Let's break this problem into clear steps: --- ## **Step 1: Resolve the Applied Force** Given: - Applied force \( F = 4\, \text{kN} = 400\, \text{N} \) - The force makes a right triangle with horizontal (4), vertical (3), hypotenuse (5). **Components:** - \( F_x = F \times \frac{4}{5} = 400 \times .8 = 320\, \text{N} \) (to the right) - \( F_y = F \times \frac{3}{5} = 400 \times .6 = 240\, \text{N} \) (downwards) --- ## **Step 2: Calculate Moment at Section \( a-a \)** Distance from force to section \( a-a \): - \( d = 300\, \text{mm} = .3\, \text{m} \) **Moment due to \( F_y \):** \[ M = F_y \times d = 240\, \text{N} \times .3\, \text{m} = 720\, \text{Nm} = 720{,}000\, \text{Nmm} \] --- ## **Step 3: Cross-section Properties** ### **Section Dimensions** - Top/bottom flange: 6 mm × 36 mm - Web: 12 mm × 6 mm ### **Area Calculations** - **Flange area (each):** \[ A_f = 6\, \text{mm} \times 36\, \text{mm} = 216\, \text{mm}^2 \] - **Web area:** \[ A_w = 12\, \text{mm} \times 6\, \text{mm} = 72\, \text{mm}^2 \] - **Total area:** \[ A_{total} = 2 \times 216 + 72 = 504\, \text{mm}^2 \] ### **Section Modulus (\(I\))** #### **Find neutral axis (ȳ):** Let the bottom of the section be \( y = \): - Bottom flange centroid: \( y_1 = 6/2 = 3\, \text{mm} \) - Web centroid: \( y_2 = 6 + 12/2 = 12\, \text{mm} \) - Top flange centroid: \( y_3 = 6 + 12 + 6/2 = 18\, \text{mm} \) \[ \bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3} \] \[ = \frac{216 \times 3 + 72 \times 12 + 216 \times 18}{504} = \frac{648 + 864 + 3888}{504} = \frac{540}{504} \approx 10.71\, \text{mm} \] #### **Moment of inertia \(I\) about neutral axis:** Use parallel axis theorem for flanges: - **Bottom flange:** \[ I_1 = \frac{1}{12} b h^3 + A (d)^2 \] \( b = 36, h = 6, d = \bar{y} - 3 = 10.71 - 3 = 7.71\, \text{mm} \) \[ I_1 = \frac{1}{12} \times 36 \times 6^3 + 216 \times 7.71^2 = 648 + 12840.8 = 13488.8\, \text{mm}^4 \] - **Top flange:** \( d = 18 - 10.71 = 7.29\, \text{mm} \) \[ I_3 = 648 + 216 \times (7.29^2) = 648 + 11472.3 = 12120.3\, \text{mm}^4 \] - **Web:** \[ I_2 = \frac{1}{12} b h^3 \] \( b = 6, h = 12, d = 12 - 10.71 = 1.29\, \text{mm} \) \[ I_2 = \frac{1}{12} \times 6 \times 12^3 + 72 \times 1.29^2 = 1036.8 + 119.8 = 1156.6\, \text{mm}^4 \] **Total:** \[ I_{total} = 13488.8 + 1156.6 + 12120.3 = 26765.7\, \text{mm}^4 \] --- ## **Step 4: Stresses at Points A & B** ### **Bending Stress (\(\sigma_b\)):** \[ \sigma_b = \frac{M y}{I} \] - For **top fiber (A)**: \( y = 18 - 10.71 = 7.29\, \text{mm} \) - For **bottom fiber (B)**: \( y = 10.71 - 3 = 7.71\, \text{mm} \) **At A:** \[ \sigma_{bA} = \frac{720{,}000 \times 7.29}{26765.7} = 196.2\, \text{N/mm}^2 \] **At B:** \[ \sigma_{bB} = \frac{720{,}000 \times 7.71}{26765.7} = 207.5\, \text{N/mm}^2 \] --- ### **Direct Stress (\(\sigma_{axial}\)):** From \( F_x \): \[ \sigma_{axial} = \frac{F_x}{A_{total}} = \frac{320}{504} = 6.35\, \text{N/mm}^2 \] --- ### **Total Normal Stress:** \[ \sigma_{total} = \sigma_b + \sigma_{axial} \] - At **A**: \( 196.2 + 6.35 = 202.6\, \text{N/mm}^2 \) - At **B**: \( 207.5 + 6.35 = 213.9\, \text{N/mm}^2 \) --- ### **Shear Stress (\(\tau\)):** Shear force \( V = F_y = 240\, \text{N} \) Shear stress at the web (at neutral axis): \[ \tau_{max} = \frac{V Q}{I b} \] - At neutral axis, \( Q = \) area above/below neutral axis × distance to centroid Let's estimate at the web (near \( B \)): - Area above neutral axis: top flange + part of web above NA - \( A_{above} = 216 + (6 \times (18 - 10.71)) = 216 + 43.74 = 259.74\, \text{mm}^2 \) - Distance from NA to centroid of above area (approx): use centroid at \( \approx 13.5\, \text{mm} \) \[ Q = A_{above} \times \text{(distance from NA to centroid of above area)} \approx 259.74 \times (13.5 - 10.71) = 259.74 \times 2.79 = 724.6\, \text{mm}^3 \] - \( b = 6\, \text{mm} \) (web thickness) \[ \tau_{max} = \frac{240 \times 724.6}{26765.7 \times 6} = \frac{173904}{160594.2} = 10.83\, \text{N/mm}^2 \] --- ## **Step 5: Principal Stresses and Maximum Shear** The state of stress at a point (A or B): \[ \sigma_x = \sigma_{total},\quad \tau_{xy} = \text{shear stress} \] - **Principal stresses:** \[ \sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} \] - **Maximum shear stress:** \[ \tau_{max} = \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2} \] For **Point A**: \[ \sigma_x = 202.6\, \text{N/mm}^2, \quad \tau_{xy} \approx \, \text{(at flange outer fiber, shear is nearly zero)} \] Thus, \[ \text{Principal stress: } \sigma_1 = 202.6\, \text{N/mm}^2,\, \sigma_2 = \] \[ \tau_{max} = \frac{202.6}{2} = 101.3\, \text{N/mm}^2 \] For **Point B** (web/neutral axis): \[ \sigma_x = 213.9\, \text{N/mm}^2, \quad \tau_{xy} = 10.83\, \text{N/mm}^2 \] \[ \sigma_{1,2} = \frac{213.9}{2} \pm \sqrt{\left(\frac{213.9}{2}\right)^2 + (10.83)^2} \] \[ = 106.95 \pm \sqrt{(106.95)^2 + 117.36} = 106.95 \pm \sqrt{11446.3 + 117.36} = 106.95 \pm \sqrt{11563.7} = 106.95 \pm 107.54 \] \[ \sigma_1 = 214.49\, \text{N/mm}^2,\quad \sigma_2 = -.59\, \text{N/mm}^2 \] \[ \tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \frac{214.49 - (-.59)}{2} = 107.54\, \text{N/mm}^2 \] --- ## **Final Answers** ### **At Point A (Top Flange Outer Fiber):** - **Principal stress:** \( \boxed{202.6\, \text{N/mm}^2} \) - **Absolute max shear stress:** \( \boxed{101.3\, \text{N/mm}^2} \) ### **At Point B (Web at Neutral Axis):** - **Principal stress:** \( \boxed{214.5\, \text{N/mm}^2} \) - **Absolute max shear stress:** \( \boxed{107.5\, \text{N/mm}^2} \) --- **Note:** All calculations are based on the provided cross-section and loading. For more precise results, consider re-checking \( I \) and \( Q \) with exact centroid locations.