1. **Problem statement:** Determine the bending stress at points A, B, and D on a T-shaped beam cross section subjected to a couple moment $M=25$ kN·m applied at an angle $\beta=15^\circ$ with the vertical axis. 2. **Given data:** - Moment magnitude: $M=25\times10^3$ N·m (converted to N·mm: $25\times10^6$ N·mm) - Angle with vertical: $\beta=15^\circ$ - Cross section dimensions: height $h=80$ mm, width $b=80$ mm, stem width $20$ mm, stem height $30$ mm 3. **Step 1: Calculate the moment components along the y and z axes.** The moment $M$ is applied at angle $\beta$ with the vertical (y-axis), so: $$M_y = M \cos\beta = 25\times10^6 \times \cos 15^\circ$$ $$M_z = M \sin\beta = 25\times10^6 \times \sin 15^\circ$$ Calculate: $$M_y = 25\times10^6 \times 0.9659 = 2.4147 \times 10^7 \text{ N·mm}$$ $$M_z = 25\times10^6 \times 0.2588 = 6.47 \times 10^6 \text{ N·mm}$$ 4. **Step 2: Find the centroid coordinates (y_c, z_c) of the T-section.** Assuming the origin at the bottom left corner of the cross section: - Flange area $A_f = 80 \times 20 = 1600$ mm², centroid at $y_f = 80 - 10 = 70$ mm, $z_f = 40$ mm (half of 80 mm width) - Stem area $A_s = 30 \times 20 = 600$ mm², centroid at $y_s = 15$ mm, $z_s = 10$ mm Calculate centroid: $$y_c = \frac{A_f y_f + A_s y_s}{A_f + A_s} = \frac{1600 \times 70 + 600 \times 15}{2200} = \frac{112000 + 9000}{2200} = 55.91 \text{ mm}$$ $$z_c = \frac{A_f z_f + A_s z_s}{A_f + A_s} = \frac{1600 \times 40 + 600 \times 10}{2200} = \frac{64000 + 6000}{2200} = 31.82 \text{ mm}$$ 5. **Step 3: Calculate moments of inertia $I_y$ and $I_z$ about centroidal axes.** Using parallel axis theorem: - Flange moment of inertia about y-axis: $$I_{y_f} = \frac{b h^3}{12} = \frac{80 \times 20^3}{12} = \frac{80 \times 8000}{12} = 53333.33 \text{ mm}^4$$ - Flange moment of inertia about z-axis: $$I_{z_f} = \frac{h b^3}{12} = \frac{20 \times 80^3}{12} = \frac{20 \times 512000}{12} = 853333.33 \text{ mm}^4$$ - Stem moment of inertia about y-axis: $$I_{y_s} = \frac{b h^3}{12} = \frac{20 \times 30^3}{12} = \frac{20 \times 27000}{12} = 45000 \text{ mm}^4$$ - Stem moment of inertia about z-axis: $$I_{z_s} = \frac{h b^3}{12} = \frac{30 \times 20^3}{12} = \frac{30 \times 8000}{12} = 20000 \text{ mm}^4$$ Calculate distances from each centroid to overall centroid: $$d_{y_f} = |70 - 55.91| = 14.09 \text{ mm}$$ $$d_{y_s} = |15 - 55.91| = 40.91 \text{ mm}$$ $$d_{z_f} = |40 - 31.82| = 8.18 \text{ mm}$$ $$d_{z_s} = |10 - 31.82| = 21.82 \text{ mm}$$ Apply parallel axis theorem: $$I_y = I_{y_f} + A_f d_{z_f}^2 + I_{y_s} + A_s d_{z_s}^2$$ $$= 53333.33 + 1600 \times 8.18^2 + 45000 + 600 \times 21.82^2$$ $$= 53333.33 + 1600 \times 66.89 + 45000 + 600 \times 476.12$$ $$= 53333.33 + 107024 + 45000 + 285672 = 490029.33 \text{ mm}^4$$ $$I_z = I_{z_f} + A_f d_{y_f}^2 + I_{z_s} + A_s d_{y_s}^2$$ $$= 853333.33 + 1600 \times 14.09^2 + 20000 + 600 \times 40.91^2$$ $$= 853333.33 + 1600 \times 198.52 + 20000 + 600 \times 1673.28$$ $$= 853333.33 + 317632 + 20000 + 1003968 = 2196933.33 \text{ mm}^4$$ 6. **Step 4: Calculate coordinates of points A, B, and D relative to centroid.** - Point A: top left corner at $(y,z) = (80,0)$ $$y_A' = 80 - 55.91 = 24.09 \text{ mm}$$ $$z_A' = 0 - 31.82 = -31.82 \text{ mm}$$ - Point B: top right corner at $(80,80)$ $$y_B' = 80 - 55.91 = 24.09 \text{ mm}$$ $$z_B' = 80 - 31.82 = 48.18 \text{ mm}$$ - Point D: bottom of stem at $(0,20)$ $$y_D' = 0 - 55.91 = -55.91 \text{ mm}$$ $$z_D' = 20 - 31.82 = -11.82 \text{ mm}$$ 7. **Step 5: Calculate bending stress at each point using:** $$\sigma = -\frac{M_y z'}{I_y} + \frac{M_z y'}{I_z}$$ Calculate stresses: - At A: $$\sigma_A = -\frac{2.4147 \times 10^7 \times (-31.82)}{490029.33} + \frac{6.47 \times 10^6 \times 24.09}{2196933.33}$$ $$= \frac{7.683 \times 10^8}{490029.33} + \frac{1.559 \times 10^8}{2196933.33} = 1568.1 + 71.0 = 1639.1 \text{ N/mm}^2$$ - At B: $$\sigma_B = -\frac{2.4147 \times 10^7 \times 48.18}{490029.33} + \frac{6.47 \times 10^6 \times 24.09}{2196933.33}$$ $$= -2373.3 + 71.0 = -2302.3 \text{ N/mm}^2$$ - At D: $$\sigma_D = -\frac{2.4147 \times 10^7 \times (-11.82)}{490029.33} + \frac{6.47 \times 10^6 \times (-55.91)}{2196933.33}$$ $$= 582.3 - 164.7 = 417.6 \text{ N/mm}^2$$ **Final answers:** - Stress at A: $\sigma_A = 1639.1$ MPa (tension) - Stress at B: $\sigma_B = -2302.3$ MPa (compression) - Stress at D: $\sigma_D = 417.6$ MPa (tension)