Question 4.SP.2: A cast-iron machine part is acted upon by the 3 kN · m coupl...
A cast-iron machine part is acted upon by the 3 \mathrm{kN} \cdot \mathrm{m} couple shown. Knowing that E=165 \mathrm{GPa} and neglecting the effect of fillets, determine (a) the maximum tensile and compressive stresses in the casting, (b) the radius of curvature of the casting.
Centroid. We divide the T-shaped cross section into the two rectangles shown and write
Centroidal Moment of Inertia. The parallel-axis theorem is used to determine the moment of inertia of each rectangle with respect to the axis x^{\prime} that passes through the centroid of the composite section. Adding the moments of inertia of the rectangles, we write
\begin{aligned} I_{x^{\prime}} & =\Sigma\left(\bar{I}+A d^{2}\right)=\sum\left(\frac{1}{12} b h^{3}+A d^{2}\right) \\ & =\frac{1}{12}(90)(20)^{3}+(90 \times 20)(12)^{2}+\frac{1}{12}(30)(40)^{3}+(30 \times 40)(18)^{2} \\ & =868 \times 10^{3} \mathrm{~mm}^{4} \\ I & =868 \times 10^{-9} \mathrm{~m}^{4} \end{aligned}a. Maximum Tensile Stress. Since the applied couple bends the casting downward, the center of curvature is located below the cross section. The maximum tensile stress occurs at point A, which is farthest from the center of curvature.
\sigma_{A}=\frac{M c_{A}}{I}=\frac{(3 \mathrm{kN} \cdot \mathrm{m})(0.022 \mathrm{~m})}{868 \times 10^{-9} \mathrm{~m}^{4}} \quad \sigma_{A}=+76.0 \mathrm{MPa}
Maximum Compressive Stress. This occurs at point B; we have
\sigma_{B}=-\frac{M c_{B}}{I}=-\frac{(3 \mathrm{kN} \cdot \mathrm{m})(0.038 \mathrm{~m})}{868 \times 10^{-9} \mathrm{~m}^{4}} \quad \sigma_{B}=-131.3 \mathrm{MPa}
b. Radius of Curvature. From Eq. (4.21), we have
\begin{aligned} \frac{1}{\rho} & =\frac{M}{E I}=\frac{3 \mathrm{kN} \cdot \mathrm{m}}{(165 \mathrm{GPa})\left(868 \times 10^{-9} \mathrm{~m}^{4}\right)} \\ & =20.95 \times 10^{-3} \mathrm{~m}^{-1} \quad \rho=47.7 \mathrm{~m} \end{aligned}
