Step 1: How to determine the sizes of Links in Drag-Link Mechanism:
In this section you will learn to actually make a 'to the scale drawing' of a Drag-Link Mechanism. If you have to design a Drag-Link-Mechanism your first step will be, to find the sizes of the linkage in order to ensure that the mechanism really works. Choosing arbitrary sizes will land you in soup and your mechanism may get stuck. This can also be used for project-work, whereby you can design the mechanism and demonstrate how it functions. The 'to-the-scale-dimensioned-drawing' of the mechanism is being posted below. The steps of construction are being articulated below the drawing:
Note: As you can see, sample dimensions have been assumed for making the above 'Drawing'.
In the Drawing, you can see, there are two circles. At first draw a centre-line almost at the centre of the drawing sheet. On the centre line draw the large circle having radius = 90mm, followed by another circle of radius = 75mm. The centre distance between the two circles has been chosen = 30mm (you are free to choose as you wish, later on, but in this case stick to the sizes shown). Now draw the line DC" (it lies on the centreline) next draw the line AB" followed by joining the ends B" and C" to get the length B"C" which is = 95.79 mm (found out by joining points B" to C" and measuring the length).
It is noteworthy that we have chosen to draw the lines DC", AB" and B"C" before drawing AB, BC and CD because we have to determine the length B"C" (key to the problem) since this the correct way of finding the length of BC which is same as B'C' or B"C".
Having determined the length B"C" you can go ahead and draw AB ( keeping the slant similar to the drawing) followed by drawing BC = 95.79 and finally joining C with point D.
For the next position of the of the linkage, take a lengths of 95.79mm and then with point C' as centre draw an arc cutting the smaller circle, the point of intersection gives you point B', join B' and C' to get B'C' = 95.79mm. Thereafter join B' to D to complete the drawing.
Having completed the drawing, measure the angle between AB' and AB" = β on the right of point A followed by measuring the angle between the same two lines on the left of point A = α .
You will find that α > β
Having completed the exercise we will go back to the post where a similar (undimensioned) diagram was posted to explain the principles behind the functioning of a Drag-Link mechanism and try to understand the same.
When Link-2 acts as the driver, rotating at a uniform speed, we will have Link-4 (the driven) moving at a non-uniform speed.
Explanation:
Suppose, starting from point C', the point, moves through an arc C'CC", i.e. the Link-4 rotates by 180° whereas the Link-2 rotates through an angle = α.
Next, the Link-4 keeps rotating from point C" and covers another 180° till point C', this time Link-2 rotates through an angle = β.
You have already measured both the angles and seen for yourself that β is much smaller than α.
As a result, it is evident that for the same amount of rotation of Link-4, rotating at a uniform speed, in the first-half covers first 180° and the second half covers second 180° ---- the 'follower' i.e. Link-2 takes much smaller time to go through the second half of rotation as compared to the first half.
Note: On the other hand if Link-2 becomes the driver and Link-4 the follower the situation will be the opposite.
No comments:
Post a Comment