This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here's the solution for Q1: To solve this, we first divide the T-section into two rectangles: a flange (Rectangle 1) and a web (Rectangle 2). We'll establish a reference axis at the bottom-left corner of the section. Step 1: Define the properties of each rectangular component. Rectangle 1 (Flange): Width (b_1) = 160 mm Height (h_1) = 40 mm Area (A_1) = b_1 × h_1 = 160 mm × 40 mm = 6400 mm^2 Centroidal x-coordinate from left edge (x_1) = 160 mm2 = 80 mm Centroidal y-coordinate from bottom reference (y_1) = 160 mm + 40 mm2 = 180 mm Rectangle 2 (Web): Width (b_2) = 40 mm Height (h_2) = 160 mm Area (A_2) = b_2 × h_2 = 40 mm × 160 mm = 6400 mm^2 Centroidal x-coordinate from left edge (x_2) = 80 mm (since it's centered) Centroidal y-coordinate from bottom reference (y_2) = 160 mm2 = 80 mm Step 2: Calculate the centroidal axes (x, y). The total area A = A_1 + A_2 = 6400 mm^2 + 6400 mm^2 = 12800 mm^2. Centroidal x-coordinate (x): Due to symmetry about the vertical axis, x is at the center of the width. x = 160 mm2 = 80 mm Centroidal y-coordinate (y): y = (A_1 y_1 + A_2 y_2)/(A_1 + A_2) y = (6400 mm^2 × 180 mm) + (6400 mm^2 × 80 mm)12800 mm^2 y = 1152000 mm^3 + 512000 mm^312800 mm^2 y = 1664000 mm^312800 mm^2 = 130 mm a) Centroidal axes: The centroid of the section is located at (x, y) = (80 mm, 130 mm) from the bottom-left corner. Step 3: Calculate the Moment of Inertia about the centroidal x-axis (I_x). We use the parallel axis theorem: I_x = (I_xi + A_i d_yi^2), where I_xi = (b_i h_i^3)/(12) and d_yi is the distance from the component's centroid to the overall centroidal x-axis. For Rectangle 1 (Flange): I_x1 = 160 mm × (40 mm)^312 = (160 × 64000)/(12) = 853333.33 mm^4 d_y1 = y_1 - y = 180 mm - 130 mm = 50 mm A_1 d_y1^2 = 6400 mm^2 × (50 mm)^2 = 16000000 mm^4 (I_x1 + A_1 d_y1^2) = 853333.33 + 16000000 = 16853333.33 mm^4 For Rectangle 2 (Web): I_x2 = 40 mm × (160 mm)^312 = (40 × 4096000)/(12) = 13653333.33 mm^4 d_y2 = y - y_2 = 130 mm - 80 mm = 50 mm A_2 d_y2^2 = 6400 mm^2 × (50 mm)^2 = 16000000 mm^4 (I_x2 + A_2 d_y2^2) = 13653333.33 + 16000000 = 29653333.33 mm^4 Total I_x: I_x = 16853333.33 mm^4 + 29653333.33 mm^4 = 46506666.66 mm^4 Step 4: Calculate the Moment of Inertia about the centroidal y-axis (I_y). Since the section is symmetrical about the y-axis, the centroidal y-axis passes through the centroid of each component, so d_xi = 0. We sum the individual moments of inertia I_yi = (h_i b_i^3)/(12). For Rectangle 1 (Flange): I_y1 = 40 mm × (160 mm)^312 = (40 × 4096000)/(12) = 13653333.33 mm^4 For Rectangle 2 (Web): I_y2 = 160 mm × (40 mm)^312 = (160 × 64000)/(12) = 853333.33 mm^4 Total I_y: I_y = 13653333.33 mm^4 + 853333.33 mm^4 = 14506666.66 mm^4 b) Moment of Inertia: The moment of inertia about the centroidal x-axis is 46.51 × 10^6 mm^4. The moment of inertia about the centroidal y-axis is 14.51 × 10^6 mm^4. Step 5: Calculate the Polar Moment of Inertia (J_z). The polar moment of inertia about the centroid is the sum of the moments of inertia about the centroidal x and y axes: J_z = I_x + I_y J_z = 46506666.66 mm^4 + 14506666.66 mm^4 J_z = 61013333.32 mm^4 c) Polar moment of inertia: The polar moment of inertia is 61.01 × 10^6 mm^4. Drop the next question!