Tutorial # 3 (Design of Couplings) Exercise 1: A rigid coupling has a bore diameter of 50 mm. Four machined bolts on a bolt circle of 125 mm diameter and fitted in reamed holes. If the bolts are made from the same material as the shaft, with an ultimate tensile strength of 550 MPa and a yield point in tension of 345 MPa, determine the necessary size of bolts to have the same capacity as the shaft in torsion. Solution: The shaft capacity is found from: D3 = (16/s)Tkt (the shaft equation for a solid shaft in torsion only). Then: s (allowable) is the smallest of: 0.18x550 = 99 MPa and 0.3x345 = 103.5 MPa and the allowance for the keyway effect is 0.75 3 Then (0.05) = (16Tkt/x0.18x550xl06)(0.75) Tkt=1.823 kNm The coupling can be designed for kt = 1.0 or Tkt can be left as a product. Using the case (1), which gives the most conservative design: Tkt = s [(/4)(d2)](0.5DBC)(n) Where: DBC = diameter of bolt circle, m d = diameter of bolt, m (shank diameter) :. 1.823x103 = 0.18x550x106( /4)(d)2(0.5x0.125)x(4) -- d = 0.00968 m Here use an M10 bolt. Using the case (2), Tkt = (3/4) s [(/4)(d2)](0.5DBC)(n) :. 1. 1.823x103 = 0.75x0.1 8x550(/4)(d)2(0.5x0.125)(4) -- d = 0.0112 m MI2 bolt may be used. _____________________________________________________________________ Exercise 2: Assume that a flange coupling has the following specifications: Number of bolts: 6. Size of bolts: 12 mm diameter Preloading of bolts: 22 KN in each bolt. Inner diameter of contact: 175 mm. Outer diameter of contact: 200 mm. Speed of rotation of coupling: 300 rpm. Coefficient of friction: 0.15 Shaft diameter: 50 mm. Shaft material, annealed steel with an ultimate tensile strength of 586 MPa and a yield point in tension of 310 MPa. The bolts are set in large clearance holes in the coupling. 1) Determine the maximum power capacity based upon friction such that slip occurs between faces of contact. 2) Compare the shaft power capacity with the friction power capacity. Assume steady load conditions and that the shaft is in torsion only. 712 kN. t determined from the above equation is very small and the difficulty of casting necessitates using a thickness much larger than calculated. _____________________________________________________________________ Exercise 3: Set up the equations or relations necessary determine: a) the hub diameter DH. N f = coefficient of friction Rf = friction radius This assumes that the pressure is uniformly distributed.75Ds to 2.7 kW The coupling has greater power capacity based on friction than the shaft capacity. s is the smallest of 0.75 x x (0. c) The thickness of flange is based upon proportions and casting requirements. .Solution: a) The torque capacity based on friction is: T= F x f x Rf Where: F = axial force caused by bolts loading. c) the flange thickness h. Where. _____________________________________________________________________ . DH ≈ (1.3x310 = 93 MPa. to Solution: a) The hub diameter (DH ) is established on the basis of proportions.Bearing of the bolts and web: T = [B x (d x t)] x (DBC /2) x n or t = 2Mt/(B x d x DBC x n) Where: B = allowable bearing stress for the bolt or web (whichever is weaker).m. Shaft power capacity = (2x T x N)/60 = (2x 1712 x 300)/60 = 53.4 kW b) Shaft torque capacity: T = s x x D3/16 = [93 x l06 x 0.0Ds) b) The minimum thickness of the web (t) is based on: 1)-Shear of the web: T = [s x (x DH x t)] x (DH/2) or t = 2Mt/(s x x DH2) **Usually. Pa of the projected area. Friction power = (2x T x N)/60 = (2x T x 300)/60= 58.'. b) the web thickness t. 2).05)3]/ 16 = 1.18x586 = 105 MPa and 0. 8 = (2 x 55 x l06) x (x db) x (0.73 mm Take Ml0 bolt.8 x (x 200)/60] x 10-3 = 20. b) Power: Power = x (x N)/60 = [981.Exercise 4: A flange coupling connects two 50 mm diameter lengths of commercial shafting. 2). The diameter of the bolt circle is 240 mm and the web thickness is 22 mm.Based upon shear: Assume uniform shear: T = s x (x db2)/4 x (DBc/2) x (n/2) 981.56 kW _____________________________________________________________________ .m.24/2) x (n/2) db = 1.Based upon bearing: T = B x (tx db) x (DBc/2) x (n/2) 981. a) Determine the minimum bolt diameter required transmitting the same torque that the shaft can transmit.8 = (55 x l06) x (x db2)/4 x (0.24/2) x (n/2) db = 9. The bolts are set in clearance holes. a) Bolts diameter: 1).69 mm Thus. Take Ml0 bolts.053)/16 = 981. b) What power may be transmitted at 200 rev/min under steady state conditions? Solution: T = (x s x D3)/16 = (x 40 x 106 x 0. The coupling webs are bolted together with four bolts of the same material as the shaft.8 N.