DESIGN OF MACHINE ELEMENTSSubmitted to School of Mechanical and Building Sciences VIT University, Vellore – 632 014 SLOT-D1 TOPIC DESIGN OF KNUCKLE JOINT SUBMITTED BY:VIKAS PORWAL(11BME0138) ASHISH JAISWAL(11BME0268) Knuckle joint :A knuckle joint is a mechanical joint used to connect two rods which are under a tensile load, when there is a requirement of small amount of flexibility, or angular moment is necessary. There is always axial or linear line of action of load. However, if the joint is guided, the rods may support a compressive load. A knuckle joint may be readily disconnected for adjustments or repairs. Applications:1.Tie rod joint of roof truss. 2.Tension link in bridge structure. 3.Link of roller chain. 4.Tie rod joint of jib crane. 5.The knuckle joint is also used in tractor. 6. Valve rod joint with eccentric rod. 7. Tension link in bridge structure. Mild steel 2. Wrought iron Dimensions of Various Parts of the Knuckle Joint .Material used for manufacturing of knuckle joint:1. The dimensions of various parts of the knuckle joint are fixed by empirical relations as given below. t1 = 0. d1 = d Outer diameter of eye. . then diameter of pin.e. 3. d2 = 2 d Diameter of knuckle pin head and collar.Knuckle pin.KNUCKLE JOINT The knuckle joint assembly consist of following major components : 1. It may be noted that all the parts should be made of the same material i.75 d Thickness of pin head.25 d Thickness of fork.5 d Thickness of single eye or rod end. d3 = 1.5 d Other dimensions of the joint are shown in Figure.Single eye. mild steel or wrought iron. 2. t = 1. If d is the diameter of rod. t2 = 0.Double eye or fork. 3. τ and σc = Permissible stresses for the joint material in tension. it is assumed that 1. d2 = Outer diameter of eye. Failure of knuckle pin in shear. 5. Failure of single eye or rod end in shear. σt . Failure of single eye or rod end in tension. Let P = Tensile load acting on the rod. 8. Failure of forked end in tension.The modes of failure of knuckle joint are : 1. 4. d = Diameter of the rod. 2. t1 = Thickness of fork. . In determining the strength of the joint for the various methods of failure. Consider a knuckle joint as shown in given Figure. Failure of forked end in shear. t = Thickness of single eye. There is no stress concentration. Failure of single eye or rod end in crushing. 6. Failure of solid rod in tension. shear and crushing respectively. 7. The load is uniformly distributed over each part of the joint. Failure of forked end in crushing. and 2. d1 = Diameter of the pin. therefore tensile strength of the rod.Due to these assumptions.Failure of solid rod in tension. Equating this to the load (P) acting on the rod. diameter of the knuckle pin (d1) is obtained. the strengths are approximate. This assumes that there is no slack and clearance between the pin and the fork and hence there is no bending of the pin. diameter of the rod ( d ) is obtained. we have t From this equation. therefore cross-sectional area of the pin under shearing 2 1 and the shear strength of the pin τ Equating this to the load (P) acting on the rod. Since the rods are subjected to direct tensile load. in . But. we have 2 P τ 1 2 1 From this equation. Failure of the knuckle pin in shear Since the pin is in double shear. however they serve to indicate a well proportioned joint. 1. 2. the eye end) and varies uniformly over the forks as shown in Figure below.actual practice. The value of maximum bending moment is given by . a load P/2 acts through a distance of t1 / 3 from the inner edge and the bending moment will be maximum at the centre of the pin. d1 = d).e. a margin of strength is provided to allow for the bending of the pin.. By making the diameter of knuckle pin equal to the diameter of the rod (i. therefore the pin is subjected to bending in addition to shearing.e. the knuckle pin is loose in forks in order to permit angular movement of one with respect to the other. it is assumed that the load on the pin is uniformly distributed along the middle portion (i. the stress due to bending is taken into account. Distribution of load on the pin Thus in the forks. In case. then increase the outer diameter of the eye (d2). Failure of the single eye or rod end in tension The single eye or rod end may tear off due to the tensile load. 3. We know that area resisting tearing = (d2 – d1) t ∴ Tearing strength of single eye or rod end = (d2 – d1) t × σt Equating this to the load (P) we have P = (d2 – d1) t × σt From this equation. the value of d1 may be obtained. the induced tensile stress (σt) for the single eye or rod end may be checked.From this expression. . In case the induced tensile stress is more than the allowable working stress. Failure of the single eye or rod end in crushing The single eye or pin may fail in crushing due to the tensile load. we have P = (d2 – d1) t × τ From this equation. We know that area resisting shearing = (d2 – d1) t ∴ Shearing strength of single eye or rod end = (d2 – d1) t × τ Equating this to the load (P). we have ∴ P = d1 × t × σc From this equation. . 5. In case the induced crushing stress in more than the allowable working stress. then increase the thickness of the single eye (t). Failure of the single eye or rod end in shearing The single eye or rod end may fail in shearing due to tensile load. the induced crushing stress (σc) for the single eye or pin may be checked.4. We know that area resisting crushing = d1 × t ∴ Crushing strength of single eye or rod end = d1 × t × σc Equating this to the load (P). the induced shear stress (τ) for the single eye or rod end may be checked. the induced tensile stress for the forked end may be checked. the induced shear stress for the forked end may be checked. Failure of the forked end in shear The forked end may fail in shearing due to the tensile load. we have P = (d2 – d1) × 2t1 × τ From this equation. the induced shear stress is more than the allowable working stress. we have P = (d2 – d1) × 2t1 × σt From this equation. 7. then thickness of the fork (t1) is increased.6. In case. Failure of the forked end in tension The forked end or double eye may fail in tension due to the tensile load. We know that area resisting tearing = (d2 – d1) × 2 t1 ∴ Tearing strength of the forked end = (d2 – d1) × 2 t1 × σt Equating this to the load (P). We know that area resisting shearing = (d2 – d1) × 2t1 ∴ Shearing strength of the forked end = (d2 – d1) × 2t1 × τ Equating this to the load (P). . . Failure of the forked end in crushing The forked end or pin may fail in crushing due to the tensile load. Note: From the above failures of the joint. we have P = d1 × 2 t1 × σc From this equation. we see that the thickness of fork (t1) should be equal to half the thickness of single eye (t / 2). But .8. We know that area resisting crushing = d1 × 2 t1 ∴ Crushing strength of the forked end = d1 × 2 t1 × σc Equating this to the load (P). the induced crushing stress for the forked end may be checked. in actual practice t1 > t / 2 in order to prevent deflection or spreading of the forks which would introduce excessive bending of pin. Thus. and σt = Permissible tensile stress for the material of the rod. a designer should consider the empirical relations in designing a knuckle joint. So fix the diameter of the pin equal to the diameter of the rod. We know that load. the diameter of rod). t where d = Diameter of the rod. 2.Design procedure of knuckle joint The empirical dimensions as discussed above have been formulated after wide experience on a particular service. First of all. 3. Other dimensions of the joint are fixed by empirical relations as discussed before. 2 P τ 1 A little consideration will show that the value of d1 as obtained by the above relation is less than the specified value (i. The following procedure may be adopted : 1. find the diameter of the rod by considering the failure of the rod in tension. After determining the diameter of the rod.e. the diameter of pin (d1) may be determined by considering the failure of the pin in shear. We know that tensile load acting on the rod. These dimensions are of more practical value than the theoretical analysis. . d2 = 2 d = 2 × 52 = 104 mm . load to be transmitted (P) =150kN t = 75MPa τ = 60MPa c =150MPa design the knuckle joint for given specifications. then the corresponding dimension may be increased.4 say 52 mm Ans. Failure of the solid rod in tension Let d = Diameter of the rod. We know that the load transmitted (P). In case the induced stress is more than the allowable stress. EXAMPLE:Let.4. 2 150 103 d ∴ d2 = 150 × 103 / 59 = 2540 or d = 50. Now the various dimensions are fixed as follows : Diameter of knuckle pin. d1 = d = 52 mm Outer diameter of eye. SOLUTION The joint is designed by considering the various methods of failure as discussed below : 1. The induced stresses are obtained by substituting the empirical dimensions in the relations as discussed in the equations of failure points. Diameter of knuckle pin head and collar.25 d = 1.25 × 52 = 65 mm Thickness of fork.75 d = 0.5 d = 0. 150 × 103 = (d2 – d1) t × τ = (104 – 52) 65 × τ = 3380 τ τ= 150 × 103 / 3380 = 44. therefore load (P). 2 150 × 103 τ 1 2 τ = 4248τ τ =150 103/4248 =35. Failure of the knuckle pin in shear Since the knuckle pin is in double shear.4 N/mm2 = 44. Failure of the single eye or rod end in tension The single eye or rod end may fail in tension due to the load. Failure of the single eye or rod end in shearing The single eye or rod end may fail in shearing due to the load.5 × 52 = 26 mm 2. 150 × 103 = (d2 – d1) t × σt = (104 – 52) 65 × σt = 3380 σt ∴ σt = 150 × 103 / 3380 = 44.4 MPa 4. d3 = 1.3 MPa 3.5 d = 1. t2 = 0. We know that load (P).75 × 52 = 39 say 40 mm Thickness of pin head.4 N / mm2 = 44. t1 = 0. We know that load (P).4 MPa . t = 1.5 × 52 = 78 mm Thickness of single eye or rod end. 5. We know that load (P). Failure of the forked end in tension The forked end may fail in tension due to the load.4 N/mm2 = 44. 150 × 103 = (d2 – d1) 2 t1× τ = (104 – 52) 2 × 40 × τ = 4160 τ τ= 150 × 103 / 4160 = 36 N/mm2 = 36 MPa . Failure of the forked end in shear The forked end may fail in shearing due to the load. We know that load (P). 150 × 103 = (d2 – d1) 2 t1 × σt = (104 – 52) 2 × 40 × σt = 4160 σt ∴ σt = 150 × 103 / 4160 = 36 N/mm2 = 36 MPa 7. Failure of the single eye or rod end in crushing The single eye or rod end may fail in crushing due to the load. 150 × 103 = d1 × t × σc = 52 × 65 × σc = 3380 σc ∴ σc = 150 × 103 / 3380 = 44.4 MPa 6. We know that load (P). therefore the joint is safe. . Failure of the forked end in crushing The forked end may fail in crushing due to the load. we see that the induced stresses are less than the given design stresses. 150 × 103 = d1 × 2 t1 × σc = 52 × 2 × 40 × σc = 4160 σc ∴ σc = 150 × 103 / 4180 = 36 N/mm2 = 36 MPa From above.8. We know that load (P). 731199 kg Volume 9.13999e-005 m^3 Material name: Description: Material Source: Material Model Type: Default Failure Criterion: Application Data: Wrought Stainless Steel Linear Elastic Isotropic Max von Mises Stress . 1 Body Name Solid Body fork(Cut-Extrude6) Material Wrought Stainless Steel Mass 0.Stress analysis of fork Units Unit system: Length/Displacement Temperature Angular velocity Stress/Pressure SI mm Kelvin rad/s N/m^2 Material Properties No. 26 7.5e+005 N using uniform distribution on 2 Face(s) apply normal force -1.1e-005 Units N/m^2 NA N/m^2 kg/m^3 N/m^2 N/m^2 /Kelvin Value Type Constant Constant Constant Constant Constant Constant Constant 19 500 W/(m. Description Load Load name Force-1 <fork> Selection set on 1 Face(s) apply normal force -1.K) J/(kg.0681e+008 1.K) Constant Constant Loads and Restraints Fixture Restraint name Fixed-1 <fork> Selection set on 2 Face(s) fixed.1702e+008 2.5e+005 N using uniform distribution Loading type Sequential Loading Description Force-2 <fork> Sequential Loading .9e+010 8000 5.Property Name Elastic modulus Poisson's ratio Shear modulus Mass density Tensile strength Yield strength Thermal expansion coefficient Thermal conductivity Specific heat Value 2e+011 0. 6719 mm) 0. 5. 17.75826e+009 N/m^2 Node: 19585 0. 10.5613 mm.5 Resultant 85702.156598 Free-body Moments Selection set Entire Body Units N-m Sum X 0 Sum Y 0 Sum Z 0 Resultant 1e-033 Study Results Default Results Name Stress1 Type VON: von Mises Stress Min 1.143353 Resultant 0. 11.59889 mm) .73517e-015 mm.5 mm.1394 mm) (31.1257 mm.543427 Sum Y -0.40795e+007 N/m^2 Node: 13770 Location (16. 16.5 Free-Body Forces Selection set Entire Body Units N Sum X 3.Reaction Forces Selection set Entire Body Units N Sum X -0.2665 mm. 19.3076 mm.973815 mm Node: 71 Location (18.03197e-005 mm. 16.6481 mm.00456444 Element: 3193 (-18.9286 mm.2316 mm) Max 1. -17.77976e-005 Element: 5953 (17.307434 Sum Z -85702.558 mm. 9.70368e-005 Sum Y -0.1602 mm) (3.1524 mm.7353 mm.6298 mm) Displacement1 URES: Resultant Displacement 0 mm Node: 549 Strain1 ESTRN: Equivalent Strain 5. 18. -11.0630307 Sum Z 0. -61. Fork-Study 1-Stress-Stress1 Fork-Study 1-Displacement-Displacement1 . Fork-Study 1-Strain-Strain1 . 731199 kg 9.9e+010 8000 5.13999e-005 m^3 3 0.Material Properties No.0681e+008 1.26 7.000115844 m^3 2 0.1702e+008 2.926753 kg Volume 0. 1 Body Name SolidBody 1(CutExtrude1) SolidBody 1(CutExtrude6) SolidBody 1(Revolve1) Material Wrought Stainless Steel Wrought Stainless Steel Wrought Stainless Steel Mass 0.9115e-005 m^3 Material name: Description: Material Source: Material Model Type: Default Failure Criterion: Application Data: Wrought Stainless Steel Linear Elastic Isotropic Max von Mises Stress Property Name Elastic modulus Poisson's ratio Shear modulus Mass density Tensile strength Yield strength Thermal expansion coefficient Thermal conductivity Specific heat Value 2e+011 0.1e-005 Units N/m^2 NA N/m^2 kg/m^3 N/m^2 N/m^2 /Kelvin Value Type Constant Constant Constant Constant Constant Constant Constant 19 500 W/(m.K) J/(kg.55292 kg 6.K) Constant Constant . Description Load Load name Force-1 <eye-1> Selection set on 1 Face(s) apply normal force -1.117023 Resultant 0.824778 Free-Body Forces Selection set Entire Body Units N Sum X -0.Loads and Restraints Fixture Restraint name Fixed-1 <fork-1> Selection set on 1 Face(s) fixed.5e+005 N using uniform distribution Loading type Sequential Loading Description Force-2 <fork-1> Sequential Loading Reaction Forces Selection set Entire Body Units N Sum X 0.198853 Sum Y -0.00129189 Sum Z -0.5e+005 N using uniform distribution on 1 Face(s) apply normal force -1.791847 Sum Z -0.00489192 Sum Y 0.317051 Resultant 0.317092 Free-body Moments Selection set Entire Body Units N-m Sum X 0 Sum Y 0 Sum Z 0 Resultant 1e-033 . 0420505 in.30129 in. -0.14457e-005 Element: 24314 (1.37843 in) Displacement1 URES: Resultant Displacement 0 mm Node: 15813 (-1. 0.95428 in) Max 7. -1. 2.253227 in.11751 in.159804 in) (0. 0.152514 mm Node: 16007 Location (0.811472 in) Strain1 ESTRN: Equivalent Strain 1.00269573 Element: 14102 (-1.182911 in) Assem2-Study 1-Stress-Stress1 . 0.229582 in. -0.378905 in) 0.33743e+008 N/m^2 Node: 28404 0.293518 in.144692 in.57449 in. -0. -0.00716 in.2048 in.37842 in.498811 in. 1. 0. -0.Study Results Name Stress1 Type VON: von Mises Stress Min 643068 N/m^2 Node: 29789 Location (1. Assem2-Study 1-Displacement-Displacement1 Assem2-Study 1-Strain-Strain1 .