Q.Design a RC rectangular beam simply supported over an effective span of 5m to support an imposed load of 20 kN/m inclusive of its self weight. Adopt M 20 grade concrete and Fe 415 steel. Provide effective depth 1.5 times the width. Use Working stress method
Q.Design a RC rectangular beam simply supported over an effective span of 5m to support an imposed load of 20 kN/m inclusive of its self weight. Adopt M 20 grade concrete and Fe 415 steel. Provide effective depth 1.5 times the width. Use Working stress method
SOLUTION:
Effective span l = 5m
Working load w = 20kN/m
Effective depth d = 1.5 b
1. PERMISSIBLE STRESSES
σcbc =7N/mm ² σst = 230 N / mm ²
2. DESIGN CONSTANTS FOR M20 CONCRETE AND Fe 415 STEEL :
• MODULAR RATIO
m = 280/3σcbc
= 280/ 3×7
= 13.33
• CRITICAL NEUTRAL AXIS FACTOR
Kc = 1/[1+(σst/m.σcbc)]
= 1/[1+(230/13.33×7)]
= 0.289
• LEVER ARM FACTOR
j = 1 – ( Kc / 3)
= 1 – ( 0.289 / 3)
= 0.904
• MOMENT OF RESISTANCE FACTOR
Q=1/2.σcbc.Kc.j
= 1/2.σcbc.Kc.[1 – ( Kc / 3)]
= 1/2 x7×0.289 x [1 – ( 0.289 /3)]
= 0.914
m= 13.33,Kc = 0.289, j = 0.904 ,Q= 0.914
3. MAXIMUM BENDING MOMENT
M = WL²/8
=20×5²/8
=62.5×10⁶ N-mm=62.5 KN/mm
4. DEPTH REQUIRED :
Equating the maximum bending moment with the moment Of Resistance (as a balance section)
M. R = Q.bd²
62.5×10⁶ = 0.914 x ×b×(1.5b)²
b = ∛(62.5×10⁶ /1.52×0.914)
= 312mm
Effective depth d= 1.5b = 1.5×312= 468 mm
Provide d= 470 mm and D= 520mm
4. AREA OF STEEL
Ast = M/(σst.j.d)
= 62.5 ×10⁶ / ( 230×0.904×470)
= 639.6 mm ²
providing 12 mm dia bars
Area of one bar ast= π*12²/4 = 113.1 mm²
No of bars Required n = Ast / ast = 639.6/113.1 = 6 bars
Provide 6 bars of 12 mm diameter, Ast Provided = 678.58 mm²
