Beams Shear Design to Eurocode 2

This article is a worked example to beam shear design as per Eurocode 2. This example discusses the method that needs to be followed when designing for shear.

Calculation procedure for beam shear capacity and shear links are discuss in detail.

The effective depth and width of the beam are 600mm and 300mm respectively.
Ultimate design shear force at face of support and at a distance ‘d ‘ are 800kN and 700kN respectively.
fck =30N/mm² and fyk = 460N/mm²

Check for maximum shear force
Consider angle as  22 degrees
VRD,max (22)   = {0.36*bw*d*(1-fck/250)*fck}/(cot22 + tan 22 )  
                  =  0.124*bw*d*(1-fck/250)*fck
                  =  0.124*300*600*(1-30/250)*30
                  =  589.25 
Hence VRD,max (22) < 700kN
There for consider angle 45 degrees
VRD,max (45)   = {0.36*bw*d*(1-fck/250)*fck}/(cot45 + tan 45 ) 
                  =  0.18*bw*d*(1-fck/250)*fck
                  =  0.18*300*600*(1-30/250)*30
                  =  855.36kN
Hence angle is between 22 degrees and 45 degrees, further Maximum shear is ok. (VRD,max (45)> 800kN)

Calculation of angle
Angle       =  0.5*Sin-1{VEf/[0.18*bw*d*(1-fck/250)*fck]}
                  =  0.5*Sin-1(VEf)/(VRD,max (45)) <= 45
                  =  0.5*Sin-1(700/855.36)
                  =  27.5 degrees 

Calculation of shear links
ASW/S     =   VED/(0.78*d*fyk*cot27.5)
                 =  700*1000/(0.78*600*460*1.92
                 =  1.69
Assume T10 bars are using for shear links              
2*113/S   =  1.69  
S             =  134mm

Proved T10@150mm c/c