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Stress and Strain tensors in cylindrical coordinates

  • #1
47
4

Homework Statement


I am following a textbook "Seismic Wave Propagation in Stratified Media" by Kennet, I was greeted by the fact that he decided to use cylindrical coordinates to compute the Stress and Strain tensor, so given these two relations, that I believed to be constitutive given an isotropic elastic medium =
$$\nabla (\textbf{u}) = \bar{\bar{\epsilon}}_{ij}$$
$$\tau_{ij}=\lambda \delta_{ij} \epsilon_{kk} + 2\mu \epsilon_{ij}$$
Given epsilon Strain and tau Stress tensors I was then surprised to see this

https://imgur.com/x4PG3iN

Equation 2.1.4, the shear stresses (crossed derivatives) and strains hold water to what I did, but the diagonals have an extra something...it looks like a divergence, multiplied by lambda, Am I missing basic calculus knowledge to solve this problem? I admit to be a bit rusty in my notation, even computing the equation 2.1.2 was tiring, but now I'm really surprised by this.
Any feedback would be greatly appreciated, thanks :)


Homework Equations


$$\nabla (\textbf{u}) = \bar{\bar{\epsilon}}_{ij}$$
$$\tau_{ij}=\lambda \delta_{ij} \epsilon_{kk} + 2\mu \epsilon_{ij}$$

The Attempt at a Solution


Stated above


P.S. [/B]= I posted it here because it is a mixture of math and physics, if it's wrong please let me know
 

Answers and Replies

  • #2
19,941
4,102
The strain tensor is the symmetric part of the displacement gradient tensor, not just the gradient (that's where the cross derivatives come in). In addition, when you are taking the gradient of the displacement vector in a curvilinear coordinate system, you also need to include the partial derivatives of the unit vectors with respect to the coordinates, which change from location to location (This is not the case in Cartesian coordinates).

Your expression for the stress tensor in terms of the strain tensor is correct.
 
  • #3
47
4
Yes you are right, I forgot to type it here. I meant to write
$$\frac{1}{2}( \nabla (\textbf{u}) + \nabla (\textbf{u})^T) = \bar{\bar{\epsilon_{ij}}}$$
And the anti symmetric tensor is zero because there are no rotations, my main issue is due to the diagonals, the rr component of Tau has more than what i'd expect it to have
 
  • #4
19,941
4,102
Yes you are right, I forgot to type it here. I meant to write
$$\frac{1}{2}( \nabla (\textbf{u}) + \nabla (\textbf{u})^T) = \bar{\bar{\epsilon_{ij}}}$$
And the anti symmetric tensor is zero because there are no rotations, my main issue is due to the diagonals, the rr component of Tau has more than what i'd expect it to have
Like I said, when you evaluate the gradient of the displacement, you need to use the product rule for differentiation, including derivatives of the unit vectors. In cylindrical coordinates, the unit vectors in the radial and circumferential directions are function of theta.
 

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