Abstract—A systematic approach to tolerance synthesisinvolves consideration of the manufacturing cost as a functionof tolerance. Allocation of manufacturing and designtolerances is an important step in product development. Thefocus of this paper is on the optimal solution of the least costtolerance design. The modified exponential cost tolerancemodel has been considered. The SA-PS algorithm ,anontraditional global optimization technique has been used asthe solution methodology for its inherent advantages.Application of methodology is demonstrated on complextolerancing product like Knuckle joint. Optimal result soobtained are compared with previous work done and foundthat individual tolerance and corresponding manufacturingcost is drastically reduced
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.
Wrought iron Dimensions of Various Parts of the Knuckle Joint .Material used for manufacturing of knuckle joint:1. Mild steel 2.
25 d Thickness of fork.KNUCKLE JOINT The knuckle joint assembly consist of following major components : 1. mild steel or wrought iron. The dimensions of various parts of the knuckle joint are fixed by empirical relations as given below. d2 = 2 d Diameter of knuckle pin head and collar. It may be noted that all the parts should be made of the same material i.5 d Other dimensions of the joint are shown in Figure. t1 = 0. If d is the diameter of rod. t = 1.5 d Thickness of single eye or rod end. 3. d3 = 1. d1 = d Outer diameter of eye. 2.Double eye or fork. .Knuckle pin.Single eye.e. then diameter of pin.75 d Thickness of pin head. t2 = 0.
There is no stress concentration. t1 = Thickness of fork. τ and σc = Permissible stresses for the joint material in tension. Failure of single eye or rod end in shear. Failure of forked end in crushing. 2.The modes of failure of knuckle joint are : 1. Failure of single eye or rod end in tension. Failure of forked end in shear. Failure of forked end in tension. it is assumed that 1. . d1 = Diameter of the pin. Failure of single eye or rod end in crushing. and 2. 6. d2 = Outer diameter of eye. Failure of knuckle pin in shear. In determining the strength of the joint for the various methods of failure. Consider a knuckle joint as shown in given Figure. 8. Failure of solid rod in tension. 4. 7. t = Thickness of single eye. Let P = Tensile load acting on the rod. d = Diameter of the rod. 3. shear and crushing respectively. 5. σt . The load is uniformly distributed over each part of the joint.
in . we have 2 P τ 1 2 1 From this equation.Due to these assumptions. Since the rods are subjected to direct tensile load. 2. Failure of the knuckle pin in shear Since the pin is in double shear. This assumes that there is no slack and clearance between the pin and the fork and hence there is no bending of the pin. therefore tensile strength of the rod. the strengths are approximate. Equating this to the load (P) acting on the rod. however they serve to indicate a well proportioned joint.Failure of solid rod in tension. diameter of the rod ( d ) is obtained. we have t From this equation. diameter of the knuckle pin (d1) is obtained. 1. But. 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.
In case..actual practice. a margin of strength is provided to allow for the bending of the pin. therefore the pin is subjected to bending in addition to shearing. Distribution of load on the pin Thus in the forks. the eye end) and varies uniformly over the forks as shown in Figure below. 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. By making the diameter of knuckle pin equal to the diameter of the rod (i. it is assumed that the load on the pin is uniformly distributed along the middle portion (i.e. The value of maximum bending moment is given by . the stress due to bending is taken into account. the knuckle pin is loose in forks in order to permit angular movement of one with respect to the other. d1 = d).e.
. the value of d1 may be obtained. Failure of the single eye or rod end in tension The single eye or rod end may tear off due to the tensile load. In case the induced tensile stress is more than the allowable working stress. 3.From this expression. the induced tensile stress (σt) for the single eye or rod end may be checked. 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. then increase the outer diameter of the eye (d2).
the induced shear stress (τ) for the single eye or rod end may be checked. In case the induced crushing stress in more than the allowable working stress. we have P = (d2 – d1) t × τ From this equation. then increase the thickness of the single eye (t). 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 know that area resisting crushing = d1 × t ∴ Crushing strength of single eye or rod end = d1 × t × σc Equating this to the load (P). we have ∴ P = d1 × t × σc From this equation. the induced crushing stress (σc) for the single eye or pin may be checked. 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). . Failure of the single eye or rod end in shearing The single eye or rod end may fail in shearing due to tensile load.4. 5.
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). 7. Failure of the forked end in tension The forked end or double eye may fail in tension due to the tensile load. . we have P = (d2 – d1) × 2t1 × σt From this equation. the induced tensile stress for the forked end may be checked. the induced shear stress is more than the allowable working stress. 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). then thickness of the fork (t1) is increased. In case.6. we have P = (d2 – d1) × 2t1 × τ From this equation. Failure of the forked end in shear The forked end may fail in shearing due to the tensile load. the induced shear stress for the forked end may be checked.
Failure of the forked end in crushing The forked end or pin may fail in crushing due to the tensile load. in actual practice t1 > t / 2 in order to prevent deflection or spreading of the forks which would introduce excessive bending of pin. 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). But . we see that the thickness of fork (t1) should be equal to half the thickness of single eye (t / 2). .8. we have P = d1 × 2 t1 × σc From this equation. the induced crushing stress for the forked end may be checked. Note: From the above failures of the joint.
Thus. t where d = Diameter of the rod. These dimensions are of more practical value than the theoretical analysis. The following procedure may be adopted : 1. the diameter of pin (d1) may be determined by considering the failure of the pin in shear. After determining the diameter of the rod. We know that load. First of all. a designer should consider the empirical relations in designing a knuckle joint. the diameter of rod). . 3. 2. Other dimensions of the joint are fixed by empirical relations as discussed before. So fix the diameter of the pin equal to the diameter of the rod.Design procedure of knuckle joint The empirical dimensions as discussed above have been formulated after wide experience on a particular service.e. find the diameter of the rod by considering the failure of the rod in tension. We know that tensile load acting on the rod. 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. and σt = Permissible tensile stress for the material of the rod.
We know that the load transmitted (P).4 say 52 mm Ans. SOLUTION The joint is designed by considering the various methods of failure as discussed below : 1. EXAMPLE:Let. d1 = d = 52 mm Outer diameter of eye. load to be transmitted (P) =150kN t = 75MPa τ = 60MPa c =150MPa design the knuckle joint for given specifications. d2 = 2 d = 2 × 52 = 104 mm . Failure of the solid rod in tension Let d = Diameter of the rod. Now the various dimensions are fixed as follows : Diameter of knuckle pin. The induced stresses are obtained by substituting the empirical dimensions in the relations as discussed in the equations of failure points. then the corresponding dimension may be increased. In case the induced stress is more than the allowable stress. 2 150 103 d ∴ d2 = 150 × 103 / 59 = 2540 or d = 50.4.
t = 1. Failure of the knuckle pin in shear Since the knuckle pin is in double shear.25 d = 1. We know that load (P). Failure of the single eye or rod end in tension The single eye or rod end may fail in tension due to the load.Diameter of knuckle pin head and collar. d3 = 1.75 d = 0. t1 = 0. 150 × 103 = (d2 – d1) t × τ = (104 – 52) 65 × τ = 3380 τ τ= 150 × 103 / 3380 = 44.25 × 52 = 65 mm Thickness of fork.4 N/mm2 = 44.5 × 52 = 26 mm 2.4 MPa .75 × 52 = 39 say 40 mm Thickness of pin head. t2 = 0.5 d = 1. 2 150 × 103 τ 1 2 τ = 4248τ τ =150 103/4248 =35.3 MPa 3. 150 × 103 = (d2 – d1) t × σt = (104 – 52) 65 × σt = 3380 σt ∴ σt = 150 × 103 / 3380 = 44.4 N / mm2 = 44.4 MPa 4.5 d = 0. We know that load (P).5 × 52 = 78 mm Thickness of single eye or rod end. therefore load (P). Failure of the single eye or rod end in shearing The single eye or rod end may fail in shearing due to the load.
We know that load (P). 150 × 103 = (d2 – d1) 2 t1× τ = (104 – 52) 2 × 40 × τ = 4160 τ τ= 150 × 103 / 4160 = 36 N/mm2 = 36 MPa .4 MPa 6. Failure of the forked end in tension The forked end may fail in tension due to the load.5. Failure of the forked end in shear The forked end may fail in shearing due to the load. We know that load (P). We know that load (P).4 N/mm2 = 44. 150 × 103 = d1 × t × σc = 52 × 65 × σc = 3380 σc ∴ σc = 150 × 103 / 3380 = 44. 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.
we see that the induced stresses are less than the given design stresses. Failure of the forked end in crushing The forked end may fail in crushing due to the load. . therefore the joint is safe. We know that load (P).8. 150 × 103 = d1 × 2 t1 × σc = 52 × 2 × 40 × σc = 4160 σc ∴ σc = 150 × 103 / 4180 = 36 N/mm2 = 36 MPa From above.
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.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 .
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.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. Description Load Load name Force-1 <fork> Selection set on 1 Face(s) apply normal force -1.K) Constant Constant Loads and Restraints Fixture Restraint name Fixed-1 <fork> Selection set on 2 Face(s) fixed.1702e+008 2.26 7.0681e+008 1.9e+010 8000 5.K) J/(kg.5e+005 N using uniform distribution on 2 Face(s) apply normal force -1.5e+005 N using uniform distribution Loading type Sequential Loading Description Force-2 <fork> Sequential Loading .
2665 mm. 5.1524 mm.70368e-005 Sum Y -0.973815 mm Node: 71 Location (18.40795e+007 N/m^2 Node: 13770 Location (16.6298 mm) Displacement1 URES: Resultant Displacement 0 mm Node: 549 Strain1 ESTRN: Equivalent Strain 5.558 mm.5 Free-Body Forces Selection set Entire Body Units N Sum X 3. 18.75826e+009 N/m^2 Node: 19585 0. 11. -17.0630307 Sum Z 0. 16.9286 mm.307434 Sum Z -85702.5 mm. -11.156598 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 1.6719 mm) 0.59889 mm) .143353 Resultant 0. -61.03197e-005 mm.6481 mm.1257 mm.77976e-005 Element: 5953 (17. 9. 19.543427 Sum Y -0.1394 mm) (31.00456444 Element: 3193 (-18. 17.2316 mm) Max 1.Reaction Forces Selection set Entire Body Units N Sum X -0.7353 mm.1602 mm) (3.5613 mm. 16. 10.3076 mm.5 Resultant 85702.73517e-015 mm.
000115844 m^3 2 0.Material Properties No.1702e+008 2.K) J/(kg.0681e+008 1.926753 kg Volume 0.9e+010 8000 5.26 7.55292 kg 6.K) Constant Constant .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.731199 kg 9.13999e-005 m^3 3 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. 1 Body Name SolidBody 1(CutExtrude1) SolidBody 1(CutExtrude6) SolidBody 1(Revolve1) Material Wrought Stainless Steel Wrought Stainless Steel Wrought Stainless Steel Mass 0.
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 -0.117023 Resultant 0. Description Load Load name Force-1 <eye-1> Selection set on 1 Face(s) apply normal force -1.Loads and Restraints Fixture Restraint name Fixed-1 <fork-1> Selection set on 1 Face(s) fixed.824778 Free-Body Forces Selection set Entire Body Units N Sum X -0.791847 Sum Z -0.317092 Free-body Moments Selection set Entire Body Units N-m Sum X 0 Sum Y 0 Sum Z 0 Resultant 1e-033 .5e+005 N using uniform distribution on 1 Face(s) apply normal force -1.00489192 Sum Y 0.317051 Resultant 0.
37843 in) Displacement1 URES: Resultant Displacement 0 mm Node: 15813 (-1. 0. -0. 0.378905 in) 0.811472 in) Strain1 ESTRN: Equivalent Strain 1.37842 in. -0. 2. 0.2048 in.182911 in) Assem2-Study 1-Stress-Stress1 .0420505 in.498811 in. 1.33743e+008 N/m^2 Node: 28404 0.95428 in) Max 7. -0.14457e-005 Element: 24314 (1.253227 in.159804 in) (0. 0.144692 in.229582 in. -0.11751 in.293518 in.57449 in.30129 in.00269573 Element: 14102 (-1. -0.152514 mm Node: 16007 Location (0.00716 in.Study Results Name Stress1 Type VON: von Mises Stress Min 643068 N/m^2 Node: 29789 Location (1. -1.