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


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:


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.


1 Answer 1


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.

  • $\begingroup$ 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

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