Help with suspension forces

[Ceboe]

Hi,

I'm the new responsible for calculating the suspension forces in our team. I had already an extensive conversation with the guy who did it last year. But thing is, he doesn't know any more how he did it. So we came up with a 'new' method. Now I would like to check our way of doing the calculations with you guys. Since we are seeing pretty large values, especially since our carbon tubes were made for 3000 N of pressure force and we would now have around 11000N max pressure force.

We started with thinking about reasonable max accelerations: 1,5 g for braking and 2 g for cornering. These would be max accelerations, so we can calculate the max force on the wishbones.

For 1,5 g deceleration we find 1,5*10 (g force to m/s²) *350 (estimated weight of the car with driver) = 5250 N

This force will be divided by 2, the know the force on a single wheel. Hence: 2625 N
This force will create a moment on the upright. 2645N*0.255(radius of rim+ tire)m = 674.5 Nm
When I calculate this back to the wishbone, the closest to the center this gives a force of: 674.5/0.08 = 8431.3 N
And a force of 674.5/0.12 = 5621 N

http://i50.tinypic.com/2u9lslh.jpg

For 2g lateral acceleration I find a force of 2*10*350 = 7000 N, divided over both wheels with a 45/55 weight distribution gives a highest force of 3850 N
This force will create a moment on the upright. 3850N*0.255m = 981.75 Nm
When I calculate this back to the wishbone the closest to the center of the wheel, this gives a force of: 981.75/0.08 = 12271,9 N.
But this would be a pulling force on the wishbone, witch is not so critical, so we calculate the push force on the wishbones in a corner at 2g:
981.75/0.12 = 8181.3 N.


http://i46.tinypic.com/2w69wde.jpg

Here we have solved the force to the individual wishbone arms by setting the moment zero and then using the perpendicular distance. I know this isn't fully correct but nobody at our college knew how to do it otherwise...

Then we have the force due to bump and weight of the car. Since we have a pullrod suspension this will cause a pushing force on our upper wishbone.

http://i47.tinypic.com/syut6w.jpg

When I put together all these cases, I found a worst case force of 11000 N that pushes on one A-arm and a max pulling force of 15000 N pulling on one A-arm.
This was done by just adding all the forces in the 3 different cases, calculated above.
Can we assume these forces are acceptable? Because for us they look pretty huge. But we can't really find an error or something..
I hope you will help us a bit, since I took my time explaining our question.

[Pete Marsh]

Hi,

Looks like you have an issue with your statics.

Take care when taking moments about a bearing, it is easy to make mistakes. It looks a lot like your wishbone loads assume some support at the bearing centre in all directions. I doubt you have this as your design. Consider treating the upright and wheel as a rigid body when appropriate.

Also, you are seeing the advantage of large mounting bases for your suspension.

Don't forget the toe links, this is where you will see the biggest loads in most designs. (and worst detrimental effects)

Pete

[BillCobb]

Try 2 g Ax and 1.6g Ay instead of what you started with. A 4g Az might also start you thinking.

[onemaniac]

This force will be divided by 2, the know the force on a single wheel. Hence: 2625 N
This force will create a moment on the upright. 2645N*0.255(radius of rim+ tire)m = 674.5 Nm
When I calculate this back to the wishbone, the closest to the center this gives a force of: 674.5/0.08 = 8431.3 N
And a force of 674.5/0.12 = 5621 N


To rephrase what Pete said, the moment should not taken about the centre of the hub bearing but the contact patch of the tire to the upper/lower wishbone mounts.

[Jay Lawrence]

Another thing to remember is that your tyres are probably not capable of giving you 2g cornering whilst also giving you 1.5 g braking (or 1.6g as BillCobb suggested). This means that it isn't really correct to 'add all the forces in the different cases' because these forces cannot all exist at once.

[Ceboe]

Thanks all for your fast responses!

I did the calculations again and indeed, you are all right. I don't know how we could see the error in the static calculations.

So now when I have a cornering acceleration of 2g there will be 3850 N and the tire contact patch, 6844 N at the lower wishbone and 3000 N at the upper wishbone.

With 1,5 g deceleration I now become 4700 N at the lower wishbone and 2100 N at the upper wishbone.

I know we can't just sum up all of the forces in the different cases, but we want our wishbones to be definitely strong enough. So I think I will just use this method. We also our looking for funding of tire data, witch we don't have now. So I don't know exactly how much g's will be left in the combined circumstances. But when I now add up the cases I still have a member of the A-arm, witch has a force of 9000N that pushes on it. Is this reasonable? This would than be the absolute max.

I also have my doubts about the was I calculate the brake force back into the wishbones. Because like I do it now, the arms of the wishbone would all be connected by hinges, but they aren't, they are mounted on a solid V-shaped end, where the rod-end will be pressed in.

And about the 4g acceleration in the z axis, do I use the weight of the car times 4g. And then divide this force by 4 (so I have a force per wheel?)

[onemaniac]

Just for estimation I guess you can manually draw a friction circle (based on 1.5G/2G guess) and solve for the resultant force that gives you the most stress on the members.

and for the 4g Az use the unsprung mass.

[Kevin Hayward]

I would recommend setting this up as a 3-d statics problem. It really doesn't take too much effort and Excel can solve the matrix quite readily.

Lets you easily look at the differences between push rods / pull rods, steering tie rod forces etc. and is no more work than what you are already doing.

Kev

[Ceboe]

I've made a calculation with a kind of tire force ellipse. This gives me a max force of 7000N on the arm of the A-arm.

I think this is fairly reasonable.
I know you guys aren't going to give me any values so if you could say, sounds pretty normal to me, then I will be satisfied. This values is still almost twice of the strength our wishbones could handle last year, but then again, our geometry wasn't very optimal.

@Kevin Hayward: I don't really know what you mean by that? I have no idea how I would start a thing that you suggest. Maybe you could give me a link or some extra info in a PM or just in this topic?

[Z]

Ceboe,

Try this "Analysis of wishbones" (http://fsae.com/eve/forums/a/tpc/f/125607348/m/98420416151/p/1) thread.

Z

[wagemd]

This might sound harsher than I mean it to, but if you don't have tire data and you're not sure what your load cases are, why are you making carbon a-arms? It seems like the time (and money - ala "looking for funding of tire data") could be FAR better spent.

The weight savings are not huge and I have seen many of them fail on several other team's cars.

It's hard to argue with the weight (or toughness) of appropriately sized, thin walled, steel tube. What happens when you driver murders a wall of cones? How bad is it if you drop a corner?

IMO, figure out the full 3d truss, get the tire data, and put the carbon somewhere else...

Excel is a great tool, so is Matlab (and Octave - free if you cant afford Matlab, but uses almost identical syntax).

[Kevin Hayward]

Ceboe,

I didn't mean to be short on info. It is a fairly straightforward statics problem, even given there are more equations to solve than most textbook problems. Just about any of the standard engineering statics textbooks will be handy. This is generally taught in the first year of mechanical engineering.

When reduced to a statics problem in a typical double wishbone system you will have 6 unknown link forces (assume pure tension/compression). You will also have the forces and moments applied to the corner via the tyre contact patch. The locations and directions of all of these forces are known through the suspension points. From this it is a simple case of the sum of forces (x,y,z) and moments (x,y,z) being zero. You will have a few cross-products to solve for the link forces to get moments, but this is not anymore difficult than first year engineering problems, in this case it is more repetitive.

Make sure you put in some decent vertical loads as well to account for hitting bumps etc. I would also look at increased lateral and longitudinal loads for similar situations. It pays to be conservative with your suspension load calcs. A failure of a suspension system can be one of the most dangerous failures in a car, coupled with the fact that it can end up being very hard to repair. A FSAE track is well contained and doesn't have a lot of things to hit if things go wrong. Not all of our test venues are the same. Will your suspension handle hitting a pothole or a grate that no-one saw until it was too late?

Once you have a matlab or excel program sorted you can very easily see how the location of the suspension points affect the link loads.

This is part of the fundamental vehicle design process. If your team is unable or unwilling to do this sort of calculation of loads then you should steer well clear of carbon wishbones and go for something much more conservative.

Out of curiosity have you looked at quality control and fatigue of the glued joints in the carbon rods? If you want any confidence in the manufactured links at all you will need to do this testing.

Kev

p.s. the thread Z mentioned is gold

[Ceboe]

Originally posted by wagemd:
This might sound harsher than I mean it to, but if you don't have tire data and you're not sure what your load cases are, why are you making carbon a-arms? It seems like the time (and money - ala "looking for funding of tire data") could be FAR better spent.

The weight savings are not huge and I have seen many of them fail on several other team's cars.

It's hard to argue with the weight (or toughness) of appropriately sized, thin walled, steel tube. What happens when you driver murders a wall of cones? How bad is it if you drop a corner?

IMO, figure out the full 3d truss, get the tire data, and put the carbon somewhere else...

Excel is a great tool, so is Matlab (and Octave - free if you cant afford Matlab, but uses almost identical syntax).

We are getting our carbon tubes sponsored, so they don't cost us any money. Steel tubes on the other hand, we have to buy..

[Ceboe]

Originally posted by Kevin Hayward:
Ceboe,

I didn't mean to be short on info. It is a fairly straightforward statics problem, even given there are more equations to solve than most textbook problems. Just about any of the standard engineering statics textbooks will be handy. This is generally taught in the first year of mechanical engineering.

When reduced to a statics problem in a typical double wishbone system you will have 6 unknown link forces (assume pure tension/compression). You will also have the forces and moments applied to the corner via the tyre contact patch. The locations and directions of all of these forces are known through the suspension points. From this it is a simple case of the sum of forces (x,y,z) and moments (x,y,z) being zero. You will have a few cross-products to solve for the link forces to get moments, but this is not anymore difficult than first year engineering problems, in this case it is more repetitive.

Make sure you put in some decent vertical loads as well to account for hitting bumps etc. I would also look at increased lateral and longitudinal loads for similar situations. It pays to be conservative with your suspension load calcs. A failure of a suspension system can be one of the most dangerous failures in a car, coupled with the fact that it can end up being very hard to repair. A FSAE track is well contained and doesn't have a lot of things to hit if things go wrong. Not all of our test venues are the same. Will your suspension handle hitting a pothole or a grate that no-one saw until it was too late?

Once you have a matlab or excel program sorted you can very easily see how the location of the suspension points affect the link loads.

This is part of the fundamental vehicle design process. If your team is unable or unwilling to do this sort of calculation of loads then you should steer well clear of carbon wishbones and go for something much more conservative.

Out of curiosity have you looked at quality control and fatigue of the glued joints in the carbon rods? If you want any confidence in the manufactured links at all you will need to do this testing.

Kev

p.s. the thread Z mentioned is gold

Hi Kev, thanks for your response.
He also have 2 companies that will supply us with the glue and they our helping us the fabrication. We have done some testing already and indeed we can look at the quality of our glued joints, also at this company.

It's just that I'm the only person in the team doing this kind of calculation, the others are working on other stuff. I have already started with the excel and it is getting there but now I am checking all calculations again because I still get pretty high forces.

Also I don't know exactly for witch force I am aiming for, I think it should be around 6000N or something for the highest loads. This is just gut feeling.

[Kevin Hayward]

Originally posted by Ceboe:

We are getting our carbon tubes sponsored, so they don't cost us any money. Steel tubes on the other hand, we have to buy..

Ceboe,

Be wary of this approach. The carbon may not cost you money, but it will cost you resources. The time taken to develop the carbon tubes will be enough that you could have earnt the money (and plenty more) working at McDonalds to pay for the steel. Heck, put on a decent BBQ and you will pay for the steel and a few other parts. Alternatively you could have put the effort into finding sponsorship and got even more.

It is much better to trade money for time than the other way around.

Kev

[Markus]

I don't think carbon tubes are that dangerous approach but as stated you want to have some fatigue testing done and a hefty safety margin everywhere and especially in the bondline (>25).

Be aware that any metal to carbon contact will corrode and this might effect the service life of the a-arms.

You might also want to study how to spread the stressess in the bondline (insert design).

[Kevin Hayward]

Markus,

I don't think there is anything wrong with carbon tubes used in suspension when properly developed. Undersized steel tubes can be just as bad as poorly developed bonded ends. Although I would be using greater safety factors and a lot more testing on the carbon.

My last post was more about the dangers of assuming something is free just because the materials have been provided.

Kev

[Markus]

I see your point and you're very right on this.