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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?

Solution Preview

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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