1. **Problem Statement:** A steel I-section has flange dimensions 200 mm by 12 mm and web dimensions 376 mm by 8 mm. It is subjected to a pure torque. The maximum shear stress allowed is 100 N/mm². We need to find the torque capacity of the cross-section and the location of the maximum shear stress. 2. **Relevant Formulas and Concepts:** - The torque capacity $T$ is related to the maximum shear stress $\tau_{max}$ by the formula: $$\tau_{max} = \frac{T c}{J}$$ where: - $c$ is the maximum distance from the center to the outer fiber (outermost point in the cross-section), - $J$ is the torsional constant (torsion constant) of the cross-section. - Rearranged to find torque: $$T = \frac{\tau_{max} J}{c}$$ - For thin-walled open sections like an I-beam, the torsional constant $J$ can be approximated by: $$J = 4 A_m^2 / \sum \left( \frac{l}{t} \right)$$ where: - $A_m$ is the median area enclosed by the median line of the cross-section walls, - $l$ is the length of each wall segment, - $t$ is the thickness of each wall segment. 3. **Calculate Median Dimensions:** - Flange median thickness line is at half thickness: 12 mm / 2 = 6 mm - Web median thickness line is at half thickness: 8 mm / 2 = 4 mm 4. **Calculate Median Perimeter and Area $A_m$:** - Flange length (median line): 200 mm (length) - Web length (median line): 376 mm (height) - Median perimeter $P_m$ = 2 (flanges) + 1 (web) = $2 \times 200 + 376 = 776$ mm - Median enclosed area $A_m$ is the area enclosed by the median lines: $$A_m = 200 \times 8 + 376 \times 4 = 1600 + 1504 = 3104 \text{ mm}^2$$ 5. **Calculate $\sum \left( \frac{l}{t} \right)$:** - For two flanges: each $l=200$ mm, $t=12$ mm - For web: $l=376$ mm, $t=8$ mm $$\sum \left( \frac{l}{t} \right) = 2 \times \frac{200}{12} + \frac{376}{8} = 2 \times 16.67 + 47 = 33.33 + 47 = 80.33$$ 6. **Calculate Torsional Constant $J$:** $$J = \frac{4 \times (3104)^2}{80.33} = \frac{4 \times 9634816}{80.33} = \frac{38539264}{80.33} \approx 479840 \text{ mm}^4$$ 7. **Calculate Maximum Distance $c$:** - The maximum distance from the centroid to the outer fiber is half the total height: $$c = \frac{376 + 12}{2} = \frac{388}{2} = 194 \text{ mm}$$ 8. **Calculate Torque Capacity $T$:** $$T = \frac{\tau_{max} J}{c} = \frac{100 \times 479840}{194} = \frac{47984000}{194} \approx 247312 \text{ Nmm} = 247.3 \text{ kNmm}$$ 9. **Location of Maximum Shear Stress:** - For thin-walled open sections under torque, the maximum shear stress occurs at the point farthest from the centroid, i.e., at the outer fibers of the flanges or web. - Here, the maximum shear stress location is at the outer edge of the flange or web, at distance $c = 194$ mm from the centroid. **Final Answers:** - Torque capacity $T \approx 247312$ Nmm - Maximum shear stress location is at $c = 194$ mm from the centroid, at the outer fibers of the cross-section.