Let's break this problem into steps as asked: ### **Given Data** - **Beam dimensions:** 380 mm (width, \(b\)), 750 mm (effective depth, \(d\)) - **Span (\(L\)):** 6 m = 600 mm - **Uniform load (\(w\)):** 60 kN/m = 60,000 N/m = 60 N/mm - **Reinforcement:** 9 bars of 16 mm dia (\(\phi\)), Fe415 at mid-span - **Concrete:** M20 - **Width of support:** 300 mm - **Bars to be curtailed:** 3 out of 9 - **No bends/hooks at curtailment** - **To be found:** Theoretical & actual cut-off points, development length checks (support/cut-off), minimum bars to be continued, and shear reinforcement design --- ## **Step (a): Theoretical Point of Cut-off** For a simply supported beam under UDL, the maximum moment is at mid-span. - Maximum BM at mid-span: \[ M_{max} = \frac{wL^2}{8} \] \[ M_{max} = \frac{60 \times 6^2}{8} = \frac{60 \times 36}{8} = \frac{216}{8} = 270 \text{ kNm} \] \[ = 270 \times 10^6 \text{ Nmm} \] - The point of zero moment is at supports, and the moment diagram is parabolic. - Theoretical cut-off point for bars is where the moment reduces to the moment that can be resisted by the curtailed reinforcement. Let the number of bars to be curtailed = 3 out of 9. Remaining bars = 6. Let \( x \) = distance from support where curtailment is proposed. - Moment at distance \( x \) from support: \[ M_x = \frac{w}{2} \left( Lx - x^2 \right) \] - Ratio of steel after curtailment: \[ \frac{A_{s,remain}}{A_{s,total}} = \frac{6}{9} = \frac{2}{3} \] - So, at curtailment, moment resisted = \(\frac{2}{3} M_{max}\) - Find \( x \) such that \( M_x = \frac{2}{3} M_{max} \): \[ \frac{w}{2}(Lx - x^2) = \frac{2}{3} M_{max} \] \[ 30(6x - x^2) = \frac{2}{3} \times 270 \] \[ 30(6x - x^2) = 180 \] \[ 6x - x^2 = 6 \] \[ x^2 - 6x + 6 = \] \[ x = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 3.464}{2} \] \[ x_1 = \frac{6 - 3.464}{2} = 1.268 \text{ m} \] \[ x_2 = \frac{6 + 3.464}{2} = 4.732 \text{ m} \] (Take the value from support: 1.27 m from the support.) ### **Answer (a):** **Theoretical cut-off point = 1.27 m from the support** --- ## **Step (b): Actual Point of Cut-off** IS 456 (Clause 26.2.3.3): Curtailment of bars should be such that the cut-off point is at a distance greater than or equal to the theoretical point plus 12 times the diameter, or effective depth, whichever is greater. - \( 12\phi = 12 \times 16 = 192 \) mm - Effective depth, \( d = 750 \) mm - So, provide extension = max(192, 750) = 750 mm - So, **actual point of cut-off**: \[ x_{actual} = x_{theoretical} + 750 \text{ mm} = 127 + 750 = 202 \text{ mm} = 2.02 \text{ m from support} \] ### **Answer (b):** **Actual point of cut-off = 2.02 m from the support** --- ## **Step (c): Check Development Length at Support** - **Development length (\( L_d \)), IS456:** \[ L_d = \frac{.87 f_y \phi}{4 \tau_{bd}} \] For M20, \(\tau_{bd} = 1.2\) N/mm² (Table 21, IS456) For Fe415, \( f_y = 415 \) N/mm², \( \phi = 16 \) mm \[ L_d = \frac{.87 \times 415 \times 16}{4 \times 1.2} = \frac{5788.8}{4.8} = 1206 \text{ mm} \] **Check: Embedding required = \( L_d \) or \( L_d/1.3 \) at support (IS456 Clause 26.2.3.3).** \[ \text{Embedment available} = \text{width of support} + \text{extension into support} \] Support width = 300 mm Extension into support = Typically, 50 mm (may be given in actual design). Let's assume full width. \[ \text{Available} = 300 \text{ mm} \] Required = \( L_d/1.3 = 1206/1.3 = 928 \) mm **Available < Required** **Hence, more bars need to be continued to satisfy development length.** --- ## **Step (d): Check Development Length at Cut-off** At the cut-off point, IS456 requires that the embedment beyond the cut-off point should be at least \( L_d \) or \( 12\phi \) or \( d \), whichever is greater. In this case, the extension after the theoretical point is already taken as 750 mm (which is less than \( L_d \)), so you may need to continue more bars or provide mechanical anchorage. --- ## **Step (e): Design of Shear Reinforcement** - **Shear force at the cut-off point:** \[ V = w \times \text{distance from support} \] \[ V = 60 \times (6/2 - 2.02) = 60 \times (3 - 2.02) = 60 \times .98 = 58.8 \text{ kN} \] - Nominal shear stress: \[ \tau_v = \frac{V}{b d} = \frac{58,800}{380 \times 750} = .206 \text{ N/mm}^2 \] - For M20, \(\tau_c\) for .5% steel is about .62 N/mm² (IS456 Table 19). Since \(\tau_v < \tau_c\), nominal links are sufficient. - **Provide 8 mm 2-legged stirrups @ 200 mm c/c** (nominal). --- ## **(f): Minimum Bars to be Continued at Support** To satisfy development length: \[ \frac{A_s}{A_{s,reqd}} = \frac{\text{Available embedment}}{\text{Required embedment}} = \frac{300}{928} = .323 \] So, minimum bars to be continued: \[ 9 \times .323 = 2.91 \implies 3 \text{ bars (minimum)} \] But, generally, at least 1/3rd of bars should be continued (IS456). **So, continue at least 3 bars to support.** --- ## **Summary Table** | Step | Result/Explanation | |------|---------------------------------------------------------------| | (a) | Theoretical cut-off point: **1.27 m from support** | | (b) | Actual cut-off point: **2.02 m from support** | | (c) | Development length at support: **Required = 928 mm, Available = 300 mm** (not sufficient) | | (d) | Development length at cut-off: **Extension after curtailment may be insufficient** | | (e) | Shear reinforcement: **8 mm 2-legged stirrups @ 200 mm c/c** | | (f) | Minimum bars to support: **3 bars (minimum required)** | --- ## **Final Answers** - **Theoretical cut-off point:** 1.27 m from support - **Actual cut-off point:** 2.02 m from support - **Development length at support:** Not sufficient, more bars must be continued to support - **Development length at cut-off:** Extension may be insufficient—bars should be continued or anchorage provided - **Shear reinforcement:** 8 mm 2-legged stirrups @ 200 mm c/c - **Minimum bars to continue to support:** 3 bars (at least 1/3rd of total) --- Let me know if you need calculations for any sub-step in more detail!