Hookes law for stress and strain

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aaaa202
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I need help understanding a passage in my textbook, where the form of hookes law in continuous elastic media is explained. It says:
"The absence of internal directions in isotopic matter tells us that there are only two tensors available to construct a linear relation between the the stress tensor and the strain tensor. One is the strain tensor itself and the other is the kronecker delta multiplied by the trace of the strain tensor. Consequently the most general strictly linear tensor relation between stress and strain is of the form:
Here follows a linear relation between the stress- and strain tensor involving the strain tensor and its trace in separate terms.
"
I don't really understand all this. Why is only the strain tensor and its trace avaible for constructing a linear relation between the two tensors? Does it have to do with the fact that we would like Hookes law to be invariant under change of directions, since the symmetry dictates that there are no internal directions?
 
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aaaa202 said:
I need help understanding a passage in my textbook, where the form of hookes law in continuous elastic media is explained. It says:
"The absence of internal directions in isotopic matter tells us that there are only two tensors available to construct a linear relation between the the stress tensor and the strain tensor. One is the strain tensor itself and the other is the kronecker delta multiplied by the trace of the strain tensor. Consequently the most general strictly linear tensor relation between stress and strain is of the form:
Here follows a linear relation between the stress- and strain tensor involving the strain tensor and its trace in separate terms.
"
I don't really understand all this. Why is only the strain tensor and its trace avaible for constructing a linear relation between the two tensors? Does it have to do with the fact that we would like Hookes law to be invariant under change of directions, since the symmetry dictates that there are no internal directions?
No. It has nothing to do with this. It's mathematical. Try to construct another tensorial form that is linear in the strain tensor.

Chet
 
I don't understand. So according to what you say we have that the most general relation is:

σij = λuij + λδijkukk

But what goes wrong if you try to put in terms proportional to other entries in the matrix u? Actually I'm not even sure if the above expression is correct.
 
aaaa202 said:
I don't understand. So according to what you say we have that the most general relation is:

σij = λuij + λδijkukk

But what goes wrong if you try to put in terms proportional to other entries in the matrix u? Actually I'm not even sure if the above expression is correct.
The above equation is the most general linear relationship (I'm assuming you are calling u the strain tensor, which I call E). If F(E) is a tensorial function of the strain tensor E, then we can use the Cayley Hamilton theorem to represent F by ##F = α I + βE+γE^2##, where α, β, and γ are functions of the invariants of E. This is linear in E only if α is a function of the trace of E (which is the only invariant that is linear in the components of E), β is a constant, and γ =0.

Chet