Design of anti roll bar

The second case is exactly the same as the first, if you take the centre of the long section (point D) as you point of reference.

E.G. Relative to point D, both cases have the tips moving up and down.

but sir in second case , as i have fixed the end (G), so reference will be (F).

Is it right to give double stroke to make possible the second case?

No no no no no! The whole point of a 'reference' point is that it is arbitrary and can be chosen to suit the problem.

Your 'Fixed point' is only fixed relative to the bench and the earth...it isn't fixed relative to the middle of the bar ... or the moon or many other reference points.

When you are driving in your car what is the reference point for the roll bar then? The middle of the bar or the chassis! Certainly not the bench or the earth. Think about what I said more carefully...

If you still don't understand... imagine the 'clamp' and everything except the bar is invisible, now imagine that you are standing on the end which is being flexed up and down very slowly... You wouldn't be able to tell which end is fixed and which is moveing..... because it doesn't matter! What matters is that they are moving relative to each other.

(Have you ever been in stationary car (or railway carriage)... you suddenly think you are moving...! But you are not ...it is the car (or train) next to you which is moving.)

I hope this explains it

Del

(PS please don't call me 'Sir'.... I think we are all equals )

I mean to say, reference for torsion will be point F in second case i.e. As point G is fixed, C and E ponts are bush clamped and we are giving up and down motion to the point A , then bar will be twisting about point F. To make second case similar to frist one , I have to give double stroke to end A as point F is becoming the reference for twisting. Correct me!

Yes. If you fix one end the stroke at the other end will now be twice the stoke that was applied to the individual ends.

(assuming that when the opposing ends were deflected the strokes were equal magnitude but opposite directions)

Del

I am also thinking in same way. Other thing can also be possible that fixing the middle of bar at D and giving the up and down motion to end A. Then in this case, stroke will be the same as in case where both ends are given to be up and down motion in opposite direction. Am I correct?

If I understand you correctly what you are saying is effectively just model half of the system as it is symmetrical. This is fair enough if you are modeling maximum deflection at either end you can look at half the syatem. However you need to think carefully about what you are trying to model. In a real situation the car may have 1 wheel relatively steady and the other may go over a hole, the limit of the suspension travel will come into play here but of course the whole of the straight section will be acting as a a torsion bar. At this point I hope I have helped with your understanding, but I feel, as you are designing it, you should be able to understand the limitations of its operation .

Regards

Del

still i am in confusion that which case is proper i.e.

firstcase : one end is fixed while other is under fatigue loading;

case second: middle section of bar is fixed and one end is under fatigue loading;

case third : both ends are under fatigue loading.

plz. tell me clearly.

Blimey...all 3 cases are fine as long as the deflection is correct...It's a matter of which is simpler to do!

E.G ... Which of these is 30?..

5+5+5+5+5+5

or 6x5

or 5x6

or (10x5)/2

It's all the same but some easier to calculate than the others.

With a big mechanical test it can make a significant difference to time and cost....

If all else fails make a small model out of some bent wire and play with it until you understand !

Understanding is not about being told, or even reading in a book...it is about appreciating actually what is happening and why, also about getting it wrong and realising your mistakes...it's called experience.

I wish some others would contribute here...as I'm no a mechanical design engineer...I do not claim to necessarily be right...I am only trying to guide and encourage. You must check what I say for yourself!

(Come on Guys....someone help me out here ...I'm flagging...)

Del

thanks for your valuable guidance. I am very appreciated by your positive reply and helping nature. still any suggestion from you, welcome sir.

plz. tell me whether i am correct or not.

as concern to you, in second case I have to fix point D which is the point of symmetry to have the similar effect similar to first case.Then there should be the same stroke to be given similar to in the first case.

Plz. help me, I need help of experts in suspension field.