The document Design Procedures For Rigid And Flexible Rubber Bushed Couplings Mechanical Engineering Notes | EduRev is a part of the Mechanical Engineering Course Machine Design.

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**Rigid Flange Coupling**

A typical rigid flange coupling is shown in **Figure- 5.1.2.1.4.2**. 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 diamete**r‘d’ based on torque transmission is given by

where T is the torque and τ_{y} is the yield stress in shear.

**(2) Hub diameter** d_{1} =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 τ_{y }is the yield stress in shear and L_{k} is the key length

This gives

If L_{k} 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 σ_{c} is crushing stress induced in the key. This gives . This gives and if σ_{c} < σ_{cy} , 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, τ_{yb} is the shear yield strength of the bolt material and d_{c} 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 t_{2} is the flange width over which the bolts make contact and σ_{cyb} is the yield crushing strength of the bolt material. This gives t_{2}. Clearly the bolt length must be more than 2t_{2} 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 t_{3} is less than t_{2}/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 d_{1} is the hub diameter and τ_{yf} is the shear yield strength of the flange material. Knowing τ_{yf} we may check if the chosen value of t_{2}is satisfactory or not. Finally, knowing hub diameter d_{1}, bolt diameter and protective thickness t_{2} we may decide the overall diameter d_{3}.

**Flexible rubber – **bushed couplings This is simplest type of flexible coupling and a typical coupling of this type is shown in Figure-5.2.2.1.

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-5.2.2.1)** 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 d_{r}= d_{b} +2t_{br} +2t_{r}

where d_{b} is the diameter of the bolt or pin , t_{br} is the thickness of the brass sleeve and t_{r} is the thickness of rubber bushing. We may now write

where d_{c }is the bolt circle diameter and t_{2} 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 d_{c} 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 d_{neck} is bolt diameter at the neck i.e at the shear plane.

**Bending Stress **

The pin loading is shown in Figure-5.2.2.2.

Clearly the bearing pressure that acts as distributed load on rubber bush would produce bending of the pin. Considering an equivalent concentrated load F= pt_{2}d 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 =

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