# Mechanics of materials

1. Jun 2, 2008

### Ssantos

I have a problem understanding what is going wrong with a simple derivation for strain energy.

Problem: Assume a prismatic bar of length L where the elastic region is assumed. If it is hanging vertically its own weight produces some strain. If all the work goes into strain energy the derivation in Gere's book (and everywhere else) for any load on a bar is:

U= $$\int$$P(x)^2dx/2EA Equation 1

Where U is the Strain energy, P(x) is the load as a function of x (where say x is the vertical axis in which teh bar is hanging), E is teh elastic modulus and A is the area of the bar.

Now, for the case of the load being the weight my procedure (that doesn't work) is teh following:

P(x)= yA(L-x) Equation 2

where P(x) is teh weight of the bar at a height x. y is the weight density. L is the length of teh bar and L-x is the length of teh bar contributing to the stran in an elementary volume at x.

Now if U= Work done by the load. We can write

W=$$\int$$P(x)ds Equation 3

where s is the displacement or deformation in the x direction.

If we introduce Eq 3 into 2 and chnage ds by dx from the following equation:

ds= dx/EA P(x) Equation 4

, we get

U= y^2AL^2/2E instead of 6E as it should.

My question. That has to do with Equation 4 right?

2. Jun 3, 2008

### Andy Resnick

I think so- your equation 4 is inappropriate. Why don't you just directly integrate equation 1, after substituting equation 2?

I get: U =y^2 A L^3/6E

3. Jun 3, 2008

### Ssantos

Thanks Andy. It just looks very tricky since they obatin the work for an element from:

W=Integral (Pds)

where first they use the fact that P=f(s) from Hooke's law, thus:

P=ks

Then they integrate and get the usual W=1/2ks^2.

The key here is the fact that insetad of leaving the formula as above they use P=ks again so: W= 1/2Ps instead of W= Ps as it would be if P was "constant". The big problem here I think is not mathematical but experimental since in fact when you apply a load to a "bar" the load is constant, that is P, so W=Integral (Pds) = PIntegral(ds). However, and I think that is the trick, correct me if Im worng, that last step would not account for the fact that P "has had to change" in the process of deformation , that is the load has to be applied slowly, then of course P would not be constant but a function of deformation, namely Hooke's law.

Am I right to argue from an experimental point of view?

Again, you are right, if taking P as avriable throughout then your answer is right, and the one shown in books. Otherwise I get the U= y^2AL^3/2E (it was a 3, sorry) as I said.

4. Jun 3, 2008

### Ssantos

Finally, Equation 4 above is, again, the equation obtained for a thin element not accounting for the slow deformation. But still I see it as a very tricky detail and I don't know to what extent it's just an experimental fact.

5. Jun 3, 2008

### Andy Resnick

I'm a little confused, but if I understand what your wrote, performing the calculation one way gives W = 0.5Ps while the other way gives W = Ps. But, this is actually not a contradiction- note that you wrote W= Ps if P is a constant. It is, if P = f(s) is replaced by the average value, which is 0.5ks.

Part of the confusion could be in differentiating between a constant load- pulling on a rope provides a constant tension along the length of the rope- and a non-constant load, in your case hanging a bar.

Or do I misunderstand?

6. Jun 4, 2008

### Ssantos

Thank again Andy.

Well, the fact is that the definition of work is Force dot product differential of displacement. The Force is, if not applied slowly and modifying it, just what you hold onto the bar, namely F or the whole load. However, if applied fast, as it would normally be if you put a weight on something, some of the work would go into increasing the kinetic energy of the "bar" and some into the Potential Energy of the bar, that is, the strain energy of the bar. However, if the load is applied slowly, so no "work" is done towards increasing the kinetic energy, then the force indeed does just the work going into strain energy. That is what I see as tricky and what I think led me to confusion. My point, I guess, is that for a force to do work all it requires is that the object moves in the direction of the force. And if you just happen to hang an object onto a bar, the weight of the object is constant so the work it does downwards is Ws where W is the weight and s is the displacement dowwards. I might be wrong though, but I think the whole thing comes down to what I say above.

Please let me know if you think otherwise since Im very interested in the theory rather than the exercise itself.

Regards.

Last edited: Jun 4, 2008
7. Jun 4, 2008

### Ssantos

One last thing arguing in favour of what I said above. If you hang a weight on a bar the work that goes into increasing the kinetic energy would set vibrations on the bar. And if the bar was to be in a vacumm, so no damping ocurred, and provided the elastic limit is not surpased, so no energy is dissipated as heat or plastic deformation, the extra work done by the weight would stay in the system (the bar load) as it happens with a spring when no damping is taken into account.

8. Jun 5, 2008

### Andy Resnick

The issue you raise involves transitioning from kinematics to dynamics. Your definition of work is time-independent; how fast a load is applied is meaningless in that formalism.

I can picture a time-dependent load initiating all kinds of waves within the bar- compression and shear. As another example, when removing the brace from a beam balance, the balance executes oscillations; these (on mine) are damped out very slowly.

9. Jun 5, 2008

### Ssantos

Thanks again. I think I more or less see what I was doing wrong (or thinking wrong). I will think a bit more about it until I get a better understanding. Thank you though!