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 5, Problem 5.7.4P

a.

To determine

The bending stress σA .

a.

Expert Solution
Check Mark

Answer to Problem 5.7.4P

The bending stressσAis 210 MPa.

Explanation of Solution

Given:

  P=12 kNL=1.25 m.bA=60 mmbB=80 mmhB=120 mmM0=10 kNm

Calculate the maximum bending stress.

  h(x)=hA(1+x3L)

  I(x)=πb(x)h( x)364S(x)=I(x)h( x)2

  S(x)=πb(x)h( x)232S(x)=[πbAhA2 ( 1+ x 3L )3]32

  M(x)=Px+M0

  σ(x)=M(x)S(x)=Px+M0[ π b A h A 2 ( 1+ x 3L ) 3 ]32=864(Px+ M 0π b A h A 2)L3( 3L+x)3

For maximum condition, we take:

  ddxσ(x)=0ddx[864( Px+ M 0 π b A h A 2 )L3 ( 3L+x )3]=0[864PπbAhA2L3 ( 3L+x )32592( Px+ M 0 π b A h A 2 )L3 ( 3L+x )4]=0[864L33PL+2Px+3M0πbAhA2 ( 3L+x )4]=0xmax=3(PL M 0)2Pxmax=0.625 in

Now, plotting the graph for σ(x) and M(x) , we get:

   Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.7.4P , additional homework tip  1Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.7.4P , additional homework tip  2

Now, the maximum bending stress is calculated.

  σmax=864(P x max+ M 0π b A h A 2)L3( 3L+ x max )3σmax=231 MPa

The largest bending stress is calculated as:

  σA=σ(0)σA=864(P( 0)+ M 0π b A h A 2)L3( 3L+( 0 ))3σA=210 MPa

Conclusion:

Therefore, the bending stress σA is 210 MPa.

b.

To determine

The bending stress σB .

b.

Expert Solution
Check Mark

Answer to Problem 5.7.4P

The bending stressσBis 221 MPa.

Explanation of Solution

The largest bending stress is calculated as:

  σB=σ(L)σB=864(P( 0)+ M 0π b A h A 2)L3( 3L+( 0 ))3σB=221 MPa

Conclusion:

Therefore, the bending stress σB is 221 MPa.

c.

To determine

The distance of maximum bending stress xmax .

c.

Expert Solution
Check Mark

Answer to Problem 5.7.4P

The distance of maximum bending stressxmaxis 0.625 in.

Explanation of Solution

The distance of maximum bending stress xmax .

  xmax=3(PL M 0)2Pxmax=0.625 in

Conclusion:

Therefore, the distance of maximum bending stress xmax is 0.625 in.

d.

To determine

The magnitude of maximum bending stress σmax .

d.

Expert Solution
Check Mark

Answer to Problem 5.7.4P

The magnitude of maximum bending stressσmaxis 231 MPa.

Explanation of Solution

The magnitude of maximum bending stress σmax .

  σmax=864(P x max+ M 0π b A h A 2)L3( 3L+ x max )3σmax=231 MPa

Conclusion:

Therefore, the magnitude of maximum bending stress σmax is 231 MPa.

e.

To determine

The magnitude of maximum bending stress.

e.

Expert Solution
Check Mark

Answer to Problem 5.7.4P

The magnitude of maximum bending stress is 214 MPa.

Explanation of Solution

Given:

  P=12 kNL=1.25 m.bA=60 mmbB=80 mmhA=90 mmhB=120 mmM0=10 kNm

Calculate the maximum bending stress.

  h(x)=hA(1+x3L)

  b(x)=bA(1+x3L)

  I(x)=πb(x)h( x)364S(x)=I(x)h( x)2

  S(x)=πb(x)h( x)232S(x)=[πbAhA2 ( 1+ x 3L )3]32

  M(x)=Px+M010Px26L

  σ(x)=M(x)S(x)=Px+M010P x 26L[ π b A h A 2 ( 1+ x 3L ) 3 ]32=288(3PxL3M0L+5Px2)L2πbAhA2( 3L+x)3

For maximum condition, we take:

  ddxσ(x)=0ddx[288(3PxL3M0L+5Px2)L2πbAhA2 ( 3L+x )3]=0[(864PL2880Px)L2πbAhA2 ( 3L+x )33(864PxL+864M0L1440Px2)L2πbAhA2 ( 3L+x )4]=0[288L29PL236PxL+5Px29M0LπbAhA2 ( 3L+x )4]=09PL236PxL+5Px29M0L=0xmax=0.105 m

Now, plotting the graph for σ(x) and M(x) , we get:

   Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.7.4P , additional homework tip  3Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.7.4P , additional homework tip  4

Now, the maximum bending stress is calculated.

  σmax=288(3PxL3M0L+5Px2)L2πbAhA2( 3L+x)3σmax=214 MPa

The largest bending stress is calculated as:

  σA=σ(0)σA=288(3PxL3M0L+5Px2)L2πbAhA2( 3L+x)3σA=210 MPa

The largest bending stress is calculated as:

  σB=σ(L)σB=288(3PxL3M0L+5Px2)L2πbAhA2( 3L+x)3σB=0 MPa

Conclusion:

Therefore, the magnitude of maximum bending stress is 214 MPa.

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Chapter 5 Solutions

Mechanics of Materials (MindTap Course List)

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