Design of Flexible Coupling.
Design of Shaft. The shaft Undergoes Torsional Shear stress . The diameter of the shaft can be calculated by
τmax=16T/πd3
Design of Key
The Dimension of the Key are set by using Standard Relation’s. Width of key Height of Key Length of Key
w=d/4 h=2/3*w or l=1.5d
h=w
The Key undergoes Direct Shear Stress as well as Crushing Stress and the Key needs to be checked for the Same. Check for Shear Stress
τd=2T(d*w*l)
Check for Shear Stress
σc=4T/(d*h*l)
Design of Hub
The Dimensions of the Hub are set using Standard Relations.
Outer Diameter “D”
Length of Hub “L” Key
D=2d
Length of Hub = Length of
The Hub undergoes Torsional Shear Stress and the Dimensions needs to be checked for the same.
Check for Torsional Shear Stress
τh=16T/πD3[1-k4] where k=d/D
For Safety
τh < τall
Design of Flange
The thickness of Flange and of Protective Flange are set by using Standard Relations.
Thickness of Flange
Thickness of Protective Flange
tf=0.5d tp=0.25d
The Flange Undergoes Direct Shear Stress and Needs to be Checked for the same. Check for Direct Shear Stress
τf=2T/πD2tf
For Safety
τf < τall
Design of Pin
The dimensions of the Pin are Set Using Standard relation’s. Number of Pins n
dp 1/2 d Dia0.5d/(n) of Enlarged portion 1 Outer Dia of Rubber Bush d2 Nominal Dia of Pin
n=3 for d less than 30mm n=4 for 30 < d ≤75 n=6 for 75< d ≤110 n=8 for 80< d ≤150
dp=
d1= db+4mm
d2=d1+16
D1 D1=D+d2+(2*8) Bolt Circle Dia
Length of Bush in Flange
Pb
Pb=2T/(nD1d2lb)
The Pin’s need to be checked for Max Shear Stress and Max Principal Stress. Max Shear Stress
τmax=[(σb/2)2+τb2]1/2
Max Principal Stress. [(σb/2)2+τb2]1/2
σmax=(σb/2)+
σb = {32*F*(lb/2)+5}/∏db3
F = T/[n*(D1/2)]
τb = F/[π * db2 /4]
For Safety
τmax < τall
Outer Diameter of Flange D2
D2=(2D1 - D)