OffRoad Vehicle Drivetrains

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Off-Road Vehicle Drivetrains

• Friction Clutches

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Macro View of a Drivetrain
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First after the flywheel.. the traction clutch

• Disconnects the prime mover from the drivetrain
• Needed for conventional transmissions
• Also needed as historically people have learned
  to stomp with their left foot in an emergency.
  Sort of.
• Must be able to transmit the maximum engine
  torque through to the driveline

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Design Issues to Be Addressed in Any Clutch

• Torque capacity how much can it transmit with
  acceptable life?
• Engagement time Heat will be generated
• Size- balance of diameter and of clutch
  engagemenet surfaces
• Duty cycle how frequently engaged
• Wet or dry?
• Material to use for friction surfaces

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Approaches to Clutch Torque Capacity Calculation
Method 1

• Consider two disks pushed together as a clutch
• Greater wear at the outer edge until the pressure
  increases near the inside
• Max pressure, pmax, then occurs at rd/2

F
F
D
r
d
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Approaches to Clutch Torque Capacity Calculation
Method 1

• This would stabilize when the work that is done
  at each radius is equal
• This occurs when pressure x radius is constant
• p x r pmax x d/2
• Solve for the pressure ,p, at any r
• ppmax x d/(2 r)

F
F
D
r
d
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Approaches to Clutch Torque Capacity Calculation
Method 1

• Consider an elemental ring, dr
• The ring has area 2p r dr
• The elemental normal force on the ring is
  pressure times area or 2p r p dr
• The total Force, F must be integrated
• Sub in the expression for p ppmax x d/(2 r)

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Max Force on the Clutch Pair

• This yields the maximum Force, F, that can be
  applied for a given maximum pressure, pmax
• Note that this is determined by the material
  properties of the friction surface
• Now we need the torque from friction

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Now for the clutch Torque

• Torque is the product of the normal force, at
  radius r, times the friction coefficient, f,
  times the radius r
• We can factor the total force F ppmaxd (D-d)/2
  out of the torque equation to get the torque
  available per friction surface as

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Clutch Torque Method 2

• What if the pressure on the clutch surfaces is
  uniform (doesnt vary with r)?
• Force between the plate surfaces is now just
  pressure x area
• F ppmax /4 x (D2 d2)
• Torque, however, must be integrated over r

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Clutch Torque Method 2

• Since the clutch pressure is uniform we use pmax
  to represent the maximum allowable pressure as
  before for the material
• We can factor out of the torque equation the
  pressure times the area (force),and are left with

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Usable Model

• Torque from this equation is for a single mating
  pair of surfaces. We have two of these (both
  sides) for each clutch disk involved
• We can design for multiple clutch disks
  sandwiched between matching rings to gain torque
  capacity

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A look at a single disk clutch

• Driven outer shell of clutch housing
• Main traction clutch disk connected to shaft to
  transmission
• Annular hydraulic cylinder relieves clamping
  force on clutch disks
• 2nd clutch disk for pto
• 2nd cylinder for pto
• Output gear for pto

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Example

• Were designing a parallel shaft transmission
  with a power shift clutch dedicated to each of
  several parallel shafts
• We need to transmit the max engine torque through
  this line
• The engine is rated at 180 kW at 2100 rpm and has
  a 35 torque reserve at 1600 rpm
• Further, by this point in the transmission we
  have cut engine speed in half

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Example

• We wish for this transmission to be reasonable
  compact and are allowed 220 mm as the maximum
  diameter of our clutch disks
• Practical inner diameter for our case is 60 mm
• The clutch material is to be filled flourocarbon
  and we will allow a maximum pressure of 2.0
  Newtons/mm2
• How many clutch disks will we require if we wish
  to have twice the minimum required torque
  capacity?

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Solution

• Determine the torque capacity required in the
  worst case
• At 180 kW T 180 x 60000/(2pne)
• at 2100 rpm we require 818 Nm
• At 1600 we require 1104 Nm
• The transmission has already halved the speed so
  we need 2208 Nm
• Double that for a margin of durability and we
  need 4416 Nm of torque through the clutch

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Solution

• The friction coefficient for filled flourocarbon
  is about 0.09
• Maximum allowable pressure on this filled
  flourocarbon is 2.0 N/mm2 ( 2 MPa)
• We will apply the clutch with an annular cylinder
  at this pressure
• The Force, F, is then p/4 x (D2-d2) x ?P
• F p/4 x (2202-602) x 2.0 70,371 N

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Solution

• Note that our diameters are mm and this would
  produce torque in N mm
• Torque 491,592 N mm or 492 N m
• We get 2 times this from each clutch disk (two
  faces on each.) or 984 N m/ disk
• We need 4.48 disks well. 5 disks

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