Question

an aluminum alloy bar fixed at its both ends is heated through 20k. find the stress developed in the bar. take modulus of elasticity and coefficient of linear expansion for the bar is 80 Gpa and 24x10^-6/k respectively.

Answer 1

Answer:

19.2 MPa

Explanation:

∆T = 20 K

Y = 80 GPa = 80×10^9 Pa

alpha = 24×10(-6) /K

thermal strain = alpha × ∆T

stress = Y × thermal strain

= 80×10^9 × 24×10(-6) × 20

= 19.2 × 10^5

= 19.2 MPa

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