FLANGED COUPLING:

What is coupling? A device that is used to connect two shafts together

for the purpose of power transmission.

General types of couplings:

Rigid: for aligned shafts Flexible: for non-aligned shaft

Rigid couplings:Rigid couplings are aligned shaft couplings that are designed to draw two shafts together so that no motion can occur between them

Types of rigid couplings:

1-Flanged 2-Sleeve/Muff Coupling

3-Clamp/Split muff/Compression Coupling

Flexible couplings:i. Used to join shafts that meet at a slight angle. ii. Angle may still change while running due to vibration

or load.

Types of flexible couplings:1-Universal Coupling2-Bushed Pin Type Coupling

3-Olbham Coupling

Flanged Coupling:Flange:

A flange is defined as a plate type device, normally round, that is attached to the end of a pipe, fitting, valve or other object to facilitate the assembly and disassembly of a piping system.

Flange coupling : Each flange is mounted

on the shaft and keyed to it. Faces are turned up at right angle to the axis of the shafts. One of the flange has a

projected portion and other has a corresponding recess.

Two flanges are coupled together by bolts and nuts.

Types of flange couplings:1-Protected type flange coupling

2-Unprotected type flange coupling

3-Marine type flange coupling

Unprotected type flange coupling:In this type of flange coupling each shaft is keyed to the boss of a flange with a counter sunk key and the flanges are coupled together by means of bolts (Generally 3,4,6 bolts are used.)Keys are staggered at right angle along the circumference of the shafts in order to divide weakening effect caused by key ways.

Unprotected type flange coupling:

The usual proportions for an unprotected type cast iron flange couplings, are shown as follows:

Let d = Diameter of shaft or inner diameter of hub,

D = Outer diameter of hub, d1 = Nominal or outside diameter of bolt, D1 = Diameter of bolt circle, n = Number of bolts, tf = Thickness of flange,

τs, τb, τk = Allowable shear stress for shaft, bolt and key material respectively τc = Allowable shear stress for the flange material i.e. cast iron,σcb, and σck = Allowable crushing stress for bolt and key material respectively

1-Design for hub:The hub is designed by considering it as a hollow shaft, transmitting the same torque (T) as that of a solid shaft. T=3.14/16×τc(D^4-d^4/D) Here D=2d and L=1.5d

2-Design for key:The key is designed with usual proportions and then checked for shearing and crushing stresses. The material of key is usually the same as that of shaft. The length of key is taken equal to the length of hub.

3-Design for flange:The flange at the junction of the hub is under shear while transmitting the torque. Therefore the torque transmitted, T= Circumference of hub × Thickness of flange × Shear stress of shaft ×Radius of hub =3.14D×tf×τc×d/2=3.14D^2/2×τc×tfHere tf=1.5d Therefore from the above relation, the induced shearing stress in the flange may be checked.

4-Design for bolts: The bolts are subjected to shear stress due to the torque transmitted. The number of bolts (n)depends upon the diameter of shaft.

And D1=3, load on each bolt=3.14/4×(d1)^2 τb Total load on all the bolts=3.14/4(d1)^2 τb×n And torque is transmitted

T=3.14/4(d1)^2τb×n×D/2

From this equation, the diameter of bolt (d1) may be obtained. Now, the diameter of bolt may checked in crushing.Area resisting crushing of all the bolts =n×d1×tfAnd crushing stress for all the bolts =( n×d1×tf) σcb Torque, T=( n×d1×tf×σcb)D1/2 From this equation, the induced crushing stress in the bolts may be checked.

CALCULATIONS:Statement: Design flange coupling with diameter of shaft 60mm and material of shaft is high carbon steel, material of key, coupling and bolt is low carbon steel with factor of safety 2. Data: d = 50mm For high carbon steel Sy = 380 MPa

For low carbon steel Sy = 300 MPa FOS = 2 Solution:For shaft: Τ all = 0.4 Sy =0.4×380 = 152 MPa Σ all = 0.9 Sy = 0.9×380 = 342 MPa nf = T all/ Td Td = 152/2 = 76 MPa Tmax = pi/16 d3 Td =pi/16 (0.06) 76= 3221.64 Nm

For hub, key,bolts,: tall = 0.4 Sy = 0.4×300 = 120 MPa σall = 0.9 Sy = 0.9×300 = 270 MPa nf = tall/td td = tall/ nf = 120/2 = 60 MPa σd = σall/ nf = 270/2 = 135 MPa 1-Design of hub: D = 2d = 2×60 = 120mm L = 1.5d = 1. ×560 =90mm T = pi/16 tc[D^4 – d^4] / D 3221640 = pi/16 tc (120^4 – 60^4 )/ 120 tc = 10.13 MPa<60MPa which is designed stress so design is safe here.2-Design of key: w= d/4 = 60/4 = 15 mm t = d/6 = 60/6 = 10mm l = 1.5d = 1.5×60 = 90m Shear of the key: T = tk×w×l(d/2)w tk = T/w×l× (d/2) = 3221640/ 15×90×30=79.54 MpaCrushing of the key: σck = 2T/ t×l× (d/2) = 6443280/ 27000

= 238.64 MPa3-Design for flange: tf = 0.5 d=30mm Resistive area=piD T=tc.pid^2/2.tf , tc =2T/piD^2tf=4.75MPaWhile for flange:

td=60MPa It means the actual shear stress(4.75MPa) in flange is less than designed(60MPa)so our design is in safe range.

4- Design for bolts: D1=3d=3(60) =180mm T=tb×pi/4×db^2×n×D1/2 db^2=8T/tb.pi.n.D1/2 db =19.49mm From Market point of view 22mm or 22M.Now, T=σcb×n×(db×tf)×D1/2 σcb = 2T/n.db.tf.D1 =13.55 MPa<135MPa which is designed stress so design is safe.

Conclusion: Coupling designed according to the above considerations will be more suitable, but the modifications according to our needs can also be made for more better results.