Beam Shear Design

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Example for beam shear design for Eurocoe 2
Effective depth and width of 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