Related Resources: mechanics machines
Shaft Couplings Design Equation and Calculator
Preview: Solid Rigid Couplings Design Calculator
The a shaft to shaft solid coupling does not allow for misalignment, except axial, but enables the extension from one piece of equipment to another. In its simplest form, the rigid coupling is nothing more than a piece of bar stock bored to receive two shafts, as shown above. Its torquehandling capacity is limited only by the strength of the material used to make the connection. The coupling is installed on one shaft before the equipment is lined up, and the mating equipment is brought into position without much chance of accurate alignment when the equipment is bolted into position.
The maximum shear stress occurs at the outer radius of the coupling and at the interface of the two bores. This stress can be derived from the torsion formula:
τ_{max} = T D_{o} / 2 J
Where J, the polar second moment of intertia is:
J = π / 32 ( D^{4}_{o}  D^{4}_{i} )
The coupling must be sized so that, typically, the stress does not exceed 10 percent of the ultimate tensile strength of the material.
Other factors to consider are the length of engagement into the coupling. The shear stress over the keyway must not exceed the allowable shear stress as given above. The centroidal radius is:
R_{c} = 0.5 * ( D_{o} / 2 + D_{i} / 2 + h )
The centroid of the bearing area is at radius (D_{i} + h) / 2. If the transmitted torque is T, then the compressive force F is 2 T / (D_{i} + h). The bearing stress σ_{b} is:
σ_{b} = F / A = 4 T / ( w L (D_{i} + h) )
The allowable compressive stress from distortion energy theory of failure is σ_{all} = τ_{all} / 0.577. Combining this equation σ_{b} with gives:
τ_{all} = [ 0.577 ( 4 ) T ] / [ w L (D_{i} + h) ]
Next, the length of key stock, for keyed shafts, must be examined to keep its shear
loading from exceeding the allowable shear stress. Referring to the equation for σ_{b} , we note
that the shear force is:
F = T / τ /2) = 2 T / D_{i} Therefore the average shear stress is
τ = F / A = ( 2 T ) / ( w L D_{i} )
Both keys must be checked, although experience has shown that small diameter shafts are more prone to failure of the key and keyway when these precautions are not followed because of their normally smaller key width and length of engagement.
Maximum Allowable Shear Stress 

Material 
Stress, psi 
Aluminum 
4000 
Cast Iron 
4500 
Ductile Iron 
6000 
Brass 
3500 
Powered Iron 
4000 
Steel 
8000 
Where:
T = Torque, lbin
D_{o} = Diameter Outer, in
D_{i} = Diameter inner, in
J = Polar Second Moment of Inertia, in^{3}
R_{c}= Centroidal radius or distance, in
F = Compressive Force, lbf
A = Cross Sectional Area (in^{2})
w = Width, in
L = Length, in
h = Keyway height, in
τ_{all} = Compresive Shear Stress, psi
τ = Shear Stress, psi
References:
ISO R773, Rectangular or Square Parallel Keys and Their Corresponding Keyways, 1969.
JISB 1301, Bore and Keyway Tolerances.
© Copyright 2000  2019, by Engineers Edge, LLC www.engineersedge.com
All rights reserved
Disclaimer
 Feedback  Advertising
 Contact
Date/Time: