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Four rods A, B, C, D of same length and material but of different radii r, 2r , 3r and 4r respectively are held between two rigid walls. The temperature of all rods is increased by same amount. If the rods do not bend, then which of these are correct:

  1. The stress in the rods are in the ratio 1 : 2 : 3 : 4.
  2. The force on the rod exerted by the wall are in the ratio 1 : 2 : 3 : 4.
  3. The energy stored in the rods due to elasticity are in the ratio 1 : 2 : 3 : 4.
  4. The strains produced in the rods are in the ratio 1 : 2 : 3 : 4.

Four rods A, B, C, D of same length and material => Same Youngs Modulus, Same coefficient of linear expansion, Same Length.

Also, The temperature of all rods is increased by same amount.

Before answering the above question I've few other questions:

  1. Suppose rods were not held between two rigid walls. Then there would have been change in length. In that case, would there be stress? Intuitively it feels like stress would be zero, as there seems to be no restoring forces developed. But Stress = Youngs modulus * strain. Strain is definitely not zero. So, stress should not be zero. Confused!

  2. No say, rods were held between two rigid walls. Assuming rods are not bending. There length will not change. So, strain would be zero. Stress = Youngs modulus * strain. So, Stress must be zero. But intuitively it seems there will be stress, because there will be restoring forces in the rod pushing walls away. Again confused!

Now coming back to the original problem. The above two confusions are causing trouble. But just going by intuition. There will be stress developed even though there is no strain. But stress = Restoring Force/Area. Here areas for all rods are different pi*r^2, because r is different. But Restoring force is same. So, the ratio must be 1/1 : 1/4 : 1/9 : 1/16. Right?

Surprisingly answers are 3,4. There is lot of confusion. Kindly clarify

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I love this question. It shows that the OP is putting thought into his formulas, not just repeating algorithms. – André Neves May 19 '14 at 1:40
up vote 5 down vote accepted

Your confusion lies within your perception of natural length in the Young's modulus formula. When we say strain=$\Delta L/L$, the $L$ refers to the natural length of the rod at a given temperature. So, if the rod is not clamped, and we increase the temperature, there is no deviation from natural length at that temperature (as we can define natural length of a rod at a temperature by calling it "the length of the rod at that temperature in the absence of any other influences"), so strain is zero. Stress is obviously zero.

If the rod is clamped, its length stays $L_0$, but it's natural length becomes $L_0(1+\alpha\Delta T)$, so the $\Delta L$ comes from the fact that its natural length has changed but its length is constant.

Summing up, in your Young's modulus formula, use strain=$\Delta L_T/L_T$, where $\Delta L_T$ is $|L_0-L_T|$, $L_T=L_0(1+\alpha\Delta T)$, and $L_0$ is length at a reference temperature.

Use this to solve the problem now.

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In the equation $ strain = {\Delta L \over L} $, you can choose what initial configuration to use. If you use the initial configuration stated in the problem, then $ \Delta L $ is zero so strain is zero for all the rods.

In the equation $$ stress = E \times strain $$ strain means mechanical strain which is not the same as the strain defined above (total strain).

$$ total \space strain = mechanical \space strain + thermal \space strain $$

so the correct equation to use for stress is

$$ stress = E \times (total \space strain - thermal \space strain) $$ $$ = E \times ({\Delta L \over L} - \alpha \Delta T) $$

Answer 1 is wrong because they all have the same stress (area doesn't appear in the above equation)

Answer 2 is wrong because the force scales with $r^2$, not $r$, as you pointed out.

Answer 3 is wrong because strain energy is proportional to $r^2$, not $r$ when length, stress, and strain are constant, which they are.

Answer 4 is wrong regardless of how you define strain. All rods have the same same total strain (0), the same thermal strain ($ \alpha \Delta T $) and the same mechanical strain (-$ \alpha \Delta T $)


EDIT: If you define the initial configuration as the rods being heated and unclamped then all the above equations still apply and all 4 answers are still wrong. The only difference is the values of the 3 kinds of strain.

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Doesn't OP say 3, 4 are right? So shouldn't that mean something is wrong with your assumptions here? – Kyle Kanos Jan 1 at 11:57
Can you see any assumption that might be wrong? I've allowed several interpretations of the type of strain and the initial configuration but none lead to 3 and 4. The author of the question probably made a mistake like swapping "radii" and "length". – user1318499 Jan 1 at 13:02
I didn't really look at your assumptions (and still haven't). What I did look at was the fact that OP says what the answers should be and the accepted answer seems to lead to a different conclusion than your own. – Kyle Kanos Jan 1 at 15:59
The accepted answer leads to the same conclusion as mine - that all 4 answers are wrong. It's a special case of my answer where you use a different initial configuration than the one implied by the problem. It seems to have a sign errors though (positive strain instead of negative) but I'm not confident enough to correct it. – user1318499 Jan 2 at 2:31

protected by Qmechanic Dec 24 '13 at 9:17

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