An aluminum bar $(E=70 \mathrm{GPa}, v=0.33)$ of diameter 20 mm is stretched by axial forces P, causing its diameter to decrease by 0.022 mm.
The load P is approximately:

Question

Quizplus · Accepted Answer

The load P can be calculated using the formula for lateral strain in the context of axial loading and Poisson's ratio. The lateral strain is the change in diameter divided by the original diameter, and Poisson's ratio ($v$) relates the lateral strain to the axial strain. The axial strain can then be related to the stress ($\sigma$) using Hooke's Law ($\sigma = E \cdot \epsilon$), and finally, the stress can be related to the force (P) using the formula $\sigma = \frac{P}{A}$, where $A$ is the cross-sectional area of the bar. Given the decrease in diameter and the properties of aluminum, the calculations lead to a load P of approximately 73 kN.

An Aluminum Bar (E=70GPa,v=0.33)(E=70 \mathrm{GPa}, v=0.33)(E=70GPa,v=0.33) Of Diameter 20 Mm Is Stretched by Axial Forces P

An Aluminum Bar $(E=70 \mathrm{GPa}, v=0.33)$ Of Diameter 20 Mm Is Stretched by Axial Forces P