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# Draw the influence lines for the forces in the members identified by an “×” of the trusses shown in Figs. P8.45–P8.52. Live loads are transmitted to the bottom chords of the trusses. FIG. PB.48

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### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

#### Solutions

Chapter
Section
BuyFindarrow_forward

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 8, Problem 48P
Textbook Problem
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## Draw the influence lines for the forces in the members identified by an “×” of the trusses shown in Figs. P8.45–P8.52. Live loads are transmitted to the bottom chords of the trusses.FIG. PB.48

To determine

Draw the influence lines for the force in member AF, BC, CF, and CG.

### Explanation of Solution

Calculation:

Influence line for the force in member AF.

Find the influence ordinate for the force in member AF at A.

The influence line for the force in member AF is determined by place a 1 k load at A.

Find the force in the member AF

Sketch the free body diagram at joint A as shown in Figure 1.

Refer Figure 1.

The support reaction (Ay) is 1 k when 1 k load applied at A.

Apply force equilibrium along vertical.

Consider upward force as positive and downward force as negative.

ΣFy=0Ay1+FAF(15)=011+FAF(15)=0FAF=0

Thus, the influence ordinate for the force in member AF at A is 0 k/k.

Find the influence ordinate for the force in member AF at B.

The influence line for the force in member AF is determined by place a 1 k load at just right of B.

The support reaction (Ay) is 34k when 1 k load applied at B.

Find the force in the member AF

Consider upward force as positive and downward force as negative.

ΣFy=0Ay+FAF(15)=034+FAF(15)=0FAF5=34FAF=1.677k

Thus, the influence ordinate for the force in member AF at B is ‑1.677 k/k.

Similarly calculate the influence ordinate for the force in member AF at different points on the beam and summarize in Table 1.

 Apply 1 k load Force in the member AF (k) Influence line ordinate for the force in member AF (k/k) A 0 0 B ‑1.677 ‑1.677 C ‑0.838 ‑0.838 D ‑0.419

Sketch the influence line diagram for ordinate for the force in member AF using Table 1 as shown in Figure 2.

Influence line for the force in member BC.

Find the force in member BC.

The expressions for the member force FBC can be determined by passing an imaginary section aa through the members BC, FC, and FG and then apply a moment equilibrium at F.

Draw the free body diagram as shown in Figure 3.

Refer Figure 3.

Apply a 1 k load just the left of B (0x16ft).

Apply moment equilibrium equation at F.

ΣMF=0

Take moment at F from E.

Consider clockwise moment as negative and counterclockwise moment as positive.

Ey(48)FBC(8)=08FCD=48EyFBC=6Ey

Apply a 1 k load just the right of B (16ftx64ft).

Apply moment equilibrium equation at F.

ΣMF=0

Take moment at F from A.

Consider clockwise moment as positive and counterclockwise moment as negative.

Ay(16)FBC(8)=0FBC=2Ay

The equation of force in the member BC,

FBC=6Ey, 0x16ft (1)

FBC=2Ay, 16ftx64ft (2)

Find the force in member BC using the Equation (1), and (2), and summarize the value in Table 2.

 x (ft) Apply 1 k load Support reaction (Ay) at A (k) Support reaction (Ey) at E (k) Influence line ordinate for the force in member BC (k/k) 0 A 1 0 0 16 B 34 14 6Ey=6(14)=1.5 32 C 12 12 2Ay=2(12)=1 48 D 14 34 2Ay=2(14)=0

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