## Step 1: Calculate the Centroid Location **Explanation:** First, we find the centroid (\( \bar{x}, \bar{y} \)) of the composite section, using the method of composite areas. Let’s break the section into three rectangles: - **A1:** Left vertical rectangle (10 mm × 60 mm) - **A2:** Middle horizontal rectangle (100 mm × 20 mm) - **A3:** Right vertical rectangle (10 mm × 60 mm) ### Areas and Centroid Positions | Area | Dimensions | \(A_i\) (mm²) | \(x_i\) (mm) | \(y_i\) (mm) | |------|---------------------|---------------|--------------|--------------| | A1 | 10 × 60 | 600 | + 5 | 60/2 + 20 | | A2 | 100 × 20 | 200 | 10 + 100/2 | 20/2 | | A3 | 10 × 60 | 600 | 110 + 5 | 60/2 + 20 | **x-coordinates:** - A1: \(x_1 = 5\) mm - A2: \(x_2 = 10 + 50 = 60\) mm - A3: \(x_3 = 110 + 5 = 115\) mm **y-coordinates:** - A1: \(y_1 = 60/2 + 20 = 30 + 20 = 50\) mm - A2: \(y_2 = 20/2 = 10\) mm - A3: \(y_3 = 30 + 20 = 50\) mm **Calculate centroid:** \[ \bar{x} = \frac{\sum A_i x_i}{\sum A_i} \] \[ \bar{x} = \frac{600 \times 5 + 200 \times 60 + 600 \times 115}{600 + 200 + 600} = \frac{300 + 120000 + 69000}{320} = \frac{192000}{320} = 60 \text{ mm} \] \[ \bar{y} = \frac{\sum A_i y_i}{\sum A_i} \] \[ \bar{y} = \frac{600 \times 50 + 200 \times 10 + 600 \times 50}{320} = \frac{30000 + 20000 + 30000}{320} = \frac{80000}{320} = 25 \text{ mm} \] **End of Step 1:** The centroid of the cross-section is at \((\bar{x}, \bar{y}) = (60, 25)\) mm. --- ## Step 2: Calculate Moments of Inertia and Principal Axes **Explanation:** Now, find the moments of inertia about the centroidal axes (\(I_{xx}, I_{yy}, I_{xy}\)), then find the principal moments and angle. ### Moments of Inertia for Each Rectangle For each rectangle, use the parallel axis theorem: \[ I_{xx} = I_{xx,centroid} + A d_y^2 \] \[ I_{yy} = I_{yy,centroid} + A d_x^2 \] \[ I_{xy} = A d_x d_y \] where \(d_x = x_i - \bar{x}\), \(d_y = y_i - \bar{y}\). #### Rectangle A1: - \(A_1 = 600\) mm², \(x_1 = 5\), \(y_1 = 50\), \(d_x = -55\), \(d_y = 25\) - \(I_{xx,1,centroid} = \frac{1}{12} b h^3 = \frac{1}{12} \times 10 \times 60^3 = 180,000\) mm⁴ - \(I_{yy,1,centroid} = \frac{1}{12} h b^3 = \frac{1}{12} \times 60 \times 10^3 = 5,000\) mm⁴ \[ I_{xx,1} = 180,000 + 600 \times (25)^2 = 180,000 + 375,000 = 555,000 \] \[ I_{yy,1} = 5,000 + 600 \times (-55)^2 = 5,000 + 1,815,000 = 1,820,000 \] \[ I_{xy,1} = 600 \times (-55) \times 25 = -825,000 \] #### Rectangle A2: - \(A_2 = 200\) mm², \(x_2 = 60\), \(y_2 = 10\), \(d_x = \), \(d_y = -15\) - \(I_{xx,2,centroid} = \frac{1}{12} \times 100 \times 20^3 = 66,667\) mm⁴ - \(I_{yy,2,centroid} = \frac{1}{12} \times 20 \times 100^3 = 1,666,667\) mm⁴ \[ I_{xx,2} = 66,667 + 200 \times (-15)^2 = 66,667 + 450,000 = 516,667 \] \[ I_{yy,2} = 1,666,667 + 200 \times ()^2 = 1,666,667 \] \[ I_{xy,2} = 200 \times \times (-15) = \] #### Rectangle A3: - \(A_3 = 600\) mm², \(x_3 = 115\), \(y_3 = 50\), \(d_x = 55\), \(d_y = 25\) - \(I_{xx,3,centroid} = 180,000\) mm⁴ - \(I_{yy,3,centroid} = 5,000\) mm⁴ \[ I_{xx,3} = 180,000 + 600 \times (25)^2 = 555,000 \] \[ I_{yy,3} = 5,000 + 600 \times (55)^2 = 1,820,000 \] \[ I_{xy,3} = 600 \times 55 \times 25 = 825,000 \] ### Sum for Total Inertia \[ I_{xx} = 555,000 + 516,667 + 555,000 = 1,626,667 \text{ mm}^4 \] \[ I_{yy} = 1,820,000 + 1,666,667 + 1,820,000 = 5,306,667 \text{ mm}^4 \] \[ I_{xy} = -825,000 + + 825,000 = \] ### Principal Moments and Angle Since \(I_{xy} = \), the centroidal axes are already principal axes. - **Principal moments:** \(I_1 = I_{xx} = 1,626,667 \text{ mm}^4\), \(I_2 = I_{yy} = 5,306,667 \text{ mm}^4\) - **Angle:** \(^\circ\) (axes are aligned with the principal axes) --- ## **Final Answer** - **The principal moments of inertia are:** - \(I_1 = 1,626,667 \text{ mm}^4\) - \(I_2 = 5,306,667 \text{ mm}^4\) - **They occur at an angle of \(^\circ\) relative to the centroidal axes (i.e., the x and y axes shown).**