1. **Problem Statement:** Determine design forces at ULS, deflection, reinforcement areas, and analyze materials for a simply supported reinforced concrete beam and steel columns in a multi-story building. 2. **Beam Load Calculations:** - Beam span $L=10$ m, width $b=0.5$ m, depth $h=1$ m. - Concrete unit weight $\gamma_c=25$ kN/m³. - Slab thickness $t_s=0.2$ m, slab width $b_s=3$ m. - Additional slab load $q_{fin}=3$ kPa, imposed load $q_{imp}=8$ kPa. 3. **Calculate self-weight of beam:** $$w_b=\gamma_c \times b \times h=25 \times 0.5 \times 1=12.5\text{ kN/m}$$ 4. **Calculate slab load:** $$w_s=\gamma_c \times t_s \times b_s=25 \times 0.2 \times 3=15\text{ kN/m}$$ 5. **Calculate finishing and imposed loads:** $$q_{fin}=3 \text{ kPa} = 3 \text{ kN/m}^2$$ $$q_{imp}=8 \text{ kPa} = 8 \text{ kN/m}^2$$ 6. **Total uniformly distributed load on beam:** $$w_{total}=w_b + w_s + q_{fin} \times b_s + q_{imp} \times b_s=12.5 + 15 + 3 \times 3 + 8 \times 3=12.5 + 15 + 9 + 24=60.5\text{ kN/m}$$ 7. **Maximum bending moment at ULS for simply supported beam:** $$M_{max}=\frac{w_{total} L^2}{8}=\frac{60.5 \times 10^2}{8}=756.25\text{ kNm}$$ 8. **Maximum shear force at ULS:** $$V_{max}=\frac{w_{total} L}{2}=\frac{60.5 \times 10}{2}=302.5\text{ kN}$$ 9. **Deflection calculation:** - Assume modulus of elasticity $E_c=25 \times 10^3$ MPa. - Moment of inertia for rectangular section: $$I=\frac{b h^3}{12}=\frac{0.5 \times 1^3}{12}=0.0417\text{ m}^4$$ - Maximum deflection for uniformly loaded simply supported beam: $$\delta_{max}=\frac{5 w_{total} L^4}{384 E_c I}$$ Convert units: $$E_c=25 \times 10^3 \text{ MPa} = 25 \times 10^3 \times 10^6 \text{ N/m}^2=2.5 \times 10^{10} \text{ N/m}^2$$ $$w_{total}=60.5 \times 10^3 \text{ N/m}$$ $$L=10 \text{ m}$$ Calculate: $$\delta_{max}=\frac{5 \times 60.5 \times 10^3 \times 10^4}{384 \times 2.5 \times 10^{10} \times 0.0417}=0.003\text{ m} = 3 \text{ mm}$$ 10. **Nominal cover and minimum beam dimensions:** - For fire resistance 1 hr, nominal cover $c=40$ mm. - Minimum beam width $b_{min}=300$ mm, depth $h_{min}=500$ mm. - Given dimensions $b=500$ mm, $h=1000$ mm are sufficient. 11. **Area of tensile reinforcement $A_s$ at ULS:** - Use design strengths: $f_{cu}=30$ MPa, $f_y=500$ MPa. - Design moment $M_u=1.5 \times M_{max}=1.5 \times 756.25=1134.4$ kNm. - Effective depth $d=0.9h=0.9$ m. - Lever arm $z=0.95d=0.855$ m. - Calculate $A_s$: $$A_s=\frac{M_u \times 10^6}{f_y \times z}=\frac{1134.4 \times 10^6}{500 \times 10^6 \times 0.855}=2653 \text{ mm}^2$$ 12. **Shear reinforcement area $A_v$ at ULS:** - Shear force $V_u=1.5 \times V_{max}=1.5 \times 302.5=453.75$ kN. - Shear link diameter $d_{link}=10$ mm. - Use formula $A_v=\frac{V_u}{f_y \times d_{link}/s}$, assuming spacing $s=200$ mm. - Calculate: $$A_v=\frac{453750}{500 \times 10^6 \times 10/0.2}=1815 \text{ mm}^2$$ 13. **Alternative materials:** - Fiber reinforced polymers, steel, timber, aluminum. - Benefits: corrosion resistance, lightweight, ease of installation. - Challenges: cost, availability, design codes. 14. **Steel beam selection:** - Maximum bending moment governs size. - Use codes for section modulus and safety factors. - Economical and safe design balances material cost and performance. 15. **Steel column design:** - Slenderness ratio $\lambda=\frac{L_e}{r}$ where $L_e$ is effective length, $r$ radius of gyration. - Effective length depends on end conditions. - High slenderness reduces load capacity. 16. **Hollow steel tube column:** - External diameter $D=250$ mm, thickness $t=50$ mm, length $L=5$ m. - Radius of gyration $r=\sqrt{I/A}$. - Calculate $I$ and $A$: $$I=\frac{\pi}{64}(D^4 - (D-2t)^4)$$ $$A=\pi \times \frac{D^2 - (D-2t)^2}{4}$$ - Calculate critical load $P_{cr}=\frac{\pi^2 E I}{(L_e)^2}$ with $L_e=2L$ (fixed-free). - Factor of safety 2 applied. 17. **I-section column safe load:** - Use Euler buckling formula with $E=2 \times 10^5$ MPa. - Calculate moment of inertia and radius of gyration. - Determine slenderness and safe load. 18. **Material comparison:** - Steel: high strength, moderate weight, corrosion prone. - Reinforced concrete: high compressive strength, heavy, durable. - Timber: lightweight, less strength, susceptible to decay. - Aluminum: lightweight, corrosion resistant, lower strength. - Recommendation: Steel for high-rise due to strength and ductility. 19. **Building Information Modelling (BIM):** - BIM integrates design, analysis, and documentation. - Enhances coordination, reduces errors, improves efficiency. **Slug:** "beam design" **Subject:** "structural engineering" **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 19