The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

Given Information:

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:

Change in length = ?

Change in diameter = ?

Answer:

Change in length = 0.01257 mm

Change in diameter = - 0.003772 mm

Explanation:

Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π (0.015) ²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = - vₚ*εlong

εlat = - 0.4*0.0006288

εlat = - 0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = - 0.0002515*15

Δd = - 0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is - 0.003772 mm