Example of the Design of Shear reinforcement in a beam

Example of the 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