0
$\begingroup$

I read following in my text book: We take an infinitesimal segment of a bar of uniform cross sectional area $A$, length of this segment starts from $x$ to $x+dx$. We apply stress $S$ in the direction of the length $x$ and $u$ is the displacement of the bar under this stress.Now if $\rho$ is mass density of the bar then from Newton's 2nd law,$$(\rho A dx)\frac{\partial^2u}{\partial t^2}=[S(x+dx)-S(x)]A$$I am having problem understanding r.h.s. of above equation because for that infinitesimal segment stress applied should be $S(x)$ only and not $[S(x+dx)-S(x)]$. Please explain how the r.h.s. in the above equation has such a form.

$\endgroup$
2
$\begingroup$

You are looking at Newton's second law $$m \vec{a} = \vec{F} $$ Now $a=\frac{\partial^{2}u}{\partial t^2}$, $m = \rho A dx$ meaning that the body under consideration is this infinitesimal volume of the bar of length $dx$. Now what is the net force on this volume? Assuming that the stress field is only along the direction of the bar, there are two forces on this volume: on the surface with area $A$ at $x$, which is $- S(x) A$, and on the surface with area $A$ at $x+dx$, which is $S(x+dx) A$. The change in sign results from the different orientations of the surfaces. The total force on the volume is just the sum of those two forces and thus we arrive at the expression you give in your question.

$\endgroup$
3
  • $\begingroup$ Your answer explained the negative sign, but I still can not understand the term $S(x+dx)$. Consider the infinitesimal segment isolated. Now, how the stress is applied at point $x+dx$ , as the stress is only applied on the cross section at $x$, i.e. we apply force from the left side of the segment only, we never touch the right side of the segment which is at $x+dx$ ? $\endgroup$ – NewStudent Nov 23 '16 at 14:52
  • $\begingroup$ Why do you assume that there is no force/stress applied at $x+dx$? The stress field gives you exactly the stress applied to the surface at any point of the material. So yes, there is a force/stress applied on both surface areas of our imaginary volume. $\endgroup$ – Sanya Nov 23 '16 at 15:00
  • $\begingroup$ After thinking for some time I understand how that r.h.s. term occures. Actually I was thinking the segment is too infinitesimal to consider the transmission of the stress from surface at $x$ surface to surface at $x+dx$. That is why the problem appeared. $\endgroup$ – NewStudent Nov 23 '16 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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