0
$\begingroup$

I found the Navier–Cauchy elastostatic equation in a lot of literature expressed in the Cartesian coordinate system. However, I didn't manage to find it anywhere in cylindrical coordinates, so I tried deriving the expression myself. The problem is, I did't get a compact final equation which makes me think I did something wrong.

The problem I am solving is axially symmetric and angular rotations do not exist. That means I am working only with the radial and longitudinal coordinates, so the angular coordinate is not taken into calculation. Also, gravity effects on the cylinder are neglected.

My starting points are the radial and longitudinal Cauchy stress equations:

$$ {{\partial \sigma_{rr}(r,z)} \over \partial r}+ {{\partial \sigma_{rz}(r,z)} \over \partial z}+ {1\over r} \sigma_{rr}(r,z) = 0 \tag1 $$

$$ {{\partial \sigma_{zz}(r,z)} \over \partial z}+ {{\partial \sigma_{rz}(r,z)} \over \partial r}+ {1\over r} \sigma_{rz}(r,z) = 0 \tag2 $$

Next, I added the two equations together to get:

$$ {{\partial \sigma_{rr}} \over \partial r}+ {{\partial \sigma_{rz}} \over \partial z}+ {{\partial \sigma_{zz}} \over \partial z}+ {{\partial \sigma_{rz}} \over \partial r}+ {1\over r} \sigma_{rr} + {1\over r} \sigma_{rz} = 0 \tag3 $$

The first four terms were easy to formulate since they are in exactly the same form as if I was dealing with the Cauchy stresses in cartesian coordinates. The short process of formulation is shown here. The result is:

$$ {{\partial \sigma_{rr}} \over \partial r}+ {{\partial \sigma_{rz}} \over \partial z}+ {{\partial \sigma_{zz}} \over \partial z}+ {{\partial \sigma_{rz}} \over \partial r}+ = (\lambda + \mu)\nabla (\nabla \cdot u) + \mu \nabla ^2 u \tag4 $$

where $u(r,z)=u_r(r,z)+u_z(r,z)$ and $\nabla = {\partial \over \partial r} + {\partial \over \partial z}$. Also, $\lambda$ is the Lamé constant for plane stress, and $\mu$ is the shear modulus.

Now comes the part I am not sure of. I want to formulate the sum of the fifth and sixth term of equation $(3)$ in terms of $\lambda$, $\mu$, and $u$. So I did this step by step.

First, I expressed the sum of stresses not under the partial derivatives in terms of dilatations:

$$ \sigma_{rr} + \sigma_{rz} = N \epsilon _{_V} + 2 \mu \epsilon _{rr} + \mu (\epsilon _{rz} + \epsilon _{zr}) \tag5 $$

Second, I expressed the dilatations in terms of displacements:

$$ \sigma_{rr} + \sigma_{rz} = N {\partial u \over \partial V} + 2 \mu {\partial u_r \over \partial r} + \mu \Bigg({\partial u_r \over \partial z} + {\partial u_z \over \partial r}\Bigg) \tag6 $$

Volume is represented with $V$ in the above equation. Next, I added and subtracted the term $\mu {\partial u_z \over \partial z}$:

$$ \sigma_{rr} + \sigma_{rz} = N {\partial u \over \partial V} + \mu {\partial u_r \over \partial r} + \mu {\partial u_r \over \partial r} + \mu {\partial u_r \over \partial z} + \mu {\partial u_z \over \partial r} + \mu {\partial u_z \over \partial z} - \mu {\partial u_z \over \partial z} \tag7 $$

Now I can group together the following terms:

$$ {\partial u_r \over \partial r} + {\partial u_r \over \partial z} + {\partial u_z \over \partial r} + {\partial u_z \over \partial z} = \nabla u \tag8 $$

This allows me to write a more compact expression of equation $(7)$ as:

$$ \sigma_{rr} + \sigma_{rz} = N {\partial u \over \partial V} + \mu \nabla u + \mu {\partial u_r \over \partial r} - \mu{\partial u_z \over \partial z} \tag9 $$

In the end, I get the following Navier–Cauchy equation:

$$ (\lambda + \mu)\nabla (\nabla \cdot u) + \mu \nabla ^2 u + {1\over r} \Bigg( N {\partial u \over \partial V} + \mu \nabla u + \mu {\partial u_r \over \partial r} - \mu{\partial u_z \over \partial z} \Bigg) = 0 \tag{10} $$

The problem in the above equation is that the terms $u_r$ and $u_z$ exist. I don't see a way I can group them together into $u$ or get rid of them in some other type of way if possible. So my questions are:

$1$. Did I do something wrong in my derivation process?

$2$. Can the Navier–Cauchy equation in cylindrical coordinates be expressed in a compact way such that it only contains terms of $u$ and does not have the hanging terms $u_r$ and $u_z$?

$3$. If yes, what is that expression and how can it be derived?

$\endgroup$

1 Answer 1

2
$\begingroup$

I'd start from the abstract form (using vector/tensor implicit notation) of the Navier-Cauchy equations

$\rho \partial_{tt} \mathbf{u} = \mathbf{f} + \nabla \cdot \boldsymbol \sigma$,

with the second order stress tensor that can related to the strain second order tensor through the fourth order elastic tensor

$\boldsymbol \sigma= \boldsymbol C : \boldsymbol \varepsilon$, with $\boldsymbol \varepsilon = \dfrac{1}{2} \left( \nabla \mathbf{u} + \nabla \mathbf{u}^T \right)$

that, for homogeneous isotropic media, can be rewritten as $\boldsymbol \sigma = \lambda (\nabla \cdot \mathbf{u}) \mathbb{I} + 2 \mu \boldsymbol \varepsilon$, so that its divergence reads

$\nabla \cdot \boldsymbol \sigma = (\lambda + \mu) \nabla(\nabla \cdot \mathbf{u}) + \mu \nabla^2 \mathbf{u}$.

Thus, the Navier-Cauchy equations in vector implicit notation in terms of displacement only reads

$\rho \partial_{tt} \mathbf{u} = \mathbf{f} + (\lambda + \mu) \nabla(\nabla \cdot \mathbf{u}) + \mu \nabla^2 \mathbf{u}$.

Next, I'd rely on the expression of the vector calculus operators using the desired system of coordinates, to get the components of the vector equation. For Cartesian, cylindrical and spherical coordinates, these expressions are available on wiki, https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.