1. **Problem Statement:** Calculate the maximum shear stresses $\tau_{max}$ at points A and B on the cross section of a cantilevered round peg subjected to a force $F=50000$ N and a torque $T=800$ N\cdot m. The cross section diameter is $d=40.00$ mm and the distance from the wall is $x=10.00$ mm. 2. **Given Data:** - Force, $F=50000$ N - Torque, $T=800$ N\cdot m - Diameter, $d=40.00$ mm = 0.04 m - Radius, $r=\frac{d}{2}=0.02$ m - Distance from wall, $x=10.00$ mm = 0.01 m 3. **Formulas and Concepts:** - Shear stress due to transverse force $F$ at the fixed end of a cantilever beam: $$\tau_F = \frac{VQ}{Ib}$$ where $V$ is the shear force, $Q$ is the first moment of area about the neutral axis, $I$ is the moment of inertia, and $b$ is the width at the point considered. - For a circular cross section, the maximum shear stress due to transverse shear force is: $$\tau_{max,F} = \frac{4F}{3A}$$ where $A=\pi r^2$ is the cross-sectional area. - Shear stress due to torque $T$ (torsion) is: $$\tau_{max,T} = \frac{T r}{J}$$ where $J=\frac{\pi d^4}{32}$ is the polar moment of inertia. - Total shear stress at a point is the sum of shear stresses from force and torque if they act in the same plane. 4. **Calculate cross-sectional area $A$:** $$A = \pi r^2 = \pi (0.02)^2 = \pi \times 0.0004 = 0.00125664\, m^2$$ 5. **Calculate polar moment of inertia $J$:** $$J = \frac{\pi d^4}{32} = \frac{\pi (0.04)^4}{32} = \frac{\pi \times 2.56 \times 10^{-7}}{32} = 2.513 \times 10^{-7}\, m^4$$ 6. **Calculate maximum shear stress due to force $F$:** $$\tau_{max,F} = \frac{4F}{3A} = \frac{4 \times 50000}{3 \times 0.00125664} = \frac{200000}{0.0037699} = 53027800\, Pa = 53.03\, MPa$$ 7. **Calculate maximum shear stress due to torque $T$:** $$\tau_{max,T} = \frac{T r}{J} = \frac{800 \times 0.02}{2.513 \times 10^{-7}} = \frac{16}{2.513 \times 10^{-7}} = 63680000\, Pa = 63.68\, MPa$$ 8. **Determine shear stresses at points A and B:** - Point A is on the outer surface where torsion shear stress is maximum. - Point B is at the center where torsion shear stress is zero. Therefore: - At A: total shear stress $\tau_{max,A} = \tau_{max,F} + \tau_{max,T} = 53.03 + 63.68 = 116.71$ MPa - At B: total shear stress $\tau_{max,B} = \tau_{max,F} + 0 = 53.03$ MPa **Final answers:** $$(\tau_{max})_A = 116.71\, MPa$$ $$(\tau_{max})_B = 53.03\, MPa$$