You are here

Knuckle Joint : Design Procedure,Problems and Questions

A) Knuckle Joint Design Procedure

Before going into detailed steps to design the dimensions of the Knuckle joint, it is essential to get oriented with all its components and their functions. Here is the diagram showing the exploded view  of Knuckle Joint.

Knuckle Joint Asssembly Exploded View

As seen in the assembly the Knuckle joint has main four parts

  1. Rods {Which are to be connected by joint }
  2. Single eye {Modified rod for assembly}
  3. Double eye or Forked end {Modified rod for assembly}
  4. Pin {Connects the two rods}
  5. Collar {to keep the pin in position}
  6. Split pin or taper pin {Not in diagram} {to prevent sliding away of pin}

Assembly Drawing and Notations used in Knuckle Joint

Knuckle Joint Crosssectional View

Notations used in design :

P  =  Tension in rod ( Load on the joint)

D =  Diameter of rod

D1= Enlarged diameter of rod

d = Diameter of pin

d1 = Diameter of pin head

d0 = Outer diameter of eye or fork

t1 = thickness of eye end 

t2= thickness of forked end (double eye)

x= distance of the Centre of fork radius R from the eye 

STEPS TO DESIGN KNUCKLE JOINT 

Step 1 : Design of Rods (D,D1)

Tensile failure of rod 

Knuckle Joint Tensile failure of rod

Using basic strength equation

Load = Stress * Area

P = (Area \quad Resisting \quad Tension) \times (Allowable \quad Stress)

P= \frac{\pi}{4} \times D^2 \times \sigma_t                          From This equation Diameter ‘D’ of rod can be found

Empirical relations

Using Empirical relations the enlarged diameter of rod D1 is determined

D_1 =1.1 \times D

Step 2 : Decide the thickness of eye end and forked end (t1,t2)

Empirical relations

Both these dimensions are decided on the basis of empirical relations,

t1= 1.25 D    and    t2= 0.75 D

Step 3 : Decide the dimensions of pin (d,d1)

Double shear Failure

The pin may get sheared off into three pieces as shown below, since the pin breaks at two places it is called double shear. Both areas are taken as resisting areas.

Knuckle Joint Double shear Failure of pin

Using basic strength equation

Load = Stress * Area

P = (Area \quad Resisting \quad Tension) \times (Allowable \quad Stress)

P= {\color{Red} 2} \times \frac{\pi}{4} \times d^2 \times \sigma_s                          

Note that 2 is because of double shear ...From This equation Diameter ‘d’ of pin can be found. But since the pin is also subjected to bending one more diameter of pin on the basis of bending is determined and the bigger of both is taken as the final size of pin

Bending failure of pin

The diameter on the basis of bending is determined using the following formula,

d =\sqrt[3]{\frac{32}{\pi \sigma_b} \times \frac{P}{2}[\frac{t_1}{4}+\frac{t_2}{3}]}           …..Calculate d from this formula

{ Discussion on how this formula is obtained is in the theory question and answer section of knuckle joint }

Empirical relation for pin head diameter

Since pin head is not subjected to any stress, its diameter is simply decided on the basis of proportionality, (it is taken 50% more than that of pin diameter )

d1=1.5 d

Step 4 : Check Stresses in Eye end 

Empirical relation for outside diameter of eye and fork

d0=2d

Tensile failure of eye end

The single eye may fail in tension as shown below { please note that when the plane of failure is perpendicular to the direction of force then the failure is either tensile or compressive}

 

Knuckle joint Tensile failure of single eye

 

Using the basic equation for stress

P = (Area \quad Resisting \quad Tension) \times (Allowable \quad Stress)

P= (d_0-d)t_1 \times \sigma_t                           ………….Using this equation find the value of \sigma_t  and check if it is less than allowable value for design to be safe.

{Note that area resisting the tension is rectangular one and not circular so its area is length time height total length is (d0-d) and height is t1.

Shear failure of eye end

The single eye may fail in shear as shown below { please note that when the plane of failure is parallel to the direction of force then the failure is Shear failure}

Knuckle Joint Shear failure of single eye

Using the basic equation for stress

P = (Area \quad Resisting \quad Tension) \times (Allowable \quad Stress)

P= 2 \times (\frac {d_0}{2}- \frac{d}{2})t_1 \times \sigma_s            simplifying this equation we get            

P= ({d_0}- d) \times t_1 \times \sigma_s………….Using this equation find the value of \sigma_s  and check if it is less than allowable value for design to be safe.

Crushing Failure of eye end

The single eye is also subjected to Crushing between pin and inner face of single eye. In case of crushing failure since the area is curved we take the projected (area which would be visible in drawing) of the cylindrical area. As we know that a cylinder appears as a rectangle in projection, hence the area will be diameter times the height of cylinder. This area is illustrated below,

Knuckle joint Crushing failure of single eye

Using the basic equation for stress

P = (Area \quad Resisting \quad Tension) \times (Allowable \quad Stress)

P= d \times t_1 \times \sigma_c                           ………….Using this equation find the value of \sigma_c and check if it is less than allowable value for design to be safe.

 

Step 5 : Check Stresses fork end

Fork end is also subjected to same failures as that of eye end, the only difference is that it has two eyes. So we get the same equations except multiplied by 2.

The equations for tensile, shear and crushing failures are given below

Tensile failure of  fork end

P={\color{Red} 2} \times (d_0-d) {\color{Red} t_2} \times \sigma_t                 {see the changes highlighted in red from the equation of single eye} Get the value of induced tensile stress from this equation and confirm that it is below allowable tensile stress.

Shear failure of  fork end

P={\color{Red} 2} \times (d_0-d) {\color{Red} t_2} \times \sigma_s                 {see the changes highlighted in red from the equation of single eye} Get the value of induced shear stress from this equation and confirm that it is below allowable shear stress.

Crushing failure of  fork end

P={\color{Red} 2} \times d \times {\color{Red} t_2} \times \sigma_s                 {see the changes highlighted in red from the equation of single eye}  Get the value of induced crushing stress from this equation and confirm that it is below allowable crushing stress.

 

Summary of Design Procedure in Tabular form

{ Student preparing for exam find the tabular form very handy and easy to remember the formulae }

Step

Failure

Equation

To find

Step 1 : Design of Rods (D,D1)

Tensile

 

Empirical

P= \frac{\pi}{4} \times D^2 \times \sigma_t

D_1 =1.1 \times D

 D=

D1=

Step 2 : Decide the thickness of eye end and forked end (t1,t2)

Empirical relations

t1= 1.25 D    

 t2= 0.75 D 

t1 =

t2 =

Step 3 : Decide the dimensions of pin (d,d1)

Double shear

Bending failure

 

 

Empirical

P= {\color{Red} 2} \times \frac{\pi}{4} \times d^2 \times \sigma_s   

d =\sqrt[3]{\frac{32}{\pi \sigma_b} \times \frac{P}{2}[\frac{t_1}{4}+\frac{t_2}{3}]}

d1=1.5 d

d =

 

d= 

{Choose higher value of both}

Step 4 : Check Stresses in Eye end 

Empirical

Tensile

Shear

Crushing

d0=2d

P= (d_0-d)t_1 \times \sigma_t

P= ({d_0}- d) \times t_1 \times \sigma_s

P= d \times t_1 \times \sigma_c

d0=

check  \sigma_t

check \sigma_s

check \sigma_c

Step 5 : Check Stresses fork end

Tensile

Shear

Crushing

P={\color{Red} 2} \times (d_0-d) {\color{Red} t_2} \times \sigma_t

P={\color{Red} 2} \times (d_0-d) {\color{Red} t_2} \times \sigma_s

P={\color{Red} 2} \times d \times {\color{Red} t_2} \times \sigma_s

check  \sigma_t

check \sigma_s

check \sigma_c

NUMERICAL PROBLEMS

 


1) It is required to design a Knuckle joint to connect two circular rods which are subjected to an axial tensile load of 50 kN. the material used for joints is 50c8, having  yield point stress 400 N/mm2. taking factor of safety as 5, find the all dimensions of Knuckle joint.


2) A Knuckle joint carries  an axial load of 100 kN.  it connects two circular rods which are made up of material 40c8, having  yield strength 360 N/mm2.  using factor of safety 4 design and draw the Knuckle joint.  Show all important dimensions in the final drawing.


3) Design a Knuckle joint to transmit a load of 50 kN, assume permissible stresses as 40 N/mm2 in 10 size 40 N/mm2 in shear and 80 N/mm2 in compression.Also draw neat dimensioned sketch of Knuckle joint showing all important dimensions designed.


4) designer Knuckle joint subjected to a pull of 30000N.  assume permissible stresses as below,

 allowable tensile stress 80 N/mm2

allowable shear stress  60 N/mm2 

 allowable crushing stress  120 N/mm2 

5) A knuckle joint is required to transmit a load of 150 kN. the rod end, fork end  and pin are made up of same material.  The design stresses may be taken as

 allowable tensile stress 75 N/mm2

 allowable shear stress 60 N/mm2

 allowable crushing stress 150 N/mm2

6) It is required to design a Knuckle joint for a tie rod of circular cross section,  the joint is subjected to maximum Pull of 70 kN.  the ultimate strength of the material of the rod against tearings  is 420  Mpa . the ultimate tensile and shearing strength of pin material or 510  Mpa and 400 Mpa respectively. Design the single eye, double eye  and pin. Take factor of safety is equal to 6

7)Design a knuckle joint required to connect two  steel bars subjected to a tensile load of 25 kN. The allowable stresses are 
75 MPa in tension, 
60 MPa in shear and 
100 MPa in crushing. Also draw the neat dimensioned sketch of the joint showing all important obtained dimensions.

8) Two MS rods are to be connected by a knuckle joint, load it has to transmit  is  120 kN. Design the joint assuming that the working stresses for both the pin and rod materials to be,

85 MPa in tension,

70 MPa in shear and

140MPa in crushing.

C) Theory Questions and answers on Knuckle joint

Theory questions are available here - Theory Questions Link

D) Objective type Questions and answers on Knuckle joint

 

 

 

 

Links to other Topics in Machine design I

Machine Design I - Introduction to Design : Theory Q&A

Machine Design -I -Design of joints : Theory Q&A

Cotter Joint : Design Procedure,Problems and Question answer

Design of turnbuckle : Design steps, Problems and Question

Design of Levers : Hand Lever, Foot Lever, Bell crank lever

Design Of Bolted and Welded Joints

Design of Shafts: Theory and Numerical Problems

Couplings : Design Procedure and Numerical problems

Design Of SPRINGS : Questions and Numerical problems

Power Screw Design

Belt drives:Theory Q&A and Selection of Flat and V belts


Comments

Best article on design of knuckle joint...

Add new comment