# Youngs and Shear modulus

Good Morning All,

I remember reading a few times that, once the Youngs modulus relaxation is measured in a uni axial test the results could be converted to Shear modulus using the relationship

G = E / 2 (1 + ni).

This would help me in a problem I have at the moment but I have a couple of doubts.

First of all it seems to me the relationship is valid only in linear elasticity, following its derivation.

Are then the statements about the relation validity based on experience, i.e. the relation work fine in spite of not being theoreticvally the most rigorous?

Thank you

Muzialis

Good points. Yes that equation is strictly only valid for linear elastic materials. For some materials the Poissons ratio is rather independent of the applied strain. In these cases this approach is still a useful approximation.

- Jorgen

Jorgen,

many thanks for your reply.

Just to close this thread, what would then the most rigorous way of determining the shear modulus relaxation be?

There are a number of shear tests I have seen mentioned (double shear,quadruple shear, etc) and I am not sure which one would be the best.

All the best

Muzialis

I dont know what is the best shear test. I personally like the double shear lap test since it keeps the specimen aligned correctly. Another nice option is torsion.

- Jorgen

Jorgen,

thanks for your reply.

My question was badly posed indeed, certainly there is not a best test.

I am just trying to understand which one replicates shear as seen on my component more faithfully.

My concern with the torsion test is that if the shear strehgth is looked at, then the result woudl be affected by the fact large deformation torsion implies noraml stresses as well.

But your reply is good food for thought.

Have a nice day

Regards

Muzialis

Jorgen,

thanks for your reply.

My question was badly posed indeed, certainly there is not a best test.

I am just trying to understand which one replicates shear as seen on my component more faithfully.

My concern with the torsion test is that if the shear strehgth is looked at, then the result woudl be affected by the fact large deformation torsion implies noraml stresses as well.

But your reply is excellent food for thought.

Have a nice day

Regards

Muzialis

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