Let's solve the problem step by step: --- ### **Step 1: Understanding the Geometry and Boom Areas** Given: - A rectangular thin-walled beam (single cell) with booms at the corners. - Boom areas: \( A_1 = 10 \, \text{cm}^2 \) (top right) \( A_2 = 5 \, \text{cm}^2 \) (top left) \( A_3 = 5 \, \text{cm}^2 \) (bottom left) \( A_4 = 10 \, \text{cm}^2 \) (bottom right) - Distance between booms: \( 50 \, \text{cm} \) both horizontally and vertically. Let's assign coordinates (assuming origin at bottom left): - \( (x_1, y_1) = (50, 50) \) - \( (x_2, y_2) = (0, 50) \) - \( (x_3, y_3) = (0, 0) \) - \( (x_4, y_4) = (50, 0) \) --- ### **Step 2: Find the Centroid** The centroid \((\bar{x}, \bar{y})\): \[ \bar{x} = \frac{\sum A_i x_i}{\sum A_i},\quad \bar{y} = \frac{\sum A_i y_i}{\sum A_i} \] Total area: \[ \sum A_i = 10 + 5 + 5 + 10 = 30 \, \text{cm}^2 \] \[ \sum A_i x_i = 10 \times 50 + 5 \times 0 + 5 \times 0 + 10 \times 50 = 500 + 0 + 0 + 500 = 1000 \] \[ \sum A_i y_i = 10 \times 50 + 5 \times 50 + 5 \times 0 + 10 \times 0 = 500 + 250 + 0 + 0 = 750 \] So, \[ \bar{x} = \frac{1000}{30} = 33.33 \, \text{cm} \] \[ \bar{y} = \frac{750}{30} = 25.00 \, \text{cm} \] --- ### **Step 3: Calculate Second Moments of Area** \[ I_{xx} = \sum A_i (y_i - \bar{y})^2 \] \[ I_{yy} = \sum A_i (x_i - \bar{x})^2 \] \[ I_{xy} = \sum A_i (x_i - \bar{x})(y_i - \bar{y}) \] Now, plug in values for each boom: #### \( I_{xx} \): - Boom 1: \( (50 - 25)^2 = 625 \), \( 10 \times 625 = 6250 \) - Boom 2: \( (50 - 25)^2 = 625 \), \( 5 \times 625 = 3125 \) - Boom 3: \( (0 - 25)^2 = 625 \), \( 5 \times 625 = 3125 \) - Boom 4: \( (0 - 25)^2 = 625 \), \( 10 \times 625 = 6250 \) \[ I_{xx} = 6250 + 3125 + 3125 + 6250 = 18,750 \, \text{cm}^4 \] #### \( I_{yy} \): - Boom 1: \( (50 - 33.33)^2 = 16.67^2 = 277.8 \), \( 10 \times 277.8 = 2778 \) - Boom 2: \( (0 - 33.33)^2 = (-33.33)^2 = 1111.1 \), \( 5 \times 1111.1 = 5555.5 \) - Boom 3: \( (0 - 33.33)^2 = 1111.1 \), \( 5 \times 1111.1 = 5555.5 \) - Boom 4: \( (50 - 33.33)^2 = 277.8 \), \( 10 \times 277.8 = 2778 \) \[ I_{yy} = 2778 + 5555.5 + 5555.5 + 2778 = 16,667 \, \text{cm}^4 \] #### \( I_{xy} \): - Boom 1: \( (50-33.33)(50-25) = 16.67 \times 25 = 416.8 \), \( 10 \times 416.8 = 4168 \) - Boom 2: \( (0-33.33)(50-25) = -33.33 \times 25 = -833.3 \), \( 5 \times -833.3 = -4166.5 \) - Boom 3: \( (0-33.33)(0-25) = -33.33 \times -25 = 833.3 \), \( 5 \times 833.3 = 4166.5 \) - Boom 4: \( (50-33.33)(0-25) = 16.67 \times -25 = -416.8 \), \( 10 \times -416.8 = -4168 \) \[ I_{xy} = 4168 - 4166.5 + 4166.5 - 4168 = 0 \] --- ### **Step 4: Shear Center Calculation** Given the symmetry, the shear center for a symmetrical rectangle with equal booms on opposite sides lies at the centroid. So, - Shear Center (SC) is at \( (\bar{x}, \bar{y}) = (33.33 \, \text{cm}, 25.00 \, \text{cm}) \) from the bottom left corner (point 3). --- ### **Step 5: Final Answer** \[ \boxed{ \text{The shear center (SC) is located at } (x, y) = (33.33\, \text{cm},\, 25.00\, \text{cm}) \text{ from the bottom left corner.} } \] --- #### **Summary Table** | Boom | \( x_i \) (cm) | \( y_i \) (cm) | \( A_i \) (cm\(^2\)) | |------|----------------|----------------|----------------------| | 1 | 50 | 50 | 10 | | 2 | 0 | 50 | 5 | | 3 | 0 | 0 | 5 | | 4 | 50 | 0 | 10 | --- If you need the detailed calculation using the equation provided in the image for a general case, let me know!