# Reinforced Concrete Design Solution (Digital Format) Unless otherwise stated, all references are from: BS811-1 --- A simply supported reinforced concrete beam carries \( G_k = 30 \) kN/m including self-weight and \( Q_k = 10 \) kN/m. Has a span of 5 m. Design the beam for bending and shear if grade 30 and 460 are to be used for concrete and all steel respectively. Take \( d = 550 \) mm and \( b = 250 \) mm. --- ## Loading **Ultimate Load** \( F = 1.4 \times 30 + 1.6 \times 10 = 58 \) kN/m --- ## Bending **Maximum moment:** \[ \text{Maximum moment} = \frac{W \times L^2}{8} = \frac{58 \times 5^2}{8} = 181.25 \text{ kN.m} \] \[ K = \frac{M}{b d^2 f_{cu}} = \frac{181.25 \times 10^6}{250 \times 550^2 \times 30} = .079 < .156 \] Therefore, no compression reinforcement is required. \[ z = d \left[.5 + \sqrt{.25 - \frac{K}{.9}}\right] \] \[ z = .903d = 496.5 \text{ mm} \] \[ A_s = \frac{M}{.95 f_{y} z} = \frac{181.25 \times 10^6}{.95 \times 460 \times 496.5} = 836 \text{ mm}^2 \] **Provided:** Provide 3T20, area provided 942.6 mm² --- ## Shear **Maximum design shear load:** \[ \text{Maximum design shear load} = \frac{W \times L}{2} = \frac{58 \times 5}{2} = 145 \text{ kN} \] **Shear stress:** \[ v = \frac{V}{b \times d} = \frac{145 \times 100}{250 \times 550} = 1.05 \text{ N/mm}^2 < .8 \times \sqrt{30} \] **Shear links (at a distance d from face of support):** \[ V = 145 - 58 \times .55 = 113.1 \text{ kN} \] \[ v = \frac{V}{b \times d} = \frac{131.1 \times 100}{250 \times 550} = .954 \text{ N/mm}^2 \] \[ v_c = .79 \left(\frac{100A_s}{bd}\right)^{1/3} \left(\frac{400}{d}\right)^{1/4} \left(\frac{f_{cu}}{25}\right)^{1/3} \] Calculate each factor: \[ \left(\frac{100A_s}{bd}\right)^{1/3} = \left(\frac{100 \times 942.6}{250 \times 550}\right)^{1/3} = (.685)^{1/3} = .88 \] \[ \left(\frac{400}{d}\right)^{1/4} = \left(\frac{400}{550}\right)^{1/4} = (.727)^{1/4} = .923 \] \[ \left(\frac{f_{cu}}{25}\right)^{1/3} = \left(\frac{30}{25}\right)^{1/3} = (1.2)^{1/3} = 1.05 \] \[ v_c = .79 \times .88 \times .923 \times 1.05 = .51 \text{ N/mm}^2 \] \[ A_{sv} = \frac{b(v - v_c)S_v}{.95 f_{yv}} \] \[ A_{sv} = \frac{250(.954 - .51) \times 200}{.95 \times 460} = .254 \] **Provided:** Provide T8 at 200 mm c/c \[ \frac{A_{sv}}{S_v} = \frac{2 \times 50.3}{200} = .5 \quad \text{OK} \] --- ## Deflection **Basic span-effective depth ratio:** \[ \frac{500}{550} = 9.1 \] **Basic ratio for simply supported rectangular beam:** 20 **Modification factor for tension reinforcement:** \[ f_s = \frac{2f_y A_{s,req}}{3A_{s,prov}} \cdot \frac{1}{\beta_b} \] \[ f_s = \frac{2 \times 460 \times 836}{3 \times 942.6} = 407.98 \] \[ \frac{M}{bd^2} = \frac{181.25 \times 10^6}{250 \times 550^2} = 2.4 \] **Modification factor:** \[ \text{Modification factor} = .55 + \frac{(477 - f_s)}{120 (.9 + \frac{M}{bd^2})} \leq 2. \] \[ = .55 + \frac{(477 - 407.98)}{120 (.9 + 2.4)} = .73 \] \[ \text{Basic ratio} = 20 \times .725 = 14.5 \] \[ \text{Actual ratio} = 9.1 < \text{basic ratio} = 14.5 \quad \text{OK} \] Deflection OK --- **Summary:** - Provide 3T20 main bars (area provided: 942.6 mm²) - Provide T8 links at 200 mm c/c - Only two 20 mm bars extend a distance d from the support face - Deflection is within allowable limits All calculations and code references are preserved as in the provided solution.