analysis of wishbones

[A]

hello guys,
i tried calculating forces on wishbone...guys correct me if im wrong..
i considered breaking+cornering situation..and drew d free body diagram..so all friction force at contact patch will be taken by lower wishbone and in front plane i applied force and moment equilibrium NEGLECTING force on tie rod link ( in front)..because that would introduce 4th unknown...am i correct???

[Dash]

Well you have to consider what type of suspension you are using ( pushrod, pullrod, direct actuation, pull damper direct actuation, leaf springs, etc) to know where the force is going to be reacted at. Also, I would thing about what degrees of freedom each link has in your system. If you have any scrub or pneumatic/mechanical trail I wouldn't neglect the tie rod. It might seem harder than it actually is. You just need to balance it in the front and side view at the same time.

[Joe_Little]

O.. If this is what you wanted to know about Adams for, it can happen quicker than I described in my other post.

If you can get the program up and running in time, email me at joseph.little[@]mscsoftware.com and I can help you churn out some results real quick. Won't take long to do a static analysis if you have hard points and spring rates ready to go. Dynamic would only take a bit longer if you have mass and damping info.

The plus side is later, with the same suspension, you could be doing a full vehicle brake lock up in a turn with TTC tires and some scaled up road frictions and look at peak loads...

[A]

we are using push rod suspension
Originally posted by Dash:
Well you have to consider what type of suspension you are using ( pushrod, pullrod, direct actuation, pull damper direct actuation, leaf springs, etc) to know where the force is going to be reacted at. Also, I would thing about what degrees of freedom each link has in your system. If you have any scrub or pneumatic/mechanical trail I wouldn't neglect the tie rod. It might seem harder than it actually is. You just need to balance it in the front and side view at the same time.

[Nick Renold]

You need to balance the forces and moments acting on the upright.

You have six suspension members (four a-arm parts, steering/toe link, and pushrod), therefore six unknowns (the tensions in these six members). Force balance in x,y,z and moment balance in x,y,z for the upright gives you six equations to solve for. Solve the equations so that they balance the force and moment induced in the upright by the force from the contact patch.

Instead of solving all these equations by hand, use your matrix math and solve the equations together using the AX=B form like you use in solving trusses.

For a more in depth explanation of the matrix math method read on:

matrix A is a 6x6 matrix where each COLUMN is composed of the x,y,z components of the unit vector representing its direction, and the x,y,z location of its attachment to the upright in that order.

vector X is a COLUMN vector consisting of the tension in each suspension member

vector B is a COLUMN vector consisting of the x,y,z components of the force at the contact patch center, and the x,y,z moments generated by the contact patch force around your origin (so that would be (r CROSS F) plus any tire self-aligning moment and overturning moments if you wish.

Now you have a handy little equation Ax=B. Left-multiply both sides by A inverse and you have just solved for all the tensions in your suspension members.


BONUS ROUND: program this into MATLAB and iterate through your expected range of forces (friction circle) to get your maximum tensions and compressions for each member.


I hope this helps many of the teams who may start out by doing crazy things like trying to solve for the moment around the steering axis to get their steering link tensions and going from there... designing suspension can make you crazy sometimes. Yes, I am guilty of this foolishness, but now I know better and so do you.

[CamMoreRon]

So.. sorry to have to dig up this thread but I've been trying to figure out how to use matrices to resolve forces and this was the most sensible thing I could find.

I'm a little confused here if I'm honest.. as I've gone through your step-by-step guide to the matrix maths and got my unit vector components and all that lovely stuff. I've used them to populate the A Matrix along with my upright attachments as you described; I've also added my force and moment components into the B Vector as you described.

The problem is that when I work out the AX=B formula in Excel (MMULT(MINVERSE(MatrixA),VectorB) I get some unrealistic numbers for my tensions - a fairly significant magnitude above what I would expect to see.

My Unit Vector maths seems to be sound as all the components add up to 1, and I've kept the same units throughout for my hardpoints (mm), so I'm struggling to find an error. The only thing I can see from writing out the formulae and matrices on paper is that there doesn't seem to be a link between the contact patch where the forces are applied and the upright points where they're reacted. Maybe I've missed something in your method that links the forces and moments at the CP into the upright.. I'm not sure.

The only place I can think to find the problem is in the B Vector, as maybe it's not as simple as just plugging in my CP forces. Say I had a 2kN braking force (and my axes were +ve x = longitudinal centreline from front to rear; +ve y = lateral from driver's left to right; +ve z = vertical from ground up) would this be input to the B Vector as: Fx=2000, Fy=0, Fz=0, Mx=0, My=0, Mz=1500 where my CP coordinate is (0,750,0)? Or is this where my error creeps in?

[Zac]

two tips:

first, start with a FBD and manually write out the equations if you haven't done so already. solving the equations the long way once to use as a check might not be a bad idea either

second, make sure that your tire forces and suspension links are represented in the same coordinated frame.

[CamMoreRon]

Zac, thanks for the response.

I've drawn the FBD and written out the equations for the sums of forces and moments and I'm happy with the force balance side of things, the problem is I'm struggling to see how to make sense of the moment balance in this method.

The thing that makes me uneasy is that if I move my origin I get different tension values, whereas logic would predict that the tensions should all stay the same no matter where your point of reference is. Since the force balance is solved with unit vectors (which don't change when the origin moves) this makes me think that there is something wrong with the way I'm calculating the moments.

Unfortunately matrices, unit vectors and r cross F were not part of my course, being only a humble BSc. While I'm starting to get my head around it, my minimal understanding is definitely putting me at a disadvantage. Nick is clearly a guy who understands this stuff in great detail, so I think maybe a key point has been lost in translation somehow. Don't get me wrong, I'm not looking to be handed the answer on a plate - where would the fun in that be? But I'd appreciate a little help in identifying where I'm getting lost so I can find my way to the correct answers.

[Zac]

The tensions should stay the same no matter where you place the origin, but only if you have the forces and moments properly represented in that coordinate system. Think about what needs to change if you sum forces and moments about the wheel center vs. the center of the contact patch.

[CamMoreRon]

At the moment I'm just solving with a force in the x-direction at the contact patch centre to represent braking, just to get to the point where I trust the numbers are sensible. I'm applying forces and therefore calculating moments from the contact patch centre as that seemed the right way to go about it. I'm also using my global coordinate system that I've used for CAD and the suspension design just for continuity, rather than a local system on the upright. I'm calculating my moments about the origin of the GCS.

[CamMoreRon]

Still struggling with this.. I'm really trying to find the logic behind how the moments are calculated here and every time I just can't see how what's been said above relates to my FBD.

To calculate moments as above, the tensions are multiplied by the x, y, and z dimensions of each attachment point. So to consider the lower wishbone outer ball joint, say it is x = 10mm, y = 650mm and z = 150mm, expanding the matrices gives me:

(0.01 x T1) + (1.01 x T2) and so on for x moments
(0.65 x T1) + (0.65 x T2) and so on for y moments
(0.15 x T1) + (0.15 x T2) and so on for z moments

If that's correct, then that means my problem is in the way I've calculated my x, y and z moments caused by the CP force, and this wouldn't be surprising due to all the confusion it's caused me! Is it going to totally screw things up if my contact patch is at x = 0mm, y = 750mm and z = 0mm? Should I be offsetting my origin so that it's below ground and I end up with non-zero values for the x and z moments?

[Z]

Originally posted by CamMoreRon:
Still struggling with this.. I'm really trying to find the logic behind how the moments are calculated...
Cam,

The easiest way to resolve this is if you post all your equations here, in fully written out matrix form, and using realistic numbers. Give some explanation for any symbols you use (eg. T1, etc.).

(BTW, you are allowed to post as much rubbish as you like on this forum, and they don't charge you anything. I know. I do it a lot... http://fsae.com/groupee_common/emoticons/icon_biggrin.gif)

At the moment I'm guessing your problem might be that you are using Newtons and millimetres, but quoting moments in Newton.metres?

Or else you are not calculating the moments correctly (ie. Mx = Fy x Rz (with correct sign +/- !), etc.)

Z

[CamMoreRon]

Sorry for the silence, I've been away from this for a little while! Ok so here's an example using some numbers I dreamt up, dimensions are in millimetres (sorry USA!) and my axes are: +ve x runs front to back down centreline, +ve y runs from driver's left to right and z is vertically up.

Point x y z
Lower WB Front Pivot -100 350 130
Lower WB Rear Pivot 250 350 150
Lower WB Outer Ball Joint -12 685 150
Upper WB Front Pivot -80 450 362
Upper WB Rear Pivot 200 450 362
Upper WB Outer Ball Joint 12 660 410
Push Rod Wishbone End -5 645 180
Push Rod Rocker End -5 485 430
Outer Track Rod Ball Joint -120 668 200
Inner Track Rod Ball Joint -125 370 176
Tyre Contact Patch Centre 0 750 0

So using the above and finding unit vectors my 6x6 matrix (MatrixA) becomes:

[-0.254, 0.616, -0.393, 0.658, -0.017, 0.000]
[-0.966, -0.788, -0.897, -0.734, -0.997, -0.539]
[-0.058, 0.000, -0.205, -0.168, -0.080, 0.842]
[-0.012, -0.012, 0.012, 0.012, -0.120, -0.005]
[ 0.685, 0.685, 0.660, 0.660, 0.668, 0.645]
[ 0.150, 0.150, 0.410, 0.410, 0.200, 0.180]

Based on the idea that each column should be: x, y and z components of unit vector of the link arm, followed by x, y and z coordinates of it's attachment point to the upright. The wishbones are A-arms with one outer balljoint each, hence 1&2 and 3&4 sharing attachment points. I've also converted the attachment point dimensions to metres.

Then if I apply a 1000N force with components x=1000N, y=0N, z=0N to represent braking, my VectorB becomes:

[1000]
[ 0]
[ 0]
[ 0] (based on 1000N force x 0m)
[ 750] (based on 1000N force x 0.75m)
[ 0] (based on 1000N force x 0m)

And then using the following Excel witchcraft: {=MMULT(MINVERSE(MatrixA),VectorB)}

VectorX i.e. the solutions for tension in the various suspension links becomes:

[-23655]
[ 26241]
[ 19881]
[-20321]
[ -267]
[ -858]

Which seems ridiculous to me, and hopefully does to you too!

Thanks for the help!

[CamMoreRon]

Right.. sorry about the mess above! I had it formatted so nicely until the forum removed all the spaces! http://fsae.com/groupee_common/emoticons/icon_rolleyes.gif

[Zac]

The formatting is fine. Pumping your numbers into matlab I get the same result so we can hopefully eliminate a simple math mistake.

The issue I see is how you've setup your B vector. Two things stick out. First, you're not reacting the braking torque into the upright properly. Does the torque come straight through the tire and into the upright? or does it get passed through several parts first? I assume your wheels roll correct?

Second, you seem to be embracing the American car trend known as "Donks." Google image search that and then look at your numbers again.

[CamMoreRon]

I thought that would be where the error creeps in!

I'm calculating all moments about the origin (which is on the front axle line and in the centre of the car at ground level) rather than about the upright - is that where I'm going wrong? Do the moments need to be resolved around the wheel centre instead? And if so, should I be using the wheel centre as the origin in this system?

The way I had read into this was that all moments about the origin should equal zero, so moments caused by reactions at the upright attachment points and moments caused by the force at the contact patch should all add up to zero.

Also, ignore the fact that the points are high up, I based them on a Lotus Shark default for a saloon car!

[Z]

Originally posted by Nick Renold:
... vector B is a COLUMN vector consisting of the x,y,z components of the force at the contact patch center, and the x,y,z moments generated by the contact patch force around your origin (so that would be (r CROSS F) ...

Originally posted by CamMoreRon:
Then if I apply a 1000N force with components x=1000N, y=0N, z=0N to represent braking, my VectorB becomes:

[1000]
[ 0]
[ 0]
[ 0] (based on 1000N force x 0m)
[ 750] (based on 1000N force x 0.75m)
[ 0] (based on 1000N force x 0m)
Cam,

I have only had a very brief look at your numbers, but the above stands out...

Your origin is at ground level, under the car centreline, and under front axle line (yes?). You have a rearward braking force of Fx=1000N at the right wheelprint (x=0, y=0.75, z=0).

Therefore,
Mx = Fz.Ry - Fy.Rz = 0x0.75 - 0x0 = 0,
My = Fx.Rz - Fz.Rx = 1000x0 - 0x0 = 0,
Mz = Fy.Rx - Fx.Ry = 0x0 - 1000x0.75 = -750Nm!!!!!!!!!!!! http://fsae.com/groupee_common/emoticons/icon_smile.gif

I have written these moments out in full (hopefully without typos!) to help you understand how they are calculated (you might also do a sketch to help make sense of it).

Your above vector B has a rearward Fx=1000N force acting on a LoA through (0,0,0.75), namely 0.75m above the origin, or just above the nose centreline.

As I said, I haven't checked any other numbers, but try the above fix....

Z

[CamMoreRon]

Z, thanks for the reply! While it hasn't got me to a solution I'm happy with it has at least done a lot to clarify RxF!

Even with the above fix, I still get differing results if I move my origin about in x or z, so there must still be something fundamentally wrong somewhere. The problem I can see, and this may just be through misunderstanding, is that this method seems to be treating the moments created by the contact patch force (i.e. those in Vector B) and moments created by the reaction points (i.e. those in Matrix A) differently. Contact patch moments are calculated with RxF as per Z's post above, but reaction point moments are calculated by multiplying the non-directional resultant by an x, y or z distance - surely this is going to give an incorrect moment?

Working out the equations for moments from Matrix A and Vector B gives:

T1, T2, T3 etc are the tensions in each member
A, B, C etc are the reaction points, so Ax is T1's attachment point x-location, By is T2's attachment point y-location etc etc

Mx = [(T1*Ax) + (T2*Bx) + (T3*Cx) + (T4*Dx) + (T5*Ex) + (T6*Fx)]
My = [(T1*Ay) + (T2*By) + (T3*Cy) + (T4*Dy) + (T5*Ey) + (T6*Fy)]
Mz = [(T1*Az) + (T2*Bz) + (T3*Cz) + (T4*Dz) + (T5*Ez) + (T6*Fz)]

So you're basically multiplying a force by a moment arm is isn't perpendicular to, and that seems totally wrong to me!

[SNasello]

Cam,

The sum of moments is not just the Force times the application point. You need to take the cross product, of the direction vector of the force (unit vector) and the moment arm. The cross product will give you the correct resultant moment, perpendicular to the force.

In the equation you wrote for the moments:

Mx = [(T1*Ax) + (T2*Bx) + (T3*Cx) + (T4*Dx) + (T5*Ex) + (T6*Fx)]
My = [(T1*Ay) + (T2*By) + (T3*Cy) + (T4*Dy) + (T5*Ey) + (T6*Fy)]
Mz = [(T1*Az) + (T2*Bz) + (T3*Cz) + (T4*Dz) + (T5*Ez) + (T6*Fz)]

Is T1, T2... a Force magnitude which is multiplied by the coordinate Ax, Bx... or a cross product of the Tension VECTOR (magnitude*unit direction vector of link) with the moment arm vector (essentially your point coordinate if you are summing moments about the origin).

[CamMoreRon]

Ok this has now completely blown my mind. http://fsae.com/groupee_common/emoticons/icon_eek.gif

It had crossed my mind that maybe I needed to use the unit vectors to find the force that was perpendicular to the moment arm, but working out how to apply it was a bit above my understanding.

Sorry to have to beat this horse to death, but I really hate to abandon something I'm struggling with!

[SNasello]

Cam,

don't abandon it, your almost there!

http://en.wikipedia.org/wiki/Cross_product

Basically the tension in the link, the moment arm, and the resultant moment are all vectors. They have magnitude and direction. The moment is a result of the cross product of the force with the moment arm vector.

The error is in how you built up your A Matrix.

[Z]

Originally posted by CamMoreRon:
Sorry to have to beat this horse to death, but I really hate to abandon something I'm struggling with!
Cam,

Perserverance is a good thing. Stick with it! http://fsae.com/groupee_common/emoticons/icon_smile.gif

I have had a closer look at the matrix equations, and, as I suspected, they are too good to be true. Namely, too simple, and hence wrong. (Too much "cogitatio caeca" = "thinking blind", which I will have to rant about sometime...)

The top three rows of AX=B, which sum forces in X, Y, and Z directions, are ok. The bottom three rows, which sum the moments, are the problem. As suggested by Stefan, the entries in A have to be fixed to reflect the "cross" (= "vector") product nature of moments. Incidentally, the link in Stefan's post above is one of the better examples of a Wiki page.
~~~o0o~~~

Anyway, here is how I would write out row 4 of the equations in full (sort of). Note I use "F1" for "force" in link 1 (save "Ts" for torques/couples), "dx1" for x-component of unit direction vector in link 1 (ie. = entry at A(1,1)), "Rx1" for x-coordinate of outer-end of link 1 (from R for radius), "*" is usual arithmetic multiply, and so on...

[dz1*Ry1 - dy1*Rz1]*F1 + A(4,2)*F2 + A(4,3)*F3 + A(4,4)*F4 + A(4,5)*F5 + [dz6*Ry6 - dy6*Rz6]*F6 = Mx.

So, entry of A at row 4, column 1 = A(4,1) = [dz1*Ry1 - dy1*Rz1], and similarly across the row.

As before, the RHS, namely B(4), is,
Mx = Fz.wp*Ry.wp - Fy.wp*Rz.wp,
with ".wp" = the force/position at the wheelprint.

You can do the entries of rows 5 and 6 of A similarly to above (with reference to my previous post for My and Mz).
~~~o0o~~~

Hope I didn't get too many typos in there... http://fsae.com/groupee_common/emoticons/icon_smile.gif

Let us know if it works...

Z

[CamMoreRon]

Gentlemen, thank you so much! http://fsae.com/groupee_common/emoticons/icon_biggrin.gif

Using cross products for the values in rows 4-6 of Matrix A has solved all my problems. I am now getting sensible answers and most importantly they stay the same no matter where I put my origin!

I'll admit I was relying on the method described initially a little too much, and I think that was definitely confusing things as it made me think I could run before I could walk, but all is good now. Gears have crunched into mesh and I definitely understand this a lot better, so I can't thank you enough for the perseverance.

[Z]

Cam,

Good to see that your perseverance has paid off! http://fsae.com/groupee_common/emoticons/icon_smile.gif (And btw thanks for the spell check... http://fsae.com/groupee_common/emoticons/icon_wink.gif)
~~~o0o~~~

In case anyone still doesn't understand the alphabet soup of the "cross product" equations, here is a brief translation.

"Mx = Fz.Ry - Fy.Rz" simply means that the moment about the longitudinal X-axis (Mx) is caused by the vertical component of force (Fz) acting along a LoA offset in the lateral direction (Ry), together with the lateral component of force (Fy) acting along a LoA offset in the vertical direction Rz. So if X is on the LHS of the equation, then only Ys and Zs on the RHS. Similarly for My and Mz.

The only tricky bit is working out the signs (+/-) on the RHS. I draw a small 3-D sketch and use the "Right Hand Rule" - right thumb points in +ve direction of axis, and if right fingers curl in same direction as the force, then +ve, else -ve.

Z

[Ceboe]

Hi, I was guided here by the other topic.

I did everything that is written here and I understand why. But my forces are still pretty high, in region of 20000N, witch is impossible.

I did the A-matrix as said above, my first three rows are like the A-matrix on the previous page. Then I calculated all the cross products. But I think there was a problem here.

For example for the A[4,1] I did:

[dZ1*Ry1 - dy1*Rz1]
witch would be concrete dZ1 = -0.058.
The value in A[1;3]. Then Ry1 I took the value of the Lower WB Outer Ball Joint witch is at -12 685 150. So Ry1 = 685 but I divided this value by 1000 to have the same units everywhere.

This gives me for A[4,1] = 0.105
Thus: (-0.058*0.685)-(-0.966*0.150)= 0.150

To expand this for the fifth row I did the following: [dx1*Rz1 - dz1*Rx1]

Witch gives the following: [(-0.254*0.150)-(-0.058*0.012)]
gives -0.039

I Think this is the correct way of doing it? No?

But when I have a 1000N of force in the X direction I yield a force of more than 16000N in the rear lower A-arm

The thing I might consider to be wrong is the R..1 value. This should maybe be the difference between the pivot point and the outer ball joint?

[SNasello]

Ceboe, why don't you post the equations that you used to derive your A matrix. Then we might be able to help better determine where you went wrong. The magnitude of your result looks similar to what was posted earlier on the previous page.

[Ceboe]

SO I used the points of the topic starter, so I can check the things I did along the way. I am making a pull-rod suspension, but that should not matter, I would just put in my points and the outcome should be correct.

So:

X Y Z
1 Lower WB front pivot -100 350 130
2 Lower WB rear pivot 250 350 150
3 Lower WB outer ball joint -12 685 150
4 Upper WB front pivot -80 450 362
5 Upper WB rear pivot 200 450 362
6 Upper WB outer ball joint 12 660 410

7 Pull Rod wishbone end -5 645 180
8 Pull Rod rocker end -5 485 430

9 Outer Track Rod Ball Joint -120 668 200
10 Inner Track Rod Ball joint -125 370 176

11 Tire Contact Patch 0 750 0

Are the points I used

This gives me three unit vectors:

X Y Z
3to1 -0,254 -0,966 -0,058
3to2 0,616 -0,788 0,000

6to4 -0,393 -0,897 -0,205
6to5 0,658 -0,734 -0,168

7to8 0,000 -0,539 0,842


9to10 -0,017 -0,997 -0,080

This seems to be correct, considering the answers given above.
So this gives me following matrix A. Where the first 3 rows are filled with X Y Z, X for row 1, Y for row 2 and Z for row 3. And then every column representing a member.
3to1, with this I mean, force from point 3 to point 1, so the front lower A-arm.

http://i50.tinypic.com/28jxngp.jpg

This is the matrix I become.
So I think the problem is in the last 3 rows.
The 0.105 of row 1 column 1, I became by solving [dz1*Ry1 - dy1*Rz1] so [(-0.058*685/1000)-(-0.966*150/1000)] witch gives 0.105
so dz1 is the value of the unit vector witch is on the third row at column 1, so -0.058. Ry1 is the Y value of the outer ball joint of the lower A-arm so 685, but this I divide by 1000 so everything is in meters. I do the same thing for dy1, witch is then -0.966 (the Y value of the unit vector) and 150/1000 with 150 being the Z value of the outer ball joint of the lower A-arm.

For the fifth row I used [dx1*Rz1 - dz1*Rx1] with dx1 = -0.254, Rz1 = 150/1000, dz1 = -0.058 and Rx1 = -12/1000, this gives a value of -0.039
And the sixt row [dy1*Rx1 - dx1*Ry1] with dy1 = -0.966, Rx1 = -12/1000, dx1 = -0.254 and Ry1 = 685/1000, witch gives a value of 0.185

I did the same thing for every column. So for example for the last column I did following:
For row 1: [dz6*Ry6 - dy6*Rz6], [(0.842*645/1000)-(-0.539*180/1000)] gives 0.640
where Ry6 is the Y value of the point where the push rod is attached to the wishbone, so the point closest to the wheel. Just like I took the point of the outer ball joint at the A arms.
But I think here might be a mistake?

[SNasello]

Ceboe,

I think you have the same problem as the previous poster. The way that you have built this up is using the coordinate of the mounting point of the link to the upright, which is essentially the moment arm about the origin.

The question is, what does the equation for your Force/moment input vector look like?

[Ceboe]

Originally posted by SNasello:
Ceboe,

I think you have the same problem as the previous poster. The way that you have built this up is using the coordinate of the mounting point of the link to the upright, which is essentially the moment arm about the origin.

The question is, what does the equation for your Force/moment input vector look like?

With the force/moment input vector, I think you mean the values of matrix B. I made this that I can just fill in the 3 forces on the tire contact patch. From these forces Mx, My, Mz will be filled in. So Mx = Fz*dz - Fy*dy where Fz and Fy are the forces on the third and second row. And dz is are the points of the contact patch. So here dz = 0 and dy = 750. So if I fill in Fx = 1000 I get a moment of -750 for Mz. Same way, a force of 1000N in the Z direction gives me an Mx of 750

But when I fill 1000N in for Fx, Fy and Fz. Then I will have a moment of 750Nm for Mx, a moment of -750Nm of Mz. My is zero.

[Ceboe]

I think I found my error. I rechecked all my values and found a couple wrong in my excel file. Some values of the upper wishbone were linked to the outer point of the lower wishbone...

Now I get some values that I think could be true.

When I enter 1000N for Fx, Fy and Fz I get following vector B:

x 1000
y -1000
z 1000
moment x 750
moment y 0
moment z -750

This then gives me following values for vector X:

Front Lower Wishbone 846 N
Rear Lower Wishbone 3058 N
Front Upper Wishbone 871 N
Rear Upper Wishbone -578 N
Steering Rod -3155 N
Pull Rod 1041 N


where a minus is pressure and a positive force is a pulling force.

Can somebody check this please?
I think this could be correct!

But a strange thing, the rear upper wishbone has a pressure force, but this is under braking and in a corner, a situation where I don't think there could be a pressure force on that member..

[Ray19]

I saw some faulty values in the last three rows of your matrix, good thing you noticed that. But, i am getting different values from the ones you are getting. Did you verify by shifting your origin and checking if the tensions remain the same?

[Ceboe]

Originally posted by Ray19:
I saw some faulty values in the last three rows of your matrix, good thing you noticed that. But, i am getting different values from the ones you are getting. Did you verify by shifting your origin and checking if the tensions remain the same?

I have again found still another error. Making Matrices at 2 in the morning and nobody to check them is'n helping very much. I have checked moving the origin by adding and subtracting a certain value at every point in x,y,z and the forces stay the same, so I guess that's ok.

So my Matrix A looks like this now and I get the following forces.

Matrix A:

http://i49.tinypic.com/30ccg7r.jpg

Vector B:

Force

x 1000
y -1000
z 1000
moment x 750
moment y 0
moment z -750

Vector X:

-707 N
2319 N
410 N
-693 N
-568 N
1046 N

[Ray19]

Yep, those are the values I am getting as well.

[Ceboe]

And so a force with minus in front of it is a pulling force on the member, a positive force pushes on the member

[CamMoreRon]

I've plugged your numbers into my Excel calculator and I get different values in the last 3 rows for several of the members. I think you might have a small error somewhere in the way you've done your RxF, as a few of them are correct but others aren't; so it might be worth going over all of them and checking you haven't got a couple of values jumbled round.

Edit: Sorry I hadn't read to the end of the topic, lazy me! I get the same answers throughout as you now, so I think (hope) you've cracked it! http://fsae.com/groupee_common/emoticons/icon_smile.gif

[Z]

Originally posted by Ceboe:
... Making Matrices at 2 in the morning and nobody to check them is'n helping very much. ...
Ceboe,

Oh, but it is!

Your efforts have greatly helped your understanding of this subject. Your examples above will also help many other FSAEers struggling with this same problem. Congratulations on your (and Cam's, etc.) perseverance! http://fsae.com/groupee_common/emoticons/icon_smile.gif
~~~o0o~~~

Now, the next thing you must do is confront your teachers and ask for your school fees back, since it seems they have been negligent in their duties. If you don't pay fees, then ask for a percentage of their wages (maybe start the negotiations at, oh, say, ~50%?). At the very least, demand that they contribute a significant sum of money towards your FSAE team, in exchange for their lack of teaching of the fundamentals of engineering!!! http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

[eilonoz]

what does your inputs represents?
the reaction at the cp or translation of it to the wheel center?

[CamMoreRon]

Originally posted by Z:
Now, the next thing you must do is confront your teachers and ask for your school fees back, since it seems they have been negligent in their duties. If you don't pay fees, then ask for a percentage of their wages (maybe start the negotiations at, oh, say, ~50%?). At the very least, demand that they contribute a significant sum of money towards your FSAE team, in exchange for their lack of teaching of the fundamentals of engineering!!! http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

Honestly, the sheer lack of fundamental content we got in our Engineering degree here in the UK never ceases to frustrate me. But ho-hum.. the Internet is the largest and greatest free university, right? Well.. once you filter out all the crap, anyway.

[Phil1988]

Hey everybody,

got some questions to you allhttp://fsae.com/groupee_common/emoticons/icon_smile.gif
I've read the Postthree times.. But i still have problems to understand this genius thing.

So far i did understand:

I have a koordinate system (0,0,0). And i have my suspension points (x,y,z). And i have forces acting on the tire contact path (x,y,z).
But all is 3D... I have problems getting the equations. the free body diagram was no problem...

is it that easy:

?Fx=0= Fx1 +Fx2+Fx3 +Fx4+ Fx5+ Fx6
?Fy=0= Fy1+ Fy2...
?Fz=0= Fz1+ Fz2...

The moments are something like a mystery^^...
I have only suspension points (x,y,z) so to get the lever arm, i need to get a Vector...
so uppright Point A - Chassis Point B= Vector BA

But how to get that in the equations for the moments? I understand the cross product to get the perpendicular lever arm.

?My=0=...

i think i have something like a undertanding problem. Hope you can help me understanding. Only taking the excel sheet would't be honourable..

regards phil

[GSpeedR]

Originally posted by Phil1988:
Hey everybody,

got some questions to you allhttp://fsae.com/groupee_common/emoticons/icon_smile.gif
I've read the Postthree times.. But i still have problems to understand this genius thing.

So far i did understand:

I have a koordinate system (0,0,0). And i have my suspension points (x,y,z). And i have forces acting on the tire contact path (x,y,z).
But all is 3D... I have problems getting the equations. the free body diagram was no problem...

is it that easy:

?Fx=0= Fx1 +Fx2+Fx3 +Fx4+ Fx5+ Fx6
?Fy=0= Fy1+ Fy2...
?Fz=0= Fz1+ Fz2...

The moments are something like a mystery^^...
I have only suspension points (x,y,z) so to get the lever arm, i need to get a Vector...
so uppright Point A - Chassis Point B= Vector BA

But how to get that in the equations for the moments? I understand the cross product to get the perpendicular lever arm.

?My=0=...

i think i have something like a undertanding problem. Hope you can help me understanding. Only taking the excel sheet would't be honourable..

regards phil

Phil, keep in mind that this Ax=B is simply a quick way to solve for your force magnitudes. You can always write out your system of equations from your FBD and solve them, which is what each row of Ax=B represents (1 row is 1 equation). I think some users are following Nick Renold's post on the first page explicitly and getting incorrect answers; the moment equations are not set up correctly as there is no mention of a cross product...so I would ignore it.

Each column of the A matrix (6x6) represents 1 of the 6 arms. The first 3 rows (1-3) represent the vector components of the arm unit vector. The last 3 rows (4-6) represent the cross product R X F of that arm.

For example,
F1 = F1*[dx1 dy1 dz1] => Force magnitude times its unit vector
R1 = [Rx1 Ry1 Rz1] => Position of arm1 connection on upright

So A(1:3,1) = [dx1; dy1; dz1]
and A(4:6,1) = [Rx1; Ry1; Rz1] X [dx1; dy1; dz1] ... Z has given you the cross product in a post above

You've got the forces right, the moments are also simple once you've set up your vectors correctly. Remember that R1 is the position vector of the arm1 connection on the upright and [dx1; dy1; dz1] is a unit vector of arm1.

Write out the matrix equation before solving...you'll notice that you have:

?Fx
?Fy
?Fz
?Mx
?My
?Mz

[Phil1988]

Thanks Gspeed for your help,

got close to the solution http://fsae.com/groupee_common/emoticons/icon_wink.gif

now i think the only problem i still have is at the beginning. Why do we need the unit vector?
All our suspension points are dimensioned from the origin (0,0,0)... I have looked the whole day to get this^^... i know how to get the unit vector but why do we need it?

Regards Phil

[M. Nader]

Given how this thread has greatly assisted me in understanding Vectors like in no way that i had understood them in my college time i will try to answer you in a simple way by explaining my process using FBD to get suspension forces.

It is true this thread is GOLD.

For everything i explain X is longitudinal, Y is lateral and Z is vertical.

The method to start is using FBD and solve it for ?Fx,?Fy,?Fz,?Mx,?My,?Mz. like you would start solving any static problem the first thing you would do with inclined angles or forces is to neutralize them to the x, y and z axis using trigonometry (sin, cos and tan) to have things perpendicular and easy to use. However using this method here isn't easy as it will cause confusion since you will solve the same wishbone in the Y-Z plane (giving you Y and Z components) then going to the X-Z plane you will have X and Z components, now you have 2 Z components!, but are these to be added or subtracted, and what about the other plane? what if your A-arms change angles will the sin and cos change? what if you want to do it dynamically with A arms angle changing constantly?

Vectors and matrices make these a lot easier!

To answer your question directly the unit vector represents how the force will be distributed along the 3 axis, so it is a resolution of the force. so by multiplying the magnitude of the force by the i component in your unit vector you get the X axis component of that force, multiply by the j and you get the y component and by the k to get the z component. you know how to get the directional vector and how to get the unit vector so i won't go into that.

now draw things in a FBD and have all the forces on the main axis; x y and z. (that means that you will multiply each force by its unit vector components). now solve that FBD using ?Fx,?Fy,?Fz,?Mx,?My,?Mz. The only problem you will now face is with the moments calculations as you have to know what the + force in each axis will cause you (+ve or -ve moment) and what the perpendicular multiplication distance is (coordinates of activation point, i.e upright mount - coordinates of origin).

Make it all symbolic so for example F1x is F1 (force in link 1) x unit vector component i. and for moments things will be (F1x)*(R1y); R1y will be the perpendicular y distance between point of action of force F1x (upright) and origin causing a moment about the z axis.


Model things easily/flexibly in Excel or Matlab and you can do all sorts of experiments you want.

Don't take Z's cross products and use them as you won't know what his directions are and force direction assumptions (although he assumes all tension). they are correct and using them will get you correct results (but you won't know tension or compression) but i recommend you do your own FBD on 2 pieces of paper and you will understand much more.


I hope this helps.

[M. Nader]

I would also like to ask something just to confirm with a dynamic simulation program i wrote.

is it normal to be seeing 10KN tension force in one of the lower wishbones in a 250 kg Formula student car cornering at about 2Gs (assuming full frontal load of 100kg on just 1 tyre due to weight transfer)? from experience and checking with Excel i think this is an OK value but would just like to make sure if others are getting similar values.

[eilonoz]

Question
if i calculate the rear sus-links forces when the car accelerates, my input need to include tire contact point reaction and the axle torque as well??

[M. Nader]

Originally posted by eilonoz:
Question
if i calculate the rear sus-links forces when the car accelerates, my input need to include tire contact point reaction and the axle torque as well??

Tyre reaction to both weight and longitudinal force causing acceleration (opposite braking direction).

[Z]

Originally posted by eilonoz:
Question
if i calculate the rear sus-links forces when the car accelerates, my input need to include tire contact point reaction and the axle torque as well??
eilonoz,

Yes, at the rear you must ALSO include the effect of driveshaft torque. There are two main options.

1. You consider the upright AND wheel as one body in the FBD.
So, the RHS "B" column vector has the forward thrust (Fx, from-road-to-wheel) acting through the ground level wheelprint centre (together with any vertical Fz and cornering Fy forces), and it ALSO has the driveshaft torque (Ty, from-diff-to-wheel, and preferably called a "couple" http://fsae.com/groupee_common/emoticons/icon_smile.gif) added to the My entry. If the driveshaft is not exactly lateral, then you should also add its other components, Tx and Tz, to the Mx and Mz entries of B. (Note the Ms also include the effects of the cross-products of the various forces, as explained before.)

2. You consider ONLY the upright in the FBD.
Now you can omit Ty by raising the forward thrust Fx upward so it acts through the upright hub-centre (ie. through the hub-bearings). In effect, Fx-through-hub-centre = Fx-through-wheelprint + Ty (with Ty acting anywhere, because couples are "free" vectors). Note this option doesn't account for a non-lateral driveshaft, so you should correct for this by adding more forces at the hub-bearings.

For inboard (diff mounted) rear brakes you solve as above (but with rearward Fx and oppositely directed Ty).

For outboard rear brakes you solve as with front wheel.

Z

[M. Nader]

@Z

at the front should we also add the brakes as a force at the contact patch and a torque? i have been in a heated discussion about this

[Z]

Originally posted by M. Nader:
at the front should we also add the brakes as a force at the contact patch and a torque? i have been in a heated discussion about this
M. Nader,

Again, you can approach this two ways (and assuming outboard front brake, as is usual).

1. Consider the upright+caliper+disc+wheel+tyre as a SINGLE body in the FBD.
This body has six forces acting on it from the suspension links (4 from wishbones + toe-link + spring-damper). These are the LHS of the AX=B "static equilibrium" equation. For equilibrium the body also has a force from-road-to-wheelprint (contact patch) acting on it (strictly speaking, a road-to-wheelprint "force screw", or "wrench"). This is the RHS of the equation, and is broken up into x,y,z force components and "moments" (see note below) about the chosen reference axes.

For front suspension analysis this is probably the easiest approach.
~o0o~

2. Consider only the upright in the FBD.
Again, six forces from the suspension links act on the upright and give the LHS of equation. For the RHS of equation you have to think about how the wheel itself exerts forces on the upright. For example, various wheel-to-upright forces act through the two roller bearings, and frictional forces from the brake-disc (connected to the wheel, so part of it) act against the caliper pads, which in turn push against the upright.

This approach is necessary if you want to do structural analysis (eg. FEA) of the upright. For example, a top or front-mounted caliper increases the loads through the bearings, while a bottom or rear-mounted caliper reduces these loads, so is generally better (ie. bearings last longer).

In a simpler approach to this 2nd case, and considering the braking forces only, you might have a rearward Fx wheel-to-upright force acting through the bearings, together with a wheel-to-upright braking torque or "couple" (which you might picture as a couple of calipers on opposite sides of the disc, exerting equal and oppositely directed forces). This braking-couple is given by,
Rearward-Fx-force-acting-at-ground-level = Rearward-Fx-force-acting-at-hub-level + Braking-couple-Ty.

Both approaches (1. = force-at-wheelprint, and 2. = force-at-wheel-hub + braking-couple), if done correctly, should give exactly the same forces in the six suspension links.
~o0o~

When learning Statics it is useful to practice moving forces away from their Lines-of-Action and then finding the couple that must be added to the picture to compensate for this move. Note again that forces are sliding vectors. They can slide anywhere along their LoAs, but MUST NOT be moved sideways from their LoAs without some compensation.

[Mini-Rant http://fsae.com/groupee_common/emoticons/icon_smile.gif] Also worth noting:
"Forces" and "Couples" are "ACTIONS", namely pushes, pulls, or twists, acting on a body.
"Moments" are the "EFFECTS" of forces and couples, as calculated (or measured) at some particular point.
"Torques" usually refer to couples acting along some kind of shaft.
Moments are most definitely NOT THE SAME as couples or torques.
Unfortunately, nobody bothers defining these things these days, so be careful...
To exist, a moment requires only a single force.
To exist, a couple requires at least two (simple linear) forces.
You cannot create a couple with a single linear force!
The moment of a given force is always the same no matter where on the force's LoA the force is, hence forces are "sliding vectors".
The moment of a given couple is always the same no matter where in space the couple is, hence couples are "free vectors".
All this is best explained in olden day books. From memory, Den Hartog's book on mechanics (= "Mechanics"?) is a good place to start.
The algebraic "cogitatio caeca" (= "thinking blind") approach to teaching Statics has all of you moving forces around willy-nilly, refering to couples as moments, believing Matlab will solve all your problems, etc. and etc., so I guess it is not surprising that it all gets a bit confusing.
Probably the best way to learn Statics is to FIRST MEMORIZE the Axioms of Statics (very simple, and in Hartog), then just lots of practice with pictorial FBDs.
[Rant Over]

Z

[M. Nader]

Yes, i understand the difference between torques and moments and i solved the front brake situation exactly as how you said (wheel and upright as one part).

But then that is (i think) the same situation as rear drive, as both the brake torque (outboard) and drive (inboard) act at the same place (the hub). so why in case 1 (Wheel and upright as 1 part) we solve for the brakes at the contact patch and for the drive we add its torque as well? i think both forces originally started at their respective places as torques and transfer to the contact patch as a force (to either move or stop the vehicle)


I guess my question is why does it differ if the force is originated inboard or outboard since both are reacted at the wheel and hub bearings? and do only 1 action at the contact patch?

[SNasello]

The difference is, that the brake torque is reacted at the UPRIGHT (for outboard brakes) and the drive torque is reacted throught the drive axle. When calculating the forces in the wishbones, what is important is the forces acting on the upright.

[Z]

M. Nader,

Firstly, this would be much easier with some pictures, or even better with a blackboard and a lot of hand waving and redrawing of vectors as they are moved about. Nevertheless, here it is in words again...

"I guess my question is why does it differ if the force is originated inboard or outboard since both are reacted at the wheel and hub bearings? ..."

1. Consider OUTBOARD brake.
Now think of [upright+caliper+disc+hub+bearings+wheel+tyre] all as a SINGLE body in the FBD. Importantly, you MUST NOT think about any forces that are internal to this body, because they will only confuse you. So DO NOT think about the brake forces "originating" at the brake-disc/caliper, or at the hub, etc. It is very important to note that they are NOT part of this picture!!! The only place that any "brake force" originates is at the wheelprint (= tyre contact patch).

So now there are six forces acting from the suspension links onto this big, composite "body" (these being the "unknowns" in this particular thread, and are the LHS of AX = B), and one ground level force at the wheelprint (possibly with x,y,z components, these being the "knowns" and RHS of AX = B)
~o0o~

2. Consider INBOARD brake.
This time think of the upright ONLY in the FBD. Can any torque/couple coming from the inboard brake, and travelling through the driveshaft, exert a torque/couple on the upright? NO!!! (<- This is important to realize.) The braking torque from the driveshaft acts only on the "wheel" body, and this is an entirely different body to the one we are dealing with in our FBD. The body in our FBD (ie. the "upright") only has forces acting on it from the suspension links and from the wheel bearings (not considering aero or inertial forces, etc.).

These wheel bearing forces can only be axial (from "cornering" Fy force, at hub height) or radial in the Fx or Fz directions. Since there are two bearings offset in the Y direction, then different radial loads at each bearing can exert Tx and Tz couples on the upright. But NO Ty couple (~braking torque) can act on the upright, from the driveshaft/axle/wheel, through the bearings.

Bottom line is that any rearward Fx "braking" force acts on the upright, from the axle, through the bearings, at wheel hub height. There is NO BRAKING TORQUE acting on the upright.
~o0o~

Perhaps also re-read my earlier posts above that give some different ways of looking at it (eg. inboard drive acting on a different FBD, which includes the wheel). Again, this is much easier to explain in person, with sketches and a lot of "No, no, no... Not that body, I mean this body...".

If anyone already has some sketches, or can word it better thanks to their previous such discussions, then please do. http://fsae.com/groupee_common/emoticons/icon_smile.gif

Z

[SJF]

Hi newbie here. I've been playing around with the numbers in this thread and decided to do a 2-D analysis in the front and side view to check with the numbers here. I'm using the same numbers as used by Cam and same boundary conditions. My FBD's are posted below (Let me know if these are correct, please.) http://imageshack.us/a/img547/8770/sideviewfbd.jpg http://imageshack.us/a/img266/6189/frontviewfbd.jpg

The equations I have for the side view are:

Sum forces X = Fx-Fax-Fbx=0
Sum forces Z = Fz-Fbz-Faz=0
Sum moments @ contact patch = Fax(410) -Faz(-12) +Fbx(150) + Fbz(12)=0

Front view equations are:

Sum forces in Y = Fd+ Fdcos(theta) - Fa=0
Sum forces in Z = Fz+ (-Fdsin(theta))=0
Sum moments @ contact patch = -Fd(L1) -Fd(cos(theta))(L1) - Fa(L2) =0

[M. Nader]

I would advice against using trigonometry here as directions will cause you a problem, + difficult to visualize effects in 3d.

Unit vectors are the way to go.

[SJF]

Originally posted by M. Nader:
I would advice against using trigonometry here as directions will cause you a problem, + difficult to visualize effects in 3d.

Unit vectors are the way to go.

My plan is to move to a 3D analysis as soon as I get my head wrapped around the 2D effects.

With that said, i'm using the following values and I want to check my numbers with someone else. This is in front view mind you.
X-Component Y-component Z-Component
Contact Patch
1.5621 M 0.5969 M 0.0000 M

Wheel Center
1.5621 M 0.5969 M 0.2515 M

Lower Control Arm Outer Joint
1.5621 M 0.5563 M 0.1207 M

Upper Control Arm Outer Joint
1.5461 M 0.5502 M 0.3747 M

Tie Rod Outer Joint
1.4994 M 0.5194 M 0.1448 M

Pushrod/Pull-rod Outer
1.5240 M 0.4826 M 0.3302 M


My forces at the contact patch are:
Fx = 188 N
Fy = 4140 N
Fz = 2179 N

My reactions are:

Upper balljoint (point A on FBD) = -2861.91 N in Y direction
Lower balljoint (point B on FBD) = 3771.42 N in Y direction

Pull-rod (point D on FBD) = 3896.68 N It is also 34 deg. from the upper a-arm

Do these values seem reasonable? Bear in mind I did this in 2D front view as that's easier for me to visualize and I don't quite have my head wrapped around the 3D method yet.

Should there be a Z component at the upper and lower balljoints? It would make my solution statically indeterminate if there were. It's also unclear to me how to resolve the force at the tie rod/steering arm.

I'd appreciate any help/input and sorry if these are noob questions in advance.

EDIT: Oops! I realize my FBD posted above is a push rod suspension.... Pretend it is a pullrod, please.

[Nesh]

Hi there,

I have a problem with inputing matrices AX=B;
Matrix A > 6x6, columns are unit vectors and upright attachments (attachments are where vectors interesect in upright; for example - lower wishbone consists of two components which are vectors, and so on. Am I right?)
Matrix X > 1x6, blank
Matrix B > 1x6, xyz of tyre contact patch components (projecting a force (vector) to xyz) and moment generated by force at origin point.
When input these, my matrix X does not show appropriate values.
What should i do?

Nebojsa,
University of Belgrade

[EvanF]

Originally posted by Nesh:
Hi there,

I have a problem with inputing matrices AX=B;
Matrix A > 6x6, columns are unit vectors and upright attachments (attachments are where vectors interesect in upright; for example - lower wishbone consists of two components which are vectors, and so on. Am I right?)
Matrix X > 1x6, blank
Matrix B > 1x6, xyz of tyre contact patch components (projecting a force (vector) to xyz) and moment generated by force at origin point.
When input these, my matrix X does not show appropriate values.
What should i do?

Nebojsa,
University of Belgrade

I am by no means an expert, but here's a quick reply.

(I recently stumbled on this forum as I work a project that involves these calculations).

Your matrices are of the correct size, however for your A Matrix the bottom three of the column are the cross product of unit vector with the moment arm (as mentioned previously in this thread). Recall that the top three rows are the summation of the forces (X,Y,Z) and the bottom three are summation of the moments (X,Y,Z) so the bottom three rows are very similar to the top three rows with the exception of the moment arm addition.

I think that's your issue as everything else looks O.K. depending on the numbers.



My question (not related to yours) for anyone is, the moment arm for the member forces is defined as what?

The center of the tire contact patch is my reference. So is it the vector from the outboard connection of a member to the patch?

Ex: X,Y,Z of Push Rod - X,Y,Z of Center Tire Patch?

Can you go directly from the tire patch to the members or...? A bit confused here.

[SJF]

Okay, fairly simple question at the expense of looking like a fool...

It seems everyone is taking the unit vector from the a-arms, push-rod and tie rod....

Yet, When I set my 3D analysis up, I did my unit vector from the contact patch to the upper and lower balljoints as well as the tie rod location.

Is it wrong if I do it that way?

[SNasello]

@SJF: The unit vector represents the direction of the resultant force you are trying to find. The contact patch (for the purpose of this analysis) is rigidly attached to the upright. YOu are looking for the forces acting on your suspension arms, from a given force applied at the contact patch. Does it make sense to you the way you set up your analysis? Draw a FBD again!


@EvanF: The moment arm is the distance from the applied force to a given fixed point. If you are applying forces and moments at the contact patch, then it makes sense to use this point.

Tip: If both of your upper wishbone members meet at the same ball joint (and same for lower) you can reduce the number of moment arms you need to calculate by 2.

[SJF]

Originally posted by SNasello:
@SJF: The unit vector represents the direction of the resultant force you are trying to find. The contact patch (for the purpose of this analysis) is rigidly attached to the upright. YOu are looking for the forces acting on your suspension arms, from a given force applied at the contact patch. Does it make sense to you the way you set up your analysis? Draw a FBD again!


@EvanF: The moment arm is the distance from the applied force to a given fixed point. If you are applying forces and moments at the contact patch, then it makes sense to use this point.

Tip: If both of your upper wishbone members meet at the same ball joint (and same for lower) you can reduce the number of moment arms you need to calculate by 2.

Thanks, I appreciate the response. After I posted that I looked at it again and thought, if I take out the wishbone and pushroad/pullrod, then I have nothing reacting the forces at the contact patch. So it is making more sense.

Aside from complaining about my statics course, I don't remember much about unit vectors if we touched on them at all. http://fsae.com/groupee_common/emoticons/icon_frown.gif

SJF

[SJF]

So i've worked out my A matrix and solved for the forces.... One thing that doesn't seem quite right with my A matrix is (comparing to Ceboe's values) that some are opposite sign to the others

lower A-arm Front
-0.253644439
-0.965578264
-0.057646464
0.105348912
-0.038738423
-0.18533338

lower A-arm Rear
0.616054625
-0.787703433
0
-0.118155515
-0.092408194
0.412544977

Front upper A-arm
-0.392760914
-0.896519478
-0.204918738
-0.232326619
0.15857295
-0.24846397

Rear upper A-arm
0.657535615
-0.734481272
-0.167881434
-0.190335575
-0.271604179
0.442787281

tie/steering rod
0.016722034
0.996633223
0.080265763
0.145709115
-0.012976298
0.130766305

Push/pull Rod
0
-0.539053696
0.842271401
-0.640294719
-0.004211357
-0.002695268


As you can see some of the signs are wrong but the numbers are correct. Which is leading me to the wrong tension values when I solve my X matrix. Any ideas??

Edit: Sorry about the formatting...

[Nesh]

Can anyone please send me an excel file? I cannot do it...

[Nesh]

I made something that seems quite normal;
Will someone please look at my results?
Car is static and there is no camber. Later I will do braking, acc, roll,...


<pre class="ip-ubbcode-code-pre"> http://i41.tinypic.com/2hrfkhe.jpg </pre>

Thanks.

[edd_16]

Seems this thread has been lying dormant for a while anyways i will try it no matter what
I am also having some problems with the calculations
For the matrix AX=B what i am in doubt is the matrix B
The moment calculations :- say my tyre contact patch is 660mm away from the car centre line now if have a force in x-direction (braking) at the wheel contact patch
hence
Mx = Fz.Ry - Fy.Rz = 0x0.75 - 0x0 = 0,
My = Fx.Rz - Fz.Rx = 1000x0 - 0x0 = 0,
Mz = Fy.Rx - Fx.Ry = 0x0 - 1000x0.66 = -660Nm!!!!!!!!!!!!
Is this the correct approach??
Should i take all my distances for moments from the tyre contact patch only?
How to take the distances when i consider lateral or vertical forces?
How will they act???
I have assumed x= longitudnal (+ve if going from frnt to rear), y=lateral (positive when to from car centerling to tire), Z= vertical(+ if upwards)?????

[Milkman]

Edd,

I put your numbers in my excel and also got -660Nm.

-Looks like the correct approach.
-Moments need two points: The point that you are taking the moment at (in your example, the middle of the front axle, on the ground) and the point where the force is applied. For your applied forces in your B matrix (which you have above), this point is the contact patch (So yes, the distance is from your first point to the tyre contact patch). However, when taking moments in your A matrix, instead of the contact patch you will need to use the point where the link connects to the upright.
-you will be able to use the same approach for lateral and normal applied forces. Assuming these are acting at the same point (contact patch), the distances will be the same.

[letthewookiewin7]

Hey everyone,

Let me start by saying that this thread has been incredibly useful and eye opening. Thank you to everyone that has contributed to this topic, as it has helped me understand what's going on so much better than any other single resource has been able to thus far.

That being said, I have a few questions regarding the FBD. What I have gathered so far is that we are defining the "body" as the tire, wheel, hub, and upright. I have defined my upright to end at the lower and upper control arm mounting points, namely the lower and upper ball joints. We are treating this as one single (call it rigid for now) piece, and are not concerned with any INTERNAL forces, only those acting EXTERNALLY on the previously defined body.

In order to solve for the forces in the individual members, we must locate where the members attach/act on the upright, as this is one part necessary in defining the unit vector. When considering the pushrod ends, one is connected to the bellcrank, and the other is connected to the lower control arm (at least in our previous design it was). Both ends are connected via rod ends, thus making them two force members. Now here is my confusion: if we are neglecting internal forces, but define the system as above (terminating at the ball joints), finding the lower control arm mounting points is easy; it would be the coordinates of the lower ball joint/spherical bearing. However, when locating the point at which the pushrod force acts on the upright, we must consider that it connects to the lower control arm, and not directly to the upright. Basically, what I am confused with is how the pushrod acts on the upright if the force must first travel through the lower control arm. The only way I see around this is to assume that the pushrod attaches at the same point that the lower control arm does, or redefine the upright to include the portion between the pushrod and ball joint. Neither solution sits right with me, but I may be looking at it wrong. I am only making this a big deal because I think the discrepancy will cause some errors when solving the system dynamically, or when I later try to apply this to a full size car where the shock mounts are a much greater distance inwards on the lower control arm.

Sorry in advance if this has been covered in a previous post, but I'll be honest I did not read the entire thread.

Thanks for your help!!

[Z]

See post above, and also,


Originally posted by Nick (...thewookie...) in a PM:
My confusion is coming from accurately drawing the FBD of the tire, wheel, hub and upright in order to determine the forces in the control arms, pushrods, and steering links. I am having trouble visualizing/figuring out how the force from the pushrod acts on the upright...

Nick,

To restate your problem, you are wondering about a pushrod (or, better yet, a direct-acting spring-damper :)) that connects to the lower-wishbone somewhere in its middle. For example, it might connect to the lower-wishbone half-way between chassis-Ball-Joints and upright-BJ, like on many production cars. This puts the lower-wishbone structure into bending, messes up some of the force directions, and thus makes you doubt whether the solution method of this thread will work.
~o0o~

Firstly, IMO there is nothing wrong with having wishbones "in bending" like this. I would probably do something like that, for a whole lot of reasons that I don't have time to cover here. BUT (!), the wishbones MUST be designed to cope with the bending!!! I note that a great many FSAE cars have wishbones loaded in bending (a little bit), but they are also quite obviously NOT designed to take those loads. Which might explain a lot of those suspension-failure DNFs...

Secondly, the "wishbones in bending" means that you CANNOT BLINDLY USE the method of this thread. Well, not without a little extra tweaking. Fortunately, this is quite simple, and, in fact, you just have to use the method twice.

The method of this thread is essentially just about finding the solution to the "static equilibrium" of a 3-D system of forces. There are six KNOWN components of force at a KNOWN wheelprint position (= the RHS = "B"), and these must "statically equilibriate" (ie. be "equal and opposite to") the six UNKNOWN forces of the suspension members (= the LHS = "AX").

In the normal use of this method, you know the positions of the six Lines-of-Action of these latter six suspension forces, because these forces are ASSUMED to act along the centrelines of the six known "ball-ended-links" that make up the 2 x wishbones + 1 x toe-link + 1 x pushrod. Normally this is a reasonable assumption, although friction in the BJs makes this a little bit untrue (but easy enough to calculate the error...). So the final bit of matrix magic is simply finding the six UNKNOWN magnitudes (= "X") of these six forces with KNOWN LoAs.

To restate your problem again, it is that because of "wishbone in bending" you DO NOT KNOW THE DIRECTION of the SINGLE 3-D force that must act through the lower-wishbone's outer-BJ (ie. the force from-Wishbone-Outer-BJ -> to-upright).

Now, you might try solving this problem by breaking this single 3-D force up into 3 components. Two of the components might act from the two Wishbone-Inner-BJs (WIBJs) to the WOBJ, in a similar way to the standard method.

But now you are left with finding the direction of the roughly vertical "pushrod" component of force. This quite clearly CANNOT be along the centreline of the pushrod, since this centreline does not intersect the WOBJ.
~o0o~

So, what to do?

I would "divide and conquer". Split the one difficult problem into two easier problems. First solve for equilibrium of the upright (+ wheel...). Then later solve for equilibrium of the lower-wishbone.

1. UPRIGHT - Start by pretending that you KNOW the centrelines of three ball-ended-links that connect to the upright-lower-BJ. Conveniently, you can just assume these links are aligned with the X, Y, and Z axes. So these three "virtual" suspension links have direction cosines of (1,0,0), (0,1,0), and (0,0,1). (Wow, you don't even need a calculator to do this stuff! :)).

Now solve your AX = B equations as before. You get two force-magnitudes at the upright-upper-BJ (ie. the two tensions/compressions in the two tubes that make up the upper-wishbone), and another force-magnitude in the toe-link. At the upright-lower-BJ you get three force-magnitudes, conveniently already in the X, Y, and Z directions. You can vectorially sum these into a single 3-D "upright-lower-BJ force", or just leave them as is.

2. LOWER-WISHBONE - You now solve for the "3-D equilibrium of forces" acting on this "lower-wishbone free-body", in pretty much the same way as above.

Here you have the KNOWN forces (= RHS = "B") acting on the WOBJ, which are taken directly from the output of step 1 above (since, yipee!, they are already in X, Y, and Z form), but with opposite signs. The KNOWN "couples" acting at the WOBJ might be taken as zero, or you can add any frictional amount you want.

You also have a KNOWN pushrod LoA, and two KNOWN points that all the other forces have to act through, namely the two Wishbone-Inner-BJs.

Importantly, there is a slight hiccup here, because the fact that you have two WIBJs, both of which can sustain forces in three orthogonal directions, means that you have an OVERCONSTRAINED linkage!!! In practice, you discover this overconstraint when you try to assemble a stiff wishbone onto a stiff chassis when the two WIBJ-separation-distances (ie. on wishbone and chassis) are slightly different.

There are several solutions to the above overconstraint problem in practice. One solution is to let one WIBJ carry three components of force, while the other WIBJ is arranged so that it can only carry two components of force (ie. its "resultant" force must lie in a plane that does NOT contain the first WIBJ). Put simply, the wishbone is built so that it is flexible in the direction between both WIBJs. Other solutions are also possible...

Anyway, you can now solve the lower-wishbone FBD by having one WIBJ with three known LoAs (say along X,Y,Z), the other WIBJ with only two known LoAs (say Y and Z), and finally the known LoA of the pushrod force. The six unknown forces (ie. all with KNOWN LoAs but UNKNOWN magnitudes, and = LHS = "AX") must equilibriate with the the six known forces at the WOBJ (= RHS = "B").

With the FBD now solved, you just have to make sure that your lower-wishbone is adequately sized to take the bending stresses!

Z

[Tim.Wright]

Importantly, there is a slight hiccup here, because the fact that you have two WIBJs, both of which can sustain forces in three orthogonal directions, means that you have an OVERCONSTRAINED linkage!!! In practice, you discover this overconstraint when you try to assemble a stiff wishbone onto a stiff chassis when the two WIBJ-separation-distances (ie. on wishbone and chassis) are slightly different.


In my opinion this is a critical point which is NEVER highlighted in the synthesis/analysis of suspension. Though in my opinion overconstrined isn't the right word - I would say its indeterminant. Its able to move (hence not overconstrained) but the system forces cannot be calculated to a single correct solution - hence its indeterminant (or indeterminate?). In fact, some computational kinematic solvers will throw an error if you try to model an A-arm with 2x spherical joints instead of a single hinge joint on the inboard side.

I stumbled across this earlier this year making an excel based wireframe suspension model of 5 link suspension. I was wondering what is the mathematical difference between an A-arm and two tension/compression links which share a common outer ball joint. the answer is that the A-arm can't be solved while the tension/compression links can. Therefore what we do with A-arms is assume that they act like 2 pin-pin links and go from there.

In my opinion this is why you need to be careful when you tell people "just do a free body diagram" because if they are smart enough they will come back to you and say its not possible. You can't solve for the forces in a wisbone which includes a spring attachment without applying a fudge.

Its the same "fudge" which unapologetically appears in the static equalibrium formulas that people use to find the CG location in plan view. If you sit down and dow a proper no BS free body diagram of the car and it's wheel forces you will see its actually not possible to calculate the CG location because you have only 3 equations but 4 unknowns. You need to fudge something (i.e. assme the CG is on the centreline) to reduce the number of unknowns.

Nothing wrong with a fudge. But I wish they were pointed out at the time. Its another thing I had to unlearn and subsequently re-teach myself and I'm still slightly bitter about that...

See you at FS-Italy.

[Nick944]

See post above, and also,

Importantly, there is a slight hiccup here, because the fact that you have two WIBJs, both of which can sustain forces in three orthogonal directions, means that you have an OVERCONSTRAINED linkage!!! In practice, you discover this overconstraint when you try to assemble a stiff wishbone onto a stiff chassis when the two WIBJ-separation-distances (ie. on wishbone and chassis) are slightly different.

There are several solutions to the above overconstraint problem in practice. One solution is to let one WIBJ carry three components of force, while the other WIBJ is arranged so that it can only carry two components of force (ie. its "resultant" force must lie in a plane that does NOT contain the first WIBJ). Put simply, the wishbone is built so that it is flexible in the direction between both WIBJs. Other solutions are also possible...

Anyway, you can now solve the lower-wishbone FBD by having one WIBJ with three known LoAs (say along X,Y,Z), the other WIBJ with only two known LoAs (say Y and Z), and finally the known LoA of the pushrod force. The six unknown forces (ie. all with KNOWN LoAs but UNKNOWN magnitudes, and = LHS = "AX") must equilibriate with the the six known forces at the WOBJ (= RHS = "B").

Z

Z, I wonder if you could expand on this. Are you suggesting to arbitrarily remove one of the ball-joint reaction forces? That method never sat right with me because the joint can react a force in that direction. Is there a special way you are deciding which force component to remove? What is the most correct way to reduce this over-constrained problem?, I agree that it should be based on making the direction between the ball-joints flexible. I don't think there is a correct way, but that might just be me giving up too early.

The next problem that people face with this analysis is when you need to solve for two springs acting on the wishbone, for example a spring and ARB acting on the lower wishbone. You need to decide how to split the force between the two components. If the spring connects to another body other then the chassis then you need to include it in the analysis also.

[Z]

In my opinion this is a critical point which is NEVER highlighted in the synthesis/analysis of suspension. Though in my opinion overconstrined isn't the right word - I would say its indeterminant...

Tim,

My understanding of the etymology:

In the idealised, abstract, world of KINEMATICS there are NO forces. There are NO masses. Everything is perfectly smooth. So there is NO friction (which is just another force anyway). And everything is PERFECTLY, PERFECTLY, RIGID. (Ahhh.... so simple... :))

If such an idealised Kinematic linkage requires two separate numbers (say, linear dimensions, or angles...) to completely specify its configuration, then it is said to have 2 Degrees-of-Freedom. If only one number is required, then 1 DoF.

If, upon assembling the linkage, it has only the one possible, unmovable, configuration, then it has zero DoFs. Think of a well designed, pinned-together, "spaceframe".

But if, in this spaceframe, you now try to squeeze in another 1.0000000 metre long strut between two of its existing "nodes" that are already CONSTRAINED to be 0.999999999 metres apart, then the extra strut will never fit. Such a design (ie. one that includes the extra strut) is said to be "Kinematically OVERCONSTRAINED". The design can never be assembled in the idealised world of Kinematics, "in theory".
~o0o~

On the other hand, in the world of STATICS there are ONLY forces. The "body" in the FBD is incidental. It acts only as a reference frame to locate the individual forces.

Doing a "Statics analysis" of the above zero-DoF spaceframe is straightforward. For a given (appropriately chosen) set of external loads, the unique system of forces that correspond to each of the struts is easily DETERMINED. For example, solve N equations in N unknowns, as explained in this thread (or, easier, solve geometrically). But squeezing in that extra strut, and adding that extra unknown, makes the whole system "Statically INDETERMINATE".
~o0o~

So, the two terms mean pretty much the same thing, but make more sense when used in their own respective fields.

In the wider field of Structural Engineering the "indeterminate" problems are solvable, but now they require a knowledge of the Elastic Modulii of the various parts. That is, you need to know how much force is required to stretch or squash the many parts of the structure so that you can squeeze that extra strut in.

And in the even wider field of Practical Engineering, the solution is achieved with a hammer. The bigger the "overconstraint", or the bigger the "indeterminancy", then the bigger the hammer you must use!

Hammers can solve a lot of problems... :)
~~~~~o0o~~~~~


Originally posted by Nick944:
... Are you suggesting to arbitrarily remove one of the ball-joint reaction forces?
Is there a special way you are deciding which force component to remove?
What is the most correct way to reduce this over-constrained problem?...

Nick,

I would guess that maybe 99% of FSAE suspensions suffer from the above overconstraint problems. And most of them use something like the last sentence of above section to achieve a solution. Well, maybe not a hammer, but some mild "massaging" of the welded V-shaped wishbone so that the second chassis-BJ-bolt can be inserted. (This massaging is OK with steel wishbones, but less so with CF.) Or one of the chassis-BJ-clevises might have slotted holes that allows adjustment of any discrepancy between the various dimensions.

I don't think there is one UNIQUE "most correct way"of doing this. But, in general, I would suggest that if you do not want to use "the hammer" too much, then it is best to design the linkage so that it has the right amount of constraint to begin with. Two examples:

1. Design the wishbone so that it has one biggish chassis-BJ, which can carry forces in three directions (ie. X, Y, and Z). This BJ is positioned to take the largest loads. The other chassis-mount in then designed as a 4 DoF joint (ie. only 2 x constraints), so that it can only carry forces in two directions (say Y and Z). This might be a BJ that can also slide freely (EITHER wrt chassis, OR wrt wishbone, but not both) along the line joining both wishbone-chassis-mounts. The theoretical analysis then follows directly from the design.

2. Design the wishbone as two separate links. Both are connected to the chassis with their own 3 DoF BJ. These two links are then connected to each other, somewhere near the upright, with a 1 DoF "revolute" joint. In practice, this can be a vertical bolt that is loosened when fitting the wishbone to chassis (to allow easy insertion of BJ bolts), and then tightened up. One of the links is the "main-link", and connects to the upright at its outer end, and to the pushrod/SD somewhere in its middle. The other link might be called the "tie-rod".

This second arrangement was extremely common on the lower wishbones of production cars in the latter half of the 1900s, both on double-wishbones and strut-type suspensions. The main-link was usually lateral to the car, and close to parallel with the steering-link. The tie-rod usually went forward/inward from-wheel-to-chassis, and connected to the chassis at a big rubber bush under the bumper-bar, thus allowing some rearward wheel movement when hitting big bumps.

More recent production cars usually have a single L-shaped lower wishbone, with one chassis-BJ connection on the axle-line, and the other more rubbery connection further back near the firewall. This arrangement puts less "strong" chassis structure forward of the front-axle-line, thus allowing for more "crush-space" in the event of accidents.
~o0o~


The next problem that people face with this analysis is when you need to solve for two springs acting on the wishbone, for example a spring and ARB acting on the lower wishbone. You need to decide how to split the force between the two components...

This is again a "Statically indeterminate" problem, thus needing an "Elastic" approach to solving the problem.

You just need to know the two spring-rates (= ~Elastic Modulii), and the amount of deflection of each spring for that particular suspension position (which includes the adjacent wheel for the ARB). The force that each spring exerts is then easily obtained.

Z

[mech5496]

Z,

a legit solution would be teh one used for years by a lot of amateur garage builders: http://images.chassisshop.com/CACHE7FB2EBFB8420184DC04293598DAB0F66/400/300/approved/C73-442/C73-442_3.jpg
Another option would be a multi-link suspension, which may have some benefits in terms of manufacturing complexity (only rods are needed) and also gives some "interesting" solutions regarding KPI/scrub etc. with vitrual steering axis but requires many more rodends and mounting points on the upright, which IMO is too much hassle. 3rd option would be to use a simple jig on which the A-arms to be manufactured with their rodends in place, and use the same jig when welding/bolting the chassis mounts for minor corrections. It works great for us!

[Z]

Harry,

Yes, your link (below) is what I meant with Example 2 above.

http://images.chassisshop.com/CACHE7FB2EBFB8420184DC04293598DAB0F66/400/300/approved/C73-442/C73-442_3.jpg

The main-link (ie. the lower one in the image) could be designed for bending, and thus take the spring loads somewhere mid-way along its length. In this case the 1 DoF revolute joint (ie. the vertical bolt connecting both links) could also be mid-way along the main-link, allowing the outboard-BJ to be closer to the wheel-centreline, whilst also allowing greater steering angles.

Think about the forces on the tie-rod-link (ie. upper one in image), especially if the revolute joint is made taller and deliberately designed for low friction. All forces acting on this link must now lie IN THE PLANE that contains the revolute-axis and the inner-BJ. It should now be quite obvious how to pick two direction cosines for the unknown force components at the tie-rod's inner-BJ.
~o0o~

Getting back to the original problem of how to analyse a single, RIGID, wishbone that attaches to chassis at two inner-BJs.

Imagine you build such a wishbone as a very stiff, triangular, structure. You measure its stiffness between the two inner-BJs as 1 ton per millimetre deflection (along the line between both BJs). You also measure the stiffness between the chassis' welded-on BJ-clevises as the same 1 ton/mm. It turns out that because of the inevitable manufacturing tolerances (it was student built! :)), the spacing between BJs on the wishbone is 1 mm more than on the chassis.

It follows that to assemble the wishbone to chassis, you must first compress the distance between BJs on the wishbone by 0.5 mm, and also stretch the chassis by 0.5 mm, in order to fit the BJ bolts. Thus, upon assembly, the two inner BJs each have equal and opposite half-ton loads acting on them. And this is even before any wheel loads start to act!

So, as before, it is better to have more flexible V-shaped wishbones, or clevises with slotted holes, or some sort of rubbery bushes, or other such fudges for Kinematic cock-ups.

Z

[Charles Kaneb]

Z,

You could also make the final holes in the wishbone clevises using the chassis attachments as the jigs to take up that 1mm difference. Lest you think that's one-off hackwork, I know of several examples of this in motorcycle and automobile production...

-CPK

[letthewookiewin7]

See post above, and also,



Nick,

To restate your problem, you are wondering about a pushrod (or, better yet, a direct-acting spring-damper :)) that connects to the lower-wishbone somewhere in its middle. For example, it might connect to the lower-wishbone half-way between chassis-Ball-Joints and upright-BJ, like on many production cars. This puts the lower-wishbone structure into bending, messes up some of the force directions, and thus makes you doubt whether the solution method of this thread will work.
~o0o~

..........

Z

Z,

Thanks very much for your reply. I wanted to take a few days to chew over your answer to make sure I get it.

The idea of breaking it up into two problems seems pretty intuitive. I was happy to hear you suggest that because I had just begun heading down that path myself, but was curious if it was correct or not. However, this does lead me to ask another question. Looking at people's previous posts that include the ball joint and pushrod locations, it appears the problem of bending stresses is almost always evident (to some degree) given that the pushrod and Lower OBJ have different coordinates. So, this begs the question: how can this cookie cutter method EVER be applied to these designs? And, perhaps more importantly (at least for my understanding): how can the equations be derived from a FBD (like the one I previously defined, including the tire, wheel, hub, and upright ending at the ball joints) if the pushrod force is not actually acting on the LOBJ? Are the bending stresses negligible if the LOBJ and pushrod coordinates are very close together?

The main thing for me is that I keep coming back to not being able to balance the moment portion of the system if I don't know where/how the pushrod force acts on the upright/LOBJ. Even IF I am correct in assuming that the method used by everyone in this thread presents some inaccuracies, I would still like to use the same method as a starter just to make sure I am 1.) Deriving the same equations, and 2.) programming it correctly. I basically want to be able to compare apples to apples first, but without being able to derive the moment equations like everyone else in this thread I am lagging behind. Maybe I am just looking at this problem through a microscope, which has been my downfall in the past.

Thanks again for the help!

[Nick944]

Hey there, Im going to say the same thing Z did, but using different words, it might help.
If the pushrod does not act directly on the upright, then you don't need to consider it in your FBD for the upright at all. Just take the upright on its own as a rigid body and consider all the external forces that act upon it. Your FBD for the upright, if the pushrod does not act directly on it, should have 1 unknown force that acts at the lower ball joint, this force could have any magnitude or direction = 3 unknowns. There is an unknown force that acts at the upper ball joint (I make the assumption that this force has no component normal to the wishbone plane), or you can consider two seperate forces each acting at the upper ball joint and directed along the wishbone links = 2 unknowns. There is a final unknown force that acts where the tierod attaches to the upright, that I assume to act along the direction of the tierod = 1 unknown. 6 unknowns, 6 equations for static equilibrium, so you can solve. And you will notice that you did not need to consider the pushrod mounting location at all. When you analyse your lower wishbone though, the pushrod location will need to be taken into account.

The cookie cutter method is drawing the FBD for all of the bodies in the system and solving them. The calculations you see in this thread make some assumptions about the forces being applied in the system and the format of the system (how the links and joints are arranged and connected).

[letthewookiewin7]

Hey there, Im going to say the same thing Z did, but using different words, it might help.
If the pushrod does not act directly on the upright, then you don't need to consider it in your FBD for the upright at all. Just take the upright on its own as a rigid body and consider all the external forces that act upon it. Your FBD for the upright, if the pushrod does not act directly on it, should have 1 unknown force that acts at the lower ball joint, this force could have any magnitude or direction = 3 unknowns. There is an unknown force that acts at the upper ball joint (I make the assumption that this force has no component normal to the wishbone plane), or you can consider two seperate forces each acting at the upper ball joint and directed along the wishbone links = 2 unknowns. There is a final unknown force that acts where the tierod attaches to the upright, that I assume to act along the direction of the tierod = 1 unknown. 6 unknowns, 6 equations for static equilibrium, so you can solve. And you will notice that you did not need to consider the pushrod mounting location at all. When you analyse your lower wishbone though, the pushrod location will need to be taken into account.

The cookie cutter method is drawing the FBD for all of the bodies in the system and solving them. The calculations you see in this thread make some assumptions about the forces being applied in the system and the format of the system (how the links and joints are arranged and connected).

Thanks for your reply, Nick.

I'm not sure if my previous questions came across correctly. I am not having trouble digesting the method that Z presented. Separating it into two bodies and using the matrix method twice is fine. My questions are in regards to NOT using two FBD's, only one like previously discussed in everyone else's posts. I fully understand that in order to accurately solve the problem of a pushrod acting on the lower control arm, you would need to use Z's approach, or something similar.

What I am still not understanding is how, after 7 or so pages of people saying that they have generated answers, is everyone else dealing with the added complexity of the pushrod not connecting to the lower control arm outer ball joint (i.e., not acting directly on the upright). If every example prior to my post had coordinates of the lower OBJ and the outer pushrod/pullrod connection sharing the same coordinates, than it would be obvious that there are no additional steps necessary, because the pushrod would be acting directly on the upright. However, that is not the case. Every example that I have seen where someone has given there coordinates, there are separate points for the LOBJ and the outer pushrod connection. That's what led me to believe that the original assumptions I made held some water, i.e. there is a force acting on the lower control arm, and causes it to see bending stresses. Now, my question to everyone is: if you have a setup that utilizes a pushrod/pullrod that DOES NOT share a joint with the lower control arm/upright, how did you solve your SINGLE FBD???

[Nick944]

They are able to solve all the wishbone and pushrod forces with a single FBD by assuming that the pushrod is attached directly to the upright, even if it is not in reality. So their analysis is most true if the physical system actually does have a pushrod that mounts to the upright, it is less true if not. You are right that there analysis is not as correct as it could be, but it may be good enough.

[letthewookiewin7]

They are able to solve all the wishbone and pushrod forces with a single FBD by assuming that the pushrod is attached directly to the upright, even if it is not in reality. So their analysis is most true if the physical system actually does have a pushrod that mounts to the upright, it is less true if not. You are right that there analysis is not as correct as it could be, but it may be good enough.

Got it. Thanks very much for your help!

[letthewookiewin7]

Importantly, there is a slight hiccup here, because the fact that you have two WIBJs, both of which can sustain forces in three orthogonal directions, means that you have an OVERCONSTRAINED linkage!!! In practice, you discover this overconstraint when you try to assemble a stiff wishbone onto a stiff chassis when the two WIBJ-separation-distances (ie. on wishbone and chassis) are slightly different.

There are several solutions to the above overconstraint problem in practice. One solution is to let one WIBJ carry three components of force, while the other WIBJ is arranged so that it can only carry two components of force (ie. its "resultant" force must lie in a plane that does NOT contain the first WIBJ). Put simply, the wishbone is built so that it is flexible in the direction between both WIBJs. Other solutions are also possible...


This makes a lot of sense. Here's anther question I have: Would it be acceptable to consider only the magnitude of the force at the WIBJ? This would reduce the number of unknowns to 2 instead of 6.

I like Z's solutions to the over-constrained problem better (allowing one inner joint to carry 3D orthogonal loads, and the other to only carry 2D) because I think it allows a more complete analysis of the arm rather than fudging it; however, because I am stuck using ball joints for the inners (mounts are already machined from last years car, parts are being reused because of budgets constraints, etc.) how can I analyse the system more completely? My justification for using magnitudes and neglecting the direction is because the suspension moves very little, and most of the load acting on the ball joint will be axial (if designed correctly). If we choose our ball joint to perform within this range of magnitude, it seems like it would do fine.

The other thing I keep coming back to from the days of statics (truss problems) was being able to solve the more complex problems by drawing individual free body diagrams of individual bodies (i.e., breaking it up into smaller parts, like already discussed). More FBD's = more equations. Couldn't you do the same thing here?

One last thing: If the upper control arm has two IBJ, how come we don't look at that as 3 unknowns each? I think what I am trying to get at is can we define the system so that the IBJ becomes an internal part and only transmits the force (much like the WOBJ on the upper control arm), and having the system terminate at the pin/bolt. This way, we can find the force that acts through the control arm and ball joint, and orient our ball joint to it's maximum loading capabilities are along that same direction (i.e. radial).

Sorry if these are redundant questions, I am still wading my way through this problem.

[Viv]

"For example, a top or front-mounted caliper increases the loads through the bearings, while a bottom or rear-mounted caliper reduces these loads, so is generally better (ie. bearings last longer)."


Hi Z,

This is Viv from the University of Toronto.

I was wondering if you could explain your comment (from a while a ago in this thread) regarding bottom mounted calipers being better for bearings compared to top-mounted calipers. When I think about the forces for a top mounted caliper, I see a rearward force from tire contact patch and a forward force from caliper acting on upright. Whereas for the bottom mounted caliper both forces are acting rearwards. Is that not worse?

Thanks,

Viv

[Viv]

Never mind that question. After a bit more thought I realized that the rearward tire contact force is pushing the wheel bearings rearwards and the forward acting caliper force for top mounted caliper is pushing the upright forward (same as pushing the wheel bearing rearwards) and is thus worse. Sorry for the random post I guess.

[DougMilliken]

Caliper location can also contribute to steering torque, see RCVD bottom of page 720, top of page 721.

[Z]

Back on pages 7 and 8 of this thread (ie. about 6 months ago), there was discussion of how to do the wishbone analysis when the pushrod, or Direct-Acting Spring-Damper, acts about midway along the lower wishbone, rather than directly at, or very close to, the wishbone's outer-BJ.

The problem was resolved by doing TWO separate FBD analyses, both using the same matrix-method explained at the beginning of this thread. So a first FBD analysis is done of the upright/wheel, and the output of that is then used as the input of a second FBD of the lower-wishbone. However, there then arose the problem of "overcontraint" of the necessarily stiff and strong one-piece lower-wishbone.

Anyway, this topic has been raised again on another thread. I thought it better to answer the questions from that thread here, to keep it all in the one place.
~~~o0o~~~

Here is a picture of a ~1972 F1 Tyrrell-006 double-wishbone front-suspension with DASD. (Note that it has inboard brakes, but it is NOT 4WD. Just something they did back then.)

This is a reasonably good example of how to build a DASD-lower-wishbone so that it can take the bending loads. Briefly, it is a folded sheet-steel fabrication that has a deepish section directly under the DASD, and the structure tapers towards its ends.

http://trackthoughts.com/wp-content/uploads/2010/11/M0401751.jpg

Keep in mind that this car would have weighed close to one ton fully fueled, and would have been braking hard from 300 kph while hitting kerbs that are much harsher bumps than anything on an FS/FSAE track.
~o0o~

This next picture shows a top wishbone (coincidentally of an ~1970 F1 Tyrrell-004) that is made in such a way that it is NOT overconstrained, so it can always be easily fitted to the chassis regardless of manufacturing tolerances (these issues discussed a few pages back).

I HIGHLY RECOMMEND this type of layout for any FS/FSAE wishbone suspensions, regardless of whether you use pushrods, pullrods, or DASDs.

The benefits of this layout are that it allows the wishbone-outer-BJs to be deep inside the wheels, for good steering-geometry and lock at front, and lesser compliance-steer at rear, while also having a very wide base at the chassis end for lower loads on the chassis BJs and chassis structure itself. Such layouts can be used as BOTH TOP AND BOTTOM wishbones.

It is worth mentioning that this layout was also common in the 1950-60s, and is, in a similar 'y'-shaped carbonfibre form, still common on many current F1 cars. This is, of course, NO reason for you to copy it. But the reasons for using it by all those decades of racecars, as briefly pointed out above, are good reasons.

If the "main-arm" of this wishbone (ie. the folded sheet-steel fabrication at the left) was made with a deeper (in Z-direction) section where the "tubular-link" connects to it, then it would be strong enough in bending for vertical spring loads to be easily carried by this section, as in the first picture above.

http://www.britishracecar.com/JohnDimmer-Tyrrell-004/JohnDimmer-Tyrrell-004-CA.jpg

This layout also helps show how to do the zero-constraint FBD analysis of a wishbone that carries spring loads somewhere in its middle. The main-arm's outer-BJ (ie. that connects to the upright) has the "given" RHS forces (ie. column vector "B") acting on it. These are found as part of the output of the upright's FBD.

The six LoAs of the LHS "A" matrix are as follows. The main-arm's inner-BJ has three X,Y,Z forces acting on it from the chassis. The main-arm also has a single pushrod/DASD force acting on it along a roughly vertical LoA somewhere near its middle (similar to first pic above).

The last two forces are on two LoAs that go through the centre of the tubular-link's chassis-BJ, and then through the top and bottom of the vertical bolt (roughly in middle of pic) that connects tubular-link to main-arm. So you might imagine, or even physically build, this tubular-link as a very narrow "V", or "hairpin", with its two connections at top and bottom of vertical bolt.
~~~o0o~~~

Hope that makes "Analysis of Tricky Wishbones" a bit more understandable. :)

BTW, both above pics were simply cut-and-pasted from the first suitable ones I found after googling "images of racecar suspensions".

Z

[Ahmad Rezq]

Erik
Great thanks for your all explanations in the whole topic :) .

[dynatune]

In our FSAE sponsorship package the DUNATUNE SUSPENSION DESIGN MODULE contains the possibility of

1) creating loadcases for the vehicle
2) enter 3D loads at the contact patch and calculate corresponding link loads
3) if you have reliable numbers for your link stiffness you can see what the link deformations do to the suspension setting.

Cheers,
dynatune