DESIGN OF MACHINE ELEMENTS

Submitted 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.

Material used for manufacturing of knuckle joint:-

1. Mild steel2. Wrought iron

Dimensions of Various Partsof the Knuckle Joint

KNUCKLE JOINT

The knuckle joint assembly consist of following major components :

1.Single eye.

2.Double eye or fork.

3.Knuckle pin.

The dimensions of various parts of the knuckle joint are fixed by empirical relations as given below. It may be noted that all the parts should be made of the same material i.e. mild steel or wrought iron.If d is the diameter of rod, then diameter of pin,d1 = dOuter diameter of eye,d2 = 2 dDiameter of knuckle pin head and collar,d3 = 1.5 dThickness of single eye or rod end,t = 1.25 dThickness of fork, t1 = 0.75 dThickness of pin head, t2 = 0.5 dOther dimensions of the joint are shown in Figure.

The modes of failure of knuckle joint are :

1. Failure of solid rod in tension.2. Failure of knuckle pin in shear.3. Failure of single eye or rod end in tension.4. Failure of single eye or rod end in shear.5. Failure of single eye or rod end in crushing.6. Failure of forked end in tension.7. Failure of forked end in shear.8. Failure of forked end in crushing.

Consider a knuckle joint as shown in given Figure, LetP = Tensile load acting on the rod,d = Diameter of the rod,d1 = Diameter of the pin,d2 = Outer diameter of eye,t = Thickness of single eye,t1 = Thickness of fork.σt , τ and σc = Permissible stresses for the joint material in tension, shear and crushing respectively.In determining the strength of the joint for the various methods of failure, it is assumed that1. There is no stress concentration, and2. The load is uniformly distributed over each part of the joint.

Due to these assumptions, the strengths are approximate, however they serve to indicate a well proportioned joint.

1.Failure of solid rod in tension.

Since the rods are subjected to direct tensile load, therefore tensile strength of the rod,

¿ π4

×d2 ×σ

Equating this to the load (P) acting on the rod, we have

P= π4

× d2× σ t

From this equation, diameter of the rod ( d ) is obtained.

2. Failure of the knuckle pin in shearSince the pin is in double shear, therefore cross-sectional area of the pin under shearing

¿2 ×π4

×d12

and the shear strength of the pin¿2 ×

π4

×d12×τ

Equating this to the load (P) acting on the rod, we haveP¿2 ×

π4

×d12×τ

From this equation, diameter of the knuckle pin (d1) is obtained. This assumes that there is no slack and clearance between the pin and the fork and hence there is no bending of the pin. But, in

actual practice, the knuckle pin is loose in forks in order to permit angular movement of one with respect to the other, therefore the pin is subjected to bending in addition to shearing. By making the diameter of knuckle pin equal to the diameter of the rod (i.e., d1 = d), a margin of strength is provided to allow for the bending of the pin.In case, the stress due to bending is taken into account, it is assumed that the load on the pin is uniformly distributed along the middle portion (i.e. the eye end) and varies uniformly over the forks as shown in Figure below.

Distribution of load on the pin

Thus in the forks, 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. The value of maximum bending moment is given by

From this expression, the value of d1 may be obtained.

3. Failure of the single eye or rod end in tension

The single eye or rod end may tear off due to the tensile load. 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 haveP = (d2 – d1) t × σt

From this equation, the induced tensile stress (σt) for the single eye or rod end may be checked.In case the induced tensile stress is more than the allowable working stress, then increase the outer diameter of the eye (d2).

4. Failure of the single eye or rod end in shearing

The single eye or rod end may fail in shearing due to tensile load. We know that area resisting shearing

= (d2 – d1) t Shearing strength of single eye or rod end∴

= (d2 – d1) t × τEquating this to the load (P), we have

P = (d2 – d1) t × τFrom this equation, the induced shear stress (τ) for the single eye or rod end may be checked.

5. 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. In case the induced crushing stress in more than the allowable working stress, then increase the thickness of the single eye (t).

6. 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 haveP = (d2 – d1) × 2t1 × σt

From this equation, the induced tensile stress for the forked end may be checked.

7. Failure of the forked end in shear

The forked end may fail in shearing due to the tensile load. 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), we haveP = (d2 – d1) × 2t1 × τ

From this equation, the induced shear stress for the forked end may be checked. In case, the induced shear stress is more than the allowable working stress, then thickness of the fork (t1) is increased.

8. Failure of the forked end in crushing

The forked end or pin may fail in crushing due to the tensile load. 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), we haveP = 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, we see that the thickness of fork (t1) should be equal to half the thickness of single eye (t / 2). But , in actual practice t1 > t / 2 in order to prevent deflection or spreading of the forks which would introduce excessive bending of pin.

Design procedure of knuckle joint

The empirical dimensions as discussed above have been formulated after wide experience on a particular service. These dimensions are of more practical value than the theoretical analysis. Thus, a designer should consider the empirical relations in designing a knuckle joint. The following procedure may be adopted :

1. First of all, find the diameter of the rod by considering the failure of the rod in tension. We know that tensile load acting on the rod,

P= π4

× d2× σ t

where d = Diameter of the rod, and σt = Permissible tensile stress for the material of the rod.

2. After determining the diameter of the rod, the diameter of pin (d1) may be determined by considering the failure of the pin in shear. We know that load,

P¿2 ×π4

×d12×τ

A little consideration will show that the value of d1 as obtained by the above relation is less than the specified value (i.e. the diameter of rod). So fix the diameter of the pin equal to the diameter of the rod.

3. Other dimensions of the joint are fixed by empirical relations as discussed before.

4. The induced stresses are obtained by substituting the empirical dimensions in the relations as discussed in the equations of failure points.In case the induced stress is more than the allowable stress, then the corresponding dimension may be increased.

EXAMPLE:-

Let, load to be transmitted (P) =150kN σ t = 75MPa τ = 60MPa σc =150MPadesign the knuckle joint for given specifications.

SOLUTION

The joint is designed by considering the various methods of failure as discussed below :1. Failure of the solid rod in tensionLet d = Diameter of the rod.We know that the load transmitted (P),

150×103¿ π4

×d2 ×σ d2

∴ d2 = 150 × 103 / 59 = 2540 or d = 50.4 say 52 mm Ans.

Now the various dimensions are fixed as follows :Diameter of knuckle pin,d1 = d = 52 mmOuter diameter of eye, d2 = 2 d = 2 × 52 = 104 mm

Diameter of knuckle pin head and collar,d3 = 1.5 d = 1.5 × 52 = 78 mmThickness of single eye or rod end,t = 1.25 d = 1.25 × 52 = 65 mmThickness of fork, t1 = 0.75 d = 0.75 × 52 = 39 say 40 mmThickness of pin head, t2 = 0.5 d = 0.5 × 52 = 26 mm2. Failure of the knuckle pin in shearSince the knuckle pin is in double shear, therefore load (P),

150 × 103 ¿2 ×π4

×d12×τ

¿2 ×π4

×522×τ = 4248τ τ =150×103/4248 =35.3 MPa

3. Failure of the single eye or rod end in tensionThe single eye or rod end may fail in tension due to the load. We know that load (P),150 × 103 = (d2 – d1) t × σt = (104 – 52) 65 × σt = 3380 σt

σ∴ t = 150 × 103 / 3380 = 44.4 N / mm2 = 44.4 MPa

4. Failure of the single eye or rod end in shearingThe single eye or rod end may fail in shearing due to the load. We know that load (P),150 × 103 = (d2 – d1) t × τ = (104 – 52) 65 × τ = 3380 τ τ= 150 × 103 / 3380 = 44.4 N/mm2 = 44.4 MPa

5. Failure of the single eye or rod end in crushingThe single eye or rod end may fail in crushing due to the load. We know that load (P),150 × 103 = d1 × t × σc

= 52 × 65 × σc

= 3380 σc

σ∴ c = 150 × 103 / 3380 = 44.4 N/mm2 = 44.4 MPa

6. Failure of the forked end in tensionThe forked end may fail in tension due to the load. We know that load (P),150 × 103 = (d2 – d1) 2 t1 × σt

= (104 – 52) 2 × 40 × σt

= 4160 σt

σ∴ t = 150 × 103 / 4160 = 36 N/mm2 = 36 MPa

7. Failure of the forked end in shearThe forked 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 ττ= 150 × 103 / 4160 = 36 N/mm2 = 36 MPa

8. Failure of the forked end in crushingThe forked end may fail in crushing due to the load. We know that load (P),150 × 103 = d1 × 2 t1 × σc

= 52 × 2 × 40 × σc = 4160 σc σ∴ c = 150 × 103 / 4180 = 36 N/mm2 = 36 MPa

From above, we see that the induced stresses are less than the given design stresses, thereforethe joint is safe.

Stress analysis of fork

Units

Unit system: SI

Length/Displacement mm

Temperature Kelvin

Angular velocity rad/s

Stress/Pressure N/m^2

Material Properties

No. Body Name Material Mass Volume

1 Solid Body fork(Cut-Extrude6)

Wrought Stainless Steel

0.731199 kg 9.13999e-005 m^3

Material name: Wrought Stainless Steel

Description:

Material Source:

Material Model Type: Linear Elastic Isotropic

Default Failure Criterion: Max von Mises Stress

Application Data:

Property Name Value Units Value Type

Elastic modulus 2e+011 N/m^2 Constant

Poisson's ratio 0.26 NA Constant

Shear modulus 7.9e+010 N/m^2 Constant

Mass density 8000 kg/m^3 Constant

Tensile strength 5.1702e+008 N/m^2 Constant

Yield strength 2.0681e+008 N/m^2 Constant

Thermal expansion coefficient

1.1e-005 /Kelvin Constant

Thermal conductivity 19 W/(m.K) Constant

Specific heat 500 J/(kg.K) Constant

Loads and Restraints

Fixture

Restraint name Selection set Description

Fixed-1 <fork> on 2 Face(s) fixed.

Load

Load name Selection set Loading type Description

Force-1 <fork> on 1 Face(s) apply normal force -1.5e+005 N using uniform distribution

Sequential Loading

Force-2 <fork> on 2 Face(s) apply normal force -1.5e+005 N using uniform distribution

Sequential Loading

Reaction Forces

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N -0.543427 -0.307434 -85702.5 85702.5

Free-Body Forces

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N 3.70368e-005 -0.0630307 0.143353 0.156598

Free-body Moments

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N-m 0 0 0 1e-033

Study Results

Default Results

Name Type Min Location Max Location

Stress1 VON: von Mises Stress

1.40795e+007 N/m^2

Node: 13770

(16.558 mm,

11.3076 mm,

19.1602 mm)

1.75826e+009 N/m^2

Node: 19585

(18.2665 mm,

16.9286 mm,

10.1394 mm)

Displacement1 URES: Resultant Displacement

0 mm

Node: 549

(3.73517e-015 mm,

17.5 mm,

16.2316 mm)

0.973815 mm

Node: 71

(31.5613 mm,

5.03197e-005 mm,

-61.6298 mm)

Strain1 ESTRN: Equivalent Strain

5.77976e-005

Element: 5953

(17.1524 mm,

-11.7353 mm,

18.6719 mm)

0.00456444

Element: 3193

(-18.1257 mm,

-17.6481 mm,

9.59889 mm)

Fork-Study 1-Stress-Stress1

Fork-Study 1-Displacement-Displacement1

Fork-Study 1-Strain-Strain1

Material Properties

No. Body Name Material Mass Volume

1 SolidBody 1(Cut-Extrude1)

Wrought Stainless Steel

0.926753 kg 0.000115844 m^3

2 SolidBody 1(Cut-Extrude6)

Wrought Stainless Steel

0.731199 kg 9.13999e-005 m^3

3 SolidBody 1(Revolve1)

Wrought Stainless Steel

0.55292 kg 6.9115e-005 m^3

Material name: Wrought Stainless Steel

Description:

Material Source:

Material Model Type: Linear Elastic Isotropic

Default Failure Criterion: Max von Mises Stress

Application Data:

Property Name Value Units Value Type

Elastic modulus 2e+011 N/m^2 Constant

Poisson's ratio 0.26 NA Constant

Shear modulus 7.9e+010 N/m^2 Constant

Mass density 8000 kg/m^3 Constant

Tensile strength 5.1702e+008 N/m^2 Constant

Yield strength 2.0681e+008 N/m^2 Constant

Thermal expansion coefficient

1.1e-005 /Kelvin Constant

Thermal conductivity 19 W/(m.K) Constant

Specific heat 500 J/(kg.K) Constant

Loads and Restraints

Fixture

Restraint name Selection set Description

Fixed-1 <fork-1> on 1 Face(s) fixed.

Load

Load name Selection set Loading type Description

Force-1 <eye-1> on 1 Face(s) apply normal force -1.5e+005 N using uniform distribution

Sequential Loading

Force-2 <fork-1> on 1 Face(s) apply normal force -1.5e+005 N using uniform distribution

Sequential Loading

Reaction Forces

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N 0.198853 -0.791847 -0.117023 0.824778

Free-Body Forces

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N -0.00489192 0.00129189 -0.317051 0.317092

Free-body Moments

Selection set Units Sum X Sum Y Sum Z Resultant

Entire Body N-m 0 0 0 1e-033

Study Results

Name Type Min Location Max Location

Stress1 VON: von Mises Stress

643068 N/m^2

Node: 29789

(1.37842 in,

1.00716 in,

-1.95428 in)

7.33743e+008 N/m^2

Node: 28404

(0.498811 in,

0.2048 in,

-0.159804 in)

Displacement1 URES: Resultant Displacement

0 mm

Node: 15813

(-1.57449 in,

-0.0420505 in,

-0.811472 in)

0.152514 mm

Node: 16007

(0.293518 in,

0.253227 in,

2.37843 in)

Strain1 ESTRN: Equivalent Strain

1.14457e-005

Element: 24314

(1.30129 in,

0.144692 in,

-0.378905 in)

0.00269573

Element: 14102

(-1.11751 in,

0.229582 in,

-0.182911 in)

Assem2-Study 1-Stress-Stress1

Assem2-Study 1-Displacement-Displacement1

Assem2-Study 1-Strain-Strain1