1. **Problem Statement:** Investigate the adequacy of a 5 m steel wide flange beam carrying a uniform load of 25 kN/m. The beam properties are: $b_f=300$ mm, $t_f=10$ mm, $d=300$ mm, $t_w=12$ mm. Material properties: $F_y=250$ MPa, $E=200000$ MPa. 2. **Step 1: Calculate the maximum moment $M_{max}$ for a simply supported beam with uniform load $w$ over length $L$:** $$M_{max} = \frac{wL^2}{8}$$ Given $w=25$ kN/m and $L=5$ m, $$M_{max} = \frac{25 \times 5^2}{8} = \frac{25 \times 25}{8} = 78.125 \text{ kN}\cdot\text{m}$$ Convert to Nmm: $$78.125 \times 10^3 \times 10^3 = 7.8125 \times 10^7 \text{ Nmm}$$ 3. **Step 2: Calculate the section modulus $S$ of the wide flange beam:** Approximate $S$ for an I-beam: $$S = \frac{I}{c}$$ where $I$ is the moment of inertia and $c$ is the distance from neutral axis to extreme fiber. 4. **Step 3: Calculate moment of inertia $I$ for the beam cross-section:** The beam has two flanges and a web. - Flange area $A_f = b_f \times t_f = 300 \times 10 = 3000$ mm$^2$ - Web area $A_w = t_w \times (d - 2t_f) = 12 \times (300 - 20) = 12 \times 280 = 3360$ mm$^2$ Neutral axis is approximately at mid-depth $d/2 = 150$ mm due to symmetry. Moment of inertia $I$ about neutral axis: $$I = 2 \times A_f \times (c - \frac{t_f}{2})^2 + \frac{t_w \times (d - 2t_f)^3}{12}$$ where $c = d/2 = 150$ mm. Calculate flange contribution: $$2 \times 3000 \times (150 - 5)^2 = 2 \times 3000 \times 145^2 = 6000 \times 21025 = 126150000 \text{ mm}^4$$ Calculate web contribution: $$\frac{12 \times 280^3}{12} = 12 \times \frac{280^3}{12} = 280^3 = 280 \times 280 \times 280 = 21952000 \text{ mm}^4$$ Total $I$: $$I = 126150000 + 21952000 = 148102000 \text{ mm}^4$$ 5. **Step 4: Calculate section modulus $S$:** $$S = \frac{I}{c} = \frac{148102000}{150} = 987346.7 \text{ mm}^3$$ 6. **Step 5: Calculate the nominal moment capacity $M_n$:** $$M_n = F_y \times S = 250 \times 987346.7 = 2.468 \times 10^8 \text{ Nmm} = 246.8 \text{ kN}\cdot\text{m}$$ 7. **Step 6: Compare applied moment $M_{max}$ with nominal capacity $M_n$:** $$M_{max} = 78.125 \text{ kN}\cdot\text{m} < M_n = 246.8 \text{ kN}\cdot\text{m}$$ Since $M_{max} < M_n$, the beam is adequate for the given load. **Final answer:** The beam is adequate to carry the uniform load of 25 kN/m over 5 m.