# Solid Elasticity, Steel Rod shortening

1. Nov 25, 2012

### Siune

1. The problem statement, all variables and given/known data
We have a steel rod with density $ρ = 7,90 kg/m^3$. When it's horizontally on the floor, it's length is $L = 6,00m$. The rods surface area $A$ is a circle with radius $r=0,04m$. Steel has Young modulus $E=2,1 \cdot 10^{11}$

Now the rod is lifted up so it's vertically straight. How much does the rod shorten $\Delta L = ?$ under it's own weight.

2. Relevant equations
Longitudinal deformation: $\frac{F}{A} = E \frac{\Delta L}{L}$

3. The attempt at a solution
Problem is easy I assume. I'm not just sure is the force that causes the shortening $F= Mg$ where $M=ρ\pi r^2 L$

I was thinking that we have infinitely small disc(s) with area $A$ and height $dL$. Now each disc would feel a force due to it's weight $dm \cdot g$.
And now intergrating through the whole rod (From 0 to $L$) we would get
$$∫dm \cdot g = ∫ (ρ \pi r^2 g) dL = ρ\pi r^2 g L = Mg$$

Now we get $$\Delta L = \frac{FL}{AE} = \frac{ρgL^2}{E} = 1,33 \cdot 10^-5 m$$.
So the shortening would be independent of the area which I hugely doubt? Especially as we are given the radius. And our teacher has never put "too much info" in the description to trick us. If anyone could expain it with words or thru math I would really appericate it.

Sincerely yours,
Siune

2. Nov 25, 2012

### Staff: Mentor

Your answer is basically correct (the radius is not relevant), except for one thing. The strain in the bar is not constant from end to end. The stress and strain in the bar are zero at the top, and are maximum at the bottom (where the entire weight is supported). On average, the stress is only half the weight divided by the area. Therefore, the change in length is only half of what you calculated.

3. Nov 25, 2012

### Siune

Okey, that makes sense. I mean I can understand it through symmetry that F is not constant and the $\Delta L$ is half. But mathematically, is it right like this:

$$\Delta L = ∫\frac{L}{AE}dF = ∫\frac{L}{\pi r^2E}gρ\pi r^2 dL = ∫\frac{ρg}{E} L dL = \frac{ρgL^2}{2E}$$

I'm really sorry, my understanding of intergrating could be better but I try to improve!

4. Nov 25, 2012

### Staff: Mentor

The local compressive stress is ρgL, where L is distance measured downward. The local compressive strain is ε =ρgL/E. The local differential of compressed length is (1 - e)dL. The local differential length shortening is εdL = (ρgL/E)dL. This agrees with the last part of your equation. So your overall final equation is correct.

5. Nov 25, 2012

### Siune

Thank you Chestermiller. That local stress were "new concept" for me, but makes sense.

Last edited: Nov 25, 2012