
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Only part e, please.
![The stress-strain behavior of a steel alloy was determine using a cylindrical test specimen having
a diameter of 15.000 mm. The stress-strain plot for this steel alloy is shown in the figure below.
Using this figure, answer the following questions:
(a) Determine the yield stress at a strain offset of 0.002 Oy = 350 MPa]
(b) What was the fracture stress? Ox=400MPa
(c) When a load of 86,000 N was applied to the test specimen did it experience elastic or
plastic deformation? Explain your answer.
(d) If the original gauge length of the test specimen was 250.00 mm, what was the increase in
length when the load of 86,000 N was applied.
Note: For part (e) below, do not used the strain value from the plot.
(e) Given that for the steel, E = 210 GPa and y = 0.3) calculate:
(i) The increase in length on a gauge length of 250.00 mm when a load of 35,000 N is
applied.
(ii) The longitudinal strain when the 35,000 N load was applied.
(iii) The new diameter of the specimen with the 35,000 N load applied
Stress (MPa)
600
500
400
-500-
✗
300
400
200
Stress (MPa)
300
200
100
100
0.000
0.002
0.004
0.006
Strain
0.00
0.04
0.08
0.12
0.16
0.20
Strain](https://content.bartleby.com/qna-images/question/d30ea730-ba6b-4d49-90df-80ffc6e77ea4/d775b602-d8b1-45a4-96cc-cf9a90e7cc9a/6r5b8y_thumbnail.jpeg)
Transcribed Image Text:The stress-strain behavior of a steel alloy was determine using a cylindrical test specimen having
a diameter of 15.000 mm. The stress-strain plot for this steel alloy is shown in the figure below.
Using this figure, answer the following questions:
(a) Determine the yield stress at a strain offset of 0.002 Oy = 350 MPa]
(b) What was the fracture stress? Ox=400MPa
(c) When a load of 86,000 N was applied to the test specimen did it experience elastic or
plastic deformation? Explain your answer.
(d) If the original gauge length of the test specimen was 250.00 mm, what was the increase in
length when the load of 86,000 N was applied.
Note: For part (e) below, do not used the strain value from the plot.
(e) Given that for the steel, E = 210 GPa and y = 0.3) calculate:
(i) The increase in length on a gauge length of 250.00 mm when a load of 35,000 N is
applied.
(ii) The longitudinal strain when the 35,000 N load was applied.
(iii) The new diameter of the specimen with the 35,000 N load applied
Stress (MPa)
600
500
400
-500-
✗
300
400
200
Stress (MPa)
300
200
100
100
0.000
0.002
0.004
0.006
Strain
0.00
0.04
0.08
0.12
0.16
0.20
Strain
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