A bar having a length of 5 in. and cross-sectional area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear elastic behavior.

The modulus of elasticity is 28.6 X 10³ ksi

Explanation:

Given -

Length, l = 5in

Force, P = 8000lb

Area, A = 0.7in²

δ = 0.002in

Modulus of elasticity, E = ?

We know,

Modulus of elasticity, E = σ / ε

Where,

σ is normal stress

ε is normal strain

Normal stress can be calculated as:

σ = P/A

Where,

P is the force applied

A is the area of cross-section

By plugging in the values, we get

σ =

σ = 11.43ksi

To calculate the normal strain we use the formula,

ε = δ / L

By plugging in the values we get,

ε =

ε = 0.0004 in/in

Therefore, modulus of elasticity would be:

Thus, modulus of elasticity is 28.6 X 10³ ksi

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