مثال محاسبه خاموتهاي برشي در تير
مثال محاسبه خاموتهاي برشي در تير
( Design of Shear reinforcement in a beam )
Design of Shear reinforcement in a beam
A reinforced cement concrete beam 300mm wide and 500mm effective depth is subjected to a shear force of 40KN at the ends. The beam is provided with 6 bars of 20mm diameter of which 3 bars are cranked at 45 degrees. Design the shear reinforcement for M20 grade concrete.
Here are the steps for the design of Shear Reinforcement in a beam:
Width of the beam = b = 300mm
Shear force = Vu = 40KN
Effective depth = d = 500mm
Area of steel, Ast = 3 x 3.14/4 x 20 x 20 = 942.47 mm2
Step one
Nominal shear stress
Tv = Vu/bd
Tv = 40 x 1000/(300 x 500) = 0.26N/mm2
Step two
Percentage of steel
Percent steel = Ast/bd x 100
Percent steel = (942.47 x 100)/ (300×500)
= 0.63%
Step three
As per IS: 456: 2000
Tc = 0.48 + (0.56-0.48)/(0.75-0.5) (0.63 – 0.5)
Tc= 0.52 N/mm2
Therefore, Tv < Tc
No shear reinforcement required.
Step four
Provide minimum shear reinforcement;
As per IS : 456 : 2000
Asv/bsv = 0.4/(0.87 fy)
Assuming 6mm diameter, 2 – legged stirrups
Asv = (2 x 3.14 x 6 x 6)/4 = 56.54 mm2
Sv = (0.87fy.Asv)/0.4b
Sv = (0.87 x 250 x 56.54)/(0.4×300) = 102.47mm say 100mm
As per IS:456:2000,
Maximum spacing = 0.75d
= 0.75 x 500
= 375mm
Provide 6mm diameter, 2-legged stirrups@100mm c/c.
Maximum spacing = 0.75d
= 0.75 x 230 = 172mm
Provide 6mm diameter 2-legged stirrups @ 100mmc/c
Shear reinforcement in a Beam
