Transcript Mechanical Concepts 101

Mechanical Concepts 101
Shannon Schnepp
Dennis Hughes
Anthony Lapp
10/29/05
Basic Concepts: Equations
 Force
= Mass * Acceleration
 Torque
= Force * Distance = Work
 Power
= Work/Time
 Power
= Torque * Angular Velocity
Basic Concepts: Traction
weight
tractive
force
torque
turning the
wheel
maximum
tractive
force
=
friction
coefficient
x
normal
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! It’s what moves the robot.
Basic Concepts: Traction
Equations
• Ffriction = m * Fnormal
• Experimentally determine m:
• Fnormal = Weight * cos(q)
• Fparallel = Weight * sin(q)
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)
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
Basic Concepts: Free Body
Diagrams
B
W A
fB
NB
fA
NA
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 = mu*N).
Basic Concepts: Weight
Distribution
less weight in front
due to fewer parts
in this area
more weight in back
due to battery and
motors
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.
Basic Concepts: Weight
Transfer
robot accelerating
from 0 mph to
6 mph
more normal force is exerted
on the rear wheels because
inertial forces tend to rotate
the robot toward the rear
inertial forces
exerted by
components
on the robot
less normal force is exerted
on the front wheels because
inertial forces tend to rotate
the robot away from the front
In an extreme case (with rear wheel drive), you pull a wheelie
In a really extreme case (with rear wheel drive), you tip over!
Basic Concepts: Gears

Gears are generally used for one of four
different reasons:
1. To reverse the direction of rotation
2. To increase or decrease the speed of
rotation (or increase/decrease torque)
3. To move rotational motion to a different
axis
4. To keep the rotation of two axes
synchronized
Basic Concepts: Gears
 The
Gear Ratio is a function of the
number of teeth of the gears
 Consecutive gear stages multiply
N2
N4
N1
N3
• Gear Ratio is (N2/N1) * (N4/N3)
• Efficiency is .95 *.95 = .90
Basic Concepts: Gears
N2
N1
N4
Wheel Diameter - Dw
Dw = Rw * 2
N3
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
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??
Lifting/Moving Objects
f
= mu*N = .25*130
 f = 65 lbs
 so…
 Work = f * dist
 Work = 65 * 10 = 650 ft lbs
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 1*Dist 1 + Force 2*Dist 2
= 10 lbs * 3 ft + 8 lbs * 6 ft
= 30 + 48 = 78 ft lbs
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.
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
Combined Motor Curves
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
Motor Calculations
 Free
Speed (rad/min) = RPM * 2 Pi
(rad/rev)
 Stall
Torque (ft*lb) = (in oz)*(1 ft/12
in)*(1 lb/16 oz)
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
Motor Calculations
Drill Motor
 Power
= Free Speed / 2 * Stall Torque /
2 *Conv. Factor
= 125664 / 2 * .48 / 2 *.02259697
= 340 W
Choosing a Motor
 Need
 Try
78 ft lbs of Torque (ex 2)
Globe Motor w/ Gearbox
 Working Torque = Stall Torque / 2
 = (15 ft lbs @ 12 V) / 2
 = 7.5 ft lbs
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!
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
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
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
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”
Hollow w/ 0.1” walls
0.8”
Solid bar
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
Structural Equations

It all boils down to 3
equations:
Bending
  Mc
I
Where:
 = Bending Stress
M = Moment (bending force)
I = Moment of Inertia of Section
c = distance from Central Axis
Tensile
 tens 
Ftens
A
Where:
 = Tensile Stress
Ftens = Tensile Force
A = Area of Section
Shear
 
Fshear
A
Where:
 = Shear Stress
Fshear = Shear Force
A = Area of Section
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?