1. **Problem Statement:** Compute the centroid and section properties of the given T-shaped cross section with dimensions: top flange 8 in. wide and 1 in. thick, vertical web 1 in. thick and 6 in. high, bottom flange 4 in. wide and 1 in. thick. 2. **Step 1: Define the coordinate system and divide the cross section into rectangles.** - Rectangle 1 (Top flange): width $b_1=8$ in, height $h_1=1$ in - Rectangle 2 (Web): width $b_2=1$ in, height $h_2=6$ in - Rectangle 3 (Bottom flange): width $b_3=4$ in, height $h_3=1$ in 3. **Step 2: Calculate the area of each rectangle.** $$A_1 = b_1 \times h_1 = 8 \times 1 = 8 \text{ in}^2$$ $$A_2 = b_2 \times h_2 = 1 \times 6 = 6 \text{ in}^2$$ $$A_3 = b_3 \times h_3 = 4 \times 1 = 4 \text{ in}^2$$ 4. **Step 3: Find the centroid $\bar{y}$ of each rectangle from the bottom of the cross section.** - Bottom flange centroid: $y_3 = \frac{1}{2} = 0.5$ in - Web centroid: $y_2 = h_3 + \frac{h_2}{2} = 1 + 3 = 4$ in - Top flange centroid: $y_1 = h_3 + h_2 + \frac{h_1}{2} = 1 + 6 + 0.5 = 7.5$ in 5. **Step 4: Calculate the overall centroid $\bar{y}$ of the T-section using the formula:** $$\bar{y} = \frac{\sum A_i y_i}{\sum A_i} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3}$$ $$= \frac{8 \times 7.5 + 6 \times 4 + 4 \times 0.5}{8 + 6 + 4} = \frac{60 + 24 + 2}{18} = \frac{86}{18} = 4.778 \text{ in}$$ 6. **Step 5: Calculate the moment of inertia $I_x$ about the centroidal axis using the parallel axis theorem:** $$I_x = \sum \left(I_{x_i} + A_i d_i^2\right)$$ where $I_{x_i} = \frac{b_i h_i^3}{12}$ is the moment of inertia of each rectangle about its own centroidal axis, and $d_i$ is the distance from each rectangle's centroid to the overall centroid. Calculate $I_{x_i}$: $$I_{x_1} = \frac{8 \times 1^3}{12} = \frac{8}{12} = 0.667 \text{ in}^4$$ $$I_{x_2} = \frac{1 \times 6^3}{12} = \frac{216}{12} = 18 \text{ in}^4$$ $$I_{x_3} = \frac{4 \times 1^3}{12} = \frac{4}{12} = 0.333 \text{ in}^4$$ Calculate distances $d_i$: $$d_1 = |7.5 - 4.778| = 2.722 \text{ in}$$ $$d_2 = |4 - 4.778| = 0.778 \text{ in}$$ $$d_3 = |0.5 - 4.778| = 4.278 \text{ in}$$ Calculate $I_x$: $$I_x = 0.667 + 8 \times (2.722)^2 + 18 + 6 \times (0.778)^2 + 0.333 + 4 \times (4.278)^2$$ $$= 0.667 + 8 \times 7.41 + 18 + 6 \times 0.605 + 0.333 + 4 \times 18.30$$ $$= 0.667 + 59.28 + 18 + 3.63 + 0.333 + 73.20 = 155.11 \text{ in}^4$$ 7. **Step 6: For part (b), given couple $M = 160$ kip-in acting in a vertical plane, calculate maximum tensile and compressive stresses using:** $$\sigma = \frac{M c}{I_x}$$ where $c$ is the distance from the neutral axis to the extreme fiber. Calculate $c$ values: - Distance to top fiber: $c_{top} = 8$ in (total height) $- \bar{y} = 8 - 4.778 = 3.222$ in - Distance to bottom fiber: $c_{bottom} = \bar{y} = 4.778$ in Calculate stresses: $$\sigma_{top} = \frac{160 \times 3.222}{155.11} = 3.32 \text{ kip/in}^2$$ $$\sigma_{bottom} = \frac{160 \times 4.778}{155.11} = 4.93 \text{ kip/in}^2$$ 8. **Final answers:** - Centroid $\bar{y} = 4.778$ in from bottom - Moment of inertia $I_x = 155.11$ in$^4$ - Maximum tensile stress $\sigma_{max} = 4.93$ kip/in$^2$ - Maximum compressive stress $\sigma_{max} = 3.32$ kip/in$^2$