10:04
Vㅏㄴㄹㅛ \( 4 G+ \)
\( 44 \% \)
Given state of soreso
\[
\begin{array}{l}
\sigma_{x}=41.37 \mathrm{mPa} \\
\sigma_{y}=6.895 \mathrm{mP} \\
\tau_{y y}=41.37 \mathrm{mPa}
\end{array}
\]
Principa stresser is given by.
\[
\begin{array}{l}
\sigma_{1,2}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-r_{y}}{2}\right)^{2}+r_{x y}^{2}} \\
\sigma_{1,2}=\frac{41.37+6.895}{2} \pm \sqrt{\left(\frac{41.37-6.895}{2}\right)^{2}+(41.37)^{2}} \\
\sigma_{1,2}=24.1325 \pm 44.8175 \\
\sigma_{1}=68.95 \mathrm{MPa}, \sigma_{2}=-20.685 \mathrm{MP},
\end{array}
\]
from Teste we get,
\( \sigma_{y p} \) for 1045 steel, \( \sigma_{y p}=310 \mathrm{MP} \)
\( \sigma_{y p} \) for lass \( 30 \mathrm{cc}, \sigma_{y p}=130 \mathrm{mPa} \)
(a) By Maximum shear theors
\[
\begin{aligned}
& \frac{\sigma_{1}-\sigma_{2}}{2} \leq \frac{\sigma_{y p}}{2\left(H_{j}\right)} \\
\Rightarrow & H_{f_{3}}=\frac{310.2}{68.95-(-20.685) .2} \\
\Rightarrow & H_{f_{3}}=3.45
\end{aligned}
\]