Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 8, Problem 54P
To determine

Draw the influence lines for the force in member AB, BG, DF, and FG.

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Explanation of Solution

Calculation:

Find the support reactions.

Apply 1 k moving load from E to G in the top chord member.

Draw the free body diagram of the member as in Figure 1.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  1

Find the reaction at A and B when 1 k load placed from E to G. (0x36ft).

Apply moment equilibrium at A.

ΣMA=01(6x)=By(24)24By=(6x)By=x2414

Apply force equilibrium equation along vertical.

Consider the upward force as positive (+) and downward force as negative ().

ΣFy=0Ay+By1=0Ay+x2414=1Ay=1x24+14Ay=54x24

Influence line for the force in member AB.

The expressions for the member force FAB can be determined by passing an imaginary section a-a through joint F and then apply a vertical force equilibrium.

Draw the free body diagram of member with section aa as shown in Figure 2.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  2

Refer Figure 2.

Find the equation of member force AB.

Apply a 1 k load at just left of F (0x18ft).

Consider the right hand portion to section a-a.

Apply moment equilibrium equation at F.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMF=0By(18)+FAB(16)=016FAB=12ByFAB=0.75By

Substitute x2414 for By.

FAB=0.75(x2414)=x32316

Apply a 1 k load at just right of F (18ftx36ft).

Consider the left hand portion to section a-a.

Apply moment equilibrium equation at F.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMF=0Ay(12)FAB(16)=016FAB=12AyFAB=0.75Ay

Substitute 54x24 for Ay.

FAB=0.75(54x24)=1516x32

Thus, the equation of force in the member AB,

FAB=x32316, 0x18ft (1)

FAB=1516x32, 18ftx36ft (2)

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

x (ft)Apply 1 k loadForce in member AB (k)Influence line ordinate for the force in member AB (k/k)
0E032316=0.18750.1875
18F1832316=0.3750.375
36G15163632=0.18750.1875

Sketch the influence line diagram for ordinate for the force in member AB using Table 2 as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  3

Influence line for the force in member BG.

The expressions for the member force FBG can be determined by passing an imaginary section a-a through the members BG, BD, and AB and then apply a moment equilibrium at F.

Draw the free body diagram of section a-a as shown in Figure 4.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  4

Refer Figure 4.

Find the force in member BG.

Apply 1 k load just left of F (0x18ft).

Consider the section EF.

The member force of EF not affected when 1 k load applied from E to F. Therefore, the influence line ordinate of member force BG is 0 k/k from E to F.

Apply a 1 k load just the right of F (18ftx36ft).

Apply moment equilibrium at F.

Consider the section right of line a-a.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMF=0By(12)+FAB(16)1662+162FBG(12)662+162FBG(16)=012By+16FAB11.24FBG5.62FBG=016.86FBG=16FAB12ByFBG=0.95FAB0.712By

Substitute 1516x32 for FAB and x2414 for By.

FBG=0.95(1516x32)0.712(x2414)=0.890.0297x0.0297x+0.178=1.0680.0594x

Thus, the equation of force in the member BG,

FBG=0, 0x18ft (3)

FBG=1.0680.0594x, 18ftx36ft (4)

Find the force in member BG using the Equation (3) and (4) and then summarize the value in Table 2.

x (ft)Apply 1 k loadForce in member BG (k)Influence line ordinate for the force in member BG (k/k)
0E00
18F00
36G1.0680.0594(36)=1.07‑1.07

Sketch the influence line diagram for ordinate for the force in member BG using Table 2 as shown in Figure 5.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  5

Influence line for the force in member DF.

The expressions for the member force FDF can be determined by passing an imaginary section bb through the members FG, DF, and AB and then apply moment equilibrium at F

Draw the free body diagram of section a-a as shown in Figure 6.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  6

Refer Figure 6.

Find the force in member DF.

Apply 1 k load just left of F (0x18ft).

Consider the section right of line bb.

Apply moment equilibrium at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMG=0FAB(16)+By(6)+45FDF(12)+35FDF(8)=016FAB+6By+9.6FDF+4.8FDF=014.4FDF=16FAB6By

Substitute x32316 for FAB and x2414 for By.

14.4FDF=16(x32316)6(x2414)14.4FDF=0.5x+30.25x+1.514.4FDF=0.75x+4.5FDF=0.75x+4.514.4

Apply 1 k load just right of F (18ftx36ft).

Consider the section left of line bb.

Apply moment equilibrium at E.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣME=0FAB(16)Ay(6)+45FDF(18)+35FDF(0)=016FAB6Ay+14.4FDF=014.4FDF=16FAB+6Ay

Substitute 1516x32 for FAB and 54x24 for Ay.

14.4FDF=16(1516x32)+6(54x24)14.4FDF=150.5x+7.50.25x14.4FDF=0.75x+22.5FDF=0.75x+22.514.4

Thus, the equation of force in the member DF,

FDF=0.75x+4.514.4, 0x18ft (5)

FDF=0.75x+22.514.4, 18ftx36ft (6)

Find the force in member DF using the Equation (5) and (6) and then summarize the value in Table 3.

x (ft)Apply 1 k loadForce in member DF (k)Influence line ordinate for the force in member DF (k/k)
0E0.31250.3125
18F‑0.625‑0.625
36G‑0.3125‑0.3125

Sketch the influence line diagram for ordinate for the force in member DF using Table 3 as shown in Figure 7.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  7

Influence line for the force in member FG.

Refer Figure 6.

Find the force in member FG.

Apply 1 k load just left of F (0x18ft).

Consider the section right of line bb.

Apply moment equilibrium at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMB=0FFG(16)=0FFG=0

Apply 1 k load just right of F (18ftx36ft).

Consider the section left of line bb.

Apply moment equilibrium at A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMA=0FFG(16)+45FDF(12)+35FDF(16)=016FFG+9.6FDF+9.6FDF=016FFG=19.2FDFFFG=1.2FDF

Substitute 0.75x+22.514.4 for FDF.

FFG=1.2(0.75x+22.514.4)FFG=0.9x2714.4

Thus, the equation of force in the member FG,

FGF=0, 0x18ft (7)

FFG=0.9x2714.4, 18ftx36ft (8)

Find the force in member FG using the Equation (7) and (8) and then summarize the value in Table 4.

x (ft)Apply 1 k loadForce in member FG (k)Influence line ordinate for the force in member FG (k/k)
0E0.31250.3125
18F‑0.625‑0.625
36G‑0.3125‑0.3125

Sketch the influence line diagram for ordinate for the force in member FG using Table 4 as shown in Figure 8.

Structural Analysis, 5th Edition, Chapter 8, Problem 54P , additional homework tip  8

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