1. **Problem Statement:** We have given load combinations and are asked to compute service load (ASD) and factored load (LRFD) given Dead Load (D) = 450 kN, Live Load (L) = 200 kN, and Roof Live Load (Lr) = 80 kN. ### Load combinations using Allowable Stress Design (ASD): 2. Calculate service loads for each combination: - 1 D + F = $450 + 0 = 450$ kN - 1 D + H + F + L + T = $450 + 0 + 0 + 200 + 0 = 650$ kN - 1 D + H + F + (Lr or R) = $450 + 0 + 0 + 80 = 530$ kN - 1 D + H + F + 0.75[L + T(Lr or R)] = $450 + 0 + 0 + 0.75 [200 + 0 \, (80)] = 450 + 0.75 \times 200 = 600$ kN - 1 D + H + F + (0.6W or E / 1.4) = $450 + 0 + 0 \times (0.6 \times 0) = 450$ kN 3. The maximum service load $S = 650$ kN (from combination 2). ### Load combinations using Strength Design or Load and Resistance Factor Design (LRFD): 4. Calculate factored loads: - 1.4 (D + F) = $1.4 \times (450 + 0) = 630$ kN - 1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lr or R) = $1.2 (450 + 0 + 0) + 1.6 (200 + 0) + 0.5 (80) = 540 + 320 + 40 = 900$ kN - 1.2 D + 1.6 (Lr or R) + (f L or 0.5 W) = $1.2 \times 450 + 1.6 \times 80 + 1.0 \times 200 = 540 + 128 + 200 = 868$ kN - 1.2 D + 1.0 W + f L + 0.5 (Lr or R) = $1.2 \times 450 + 1.0 \times 0 + 1.0 \times 200 + 0.5 \times 80 = 540 + 0 + 200 + 40 = 780$ kN - 1.2 D + 1.0 F + f L = $1.2 \times 450 + 1.0 \times 0 + 1.0 \times 200 = 540 + 0 + 200 = 740$ kN - 0.9 D + 1.0 W + 1.6 H = $0.9 \times 450 + 1.0 \times 0 + 1.6 \times 80 = 405 + 0 + 128 = 533$ kN - 0.9 D + 1.0 E + 1.6 H = Same as above $= 533$ kN 5. The maximum factored load $U = 900$ kN (from combination 2). ### Moment of inertia and centroid calculations for the composite section: 6. Calculate moment of inertia $I$ for rectangular section: $$I = \frac{bh^3}{12} = \frac{350 \times 600^3}{12} = 6.3 \times 10^9 \, mm^4$$ 7. Compute centroid $\bar{y}_T$ for composite section: $$\bar{y}_T = \frac{350 \times 800 \times \frac{300}{2} + 500 \times 100 \times (300 + \frac{100}{2})}{350 \times 300 + 500 \times 100} = 214.52 \, mm$$ 8. Calculate total moment of inertia $I_T$ about centroid: $$I_T = \frac{350 \times 800^3}{12} + 350 \times 300^3 (214.52 - 150)^2 + \frac{500 \times 100^3}{12} + 500 \times 100 (400 - 214.52 - 50)^2 = 2.184 \times 10^9 \, mm^4$$ ### Stress calculations in reinforced concrete beam: 9. Given: - $b = 350 \, mm$, $h = 600 \, mm$ - Reinforcement area $A_s = 4 \times \frac{\pi}{4} (28)^2 = 784 \pi \, mm^2$ - Moduli and loads: $f_y=420$, $f_c=28$, $f_r=3.28$, $n=8$ - Moment $M = 60 \times 10^6 \, Nmm$ 10. Compute updated $I_T$ including reinforcement: $$I_T = 7.1429 \times 10^9 \, mm^4$$ 11. Calculate tensile and compressive stresses: - Tensile stress in concrete at bottom fiber: $$f_{ct} = \frac{M (600 - y_1)}{I_T} = 2.37 \, MPa$$ - Compressive stress in concrete: $$f_{cc} = \frac{M y_1}{I_T} = 2.67 \, MPa$$ - Stress in steel reinforcement: $$f_s = \frac{M (530 - y)}{I_T/8} = 14.28 \, MPa$$ 12. Since $f_{ct} = 2.37 < f_r = 3.28$, the concrete is uncracked under this loading. **Final answers:** - Maximum service load (ASD): $650$ kN - Maximum factored load (LRFD): $900$ kN - Moment of inertia $I_T$: approx $7.14 \times 10^9 \, mm^4$ - Concrete tensile stress $f_{ct} = 2.37$ MPa (uncracked) - Concrete compressive stress $f_{cc} = 2.67$ MPa - Steel reinforcement stress $f_s = 14.28$ MPa