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How to calc force required to steer an articulated vehicle?

  1. Apr 11, 2015 #1
    Ok so for a project of mine I need to calc. the force required to steer an articulated heavy on road vehicle and to overcome friction (so I need to calc frictional forces too). I understand I need to draw FBDs but it has been a while since i have done that and I have never worked with an articulated vehicle and it is very confusing for me.
    I have been told that this can be worked by using only one side (left or right) of the vehicle as it is a dual-cylinder setup and calculating moment about center pivot. Is this moment the turning force? Actually, on second thought it would not be as it is a torque but how do I use that to get the turning force?

    So what I know:
    - vehicle's front axle and rear axle weights (both are single-axle but with 4 tyres front and rear)
    -I know the hydraulic working pressure and area of the pistons and hence I have calculated the force required to extend and retract the cylinders (so the max. force that can be provided through the hydraulics to steer the vehicle)
    -max articulation angle either side of center line (unsure if important to know)
    -angle and length from center pivot to front cyl mount
    -length mount to mount of cylinder
    -length at 90 degrees from center pivot to cylinder
    -length from center pivot to rear cyl mount
    -offset angle as to which cylinder is mounted from the 90 degree position when steering is centered (value is 0.31 rad to the left of 90 degrees when looking at right hand side..so cylinders point inwards at front)

    What do I do now, how do I go about calulting the turning force and the friction? Any help would be appreciated sorry for the length of this..
     
  2. jcsd
  3. Apr 12, 2015 #2

    jack action

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    What will happen is the following: The midpoints of the front and rear axles will stay in place, the axles rotating about their respective midpoint as your articulate the vehicle.

    This means that the wheels are in pure rolling. Hence the only resistive force is rolling resistance. You can get some rolling resistance coefficients values here as well.

    Because - on the same axle - one wheel turn one way and the other, the other way, this will create a torque on your FBD. The other axle will create another torque, but in the other direction. So the input torque needs to be as high as the sum of these torques to initiate/maintain motion. Higher torque will create a greater rotational acceleration. So you will also have to define how fast you want this motion to happen to determine the turning force needed.
     
  4. Apr 12, 2015 #3
    OK thanks for that starting to understand it a bit better. Using that equation for rolling resistance I am getting a force of 32 N for front right and 48 N for rear right tyres...for a 20 tonne vehicle seems too little? I also found another equation which yields closer to 100 N... F=Crr * N/R where R is wheel radius...unsure which to use

    and with calculating the moment do i disregard the left side (as they would cancel)? and then only use half the axle lengths so moment about the centre of axle = roll resist. force *half axle length?? And then I just add this to that found for the other axle and this should give me my required torque? Or am i supposed to use these moments to then do another fbd to calculate the torque at the centre pivot and then add these together?

    Because I have been given a max torque provided by cylinders which gives around 40 000 Nm at centre pivot...seem like my torques from my calcs are too small
     
  5. Apr 12, 2015 #4
    actually come to think of it..since both moments on the same axle act at the center (thererfore spinning the axle in same direction) they would cause the moment to double??so calculate the moment on that axle for one side and double the answer to get total moment of axle??
     
  6. Apr 12, 2015 #5
    also another thing...my "axle length" is actually length from centre of tyre to centre of opposite tyre on the axle...is that ok?

    *Except im using half that*
     
    Last edited: Apr 12, 2015
  7. Apr 12, 2015 #6
    lol hold up i was using kg not N

    changed it and only gives me ~430 Nm for rear axle moment about centre of the axle...
     
    Last edited: Apr 12, 2015
  8. Apr 13, 2015 #7

    jack action

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    After looking at it some more, there will be some lateral displacement of the axles too. This introduces lateral friction which can be defined with the coefficient of friction of the tires.

    After that, you need to do the FBD for both parts of the vehicle, each having its own mass and moment of inertia. For example:

    art-veh.png

    RED: rolling resistance and lateral friction;
    GREEN: Reaction at pin (same for both);
    PURPLE: Input force from hydraulic cylinder (same for both; I've put only 1 cylinder);
    BLUE: acceleration (linear & rotational) at CG.

    I would start with something like that.
     
  9. Apr 13, 2015 #8
    ok thanks for the help

    So what I did was use the lateral force with mu=0.7

    i then calculated the moments of both axle centres (this uses the small values of rolling resistance forces)...turns out you use the whole axle as the opposing directions of the rolling resistance forces actually double the moment not cancel it.

    I then noted down these moment values...one was negative the other positive.

    then I 'removed' the axles and treated the rest as a single beam with pivot at centre..i then calculated the moment about the centre summing the axle moments with the frictional forces*dist to pivot.

    This gave me a reasonable answer of 30 000 Nm

    The hydraulics provide 48 000 Nm at the pivot which is more then enough to overcome friction.

    But...The 30 000 Nm is the torque required to only just get the vehicle turning right? But then this should be the max value since it is overcoming static friction and once turning at a stand still (and also while moving at speed) the torque required should be less

    So, the extra torque supplied by hydraulics could be some sort of safety factor, having more than what is needed? Or would the extra value be there to get a certain turning velocity??
     
    Last edited: Apr 13, 2015
  10. Apr 14, 2015 #9

    jack action

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    Yes, whatever force/torque that will exceed what is needed will translate into an acceleration of the motion; F=ma, T=Iα.
     
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