1
$\begingroup$

The usual formula for Young's modulus:

$$ Y=\frac{FL}{A\Delta l}, $$ (where $F$, $L$, $A$ and $Δl$ are load, original length, and area over which force is applied respectively) can be rearranged to give

$$Δl=\frac{FL}{AY}.$$

The assumption is that the cross-sectional area remains effectively constant, however, this is not true for longer elongations. If area changes with change in length, then $Δl$ becomes $dl$ since $Δl$ becomes a function of area (which changes with $l$) and the change in length can be only evaluated within the tiny interval within which area is constant. For constant volume, the following relation is true

$$A=\frac{V}{l},$$ where $l$ is the length at a given instant.

Plugging this into the previous equation we get:

$$dl=\frac{FLl}{YV}$$

I'm unable to proceed beyond this point. Here, how can I evaluate the total change in length $l$? $l$ varies from $L$ to $L+Δl$ where $Δl$ is the sum of $dl$. I'm very confused. Feel free to correct any wrong steps.

$\endgroup$

1 Answer 1

2
$\begingroup$

Here, how can I evaluate the total change in length l?

You integrate; relabeling the initial length $L$ as $L_0$ for clarity and defining the initial area $A_0=V/L_0$, we have:

$$\int_{L_0}^{L_0+\Delta L}\frac{dl}{l}=\int_0^F\frac{df}{YA_0};$$

$$\ln\left(\frac{L_0+\Delta L}{L_0}\right)=\frac{F}{YA_0};$$

$$\Delta L=L_0(e^{F/YA_0}-1).$$

Note that this carries all your original assumptions: constant volume $V$, constant stiffness $Y$. You can verify that for small $\Delta L$ (where we're more likely to encounter near-constant stiffness), the relation simplifies to

$$\ln\left(\frac{L_0+\Delta L}{L_0}\right)\approx \frac{\Delta L}{L_0}=\frac{F}{YA_0}\longrightarrow\Delta L=\frac{FL_0}{YA_0},$$

which is Hooke's Law.

$\endgroup$
2
  • $\begingroup$ Gee! Thanks so much, your insight was that F should be replaced with df, which I'd be glad if you helped me understand better, btw, is this a formal formula? why can't I find a formula for this anywhere? I've searched the internet as thoroughly as possible. $\endgroup$ Jun 30, 2021 at 15:26
  • $\begingroup$ Well, you can't have only one differential in an equation—it's too small and would evaluate to zero. You must have (at least) two, which then gives a meaningful ratio. You obtain a very small displacement dl from a very small force, so this was the motivation to replace F with df. Many constitutive equations start with differentials, and then if their ratio is constant, one can integrate to obtain, e.g., V=IR, F=ma, F=-kx, σ=Eε. $\endgroup$ Jun 30, 2021 at 15:54

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.