0
$\begingroup$

We have: $\sigma = E \epsilon$ and $\epsilon = \Delta L/L_0$. This seems to imply that for a constant strain, we will achieve a proportional change in length.

However, it doesn't rest well with me that this relation is not dependent on $L_0$. For instance, say I have some material that has an $E = 1$ MPa and an area of $1$ m$^2$, and I put a $1$ N force on it. Then it would have a strain of $\epsilon = 1\times 10^{-9}$.

So if my bar is 1m long, then there would be a millimetre's difference in the length. But if my bar was in space and a kilometre long, then would I expect to see the bar shrink by a full metre?

Or do I have an overly simplistic understanding of stress and strain?

$\endgroup$
1
  • $\begingroup$ $E$ is usually in hundreds of GPa so your value isn't realistic at all. $\endgroup$ – John Alexiou Sep 6 '20 at 1:12
2
$\begingroup$

So in the context of overhead powerlines which can be as long as several miles on river crossings the amount of deformation can be of the order of 0.5% of length which is several feet in length.

fig

The example above is a 5280 ft length of cable, tensioned to 28,000 lbf that stretches it do 5304.3 ft (an elongation of 24.3 ft).

So yes, elongation of crazy proportions is real and important part of a well-engineered power grid.

$\endgroup$
2
$\begingroup$

Yes, you would have the equivalent of 1000 rods of one meter each joined end to end. Each would change by 1 mm, and the total change for all of them would be 1000 mm. Although I would stretch it rather than shrink it to avoid buckling.

$\endgroup$

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.