft.

Calculation:

Influence line for the force in member GH.

Find the force in member GH.

The expressions for the member force FGH can be determined by passing an imaginary section aa through the members GH, CH, and CD and then apply a moment equilibrium at G.

The appropriate support reaction is first determined by proportions and then influence line ordinate can be found

Place 1 k load successively at the points A, B, C, D, and E.

Draw the free body diagram of section aa as shown in Figure 1.

Refer Figure 1.

Apply a 1 k load just the left of C (0x40ft).

Apply moment equilibrium equation at C.

ΣMC=0

Take moment at C from E.

Consider clockwise moment as negative and counterclockwise moment as positive.

Ey(40)+FGH(15)=015FGH=40EyFGH=83Ey

Apply a 1 k load just the right of C (40ftx80ft).

Apply moment equilibrium equation at C.

ΣMC=0

Take moment at C from A.

Consider clockwise moment as positive and counterclockwise moment as negative.

Ay(40)+FGH(15)=015FGH=40AyFGH=83Ay

The equation of force in the member GH,

FGH=83Ey, 0x40ft        (1)

FGH=83Ay, 40ftx80ft        (2)

Find the force in member GH using the Equation (1) and (2) and summarize the value in Table 1

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