**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.

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

- Rods {Which are to be connected by joint }
- Single eye {Modified rod for assembly}
- Double eye or Forked end {Modified rod for assembly}
- Pin {Connects the two rods}
- Collar {to keep the pin in position}
- 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

D_{1}= Enlarged diameter of rod

d = Diameter of pin

d_{1} = Diameter of pin head

d_{0} = 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,D**_{1})

_{1})

**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 D_{1} is determined

**Step 2 : Decide the thickness of eye end and forked end (t**_{1},t_{2})

_{1},t

_{2})

**Empirical relations**

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

t_{1}= 1.25 D and t_{2}= 0.75 D

**Step 3 : Decide the dimensions of pin (d,d**_{1})

_{1})

**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 )

d_{1}=1.5 d

**Step 4 : Check Stresses in Eye end **

**Empirical relation for outside diameter of eye and fork**

d_{0}=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 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}

Using the basic equation for stress

simplifying this equation we get

………….Using this equation find the value of 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,

Using the basic equation for stress

………….Using this equation find the value of 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**

{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
Empirical | D= D1= | |

| Empirical relations | t t | t1 = t2 = |

| Double shear Bending failure
Empirical |
d | d =
d= {Choose higher value of both} |

| Empirical Tensile Shear Crushing | d | d check check check |

| Tensile Shear Crushing | check check check |

**NUMERICAL PROBLEMS**

**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/mm^{2} in shear and 80 N/mm^{2} 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/mm**^{2}
**allowable shear stress 60 N/mm**^{2 }
** allowable crushing stress 120 N/mm**^{2 }

**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/mm**^{2}
** allowable shear stress 60 N/mm**^{2}
** allowable crushing stress 150 N/mm**^{2}

**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**

## Comments

## Mechanical

## Praise

## DESIGN OF MACHINE ELEMENTS

## Machine design

## solutions

## machine design

## Add new comment