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 21P
To determine

Draw the influence lines for the shear and bending moment at point C.

Draw the influence lines for the shear at internal hinge D.

Expert Solution & Answer
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Explanation of Solution

Calculation:

Equation of influence line ordinate of support B:

Apply a 1 k unit moving load at a distance of x from left end D.

Sketch the free body diagram of frame as shown in Figure 1.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  1

Refer Figure 1.

Consider the unit load at a variable position x to the left hinge D. (placed portion AC of the beam 0x<36ft).

Find the vertical support reaction (By) at B for portion AD.

Take Moment at hinge D from left end A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMDAD=0By(24)1(36x)=024By=36xBy=1.5x24

Consider the unit load at a variable position x to the right hinge D. (Place portion DG of the beam 36ftx72ft).

Find the vertical support reaction (By) at D for portion DG.

Take Moment at hinge D from left end A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMDAD=0By(24)=0By=0

Thus, the equations of the influence line ordinate for By are,

By=1.5x24, 0x<36ft        (1)

By=0, 36ftx72ft        (2)

Equation of influence line ordinate of support E:

Refer Figure 1.

Find the equation of influence line ordinate for the vertical reaction (Ey) using equilibrium equation.

Apply moment equilibrium at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMG=0By(60)+Ey(24)1(72x)=060By+24Ey72+x=024Ey=72x60ByEy=3x242.5By        (3)

Find the influence line ordinate of vertical reaction (Ey) for portion AD (0x<36ft).

Substitute 1.5x24 for By in Equation (3).

Ey=3x242.5(1.5x24)=3x243.75+2.5x24=1.5x240.75

Find the influence line ordinate of vertical reaction (Ey) for portion DG (36ftx72ft).

Substitute 0 for By in Equation (3).

Ey=3x242.5(0)=3x24

Thus, the equations of the influence line ordinate for Ey are,

Ey=1.5x240.75, 0x<36ft        (4)

Ey=3x24, 36ftx72ft        (5)

Equation of influence line ordinate of support G:

Refer Figure 1.

Find the equation of influence line ordinate for the vertical reaction (Gy) using force equilibrium equation.

Consider the vertical forces equilibrium condition, take the upward force as positive (+) and downward force as negative ().

ΣFy=0By+Ey+Gy=1Gy=1ByGy        (6)

Find the influence line ordinate of vertical reaction (Gy) for portion AD (0x<36ft).

Substitute 1.5x24 for By and 1.5x240.75 for Ey in Equation (6).

Gy=1(1.5x24)(1.5x240.75)=11.5+x241.5x24+0.75=0.250.5x24

Find the influence line ordinate of vertical reaction (Gy) for portion DG (36ftx72ft).

Substitute 0 for By and 3x24 for Ey in Equation (6).

Gy=1(0)(3x24)=13+x24=x242

Thus, the equations of the influence line ordinate for Gy are,

Gy=0.250.5x24, 0x<36ft        (7)

Gy=x242, 36ftx72ft        (8)

Influence line for shear at point C.

Find the equation of shear at C of portion AC (0x<24ft).

Apply 1 k load just left of C.

Consider the portion AC.

Sketch the free body diagram of the section AC as shown in Figure 2.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  2

Refer Figure 2.

Apply 1 k load just right of C.

Apply equilibrium equation of forces.

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

ΣFy=0

By1SC=0SC=By1

Substitute 1.5x24 for By.

SC=(1.5x24)1=0.5x24

Find the equation of shear at C of portion CG (24ft<x72ft).

Consider the portion AC.

Sketch the free body diagram of the section CG as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  3

Refer Figure 3.

Apply equilibrium equation of forces.

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

ΣFy=0

SC+By=0SC=By        (9)

Find the influence line ordinate of shear at C(SC) for portion CD (24ftx36ft).

Substitute 1.5x24 for By in Equation (9).

SC=1.5x24

Find the influence line ordinate of shear at C(SC) for portion DG (36ftx72ft).

Substitute 0 for By in Equation (9).

SC=0

Thus, the equations of the influence line ordinate for SC are,

SC=0.5x24, 0x24ft        (10)

SC=x241.5, 24ftx36ft        (11)

SC=0, 36ftx72ft        (12)

Find the influence line ordinate of SC using Equation (10), (11) and (12) and summarize the values in Table 1.

x (ft)PointsInfluence line ordinate of SC(k/k)
0A0.5
12B0
24C0.5
24C+0.5
36D0
48E0
60F0
72G0

Sketch the influence line diagram for the shear at point C as shown in Figure 4.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  4

Influence line for moment at point B:

Refer Figure 2.

Find the equation of bending moment at B of portion AC(0x<24ft).

Take moment at C.

Consider clockwise moment as positive and anticlockwise moment as negative.

MC=By(12)1(24x)

Substitute 1.5x24 for By.

MC=(1.5x24)(12)1(24x)=18x224+x=x26

Refer Figure 3.

Find the equation of bending moment at C of portion CG.

Take moment at C.

Consider clockwise moment as negative and anticlockwise moment as positive.

MC=By(12)        (13)

Find the influence line ordinate of moment at C(MC) for portion CD(24ftx36ft).

Substitute 1.5x24 for By.

MC=(1.5x24)(12)=180.5x

Find the influence line ordinate of moment at C (MC) for portion DG(36ftx72ft).

Substitute 0 for By in Equation (13).

MC=0(12)=0

Thus, the equations of the influence line ordinate for MB are,

MC=x26, 0x<24ft        (14)

MC=180.5x, 24ftx36ft        (15)

MC=0, 36ftx72ft        (16)

Find the influence line ordinate of MC using Equation (14), (15), and (16) and summarize the values in Table 2.

x (ft)PointsInfluence line ordinate of MC(k-ft/k)
0A6
12B0
24C6
36D0
48E0
60F0
72G0

Sketch the influence line diagram for the bending moment at point C as shown in Figure 5.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  5

Influence line for shear at hinge D:

Find the equation of shear at D of portion AC(0x<36m).

Sketch the free body diagram of the section AD as shown in Figure 6.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  6

Refer Figure 6.

Apply equilibrium equation of forces.

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

ΣFy=0

By1SD=0SD=By1

Substitute 1.5x24 for By.

SD=(1.5x24)1=0.5x24

Find the equation of shear at C of portion DG (36ft<x72ft).

Sketch the free body diagram of the section DG as shown in Figure 7.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  7

Refer Figure 7.

Apply equilibrium equation of forces.

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

ΣFy=0

BySD=0SD=By

Substitute 0 for By.

SD=0

Thus, the equations of the influence line ordinate for SD are,

SD=0.5x24, 0x<36ft        (17)

SD=0, 36ft<x72ft        (18)

Find the influence line ordinate of SD using Equation (17) and (18) and summarize the values in Table 3.

x (ft)PointsInfluence line ordinate of SD(k/k)
0A0.5
12B0
24C0.5
36D1
48E0
60F0
72G0

Sketch the influence line diagram for the shear at point D as shown in Figure 8.

Structural Analysis, 5th Edition, Chapter 8, Problem 21P , additional homework tip  8

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