Mechanical Concepts 101

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Mechanical Concepts 101

• Shannon Schnepp
• Dennis Hughes
• Anthony Lapp
• 10/29/05

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Basic Concepts Equations

• Force Mass Acceleration
• Torque Force Distance Work
• Power Work/Time
• Power Torque Angular Velocity

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Basic Concepts Traction
maximum tractive force
normal force
friction coefficient
x

torque turning the wheel
weight
tractive force
normal force
The friction coefficient for any given contact
with the floor, multiplied by the normal force,
equals the maximum tractive force can be applied
at the contact area. Tractive force is
important! Its what moves the robot.
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Basic Concepts Traction Equations

• Ffriction m Fnormal
• Experimentally determine m
• Fnormal Weight cos(q)
• Fparallel Weight sin(q)

When Ffriction Fparallel, no slip Ffriction m
Weight cos(q) Fparallel Weight sin(q) m
Weight cos(q) m sin(q) / cos(q) m
tan(q)
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Basic Concepts Coefficient of Friction

• Materials of the robot wheels (or belts)
• High Friction Coeff soft materials, spongy
  materials, sticky materials
• Low Friction Coeff hard materials, smooth
  materials,shiny materials
• Shape of the robot wheels (or belts)
• Want the wheel (or belt) surface to interlock
  with the floor surface
• Material of the floor surface
• Surface conditions
• Good clean surfaces, tacky surfaces
• Bad dirty surfaces, oily surfaces

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Basic Concepts Free Body Diagrams
W
A
B
fA
fB
NA
NB
The normal force is the force that the wheels
exert on the floor, and is equal and opposite to
the force the floor exerts on the wheels. In the
simplest case, this is dependent on the weight of
the robot. The normal force is divided among the
robot features in contact with the ground. The
frictional force is dependent of the coefficient
of friction and the normal force (f muN).
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Basic Concepts Weight Distribution
less weight in front due to fewer parts in this
area
more weight in back due to battery and motors
EXAMPLE ONLY
front
more normal force
less normal force
The weight of the robot is not equally
distributed among all the contacts with the
floor. Weight distribution is dependent on where
the parts are in the robot. This affects the
normal force at each wheel.
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Basic Concepts Weight Transfer
EXAMPLE ONLY
In an extreme case (with rear wheel drive), you
pull a wheelie In a really extreme case (with
rear wheel drive), you tip over!
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Basic Concepts Gears

• Gears are generally used for one of four
  different reasons
• To reverse the direction of rotation
• To increase or decrease the speed of rotation (or
  increase/decrease torque)
• To move rotational motion to a different axis
• To keep the rotation of two axes synchronized

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Basic Concepts Gears

• The Gear Ratio is a function of the number of
  teeth of the gears
• Consecutive gear stages multiply

• Gear Ratio is (N2/N1) (N4/N3)
• Efficiency is .95 .95 .90

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Basic Concepts Gears
Wheel Diameter - Dw Dw Rw 2
Fpush

• Gear 4 is attached to the wheel
• Remember that T F Rw
• Also, V w Rw
• T4 T1 N2/N1 N4/N3 .95 .95
• w4 w1 N1/N2 N3/N4
• F T4 / Rw
• V w4 Rw

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Lifting/Moving Objects

• Example 1
• A box weighs 130 lbs and must be moved 10 ft. The
  coefficient of friction between the floor and the
  box is .25.
• How much work must be done??

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Lifting/Moving Objects

• f muN .25130
• f 65 lbs
• so
• Work f dist
• Work 65 10 650 ft lbs

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Lifting/Moving Objects

• Example 2 The arm weighs 10 lbs and moves 3 ft
  vertically. The mechanism that contains the balls
  weighs 5 lbs. The balls weigh 3 lbs. The
  mechanism and balls move 6 ft vert.
• Work Force 1Dist 1 Force 2Dist 2
• 10 lbs 3 ft 8 lbs 6 ft
• 30 48 78 ft lbs

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Lifting/Moving Objects

• Example 2A
• Desire this motion to be completed in 10 seconds.
• Power 78 ft lbs / 10 seconds (60sec/1min)
  .02259697
• 10.6 Watts
• Note There is only a certain amount
• of power available.

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Lifting/Moving Objects

• Example 2B
• Desire this motion to be completed in 3 seconds.
• Power 78 ft lbs / 3 seconds (60sec/1min)
  .02259697
• 35.3 Watts

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Combined Motor Curves
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Motor Calculations

• Motor Power Power Available
• Free Speed / 2 Stall Torq. / 2 C.F.
• Where
• Free Speed is in rad / min
• Stall Torque is in ft lbs
• Conversion Factor .02259697

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Motor Calculations

• Free Speed (rad/min) RPM 2 Pi (rad/rev)
• Stall Torque (ftlb) (in oz)(1 ft/12 in)(1
  lb/16 oz)

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Motor Calculations

• Drill Motor
• Free Speed 20000(rev/min)2PI(rad/rev)
• 125664 rad/min
• Stall Torque 650 (Nmm)(1 lb/4.45 N) (1 in/
  25.4mm)(1 ft/12 in)
• .48 ft lbs

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Motor Calculations

• Drill Motor
• Power Free Speed / 2 Stall Torque /
• 2 Conv. Factor
• 125664 / 2 .48 / 2 .02259697
• 340 W

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Choosing a Motor

• Need 78 ft lbs of Torque (ex 2)
• Try Globe Motor w/ Gearbox
• Working Torque Stall Torque / 2
• (15 ft lbs _at_ 12 V) / 2
• 7.5 ft lbs

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Gear Ratios

• Gear Ratio Torque Needed / Torque Available
• 78 ft lbs / 7.5 ft lbs
• 10.4 1
• Now time to find the gear train that will work!

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Choosing a Motor

• In Summary
• All motors can lift the same amount (assuming
  100 power transfer efficiencies) - they just do
  it at different rates
• BUT, no power transfer mechanisms are 100
  efficient
• If you do not account for these inefficiencies,
  your performance will not be what you expected

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Materials

• Steel
• High strength
• Many types (alloys) available
• Heavy, rusts,
• Harder to processes with hand tools
• Aluminum
• Easy to work with for hand fabrication processes
• Light weight many shapes available
• Essentially does not rust
• Lower strength

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Material

• Lexan
• Very tough impact strength
• But, lower tensile strength than aluminum
• Best material to use when you need transparency
• Comes in very limited forms/shapes
• PVC
• Very easy to work with and assemble prefab shapes
• Never rusts, very flexible, bounces back (when
  new)
• Strength is relatively low

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Structure

• Take a look at these two extrusions - both made
  from same Aluminum alloy
• Which one is stronger?
• Which one weighs more?

1.0
0.8
1.0
0.8
Hollow w/ 0.1 walls
Solid bar
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Structure

• The solid bar is 78 stronger in tension
• The solid bar weighs 78 more
• But, the hollow bar is 44 stronger in bending
• And is similarly stronger in torsion

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Structural Equations

• It all boils down to 3 equations

Bending
Tensile
Shear
Where ? Bending Stress M Moment (bending
force) I Moment of Inertia of Section c
distance from Central Axis
Where ? Tensile Stress Ftens Tensile Force A
Area of Section
Where ? Shear Stress Fshear Shear Force A
Area of Section
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Stress Example

• Let's assume we have a robot arm (Woo hoo!)
  that's designed to pick up a few heavy weights.
  The arm is made out of Al-6061, and is 3/8" tall,
  1" wide, and 3 feet long. The yield strength is
  about 40,000 PSI. In the competition they are
  hoping to to pick up 3 boxes of 15 lbs each. Will
  this arm be strong enough?

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Woo Hoo! You Made It!