1. **Stating the problem:** Calculate service load and factored load for a structure given Dead Load (D) = 450 kN, Live Load (L) = 200 kN, Roof Live Load (Lr) = 80 kN. 2. **Load combinations using Allowable Stress Design (ASD):** - 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)] = $450 + 0 + 0 + 0.75 (200 + 0) = 600$ kN - 1 D + H + F + (0.6W or E/1.4) = $450 + 0 + 0 + 0 = 450$ kN Therefore, service load $S = 650$ kN (maximum). 3. **Load combinations using Load and Resistance Factor Design (LRFD):** - $1.4 (D + F) = 1.4 (450 + 0) = 630$ - $1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lr or R) = 1.2 (450 + 0 + 0) + 1.6 (200 + 0) + 0.5 (80) = 900$ - $1.2 D + 1.6 (Lr \, or \, R) + f L = 1.2 (450) + 1.6 (80) + 1.0 (200) = 868$ - $1.2 D + 1.0 W + f_1 L + 0.5 (Lr \, or \, R) = 1.2 (450) + 0 + 1.0 (200) + 0.5 (80) = 780$ - $1.2 D + 1.0 E + f_1 L = 1.2 (450) + 0 + 1.0 (200) = 740$ - $0.9 D + 1.0 W + 1.6 H = 0.9 (450) + 0 + 1.6 (80) = 405$ - $0.9 D + 1.0 E + 1.6 H = 0.9 (450) + 0 + 1.6 (80) = 405$ Hence, the factored load $U = 900$ kN (maximum). 4. **Moment of inertia $I$ for rectangular section:** Given width $b=350$ mm and height $h=600$ mm, $$I = \frac{b h^3}{12} = \frac{350 \times 600^3}{12} = 6.3 \times 10^9 \text{ mm}^4$$ 5. **Composite centroid $y_T$ calculation:** $$y_T = \frac{350 \times 800 \times (300/2) + 500 \times 100 \times (300 + 100/2)}{350 \times 300 + 500 \times 100} = 214.57 \text{ mm}$$ 6. **Total moment of inertia $I_T$ for composite section:** $$I_T = \frac{350 \times 800^3}{12} + 350 \times 300^3 (y_T - 150)^2 + \frac{500 \times 100^3}{12} + 500 \times 100 (400 - y_T - 50)^2 = 2.184 \times 10^9 \text{ mm}^4$$ 7. **For beam with width 350 mm, depth 600 mm, and effective depth 530 mm, with 4 tension bars of diameter 28 mm:** - Area of steel $A_s = 4 \times \frac{\pi}{4} \times 28^2 = 784 \pi$ mm$^2$ - Given strengths: $f_y=420$ MPa, $f_c'=28$ MPa, $f_r=3.28$ MPa, modular ratio $n=8$ - Bending moment $M=60$ kN·m = $60 \times 10^6$ N·mm 8. **Moment of inertia $I_T$ including reinforcement:** $$I_T = \frac{350 \times 600^3}{12} + 350 \times 600 (y_T - 300)^2 + (8-1) \times 784\pi (530 - y_T)^2 = 7.143 \times 10^9 \text{ mm}^4$$ 9. **Calculate stresses:** - Concrete tension: $$f_{ct} = \frac{M (600 - y_T)}{I_T} = \frac{60 \times 10^6 (600 - 214.57)}{7.143 \times 10^9} = 2.37 \text{ MPa}$$ - Concrete compression: $$f_{cc} = \frac{M y_T}{I_T} = \frac{60 \times 10^6 (214.57)}{7.143 \times 10^9} = 2.67 \text{ MPa}$$ - Steel stress: $$\frac{f_s}{n} = \frac{M (530 - y_T)}{I_T} \Rightarrow f_s = 8 \times \frac{60 \times 10^6 (530 - 214.57)}{7.143 \times 10^9} = 14.28 \text{ MPa}$$ 10. **Compare tensile concrete stress with fracture strength:** $$2.37 < 3.28 \Rightarrow \text{Uncracked section}$$ Hence, stresses are within safe limits as per the given parameters and loads. **Final answers:** - Service load $S = 650$ kN - Factored load $U = 900$ kN - Moment of inertia: $I = 6.3 \times 10^9$ mm$^4$ - Composite moment of inertia: $I_T = 7.143 \times 10^9$ mm$^4$ - Concrete compression stress $f_{cc} = 2.67$ MPa - Concrete tension stress $f_{ct} = 2.37$ MPa - Steel stress $f_s = 14.28$ MPa These demonstrate the load calculations and structural stress analysis based strictly on the given data and illustrated formulas.