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Structural Analysis

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

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 9, Problem 15P
Textbook Problem
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For the beam of Problem 8.23, determine the maximum positive bending moment at point D due to the wheel loads of the moving HS15-44 truck shown in Fig. P9.15.

FIG. P9.15, P9.21

Chapter 9, Problem 15P, For the beam of Problem 8.23, determine the maximum positive bending moment at point D due to the , example  1

FIG. P8.23, P8.25

Chapter 9, Problem 15P, For the beam of Problem 8.23, determine the maximum positive bending moment at point D due to the , example  2

To determine

Find the maximum positive bending moment at point D.

Explanation of Solution

Calculation:

Apply 1 k load at D.

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

Find the vertical reaction (Ay) at A.

Consider moment at B from A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMB=0Ay×20=0Ay=0

Find the vertical reaction (Gy) at G.

Consider moment at F from G.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMF=0Gy×20=0Gy=0

Find the vertical reactions Cy and Ey.

Consider moment at B from G.

Consider clockwise moment as negative and anticlockwise moment as positive.

ΣMB=0Gy(110)+Ey(80)1(50)+Cy(20)=0

Substitute 0 for Gy.

(Gy)(110)+Ey(80)1(50)+Cy(20)=080Ey+20Cy=50        (1)

Consider moment at F from A.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMF=0Ay(115)+Cy(75)1(45)+Ey(15)=0

Substitute 0 for Ay.

(0)(115)+Cy(75)1(45)+Ey(15)=075Cy+15Ey=45        (2)

Solve Equations (1) and (2),

Cy=0.5kEy=0.5k

Insert a hinge at D to obtain the released structure of the beam. Next apply a small rotation θ in positive direction of MD at D in the released structure. The rotation θ rotate the portion CD in counterclockwise and the portion DE in clockwise to obtain the deflected shape.

Sketch the deflected shape of the released beam when insert a hinge at D as shown in Figure 2.

The general shape of the influence line for bending moment (MD) at D is the similar of deflected shape.

Sketch the general shape of the influence line for bending moment (MD) at D as shown in Figure 3.

Find the numerical value of influence line ordinate for MD at D.

Refer Figure 3.

Find the moment at D.

Consider clockwise moment as positive and anticlockwise moment as negative.

MD=Ay(70)+Cy(30)

Substitute 0 for Ay and 0.5 k for Cy.

MD=(0)(70)+(0.5)(30)=15k-ft

Thus, the value of influence line ordinate at D is 15k-ft/k.

The remaining values of influence line ordinate at different points on the beam determined by geometry (similar triangle) of the influence line and summarize the values as in Table 1.

x (ft)PointsInfluence line ordinate of Cy (k/k)
0A0
20B10
40C0
70D15
100E0
115F7.5
130G0

Sketch the influence line diagram for moment at point D as shown in Figure 4.

Refer Figure 4.

The slope AB of influence diagram is equal to slope BC of influence diagram.

Find the slope (ϕAC) of influence line diagram of moment at D in the portion AC.

ϕAC=10LAB

Here, LAB is the length of the beam from A to B.

Substitute 20 ft for LAB.

ϕAC=1020=12

The slope CD of influence diagram is equal to slope DE of influence diagram.

Find the slope (ϕCE) of influence line diagram of moment at D in the portion CE.

ϕCE=15LCD

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

Substitute 30 ft for LCD.

ϕCD=1530=12

The slope EF of influence diagram is equal to slope FG of influence diagram

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