0
$\begingroup$

Consider a situation where we have a frustum between 2 walls. We heat it , so it wants to expand but the walls prevent it from doing so.

Does the stress in the frustum remain constant or not ??

Case 1 Say we consider a small element of "$dx$" length so that we can use equation of stress and strain

Because we know $$\text{Stress} = \frac{\text{Force}}{\text{Area}}$$ and since Force from both sides is same and A varies , stress should vary.

Case 2 But also, since $$\text{ Stress} =\text{ Young's Modulus} * \text{Strain} $$ And $$ \text{Strain} =\frac{\Delta L}{\Delta x}$$ which comes out to be $$ \text{Strain} = \text{(coeff of linear expansion)}* T $$ , which is constant ,thus stress should be constant.

Then which of the following is true ???

$\endgroup$
  • $\begingroup$ The strain is not constant. The strain is the sum of a thermal part and a compressive part. The thermal part is constant, but the compressive part is not. $\endgroup$ – Chet Miller May 18 at 16:50
  • $\begingroup$ @ChetMiller Please can you elaborate , what do you mean by compressive part ?? $\endgroup$ – user232243 May 18 at 16:53
1
$\begingroup$

Suppose that, rather than heating the frustum between two rigid walls, you did the process in two steps. In the first step, you heated it without any constraint on its ends so that it could expand freely. Here, the strain would be uniform ($\epsilon_1=\alpha\Delta T=\frac{du_1}{dx}$). Then, in the second step, you compressed it back to its original length. In the second step, the stress and strain within the frustum would vary along its length: $$\sigma(x)=-\frac{F}{A(x)}=E\epsilon_2(x)=E\frac{du_2}{dx}$$where $u_2(x)$ is the axial displacement in step 2. This would be subject to the constraint that $u_2(l)=-u_1(l)$.

$\endgroup$
  • $\begingroup$ Thank You , Gr8 explanation $\endgroup$ – user232243 May 18 at 20:21
0
$\begingroup$

The force is constant. The stress is $\frac{F}{A}$. Since the cross sectional area, $A$, varies across the object, the stress also varies.

Since the stress varies across the object, the strain also varies. The value $\frac{\Delta L}{\Delta X}$ is the average strain across the object.

$\endgroup$
  • $\begingroup$ But what about a small elemental pieces of length dx , they will have the same elongations of dl as... ? $\endgroup$ – user232243 May 18 at 18:33

Your Answer

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