The cross-section of a thin-walled beam with uniform wall thickness \(t\), shown in the figure, is subjected to a bending moment \(M_x = 10 \, \text{Nm}\). If \(h = 1 \, \text{m}\) and \(t = 0.001 \, \text{m}\), then the magnitude of the maximum normal stress in the cross-section is _______ N/m\(^2\) (answer as an integer).
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For thin-walled sections, always approximate flange inertia using parallel axis theorem \(\text{Area} \times (\text{Distance})^2 \), since local inertia about centroidal axis of a thin plate is negligible.
Step 1: Identify the section geometry. The cross-section resembles a \(\,T\)-shaped thin-walled section: - Vertical stem: height \(h\), thickness \(t\). - Two horizontal flanges: each of length \(\tfrac{h}{2}\) and thickness \(t\). Given: \[ h = 1.0 \, \text{m}, \quad t = 0.001 \, \text{m} \] Step 2: Locate neutral axis and compute moment of inertia. Since the section is symmetric about the $y$-axis, the neutral axis coincides with the vertical centroidal axis. The bending moment is about the $x$-axis, hence we require the second moment of area about $x$-axis (\(I_x\)). Contribution: 1. Vertical stem: Rectangle dimensions: \(b = t, \; h = h\). \[ I_{x,\text{stem}} = \frac{1}{12} b h^3 = \frac{1}{12}(0.001)(1^3) = 8.33 \times 10^{-5} \, \text{m}^4 \] 2. \underline{Two flanges:} Each flange: \(b = h/2 = 0.5 \, \text{m}, \; h = t = 0.001 \, \text{m}\). Moment of inertia about its own centroidal horizontal axis: \[ I_{x,\text{flange-local}} = \frac{1}{12} b h^3 = \frac{1}{12}(0.5)(0.001^3) \approx 4.17 \times 10^{-11} \, \text{m}^4 \] This is negligible compared to the parallel axis term. Distance of flange centroid from neutral axis (x-axis passes through mid-height of vertical web): \[ d = \frac{h}{2} = 0.5 \, \text{m} \] Area of each flange: \[ A = b \cdot h = (0.5)(0.001) = 0.0005 \, \text{m}^2 \] Parallel axis contribution: \[ I_{x,\text{flange}} \approx A d^2 = (0.0005)(0.5^2) = 1.25 \times 10^{-4} \, \text{m}^4 \] Since there are two flanges: \[ I_{x,\text{flanges total}} = 2(1.25 \times 10^{-4}) = 2.5 \times 10^{-4} \, \text{m}^4 \] Step 3: Total moment of inertia. \[ I_x = I_{x,\text{stem}} + I_{x,\text{flanges total}} \] \[ I_x = (8.33 \times 10^{-5}) + (2.5 \times 10^{-4}) = 3.33 \times 10^{-4} \, \text{m}^4 \] Step 4: Bending stress. Bending stress is: \[ \sigma = \frac{M_x y}{I_x} \] Maximum distance from neutral axis to extreme fiber: \[ y_{\text{max}} = \frac{h}{2} = 0.5 \, \text{m} \] Given bending moment: \[ M_x = 10 \, \text{Nm} \] So, \[ \sigma = \frac{(10)(0.5)}{3.33 \times 10^{-4}} = \frac{5}{3.33 \times 10^{-4}} \] \[ \sigma = 15000 \, \text{N/m}^2 \] Step 5: Rounding as per requirement. The nearest integer is: \[ \boxed{15000 \, \text{N/m}^2} \]
