a) determine the principal stresses at point B

b)determine the magnitude of the maximum in-plane shear stress at point __B__

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\begin{array}{ll}d=190 \mathrm{~mm} & A=\frac{\pi}{4}(190)^{2} \\F_{a}=12.5 \mathrm{kN} & I=\frac{\pi}{64}(190)^{4} \\M=1.2 k \mathrm{~N}-\mathrm{m} & J=\frac{\pi}{32}(190)^{4}\end{array}Shaft Cross-sectionBending moment M=1.2 \mathrm{kN}-\mathrm{m}, acts at A and does not act at 'B', as can be seen from the obove diagramAt B\text { Axial stress }=\sigma_{a}=\frac{F_{a}}{A}=\frac{12.5 \times 10^{3} \mathrm{~N}}{\frac{\pi}{a} \times 190^{2}}=-0.441 \mathrm{MPa}Due to torsional moment, twisting Shear stress geverated at B\tau=\frac{T(d / 2)}{J}=\frac{2.25 \times 10^{3} \mathrm{~N}(0.19 \mathrm{~m}) \times 10^{6} \mathrm{MPa}}{\frac{\pi}{32}(190)^{4} \mathrm{~mm}^{4}}\tau=3.34 \mathrm{MPa}Principal stress formula\begin{aligned}\sigma_{1,2} & =\frac{1}{2}\left[\sigma_{x}+\sigma_{y} \pm \sqrt{\left(\sigma_{x}-\sigma_{y}\right)^{2}+4 \tau^{2}}\right] \\& =\frac{1}{2}\left[-0.441+0 \pm \sqrt{(-0.441-0)^{2}+4(3.34)^{2}}\right]\end{aligned}(a)\begin{array}{l}\sigma_{1}=3.128 \mathrm{mPa} \text { and } \\\sigma_{2}=-3.57 \mathrm{MPa}\end{array}(B) Maximum in-plane shear stress at B\begin{aligned}\tau_{\text {max }} & =\left|\frac{\sigma_{1}-\sigma_{2}}{2}\right| \\& =\left|\frac{3.128-(-3.57)}{2}\right| \\\tau_{\text {max }} & =3.348 \mathrm{MPa}\end{aligned}Mons circle ...