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A Question and solution concerning linear thermal expansion

1) (a) If a rod of 1025 steel 19.7 in. long is heated from 68 to 176ï?°F while its ends remain fixed, determine the type and magnitude of stress that develops. Assume that at 68ï?°F the rod is stress free. (b) What will be the stress magnitude if a rod 39.4 in. long is used? (c) If the rod in part (a) is cooled from 68 to 14ï?°F, what type and magnitude of stress will result? (d) What is the magnitude of the stress in part (a) if only one end of the rod is fixed?

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See attached file.

Use the definition of strain in terms of linear change in length ΔL over original length L

Strain = Delta(L)/L (1)

Change in length of a material due to temperature change Delta(T) is given by

Delta(L) = Alpha*L*Delta(T) (2)

where Alpha is the coefficient of linear expansion for the material in question

Thus we can say that the strain can be given by combining (1) & (2) thus

Strain = Dleta(L)/L = Alpha*Delta(T) (3)

Young Modulus Y for a material relates the Stress (Sigma) to the Strain in the elastic regime for the material

Thus

Y = Stress/Strain = Sigma/Strain (4)

Combining (4) with (3) we get a relationship between Stress in a material and other ...

Solution Summary

The thermal linear expansion in a material (1025 steel) is considered and a formula for the linear stress developed from the formulas for strain, Youngs Modulus and linear expansion for the material. The stress in a steel bar is determined from this analysis

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