A 12-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in length of the wire must not exceed 25 mm.

The smallest wire diameter that can be used is 1 cm

Explanation:

First, we find the smallest diameter using the criterion of maximum normal stress:

Max. Stress = 150 x 10^6 Pa = F/A

150 x 10^6 Pa = 12000 N/(πd²/4)

d² = (12000 N)(4)/(150 x 10^6 Pa)(π)

d = √1.0185 x 10^-4 m²

d = 0.010 m = 1 cm

Now, we find the smallest diameter using the criterion of maximum strain:

Max. Strain = Max. Change in Length/Original Length = 0.025 m/50 m

Max. Strain = 5 x 10^-4 mm/mm

Now,

Max. Strain = Stress/E = (F/A)/E = F/AE

using values:

5 x 10^-4 mm/mm = (12000 N)/(200 x 10^9 Pa)(πd²/4)

d =√(12000 N)(4)/(5 x 0^-4)(200 x 10^9 Pa)(π)

d = 0.012 m = 1.2 cm

Now, by comparison in both cases it can be noted that the smallest value of the diameter is 1 cm, which is limited by maximum stress.

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: