11. Two short shafts having identical diameters of 44.45 mm and rotating at 420 rpm are Stress of the bolt is 13 N/mm² and design compressive stress of the flange is 17 N/mm².What is connected by a flange coupling having 4 bolts with a 120 mm bolt circle. The design shearing the power transmitted by the short shaft in KW? 12. From the Previous Problem, What diameter of bolt should be used? 13. From the Previous Problem, How thick should be the flange in mm? Definition COUPLINGS Coupling a mechanical device which is used to connect lengths of shafting permanently Types of Couplings 1. Rigid Couplings couplings that do not allow angular, axial or rotational used with collinear shafts. flexibility A. Flange Coupling type of rigid coupling which consists of two halves of flanges c to each other by bolts. connema B. Sleeve or Collar Coupling rigid coupling which is a cylindrical collar pressed over th ends of two collinear shafts 2. Flexible Couplings couplings which allow angularity to take care of misalignment of th shafts. Oldham coupling, chain coupling, flexible disk coupling, flexible gear type coupling, hydra coupling, universal joints, are examples of flexible couplings. esses in Flang P-2RTN Ftotal transmitted load on bolts Torque %2 force per bolt = no. of be Sshear stress in bolts = Scompressive stress on flan Flange Coupling: Flange Coupling Sleeve Coupling Oldham Coupling Chain Coupling Flexible Disc Coupling Universal Joint where: D diameter of the bolt circ D, diameter of the shaft t=thickness of the flange d diameter of the bolt Torque Capacity of Couplin T-FRn F n Torque Capacity of Coupli T=F₂Rana +FaRanz F Hydraulic Coupling Flexible Gear Type Coupling Flexible Jaw Type Coupling eg permanently. 2RTN total transmitted load on bolts Torque Flange Coupling al flexibility and flanges connected pressed over the salignment of the force per bolt= 6- no. of bolts shear stress in bolts = I F Fr F upling, hydraulic compressive stress on flange td Flange Coupling: www 10 ° о where: D= diameter of the bolt circle diameter of the shaft t-thickness of the flange d=diameter of the bolt Torque Capacity of Coupling (One Concentric Row): T=FRn = F-n 2 Torque Capacity of Coupling (Two Concentric Rows) R+F₂m TFaRan +FaRanz F- 100 R₂ R₁ ang permanently. nal flexibility and of flanges connected ar pressed over the 6x esses in 2xTN total transmitted load on bolts Torque D %2 force per bolt F no. of bolts shear stress in bolts = F Flange Coupling isalignment of the Oupling, hydraulic & compressive stress on flange : F td Flange Coupling: F-3 ° where: D= diameter of the bolt circle 0 diameter of the shaft thickness of the flange d diameter of the bolt Torque Capacity of Coupling (One Concentric Row): T=FRn=F D 2 n Torque Capacity of Coupling (Two Concentric Rows): TFaRana +FaRana = RDm+F2D2m2 ° 100 R R₂ F-4 Where: T-torque capacity of coupling or torque transmitted by shaft Fshearing force of one bolt R=radius of bolt circle D diameter of bolt circle number of bolts Relation of Shear Strain, Radial Distance and Shear Stress: Relation of Shear Strain and Radial distances from the axis of the shaft: nn R₁ R₂ Relation of Shear Stress and Radial distances from the axis of the shaft: SS2 GR₁ G₂R₂ Relation of Shearing Force and Radial distances from the axis of the shaft: R₁ R₂ Where: y shear strain R radius of bolt circle S, shear stress G = modulus of rigidity SOLVED P 1. A flanged bolt coupl 415 mm bolt circle. Det stress in the bolt is 50 N A 59.95kN-m 8. 52.6kN-m⭑ Si Se shearing stress pe S-50 MN/m² 50, 50,000- F T (0.025 Fforce per bolt F force on all bolts = T torque Fx boll 2. A flange coupling There are four 16 mm the torsional stress in bolts if uniformly distr A. 8.5 N/mm² B. 5.8 N/mm² Solution 167 Shearing sta 26,000- 167 (0.40) T=0.3267256 kN-m force on bolts F= 4.6675 kN 4,E 4.667.5 =1,7 4 COUPLINGS SOLVED PROBLEMS. F-5 425 mm bolt circle. Determine the torque capacity of the connection if the allowable shearing 1 A flanged bolt coupling has ten (10) steel 25.4 mm diameter bolts evenly tighten around a stress in the bolt is 50 MN/m. (ME Bd. Oct 97). A 59.95kN-m 52.6kN-m" C. 46.15 kN-m D. 43.8 kN-m Sashearing stress per bolt = F Ab S=50 MN/m² = 50,000 kN/m² Fr 50,000= (0.0254) 6 force per bolt 25.335 kN force on all bolts 25.335 x 10-253.35 kN 1 torque Fx bolt circle radius=253.35 x 0415 = 52.57 kN-m 2. A flange coupling has an outside diameter of 200 mm and connects two 40 mm shafts. There are four 16 mm bolts on a 140 mm bolt circle. The radial flange thickness is 20 mm. If the torsional stress in the shaft is not to exceed 26 MPa, determine the shearing stress in the bolts if uniformly distributed. (ME Bd. Oct 90). A. 8.5 N/mm² 8.5.8 N/mm² C. 6.5 N/mm² D. 7.5 N/mm² Slatin 16T shearing stress of the shaft 167 26,000- (0.40)³ T 0.3267256 kN-m Fa force on bolts= F= 4.6675 kN 4,667.5 Torque 0.3267256 D/ 0.140 = 4,667.5 N = 1,167 N F-6 MACHINE DESIGN Shearing Stress on bolt- F 1,167 = 5.8 N/mm² a fac 3. A flange coupling connects two 2" diameter shafts. The flanges are fitted with 6 bolts of Sa of safety of 5, ultimate tension of 70,000 psi, and ultimate shear of 55,000 psi.What is th 1040 steel on a 7" bolt circle. The shaft runs at 300 rpm and transmits 45 Hp. Assume a torque transmitted? Salatine Power 2xTN 45(33,000) 2T (300) T=787.817 ft-lbs 9,453.8 in-lbs 7 in 2 in 4. From the Previous Problem, Determine the diameter of bolts required. Solutio T = F(r) F-9453.8/3.5 2701 lbs F per bolt 2701/6=450.17 lbs S=F/A 55,000 45017 5 (/4)2 d = 0.228 in. 5. From the Previous Problem, How thick should the flange be? Selation S = F/A 70,000 45017 5 (0.228) t=0.141 in ' 6. A flange coupling is to connect two 57 mm shafts. The hubs of the coupling are each 1 mm in diameter and 92 mm thick and the flange webs are 19 mm thick. Six 16 mm bolts in thickness and key is 14 mm x 14 mm. Coupling is to transmit 45 KW at 160 rpm. For all parts 165 mm diameter circle connect the flanges. The keyway is 6 mm shorter than the hub yield point value in shear is one-half the yield point in tension or compression which is 4 MPa. Find the stress and factors of safety based on yield point shear in key. F F SH with 6 bolts of S Assume a fact psi. What is th Length of key-92-6-86 mm Power=2xTN 5 2xT (160/60) T-2.686kN-m COUPLINGS 111 mm 92 mm 16 mm 57 2 =28.5 mm 57 mm 165 mm F-7 2n radius 19 mm 14 mm 86 mm 7 in Force on shafts- torque 2.686 14 mm = =94.246 kN radius 0.0285 force 94.246 Shear in key shear area 0.0140.086) =78,277 KPa =78.277 MPa 224 =2.86 FS= 78.277 1. From the Previous Problem, Find the stress and factors of safety based on yield point bearing in the key. Si Bearing stress in key= force 94.246 bearing area 0.007(0.086) -156,555 KPa = 156.555 MPa FS. 448 156555 -2.86 & Find the stress and factors of safety based on yield point shear in bolts. Solution torque Force on coupling= radius where: radius= 165 82.5 mm 0.0825 m 2 torque 2.686 -32.557 kN radius 0.0825 ng are each 111 6 mm bolts in a than the hub's m. For all parts On which is 448 Force on coupling= Force per bolt 32.557/6=5.426 KN Shear in bolts 224 force 5.426 shear area (/4) (/4)(0.016) FS. = 26.987 =8.3 5.426 =26,987 KPa = 26.987 MPa F-9 dameter bolts are evenly spaced. Determine the torque transmitted by the coupling if the 2. A flanged bolt coupling has a bolt circle 360 mm in diameter where eight steel 25-mm design S n shearing stress in the bolts is 60 MPa. T-torque transmitted by the coupling =PR= =d² (Sa XRX(n) = (0.7854) (25) (60) 320) (8). T-42.411 kN-m 4241.15 x 10 N-mm 13. A solid circular shaft 90 mm in diameter is connected by a rigid coupling to a hollow shaft 100 mm in outside diameter, det mm in inside diameter. If the allowable stress in shear in the shafts and bolts is 70 MPa, determine the number of 12-mm-diameter steel bolts to be ed on a 240-mm-diameter bolt circle so that the coupling will be as strong as the weaker shaft for the hollow shaft: 100 2 Je =50 mm (OD* - ID*) = (100-90*) = 337.623 x 10" mm* torque that could be transmitted by the hollow shaft 60337.623 x 104 50 For the solid shaft: 790/2 = 45 mm (90) 32 =405.1476 x 10' N-mm = 4.0515 kN-m =644.125 x 10mm* 1-torque that can be transmitted by the solid shaft. 60644125 x 104) 45 Hollow shafting is weaker. As area of one! 7.7295 x 10 N-mm -7.7295 kN-m bolt 0.7854 (12) 113.1 mm² Solving for the force per bolt: FS (A)=70(113.1)=7917 N 405.1476x10 Total force on bolts 120 = 33,762.29 N No, of bolts 33,76229 7,917 4.26 Use 5 bolts diameter steel bolts on a concentric bolt circle 100 mm in radius were used in a rigid coupling 14. Eight 10-mm-diameter steel bolts on a bolt circle 150 mm in radius and six 20mm) If the design stress in the bolts is 60 MPa, determine the torque capacity of the coupling Sabali Where the subscripts 1 and 2 refer to the bolts in the outer and inner circles: 20 100 F₂= -2.667F 10 150 FA, Ss, -0.7854/10 (60)-4712.4 N T-torque capacity of coupling T=F,Run + F₂Ron=4712.4(150)(8) +(2.66)(4712.4)(100)(6) =13.175 x 10' N-mm =13.175 kN-m Ο Ο R=150 mm ° R₁ =100 mm 15. A flanged coupling having an outside diameter of 190 mm connects two 40-mm sha There are three 16-mm bolts on a 134-mm bolt circle. The radial flange thickness is 20 mm. the torsional stress in the shaft is not to exceed 30 MPa, determine the power that can be transmitted at 900 rpm. Solution D'Ss 40 (30) T-Torque transmitted = = 376.991 x 10' N-mm = 376.991 N-m 16 16 Power transmitted, P=2RTN=2(376991) 900 60 =35,530.56 W = 35.53 kw 16. From the Previous Problem, determine the shearing stress in the bolts if uniformly distributed. Solution T Fshearing force per bolt= (376991) = 1875.57 N De 2 134 2 Ss shearing stress in the bolts == F 187557 = 9.3 MPa A (16 six 20-mm aid coupling pling. F-11 17. From the Previous Problem, determine the maximum shearing stress induced in the bolts. Spo (9.3) 12.43MPa 18. From the Previous Problem, determine the bearing pressure in the bolts. Slatine Bearing pressure= Shearing force 187557 Projected area of bolt 16 x 20 5.8 MPa 13. Two 38.1 mm shafts are connected by a flanged coupling. The flanges are fitted with 6 bolts of SAE 1020 steel on a 152.4 mm bolt circle. The shafts run at 260 rpm and transmit a torque of 850,000 N-mm. Assuming a factor of safety of 5, ultimate tension, 430 MPa, and ultimate shear, 330 MPa, what power is transmitted? m Salati P=2RTN=2(850) 260 60 =23,143 W 23.1 kw nm m shafts. 20 mm. t can be 20. From the Previous Problem, Determine the diameter of the bolts required. Solution number of bolts 6 D= diameter of bolt circle 152.4 mm Fa shearing force per bolt= T 30. m Fa (850000) = 1859.14 N 1524 2 Design stress, S= 330 = 66MPa 5 F= A(S) 1859.14= d²(66) d=5.98 mm use 6 mm bolt 21. From the Previous Problem, How thick should the flange be? Salation 430 = 86 MPa 5 F-12 MACHINE DESIG F=1859.14 N per bolt F= d(t) (S) 185914 6x86 3.6 mm 22. Two solid shafts 120 mm in diameter are coupled by bolts 30 mm in diameterwith 120 mm from the axis. How many bolts are necessary? Salatin d-diameter of bolts 30 mm De diameter of bolts circle 120 x 2=240 D=shaft diameter = 120 mm n=number of bolts T-torque transmitted by bolts = T-torque of shaft n = D'S, 16 (120) 1200.7854:30² (16) Use 4 bolts 4 о O D. 240 mm Thollow SJ Use the smaller va d²S, Rn T = 4 5187.67= n = 5.95 or (0.0 0-30 mm 25. A flanged bot in diameter, and s diameter. What t bolts? PR R₂ 2/3 23. A flanged bolt coupling consists of eight steel 25-mm diameter bolts spaced evenly a a bolt circle 320 mm in diameter. Determine the torque capacity of the coupling if the allowable shearing stress in the bolts is 50 MPa. Solution T = d²S,Rn 4 T = (0.025) (50x100.160)(8) T 31,415.9 N-m shaft 90 mm in outside diameter and 75 mm in inside diameter. If the allowable shearing 24. A flanged bolt coupling is used to connect a solid shaft 75 mm in diameter to a hollow stress in the shafts and bolts is 70 MPa, how many 12-mm diameter steel bolts must be on a 220-mm diameter bolt circle so that the coupling will be as strong as the weaker sha Solution Tod= S, 70x10(0.075) 16 16 5798.44 N-m T P₁Ran +P T=PR T = R(RM (0.012) T = 10,291.85 26. From the F must be used o Solution T = P₂R₁₂+ T= PRn 14,000 ₁ = 10.01 USE 10 bolts terwith centers mm Telow it SJ COUPLINGS 0.09-0.075 70x10 0.09 2 Use the smaller value of T: 1 Ed's, Rn 5187.67 5.95 or (0.012) (70x100.220 USE n 6 bolts 5187.67 N-m F-13 25 A flanged bolt coupling consists of six 12-mm diameter steel bolts on a bolt circle 300 mm in diameter, and four 12-mm diameter steel bolts on a concentric bolt circle 200 mm in dameter. What torque can be applied without exceeding a shearing stress of 70 MPa in the bots? AR ARK evenly around if the Da hollow shearing must be used eaker shaft? T= PR₂n + P₂Ranz TRR₂+ IN -품(0012270x100 [01(6)+] (02)] T10,291.85 N-m 25. From the Previous Problem, determine the number of 12-mm diameter steel bolts that must be used on the 300-mm bolt circle to increase the torque capacity to 15 kN-m. Salation TPRana +PaRana 1 PR 14.000 *품 (0.0122(100100 [02(7)+] (224)] 10.01 USE 10 bolts F-14 MACHINE DESIGN REVIEWER BY JAS TORDILLO aroun 27. A flanged bolt coupling consists of six 12-mm diameter steel bolts evenly spaced a bolt circle 300 mm in diameter, and the four 25-mm diameter aluminum bolts on a conce stress of 60 MPa in the steel or 40 MPa in the aluminum? Use G of steel = 83 GPa and G bolt circle 200 mm in diameter. What torque can be applied without exceeding a shearing aluminum 28 GPa Saladin ES E Rs Ra PL SPS 8= LL Ss Sa ESRs EaRa Ss AE E Sa 83x10° (0.15) 28x10° (0.1) S = 4.44642857 S, о 00 0 If S, 40 MPa S. 177.86 MPa (overstressed) If S, 60 MPa Ss 13.49 MPa (Governs) T T T = P,Ran,+PR₂ns AS, Ron,+As Ss Rsns 0 O. (0.025) (0.14)13.49x10)+(0.012) (0.15)(6)60x106) T = 8756 N-m 150 mm 100 mm SUP 1. A flanger 400 mm bc stress in th 2. A flang There are the torsion bolts if uni 3. A flang 430 mm b stress in th 4. A fland There are the torsic bolts if un 5. A flang 1040 stee of safety torque tr 6. From 1 7. From 8. A flar mm in c 165 mm thicknes yield po MPa. F 9. From bearing 10. Find 11. Tw connec ith 6 bolts of Sa Assume a fac psl. What is the Length of key-92-6-86 mm Power 2xTN 15-2xT (160/60) T-2.685 kN-m COUPLINGS 92 mm 111 mm 57 16 mm radus =28.5 mm 57 mm 165 mm F-7 2n 19 mm 14 mm 86 mm 7 in Force on shafts torque 2.686 14 mm radius 0.0285 -94.246 kN force 94.246 Shear in key shear area 0.0140.086) =78,277 KPa 78.277 MPa 224 =2.86 15.- 78.277 1. From the Previous Problem, Find the stress and factors of safety based on yield point bearing in the key. Bearing stress in key = force bearing area 0.007(0.086) 94.246 156,555 KPa 156.555 MPa 448 F.S.= = 2.86 156555 & Find the stress and factors of safety based on yield point shear in bolts. Solation Force on coupling= torque radius ng are each 111 6 mm bolts in a than the hub's m. For all parts On which is 448 where 165 adus- 2 82.5 mm 0.0825 m. Force on coupling torque 2.686 = 32.557 kN radius 0.0825 Force per bolt 32.557/6=5.426 KN Shear in bolts - 13.= 224 26.987 force 5.426 5.426 shear area (/4) (/4) (0.016) 8.3 =26,987 KPa 26.987 MPa