Let's break down the solution step by step: --- ## **Given:** - \( a = 100 \) mm - \( b = .8a = 80 \) mm - \( t = 16 \) mm --- ## **Step 1: Section Properties** ### **Identify Subsections** - **Bottom horizontal:** \( a \times t \) - **Left vertical:** \( t \times b \) - **Right vertical:** \( t \times b \) --- ## **Step 2: Calculate the Centroid (ȳ)** Let's set the bottom of the section as \( y= \), and measure upward. | Section | Area \(A\) | y (to centroid of each) | |-------------------|--------------------|------------------------| | Bottom | \( a \times t \) | \( t/2 \) | | Left vertical | \( t \times b \) | \( t + b/2 \) | | Right vertical | \( t \times b \) | \( t + b/2 \) | - **Areas:** - \( A_1 = a t = 100 \times 16 = 160 \) mm² - \( A_2 = t b = 16 \times 80 = 128 \) mm² (each vertical) - Total Area: \( A = 160 + 128 + 128 = 416 \) mm² - **Centroid:** \[ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \] \[ \bar{y} = \frac{160 \times 8 + 128 \times (16+40) + 128 \times (16+40)}{416} \] \[ \bar{y} = \frac{160 \times 8 + 128 \times 56 + 128 \times 56}{416} \] \[ = \frac{12800 + 71680 + 71680}{416} \] \[ = \frac{156160}{416} = 37.56 \text{ mm (rounded to 2 decimal places)} \] --- ## **Step 3: Moments of Inertia About the Centroid (Parallel Axis Theorem)** ### **For each section:** #### **1. Bottom (horizontal)** - \( I_{x1} = \frac{1}{12} a t^3 \) - Distance from centroid: \( d_1 = \bar{y} - t/2 = 37.56 - 8 = 29.56 \) mm \[ I_{x,1}^{total} = \frac{1}{12}(100)(16^3) + 160 \times (29.56^2) \] #### **2. Left/Right (vertical)** - \( I_{x2} = \frac{1}{12} t b^3 \) - Distance from centroid: \( d_2 = |(t + b/2) - \bar{y}| = |16 + 40 - 37.56| = 18.44 \) mm \[ I_{x,2}^{total} = 2 \left[ \frac{1}{12}(16)(80^3) + 128 \times (18.44^2) \right] \] --- ### **Calculate \( I_x \):** #### **Bottom:** - \( \frac{1}{12} \times 100 \times 16^3 = \frac{1}{12} \times 100 \times 4096 = 100 \times 341.33 = 34,133 \) mm⁴ - \( 160 \times (29.56^2) = 160 \times 873.86 = 1,398,176 \) mm⁴ Total bottom: \( 34,133 + 1,398,176 = 1,432,309 \) mm⁴ #### **Verticals:** - \( \frac{1}{12} \times 16 \times 80^3 = \frac{1}{12} \times 16 \times 512,000 = 16 \times 42,666.67 = 682,666.67 \) mm⁴ - \( 128 \times (18.44^2) = 128 \times 340.17 = 435,417.6 \) mm⁴ Sum per vertical: \( 682,667 + 435,418 = 1,118,085 \) mm⁴ Total verticals (both): \( 2 \times 1,118,085 = 2,236,170 \) mm⁴ --- ### **Total \( I_x \):** \[ I_x = 1,432,309 + 2,236,170 = 3,668,479 \text{ mm}^4 \] --- ### **Now \( I_y \) (centroidal):** - Bottom: \( I_{y1} = \frac{1}{12} t a^3 \) - Verticals: Each \( I_{y2} = \frac{1}{12} b t^3 + A d^2 \) (distance from centroid is half the distance between vertical centers and centroid, which is symmetrical.) #### **Bottom:** - \( \frac{1}{12} \times 16 \times 100^3 = \frac{1}{12} \times 16 \times 1,000,000 = 16 \times 83,333.33 = 1,333,333 \) mm⁴ #### **Verticals:** - \( \frac{1}{12} \times 80 \times 16^3 = \frac{1}{12} \times 80 \times 4096 = 80 \times 341.33 = 27,306.67 \) mm⁴ Centroid x-location: Section is symmetric, the centroid is at \( x = a/2 = 50 \) mm. - Distance from centroid for each vertical: \( |x_{vertical} - x_{centroid}| = |t/2| = |8 - 50| = 42 \) mm - \( A d^2 = 128 \times 42^2 = 128 \times 1764 = 2,258,880 \) mm⁴ (each) Total per vertical: \( 27,307 + 2,258,880 = 2,286,187 \) mm⁴ Both: \( 2 \times 2,286,187 = 4,572,374 \) mm⁴ --- ### **Total \( I_y \):** \[ I_y = 1,333,333 + 4,572,374 = 5,905,707 \text{ mm}^4 \] --- ## **Step 4: Product of Inertia \( I_{xy} \)** Since the section is symmetric about the y-axis, \( I_{xy} = \). --- ## **Step 5: Principal Moments and Angles** ### **Formulas:** \[ I_{max/min} = \frac{I_x + I_y}{2} \pm \sqrt{ \left( \frac{I_x - I_y}{2} \right)^2 + I_{xy}^2 } \] \[ \tan 2\theta = \frac{2I_{xy}}{I_x - I_y} \] Given \( I_{xy} = \): \[ I_{max} = \max\{I_x, I_y\} \] \[ I_{min} = \min\{I_x, I_y\} \] ### **Plug in values:** - \( I_x = 3,668,479 \) - \( I_y = 5,905,707 \) \[ I_{max} = 5,905,707 \text{ mm}^4 \text{ at } ^\circ \] \[ I_{min} = 3,668,479 \text{ mm}^4 \text{ at } 90^\circ \] --- ## **Step 6: Final Answers** | Quantity | Value | Angle | |-----------|------------|--------------| | \( I_{max} \) | **5,905,707 mm⁴** | **.°** | | \( I_{min} \) | **3,668,479 mm⁴** | **90.°** | --- ### **Summary Table** | \( I_{max} \) | \( I_{min} \) | Angle for \( I_{max} \) | Angle for \( I_{min} \) | |:-------------:|:-------------:|:-----------------------:|:-----------------------:| | 5,905,707 mm⁴ | 3,668,479 mm⁴ | .° | 90.° | --- Let me know if you need further explanation or calculations!