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Thread: analysis of wishbones

  1. #1
    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???

  2. #2
    Senior Member
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    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.
    --Dash Robinson
    --Mississippi State University

  3. #3
    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...
    Joseph Little
    Chrysler Group LLC
    CAE/Vehicle Dynamics Tools Development
    Mississippi State Motorsports Alum 2008-2010
    Get Adams/Car, the FSAE Database, and Training Materials:
    www.mscsoftware.com/fsae

  4. #4
    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.

  5. #5
    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.
    _______________________
    "It doesn't get easier, you just go faster." - Greg Lemond

    Nick Renold
    Northwestern Formula Racing - Suspension Lead '11

  6. #6
    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?

  7. #7
    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.

  8. #8
    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.

  9. #9
    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.

  10. #10
    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.

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