Question

An 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine (a) whether the wire will plastically deform; and (b) whether the wire will experience necking.

Answer 1

Answer:

a)Yes will deform plastically

b) Will NOT experience necking

Explanation:

Given:

- Applied Force F = 850 lb

- Diameter of wire D = 0.15 in

- Yield Strength Y=45,000 psi

- Ultimate Tensile strength U = 55,000 psi

Find:

a) Whether there will be plastic deformation

b) Whether there will be necking.

Solution:

Assuming a constant Force F, the stress in the wire will be:

                       stress = F / Area

                       Area = pi*D^2 / 4

                       Area = pi*0.15^2 / 4 = 0.0176715 in^2

                       stress = 850 / 0.0176715

                       stress = 48,100.16 psi

      Yield Strength < Applied stress > Ultimate Tensile strength

                        45,000 < 48,100 < 55,000

Hence, stress applied is greater than Yield strength beyond which the wire will deform plasticly but insufficient enough to reach UTS responsible for the necking to initiate. Hence, wire deforms plastically but does not experience necking.

Answer 2

The wire will deform plastically but will not experience necking.

Stress on the nickel wire

The stress on the nickel wire due to the applied force is calculated as follows;

Stress = Force / Area

Area of the wire = πd² / 4

                           = π x (0.15)² / 4 = 0.0177 in²

Stress = 850 / 0.0177

Stress = 48,022.6 psi

45,000 < 48,022.6 < 55,000

From the data above, the stress of the wire is greater than the yield strength but less than the tensile strength. Thus, the wire will deform plastically but will not experience necking.

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