Let's analyze the problem carefully. **Given dimensions for the I-section:** - Flange width = 50 mm - Web height = 300 mm - Web thickness = 50 mm - Flange thickness = 50 mm - Overall height = 200 mm --- ### Step 1: Clarify the section geometry Assuming the I-section is symmetric with: - Total height \( H = 300\,mm \) - Flange width \( b_f = 50\,mm \) - Web thickness \( t_w = 50\,mm \) - Flange thickness \( t_f = 50\,mm \) Note: The overall height given is 200 mm in the problem statement, but the web height is 300 mm. Likely, the 300 mm is the web height, and the total height might include flanges. **Potential correction:** Suppose: - Total height \( H = 300\,mm \) - Flange width \( b_f = 50\,mm \) - Flange thickness \( t_f = 50\,mm \) - Web height \( h_w = 200\,mm \) - Web thickness \( t_w = 50\,mm \) which matches the given data: "50mm 300mm 50mm 50mm 200mm". Assuming the sequence: - Flange width = 50 mm - Total height = 300 mm - Flange thickness = 50 mm - Web thickness = 50 mm - Web height = 200 mm --- ### Step 2: Calculate the centroid Since the section is symmetric about the vertical axis, the centroid \( y_c \) is at the mid-height, i.e., at \( H/2 = 150\,mm \). --- ### Step 3: Calculate \( I_y \) (moment of inertia about centroidal vertical axis) For an I-beam, the moment of inertia about the y-axis (vertical axis through centroid) is obtained by summing the moments of inertia of individual rectangles (flanges and web), applying the parallel axis theorem where needed. --- ### Step 4: Calculate areas and moments of inertia of individual parts **Flanges:** - Area of flange \( A_f = b_f \times t_f = 50\,mm \times 50\,mm = 2500\,mm^2 \) - Distance from centroid to flange centroid: - Top flange centroid is located at \( y_{top} = H - t_f/2 = 300 - 25 = 275\,mm \) - Bottom flange centroid at \( y_{bottom} = t_f/2 = 25\,mm \) - Since the centroid is at 150 mm, the distances are: \( d_{top} = 275 - 150 = 125\,mm \) \( d_{bottom} = 150 - 25 = 125\,mm \) **Web:** - Area \( A_w = h_w \times t_w = 200\,mm \times 50\,mm = 10,000\,mm^2 \) - The web centroid is at \( y_w = H/2 = 150\,mm \), coinciding with the centroid, so the web's contribution to \( I_y \) is about its own centroid. --- ### Step 5: Moment of inertia calculations **(a) Flanges:** Moment of inertia of each flange (rectangle about its own centroid): \[ I_{f,local} = \frac{b_f t_f^3}{12} = \frac{50 \times 50^3}{12} = \frac{50 \times 125,000}{12} \approx 520,833\,mm^4 \] But this is about the flange's own centroid (horizontal axis). For the vertical axis \( y \), the flange's inertia about its centroid is: \[ I_{f,about\, centroid} = I_{f,local} + A_f d^2 \] where \( d = 125\,mm \). Calculate: \[ I_{f,about\, centroid} = 520,833 + 2500 \times (125)^2 \] \[ = 520,833 + 2500 \times 15,625 = 520,833 + 39,062,500 = 39,583,333\,mm^4 \] for each flange, so total for both flanges: \[ I_{flanges} = 2 \times 39,583,333 = 79,166,666\,mm^4 \] --- **(b) Web:** Since the web's centroid coincides with the overall centroid: \[ I_{w} = \frac{h_w t_w^3}{12} = \frac{200 \times 50^3}{12} \] Calculate: \[ 50^3 = 125,000 \] \[ I_w = \frac{200 \times 125,000}{12} = \frac{25,000,000}{12} \approx 2,083,333\,mm^4 \] --- ### **Total \( I_y \):** \[ I_y = \text{flanges} + \text{web} = 79,166,666 + 2,083,333 \approx 81,250,000\,mm^4 \] Expressed in scientific notation: \[ I_y \approx 8.125 \times 10^7\,mm^4 \] --- ### **Compare with options:** Options are: 1. \( I_{y} = 102 \times 10^6 \times m \times m^4 \) 2. \( I_{x} = 82 \times 10^4 \times m \times n^4 \) 3. \( I_{y} = 69.8 \times 10^6 \times m \times m^4 \) 4. \( I_{y} = 40.5 \times 10^6 \times rn \times m^4 \) Our calculation (~81.25 million \(mm^4\)) is closest to **Option 1**: \(102 \times 10^6\,mm^4\), but slightly lower. Given the approximate nature and the options, **Option 1** seems the most appropriate. --- ## **Final conclusion:** **The correct option is:** ### **I_{y} = 102 \times 10^6 \times m \times m^4** **Thus, the first option is the best match.**