A 120 mm x 120 mm x 40 mm thick sheet of copper is stretched biaxially as shown.

Question

A 120 mm x 120 mm x 40 mm thick sheet of copper is stretched biaxially as shown. If the normal stress is 40 MPa in the y direction and the new thickness of the plate in the z direction is 39.98 mm, determine the normal stress in the x direction. Copper has a modulus of elasticity of 110 GPa and a Poisson

A 120 mm x 120 mm x 40 mm thick sheet of copper is stretched biaxially as shown. If the normal stress is 40 MPa in the y direction and the new thickness of the plate in the z direction is 39.98 mm, determine the normal stress in the x direction. Copper has a modulus of elasticity of 110 GPa and a Poisson's ratio of 0.35

Explanation / Answer

strain in z = -0.02/40

strain in z = [ -mew *( stress in x + stress in y)]/E

mew = 0.35

E = 110*10^9 Pa

stress in x + stress in y = 157.143 MPa

stress in x + 40 = 157.43

stress in x = 157.43 -40 = 117.143 MPa .....(ans)

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