A) DESIGN PROCEDURE OF KNUCLE JOINT
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.
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
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
Using basic strength equation
Load = Stress * Area
From This equation Diameter ‘D’ of rod can be found
Empirical relations
Using Empirical relations the enlarged diameter of rod D1 is determined
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.
Using basic strength equation
Load = Stress * Area
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,
…..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}
Using the basic equation for stress
………….Using this equation find the value of
less than allowable value for design to be safe.
and check if it is
{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}
Using the basic equation for stress
simplifying this equation we get
………….Using this equation find the value of
allowable value for design to be safe.
and check if it is less than
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,
Using the basic equation for stress
………….Using this equation find the value of
than allowable value for design to be safe.
and check if it is less
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
{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
{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
{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
Tensile
D=
Step 1 : Design of Rods (D,D1)
D1=
Empirical
Step 2 : Decide the thickness
Empirical
of eye end and forked end (t1,t2) relations
t1= 1.25 D
t1 =
t2= 0.75 D
t2 =
Double shear
d=
Bending failure
Step 3 : Decide the dimensions
of pin (d,d1)
d=
Empirical
Empirical
Step 4 : Check Stresses in Eye
end
d1=1.5 d
{Choose higher
value of both}
d0=2d
d0=
Tensile
check
Shear
check
Crushing
check
Step 5 : Check Stresses fork end Tensile
check
Shear
check
Crushing
check
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/mm 2 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.
A) DESIGN PROCEDURE OF COTTER JOINT
Cotter Joint Introduction
Before going into detailed steps to design and find dimensions of cotter joint, it is necessary to
understand clearly the various components, their functions and assembly of cotter joint. Here is the
exploded view of cotter joint.
Exploded View of Cotter joint assembly
Let us see the animation of each component so as to get clear idea of its details and dimensions.
Spigot is the male part of the joint , it has a rectangular slot for passing the cotter through it. Spigot has
a collar which rests against the socket end.
Socket is the female part of the joint, it also has a rectangular slot for passing the cotter through it. It
has a circular hole in which spigot fits.
Cotter is a wedge shaped piece of metal which actually connects two parts which are non
rotating. Cotter is fitted in the slot and remains in its position by wedge action.
Taper is provided to, i) With taper it is easy to remove the cotter and dismantle the joint ii) It ensures
tightness of the joint in operation and prevents loosening of the parts.
Cotter is driven in or out using the hammer. Value of taper on cotter is 1 in 48 to 1 in 24 .
DESIGN PROCEDURE OF COTTER JOINT
Notations used in design are as follows,
P= Load on the joint or pull acting on rods, d= Diameter of the rod, d1= outer diameter of socket,d2=
Diameter of spigot or inside diameter of socket, d3= Outside diameter of spigot collar,d4= Diameter of
socket collar, t1= Thickness of spigot collar, a = Distance from the end of the slot to end of spigot, c=
thickness of socket collar, b,t,l= width , thickness and length of cotter
Step 1 : Design of rod( ‘d’)
Tensile failure of rod
Since the area resisting is circular using the basic equation of design
Load = Stress * Area
...From this equation diameter d of the rod is determined.
Step 2 : Design of Spigot { ‘d2’,’d3’,t1’,’a’}
Tensile failure of Spigot
The spigot end may fail under tension into two parts as shown in figure below, Since the plane of failure
is perpendicular to the direction of force it is tensile failure,
the area resisting the failure is shown by red in the diagram, since it is circular minus a rectangle the
equation will be,
………….From this equation the diameter of spigot end d2 can be found.
If you observe clearly it will form a quadradic equation .
Emperical Relations
{Other three dimensions of Spigot are decided on the basis of emperical relations and then checked
whether the stress induced is withing safe limits or not}
Crushing failure of Spigot ( Check crushing stress induced)
……...Find
within safe limits or not .
Using this equation and check if it is
Shear failure of Spigot ( Check Shear stress induced)
……...Find
Using this equation and check if it is
within safe limits or not .
Diagram showing the Shear failure of Spigot end
Step 3 : Design of Socket { ‘d1’,’d4’,’c’}
Following dimensions of socket are to be found and stresses induces must be checked for safe limit.
Tensile failure of Socket
The socket may also fail under tension into two parts as shown in figure below, since the plane of failure
is perpendicualr to direction of force it is the tensile failure.
Area resisting is a hollow square minus a rectangular strip having length (d1-d2) and thickness t, so the
strength equation becomes,
………...From this equation diameter d1 can be
determined, again there is quadradic equation formation. Refer to numerical problems section to view
the detailed solution.
Emperical Relations
{Other three dimensions of Socket are decided on the basis of emperical relations and then checked
whether the stress induced is withing safe limits or not}
Crushing failure of Socket( Check crushing stress induced)
……...Find
Using this equation and
check if it is within safe limits or not .
Shear failure of Socket ( Check Shear stress induced)
……...Find
it is within safe limits or not .
Shear Failure of socket end is shown below
Using this equation and check if
Step 4 : Design of Cotter { ‘t’, ‘b’ and ‘l’ of cotter}
Empirical relation
The cotter thickness is determined using an empirical relation, and is proportional to the diameter of
rod,
Calculate the width of b of the cotter by sharing and bending consideration
Double shear of Cotter
{Note that 2 is because of double shear failure}
Bending failure of cotter
{This formula is based on the bending stress induced in the cotter, the derivation of this formula is
described in the theory section of this chapter}
Design procedure of cotter joint in tabular form {Grouped for easy to remember steps}
tep Component
Rod
Criterion
Design equation
To find
Tensile Failure
d= ‘
Tensile Failure
d2 =’
Emperical relation
d3= ‘ ‘
‘
‘
t1 = ‘ ‘
a=‘ ‘
Spigot
Crushing Failure
Shear Failure
Tensile Failure
d1 = ‘
‘
Emperical relations
d4 =’ ‘
Socket
c=‘ ‘
Crushing Failure
=’ ‘
Shear Failure
=‘ ‘
t=‘ ‘
Emperical Relation
Cotter
Double shear
Bending failure
b= ‘ ‘
b= ‘ ‘
{Choose higher b
of both above}
B) Numerical Problems
1) Design and draw a detailed sketch of cotter joint which is required to withstand a load of 30 kN in
tension and compression. The material used for the cotter joint has following allowable stress values
Allowable stress in tension =50 MPa,
Shear Stress =35 Mpa
Crushing stress= 90 Mpa
2) Design and draw cotter joint required to withstand a load of 45 kN, The cotter joint is made up of
material having following allowable stresses
Allowable stress in tension =55 MPa
Allowable shear stress = 40 MPa
Allowable Crushing stress = 70 MPa
3) It is required to design a cotter joint which is used to connect two steel rods having same diameter.
The maximum axial load on the joint is 50 kN. The material used for joint is 30C8 ,ultimate tensile
stress for the material is 400 MPa. Using factor of safety 6. Design all dimensions of the joint. Take
ultimate shear stress half that of ultimate tensile stress and Ultimate Crushing stress double that of
ultimate tensile stress. Draw the neat dimensioned sketch of the joint designed showing all important
dimensions.
4) A cotter joint carries a load of 150 kN, is made up material 45C8. The ultimate tensile stress is 500
MPa. Using factor of safety of 4 design the joint.Take ultimate shear stress half that of ultimate tensile
stress and Ultimate Crushing stress double that of ultimate tensile stress. Draw the neat dimensioned
sketch of the joint designed showing all important dimensions.
5) Design and draw cotter joint required to withstand a load of 200 kN, The cotter joint is made up of
material having following allowable stresses
Allowable stress in tension =65 MPa
Allowable shear stress = 45 MPa
Allowable Crushing stress = 125 MPa
6) Design and draw cotter joint required to withstand a load of 120 kN, The cotter joint is made up of
material having following allowable stresses
Allowable stress in tension =95 MPa
Allowable shear stress = 65 MPa
Allowable Crushing stress = 186 MPa