Tensile Tests Load-Extension into Stress-Strain

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SUMMARY

The discussion centers on calculating strain from tensile test data, specifically when using Load-Extension tables for materials like brass, copper, nylon, and polystyrene. The key formula for strain is defined as ε = (l_i - l_o) / l_o, where l_o represents the initial gauge length. It is established that the gauge length must be marked on the sample before testing to accurately measure elongation and calculate strain, as the overall length of the sample is not necessary for this calculation. The conversation emphasizes that strain is a dimensionless variable and is solely dependent on the ratio of elongation to the initial gauge length.

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  • Understanding of tensile testing principles
  • Familiarity with stress and strain calculations
  • Knowledge of material properties (e.g., plasticity, elasticity)
  • Experience with measuring and marking gauge lengths on test samples
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ajwinkworth
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Homework Statement



Via tensile testing of various materials I have been supplied with Load-Extension tables. I must calculate and show the Strain-Stress Graph. I know from this it is fairly easy to calculate the stress by force/area. What I am having trouble with is strain.

I have a gauge length of 50mm but the actual piece of material was longer. When putting the broken sample back together I have a measurement of the elongated sample (plasticity deformed). The problem I seem to be having is obtaining the strain for between the gauges from extension of the overall sample. I do not know and was never instructed to record the overall length of the sample. The Samples were dog bone shape with materials of brass, copper, nylon and polystyrene.

Is there some way to convert from the extensions of the overall piece to strain between the gauge? Seems like some information is missing for me to be able to do this? Or is the gauge length not important for this? if i just go by the extensions then the original length is l_o = 0 so i would get a divide by zero out of the equation for strain.


Homework Equations


Strain:
\epsilon = \frac{l_i-l_o}{l_o}

l_o = ? Can that be gauge length? It just doesn't seem right?

Stress:
\sigma = \frac{F}{A_o}

Strain involving gauge lengths:
missing?

The Attempt at a Solution



After a pointer to a formula or theory rather then a solution. (Or a D.N.E)
 
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The strain is based on the gauge length alone, because, in testing errorss creep in principally due to the jaws of the machine slipping. So you can't base the strain on the distance apart of the jaws. I imagine you are looking at strain just in the elastic region? Why do you say that ell nought cannot be the gauge length?
 
I know I'm a little late and you don't care about the answer, but someone else may someday. Basically, strain is a dimensionless variable (units of distance/distance). Thus, it doesn't matter how long your sample is, only the ratio of how much the sample elongated (final length - initial length) divided by the initial length. Thus, to do the experiment, you need to make markings on your dogbone that is your gauge length. You will then elongate it in the machine (usually until fracture). Once fractured (and even if not), you will measure the new distance between the markings you originally made for your gauge length (this length will be larger than the gauge length). To get strain, you take the difference between the final and initial length and divide that by the initial length.

Boom.
 

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