Recent Updates

Wednesday, 7 January 2015

FOUR BAR LINKAGE MECHANISM


In this Post, we will have to discuss about the four bar linkage mechanism. Four bar linkage mechanism has some important features as mentioned below 

1. Four bar linkage is the simplest form of all closed loop types of linkage mechanism.
2. There will be one fixed link, 3 moving link and four pin joints.

Objective

To analysis four bar linkage mechanism in real life mechanism, let us discuss the real world example of four bar mechanism that we generally use in our engineering life and it is locking pliers. 

Locking pliers are pliers that using over center mechanism and 4 bar mechanism to enable users to maintain the pliers in locked position.

Let us consider about which four bar we are talking, in following figure we have mentioned the links. 

We have taken measurement of links from our workshop locking pliers and for easy understanding we are just indicating the links below.


There are three moving links and one fixed link, lets understand about the links 
DC form one link, AD form second link, AB form third link and BC form fourth link
As we have mentioned above, DC link is fixed

AB = 80 mm,              BC= 55 mm
AD= 35 mm,               DC= 25 mm

We have fixed the link DC, the bottom one which is quite easy to make it fix and we may draw the movement of instantaneous center throughout the reason. 

Before that we must understand the concept of instantaneous center and we can understand about the instantaneous center by considering the below figure.

In order to calculate the movement of instantaneous center we must calculate the displacement equations. 

The displacement equation of four bar linkage might be secured by considering a rectangular coordinate system as shown in figure. 


For A,                                                                    For, B
XA= -a1Cosᶲ                                                   XB= a- a3Cos ψ
YA= a1Sinᶲ                                                     YB= a3Sin ψ   
A= Sin ᶲ,             B = (a4/ a1) +Cosᶲ
C= (a4/ a3) Cosᶲ + {a1– a22+ a32+ a42}/ (2a1a3)

For one rotation, I am calculating and rest could be calculated as I have given the above formulas
For, ᶲ= 20 degree, we will have from above formulas, then right link will move 65.81 degree.

There will be limitation also in changing the angle and changing of instantaneous center and its depends on the mechanism for example if we see the pliers mechanism then when we press at one end then there will be one stroke and for that one stroke only link will change its, angle and instantaneous center will be changed.

Let us consider various positions in single diagram.
There will be two basic constraints equations as mentioned below. 


Let us move for new post based on “Fluid Coupling "

Also read:


Do you have suggestions? Please write in comment box