Answer
$$D_{thread} = 0.564 \space mm$$
$$\sigma_{avg} = 36.3 \space MPa$$
Work Step by Step
Young's Law:
$\sigma = E \epsilon$
1. Substituting known values find tensile stress in string:
$\sigma=3.3*10^9 N/m^2 * 0.011 \space m/m = 36.3*10^6 N/m^2 = \boxed{36.3 \space MPa}$ $\leftarrow ANS (b)$
2. Average tensile Stress:
$\sigma = \frac{P}{A}$
Assuming:
$\bullet$ thread cross section is circular and constant
$\bullet$ transverse contraction is negligible (ignoring Poisson's effect)
$A = \frac{\pi D^2}{4}$
After substituting known values into tensile stress equation and solving for D:
$\ 36.3*10^6 N/m^2= \frac{4 * 8.5 N}{\pi D^2} \Rightarrow D = 5.46*10^{-4} \space m = \boxed{0.546 \space mm}$ $\leftarrow ANS (a)$
