# Find elongation of a rod of constant volume given Young's modulus, original length and load without assuming that cross-sectional area is constant

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.

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.

• 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. Jun 30, 2021 at 15:26
• 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. Jun 30, 2021 at 15:54