mechanical concepts 101
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Mechanical Concepts 101
Shannon Schnepp
Dennis Hughes
Anthony Lapp
10/29/05
Basic Concepts: EquationsForce = Mass * Acceleration
Torque = Force * Distance = Work
Power = Work/Time
Power = Torque * Angular Velocity
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.
normalforce
tractiveforce
torqueturning the
wheel
maximumtractive
force
normalforce
frictioncoefficient= x
weight
Basic Concepts: Traction
Basic Concepts: Traction Equations
• Ffriction = * Fnormal
• Experimentally determine :
• Fnormal = Weight * cos()
• Fparallel = Weight * sin() Fnormal
Ffriction
Weight
Fparallel
When Ffriction = Fparallel, no slip
Ffriction = * Weight * cos()
Fparallel = Weight * sin() = * Weight * cos()
= sin() / cos() = tan()
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
W
fA
NA
fB
NB
AB
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
more weight in backdue to battery andmotors
front
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.
morenormalforce
lessnormalforce
less weight in frontdue to fewer partsin this areaEXAMPLE
EXAMPLEONLYONLY
Basic Concepts: Weight Transfer
EXAMPLEEXAMPLEONLYONLY
robot acceleratingfrom 0 mph to6 mph
inertial forcesexerted bycomponentson the robot
more normal force is exertedon the rear wheels becauseinertial forces tend to rotatethe robot toward the rear
less normal force is exertedon the front wheels becauseinertial forces tend to rotatethe robot away from the front
In an extreme case (with rear wheel drive), you pull a wheelieIn 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: GearsThe Gear Ratio is a function of the
number of teeth of the gearsConsecutive gear stages multiply
N1
N2
N3
N4
• Gear Ratio is (N2/N1) * (N4/N3)• Efficiency is .95 *.95 = .90
Basic Concepts: Gears
N1
N2
N3
N4
• Gear 4 is attached to the wheel• Remember that T = F * Rw
• Also, V = * Rw
• T4 = T1 * N2/N1 * N4/N3 * .95 * .95• 4 = 1 * N1/N2 * N3/N4
• F = T4 / Rw
• V = 4 * Rw
Wheel Diameter - Dw
Dw = Rw * 2
Fpush
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 lbsso…Work = f * distWork = 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 amountof power available.
Lifting/Moving ObjectsExample 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 CalculationsMotor 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 MotorFree 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 MotorNeed 78 ft lbs of Torque (ex 2)
Try Globe Motor w/ GearboxWorking Torque = Stall Torque / 2= (15 ft lbs @ 12 V) / 2= 7.5 ft lbs
Gear RatiosGear 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”
1.0” 0.8”
0.8”
Hollow w/ 0.1” walls Solid bar
StructureThe 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:
IMc
A
Ftens
tens
A
Fshear
Where: = Bending StressM = Moment (bending force)I = Moment of Inertia of Sectionc = distance from Central Axis
Where: = Tensile StressFtens = Tensile ForceA = Area of Section
Where: = Shear StressFshear = Shear ForceA = Area of Section
Bending Tensile Shear
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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