A 9-m length of 6-mm-diameter steel wire is to be used in a hanger. The wire stretches 18mm when a tensile force P is applied. If E = 200 GPa, determine the magnitude of the force P, and the normal stress in the wire.

The magnitude of the force is 11.3KN and the normal stress is 400 MPa

Given that length (L) = 9 m, diameter (d) = 6 mm = 6 * 10⁻³ m, extension (δ) = 18 mm = 18 * 10⁻³ m, E = 200 GPa = 200 * 10⁹ Pa

The area of the wire (A) is:

The magnitude of the force is 11.3KN and the normal stress is 400 MPa

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Force P is 11304 N and normal stress is 400 N/mm²

Explanation:

Given-

Length, l = 9 m = 9000 mm

Diameter, d = 6 mm

Radius, r = 3 mm

Stretched length, Δl= 18 mm

Modulus of elasticity, E = 200 GPa = 200 X 10³MPa

Force, P = ?

According to Hooke's law,

Stress is directly proportional to strain.

So,

σ ∝ ε

σ = E ε

Where, E is the modulus of elasticity

We know,

ε = Δl / l

So,

σ = E X Δl/l

σ =

We know,

σ = P/A

And A = π (r)²

σ = P / π (r)²

Therefore, Force P is 11304 N and normal stress is 400 N/mm²

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answer:

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explanation: