1. **Problem Statement:** Investigate the adequacy of a 3.5 m steel channel beam (Beam 2) carrying a uniform load of 20 kN/m. The beam has dimensions: $b_f=300$ mm, $t_f=12$ mm, $d=325$ mm, $t_w=10$ mm. Material properties: $F_y=250$ MPa, $E=200000$ MPa. 2. **Step 1: Calculate the maximum bending moment $M_{max}$ for a simply supported beam with uniform load $w$:** $$M_{max} = \frac{wL^2}{8}$$ Given $w=20$ kN/m and $L=3.5$ m, $$M_{max} = \frac{20 \times 3.5^2}{8} = \frac{20 \times 12.25}{8} = 30.625 \text{ kN}\cdot\text{m}$$ Convert to N·mm: $$M_{max} = 30.625 \times 10^3 \times 10^3 = 3.0625 \times 10^7 \text{ N}\cdot\text{mm}$$ 3. **Step 2: Calculate section modulus $S$ of the channel beam:** Approximate $S$ by calculating the moment of inertia $I$ and distance from neutral axis to extreme fiber $c$. 4. **Step 3: Calculate moment of inertia $I$ for the channel cross-section:** The channel consists of two flanges and one web. - Flange area $A_f = b_f \times t_f = 300 \times 12 = 3600$ mm$^2$ - Web area $A_w = t_w \times (d - 2t_f) = 10 \times (325 - 24) = 10 \times 301 = 3010$ mm$^2$ Calculate centroid $\bar{y}$ from bottom: $$\bar{y} = \frac{2 A_f y_f + A_w y_w}{2 A_f + A_w}$$ where $y_f$ is centroid of flange from bottom (flange thickness/2 = 6 mm for bottom flange, and $d - t_f/2 = 325 - 6 = 319$ mm for top flange), and $y_w$ is centroid of web (at $d/2 = 162.5$ mm). Calculate: $$\bar{y} = \frac{3600 \times 6 + 3600 \times 319 + 3010 \times 162.5}{3600 + 3600 + 3010} = \frac{21600 + 1148400 + 489125}{10210} = \frac{1610125}{10210} \approx 157.7 \text{ mm}$$ 5. **Step 4: Calculate $I$ using parallel axis theorem:** - $I_{flange} = 2 \times \left( \frac{b_f t_f^3}{12} + A_f (y_f - \bar{y})^2 \right)$ - $I_{web} = \frac{t_w (d - 2 t_f)^3}{12} + A_w (y_w - \bar{y})^2$ Calculate each term: - $I_{flange} = 2 \times \left( \frac{300 \times 12^3}{12} + 3600 (y_f - 157.7)^2 \right)$ - Bottom flange: $y_f = 6$ mm, $(6 - 157.7)^2 = 23198$ - Top flange: $y_f = 319$ mm, $(319 - 157.7)^2 = 25922$ Average for two flanges: use both separately and sum. Calculate bottom flange: $$I_{bf} = \frac{300 \times 12^3}{12} + 3600 \times 23198 = 43200 + 83512800 = 83556000$$ Calculate top flange: $$I_{tf} = \frac{300 \times 12^3}{12} + 3600 \times 25922 = 43200 + 93319200 = 93362400$$ Sum flanges: $$I_{flange} = I_{bf} + I_{tf} = 83556000 + 93362400 = 176918400 \text{ mm}^4$$ Calculate web: $$I_{web} = \frac{10 \times 301^3}{12} + 3010 \times (162.5 - 157.7)^2 = \frac{10 \times 27270901}{12} + 3010 \times 23.04 = 22725742 + 69362 = 22795104$$ 6. **Step 5: Total moment of inertia:** $$I = I_{flange} + I_{web} = 176918400 + 22795104 = 199713504 \text{ mm}^4$$ 7. **Step 6: Calculate section modulus $S$:** $$S = \frac{I}{c}$$ where $c = \max(\bar{y}, d - \bar{y}) = \max(157.7, 325 - 157.7) = 167.3$ mm $$S = \frac{199713504}{167.3} \approx 1.194 \times 10^6 \text{ mm}^3$$ 8. **Step 7: Calculate allowable bending moment $M_{allow}$:** $$M_{allow} = F_y \times S = 250 \times 1.194 \times 10^6 = 2.985 \times 10^8 \text{ N}\cdot\text{mm} = 298.5 \text{ kN}\cdot\text{m}$$ 9. **Step 8: Compare $M_{max}$ and $M_{allow}$:** $$M_{max} = 30.625 \text{ kN}\cdot\text{m} < M_{allow} = 298.5 \text{ kN}\cdot\text{m}$$ **Conclusion:** The beam is adequate for the given load since $M_{max} < M_{allow}$.