Rigid Flange Coupling
A typical rigid flange coupling is shown in Figure- 188.8.131.52.4.2. If essentially consists of two cast iron flanges which are keyed to the shafts to be joined. The flanges are brought together and are bolted in the annular space between the hub and the protecting flange. The protective flange is provided to guard the projecting bolt heads and nuts. The bolts are placed equi-spaced on a bolt circle diameter and the number of bolt depends on the shaft diameter d. A spigot ‘A’ on one flange and a recess on the opposing face is provided for ease of assembly. The design procedure is generally based on determining the shaft diameter d for a given torque transmission and then following empirical relations different dimensions of the coupling are obtained. Check for different failure modes can then be carried out. Design procedure is given in the following steps:
(1) Shaft diameter‘d’ based on torque transmission is given by
where T is the torque and τy is the yield stress in shear.
(2) Hub diameter d1 =1.75d +6.5mm
(3) Hub length L = 1.5d
But the hub length also depends on the length of the key. Therefore this length L must be checked while finding the key dimension based on shear and crushing failure mode
(4) Key dimensions: If a square key of sides b is used then b is commonly taken as d/4 . In that case, for shear failure we have
where τy is the yield stress in shear and Lk is the key length
If Lk determined here is less than hub length L we may assume the key length to be the same as hub length. For crushing failure we have
where This gives σc is crushing stress induced in the key. This gives . This gives and if σc < σcy , the bearing strength of the key material , the key dimensions chosen are in order.
(5) Bolt dimensions : The bolts are subjected to shear and bearing stresses while transmitting torque. Considering the shear failure mode we have
where n is the number of bolts, db the nominal bolt diameter, T is the torque transmitted, τyb is the shear yield strength of the bolt material and dc is the bolt circle diameter. The bolt diameter may now be obtained if n is known. The number of bolts n is often given by the following empirical relation:
where d is the shaft diameter in mm. The bolt circle diameter must be such that it should provide clearance for socket wrench to be used for the bolts. The empirical relation takes care of this Considering crushing failure we have
where t2 is the flange width over which the bolts make contact and σcyb is the yield crushing strength of the bolt material. This gives t2. Clearly the bolt length must be more than 2t2 and a suitable standard length for the bolt diameter may be chosen from hand book.
(6) A protecting flange is provided as a guard for bolt heads and nuts. The thickness t3 is less than t2/2 . The corners of the flanges should be rounded.
(7) The spigot depth is usually taken between 2-3mm. (8) Another check for the shear failure of the hub is to be carried out. For this failure mode we may write
where d1 is the hub diameter and τyf is the shear yield strength of the flange material. Knowing τyf we may check if the chosen value of t2is satisfactory or not. Finally, knowing hub diameter d1, bolt diameter and protective thickness t2 we may decide the overall diameter d3.
Flexible rubber – bushed couplings This is simplest type of flexible coupling and a typical coupling of this type is shown in Figure-184.108.40.206.
In a rigid coupling the torque is transmitted from one half of the coupling to the other through the bolts and in this arrangement shafts need be aligned very well. However in the bushed coupling the rubber bushings over the pins (bolts) (as shown in Figure-220.127.116.11) provide flexibility and these coupling can accommodate some misalignment. Because of the rubber bushing the design for pins should be considered carefully.
(1) Bearing stress
Rubber bushings are available for different inside and out side diameters. However rubber bushes are mostly available in thickness between 6 mm to 7.5mm for bores upto 25mm and 9mm thickness for larger bores. Brass sleeves are made to suit the requirements. However, brass sleeve thickness may be taken to be 1.5mm. The outside diameter of rubber bushing dr is given by dr= db +2tbr +2tr
where db is the diameter of the bolt or pin , tbr is the thickness of the brass sleeve and tr is the thickness of rubber bushing. We may now write
where dc is the bolt circle diameter and t2 the flange thickness over the bush contact area. A suitable bearing pressure for rubber is 0.035 N/mm2 and the number of pin is given by where d is in mm. The dc here is different from what we had for rigid flange bearings. This must be judged considering the hub diameters, out side diameter of the bush and a suitable clearance. A rough drawing is often useful in this regard. From the above torque equation we may obtain bearing pressure developed and compare this with the bearing pressure of rubber for safely.
(2) Shear stress
The pins in the coupling are subjected to shear and it is a good practice to ensure that the shear plane avoids the threaded portion of the bolt. Unlike the rigid coupling the shear stress due to torque transmission is given in terms of the tangential force F at the outside diameter of the rubber bush. Shear stress at the neck area is given by
where dneck is bolt diameter at the neck i.e at the shear plane.
The pin loading is shown in Figure-18.104.22.168.
Clearly the bearing pressure that acts as distributed load on rubber bush would produce bending of the pin. Considering an equivalent concentrated load F= pt2d the bending stress is
Knowing the shear and bending stresses we may check the pin diameter for principal stresses using appropriate theories of failure. We may also assume the following empirical relations:
Hub diameter = 2d
Hub length = 1.5d
Pin diameter at the neck =