Let's break down the solution step by step as requested. --- ### **Step 1: Determine d so the centroid is at the origin (Part a)** #### **Divide the shape into simple rectangles:** - **Top flange:** 1 m (width) × 1.5 m (height) - **Stem:** 0.5 m (width) × 4 m (height) - **Bottom flange:** 1 m (width) × 1 m (height) #### **Find the centroid for each part:** Assume the bottom of the figure as the reference (y = 0). - **Top flange:** Area \(A_1 = 1 \times 1.5 = 1.5\,m^2\) Centroid \(y_1 = d + 4 + \frac{1.5}{2} = d + 4 + 0.75 = d + 4.75\,m\) - **Stem:** Area \(A_2 = 0.5 \times 4 = 2\,m^2\) Centroid \(y_2 = d + \frac{4}{2} = d + 2\,m\) - **Bottom flange:** Area \(A_3 = 1 \times 1 = 1\,m^2\) Centroid \(y_3 = d + \frac{1}{2} = d + 0.5\,m\) #### **Calculate the centroid (ȳ):** \[ \bar{y} = \frac{\sum A_i y_i}{\sum A_i} \] \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3} \] \[ = \frac{1.5(d+4.75) + 2(d+2) + 1(d+0.5)}{1.5+2+1} \] \[ = \frac{1.5d + 7.125 + 2d + 4 + d + 0.5}{4.5} \] \[ = \frac{(1.5d + 2d + d) + (7.125 + 4 + 0.5)}{4.5} \] \[ = \frac{4.5d + 11.625}{4.5} \] Set \( \bar{y} = 0 \) (since O is at the centroid): \[ 0 = \frac{4.5d + 11.625}{4.5} \] \[ 4.5d + 11.625 = 0 \] \[ d = -\frac{11.625}{4.5} \] \[ d = -2.583\,m \] **Explanation:** We summed up the moments of each rectangle about the reference axis, divided by the total area, and set it to zero to solve for \(d\), ensuring the centroid is at the origin. --- ### **Step 2: Calculate the area moment of inertia about the x-axis (Part b)** - **Parallel axis theorem:** \(I_x = I_{x,\text{centroid}} + A \cdot (y')^2\) - For each part, \(y'\) is the distance from the centroid of each rectangle to the centroid of the entire section (which is at \(y = 0\)). #### **For each part:** **Top flange:** \[ I_{x1} = \frac{1 \times (1.5)^3}{12} + 1.5 \times (d + 4.75)^2 \] \[ I_{x1} = \frac{1 \times 3.375}{12} + 1.5 \times (2.167)^2 \] \[ I_{x1} = 0.28125 + 1.5 \times 4.697 \] \[ I_{x1} = 0.28125 + 7.0455 = 7.32675\,m^4 \] **Stem:** \[ I_{x2} = \frac{0.5 \times (4)^3}{12} + 2 \times (d + 2)^2 \] \[ I_{x2} = \frac{0.5 \times 64}{12} + 2 \times (0.417)^2 \] \[ I_{x2} = \frac{32}{12} + 2 \times 0.174 \] \[ I_{x2} = 2.6667 + 0.348 = 3.0147\,m^4 \] **Bottom flange:** \[ I_{x3} = \frac{1 \times (1)^3}{12} + 1 \times (d + 0.5)^2 \] \[ I_{x3} = \frac{1}{12} + 1 \times ( -2.083 )^2 \] \[ I_{x3} = 0.0833 + 4.34 = 4.4233\,m^4 \] Now sum: \[ I_x = I_{x1} + I_{x2} + I_{x3} \] \[ = 7.32675 + 3.0147 + 4.4233 \] \[ = 14.76475\,m^4 \] **Explanation:** We used the parallel axis theorem to compute the moment of inertia for each rectangle about the x-axis passing through the centroid, and summed all contributions. --- ## **Final Answers:** - **(a) d = -2.583 m** - **(b) Area moment of inertia about the x-axis, \(I_x = 14.765\,m^4\) (rounded to three decimals)** --- If you need Step 3 (moment of inertia about the y-axis), please let me know!

Certainly! Based on the provided diagram and problem statement, here's a detailed two-step solution with explanations: --- ### **Step 1: Find the distance \(d\) so that the centroid is at the origin** **Understanding the problem:** - The figure is composed of three parts: top flange, stem, and bottom flange. - The coordinate system is set such that the bottom of the figure is at \(y = 0\). - The variable \(d\) represents the vertical distance from the bottom of the figure to the reference point \(O\). **Step-by-step:** 1. **Calculate the area of each part:** - Top flange: width = 1 m, height = 1.5 m \[ A_1 = 1 \times 1.5 = 1.5\,m^2 \] - Stem: width = 0.5 m, height = 4 m \[ A_2 = 0.5 \times 4 = 2\,m^2 \] - Bottom flange: width = 1 m, height = 1 m \[ A_3 = 1 \times 1 = 1\,m^2 \] 2. **Determine the centroid \(y_i\) of each part relative to the bottom:** - For the **bottom flange**: \[ y_3 = d + \frac{1}{2} = d + 0.5\,m \] - For the **stem**: \[ y_2 = d + 1 + \frac{4}{2} = d + 1 + 2 = d + 3\,m \] - For the **top flange**: \[ y_1 = d + 1 + 4 + \frac{1.5}{2} = d + 1 + 4 + 0.75 = d + 5.75\,m \] 3. **Calculate the combined centroid \( \bar{y} \):** \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3} \] Substituting: \[ \bar{y} = \frac{1.5(d + 5.75) + 2(d + 3) + 1(d + 0.5)}{1.5 + 2 + 1} = \frac{1.5d + 8.625 + 2d + 6 + d + 0.5}{4.5} \] Simplify numerator: \[ (1.5d + 2d + d) + (8.625 + 6 + 0.5) = 4.5d + 15.125 \] So: \[ \bar{y} = \frac{4.5d + 15.125}{4.5} \] 4. **Set the centroid at the origin (i.e., \(\bar{y} = 0\)) to find \(d\):** \[ 0 = \frac{4.5d + 15.125}{4.5} \] \[ 4.5d + 15.125 = 0 \] \[ d = -\frac{15.125}{4.5} \approx -3.36\,m \] **_Explanation:_** By calculating the centroid position and setting it at zero, we solve for \(d\). The negative value indicates that the reference point \(O\) is below the actual centroid of the shape. --- ### **Step 2: Calculate the area moment of inertia \(I_x\) about the x-axis passing through the centroid** **Understanding:** - Use the parallel axis theorem to find each part's \(I_x\) about the centroidal axis. **Step-by-step:** 1. **Determine the centroidal distance \(y_i'\) of each part from the overall centroid:** \[ y_i' = y_i - \bar{y} \] Since the overall centroid is at \(y=0\), and from previous step, the centroid is at \(\bar{y} = 0\), the distances are just \(y_i\). 2. **Moment of inertia for each rectangle about its own centroid (about x-axis):** \[ I_{x,i} = \frac{b_i h_i^3}{12} \] where \(b_i\) and \(h_i\) are width and height. - Bottom flange: \[ I_{x3} = \frac{1 \times (1)^3}{12} = \frac{1}{12} \approx 0.0833\,m^4 \] - Stem: \[ I_{x2} = \frac{0.5 \times (4)^3}{12} = \frac{0.5 \times 64}{12} = \frac{32}{12} \approx 2.6667\,m^4 \] - Top flange: \[ I_{x1} = \frac{1 \times (1.5)^3}{12} = \frac{1 \times 3.375}{12} \approx 0.2813\,m^4 \] 3. **Apply the parallel axis theorem to shift each \(I_x\) to the overall centroid:** \[ I_{x,\text{total}} = \sum \left[ I_{x,i} + A_i (y_i - \bar{y})^2 \right] \] Since \(\bar{y} = 0\), \(y_i - \bar{y} = y_i\). - Bottom flange: \[ I_{x3,\text{total}} = 0.0833 + 1 \times (0.5)^2 = 0.0833 + 0.25 = 0.3333\,m^4 \] - Stem: \[ I_{x2,\text{total}} = 2.6667 + 2 \times (3)^2 = 2.6667 + 2 \times 9 = 2.6667 + 18 = 20.6667\,m^4 \] - Top flange: \[ I_{x1,\text{total}} = 0.2813 + 1.5 \times (5.75)^2 = 0.2813 + 1.5 \times 33.0625 \approx 0.2813 + 49.59375 = 49.875\,m^4 \] 4. **Sum all contributions:** \[ I_x = 0.3333 + 20.6667 + 49.875 \approx \boxed{70.875\,m^4} \] **_Final answer:_** - **Distance \(d \approx -3.36\,m\) (to position the centroid at the origin).** - **Moment of inertia about the x-axis: \(I_x \approx 70.88\,m^4\).** --- **Note:** The negative \(d\) indicates the reference point is below the centroid; in some contexts, you might redefine the reference or interpret accordingly. --- If you need the moment of inertia about the y-axis or further clarification, please ask!

Certainly! Here's the detailed two-step solution with explanations for calculating the area moment of inertia about the y-axis for the given shape. --- ### **Step 1: Find the centroid \( \bar{x} \) along the y-axis** **Understanding:** - The shape is composed of three rectangular parts: bottom flange, stem, and top flange. - To compute the moment of inertia about the y-axis, we need the position of the centroid along x-axis (horizontal direction). **Assumptions:** - The shape is symmetric about its vertical centerline. - The bottom of the shape is at \( x = 0 \). **Step-by-step:** 1. **Calculate the area of each part:** - Bottom flange: width = 1 m, height = 1 m \[ A_3 = 1 \times 1 = 1\,m^2 \] - Stem: width = 0.5 m, height = 4 m \[ A_2 = 0.5 \times 4 = 2\,m^2 \] - Top flange: width = 1 m, height = 1.5 m \[ A_1 = 1 \times 1.5 = 1.5\,m^2 \] 2. **Determine the centroid \(x_i\) of each part:** - For all parts, assume the centroid along x is at the center of each rectangle: - Bottom flange: \( x_3 = 0.5\,m \) - Stem: \( x_2 = 0.25\,m \) - Top flange: \( x_1 = 0.5\,m \) 3. **Calculate the overall centroid \( \bar{x} \):** \[ \bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A_1 + A_2 + A_3} \] \[ \bar{x} = \frac{1.5 \times 0.5 + 2 \times 0.25 + 1 \times 0.5}{1.5 + 2 + 1} = \frac{0.75 + 0.5 + 0.5}{4.5} = \frac{1.75}{4.5} \approx 0.389\,m \] **_Explanation:_** The centroid along the x-axis is approximately 0.389 meters from the left edge, calculated using the areas and centroid positions of each part. --- ### **Step 2: Compute the area moment of inertia \(I_y\) about the y-axis** **Understanding:** - Use the sum of each part's moment of inertia about its own centroid plus the parallel axis term to shift to the overall centroid. **Step-by-step:** 1. **Moment of inertia for each rectangle about its own centroid (about y-axis):** \[ I_{y,i} = \frac{h_i \times b_i^3}{12} \] where \(b_i\) is along x, \(h_i\) along y. - Bottom flange: \(b=1\,m\), \(h=1\,m\) \[ I_{y3} = \frac{1 \times 1^3}{12} = \frac{1}{12} \approx 0.0833\,m^4 \] - Stem: \(b=0.5\,m\), \(h=4\,m\) \[ I_{y2} = \frac{4 \times 0.5^3}{12} = \frac{4 \times 0.125}{12} = \frac{0.5}{12} \approx 0.0417\,m^4 \] - Top flange: \(b=1\,m\), \(h=1.5\,m\) \[ I_{y1} = \frac{1.5 \times 1^3}{12} = \frac{1.5}{12} = 0.125\,m^4 \] 2. **Calculate the parallel axis term for each part:** \[ I_{y,i}^{\text{total}} = I_{y,i} + A_i (x_i - \bar{x})^2 \] - Bottom flange: \[ (x_3 - \bar{x}) = 0.5 - 0.389 = 0.111\,m \] \[ I_{y3}^{\text{total}} = 0.0833 + 1 \times (0.111)^2 = 0.0833 + 0.0123 = 0.0956\,m^4 \] - Stem: \[ (x_2 - \bar{x}) = 0.25 - 0.389 = -0.139\,m \] \[ I_{y2}^{\text{total}} = 0.0417 + 2 \times (0.139)^2 = 0.0417 + 2 \times 0.0193 = 0.0417 + 0.0386 = 0.0803\,m^4 \] - Top flange: \[ (x_1 - \bar{x}) = 0.5 - 0.389 = 0.111\,m \] \[ I_{y1}^{\text{total}} = 0.125 + 1.5 \times (0.111)^2 = 0.125 + 1.5 \times 0.0123 = 0.125 + 0.0185 = 0.1435\,m^4 \] 3. **Sum all contributions:** \[ I_y = 0.0956 + 0.0803 + 0.1435 \approx \boxed{0.319\,m^4} \] --- ## **Final answer:** **The area moment of inertia about the y-axis passing through the centroid is approximately \(\boxed{0.319\,m^4}\).** --- **Note:** - Calculations are approximate; rounding may slightly vary. - The shape's symmetry simplifies some calculations, especially if symmetry is considered, but here we used the detailed approach for accuracy. Let me know if you'd like further clarification!