A cylindrical rod of a metal alloy is stressed elastically in tension. The original diameter was 11 mm and a force of 55 kN was applied. Compute the diameter of the rod upon loading, in millimeters, if the alloy has a modulus of elasticity is 125 GPa and a Poisson's ratio of 0.31. Answer Format: X. XXX Unit: mm

10.984mm

Explanation:

by elastic modulus

stress=modulus of elasticity*strain

stress=loading/area                          area" cross-section"

11mm=0.011m

area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter

stress=55000/(9.503*10^-5)=578.745 MPa

convert MPa and GPa to pascal.

strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463axial strain

v=Poisson  ratio

lateral strain=(-v)*axial strain= -0.31*0.00463

lateral strain= -1.4353*10^-3=change in diameter/ original diameter

change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m

negative indicates decrease in diameter.

decrease in dia.=0.01578mm

new diameter=11-0.01578= 10.984mm

answer:

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explanation: