Shaft. keys and cuopling

Shaft. keys and cuopling

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  Submitted by-: Nameof Faculty-: Svabhiman Singh Mr. Umed Singh Sir (131MC00230)
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  Content Sr. No. Name 9.23Bushed-Pin Flexible Coupling 9.24 Design Procedure for Flexible Coupling
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  (9.23) Bushed-Pin FlexibleCoupling-: Rigid Coupling Can be used only when there is perfect alignment between the two axes of two shafts and motion is free from vibrations and shocks.  In practice, it is impossible to obtain prefect alignment of shafts. Misalignment exists due to the following reasons: (i) Deflection of shafts due to lateral forces; (ii) Error in shaft mounting due to manufacturing tolerances; (iii) Use of two separately manufactured units such as an electric motor and a worm gear box; (iv) Thermal expansions of parts. To over these problems, Flexible coupling is used. A flexible coupling employs a flexible like a rubber bush between the driving and the driven flanges. This flexible bush not only accommodates the misalignment but also absorb shocks and vibrations.
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  The basic typesof misalignment between axes of the input and output shafts are shown :
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  The coupling havingtwo flanges, one keyed to input shaft and other to output shaft. The tow flanges are connected together by means of four or six pins. At one end, the pin is fixed to the put flange by means a nut. The diameter of the pin is enlarged in the input flange where a rubber bush is mounted over the pin. The rubber bush is provided with brass lining at the inner surface. The lining reduces the wear of the inner surface of the rubber bush. Power transmission-: Input shaft key Input flange rubber bush Output flange Pin Output shaft Advantages of bush-pin flexible-: (i) It can tolerate 0.5mm lateral or axial misalignment and 1.5̊ of angular misalignment. (ii) It prevents transmission of shock from one shaft to other and absorb vibrations. (iii) It can be used for transmitting high torques. It is simple in construction and easy to assemble and disassemble and also easy to design and manufacturing of coupling.
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   Disadvantages ofbush-pin flexible-: (i) Cost of flexible is more than that of rigid coupling due to additional parts. (ii) It requires more radial space compared with other types of coupling. The basic difference between the rigid coupling and flexible coupling is; In rigid coupling two flanges are identical except for the provision of spigot and recess ,In flexible coupling input flanges accommodates the rubber bushes comparatively large diameter than the diameter of pins accommodated in output flanges as shown in fig;
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  Where, dh= outsidediameter of hub (dh = 2d) lh= length of hub or effective length of key(lh= 1.5d) D= pitch circle diameter of pins (D= 3d to 4d) t = thickness of output flange (t = 0.5d) t1 = thickness of protective rim (t1= 0.25d) d1 = diameter of pin (d1= 0.5d ) √ N N= number of pins d = shaft diameter. As the flange on the input shaft rotates, it exerts a force P on each rubber bush and the resisting forces on the rubber bushes are shown in fig-:
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  M= P*D*N (a) 2 Where, M= torque transmitted by coupling P=force acting on each rubber or pin D= pitch circle diameter of bushes or pins N= number of bushes or pins.
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  Projected area ofrubber of rubber bush is shown in fig; the force P is equal to the projected area and intensity of pressure, so P = (Db*lb)*pm (b) where, Db= outer diameter of bush lb = effective length of bush in contact with input flange pm= permissible intensity of pressure between flange & rubber bush. From , (a) & (b), Mt = 1*pm*Db*lb*DN (c) 2 The permissible intensity of pressure between the rubber bush and the cast iron flange is usually (1 N/mm2). The ratio of length to the outer diameter for the rubber bush usually assumed as 1.
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  So, lb Db (d) andpm = 1N/mm2 After solving (c) & (d) , then Mt = 1*Db^2*DN 2 The pin is subjected to direct shear stress due to force P , so direct stress in pin given by t(tou) = P π*(d1)^2 (d) 4 From (a) & (d), t(tou) = 8Mt (e) π*(d1)2*DN According to the Indian standard the allowable shear stress for pins is 35N/mm2 . The maximum allowable peripheral speed of the coupling is 30m/s. There are two important features of flexible bush coupling , which are different than rigid flange coupling are -: (i) there is a gap between driving and driven flanges of flexible bush coupling. This gap is essential for taking care of angular misalignment between two shafts.
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  There is nosuch clearance between flanges of rigid coupling. There fore, rigid coupling can’t be tolerate any misalignment. (ii) In case of rigid coupling, torque is transmitted by means bolts. These bolts are made of steel resisting shear or tensile stress are high. Therefore, diameter of bolts are pitch circle diameter of bolts is competitively less than that of flexible bush coupling. On the hand, torque is transmitted by means of a force passing through a rubber bush in case of flexible bush coupling. The permissible pressure between rubber bush and cast iron flange is only 1N/mm2. So, the diameter of the pin or pitch circle diameter of pins is competitively large than that of rigid coupling. (9.24) Design Procedure for Flexible Coupling-: The basic procedure for finding out dimensions of bushed pin type flexible coupling consists of the following steps-: (i) Shaft Diameter-: Calculate the shaft diameter by using the following equations – 60*10^6(kw) and 16Mt Mt 2 πn t(tou) πd^3 (ii)Dimensions of Flanges-: Calculate the dimensions of flanges by the following empirical relations-
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  dh = 2d lh= 1.5d D = 3d to 4d t = 0.5d t1 = 0.25d •To calculate torsional shear stress in hub we consider that a hollow shaft subjected to torsional moment Mt. The inner and outer diameters of hub are d and dh respectively, then torsional shear stress - Mt*r and π*(dh ^4- d^4 ) t(tou) J J 32 r = (dh/2) •The shear stress in the flange at the junction with the hub calculated by; 1* π* dh ^2*t*t(tou) M 2 (iii)Diameter of pins-: The number of pins is usually 4 or 6. The diameter of pins is calculated by following relationship- 0.5*d d1 √ N Determine shear stress in pins is; 8Mt t(tou) π*d1 ^2*DN The shear stress is calculated by above equation should be less than 35 N/mm2.
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  (iv) Dimensions ofBushes-: Calculate the outer diameter of the rubber bush equation is 1*Db^2*DN Mt 2 • Calculate the effective length of the rubber bush by the following relationship, lb = Db (v) Dimensions of keys -: Determine the standard cross-section of flat key(Table9.3). The length of the key in each section is lh. Therefore, l = lh With the above dimensions of key, check the shear and compressive stresses in the key by the equations { 2Mt (9.27)} and { 4Mt (9.28) } { t(tou) dbl } {(sigma)c dhl } respectively. 2Mt and 4Mt t(tou) dbl (sigma)c dhl