Q.Design a RC rectangular beam simply supported over an effective span of 6 m to support an imposed load of 24 kN/m inclusive of its self weight. Use M 25 grade concrete and Fe 415 steel. Provide width of the beam equal to half of the depth. Use working stress method
Q.Design a RC rectangular beam simply supported over an effective span of 6 m to support an imposed load of 24 kN/m inclusive of its self weight. Use M 25 grade concrete and Fe 415 steel. Provide width of the beam equal to half of the depth. Use working stress method
SOLUTION:
Effective span l = 6m
Working load w = 24kN/m
Effective depth d = 2 b or b= d/2
1. PERMISSIBLE STRESSES
σcbc =8.5N/mm ² σst = 230 N / mm ²
2. DESIGN CONSTANTS FOR M20 CONCRETE AND Fe 415 STEEL :
• MODULAR RATIO
m = 280/3σcbc
= 280/ 3×8.5
= 10.98
• 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 x8.5×0.289 x [1 – ( 0.289 /3)]
= 1.11
m= 10.98,Kc = 0.289, j = 0.904 ,Q= 1.11
3. MAXIMUM BENDING MOMENT
M = WL²/8
=24×6²/8
=108×10⁶ N-mm=108 KN/mm
4. DEPTH REQUIRED :
Equating the maximum bending moment with the moment Of Resistance (as a balance section)
M. R = Q.bd²
180×10⁶ = 1.11 x ×b×(2b)²
b = ∛(108×10⁶ /1.11×2²)
= 290mm
Effective depth d= 2b = 2×290= 580 mm
Provide d= 580 mm and D=630mm
4. AREA OF STEEL
Ast = M/(σst.j.d)
= 108 ×10⁶ / ( 230×0.904×580)
= 895.6mm ²
providing 16 mm dia bars
Area of one bar ast= π*16²/4 = 201.1mm²
No of bars Required n = Ast / ast = 895.6/201.1 = 5 bars
Provide 5 bars of 16 mm diameter, Ast Provided = 1005.3 mm²
