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?