A steel railroad track (Es=200 GPA, as=11.7x10-6/oC) was laid out at a temperatu

Question

A steel railroad track (Es=200 GPA, as=11.7x10-6/oC) was laid out at a temperature of 6/oC. Determine the normal stress in the rails when the temperature reaches 48oC assuming that the rails (a) are welded to form a continuous track (b) are 10m long with 3 mm gaps between them.



I found (a). s=98.28MPa ...That one was easy but I can't figure out how to solve for (b) without the Area. I've tried to manipulate the problem to cancel out the Area since I don't have it but then it also cancels out the length so I get the same answer as (a), which isn't correct. I know the answer should be -38.3MPa...

Explanation / Answer

Thermal expansion = (alpha)(temperature change)(length)

= (11.7x10-6 /deg.C)(48-6 deg.C)(10m) = 4.914mm

Thermal Expansion = Gap + Stress Expansion

4.914mm = 3mm + (PL/AE)

normal stress = P/A (It is actually (-Normal Stress) because the rails are in compression)

4.914mm = 3mm + (P/A)(L/E)

4.914mm = 3mm + (-Normal Stress MPa)(10m / 200,000 MPa)

Normal Stress = -38.28 MPa

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