Thermal expansion stress and strain [closed]

I'm having trouble with this question: https://i.imgur.com/rWK0RNI.png

and this is the work i've done so far: https://i.imgur.com/wy0IeTQ.jpg

Am I going along the right lines? It seems its a 3 variable simultaneous equation to solve for the stress but it seems like its too long for the 15 marks so am I missing something?

closed as off-topic by JMac, John Rennie, stafusa, Kyle Kanos, Jon CusterJan 9 '18 at 15:24

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I don't think you need to solve a simultaneous system of equations. You can do this serially. Instead of the process stated, imagine that first the tubes were heated, expanding to three different lengths, and then they were stretched/compressed as much as needed to bring them into alignment. The length of each component after heating is $L_n=L_0+L_0\alpha_n\Delta T$. When you bring all the components into alignment (at length $L$), there is no net force, so you have $\sum A_nE_n(L_n-L)/L_n=0$. This is a single equation in a single unknown ($L$). Once you have solved for $L$ you can find the stress in each component easily enough.

• Reason for downvote? – Ben51 Jan 9 '18 at 14:11
• How did you get A_n E_n (L_n - L) from the first equation? – Johnathon Jan 10 '18 at 17:20
• Area times modulus times strain is force. Sum of the three forces is zero. I notice that I made an error though, and wrote not strain but change in length. I will edit the answer to remedy. – Ben51 Jan 10 '18 at 17:27
• Wouldn't that just give the strain as 0? – Johnathon Jan 11 '18 at 15:37
• I'm not sure I follow. Strain is fractional change in length. The unstressed length of each component is $L_n$. Each is brought to a length $L$. So the strain is $(L_n-L)/L_n$. Is there a problem with this? – Ben51 Jan 11 '18 at 15:41