Design of a Three-Span Continuous Beam

Given Data:
- UDL = 35 kN/m
- Clear Span = 4.5 m
- Support Width = 300 mm
- Concrete Grade = M25
- Steel Grade = Fe500

Step 1: Calculation of Dead Load
- Self-weight of beam = (0.23 x 1 x 4.5) + (0.23 x 2 x 4.5) = 2.07 kN/m
- Self-weight of slab = 0.1 x 25 = 2.5 kN/m
- Total Dead Load = (2.07 + 2.5) = 4.57 kN/m

Step 2: Calculation of Live Load
- Ultimate Load = 1.5 x (DL + LL) = 1.5 x (4.57 + 35) = 62.355 kN/m
- Factored Load = 1.2 x Ultimate Load = 1.2 x 62.355 = 74.826 kN/m
- Maximum Bending Moment = (wL^2)/8 = (74.826 x 4.5^2)/8 = 126.79 kNm

Step 3: Selection of Beam Section
- Adopt a rectangular section
- Depth of beam = clear span/10 = 4.5/10 = 0.45 m
- Width of beam = support width + effective depth of slab = 0.3 + 0.05 = 0.35 m
- Effective Depth = 0.45 - 0.05 = 0.4 m
- Area of Steel Required = (0.87 x fy x Ast x d)/fck = (0.87 x 500 x Ast x 0.4)/25
- Ast = 0.00314 m^2/m
- As per IS 456:2000, minimum reinforcement required = 0.85 x (0.18/100) x b x D = 0.85 x (0.18/100) x 350 x 400 = 21.21 mm^2/m
- Provide 4 bars of 12 mm diameter
- Provide a beam of size 350 mm x 450 mm

Step 4: Check for Deflection
- Limiting deflection = span/250 = 4.5/250 = 0.018 m
- Effective span = clear span + effective depth = 4.5 + 0.4 = 4.9 m
- Modular ratio (m) = Es/Ec = 2.5 x 10^5/5000 = 50
- Permissible deflection = (0.83 x span x LL)/m = (0.83 x 4.9 x 35)/50 = 1.145 mm
- Permissible deflection is less than the limiting deflection, hence the beam is safe in deflection.

Step 5: Check for Shear
- Shear force at the end support = wL/2 = 35 x 4.5/2 = 78.75 kN
- Shear force at mid-span = wL/4 = 35 x 4.5/4 = 39.375 kN
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