1. **Problem Statement:** Calculate various moments and strengths for a rectangular reinforced concrete beam with given dimensions and material properties. 2. **Given Data:** - Width $b = 300$ mm - Effective depth $d = 525$ mm - Total depth $h = 600$ mm - Steel yield strength $f_y = 414$ MPa - Concrete compressive strength $f_c' = 27.6$ MPa - Reinforcement: 3 bars of 32 mm diameter 3. **Step (a): Moment causing first cracking using $I_g$ (gross moment of inertia):** - Calculate area of concrete section: $A_c = b \times h = 300 \times 600 = 180000$ mm$^2$ - Calculate moment of inertia of gross section about bottom fiber: $$I_g = \frac{b h^3}{12} = \frac{300 \times 600^3}{12} = 5.4 \times 10^9 \text{ mm}^4$$ - Calculate modulus of rupture $f_r$ (approximate cracking stress): $$f_r = 0.62 \sqrt{f_c'} = 0.62 \sqrt{27.6} = 3.26 \text{ MPa}$$ - Calculate distance from neutral axis to bottom fiber $c = \frac{h}{2} = 300$ mm - Calculate cracking moment: $$M_{cr} = f_r \times \frac{I_g}{c} = 3.26 \times \frac{5.4 \times 10^9}{300} = 5.87 \times 10^7 \text{ Nmm} = 58.7 \text{ kNm}$$ 4. **Step (b): Moment causing first cracking using $I_u$ (uncracked transformed section):** - Calculate steel area $A_s$: $$A_s = 3 \times \pi \times \left(\frac{32}{2}\right)^2 = 3 \times \pi \times 16^2 = 2413 \text{ mm}^2$$ - Modular ratio $n = \frac{E_s}{E_c} \approx \frac{200000}{4700 \sqrt{f_c'}} = \frac{200000}{4700 \times 5.25} = 8.1$ - Calculate transformed steel area $n A_s = 8.1 \times 2413 = 19545$ mm$^2$ - Calculate neutral axis depth $x$ from top by solving: $$b x = n A_s (d - x) \Rightarrow 300 x = 19545 (525 - x)$$ $$300 x = 19545 \times 525 - 19545 x$$ $$300 x + 19545 x = 19545 \times 525$$ $$19845 x = 10255125$$ $$x = \frac{10255125}{19845} = 517 mm$$ - Calculate moment of inertia of transformed section about bottom fiber: $$I_u = \frac{b x^3}{3} + n A_s (d - x)^2 = \frac{300 \times 517^3}{3} + 19545 \times (525 - 517)^2$$ $$= 1.37 \times 10^9 + 19545 \times 64 = 1.37 \times 10^9 + 1.25 \times 10^6 = 1.37125 \times 10^9 \text{ mm}^4$$ - Calculate cracking moment: $$M_{cr,u} = f_r \times \frac{I_u}{c} = 3.26 \times \frac{1.37125 \times 10^9}{300} = 1.49 \times 10^7 \text{ Nmm} = 14.9 \text{ kNm}$$ 5. **Step (c): Maximum moment without exceeding stress limits:** - Maximum concrete compressive stress $= 0.45 f_c' = 0.45 \times 27.6 = 12.42$ MPa - Maximum steel stress $= 0.40 f_y = 0.40 \times 414 = 165.6$ MPa - Calculate neutral axis depth $c$ using equilibrium: $$C = 0.85 f_c' b a, \quad a = \beta_1 c, \quad \beta_1 = 0.85$$ - Steel stress limit governs, so calculate $c$ from steel stress: $$f_s = E_s \varepsilon_s \leq 165.6 \text{ MPa}$$ - Approximate $c$ by iteration or assume balanced condition: $$c = \frac{A_s f_s}{0.85 f_c' b} = \frac{2413 \times 165.6}{0.85 \times 27.6 \times 300} = 57.3 \text{ mm}$$ - Calculate moment arm $z = d - a/2 = d - \beta_1 c / 2 = 525 - 0.85 \times 57.3 / 2 = 525 - 24.35 = 500.65$ mm - Calculate nominal moment capacity: $$M_{max} = A_s f_s z = 2413 \times 165.6 \times 500.65 = 2.0 \times 10^8 \text{ Nmm} = 200 \text{ kNm}$$ 6. **Step (d): Nominal and design flexural strength:** - Nominal flexural strength $M_n = M_{max} = 200$ kNm - Design strength $\phi M_n$ with $\phi = 0.9$ (typical): $$M_u = 0.9 \times 200 = 180 \text{ kNm}$$ **Final answers:** - (a) $M_{cr} = 58.7$ kNm - (b) $M_{cr,u} = 14.9$ kNm - (c) $M_{max} = 200$ kNm - (d) $M_n = 200$ kNm, $M_u = 180$ kNm