This article is a practical example of shear beam design according to Eurocode 2. This example explains the method to be followed in shear design.
The procedure for calculating the shear capacity of beams and connections is explained in detail.
The effective depth and width of the beam are 600 mm and 300 mm, respectively.
The maximum shear force on the support surface and at distance “d” is 800 kN and 700 kN, respectively.
fck = 30 N/mm 2 and fyk = 460N/mm 2
Check maximum shear force
Consider the angle as 22 degrees
VRD,max (22) = {0.36*bw*d*(1-fck/250)*fck}/(cot22 + tan 22 )
= 0.124*b*d*(1-fck/250)*fck
= 0.124*300*600*(1-30/250)*30
= 589.25
Therefore VRD,max (22) <700kN
An angle of 45 degrees must be taken into account.
VRD,max (45) = {0.36*bw*d*(1-fck/250)*fck}/(cot45 + tan 45 )
= 0.18*b*d*(1-fck/250)*fck
= 0.18*300*600*(1-30/250)*30
= 855.36 kN
So the angle is between 22 degrees and 45 degrees, and the maximum shear is ok. (VRD,max. (45)>800kN)
Angle calculation
Angle = 0.5*Sin-1{VEf/(0.18*bw*d*(1-fck/250)*fck)}
= 0.5*Sen-1(VEf)/(VRD,max (45)) <= 45
= 0.5*Sin-1(700/855.36)
= 27.5 degrees
Calculation of shear elements
ASW/S = VED / (0.78*d*fyk*cot27.5)
= 700*1000/(0.78*600*460*1.92
= 1.69
Suppose T10 bars are used for shear connections
2*113/ S = 1.69
S = 134mm
Proven T10 with 150mm center distance
