Let's solve each question step by step: --- ### **1. Minimum Required Number of 25-mm Bars at End H of Beam GH** #### **Step 1: Calculate Required Area of Steel (As,req)** The required area of steel at the end (support) is given by: \[ A_{s,req} = 0.5 \times A_{s,bal} \] But since details about moment at support are not given, a practical minimum based on ACI 318 or NSCP is (for beams): \[ A_{s,min} = 0.25\% \times b \times d \] Where: - \( b = 250 \) mm (width) - \( d = h - \text{cover} = 400 - 75 = 325 \) mm \[ A_{s,min} = 0.0025 \times 250 \times 325 = 203.125 \, \text{mm}^2 \] #### **Step 2: Area Provided by One 25-mm Bar** \[ A_{bar} = \frac{\pi}{4} \times (25)^2 = \frac{3.1416}{4} \times 625 = 490.87 \, \text{mm}^2 \] #### **Step 3: Number of Bars Required** \[ \text{Number of bars} = \frac{A_{s,min}}{A_{bar}} = \frac{203.125}{490.87} \approx 0.414 \] So, minimum 1 bar is enough, but code minimum for beams is at least **2 bars** at support. #### **Final Answer:** **C. 2** --- ### **2. Shear Force at End H of Beam GH** #### **Step 1: Calculate Factored Load (U)** \[ U = 1.4D + 1.7L \] - Dead Load (D) = 2.6 kPa - Live Load (L) = 3.6 kPa - Slab Thickness = 0.1 m, Unit Weight = 23.6 kN/m³ **Slab Weight:** \[ w_{slab} = 0.1 \times 23.6 = 2.36 \, \text{kN/m}^2 \] **Total Dead Load:** \[ D_{total} = 2.6 + 2.36 = 4.96 \, \text{kN/m}^2 \] **Factored Load:** \[ U = 1.4 \times 4.96 + 1.7 \times 3.6 = 6.944 + 6.12 = 13.064 \, \text{kN/m}^2 \] #### **Step 2: Tributary Width for Beam GH** \[ \text{Tributary width} = S = 2.6 \, \text{m} \] \[ w_{beam} = U \times S = 13.064 \times 2.6 = 33.97 \, \text{kN/m} \] #### **Step 3: Shear at End H (Simply Supported, max at end):** \[ V = \frac{w_{beam} \times L}{2} = \frac{33.97 \times 5.3}{2} = \frac{180.04}{2} = 90.02 \, \text{kN} \] Closest answer: **C. 78.4** But, let's check if a portion of the load is distributed differently. If the beam only takes half the span (assuming slab distributes load equally), then: \[ V = w_{beam} \times \frac{L}{2} = 33.97 \times 2.65 = 90.02 \, \text{kN} \] But none of the answers match exactly; **C. 78.4** is the closest. #### **Final Answer:** **C. 78.4** --- ### **3. Moment Capacity of Beam at H with 3-25 mm Bars (Top and Bottom)** #### **Step 1: Calculate Area of Steel Provided** \[ A_s = 3 \times 490.87 = 1,472.6 \, \text{mm}^2 \] #### **Step 2: Use Flexural Formula** \[ M_n = A_s f_y \left( d - \frac{a}{2} \right) \] Where \( a = \frac{A_s f_y}{0.85 f'_c b} \) - \( f_y = 415 \) MPa - \( f'_c = 27 \) MPa - \( b = 250 \) mm - \( d = 325 \) mm Find \( a \): \[ a = \frac{1472.6 \times 415}{0.85 \times 27 \times 250} = \frac{611,929}{5,737.5} = 106.7 \, \text{mm} \] \[ M_n = 1472.6 \times 415 \times \left( 325 - \frac{106.7}{2} \right) \times 10^{-6} \, \text{kN-m} \] \[ = 1472.6 \times 415 \times (325 - 53.35) \times 10^{-6} \] \[ = 1472.6 \times 415 \times 271.65 \times 10^{-6} \] \[ = 611,929 \times 271.65 \times 10^{-6} \] \[ = 166,319,069 \times 10^{-6} \] \[ = 166.3 \, \text{kN-m} \] Closest answer: **D. 168** --- ## **Final Answers** 1. **C. 2** 2. **C. 78.4** 3. **D. 168** --- If you need further clarification for any step or want the calculations shown in more detail, let me know!