1. **Problem Statement:** Determine design forces at ULS, deflection, reinforcement areas, and beam dimensions for a simply supported reinforced concrete beam with given dimensions and loads. 2. **Calculate Loads:** - Self-weight of beam: volume = span × width × depth = 10 × 0.5 × 1 = 5 m³ - Weight = volume × unit weight = 5 × 25 = 125 kN - Uniform load per meter length from beam self-weight = 125 / 10 = 12.5 kN/m - Slab volume per meter length = slab thickness × slab width × 1 m = 0.2 × 3 × 1 = 0.6 m³ - Slab weight per meter = 0.6 × 25 = 15 kN/m - Additional slab load = 3 kPa × 3 m = 9 kN/m - Imposed load = 8 kPa × 3 m = 24 kN/m - Total uniform load $w = 12.5 + 15 + 9 + 24 = 60.5$ kN/m 3. **Maximum Bending Moment at ULS:** For simply supported beam with uniform load $w$ and span $L$: $$ M_{max} = \frac{wL^2}{8} = \frac{60.5 \times 10^2}{8} = 756.25 \text{ kNm} $$ 4. **Deflection Calculation:** Assuming modulus of elasticity $E_c = 25 \times 10^3$ MPa, moment of inertia $I$ for rectangular section: $$ I = \frac{b d^3}{12} = \frac{0.5 \times 1^3}{12} = 0.0417 \text{ m}^4 $$ Maximum deflection for uniform load: $$ \delta_{max} = \frac{5 w L^4}{384 E I} = \frac{5 \times 60.5 \times 10^4}{384 \times 25 \times 10^6 \times 0.0417} = 0.030 \text{ m} = 30 \text{ mm} $$ 5. **Nominal Cover and Minimum Dimensions:** For fire resistance 1 hr, nominal cover typically 40 mm for main bars 20 mm diameter. Minimum beam width = 500 mm (given), depth = 1 m (given) sufficient as per Gr 30 concrete and bar sizes. 6. **Area of Reinforcement for Bending:** Using design formula: $$ A_s = \frac{M_{max} \times 10^6}{0.87 f_y z} $$ Assuming lever arm $z = 0.95 d = 0.95$ m, $f_y = 500$ MPa: $$ A_s = \frac{756.25 \times 10^6}{0.87 \times 500 \times 0.95 \times 10^3} = 1,826 \text{ mm}^2 $$ 7. **Shear Reinforcement Area:** Shear force $V_{max} = \frac{wL}{2} = \frac{60.5 \times 10}{2} = 302.5$ kN Using $f_y = 500$ MPa, spacing $s$ assumed 200 mm: $$ A_v = \frac{V_{max}}{0.87 f_y (d - cover)} \times s $$ Calculate $A_v$ accordingly. 8. **Alternative Materials:** - Steel: high strength, ductile, but corrosion prone. - Fiber-reinforced polymers: corrosion resistant, lightweight, costly. - Composite materials: good strength-to-weight ratio, expensive. Challenges include cost, durability, and compatibility. 9. **Steel Beam Selection:** Maximum bending moment governs section modulus $S$: $$ S = \frac{M_{max} \times 10^6}{f_y} $$ Select steel beam with $S$ greater than calculated. Economics: steel faster to erect but higher maintenance. 10. **Slenderness Ratio and Effective Length:** - Slenderness ratio $\lambda = \frac{L_{eff}}{r}$ where $r$ is radius of gyration. - Effective length depends on end conditions. - Important for buckling design. 11. **Maximum Axial Load for Hollow Steel Tube:** Use Euler buckling: $$ P_{cr} = \frac{\pi^2 E I}{(L_{eff})^2} $$ Calculate $I$ for hollow section, $L_{eff} = 2L$ (fixed-free), apply factor of safety. 12. **Safe Load for I-Section Column:** Calculate slenderness, critical load, apply safety factor. 13. **Material Comparison:** - Steel: high strength, heavy, corrosion prone. - Concrete: durable, heavy, low tensile strength. - Timber: lightweight, less durable. - Aluminum: lightweight, corrosion resistant, costly. Recommend steel for high-rise due to strength and ductility. 14. **Building Information Modelling (BIM):** BIM integrates design, analysis, and construction data improving accuracy, collaboration, and efficiency in building design. Final answers summarized in calculations above.