1. **Problem Statement:** Find the moments of inertia $I_x$ and $I_y$ about the centroidal axes for the given composite shape consisting of three rectangles and one circular hole. 2. **Step 1: Find the centroid $(\bar{x}, \bar{y})$ of the composite shape** - Use the bottom-left corner as origin $(0,0)$. - Calculate area $A_i$, centroid coordinates $(x_i, y_i)$ for each component. - For rectangles: $A = b \times h$, centroid at center. - For circle (hole): $A = -\pi r^2$ (negative area), centroid at center. 3. **Component areas and centroids:** - Shape 1 (Base rectangle): $b=250$, $h=20$, $A_1=250 \times 20=5000$ mm$^2$ Centroid: $x_1=125$, $y_1=10$ - Shape 2 (Vertical stem): $b=50$, $h=75$, $A_2=50 \times 75=3750$ mm$^2$ Centroid: $x_2=225$, $y_2=57.5$ - Shape 3 (Top flange): $b=150$, $h=50$, $A_3=150 \times 50=7500$ mm$^2$ Centroid: $x_3=125$, $y_3=120$ - Shape 4 (Circular hole): $r=20$, $A_4=-\pi \times 20^2 = -1256.64$ mm$^2$ Centroid: $x_4=225$, $y_4=10$ 4. **Calculate total area:** $$A_T = 5000 + 3750 + 7500 - 1256.64 = 14993.36 \text{ mm}^2$$ 5. **Calculate $\bar{x}$ and $\bar{y}$:** $$\bar{x} = \frac{\sum A_i x_i}{A_T} = \frac{5000 \times 125 + 3750 \times 225 + 7500 \times 125 - 1256.64 \times 225}{14993.36}$$ Calculate numerator: $$= 625000 + 843750 + 937500 - 282243 = 2120007$$ $$\bar{x} = \frac{2120007}{14993.36} \approx 141.36 \text{ mm}$$ $$\bar{y} = \frac{\sum A_i y_i}{A_T} = \frac{5000 \times 10 + 3750 \times 57.5 + 7500 \times 120 - 1256.64 \times 10}{14993.36}$$ Calculate numerator: $$= 50000 + 215625 + 900000 - 12566.4 = 1152058.6$$ $$\bar{y} = \frac{1152058.6}{14993.36} \approx 76.85 \text{ mm}$$ 6. **Step 2: Calculate moments of inertia about each component's centroidal axes:** - Rectangle about centroidal axes: $$I_x' = \frac{b h^3}{12}, \quad I_y' = \frac{h b^3}{12}$$ - Circle about centroidal axes: $$I_x' = I_y' = \frac{\pi r^4}{4}$$ Calculate for each shape: - Shape 1: $$I_{x1}' = \frac{250 \times 20^3}{12} = \frac{250 \times 8000}{12} = 1666667 \text{ mm}^4$$ $$I_{y1}' = \frac{20 \times 250^3}{12} = \frac{20 \times 15625000}{12} = 26041667 \text{ mm}^4$$ - Shape 2: $$I_{x2}' = \frac{50 \times 75^3}{12} = \frac{50 \times 421875}{12} = 1757812.5 \text{ mm}^4$$ $$I_{y2}' = \frac{75 \times 50^3}{12} = \frac{75 \times 125000}{12} = 781250 \text{ mm}^4$$ - Shape 3: $$I_{x3}' = \frac{150 \times 50^3}{12} = \frac{150 \times 125000}{12} = 1562500 \text{ mm}^4$$ $$I_{y3}' = \frac{50 \times 150^3}{12} = \frac{50 \times 3375000}{12} = 14062500 \text{ mm}^4$$ - Shape 4 (circle hole): $$I_{x4}' = I_{y4}' = \frac{\pi \times 20^4}{4} = \frac{3.1416 \times 160000}{4} = 125664 \text{ mm}^4$$ 7. **Step 3: Calculate distances $d_x = |x_i - \bar{x}|$, $d_y = |y_i - \bar{y}|$ for parallel axis theorem:** - Shape 1: $$d_{x1} = |125 - 141.36| = 16.36, \quad d_{y1} = |10 - 76.85| = 66.85$$ - Shape 2: $$d_{x2} = |225 - 141.36| = 83.64, \quad d_{y2} = |57.5 - 76.85| = 19.35$$ - Shape 3: $$d_{x3} = |125 - 141.36| = 16.36, \quad d_{y3} = |120 - 76.85| = 43.15$$ - Shape 4: $$d_{x4} = |225 - 141.36| = 83.64, \quad d_{y4} = |10 - 76.85| = 66.85$$ 8. **Step 4: Apply parallel axis theorem to find $I_x$ and $I_y$ about centroidal axes:** $$I_x = \sum \left(I_x' + A d_y^2\right)$$ $$I_y = \sum \left(I_y' + A d_x^2\right)$$ Calculate each term: - Shape 1: $$I_{x1} = 1666667 + 5000 \times 66.85^2 = 1666667 + 5000 \times 4470.92 = 1666667 + 22354600 = 24021267$$ $$I_{y1} = 26041667 + 5000 \times 16.36^2 = 26041667 + 5000 \times 267.62 = 26041667 + 1338100 = 27379767$$ - Shape 2: $$I_{x2} = 1757812.5 + 3750 \times 19.35^2 = 1757812.5 + 3750 \times 374.42 = 1757812.5 + 1404080 = 3161892.5$$ $$I_{y2} = 781250 + 3750 \times 83.64^2 = 781250 + 3750 \times 6994.5 = 781250 + 26229375 = 27010625$$ - Shape 3: $$I_{x3} = 1562500 + 7500 \times 43.15^2 = 1562500 + 7500 \times 1862.9 = 1562500 + 13971750 = 15534250$$ $$I_{y3} = 14062500 + 7500 \times 16.36^2 = 14062500 + 7500 \times 267.62 = 14062500 + 2007150 = 16069650$$ - Shape 4 (hole): $$I_{x4} = 125664 - 1256.64 \times 66.85^2 = 125664 - 1256.64 \times 4470.92 = 125664 - 5619990 = -5494326$$ $$I_{y4} = 125664 - 1256.64 \times 83.64^2 = 125664 - 1256.64 \times 6994.5 = 125664 - 8789930 = -8664266$$ 9. **Sum all components:** $$I_x = 24021267 + 3161892.5 + 15534250 - 5494326 = 37233083.5 \text{ mm}^4$$ $$I_y = 27379767 + 27010625 + 16069650 - 8664266 = 61885276 \text{ mm}^4$$ 10. **Final answer:** $$\boxed{I_x = 3.72 \times 10^7 \text{ mm}^4, \quad I_y = 6.19 \times 10^7 \text{ mm}^4}$$