Chapter 3 Review

Question 1

A brass rod of length L = 0.75 m is twisted by torques T until the angle of rotation between the ends of the rod is 3.5°. The allowable shear strain in the copper is 0.0005 rad.

Determine The maximum permissible diameter of the rod:

Answer 1

We will leverage the Shear Strain formula to solve this problem: γ = (d/2)Φ/L

d = 2(0.75 m)(0.0005 rad) / (0.061 rad)

d = 12.8 mm

Question 2

A brass bar twisted by torques T acting at the ends has the following properties:
L = 2.1 m
d =38 mm
G = 41 GPa

Determine the torsional stiffness of the bar.

Answer 2

We can leverage the Stiffness Equation: k = GＩₚ/ L

k = (41x10⁹ Pa)(π/32)(38x10⁻³ m) / (2.1 m)

k = 3667 N M