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Young's modulus vs tangent modulus vs secant modulus 1

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elogesh

Mechanical
May 10, 2002
187
Hai,

1)What is the difference between young's modulus, tangent modulus and secant modulus?

The understanding I have is,

Young's modulus represents the ratio of stress versus strain(slope) in the elastic range.

Tangent modulus and secant modulus represents the ratio of stress versus strain in the inelastic range and it has to be specified based on the value of stress value.

2)My second question, I have the values of secant modulus and tangent modulus, is there any way to find young's modulus of the material from these values.

If there is anything technically wrong with my question, please correct me.

With advanced Thnaks and Regards,
Logesh.E
 
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Usually finding the Young's Modulus of the material you want is relatively simple. Knowing the inelastic behaviour is more difficult for some materials. If you have the tangent modulus, ie. the slope in the inelastic region, then I'm not sure what the problem is other than perhaps you don't know what the material is and hence don't know what the Young's modulus is. Knowing the slope in the inelastic region won't, unfortunately, tell you how it behaves elastically.
 
The tangent modulus is the instantaneous modulus
(d(sig)/d(e)) of the stress-strain curve, while the secant modulus (if my memory is correct) is the modulus from 0,0 to e(i), sig(i). By that definition (if I've gotten it correct), the secant modulus at a low strains approaches the Young's modulus, since Young's modulus is the linear modulus and at linear strains the secant modulus would directly be E.

Brad

Sorry--sig being stress, e being strain
 
It's amazing.

Every time bradh posts a reply, I learn something!

Thanks for another great response, Brad, even though it wasn't my question...
 
The response from Bradh is more meaningful if you look at the properties of a thermoplastic. Here the linear portion of the stress strain curve hardly exists. Sharing knowledge is a way to immortality
 
Hai,

A star for your answer.

Thanks for your reply.

Regards,
Logesh.E
 
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