glra2222
Hello,

I am currently working on a uni project which involves getting material properties of different materials (steel, rubber, foam, etc.).
Now I have always thought that getting Young's modulus was simply from doing a tensile test, getting the stress-strain graph from stress = P/A and strain = deltaL/L
and then E = stress/strain.
However, after doing a compression (flexural bending) test on a steel rebar 10x10 mm cross-sectional area and 475 mm length -> I get the pre-yield part of the graph going up to 6 MPa stress and strain of 0.02. 6/0.02 = 300MPa which is about 1000x smaller than the E = 200GPa that I should be getting.
I've already checked the units - 400 N of load gives me about 5mm of deflection. The equation deflection = 1/48*PL^3/(EI) shows this to be correct. P/A = 400/100 = 4 N/mm^2 = 4 MPa and 5 mm/475 mm = approx. 0.01 strain.
I have also done a tensile test which gives me a graph within the similar range.
What am I doing wrong?
glra2222
Okay, ive finally figured out what I had done wrong.
The strain in the stress-strain graph is not the change in length of the bar divided by the length of the bar. Strain can only be obtained through a strain gauge.
Now the only thing is, I'm not sure why? I thought that the strain gauge also measured change in length over length - unless there is a specific length that you can only test it on.
rpenner
When you bend a beam, you get compression and deflection. As such the stress is of two types and not all the area of the beam contributes.

I suggest you re-read the sections on second moment of area and deflection of a cantilever. And if you have no textbook, here's Wikipedia.

http://en.wikipedia.org/wiki/Deflection_(engineering)

http://en.wikipedia.org/wiki/Area_moment_of_inertia
Enthalpy
With the right units, the figures you give are consistent. Except that 5mm deflection and 475mm length do not combine simply to give 0.01 strain (which would already be a lot for steel).

By the way, bending combines tension and compression, yes... But near to another, and then the upper and lower part of the beam aren't allowed to expand freely in the direction of the width. Then, some form of Poisson's factor must improve the formula a bit, like (1-n^2), but in a complicated manner that depends on the thickness-to-width ratio. With 10mm*10mm, this correction isn't attainable by a hand computation.

You mention rubber: bad luck. All elastomers are extremely nonlinear, most of them have hysteresis, as so computations with Young's and sheer modulus and strength, which work with metals, don't nearly work with elastomers. They work so little that elastomer manufacturers generally refuse to give such numbers; you just get Shore hardnesses, oh good.
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