Lecture - 14 ( double slider crank mechanism)

DOUBLE SLIDER CRANK MECHANISM 

Double slider crank mechanism is a combination of four links having two turning pairs and two sliding pairs. Diagram of this mechanism is shown below-

This mechanism consists of three inversions as listed below-
  • First inversion (fixed slotted plate)
  • Second inversion (fixed slider)
  • Third inversion (fixed connecting rod)

First inversion  

When slotted plate is fixed then it become the first inversion of double slider crank mechanism. Its example is elliptical trammel. Diagram of elliptical trammel is shown below- 

An electrical trammel is a simple mechanism which can trace exact elliptical path. It consists of two shuttles which are are confined to perpendicular channels or Rails, and a rod which is attached to shuttles by pivots at fixed position along the rod. As shuttle moves back and forth, each along its channel, the end of the rod moves in an elliptical path. Elliptical trammel is used in automatic tool changer in a machining.
Consider the following diagram of elliptical trammel, here point P lies between point A and B on a straight line. The straight line forms angle 'θ' with respect to x-axis. The point P is X distance away from y-axis and Y distance away from x-axis. 
From the above derivation we get the equation of ellipse. Therefore it can be concluded that each locus on line AB will draw an ellipse.
Different sizes of ellipse can be generated with the help of elliptical trammel just by adjusting AP and BP.

Always remember that each locus on line AB  will draw ellipse accept midpoint, midpoint will always generate a circle.
Let us consider point P lies in middle of line AB such that AP and PB are equal.

Second inversion 

When anyone of the slider is fixed in the mechanism then it become second inversion of double slider crank. Example of second inversion is scotch yoke mechanism. In this mechanism rotation is converted into reciprocation motion or vice-versa. Diagram of scotch yoke mechanism is shown below-
Scotch yoke mechanism is used in power hexa machine.

Third inversion

When link connecting slider is fixed then it become the third inversion of double slider crank mechanism. 


It's example is Oldham's Coupling. Oldhams coupling is used to connect the shaft having lateral misalignment. Its diagram is shown below-

Comments

Post a Comment

Popular posts from this blog

Lecture - 11 (Transmission angle in four bar mechanism)

Image
The angle between coupler link and output link in four bar mechanism is known as transmission angle. It is represented as ' μ'.   In the given figure Let us consider angle between link AO and link OC (fixed link) is represented by ' θ'.   Let us consider length of link OC is 'a', length of link OA is 'b', length of AB is 'c' and length of BC is 'd'.  According to law of cosine we get -  Now  μ will be different for different values of  θ. μ will be maximum for  θ = 180 degrees  & minimum for  θ = 0 degrees.   Problem 1-  Calculate maximum and minimum value of transmission angle for the following mechanism. Length of link OC is 5cm, length of link AO is 1cm, length of link AB is 4cm and length of link BC is 3cm. Solution 1 -   To calculate minimum value of transmission angle observe the mechanism when  θ = 0 degrees   To calculate maximum value of transmission angle observe the mechanism when  θ = 180 degr
Read more

Lecture - 8 ( problems on degrees of freedom)

Image
PROBLEM 1  - For the mechanism shown in figure, calculate it's degrees of freedom (F) SOLUTION 1  -  number of links (L) = 8 Number of binary joints (J) = 10 Number of higher pair (H) = 0 Number of those motions which are not the part of mechanism (N) = 0 KUTZBACH equation: F=3(L-1) - 2(J) - H - N F = 3(8-1) - 2(10) - 0 - 0 = 1 F = 1  PROBLEM 2 -  Calculate degrees of freedom for given mechanism SOLUTION 2 -   L = 4 ; J = 3 ; H = 1 ; N = 1 F = 3(L-1) - 2J - H - N = 3(4-1) -2(3) -1 -1 = 1 F = 1    PROBLEM 3 -  Calculate degrees of freedom for given mechanism.  SOLUTION 3 -   L = 5 ; H = 1 ; J = 5 ; N = 0 F = 3(5-1) -2(5) - 1 - 0 = 1  F = 1 PROBLEM 4 -  Calculate degrees of freedom for given mechanism SOLUTION 4 -    Here one link is spring. Spring is the link of flexible length. Spring can be replaced by 2 links connected with turning pair. L = 5 ; J = 5 ; H = 1 ; N = 0 F = 3(5-1) -2(5) - 1 = 1  F = 1
Read more

Lecture - 15 (Mechanical advantage, mechanism efficiency, toggle mechanism)

Image
Mechanical advantage   It is defined as the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of machine.  The combination of pulley system is shown in the below figure-  Here for lifting download F force is required but the effort given to the system is 8 times less, which is the good example of mechanical advantage. For the above system mechanical advantage is 8 Mechanical advantage = load / effort  Mechanical advantage for linear motion is -  Mechanical advantage for angular motion is -  Efficiency of system or mechanism Efficiency of the system is defined as the ratio of output power to the input power. PROBLEM - Find out mechanical advantage of the given mechanism having 100% efficiency. Solution -  In the sense of angular velocity It means that above mechanism is toogle mechanism.  Toggle mechanism   Those mechanism whose mechanical advantage is so high that it approaches
Read more