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Thermal Stress on Two Rods 
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#1
Jan2508, 12:18 AM

P: 54

1. The problem statement, all variables and given/known data
A steel rod 0.350 m long and an aluminum rod 0.250 m long, both with the same diameter, are placed end to end between rigid supports with no initial stress in the rods. The temperature of the rods is now raised by 60.0 degrees C. What is the stress in each rod? 2. Relevant equations Thermal Stress equation: F/A = Y(([tex]\Delta[/tex]L / L)  ([tex]\alpha[/tex] [tex]\Delta[/tex] T)) 3. The attempt at a solution Since the two rods are in equilibrium when they are finished expanding, the net force acting where the they touch must be zero; the two contact forces are equal. Using the thermal stress equation, I found the force for both rods and set them equal to each other. Then I solved for the change in length and substituted it back into the equation to find the stress. For some reason I'm not getting a good answer. 


#2
Jan2508, 04:35 AM

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P: 1,979

A simple way to think would be to let a rod expand, and then bring it back to its original size by applying compression. Can you do that?



#3
Jan2508, 04:06 PM

P: 54

Yes. That's essentially where the thermal stress equation comes from. However, since the two rods do change their length, I think you have find the change in length before you find the thermal stress on either of them. To find the change in length, you set the two forces equal to each other and solve for it....I think. But then again I'm not sure because I keep getting a wrong answer. 


#4
Jan2508, 05:35 PM

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P: 1,979

Thermal Stress on Two Rods
Due to change in temp 't', the rod of length l changes to (l+delta_l), and we know the eqn relating delta_l to alpha.
Now, due to compressive force F, the rod of length (l+delta_l) changes to l, and we know the eqn relating delta_l to F, Y etc. Eliminate delta_l to get the final F/A. (There's no subtraction of terms.) 


#5
Jan2608, 04:17 AM

P: 54

All I want to know is if I'm approaching this problem the wrong way. It doesn't seem to me like there are many alternative approaches.



#6
Jan2608, 06:37 AM

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P: 1,979

Yes, somewhat wrongly.
Which portion of post #4 do you not understand? To start you on your way, the first eqn would be: delta_l = atl, where a is alpha, t = change in temp and l is the length. For the next part, write the strain delta_l/l for compressing a rod of length l by delta_l. Equate the two, as I've said before. (Note: since delta_l is very small, we can neglect it's square. That's why in the 2nd case too, we'll take take the length to be l.) Your correct ans for one rod would be F/A = atY. 


#7
Jan2608, 03:57 PM

P: 54

The equation you derived is assuming that the length of each rod does NOT change. This is not true. The length of each rod does change, but the total length remains the same. This is why the problem is difficult....the rods each change in length, but less than they would if they were free to expand. The only way I found to solve for delta l is to use the fact that the sum of the forces at the junction between the two rods is 0 and equate them. Once you have delta l, you can find the stress on each, but it's not working for some reason. 


#8
Jan2708, 08:11 AM

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P: 1,979

The force F due to stress is constant throughout the rod. That is one point. The other information to use is that the sum of the lengths of the rods are same. The crosssectional area of the rods will change due to thermal expansion. Initially both were A, and at the final temp, let them be A1 and A2. So, total expansion of one rod = total contraction of the other rod. Total expansion of 1st rod = thermal expansion  contraction due to stress = a1*t*L1  F*L1/(A1*Y1). Same for the 2nd rod. Then, a1*t*L1  F*L1/(A1*Y1) = F*L2/(A2*Y2)  a2*t*L2. You can now solve for F and find F/A1 and F/A2. 


#9
Jan2708, 05:01 PM

P: 54

Interesting. I used exactly the same information as you, but I proposed a different method in which you solve for the change in length by equating the two forces, rather than vice versa. I think I must be making a mathematical mistake or something, because both ways should work. 


#10
Jan2708, 05:43 PM

P: 54

Got it. Thanks.
Yeah....I have no clue why my way wasn't working. Logically, the problem could be solved by equating the forces and then solving for change in length rather than equating the change in length and then solving for forces. I think it's probably a little messier the way I was doing it, and I probably kept making sign errors or something. 


#11
Jan1610, 12:20 AM

P: 1

Don't we need to have A first to find F?



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