# Mechanical Systems Depth Question 520



## GTPE2B (Apr 22, 2009)

Can someone explain this one to me? I can follow their logic through the problem. However, I don't understand why you can't calculate the thermal deflection and then, using that x, calculate force as -kx. The answer is close, but not close enough to chalk up to rounding or significant digits. I'm probably missing something obvious due the frazzled state of my brain at the moment. Please enlighten me.


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## bph (Apr 22, 2009)

GTPE2B said:


> Can someone explain this one to me? I can follow their logic through the problem. However, I don't understand why you can't calculate the thermal deflection and then, using that x, calculate force as -kx. The answer is close, but not close enough to chalk up to rounding or significant digits. I'm probably missing something obvious due the frazzled state of my brain at the moment. Please enlighten me.


I solved it by a slightly different method, but the results are the same 628 lbs, and the theory is the same.

The key is that the stiffness of the spring is high, and on the order of the stiffness of the bar, so you effectively have two springs in series which you must account for.

The combined spring constant is calculated by adding the springs in series K overall = (1/K1 + 1/k2)^-1

Where k1 is the spring constant of the spring and k2 is the spring constant of the rod = (A * E)/L

Since the force is the same on both springs, and you know the total deflection of both springs together is just the thermal expansion of the rod, you can then get:

F= KX, where X is the thermal expansion, and K is the combined spring constant.

So,

F = a * L * delta T * (1/K1 + 1/k2)^-1

The result is 628 lbs.

Were a is the coef of thermal expansion

Does that make sense?

I noticed that even if you did not pick this up, you would still likely guess correctly.

BPH


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## GTPE2B (Apr 23, 2009)

bph said:


> I solved it by a slightly different method, but the results are the same 628 lbs, and the theory is the same.The key is that the stiffness of the spring is high, and on the order of the stiffness of the bar, so you effectively have two springs in series which you must account for.
> 
> The combined spring constant is calculated by adding the springs in series K overall = (1/K1 + 1/k2)^-1
> 
> ...


BPH, thanks for the reply. I think I understand where I went wrong. I didn't consider the bar as a spring and assumed that it would expand to the full length indicated by the thermal expansion calculation alone. However, after reading your reply and looking up further information in the MERM, I see that if the bar is restrained in anyway, it will develop internal stresses just as if you tried to compress it.


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## bph (Apr 23, 2009)

GTPE2B said:


> ...I see that if the bar is restrained in anyway, it will develop internal stresses just as if you tried to compress it.


Yes, the key point is the bar IS a spring. No different than the coils spring, just a linear spring. You have to account for both springs. Don't get confused with "internal pressure", it's just another spring.

BPH


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## Fitz (Apr 3, 2011)

I see how you are explaining above how you treated the problem as two spring however agreeing with the OP I don't follow the logic to the given solution in the back of the practice test. I think I understand most of it but on thing that is really throwing me off is why they are using PI to calculate the area when it is given as .375in^2 in the problem statement. Because of this, I am coming up with a slightly different answer. I want to make sure I am not missing something.


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