A pipe is subjected to a tension force of P = 50.00 kN. The pipe outside diameter is 32.00 mm, the wall thickness is 6.00 mm, and the elastic modulus is E = 150.00 GPa. Determine the normal strain in the pipe. Express the strain in mm/mm.

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hamzaahmeds hamzaahmeds · Answer 1 · 2020-02-16T15:41:29+01:00

Answer:

Strain = 6.8 x 10^-4 mm/mm

Explanation:

First we need to calculate normal stress on the pipe. For that we have formula:

Normal Stress = σ = F/A

where,

F = Normal Force = 50 KN = 50000 N

A = Area normal to the force = Area of outer circle - Area of inner circle = Ao - Ai

Now, we find outer area and inner area:

Ao = π (outside radius)²

Ao = π (0.032m/2)²

Ao = 8.04 x 10^-4 m²

Ai = π (inside radius)²

Ai = π (0.032m/2 - 0.006 m)²

Ai = 3.14 x 10^-4 m²

Now, the Area normal to force is calculated as:

A = Ao - Ai = 8.04 x 10^-4 m² - 3.14 x 10^-4 m²

A = 4.9 x 10^-4 m²

Now the stress can be calculated by given formula:

σ = 50000 N/4.9 x 10^-4 m²

σ = 0.102 GPa

Now, for the strain we have formula:

Elastic Modulus = E = σ/strain

Strain = σ/E

Strain = (0.102 GPa)/(150 GPa)

Strain = 6.8 x 10^-4 mm/mm