moments at contact tyre patch and wheel spindle

[shott92]

hi guys this is an image of the rear wheel of a push bike going into a corner as well as what i think are relevant forces,
i need to calculate the moments or torque at A (the contact patch) and B (the hub centre)
this is to calculate the force on the left hand side of the frame and right hand side of the frame while in a corner
Fsr = the force going through the suspension
Fc = lateral cornering forcer on the rear wheel
phi = lean angle
MG = mass & gravity
t = the cross sectional radius of the tyre

i can and have calculated the above parameters through lot of prior equations that take into account the wheelbase, velocity, friction, radius of the turn ect. but i think that the above are the require ones to calculate this... if there is an easier way please let me know

im sure its quite simple really but i have either over thought it are just been daft

cheers for any help guys but could you please explain your reasoning behind your answer so that i can verify it please

https://s.yimg.com/hd/answers/i/7c911b9cff554eabad1ca29c0f9e8161_A.jpeg?a=answers&mr=0&x=1398491203&s=49df46fb82b50fa73242ec5bd28516c4

Edited April 26, 2014 by shott92

[studiot]

You need to work on your diagram before doing any calculations.

 

Do you understand that the bike is not in equilibrium so you cannot use an equilibrium analysis directly?

 

How is F_sr applied to the wheel?

 

What is F_c ? Friction or circular motion forces?

 

On the subject of the curvilinear motion, are you considering accelerations (Newton's solution) or Quasi Equilibrium (D'Alembert's solution) ?

[shott92]

hi well that is a little more advanced than i was hoping,
the frictional / tractive for would be the force that is opposing Fc i assumed that tractive force would have to be higher but

 

V = Sqrt((mu)gr) for this instance i am using (mu) the traction co efficient as 1.3

Fc = (sprung mass *v^2)/®*(wheelbase-CoG(X))/wheelbase)cos(r/(sqrt(CoG^2+r^2))

tractive force = 1/2 Cog(x)(sqrt(g^2)+(V^2/r))

Fsr = mg(wheelbase-CoG(X))/wheelbase)+Frontal area*coeffiecient of drag*(Cog(y)/wheelbase)

the reason i didnt put these equations in is becuase of the readability

 

and i am not sure i can assume it would have been newtons solution as thats all we have been taught
and i was planning or attempting to work the moments out as simply as possible would there be an easier way if we was to assume a constant state? this is what i was trying to do as i would be able to find for Vmax and a specific lean angle hoping that it would give a useable single point answer rather than finding for a certain displacement or deriving for a point along the corner

 

i hope this provides enough information for you to be able to help me (also if i have mucked up one of the above equations then let me know but i have reworked the a few times they seem accurate) thanks very much the help is hugely appriciated

[studiot]

 

Fc = (sprung mass *v^2)/®*(wheelbase-CoG(X))/wheelbase)cos(r/(sqrt(CoG^2+r^2))

 

This is the first time I have ever seen a formula that includes a registered trade mark!

 

But I agree that is way too complicated to start off with.

 

But I was not asking for this, I was asking about the mechanics of your diagram.

 

Your diagram should be a free body diagram or rather set of diagrams since it is a 3 dimensional situation.

 

You are asking about forces and moments on the tyres so I would suggest the free body should be the wheel/tyre assembly.

 

Given this, the answer to my question where does F_sr act is "F_sr acts on the hub" assuming it the the froce applied by the forks to the wheel.

 

From what I can see (if you are saying F_c is the frictional force) F_c acts perpendicular to the plane of your diagram not in it.

 

Also as you have drawn it the velocity is perpendicular to the plane of your diagram.

 

If you have been taught Newton you need to introduce the centripetal acceleration, which is in the plane of your diagram.

 

You did not answer my first question about equilibrium analyses.

 

Incidentally I cannot see whatever you have linked to in the first post.

Please post the necessary information in the thread, this is also forum policy.

Edited April 26, 2014 by studiot

[shott92]

hi the cheers
sorry when i typed ( r ) without the spaces it automatically made it to a ® :/

the Fc is the lateral force from the center of the corner outwards (the centripetal force) as such i am basing this not on centrpetal acceleration just the given force in Newtons, the velocity is perpendicular to the plane of the diagram but technically irrelevant as this should be able to be solved in 2D, the velocity and radius of the corner though give the lateral force, friction is the centrifugal force or the force that allows the bike not to slip hence the friction must be higher than Fc or slip would occur, and for the given point in time the bike is in equilibrium as the lean angle is definite as are all the other factors the bike is only in forward motion for this practice the forward velocity as far as i can tell is irrelevant

this scenario could be mimicked by the bike being stationary but leant over to a certain angle then a force pushing onto the seat (down through the CoG to the contact patch) the wheel would still want to twist to an upright position at both A (the contact patch) and B the spindle or hub on the forks, as i said this is what im trying to figure i'm sure there should be a simple moment around a point equation to calculate this but i cant figure what value it is i actually need :/

sorry for being awkward but i've spent the last 6 months calculating many force equations for most of the degrees of freedom on the bike across the x,y and z axis im now trying to calculate the rotational forces and having abit of difficulty here im coming to the end of my 2nd year at university and this is a project i set myself as such there is no help sheets that i can refer to and the magnitude of books i have dont have the equations i need for this task, please accept my appolagies for any unclarity but this unclarity is why i am asking for help, and your time is much appreciated so thank you

[studiot]

From what you say, this is not really a homework question.

 

So I can post a more complete answer.

 

I think you are trying to make the issue too complicated by concentrating on the patch of contact between the tyre and the road.

The crux of the analysis is that the reaction between the tyre and the road has to pass through the centre of gravity of the bike or it would fall over.

 

Consider a bike plus rider, of weight W, travelling around a horizontal curve or radius r as shown with velocity v.

 

 

It is subject to a centripetal acceleration v²/r

 

There is one point (or line) of contact with the road, so the total reaction here R, is comprised of the vertical reaction, Q and the frictional force F.

 

F is the only force that can supply the horizontal centripetal force necessary to exert the centripetal acceleration, thus

 

[math]F = \frac{W}{g}\frac{{{v^2}}}{r}[/math]

If the cyclist leans over at an angle theta the line from the point of contact with the road must pass through the centre of gravity, so R must pass through the centre of gravity, G. (W passes through G and also has no moment about it)

 

Further since there is zero vertical movement

 

[math]Q = W[/math]

 

So

[math]\tan (\theta ) = \frac{Q}{F} = \frac{W}{F}[/math]

 

The analysis of four wheeled vehicles is different.