0
$\begingroup$

On page $519$ of the book Engineering Vibration (which can be downloaded from here), the following wave equation of a variable cross-section rod is derived:

$${ {\partial}\over{\partial}x} \Big( EA(x){ {\partial}w(x,t)\over{\partial}x} \Big)={\rho}A(x) { {\partial}w^2(x,t)\over{\partial}t^2}\tag{1}$$

where:

  • $x$ is the spatial coordinate
  • $t$ is time
  • $E$ is the Young's modulus of the rod
  • $A(x)$ is the variable cross-section function of the rod
  • $\rho$ is the density of the rod
  • $w(x,t)$ is the wave function

Because some of the steps are missing from the derivation process, I tried deriving the same equation myself but failed. Please point out my mistake.

enter image description here

For the variable cross-section rod in the picture above, I wrote the following dynamics equation: $$\Big(F(x,t)+dF(x,t)\Big) - F(x,t) = \Big(dm(x)\Big){ {\partial}w^2(x,t)\over{\partial}t^2}\tag{2}$$ where $dm(x)$ is the mass of the Infinitesimal element of the rod. Next, I crossed out the forces $F(x,t)$ and defined the mass $dm(x)$ as a product of volume $dV(x)$ and density: $$dF(x,t) = {\rho}\Big(dV(x)\Big){ {\partial}w^2(x,t)\over{\partial}t^2}\tag{3}$$ Since the volume $dV(x)$ is: $$dV(x) = A(x)dx\tag{4}$$ equation $(3)$ can be rewritten as: $$dF(x,t) = {\rho}A(x)dx{ {\partial}w^2(x,t)\over{\partial}t^2}\tag{5}$$ After this, I defined $dF(x,t)$ as: $$dF(x,t) = d\Big(P(x,t)A(x)\Big)=\Big(dP(x,t)\Big)A(x)+ P(x,t)\Big(dA(x)\Big)\tag{6}$$ where $P(x,t)$ is the pressure. By expanding the total derivatives, I obtained: $$dF(x,t) = \Big( { \partial P(x,t) \over \partial x}dx + { \partial P(x,t) \over \partial t}dt \Big)A(x) + P(x,t){ \partial A(x) \over \partial x}dx \tag{7}$$ Hook's law states: $$P(x,t) = E{ {\partial}w(x,t)\over{\partial}x}\tag{8} $$ and because of that, I rewrote equation $(7)$ as: $$dF(x,t) = E\Bigg(\Big( { \partial ^2 w(x,t) \over \partial x^2}dx + { \partial ^2 w(x,t) \over \partial x \partial t}dt \Big)A(x) + w(x,t){ \partial A(x) \over \partial x}dx\Bigg) \tag{9}$$ In the end, I derived the expression: $$E\Bigg(\Big( { \partial ^2 w(x,t) \over \partial x^2}dx + { \partial ^2 w(x,t) \over \partial x \partial t}dt \Big)A(x) + w(x,t){ \partial A(x) \over \partial x}dx\Bigg) = {\rho}A(x)dx{ {\partial}w^2(x,t)\over{\partial}t^2}\tag{10}$$ In equation $(10)$ there is an extra term. Namely, the ${ \partial ^2 w(x,t) \over \partial x \partial t}dt$ term. If that did not exist, I would get equation $(1)$. So, my question is, why does that term not exist in equation $(1)$?

$\endgroup$
1
  • $\begingroup$ The only explanation is that the pressure on a line x=cst is constant, $\partial p(x,t)/\partial t |_{x=cst}=0$ $\endgroup$
    – The Tiler
    Commented Sep 26, 2022 at 19:17

1 Answer 1

1
$\begingroup$

The problem is in equation 2, it should be the increment in the $x$ direction, the part in parenthesis should be

$$F + \frac{\partial F}{\partial x} dx\, .$$

Also, I would probably take finite values for the size of your element: $\Delta x$. Then you can take the limit, and new derivatives appear.

$\endgroup$
1
  • $\begingroup$ You are correct. I found the solution in a nother piece of literature. Thank you Nicoguaro. $\endgroup$ Commented Oct 2, 2022 at 17:14

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.