Vector Mechanics for Engineers: Statics
Vector Mechanics for Engineers: Statics
12th Edition
ISBN: 9781260501735
Author: BEER
Publisher: MCG
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Chapter 7.2, Problem 7.61P

Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M|max. (See the hint for Prob. 7.55.)

Chapter 7.2, Problem 7.61P, Knowing that P = Q = 150 lb, determine (a) the distance a for which the maximum absolute value of

Fig. P7.60

7.61 Solve Prob. 7.60 assuming that P = 300 lb and Q = 150 lb.

(a)

Expert Solution
Check Mark
To determine

The distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest.

Answer to Problem 7.61P

The distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest is 46.8in.

Explanation of Solution

Refer Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 7.2, Problem 7.61P , additional homework tip  1

Write an expression to calculate the counter clockwise moment at point A.

ΣMA=0                                       

Here, ΣMA is the moment about point A.

Write an expression to calculate the counter clockwise moment at point A.

ΣMC=0                                       

Here, ΣMC is the moment about point A.

Write an expression to calculate the counter clockwise moment at point A.

ΣMD=0                                        

Here, ΣMD is the moment about point A.

Conclusion:

Refer Figure 1:

Calculate the moment about point A.

ΣMA=Da(P)(AC)(Q)(AB)=0

Here, P is the weight at point A and a is the distance between points A and D, Q is the weight at Q.

Rearrange the equation to calculate the D.

D=(P)(AC)+(Q)(AB)a

Substitute 300lb for P, 30in for AC, 150lb for Q and 60in for AB to find D.

D=(300lb)(30in)+(150lb)(60in)a=18,000lbina

Refer Figure 2.

Vector Mechanics for Engineers: Statics, Chapter 7.2, Problem 7.61P , additional homework tip  2

Calculate the moment about point C.

ΣMC=MC(P)(AC)+(D)(aAC)=0

Rearrange the equation to calculate the MC.

MC=(P)(AC)+(D)(aAC)=0

Substitute 150lb for P, 30in for AC, 18,000lbina for D and 60in for AB to find D.

MC=(150lb)(30in)+(18,000lbina)(a30in)=13,500lbin(140a) (I)

Refer Figure 2.

Calculate the moment about point D.

ΣMD=MD(DB)(ABa)=0

Rearrange the equation to calculate the MD.

MD=(DB)(ABa)

Substitute 150lb for DB and 150lb for Q to find MD.

MD=(150lb)(60ina) (II)

The magnitude of the maximum moment is equal to the magnitude of the minimum moment.

Mmax=|Mmin|MC=|MD|

Substitute (I) and (II) in above equation to find a.

(13,500lbin)(140a)=(150lb)(60ina)

Rearrange the equation to find a.

a=30in+15,300in22=46.8in

Thus, the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam AB is the smallest is 46.8in.

(b)

Expert Solution
Check Mark
To determine

The value of Mmax

Answer to Problem 7.61P

The value of Mmax is 1,973lbin.

Explanation of Solution

Refer Figure 4.

Vector Mechanics for Engineers: Statics, Chapter 7.2, Problem 7.61P , additional homework tip  3

The magnitude of the maximum moment is equal to the magnitude of the minimum moment.

Mmax=|Mmin|=|MD|

Conclusion:

Substitute 46.8in for a in equation (II) to calculate the value of Mmax .

Mmax=(150lb)(60in(46.8in))=1,973lbin

Thus, the value of Mmax is 1,206lbin.

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Chapter 7 Solutions

Vector Mechanics for Engineers: Statics

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