Determing force on anti overrol construction

[Mark23]

Hi there,

I am going to do the sine with dwell test on a vehicle. The vehicle can turn over on this test. For the safety of the driver, there must be an construction so that the vehicle can't turn over. The vehicle speed with this test is 80 km/h around 55 miles per hour. The steering movement can be seen in the following link.

http://www.google.nl/imgres?imgurl=...=nl&sa=N&tbs=isch:1&ei=u8ViTZGWMMb5sgbsxPm4CA

I use the som of moments to solve this. In the figure you can see mine VLS of the vehicle. I want to know force B, wich is the force on the anti overrol construction when the vehicle turns over. I think that B = (F*1100)/1500. And I think that F is m *a, and a is the lateral acceleration from 0.9 g, so this is around 8.3 m/s^2. Whith the vehicle mass of 3500 kg the force B would be 22661 N.

So the force on this contact surface from the anti overrol construction is 22661 N. Can anybody confirm if this is wright or wrong?

Next to this force, there is working another force on the contact surface from the anti overrol construction. This is the friction force. This is force B * the frictioncoefficient.

I am not sure of this all. I appreciate it if anybody can confirm this or when i am wrong correct me. Many thanks.

p.s. sorry for my bad English, i am from Netherland and don't have much English lessons.



Naast deze kracht werkt er naar mijn idee ook een wrijvingskracht op dit contactoppervlak van kunststof. Dit is de wrijvingscoefficient maal kracht B als ik het goed heb.

Kan iemand mijn verhaal bevestigen of juist ontkrachtigen als het niet goed is.

 

[GregLocock]

The overturning moment on the vehicle is Mx=cgz*m*a

This is created by a 'weight' transfer across the track t of WT=Mx/t

However, this number is remarkably useless to you I suspect, unless you are trying to predict wheel lift off.





Cheers

Greg Locock


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[Mark23]

Thanks for your reply, I am think i must calculate the force when the right wheel lifts of the ground. Thats why I didnt take the reaction force from this wheel in my calculation.

But how I see it, the moment from the anti overrol construction must be the same as the overturning moment. Am i correct?

With the formula you given i can calculate the maximum acceleration and the maximum overturning moment, because all the weight of the right wheels where transfert to the left wheels. Then it must be correct.

But anybody know something about the friction force? If the contact surface from the anti overrol construction hits the ground there is a friction force i guess. Ffriction = B * frictioncoeffienct. Is this correct?

Thanks

 

[GregLocock]

Yes sort of. Trouble is that the friction coefficient changes with Fz.

Cheers

Greg Locock


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[GregLocock]

Oh and also one axle may lift before the other.

Cheers

Greg Locock


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[Mark23]

What do you mean by Fz, is that the force on what I called B in my free body diagram in the attachment? That the friction force is depending from this force I know.

Do you know how I deal with the fact that one axle lifts before the other? I am now doing an internship and I still niet to learn a lot. Thanks

 

[GregLocock]

Since you haven't told me what you are really trying to do, no, I don't know how you need to account for the different roll stiffness fron to rear.

I haven't seen an FBD, ah ok.

B*2.5=1.1*m*latacc at rollover.

Cheers

Greg Locock


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[Mark23]

I must design the anti overrol frame that is placed on the vehicle during the test. I must know the reaction force on the globe wich you can see in the attachment of this message. That is what I want to know.

Do i need the wheigt transfer of both axles for this calculation or am i going in the wrong way.

 

[GregLocock]

If the sliding puck is a single piece then no you don't need to worry about which axle lifts first.

So, how are you going to determine the friction coefficient? and what happens to to the forces when the puck digs in?

Cheers

Greg Locock


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[Mark23]

Oke thanks, but why I don't need to worry if the sliding piece is a single piece? Because I don't understand why that is so and I want to understand.

The friction coefficient is the next task. I think I must determine this experimental. I dont know if I need the static of kinematic friction. I think the kinematic.

Also the forces when the puck dugs in are unknown to me. I dont know how to determine this. Because I cant imagine it, how this will go. You have an idee?

 

[GregLocock]

I'd estimate it as above using mu of 1, and then use a factor of safety of 4 or more.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[Mark23]

Oke a mu of 1 is high, but that doesnt matter, because there goes a safety factor over of 3 or 4. What I thought by now, is that I maybe forgot the vehicle mass * 9.81 from the gravitation force in the center of gravity. I quote: I haven't seen an FBD, ah ok. Is this the force you where missing?