How Can I Determine the Velocity of Points C and F in a 4-Bar Linkage System?

[blue-steel]

Hi all,

(see the attached images)

working on a dynamics project about the suspension system in a bike, we have been given a 4 bar linkage - dynamic system, and I am trying to find the velocity of POINT C and POINT F, but given in terms of the angular velocity of bar AB in the STATIC POSITION the solid grey position, the other 'shadow' image is the linkage at it's deflected image,

the 2 images attached show a diagram of the system, Point C represents the centre of the back wheel, all black circles are pivot points, the second gives the coordinates of the points in the two positions

what I am workin on so far is the relative velocity relationship, Vc = Vb + Vc/b to find Vc, only problem is this (ultimately) involves omega_BC when the final value for Vc should only involve omega_AB

should look something like this
Vc = somenumber.omegaAB.i + somenumber.omegaAB.j

anyone familiar with similar problem or have any advice on how to go about it would greatly appreciate it

Cheers

blue steel

 

[MikeLizzi]

I suppose there are lots of ways to approach this problem. Here's mine.

I would start by redrawing the picture. Leave out the spring and E-G. Redraw B-C-D as a straight bar B-D. (You loose point C but that can be added later.) Now you have four unequal bars pined at their four corners. A-E is vertical. A-B is horizontal. Now consider a small rotation of bar A-B while holding A-E fixed. Can you calcualte the required rotation of E-D? Its messy geometry but you should be able to get that angule of E-D as a function of the angle of A-E.

Once you’ve done that you can calculate the angular velocity of E-D as a function of the angular velocity of A-E. Then you can calculate the linear velocity of any point in bar E-D. That gives you the velocity of point F and D.

Now replace bar B-D with the original B-C-D. You know the velocity at points B and D. Do you know how to use that info to calculate the velocity at C?

 

[blue-steel]

Thanks a lot, the thing I was getting confused with most was the 'bent link' BCD but since all points on it travel with the same angular velocity the simplification BD worked

Cheers