Cross Sectional Area, Stress and Strain confusion

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SUMMARY

The discussion centers on calculating the cross-sectional area (CSA), stress, and strain for materials with a rectangular cross-section of 20mm by 1mm. The CSA is determined using the formula CSA = Length * Width, resulting in 0.00002 m². Stress is calculated as Force divided by Area, yielding values that appear unusually high compared to other materials. The original length for strain calculations should be clarified, as the problem presents two lengths: 20mm and 375mm, with the latter being the correct choice for accurate strain measurement.

PREREQUISITES
  • Understanding of stress and strain concepts in material science.
  • Familiarity with the formula for cross-sectional area (CSA = Length * Width).
  • Knowledge of unit conversions, specifically from millimeters to meters.
  • Basic principles of tensile testing and material properties.
NEXT STEPS
  • Research the significance of using original length in strain calculations.
  • Learn about stress-strain curve graphing techniques and best practices.
  • Study the modulus of elasticity and its calculation from stress-strain graphs.
  • Explore the differences between stress units (Pa vs. MPa) and their applications in material science.
USEFUL FOR

Students in engineering or material science, particularly those involved in mechanical testing and analysis of materials, will benefit from this discussion.

joe465
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Homework Statement




For each material calculate the cross sectional area in m². Each sample is of rectangular cross section 20mm by 1mm.

Produce Stress and Strain tables for the data.

TEST DATA FOR X
(N) EXTENSION (mm)
1000 0.48
2000 0.60
3000 0.70
4000 0.80

Homework Equations



CSA = Length * Width

mm to m = divide by 1000

Stress = Force / Area

Strain = Extension / original length

The Attempt at a Solution



First convert mm to m by dividing by 1000.

20mm * 1mm = 0.020m * 0.001m = 0.00002m^2

Area = 0.00002m^2

Now i presume because it states all 3 materials have the same cross section they have the same Cross sectional area?


Now for the stress and strain data.

There are more numbers but i have only provided a few for an example.

Stress = Load (N) / Area (m^2)

(N) (mm ext) Stress(N/m^2) Strain (m/m)
1000 0.48 50000000 0.024
2000 0.60 100000000 0.03
3000 0.70 150000000 0.035
4000 0.80 200000000 0.04

The values for stress seems strangely high when compared to another materials values where stress is 70, 100, 150, 200 and 220. (these are pre-given values). When calculating strain i used the length given by the first question 20mm as the original sample length, however underneath the tables for the test data giving the Load and extension data it gives an additional value ''original length of sample X: 375mm''. Which figure should i use, the original sample length as given in question 1 or the additional information?

This baffles me and without knowing which figure to use will affect the rest of the results and observations?

Thanks for any help given in advance,

Joe
 
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Cross sectional area can also be width and height (the terms length, width, and height are somewhat generic). For strain calculations, you must use the axial length of the sample (the direction in which the strain is produced by the application of the load).

For the stress values of the other materials (70, 100, 150, 200, and 220) no units are given in the OP. Could these units perhaps be in MPa rather than plain old Pa?
 
Thankyou for your reply,

The lower stress values indicated in the table (70, 100, 150, 200, and 220) are under the heading of Stress (SI Units).

The SI units of stress are N/m^2 or Pa so these must be correct, just seems unusual that there is a big difference between the other materials.

As for the Original length query, I'm still puzzled to why they would give another value. It says original length of sample, the only thing i can think of is that was the length before a sample was taken.

Cheers,

Joe
 
You don't say what these other materials are which are in the table. Although SI units for stress are Pascals, 1 pascal is a relatively small magnitude (1 N/m^2). Stresses on the order of 1 MPa (10^6 Pa) are encountered frequently. Since recording stresses in Pa entails writing a lot of zeroes, MPa is usually used to eliminate the zeroes.

It's not clear if the formula for cross sectional area was included in the original problem statement, but this same formula could also have been written CSA = width*height, which would eliminate the confusion with the length of the sample. BTW, a sample with a length of 20 mm is much too short to obtain a reasonable tensile test. A length of 375 mm allows the sample to be mounted into the tensile tester and, when pulled, it readily shows the reduction in the CSA due to the stress in the sample.
 
Thankyou, i just double checked and it does not state the formulae, therefore i will take CSA = width * height. This relieves the confusion for the length.

The materials are not given because the idea of the assignment is to draw stress - strain graphs, calculate the modulus of elasticity from the graphs and then use relevant research to determine what the materials are and provide evidence.

Still unsure about the units, since it specifically states SI Units it wouldn't make sense to not say if it used difference units. i.e MPa
 
I have now converted the values to MPa and they all seem to fit, thankyou for your help.

One final question is with reference to drawing, Stress - Strain curve graphs, should i include a 0 stress 0 strain data because my line is starting from thin air otherwise or do you suggest ammending the starting point for each axis on the graph so it coincides with the first piece of data?
 
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