Design a rectangular R.C.C beam to resist a bending moment of 65 kN-m. Use M 20 grade concrete and Fe 415 steel. Assume b = 0.5d

Q.Design a rectangular R.C.C beam to resist a bending moment of 65 kN-m. Use M 20 grade concrete and Fe 415 steel. Assume b = 0.5d

SOLUTION:

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. DEPTH REQUIRED :

M. R = Q.bd²

65×10⁶ = 0.914 x 0.5d×d²

d = ∛(65×10⁶ /0.5×0.914)

= 522mm

Provide  d= 530 mm and D= 570mm

4. AREA OF STEEL

Ast = M/(σst.j.d)

= 65×10⁶ / ( 230×0.904×530)

= 589.8 mm ²

providing 16mm dia bars

Area of one bar ast= π*16²/4 = 201.06 mm²

No of bars Required  n = Ast / ast = 589.8/201.06 = 3 bars

Provide 3 bars of 16 mm diameter, Ast Provided = 603.2 mm² 

Thank u