Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 6, Problem 6.2P

(a)

To determine

The change in radius of specimen.

(a)

Expert Solution
Check Mark

Answer to Problem 6.2P

The change in radius of specimen is not significant and the final radius of specimen is 0.005 m .

Explanation of Solution

Given:

Poisson’s ratio is 0.29 .

Radius of specimen is 0.005 m .

Elongation in steel rod is 0.616×102 m .

Write the expression for tensile stress in rod.

  σ=FA0

Here, σ is the tensile stress in rod, F is the applied force and A0 is the original cross-sectional area of specimen.

Substitute πr02 for A0 in above expression.

  σ=Fπr02 .......... (1)

Here, r0 is the original radius of specimen.

Write the expression for axial strain in steel rod.

  εx=Δll0 .......... (2)

Here, εx is the axial strain, Δl is the elongation in rod and l0 is the original length of rod.

Write the expression for axial strain in y direction.

  εy=vεx .......... (3)

Here, εy is the axial strain in y direction and v is Poisson’s ratio.

Write the expression for final radius of specimen.

  r=r0+εyr0 .......... (4)

Here, r is the final radius of steel rod.

Substitute 0.005 m for r0 and 0.250×105N for F in equation (1).

  σ=0.250× 105π ( 0.005 )2=3.183×108N/m2

Substitute 0.616×102m for Δl and 4.000m for l0 in equation (2).

  εx=0.616× 10 24.000=1.54×103

Substitute 0.29 for v and 1.54×103 for εx in equation (3).

  εy=(0.29)(1.54× 10 3)=0.045×102

Substitute 0.045×102 for εy and 0.005m for r0 in equation (4).

  r=0.005+(0.045× 10 2)(0.005)=0.0052.25×106=0.0050.000002250.005

Thus, the change in radius of specimen is not significant and the final radius of specimen is 0.005 m .

(b)

To determine

The difference between true stress and engineering stress.

(b)

Expert Solution
Check Mark

Answer to Problem 6.2P

There is no difference between true stress and engineering stress for this specimen.

Explanation of Solution

As the change in radius of specimen is not significant, the actual and original radius of specimen can be considered as same. The true stress is evaluated on the basis of actual cross-sectional area of specimen and engineering stress is calculated on original cross-sectional area of specimen. As the value of area is constant for same final radius there will be no difference in Engineering stress and true stress.

The calculate engineering stress or true stress of specimen is 3.183×108N/m2 .

Thus, there is no difference between true stress and engineering stress for this specimen.

(c)

To determine

The value of true strain and compare it with engineering strain.

(c)

Expert Solution
Check Mark

Answer to Problem 6.2P

The true strain for the specimen is 0.154×102 .

Explanation of Solution

Write the expression for true strain of specimen.

  εt=ln(1+ε) .......... (5)

Here, εt is true strain.

Substitute 1.54×103 for ε in equation (5).

  εt=ln[1+(1.54× 10 3)]=0.154×102

The engineering strain and true strain will remain same for the specimen.

Thus, the true strain for the specimen is 0.154×102 .

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