Mass conservation in a deformed membrane in cylindrical coordinates

Question

This is clearly an obvious question but here is my issue.

Context :

We assume an axisymmetric deformation of a membrane, and work with cylindrical coordinates $(r; \phi; y)$. At time $t = 0$ we let $r$ and $\phi$ be the initial radial and circumferential location of a fabric point in the infinite, untensioned membrane lying at rest in the level ground plane $y = 0$. At time $t$ after impact, we let $u$ and $v$ be the in-plane and normal, out-of-plane components, respectively, of the displacement of this point, but its cylindrical angle $\phi$ remains the same. The new coordinates of the fabric point now become $(r +u; \phi; 0 + v)$

Visually it gives this :

Problem :

If we consider a membrane element initially of volume $dV_{0} = hr drd\phi$, where $h$ is its initial thickness and $\rho$ is its initial density then after the deformation the volume is written : $dV = h(r+u)(1+\epsilon_t)drd\phi$ with $\epsilon_t$ its local in-plane strain oriented radially. Then conservation of mass should be written : $\rho h r dr d\phi = \rho h (r+u)(1+\epsilon_t)drd\phi$

Question :

I understand this expression and the fact that the new "deformed" length is no longer $dr$ but $(1+\epsilon_t) dr$. However, I am trying to demonstrate it in a more formal way and to do that I wrote the displacement vector $\vec{u} = (u,0,v)$ in the cylindrical space and that's in order to obtain the expression of the tangent linear operator $F$ as $F = I + Grad(\vec{U})$ . Then, we know that there is a direct link between the variation of a volume element and the determinant of this tensor such as : $dV = J dV_0$ with $J = det(F)$. Using this : I am unable to recover the expression $(1+\epsilon_t)$ that appears in the expression presented before...

Am I doing something wrong ? I don't understand how I can't express this volume variation like suited.

Thanks in advance for any help.

Best regards.

P.S : The expression of $\epsilon_t$ is $\epsilon_t = \sqrt{(1 + \frac{\partial u}{\partial r})^2+(\frac{\partial v}{\partial r})^2} - 1$

Answer 1

The position vector $~\mathbf R~$ to a point on the curve surface is

$$\vec R= \left[ \begin {array}{c} r\cos \left( \varphi \right) \\ r\sin \left( \varphi \right) \\ f \left( r,\varphi ,y \right) \end {array} \right] $$

thus the Jacobian $~\mathbf J~$ $$\mathbf J_{ij}= \frac{\partial\vec R_i}{\partial \mathbf q_j}\quad, \text{where}\quad \mathbf q=\left[ \begin {array}{c} r\\ \varphi \\ y\end {array} \right] $$

from here the determinate of the Jacobian

$~\rm det(\mathbf J)=r{\frac {\partial }{\partial y}}f \left( r,\varphi ,y \right)~$

and the volume element $~dV=\rho\,r{\frac {\partial }{\partial y}}f \left( r,\varphi ,y \right)\,dr\,d\varphi\,dy~$

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instead of r and y you can use u and v