Traction and Soil

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Traction and Soil
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Questions We Would Like to Answer

• How much pull (draft) can we produce with this
  vehicle?
• Skid steer, tractor, tank, tracked trencher
• What determines that draft?
• Which parameters matter most?
• Which ones do we have any control over?
• If designing a vehicle what shall we do?

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Soil !

• Consider what draft you might expect to generate
  on the following surfaces
• two inches of dry sand over firm ground
• saturated clay
• freshly worked silt loam
• dry silt loam no till
• hard packed moist clay
• Whats the difference?
• The strength of the soil..

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Normal Force
Horizontal Force
Displacement Of Plate
Lugs engage soil
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Maximum Force for any given normal loading
Different Normal Loads
Horizontal Force Sustained
A given normal load is applied. The plate is
displaced to the right. If we had a sensor
connected we would measure the horizontal force
as the plate is displaced.
Horizontal Displacement of our grouser plate
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Soil Maximum Shear Strength
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Soil Maximum Shear Strength

• Lets look at this as you did with other
  materials in Mechanics of Materials
• Remember Mohrs Circle of Stress Analysis
• Applied Normal stress
• Applied Shear stress
• Maximum Principle Stress?
• Now back to soil

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Classic Mohr Coulomb Model

• Force developed at maximum soil strength with our
  shear plate was
• Fmax Ac Wtanf Note A is area, W is weight,
  c is soil cohesion and f is internal friction
  angle of the soil
• Divide by the Area of the plate, A
• tmax c W/A tanf c p tan f

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Soil Maximum Shear Strength
tmax c ? tan f
t1 t2 t3
Maximum Shear Stress
c
?1 ?2 ?3
Normal Stress, ?
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Max soil shear strength

• Note that this strength function is linear (2
  parameters.)
• Intercept is c, the soil cohesion
• Slope is ?t/??
• We call this the tangent of the soils internal
  friction angle and represent internal friction
  angle with f
• Memorize the meaning of these c and f

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What does it mean??

• Soil shear strength is a function of the
  compressive (normal) load on it
• Greater soil strength allows greater tangential
  loads before soil failure
• How do we get higher normal loads?
• Heavier vehicle
• Ballast
• So we ballast to increase soil strength to allow
  greater tractive effort

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Lets look at a Track Model

• Track lugs enter the soil (similarity to test
  plate)
• However, soil displacement varies with lug

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Less than Max Soil Strength Case

• For small lug displacements we wont reach the
  max soil strength
• Then we need a function to indicate the soil
  strength as a function of displacement
• We need an equation for the curves at right

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Less than Max Soil Strength Case

• Curves look exponential
• t (c p tanf)(1 e-j/K)
• p is normal pressure
• j is soil displacement
• K is a soil related constant
• Now we have a curve, but where are we on it?
• How do we know what j, the displacement is ?

K
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Track/Soil Displacement
Vs

• Vehicle has a theoretical, no slip, velocity, Vt
• Vehicle has an actual velocity, Va
• Vt Va Vs , the slip velocity
• The bottom of the track moves at Vs rearward
  relative to the soil. See above

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Relate slip velocity to track displacement
x

• Denote x the position of a point on the contact
  patch (not the track), measured from its leading
  edge
• x Vtt (theoretical track velocity times time)
• Recall, Vt Va Vs , the slip velocity
• Displacement, j, Vst (velocity times time)
• Then j Vst Vs(x/Vt) sx

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The contact patch
b
x
l
0
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What does THAT mean?

• Shear displacement, j, increases linearly from
  the entry point to the exit point of the track
  and is dependent upon the slip
• How do we define slip?
• Slip s (Vt Va) / Vt Vs/ Vt
• Slip can range from near zero (on dry concrete)
  to 1, or 100, (stuck.)

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Now Model Track Thrust, F

• We have a model for soil shear strength, and for
  shear force, and one for soil shear displacement
  so
• Combine them
• t (c ptanf)(1 e-j/K ) tmax(1-e-sx/K)

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Track Thrust Model

• Our model predicts the thrust that the track will
  generate
• It includes varying soils strength as the soil is
  progressively sheared under the track
• Note that it assumes a constant pressure, p, upon
  the ground from the track
• Hmmmmmmm.

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Ground Pressure Under a Track

• Pressure under an Abrams tank might be relatively
  constant with LOTS of bogies
• Pressure under a track for a vehicle pulling an
  implement (tractor.) might see
• Variation under bogie or roller wheels
• Increase front to rear as weight transfer from
  draft force puts more downpressure on the rear
  drive wheel

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Bogies or Rollers
Rear Drive Wheel
Front Tension Wheel
Pressure under a track
Elapsed Time (represents position under track)
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Incorporating Pressure Variation

• Note that our earlier model is integrated over
  the contact patch length
• If we can come up with a function for the
  pressure over that same length we can add that
  function to the integral
• Why?
• Because soil strength is a function of pressure

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Lets try an example

• Givens
• Track contact length 7 or 84 inches
• Track width 20.8 inches
• Weight on each track is 9082 lbs
• Soil cohesion 3.8 psi
• Soil internal friction angle 24 degrees
• Soil shear deformation modulus 2.2 inches
• Slip 6 or 0.06
• Track pressure is UNIFORM (not but well assume
  it is for this example)

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Thrust force Equation
I used a website for integrals to find the
integral for the hard part
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Max Shear Strength tmax

• Since we are treating the ground pressure as a
  constant we have a constant value of the maximum
  soil shear strength

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Now we calculate the Track Force, F

• Slip0.06, l 84 inches, K2.2 inches
• At x l 84 and b20.8, t 6.11
• At x 0
• The Track Force is 11,147 - 4,6606,487 lb

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Some Discussion

• Note that the traction force, F, is less than we
  might expect for the contact patch failing at
  tmax, and yet we are using tmax
• F c (bl) W tanf 10,683 lb , but we got
  4,660 lb
• Why is this so?
• What are the units on the bracketed term below?

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Some Answers

• The effective track length is less than 84
  inches because the soil does not develop its full
  strength until the shear displacement reaches a
  value well above 2.2 inches
• Only the soil under the back portion of the track
  is providing near full strength
• This would be even more pronounced with a
  situation where the loading was heavier at the
  rear of the track

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Question

• How would you approach the problem if you had a
  non uniform soil pressure from front to rear
  under the track?
• Consider how you would describe pressure
• Consider K .
• The integral could get messy

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More Discussion

• What have we left out?
• Weve modeled the thrust that can be generated as
  the soil is displaced and failed
• Consider operating in very soft soil or mud
• As we dig a hole we may encounter some soil
  resistance
• We may push a wave of soil ahead of us
• It takes energy to compress the soil or
  continually climb out of our trench

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Motion Resistance in a Track-Soil System

• We have outlined a model to calculate the thrust
  force from the soil on a track given
• Soil strength model (c, f, K)
• Track width and contact length
• Track slip
• The track thrust is forward on the track
• What about resistance to track motion?
• Think of towing a tracked vehicle in neutral

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Track Rut and Bow Wave

• Soil level prior to passage
• Vehicle track
• Bottom of rut and bow wave
• Logarithmic soil fracture characteristic

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What causes motion resistance?

• Motion resistance depends upon how deep we sink
  in as we move along
• This sinkage is dependent upon the soil
• We need a model to predict the sinkage
• Would help if it was a function of stuff that we
  know about the vehicle already
• Ground pressure
• Track dimensions

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Soil Bearing Strength Model

• Bekker gives a soil sinkage model
• P is pressure (psi or Pa)
• b is track or test plate width (in. or m)
• Z is sinkage (in. or m)
• Kc , kf , and n are constants determined from
  load-sinkage tests with two or more different
  size plates

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Work Done in Sinkage of the Track

• Assume sinkage occurs linearly from track entry
  to exit and is governed by pressure
• Express sinkage, z, as a function

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Integrate over rut depth

• Work integral
• But P is related to z
• Sub P into the integral
• Or.
• Now substitute z from above into this
• This should be the same as work done pulling the
  loaded track a distance l

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Motion Resistance Force

• Let Rmr be the motion resistance force
• Rut work Rmr x l
• Divide the soil model work (previous slide) by
  track length, l, to get the force, Rmr
• Or. Reorganized slightly

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What about that Bow Wave

• This can be modeled with the Universal
  Earthmoving Equation
• Complex topic for this class but it would look
  like this for our case

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Bow Wave Term Definitions

• Rb is motion resistance
• Kpc is a coefficient related to cohesion
• Kp? is a coefficient related to inertial forces
• ? is the soil specific weight
• b, c, zo are track width, soil cohesion, and
  track sinkage respectively

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One more thing.

• Why is Bobcat interested in the track testing
  station that last years design team built?
• Parasitic forces involved in flexing a relatively
  stiff track around the front and rear wheels
• Our soil model does not include these and they
  will be specific to each track design

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More Discussion

• We will leave tracks at this point. It is
  possible to model their likely performance with
  classic soil strength theory as we have
  introduced. To be realistic, non-uniform ground
  pressure and motion resistance would have to be
  treated before we could begin to predict net
  tractive performance and available draft.