How to Plot a Stress-Strain Graph for Steel Sheet?

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To plot a stress-strain graph for a steel sheet, the stress can be calculated using the formula F/A, where F is the load and A is the cross-sectional area. For a steel sheet with a width of 12.5mm and a thickness of 3mm, the correct cross-sectional area is 12.5mm x 3mm. The discussion clarifies that this formula is applicable to sheets as long as the thickness and width are small compared to the gauge length. The maximum proposed stress for the new design is 670MPa, which is relevant for evaluating the tensile strength of the material. Understanding these parameters is essential for accurately plotting the stress-strain graph.
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Homework Statement


The maximum proposed stress of their new design is 670MPa. You have measured the tensile strength of pieces of their steel, using an appropriate "standard" to guide you. (What does this even mean?) The load vs displacement graph is given. The steel sheet is 3mm thick. The width of the sample is 12.5mm and the gauge length sample is 50mm. Plot the stress strain graph.

Homework Equations


mm load, N
7.52E-03 1.32E+03
1.69E-02 3.30E+03
2.61E-02 4.85E+03
3.68E-02 6.53E+03
5.02E-02 8.48E+03
6.34E-02 1.05E+04
7.74E-02 1.25E+04
9.42E-02 1.44E+04
1.04E-01 1.63E+04
1.13E-01 1.81E+04
1.27E-01 1.98E+04
1.44E-01 2.14E+04
1.80E-01 2.27E+04
4.48E-01 2.34E+04
8.10E-01 2.40E+04
1.13E+00 2.47E+04
1.47E+00 2.52E+04
1.80E+00 2.57E+04
2.16E+00 2.62E+04
2.50E+00 2.65E+04
2.81E+00 2.69E+04
3.09E+00 2.75E+04
3.42E+00 2.80E+04
3.76E+00 2.83E+04
4.18E+00 2.80E+04
4.63E+00 2.74E+04
5.12E+00 2.72E+04
5.47E+00 2.67E+04
5.71E+00 2.59E+04
5.71E+00 2.49E+04
5.71E+00 2.37E+04
5.70E+00 2.21E+04
5.30E+00 2.03E+04

The Attempt at a Solution



How do I find the stress for each load? I know that it is F/A, but I thought this only applied to rods and not a metal sheet? Or am i allowed to use 12.5x5 as the cross sectional area?
 
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ipocoyo said:

Homework Statement


The maximum proposed stress of their new design is 670MPa. You have measured the tensile strength of pieces of their steel, using an appropriate "standard" to guide you. (What does this even mean?) The load vs displacement graph is given. The steel sheet is 3mm thick. The width of the sample is 12.5mm and the gauge length sample is 50mm. Plot the stress strain graph.

Homework Equations


mm load, N
7.52E-03 1.32E+03
1.69E-02 3.30E+03
2.61E-02 4.85E+03
3.68E-02 6.53E+03
5.02E-02 8.48E+03
6.34E-02 1.05E+04
7.74E-02 1.25E+04
9.42E-02 1.44E+04
1.04E-01 1.63E+04
1.13E-01 1.81E+04
1.27E-01 1.98E+04
1.44E-01 2.14E+04
1.80E-01 2.27E+04
4.48E-01 2.34E+04
8.10E-01 2.40E+04
1.13E+00 2.47E+04
1.47E+00 2.52E+04
1.80E+00 2.57E+04
2.16E+00 2.62E+04
2.50E+00 2.65E+04
2.81E+00 2.69E+04
3.09E+00 2.75E+04
3.42E+00 2.80E+04
3.76E+00 2.83E+04
4.18E+00 2.80E+04
4.63E+00 2.74E+04
5.12E+00 2.72E+04
5.47E+00 2.67E+04
5.71E+00 2.59E+04
5.71E+00 2.49E+04
5.71E+00 2.37E+04
5.70E+00 2.21E+04
5.30E+00 2.03E+04

The Attempt at a Solution



How do I find the stress for each load? I know that it is F/A, but I thought this only applied to rods and not a metal sheet? Or am i allowed to use 12.5x5 as the cross sectional area?
Yes, you are, although it should be 12.5x3. What made you think that it only applies to a rod?

It certainly applies to a sheet if the thickness and width are small compared to the gauge length.

Chet
 
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Chestermiller said:
Yes, you are, although it should be 12.5x3. What made you think that it only applies to a rod?

It certainly applies to a sheet if the thickness and width are small compared to the gauge length.

Chet
Thank you for clearing up the confusion!
 

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