   # For the truss of Problem 8.49, determine the maximum compressive axial force in member GH due to the series of four moving concentrated loads shown in Fig. P9.14. FIG. P9.14, P9.16, P9.19, P9.23 FIG. P8.49

#### Solutions

Chapter
Section
Chapter 9, Problem 16P
Textbook Problem
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## For the truss of Problem 8.49, determine the maximum compressive axial force in member GH due to the series of four moving concentrated loads shown in Fig. P9.14. FIG. P9.14, P9.16, P9.19, P9.23 FIG. P8.49

To determine

Find the maximum compressive axial force in member GH.

### Explanation of Solution

Calculation:

Find the equation of support reactions using equilibrium.

Apply 1 k load from A to E.

Sketch the free body diagram as shown in Figure 1.

Refer Figure 1.

Find the equation of support reaction at D using moment equilibrium at A.

Consider clockwise moment as positive.

ΣMA=0(Dy)(60)+1(x)=0Dy=x60        (1)

Find the equation of support reaction at A using force equilibrium at A.

ΣFy=0Ay+Dy=1Ay+x60=1Ay=1x60        (2)

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, CG, and BC and then apply a moment equilibrium at G.

Draw the free body diagram as shown in Figure 2.

Refer Figure 3.

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.

Dy(20)+FGH(15)=015FGH=20DyFGH=43Dy

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=43Dy, 0x40ft        (3)

FGH=83Ay, 40ftx80ft        (4)

Find the support reactions using Equations (1) and (2), force in member GH using the Equations (3) and (4) and summarize the value in Table 1.

 x (ft) Apply 1 k load Support reaction (Ay) at A (k) Support reaction (Dy) at D (k) Influence line ordinate for the force in member GH (k/k) 0 A 1 0 0 20 B 23 13 −43Dy=−43(13)=−0.444 40 C 13 23 −43Dy=−43(23)=−0.889 60 D 0 1 −83Ay=0 80 E −13 43 −83Ay=−83(−13)=0.889

Sketch the influence line diagram for the force in member GH using Table 1 as shown in Figure 3.

Refer Figure 3.

Find the slope (θAC) of influence line diagram for the force in member GH.

θAC=0.889LAC

Here, LAC is the length of the truss from A to C.

Substitute 40 ft for LAC.

θAC=0.88940=145

Find the slope (θCD) of influence line diagram for the force in member GH.

θCD=0.889LCD

Here, LCD is the length of the truss from C to D.

Substitute 20 ft for LCD.

θCD=0.88920=122.5

Find the slope (θDE) of influence line diagram for the force in member GH.

θDE=0.889LDE

Here, LDE is the length of the truss from D to E.

Substitute 20 ft for LDE.

θDE=0.88920=122.5

Find the maximum compressive axial force in member GH.

Sketch the loading position on the truss when the load 1 placed at just right of C as shown in Figure 5.

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