A cylindrical rod of nickel is stressed elastically in tension. The original dia

Question

A cylindrical rod of nickel is stressed elastically in tension.

The original diameter was 25 mm and a force of 52 kN was applied.

Compute the diameter of the rod upon loading, in millimeters.

Answer Format: X.XXX

Unit: mm

Values Are For Room Temperature, 1 atm Ambient Pressure, and Annealed Microstructures CUnless Noted Density g/m 87 .85 8.00 Material Elastic Modulus, GPa Yield Point, MPa Ultimate Tensile Strength, Mpa Poisson's Ratio Tensile Strain at Failure S 415 290 17.7 50 4140 1020 0.29 316 Stairless Steel Nckel Tungsten 193 207 407 580 317 960 0.27 0.31 0.28 7075 Aluminum-T6 T 8.88 19.3 760

Explanation / Answer

Area of cross section A = 0.7855 x (25) ^2 = 490.9375 mm2

Load P = 52000 N

Given elastic modulus E = 200 GPa = 200 x 10^3 N/mm2

Stress S = P/A = 52000/490.9375 = 105.92 N/mm2 or MPa < Sy= 130 MPa

Hence the material loading is within the elastic limit.

Strain e = S/E = 105.92/(200 x 10^3 ) = 5.3 x 10 ^ -4

We know the poison ration v = Transverse strain / Longitudinal strain = et / e

So, 0.31 = et / 5.3 x 10 ^ -4

et = 0.31 x 5.3 x 10 ^ -4 = 1.642 x 10 ^ -4

but, d / D = et = change in dia / original dia

d = D x 1.642 x 10 ^ -4

d = 25 x 1.642 x 10 ^ -4

d = 0.0041 mm

new dia D’ = D-d = 25-0.0041 = 24.9959 mm

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