Q. Design a R.C.C beam 230 mm wide to resist a bending moment of 30 kN-m. Use M 20 grade concrete and Fe 415 steel.
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
30×10⁶ = 0.914 x 230×d²
d= √(30×10⁶ /230×0.914)
= 377.8mm
Provide d= 380 mm and D= 420mm
4. AREA OF STEEL
Ast = M/(σst.j.d)
= 30×10⁶ / ( 230×0.904×380)
= 379.7 mm ²
providing 12mm dia bars
Area of one bar ast= π*12²/4 = 113.1 mm²
No of bars Required n = Ast / ast = 379.7/113.1 = 4bars
Provide 4 bars of 12 mm diameter, Ast Provided = 452.4 mm²
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