Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 4, Problem 4.5.27P

The simple beam ACE shown in the figure is subjected to a triangular load of maximum intensity q0= 200 lb/ft at a = 8 ft and a concentrated moment M = 400 Ib-ft at A.

  1. Draw the shear-force and bending-moment diagrams for this beam,
  2. Find the value of distanced that results in the maximum moment occurring at L/2. Draw the shear-force and bending-moment diagrams for this case.
  3. Find the value of distance a for which Mmaxis the largest possible value.

  Chapter 4, Problem 4.5.27P, The simple beam ACE shown in the figure is subjected to a triangular load of maximum intensity q0=

(a).

Expert Solution
Check Mark
To determine

To draw: Shear force and bending moment diagrams for simply supported beam.

Answer to Problem 4.5.27P

The SFD and BMD are shown in explanation part.

Explanation of Solution

Given Information:

Max load q0=200lb/ft

Distance a=8 ft

Moment M=400 lb-ft at A

Length L=10 ft

Concept Used:

Shear forces and bending moments at various points shall be calculated.

Calculation:

Draw free body diagram

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  1

From equilibrium

  RA+RB=12(200×8+200×2)=1000 lb  ............(1)

Also,

MB = 0

RA×10+400[12×200×8×(83+108)]12×200×2×2×23=0RA×10+4003733.3333266.6667=0RA=360 lb

From equation (1)

  RA+RB=1000360+RB=1000RB=640

Shear Force calculation

  SFB=RB=640SFCBx=0.5x×100xRB=50x2640SFC=[50x2640]x=2=50×22640=440SFA=RA=360

SFD

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  2

Bending Moment calculation

  BMB=0BM between C and B when x from rightBMCBx=RB×x12×x×100x×x3=640x503x3BMC=[640x503x3]x=2=640×2503×23=1146.67BM between A and C when x from leftBMACx=RA×x+40012×x×25x×(x3)=360x+4004.167x3BMA=400

BMD

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  3

Conclusion:

The SFD and BMD are shown in explanation part.

(b).

Expert Solution
Check Mark
To determine

To find: The value of a for maximum bending moment at L/2

Answer to Problem 4.5.27P

The distance is a=5 ft

Explanation of Solution

Given Information:

Max load q0=200lb/ft

Distance =a

Moment M=400 lb-ft at A

Length L=10 ft

Concept Used:

Shear forces and bending moments at various points shall be calculated.

Calculation:

The free body diagram is as follows:

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  4

From equilibrium

  RA+RB=12[200×a+200×(10a)]=1000 lb  ............(1)

Also,

MB = 0

RA×10+400[12×200×a×(100.67a)]12×200×(10a)×(10a)×23=0RA×10+4001000a+67a26666.67+1333.33a66.67a2=0RA=33.333a+626.667 lb

From equation (1)

  RA+RB=100033.33a+626.67+RB=1000RB=33.33a+373.33 lb

Bending moment

BML/2=RA×0.5L+40012×0.5L×200a×0.5L×(0.5L3)BML/2=(33.33a+626.67)×5+40012×5×200a×5×(53)BML/2=166.65a+3133.354166.67/aFor max momentdda(BML/2)=0dda(166.65a+3133.354166.67/a)=0166.65+4166.67/a2=0166.65a2+4166.67=0

On solving above equation, we get:

a=±5a=5 ft

Shear Force calculation:

  SFB=RB=(33.33a+373.33)=540 lbSFCBx=0.5x×2005xRB=20x2540SFC=[20x2540]x=5=20×52540=40 lbSFA=RA=33.333a+626.66=460

SFD

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  5

Bending Moment calculation:

  BMB=0BM between C and B when x from rightBMCBx=RB×x12×x×40x×x3=540x203x3BMC=[540x203x3]x=5=540×5203×53=1866.67BMA=400

BMD

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  6

Conclusion:

The distance is a=5 ft .

(c).

Expert Solution
Check Mark
To determine

To find: The value of a for maximum bending moment.

Answer to Problem 4.5.27P

The distance is a=4.7005 ft .

Explanation of Solution

Given Information:

Max load q0=200lb/ft

Distance =a

Moment M=400 lb-ft at A

Length L=10 ft

Concept Used:

Bending moment shall be calculated.

Calculation:

The free body diagram:

Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.27P , additional homework tip  7

From part b we get,

RA=33.333a+626.667 lb

  RB=33.33a+373.33 lb

Bending moment

  BM between C and B when x from right:BMCBx=RB×x12×x×20010ax×x3=(33.33a+373.33)x33.3310ax3But x=10-aBMCBx=[33.33(10x)+373.33]x33.33x2BMCBx=[333.3x33.33x2+373.33x33.33x2]=66.66x2+706.63xfor max bending moment :ddx(BMCBx)=0ddx(66.66x2+706.63x)=0133.33x706.63=0x=5.2995a=10xa=105.2995=4.7005

And maximum bending moment is Mmax=1872.67

Conclusion:

The distance is a=4.7005 ft .

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Chapter 4 Solutions

Mechanics of Materials (MindTap Course List)

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