Let's break down the problem step by step: --- ### **Given Data:** - Cross-section: Symmetric, open thin-walled section ("C" shape) - Dimensions (centerline to centerline): - Height = \(200\, \text{mm}\) - Width = \(100\, \text{mm}\) - Wall thickness, \( t = 2\, \text{mm} \) --- ## a) **Moment of Inertia \( I_y \)** We are to find the moment of inertia about the **y-axis** (vertical centroidal axis). ### **Step 1: Centroid Location** The section is symmetric about the \( y \)-axis, so the centroid is at the mid-width horizontally. ### **Step 2: Moment of Inertia Calculation** For thin-walled sections: - Ignore the corner radii (since \( t \ll \text{lengths} \)) - Treat each wall as a rectangle #### **1. Vertical Walls (Left and Right)** - Height: \(200\, \text{mm}\) - Thickness: \(2\, \text{mm}\) - Distance from y-axis: \( \frac{100}{2} = 50\, \text{mm} \) Moment of inertia of one wall (parallel axis theorem): \[ I_{y, \text{vertical}} = A \cdot d^2 = (200 \times 2) \cdot (50^2) = 400 \times 2500 = 1,000,000\, \text{mm}^4 \] There are two such walls (left and right), so total: \[ I_{y, \text{vertical (total)}} = 2 \times 1,000,000 = 2,000,000\, \text{mm}^4 \] #### **2. Horizontal Wall (Top)** - Width: \(100\, \text{mm}\) - Thickness: \(2\, \text{mm}\) - Distance from y-axis: 0 (since it lies along the y-axis) Moment of inertia about y-axis (through its own centroid): \[ I_{y, \text{horizontal}} = \frac{1}{12} b h^3 = \frac{1}{12} \times 100 \times 2^3 = \frac{1}{12} \times 100 \times 8 = 66.67\, \text{mm}^4 \] But for a thin strip, this is very small compared to the vertical walls, so **often neglected**. But let's include it for completeness. #### **Total Moment of Inertia:** \[ I_y \approx 2,000,000 + 66.67 \approx 2,000,067\, \text{mm}^4 \] --- ## b) **Shear Flow Distribution for \( V_y = 1\, \text{kN} \)** Shear force \( V_y = 1\, \text{kN} = 1000\, \text{N} \) applied vertically. Shear flow at distance \( s \) from the bottom (start at bottom left, go up, then right, then down): \[ q(s) = \frac{V_y}{I_y} \int_{\text{cut}}^s t \cdot y \, ds \] - For left vertical wall (from bottom up to \( y \)): \[ q_{\text{left}}(y) = \frac{V_y}{I_y} t \int_0^y y' dy' = \frac{V_y t}{I_y} \frac{y^2}{2} \] Where \( y \) is measured from the bottom. - For top wall (from left to right, at \( y = 200\, \text{mm} \)): \[ q_{\text{top}}(x) = q_{\text{left}}(200) \] (Shear flow constant along the top wall, since \( y \) is constant.) - For right vertical wall (from top to bottom, \( y \) decreases from 200 to 0): \[ q_{\text{right}}(y) = q_{\text{top}} - \frac{V_y t}{I_y} \int_{200}^y y' (-dy') \] (You can write the full expression if needed.) #### **Shear Flow Sketch:** - Starts at 0 at bottom left - Increases quadratically up the left wall - Constant along the top wall - Decreases down the right wall back to 0 --- ## c) **Shear Center** For a symmetric open section like this, the **shear center** lies on the axis of symmetry, halfway along the width (i.e., at the midpoint of the base). **For a channel section:** - If walls are of equal thickness and length, the shear center is on the axis of symmetry, outside the section, a certain distance from the web. But in this specific geometry, with open C-section (not U), the shear center is **at the centroid**, i.e., halfway along the base width. --- ### **Summary of Answers** #### a) **Moment of inertia \( I_y \):** \[ I_y \approx 2,000,067\, \text{mm}^4 \] #### b) **Shear flow \( q \) distribution:** \[ q_{\text{left}}(y) = \frac{V_y t}{I_y} \frac{y^2}{2} \] \[ q_{\text{top}} = q_{\text{left}}(200) \] (Sketch: quadratic up left, constant on top, quadratic down right.) #### c) **Shear center:** At the centroid, halfway along the base width. --- **Let me know if you'd like detailed calculations, a sketch, or further explanation for any part!**