A bar of diameter d = 18 mm and length L = 0.75 m is loaded in tension by forces P. The bar has modulus E = 45 GPa and allowable
Normal stress of 180 MPa. The elongation of the bar must not exceed
2.7 mm. The allowable value of forces P is approximately:
A) 41 kN
B) 46 kN
C) 56 kN
D) 63 kN

Question

Quizplus · Accepted Answer

The allowable force P can be calculated using the formula for stress ($\sigma = \frac{P}{A}$) and the formula for elongation ($\Delta L = \frac{PL}{AE}$), where $A$ is the cross-sectional area of the bar, $E$ is the modulus of elasticity, $L$ is the length of the bar, and $\Delta L$ is the allowable elongation. Given the allowable stress of 180 MPa and the diameter $d = 18$ mm, the cross-sectional area $A = \frac{\pi d^2}{4}$. Substituting the given values and solving for $P$ yields an allowable force of approximately 41 kN.

A Bar of Diameter D = 18 Mm and Length