Hooke's Law: A Beginner's Guide to Tensor Form and Matrix Form

[sol66]

Ok I wan't to start by saying I'm in a ridiculous solid state physics class where the stuff we are learning is either poorly explained by our textbook or even non-existent in the text. My teacher asked me the following question ... A single cube-shaped crystal of a simple cubic metal with face normals [100][010][001] has a value of the elastic stiffness constant c11 = cxx = 293 GPa. Write down the generalized form of Hooke's Law, in the tensor form and in the reduced index (matrix) form.

So I just want to say, outside of this class I've never seen a tensor in my life except in classical dynamics which made sense for rotations. We went over tensors for like 20 minutes, where the basic idea of a tensor was described as being an object that calls a particular value a number of times. I have a somewhat ok idea of what a tensor is, though never in my life have I ever had to use them. I have no idea what my teacher is talking about and I can't find good information on the web or in my horrid little Intro to solids Charles Kittel book.

Apparently hookes law is a fourth rank tensor and I have no freakin clue why that is.

Thanks for the replies

 

[Mapes]

Stiffness is represented by a fourth-rank tensor because it couples (relates) two second-rank tensors, stress and strain. And these are tensors because they each couple two vectors (aka first-rank tensors): the stress tensor relates a force vector to a vector representing the direction normal to an area, and the strain tensor relates an initial undeformed vector to the deformed version. Does this help?

http://en.wikiversity.org/wiki/Introduction_to_Elasticity/Constitutive_relations" might be helpful.

 

[turin]

Hi, Mapes. Thanks for the link. I was wondering if you could confirm for me a factor-of-two error in that article. Here is the equation sequence that I believe has the error:

$$\sigma_{ij}=\frac{\partial{w}}{\partial{\epsilon_{ij}}}=C_{ijkl}\ \epsilon_{kl}$$

$$w=C_{ijkl}\ \epsilon_{ij}\ \epsilon_{kl}$$

where C is the stiffness tensor, ε is the strain tensor, σ is the stress tensor, and w is the strain energy density. I would replace the last equation with:

$$w=\frac{1}{2}\ C_{mnkl}\ \epsilon_{mn}\ \epsilon_{kl}$$

which leads to:

$$\frac{\partial{w}}{\partial\epsilon_{ij}}<br /> =\frac{1}{2}\ C_{mnkl}\ \frac{\partial\left(\epsilon_{mn}\ \epsilon_{kl}\right)}{\partial\epsilon_{ij}}<br /> =\frac{1}{2}\ C_{mnkl}\ \frac{\partial\epsilon_{mn}}{\partial\epsilon_{ij}}\ \epsilon_{kl}\ +\ \frac{1}{2}\ C_{mnkl}\ \epsilon_{mn}\ \frac{\partial\epsilon_{kl}}{\partial\epsilon_{ij}}<br /> =\frac{1}{2}\ C_{ijkl}\ \epsilon_{kl}\ +\ \frac{1}{2}\ C_{mnij}\ \epsilon_{mn}$$

and then using the symmetry of C and renaming the dummy indices mn → kl:

$$\frac{\partial{w}}{\partial\epsilon_{ij}}=C_{ijkl}\ \epsilon_{kl}$$

Maybe I don't understand their notation.

 

[Mapes]

Yep, looks like a missing 1/2 to me too.

 

[Modey3]

turin,

Nothing is wrong with the final expression since $$C_{ijkl} = C_{klij}$$ due to symmetry. After summing on the indices from 1 to 3 the two terms end up being equal. Intuitively, it makes sense because the first derivative of energy wrt object-position gives the object-force or in this case the stress on a unit cell. Which is what you end up with.

modey3

 

[Modey3]

Ok I wan't to start by saying I'm in a ridiculous solid state physics class where the stuff we are learning is either poorly explained by our textbook or even non-existent in the text. My teacher asked me the following question ... A single cube-shaped crystal of a simple cubic metal with face normals [100][010][001] has a value of the elastic stiffness constant c11 = cxx = 293 GPa. Write down the generalized form of Hooke's Law, in the tensor form and in the reduced index (matrix) form.

So I just want to say, outside of this class I've never seen a tensor in my life except in classical dynamics which made sense for rotations. We went over tensors for like 20 minutes, where the basic idea of a tensor was described as being an object that calls a particular value a number of times. I have a somewhat ok idea of what a tensor is, though never in my life have I ever had to use them. I have no idea what my teacher is talking about and I can't find good information on the web or in my horrid little Intro to solids Charles Kittel book.

Apparently hookes law is a fourth rank tensor and I have no freakin clue why that is.

Thanks for the replies

sol66,

The best place to learn tensor applications in mechanics is by reading a continuum mechanics book. I recommend getting the Schaums Outline for continuum mechanics. That book has served me well in my graduate career.

modey3

 

[Mapes]

turin,

Nothing is wrong with the final expression since $$C_{ijkl} = C_{klij}$$ due to symmetry. After summing on the indices from 1 to 3 the two terms end up being equal. Intuitively, it makes sense because the first derivative of energy wrt object-position gives the object-force or in this case the stress on a unit cell. Which is what you end up with.

modey3

Modey3, just to be clear, turin suggests that the "Wikiversity" equation $\frac{1}{2}\bold{\epsilon C\epsilon}= C_{ijkl}\epsilon_{ij}\epsilon_{kl}$ should be $\frac{1}{2}\bold{\epsilon C\epsilon}= \frac{1}{2}C_{ijkl}\epsilon_{ij}\epsilon_{kl}$; is this what you're disagreeing with?

 

[Modey3]

Modey3, just to be clear, turin suggests that the "Wikiversity" equation $\frac{1}{2}\bold{\epsilon C\epsilon}= C_{ijkl}\epsilon_{ij}\epsilon_{kl}$ should be $\frac{1}{2}\bold{\epsilon C\epsilon}= \frac{1}{2}C_{ijkl}\epsilon_{ij}\epsilon_{kl}$; is this what you're disagreeing with?

I'm not disagreeing with turin. The guy who wrote the Wiki forgot to add the 1/2.

modey3