Stress and Strain tensors in cylindrical coordinates

In summary: That's why you have extra terms in the diagonal components. In summary, the conversation discusses the use of cylindrical coordinates to compute the stress and strain tensors in an isotropic elastic medium. The strain tensor is the symmetric part of the displacement gradient tensor and the stress tensor can be expressed in terms of the strain tensor. However, when evaluating the stress and strain tensors in cylindrical coordinates, additional terms arise in the diagonal components due to the product rule for differentiation and the dependence of unit vectors on theta.
  • #1
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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 aboveP.S. [/B]= I posted it here because it is a mixture of math and physics, if it's wrong please let me know
 
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  • #2
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
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
Remixex said:
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.
 

1. What are stress and strain tensors in cylindrical coordinates?

Stress and strain tensors in cylindrical coordinates are mathematical representations of the distribution of forces and deformations within a cylindrical object. They are used to describe the internal stresses and strains experienced by materials subjected to external forces, and are important in understanding the mechanical behavior of cylindrical structures.

2. How are stress and strain tensors calculated in cylindrical coordinates?

Stress and strain tensors in cylindrical coordinates are calculated using equations derived from the basic principles of mechanics and the geometry of cylindrical systems. These equations take into account the cylindrical shape of the object and the direction and magnitude of the applied external forces.

3. What is the significance of stress and strain tensors in cylindrical coordinates?

Stress and strain tensors in cylindrical coordinates provide a comprehensive description of the mechanical behavior of cylindrical structures, including the distribution of stresses and strains along different directions. They are essential for analyzing the stability, strength, and failure of cylindrical objects.

4. How do stress and strain tensors in cylindrical coordinates differ from those in other coordinate systems?

The main difference between stress and strain tensors in cylindrical coordinates and those in other coordinate systems, such as Cartesian or spherical coordinates, is the way in which the tensors are defined and calculated. This is due to the unique cylindrical geometry of the object and the specific directions of the applied external forces.

5. What are some real-world applications of stress and strain tensors in cylindrical coordinates?

Stress and strain tensors in cylindrical coordinates have numerous applications in engineering and science, including in the design and analysis of cylindrical structures such as pipelines, pressure vessels, and rotating machinery. They are also used in geomechanics to study the behavior of cylindrical geological formations, and in biomedical engineering to understand the mechanics of cylindrical tissues and organs.

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