## Cantilever Beam Design Solution ### **Given Data** - **Length of cantilever, \( L \) = 2.25 m** - **Uniformly distributed load (UDL), \( w \) = 20 kN/m** - **Point load at free end, \( P \) = 6 kN** - **Width of beam, \( b \) = 230 mm** - **Width of support = 300 mm** - **Concrete grade: M15 (\( f_{ck} = 15 \) MPa)** - **Steel grade: Fe415 (\( f_y = 415 \) MPa)** --- ### **Step 1: Calculate Maximum Moments** #### **A. Due to UDL:** \[ M_{UDL} = \frac{w L^2}{2} \] \[ = \frac{20 \times (2.25)^2}{2} = \frac{20 \times 5.0625}{2} = \frac{101.25}{2} = 50.625 \text{ kNm} \] #### **B. Due to Point Load:** \[ M_{P} = P \times L = 6 \times 2.25 = 13.5 \text{ kNm} \] #### **C. Total Maximum Moment at the Face of Support:** \[ M_{max} = M_{UDL} + M_{P} = 50.625 + 13.5 = 64.125 \text{ kNm} \] \[ = 64,125 \text{ Nm} = 64,125,000 \text{ Nmm} \] --- ### **Step 2: Assume Depth of Beam (\( d \))** Let’s assume an effective depth (\( d \)): \[ d = \frac{M_{max}}{.138 f_{ck} b} \] But first, let's estimate \( d \): Try \( d = 450 \) mm (typical for cantilever beams, but check adequacy). --- ### **Step 3: Check Depth for Flexure** The moment of resistance for a singly reinforced beam: \[ M_{u,lim} = .138 f_{ck} b d^2 \] Let’s use: - \( f_{ck} = 15 \) MPa - \( b = 230 \) mm - \( M_{u,lim} = 64,125,000 \) Nmm \[ d = \sqrt{\frac{M_{u}}{.138 \times f_{ck} \times b}} \] \[ d = \sqrt{\frac{64,125,000}{.138 \times 15 \times 230}} \] \[ = \sqrt{\frac{64,125,000}{476.1}} \] \[ = \sqrt{134,697} \] \[ = 367 \text{ mm} \] Adopt **effective depth \( d = 370 \) mm**. Assume overall depth \( D = d + 50 \) mm (cover and bar diameter) \( \approx 420 \) mm. --- ### **Step 4: Area of Tension Reinforcement (\( A_{st} \))** \[ M_u = .87 f_y A_{st} d \left[1 - \frac{A_{st} f_y}{f_{ck} b d}\right] \] For preliminary design, ignore the bracketed term: \[ A_{st} = \frac{M_u}{.87 f_y d} \] \[ A_{st} = \frac{64,125,000}{.87 \times 415 \times 370} \] \[ = \frac{64,125,000}{133,459.5} \] \[ = 480.3 \text{ mm}^2 \] --- ### **Step 5: Provide Reinforcement** Use 2 bars of 16 mm diameter: \[ \text{Area of 1 bar} = \frac{\pi}{4} \times 16^2 = 201 \text{ mm}^2 \] \[ \text{Total for 2 bars} = 402 \text{ mm}^2 \] Use 2 bars of 20 mm diameter: \[ \text{Area of 1 bar} = \frac{\pi}{4} \times 20^2 = 314 \text{ mm}^2 \] \[ \text{Total for 2 bars} = 628 \text{ mm}^2 \] **Provide 2 bars of 20 mm diameter as main tension reinforcement.** --- ### **Step 6: Curtailment of Tension Reinforcement** For cantilever beams, the main reinforcement should extend up to the free end; no curtailment is recommended. Provide minimum distribution steel as well. --- ### **Step 7: Check for Shear** \[ V_{max} = w \times L + P = 20 \times 2.25 + 6 = 45 + 6 = 51 \text{ kN} \] Shear strength of concrete (\( \tau_{c} \)) for M15 and .5% steel = .36 N/mm² (from IS 456 Table 19). \[ V_{c} = \tau_{c} \times b \times d = .36 \times 230 \times 370 = 30,636 \text{ N } = 30.6 \text{ kN } \] Since \( V_{max} > V_c \), design shear reinforcement (stirrups) is required. --- ### **Step 8: Provide Shear Reinforcement** Use 2-legged 8 mm stirrups @ 150 mm c/c near support. --- ## **Summary of Design** - **Beam size:** 230 mm (width) × 420 mm (overall depth) - **Main reinforcement:** 2 bars of 20 mm diameter (Fe415) - **Shear reinforcement:** 2-legged 8 mm stirrups @ 150 mm c/c near support - **Concrete:** M15 - **Curtailment:** Main bars to be continued to the free end (no curtailment) --- If you need detailed drawings or further checks (deflection, crack width, etc.), let me know!