Independent Research Project
Knuckle Joint
Problem Statement: -
A pull of 70KN is transmitted from a bar X to the Y – through a pin. If the maximum permissible tensile stress in the bar is 100N/mm^2 and the permissible shear stress in the pin is 60N/mm^2. Find the diameter of bars and the pin.
Aim: -Write a python program for Knuckle joint.
Diagram of Knuckle joint
Problem diagram
Knuckle Joint: -
A knuckle joint is a type of joint which connects two cylindrical rods whose axes lies on the same plane and are under a tensile load. It is used to transmit axial tensile force.
This joint allows angular movement of the two cylindrical rods in one plane.
Knuckle joint is named so as it uses a knuckle pin to join the two cylindrical rods.
Parts of knuckle joint :-
1) Two rods which are to be connected
2) Eye end
3) Double Eye End or Fork End
4)Knuckle Pin
5) Collar
6) Taper Pin or Lock Pin or Split Pin
Construction of knuckle joint and Functions of different parts :-
Knuckle joint is used to connect two rods in which one rod has one eye end and the other rod has two eye end or the fork end. This ends are named one and two eye ends as one of these has one hole and the other one has two holes.
One eye end is placed between the two eyes of the two eye end or fork end. After that a knuckle pin is inserted in the holes of the eye end and fork end. Knuckle Pin is used to hold fork end and eye end together
.
Some part of knuckle pin comes outside of the lowermost hole of the fork end. Knuckle pin has a small hole near its bottom. A collar and a tapper pin is used to lock the knuckle pin in the eye and fork end. The collar has two holes in it and the collar is placed in such a way that hole of knuckle pin coincides with holes of the collar and after that a tapper pin is inserted in the holes of the collar through the hole of knuckle pin. In this way a knuckle joint is formed.
Advantages of using knuckle joint :-
1) It can withstand large tensile loads.
2) It has good mechanical rigidity.
3) It can be easily dismantled and assembled.
Disadvantage of using knuckle joint :-
1) This joint cannot withstand large compressive loads.
2) Less flexible than universal joints
3) Allow movement in only one plane.
Applications of knuckle joint in different machines :-
1) In automobile a knuckle is supporting skeleton for wheel assembly.
2) Joint between link of Suspension bridge.
3) Joint between tie bar and roof trusses.
4) Tie rod joint of jib crane.
4) Chains Of bicycle
5) Wipers of vehicle
6) Chain straps Of watches
7) Earth Movers
Axial load
An Axial load is a force administered along the lines of an axis. Axial loading occurs when an object is loaded so that the force is normal to the axis that is fixed, as seen in the figure. Taking statics into consideration the force at the wall should be equal to the force that is applied to the part.
Maximum permissible tensile stress: -
Permissible stress means the allowable stress or allowable strength is the maximum stress (tensile, compressive or bending) that is allowed to be applied on a structural material
Maximum permissible shear stress: -
Permissible shear stress of a beam is the highest value of shear stress developed in a beam at which it reaches the permissible limit of the shear stress or at which, crack propagation starts in the beam. Please note that this shear stress is generated because of the transverse load.
Tensile load and Shear load
Tensile Load is the capability of a material to withstand a pulling force. It is usually measured in units of force per cross-sectional area.
Shear load is a force that causes shear stress when applied to a structural element. Shear stress, which is a force per unit area, occurs in the plane perpendicular to normal stress; it is created when two planes of the same object are trying to slide past one another.
Program: -
Case1( For given problem)
#Python program for find out diameter of Knuckle pin
import math
import matplotlib.pyplot as plt
#Force due to axial Load (N)
P = 70000
#Maximum Permissible tensile stress (MPA)
St = 100
#Maximum Permissible shear stress (MPA)
Ss = 60
#For tensile load
TL = math.sqrt(P/(St*(3.14/4)))
print("Diameter of pin due to tensile load", round(TL),"mm")
#For shear load
SL = math.sqrt(P/(Ss*(3.14/4)))
print("Diameter of pin due to shear load", round(SL),"mm")
if(TL>SL):
print("FINAL Diameter of pin (TL)", round(TL),"mm")
else:
print("FINAL Diameter of pin (SL)", round(SL),"mm")
Case 2
#Python program for find out diameter of Kunckel pin
import math
import matplotlib.pyplot as plt
#Force due to axial Load (N)
P = 70000
#Maximum Permissible tensile stress (MPA)
St = 100
#Maximum Permissible shear stress (MPA)
Ss = 110
#For tensile load
TL = math.sqrt(P/(St*(3.14/4)))
print("Diameter of pin due to tensile load", round(TL),"mm")
#For shear load
SL = math.sqrt(P/(Ss*(3.14/4)))
print("Diameter of pin due to shear load", round(SL),"mm")
if(TL>SL):
print("FINAL Diameter of pin (TL)", round(TL),"mm")
else:
print("FINAL Diameter of pin (SL)", round(SL),"mm")
Result screenshot: -
Case1
In this case TL<SL therefore else condition of diameter of pin (SL) is print.
Case2
In this case TL>SL therefore if condition of diameter of pin (TL) is print.
Program Explanation: -
- Import the math in a core library
- Import matplotlib.pyplot as plt to make python as matlab.matplot.pyplot is collection.
- To enter the value of axial load(N), maximum permissible tensile stress(MPA) and maximum permissible shear stress(MPA).
- Using the print command to print the value
- Using an if and else statement this statement contains the block of code that execute if condition expression in the if statement resolve to 0 or false value.
Conclusion
We are above the program conclude that when changing the value of maximum permissible shear stress out put result are change, comparison given in above screen short.