Foundations of Materials Science and Engineering
Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 6.13, Problem 51AAP

A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter decreases to 0.210 cm, determine (a) the engineering stress and strain at this load and (b) the true stress and strain at this load.

a)

Expert Solution
Check Mark
To determine

Engineering stress and strain has to be determined.

Answer to Problem 51AAP

The magnitude of engineering stress is 1019 MPa.

The magnitude of engineering strain is 0.417.

Explanation of Solution

Write the expression for initial area of the bar:

A0=π4d02 (I)

Here, initial diameter of the bar is d0.

Write the expression for initial area of the bar:

Af=π4df2 (II)

Here, final diameter of the bar is df.

The volume of the bar before and after deformation is same.

Vi=Vf

A0l0=Aflf

A0Af=lfl0 (III)

Write the expression for the engineering stress as given as follows.

σ=FA0 (IV)

Here, applied force is F and area of the bar is A0.

Write the expression for the engineering strain as given as follows.

ε=lfl0l0

ε=lfl0l0l0 (V)

Substitute equation (III) in equation (V).

ε=A0Af1 (VI)

Conclusion:

Substitute 0.25 cm for d0 in equation (I).

A0=π4(0.25 cm)2=0.04909 cm2×(1 m100 cm)2=4.909×106 m2

Substitute 0.21 cm for df in equation (II).

Af=π4(0.21cm)2=0.03464 cm2×(1m100cm)2=3.464×106 m2

Substitute 5000N for F and 4.909×106 m2 for A0 in equation (IV).

σ=5000N4.909×106 m2=1019×106 Pa×1MPa106 Pa=1019MPa

Thus, the magnitude of engineering stress is 1019 MPa.

Substitute 4.909×106 m2 for A0 and 3.464×106 m2 for Af in equation (VI).

ε=4.909×106 m23.464×106 m21=1.4171=0.417

Thus, the magnitude of engineering strain is 0.417.

b)

Expert Solution
Check Mark
To determine

true stress and strain has to be determined.

Answer to Problem 51AAP

The magnitude of true stress is 1443 MPa.

The magnitude of true strain is 0.349.

Explanation of Solution

Write the expression for the true stress as given as follows:

σT=FAf (VII)

Write the expression for the true strain as given as follows:

ε=ln(lfl0)

ε=ln(A0Af) (VIII)

Conclusion:

Substitute 5000 N for F and 3.464×106 m2 for Af in equation (VI).

σT=5000 N3.464×106 m2=1443×106 Pa×1 MPa106 Pa=1443 MPa

Thus, the magnitude of true stress is 1443 MPa.

Substitute 4.909×106 m2 for A0 and 3.464×106 m2 for Af in equation (VII).

εT=ln(4.909×106 m23.464×106 m2)=ln(1.418)=0.349

Thus, the magnitude of true strain is 0.349.

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Foundations of Materials Science and Engineering

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