1. **Problem Statement:** A square footing supports a 300mm x 500mm RC column with dead load 770 kN and live load 420 kN. The footing bottom is 1.5m below ground. Soil weight is 18.7 kN/m³, allowable soil bearing pressure 130 kPa. Footing thickness is 500mm, concrete cover 75mm, concrete strength $f'c=21$ MPa, steel yield strength $f_y=415$ MPa, reinforced with 20mm bars, concrete unit weight 23.5 kN/m³. Find the design strength for wide beam shear and check safety. 2. **Step 1: Calculate total load on footing** $$P_u = 1.2 \times \text{Dead Load} + 1.6 \times \text{Live Load} = 1.2 \times 770 + 1.6 \times 420 = 924 + 672 = 1596\text{ kN}$$ 3. **Step 2: Determine footing size** Allowable soil pressure $q_{allow} = 130$ kPa = 130 kN/m². Required area: $$A = \frac{P_u}{q_{allow}} = \frac{1596}{130} = 12.2769\text{ m}^2$$ Since footing is square: $$\text{Side length } = \sqrt{12.2769} = 3.505\text{ m}$$ 4. **Step 3: Calculate effective depth $d$** Thickness $h = 500$ mm = 0.5 m Concrete cover = 75 mm = 0.075 m Bar diameter = 20 mm = 0.02 m Assuming effective depth: $$d = h - \text{cover} - \frac{\text{bar diameter}}{2} = 0.5 - 0.075 - 0.01 = 0.415\text{ m}$$ 5. **Step 4: Calculate factored shear force $V_u$ at critical section** Critical section for wide beam shear is at distance $d$ from face of column. Column dimensions: 0.3 m x 0.5 m Length of critical section perimeter: $$b_0 = 2(0.3 + 0.5) + 4d = 2(0.8) + 4(0.415) = 1.6 + 1.66 = 3.26\text{ m}$$ Area of critical section: $$A_v = b_0 \times d = 3.26 \times 0.415 = 1.3529\text{ m}^2$$ Weight of footing: $$W_f = \text{footing volume} \times \text{concrete unit weight} = (3.505)^2 \times 0.5 \times 23.5 = 12.28 \times 0.5 \times 23.5 = 144.19\text{ kN}$$ Weight of soil above footing: $$W_s = \text{footing area} \times \text{depth} \times \text{soil unit weight} = 12.28 \times 1.5 \times 18.7 = 344.6\text{ kN}$$ Total vertical load at critical section: $$V_u = P_u + W_f + W_s = 1596 + 144.19 + 344.6 = 2084.79\text{ kN}$$ 6. **Step 5: Calculate nominal shear strength $V_c$ of concrete** Using ACI 318 for wide beam shear: $$V_c = 0.17 \sqrt{f'_c} b_0 d$$ Where $f'_c$ in MPa, $b_0$ and $d$ in mm: Convert $b_0$ and $d$ to mm: $$b_0 = 3.26\text{ m} = 3260\text{ mm}, \quad d = 415\text{ mm}$$ Calculate: $$V_c = 0.17 \times \sqrt{21} \times 3260 \times 415 = 0.17 \times 4.583 \times 3260 \times 415$$ $$= 0.17 \times 4.583 \times 1,352,900 = 0.17 \times 6,198,000 = 1,053,660\text{ N} = 1053.66\text{ kN}$$ 7. **Step 6: Check design shear strength** Design shear strength: $$\phi V_c = 0.75 \times 1053.66 = 790.25\text{ kN}$$ Compare with factored shear $V_u = 2084.79$ kN Since $V_u > \phi V_c$, the design is **not safe** for wide beam shear. **Final answer:** Design strength for wide beam shear is approximately 790 kN, which is less than the factored shear force 2085 kN. Therefore, the design is not safe for wide beam shear and requires redesign or additional reinforcement.