# Design5x - MIT

Urban and Civil

Nov 15, 2013 (4 years and 8 months ago)

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1

Mechanical Elements and Systems Design Quals

Jan 2010

Robert Panas

Statistics

1
1
N
i
i
x x
N

average

2
1
1
1
N
i
i
s x x
N

 

standard deviation

Least squares fit
, y = ax

o

x
y
as
r
s

Chance σ>s: (stress greater than limit v
alue), want σ<s chance 0.99.

o

y i
m s s
 

define margin of safety variable
, m (+ if safe), y is yield, i is stress of
interest

o

2 2 2
m y i
m y i
  
  
 
 

o

m
m
deviation from average
m
z
standard deviation

o

Failure is when m is negative. Ideally it is largely
positive. Question is, what %
of m<0? Calculate p(m>0)

o

2 2
0
( 0)
y i
m
m
y i
p m
 

 

   

o

( 0) 1 ( 0)
p m p m
   

Stress

Flexures

o

4
4
( )
dV d y x
q EI
dx dx
 

derivative of shear (positive upwards)

3
3
( )
dM d y x
V EI
dx dx
 

shear is derivative
of moment
(positive downwards)

2
2
( ) ( )
( )
d x d y x
M x EI EI
dx dx

 

for small
deflections

( )
( )
dy x
x
dx

slope of beam

( )
y x

vertical deflection of beam

Shear/Moment directions

+x
+
V(x
)
+
M(x
)
+x
+
V(x
)
+
M(x
)
+y
+x
+Moment
doublet
+
θ

2

o

Singularity functions

Moments:
doublet

unit doublet (
-
2) is in NEGATIVE direction for moments

(if we
assume Z axis out of page, towards reader)
, so a positive moment
M is
2
x a
M M x a

  
, correct + rotation shown in fig above

step functions

higher polynomials: use
ramp, parabolic, etc…

when integrating, bring
exponent down for
everything above 0

o

Solving process

Use singularity functions to
define q(x) including

integrate from q four
times
to get deflection
(to
V,M,θ,y)
integration constant each
time
. When go from M to
θ, multiple M(x)*1/EI
before integration

use boundary conditions to
define the four constants

and any other unknowns (reaction loads).
Common

BC, up to 8 possibl
e at edges:

shear V(0
-
)=0,
and moment M(0
-
)=0 before beam starts

shear V(L+)=0
, and moment M(L+)=0, after beam ends

slope

ydot(0
)
=0, slope ydot(L)=0 if clamped guided

y(0)=0, y(L)=0 if clamped clamped, y(x)

may want to do
this as each
condition is reached during
integration (shear = 0 after
first integration, moment = 0
after 2
nd

integration, etc)

no BC on q(x), but can have
on V, M, theta, y

o

Examples:

o

Derivation

dx yd

compressio
n/expansion off
neutral axis (y = dist off axis)

Shear/Moment directions

Differential bending segment

3

2
2
3
2
2
1
1
d y
d
d x
ds
dy
dx

 
 
 

 
 
 
 
 

where ρ is radius of curvature, Φ is angle and s is
displacement along beam center

(neutral axis)

dx yd y
ds ds

    
,

strain is ε

2
1 1
y
M ydA E ydA E y dA EI

  
 
   
 
 
  

where M is moment at s,
I is moment of inertia, E is youngs modulus. Derived from forcing
internal moment (integral sigma*y*da) to equal external applied moment

E M
E y y
I
 

  
stress proportional to moment, distance from neutral
axis and momen
t of inertia

For torsion, shear stress
Tr
J

where T is torque, r is radius and J
is the polar moment of inertia

Curved flexures

Neutral axis moved to average radius of curvature
1 1 1
n
dA
A
 

Common relations

Common
deflection relations

Stress concentrations

L
F
F
F
w
w
w
δ
max
θ
max
EI
FL
3
3
max

EI
FL
2
2
max

EI
wL
8
4
max

EI
FL
48
3
max

EI
wL
384
5
4
max

EI
FL
192
3
max

EI
wL
384
4
max

EI
wL
6
3
max

EI
FL
32
2
max

EI
wL
24
3
max

??
2
2
max
EI
FL

EI
wL
3000
2
max

2
max
16
FL
EI

2
max
64
FL
EI

L
F
F
F
w
w
w
δ
max
θ
max
EI
FL
3
3
max

EI
FL
2
2
max

EI
wL
8
4
max

EI
FL
48
3
max

EI
wL
384
5
4
max

EI
FL
192
3
max

EI
wL
384
4
max

EI
wL
6
3
max

EI
FL
32
2
max

EI
wL
24
3
max

??
2
2
max
EI
FL

EI
wL
3000
2
max

2
max
16
FL
EI

2
max
64
FL
EI

2
16
FL
EI
4
64
d E
k
DN

  
 
2
16
FL
EI
4
64
d E
k
DN

  
 

4

o

2
1
t
b
K
a
 

theoretical stress concentration factor for elliptical hole where b is
half width and a is half height and width of part is infinite. Stress in height
direction

o

Usually less than 3

Thi
n walled vessel

o

If wall radius <0.05*r, thin walled.

o

2 2
2 2 2
1
i i o
r i
o i i
r p r
p
r r r

 
   
 

 

same as outer radius ro, with internal pressure pi.

o

2
pd t

pressure trying to separate cylinder in half integrated over inner
surface gives p*d, hoop stress resisting is wall area (2*t) times stress, σ.

o

4
z
pd
t

longitudinal stress (z direction in cylindrical coordinates) due to caps at
end of c
ylinder

Hertzian
Contact Stresses

o

2
cylinders, check hale thesis for other choices

1 1 1
e 1 2
R R R
 

surfaces, such that sphere has 2 of equal

size, cylinder has one finite and
one infi

2 2
1 2
1 2
1 1
1
e
E E E
 
 
 

eff
ective youngs modulus of full interface

1 2
3 3
1 3
4
e e
F
R E

   

   
   

deflection between points far from contact area

1
3
3
4
e
e
FR
a
E
 

 
 

,

a,
between two spheres with
po
isson’s ratio, E
and diameter d for given force F

1
2
3
2 2
6
3 1
2
e
e
FE
F
p
a R
 
 
 
 
 

maximum surface pressure, found in exact center
of contact area, 1.5x the average pressure

Max shear stress
0.31
p

at
0.48
z a

this is where yield occurs,
whe
n this stress stress is about 58% of tensile strength

Max tensile stress

1 2
3
p
 
 

at
r c

Stress/Strain Relations

5

o

Von Mises:

2
2 2 2
2 2 2
11 22 22 33 11 33 12 23 31
6
2
v
        

       

where the von mises stress indicates failure when it reaches material yield

If all shears turned into stresses (‘principle stresses’), then von mises
criterion reduces to:

2 2 2
2
11 22 22 33 11 33
2
v
      
     

o

Tresca criterion is:

1 3
t
  
 

using pr
inciple stresses, ordered from largest
(1) to smallest (3)

o

1
i i j k
T
E
    
 
    
 

general strain equation

o

lateral
axial

 

o

o

0
2
2
,
3
2
2
2
1

xy
Y
x
Y
x

Mohr’s circle

o

T
E

 
  

thermal strain

o

2 1
E
G

Deflection and Strain

Simple spring calculations

o

AE
k
l

linear

o

GJ
k
l

rotational shear

o

r
GA
k
l

linear shear

Inertias

o

1
st

moment
x
I ydA

over surface of beam, etc.

o

2
nd

moment
2
xx
I y dA

over surface of beam, etc.
, neutral plane (linear
deflection)

Parallel Axis theorem
2
new old
 

this is true when Iold is the 2
nd

moment around the center of mass

o

Polar moment
2
J r dA

, neutral axis (rotational deflection)
, only holds for
circular cross
-
sections. For rectangular cross
-
sections, is not a clear simple
expression. See Roarkes Formulas for Torsion for details.

o

Mass moment
2
m
I r dm

6

Parallel Axis theorem
2
I
mnew mold
I md
 

this is true when Iold is the 2
nd

moment around the center of mass

Second moment of inertias

Strain Energy

o

2
2
M
U dx
EI

for bending

o

2
2
F
U dx
EA

for tension

o

2
2
T
U dx
GJ

for torsion

o

2
2
V
U dx
GA

for shear

o

2
2 2
U dx dV
E
 
 
 

for generic volume

Castigliano’s Theorem

o

i
i
U
F

displacement or force calculated as partial derivative of energy with
respect to other variable

o

Can be done for fictitious force by adding in this fake force, Q, taking derivative,
then setting Q=0

o

Can enforce zero displacement with this method too

Buckli
ng

o

Equations

4
3
4
1 1
0.21 1
3 12
is long side, is short side
b b
ab
a a
a b
 
 
 
 
 
 
 
3 4
4
1
1 0.63 1
3 12
ab b b
a a
a b
 
 
 
 
 
 
 

2 3 4
2
3
1 0.6095 0.8865 1.8023 0.9100
T b b b b
ab a a a a
a b
 
     
   
 
     
     
 
 

At the midpoint of each longer side (a)

7

2
2
crit
C EI
P
l

critical
am

2
2
/
crit
crit
P
C E
A
l k

 

where
l/k is the slenderness ratio

2
I
k
A

definition of k as
the radius of gyration, with I
being 2
nd

moment and A is
cross sectional area

o

Constant:

A 4 fixed fixed

B 2 fixed pinned

C 1 fixed guided

D 1 pinned pinned

E ¼ fixed free

F ¼ pinned guided

o

Regimes

When low, buckling very high, failure at yield stress

When high, buckling first f
ailure mode, failure at crit stress

Crossover when yield=crit stress. Above this, critical stress (buckling) is
failure, below, yield is failure.

This allows calculation of l/k at which buckling becomes an issue

Usually calculate for buckling failure when

l/k ratio such that crit stress <
yield/2, when around crit=yield, use intermediary, when crit>>yield, use
yield

Damping

o

Constant orifice snubber: Classically linear with velocity

o

Conventional snubber: Can made dampers that put out nearly constant force f
or
any velocity

o

Progressive snubber: Increasing force for lower velocities

Mechanical Springs

For helical spring,
4
3
8
d G
k
D N

, d is wire diameter, D is helix diameter, N number of turns,
k is linear stiffness

Light service (less than 10
4
),
average, (10
5
-
6
), severe (>10
6
)

Torsion springs
-

helical springs with different ends

Materials

Yield= 0.2% deflection from linear

UTS = maximum engineering stress observed

o

UTS (
Su
)
~3
.5
HB
[MPa] for steel, where HB is brinell hardness number

Failure Regimes

Boundary Conditions

Stress
σ
Slenderness Ratio
l/k
Yield
Buckle
σ
y
Transition
Buckling
Yield
Stress
σ
Slenderness Ratio
l/k
Yield
Buckle
σ
y
Transition
Buckling
Yield
C=
4
2
1
1
1/4
1/4
C=
4
2
1
1
1/4
1/4

8

Low cyle (<1000
)

Infinite life region (>10
6
) for steels

Fracture stress
-

stress value at which material fails

True strain:
1
ln( )
i
o
l
i
o
l
l
dl
l l

 

so
ln( 1)
true eng
 
 

In true stress/strain, fracture stress> UTS

Plastic stress
0
m
 

Plastic deformation in cold working moves elastic limit up. Return path not to 0, but to
new equilibrium with yield stress being a greater value
-

larger elastic limit, same E,
reduced amount of effort r
equired after this to reach UTS,
less ductility

Manufacturing Processes

o

casting

Sand casting

Shell casting

Investment casting

o

Powder metallurgy

o

Hot working processes

Rolling

Forging

Extrusion

o

Cold Working

Drawing

Blanking

F
orming

Breaks grains
-

get smaller

o

Annealing

Heated above
transition temp, relieve residual stresses, lower yield, greater
ductility

Grain growth

o

Quenching
-

can change crystal structure of steel this way

o

Plastics

Thermoplastic
-

can reheat/reshape

Thermoset
-

once set, cannot be changed, chemically set

Friction

o

Tef

o

Ice 0.1

o

Steel/aluminum 0.25

o

o

Plastic 0.1
-
0.2

Cost

o

Steel

1x \$0.5/lb

o

Alloy steel

2x

o

Stainless steel

3x

o

Tool steel

13x

o

Aluminum

4x

9

o

Brass

2x

Screws, Fastener, and the Design of Nonpermanent Joints

o

Pitch
-

distance b

o

Minor diameter
-

di
a of bot

o

Major diameter
-

dia

o

Lead is distance nut moves along screw axis
when it is rotated 360 degrees

1
2
m m
m
Fd fd
T
d fL

 

 

 

torque required to raise a load
with F load, dm is mean diameter,
L

is height gained
from one rotation, f is coefficient of friction,

o

Self locking if tan(λ)

, where λ is lead angle
tanλ = L/(πd
m
)
, μ is friction coefficient

o

Reduces to:
m
L
d

With no friction,
2
T FL

 
, torque T to turn 1 rotation is energy required to apply force
F over length L

,
-
1.5 D wort

50% of energy in joint into bolt head friction, 40% into thread friction, 10% into elastic
deformation

S
tiffness
:

o

Material stiffness found
through assuming pressure cone of about 30deg off
vertical

Assuming this angle and diameter about 1.5 times hole,

tan
tan
2ln
tan
w w
w w
Ed
k
l d d d d
l d d d d
 

 
  
 
  
 

where d is diameter of sc
rew hole,
dw

is diameter of

washer at top of conic section, l is depth of screw hole
thro
ugh material

o

Proof strength
-

o

Supposed to fail just under head

o

o

i
T KFd

torque

T

Fi

relation, K roughly 0.2 (when lubrication about f
= 0.15),
d is mean diameter,

o

, F

o

, F

Pins/Keys

Model of a bolted joint

10

o

Compressive stresses from interference fit:

2
E
p
R

, R is nominal pin size, d is
the difference in pin and hole radii, p is pressure

Want threads to fail (shear) just as shaft breaks

Welding, Brazing, Bonding, and the Design of Permanent Joints

Welding

o

Often case that weld metal is strongest part of joint

Resistance Welding (spot welding)

o

Can be more resistant to failure

than mechanical joints

Mismatch of thermal expansion coefficients can lead to stresses

Dowel pin

Spiral pin
-

rolled cylinder of material, expands in hole

Spring pin
-

slotted cylinder, also expands in hole

rivets

Bearings

Types

o

Sliding

o

Roller/ball

o

Magneti
c

o

Flexural

o

Hydrostatic

Journal

Rotary thrust

Conical journal/thrust

o

Aerostatic

Rolling Contact Bearings

Components

o

Outer ring

-

surface either circular arc or gothic
, gothic allows 4 pt contact,
but more wear/friction

o

Inner ring
-

surface either circular arc or gothic

o

Balls/rolling element

o

Separator
-

omitted in cheap bearings

11

Ball Types

o

B,C,D,E
Single row, deep groove take radial and some thrust

, most
common

o

B,C deep groove has little moment resistance, does not resist angular
misalignment or thrust greatly

o

D 4 point contact good at resisting moment

and thrust

o

E angular contact good at resisting thrust in one direction

o

F double row instant centers are far apart, so very good at resisting moments
and thrust, also able to handle thermal growth of axle

o

G

full freedom in angular alignment
-

no moment resi
stance, moderate thrust

Roller Types:

o

H
Cylindrical roller

no thrust
, very good at radial, much higher than ball
bearing

o

I Double Cylindrical roller, same as H, just twice as much load

o

J Tapered roller

heavy axial
one direction, advantages of ball and straight rollers

o

K Needle roller (d)
-

o

L
Spherical roller

(b,c)

self aligning, good with heavy loads and
misalignment, increase

Bearing Life

o

Measure is number of hours at standard rotational speed or number of
revolutions, rating is that at which 10% have failed, called L
10

o

Ideal failure is metal fatigue in race

o

1/
a
FL K

, F is load, L is li
fe (# of revolutions), a is constant= 3 for ball
bearings, 10/3 for roller bearings

o

Manufacturer chooses standard life L
10

= 10
6

then

rate each bearing with the
10

such that 10% fail by that lifetime. Gives value of constant
1/1/
10 10
a a
C L K FL
 

o

0
0
b
x x
x
R e

 

 

 

For reliabilities R other than 90% (10% failure), where
x=L/L
10
,
θ is std dev of distribution, x0 is guaranteed or minimum value of the
variate, b is shape parameter of
skewness
.

Weibull distribution

L
50

= 5*L
10

for 50% reliability

L
5

= 0.62*L
10

for 95% reliability

L
1

= 0.21*L
10

for 99% reliability

o

a
i i
D F L

Damage D occurs as sum of load to the a power times number
of revolutions, sum up all segments

1/
a
a
i i
tot
i
F L
F
L
 

 
 
 

equivalent load due to damage theory

A B C D E F G H I J

K L

Types of Bearings

12

One way to c
heck where this lies
:
,
calculate K for bearing, using Ftot,
gives an L. calc x = L/L
10
, then determine R for that point, gives
chance of failure at that point.

o

When bearing is
mis
-
aligned, life drops

o

Friction around 0.01

Tapered Roller Bearings

o

components

Cone (inner ring)

Cup (outer ring)

Tapered rollers

Cage (space retainer)

o

0.47*
r
a
F
F
K

constant K

use

two bearing sets pointed opposite directions to cancel out

Mounting and Enclosure

o

Mounting plan: often axially constrain only one set of bea
rings, so not
overconstrained.

o

Duplexing
-

pairs of angular contact ball bearings set such that when clamped
togethe

o

Preloaded desired to remove internal clearances, increase fatigue life, decrease
shaft slope

Inner or outer interference fit

Ensure all of ball is under compression

Med

o

Seals should be used to keep out grease/dirt

Felt (rubbing on surface) cheap, good for low speed

Rubber seal (rubbing on surface) good for low speed

Labyrinth

non contact, good for high speed, no direct path through
gap

Lu
brication and Journal Bearings

o

Revolute pair
-

(journal bushing)
-

allows
rotation while constraining other motion

o

Lubrication

Hydrodynamic
-

su
rfaces separated by thi
n

film, fluid mechanics physics

du U R
dx h h

   
  

fluidic shear stress, τ, with relative
velocity U and gap h, dynamic viscosity μ

(10
-
3

water, 10
-
5

air),
radius R and rotational speed ω

Petroff’s equation:
2
2
N r
f
P h

, coeff of friction f; rev per sec,
N; radius r; P = W/(2rl)

13

Hydrostatic

pressurized air/water into gap, no need for motion of
surfaces

Elastohydrodynamic

between surfaces in rolling contact (gears)

Boundary

very thin film, few molecules, viscosity less important than
chemical composition

Oily chemicals bond to surface v
ery strongly

Solid film

graphite as example, extreme temperatures

Gears

Can be up to 95% efficient

o

Generally used to slow down rotation, increase torque, because velocity easy to
generate
with voltage, torque requires current

Types:

o

Spur gears, teeth
parallel to axis of rotation

Used to transmit torques between parallel axes

o

Helical gears, teeth sheared from spur gear alignment

Less noisy
-

Develops thrust loads and bending, fixed with double helica
l

o

Bevel gear, conical shaped surface for gears

Between intersecting shafts

Spiral bevel, like helical

o

Hypoid gear

Helical gear for non
-
intersecting shafts

o

Worm gear

Used when speed ratio >3 between shafts

Non
-
intersecting shafts

Worm gear, ~ 30
-
50%
efficient (sliding contact)

o

Planetary Gear

Sun in center

Ring gear outside

Planet rolls between these

TR

from sun to arm which is
holding planets:

must be higher than 2
(corresponds to 0 size
planets)

lowest practical: 3:1

highest practical: 10:1,
beyond this, sun is so
small planets hit

Usually big speed
reduction

Planetary Gear Transmission Ratios

Memorization Trick for Planetary TR

!
!

14

arm ring
sun
ring sun arm
N
N
 
 

sun is inner gear in planetary arrangement, ring
is the ou
ter ring with gears on it’s inner surface, planet moves between
these

Nomenclature

o

Pitch circle:
theoretical gear
diameter for calc

Pitch diameter

o

Pinion:
smaller gear

o

Gear: larger of the
two

o

Circular pitch

p
:
distance between
gears on pitch circle

d
p
N

Sum of tooth
thickness and
spacing between gears

o

Module
pitch
d
m
N

ratio pitch diameter to number of teeth

o

Diametral pitch
P
= 1/m
,
N
P
d

o

o

Dedendum: radial distance from pitch circle to bottom of teeth

o

Clearance circle: height of other gears teeth tips

Teeth

o

Constant angular velocit
y ratio during meshing: conjugate action

o

Common form: involute profile

o

Pressure angle
cos
b
r
r

, rb is radius of clearance circle, r is radius of pitch
circle

Spur Gear

Helical Ge
a
r (double)

Spiral Bevel Gear

Worm Gear

Gear Details

15

o

Contact ratio:
t
c
q
m
p

p is circular pitch, qt is the arc of action (arc over which
gears touch)

Can be more than 1, indicating multiple teeth in contact, generally <1.2

o

To avoid teeth causing interference, must choose number of teeth carefully
-

too
few or too many and will hav
e problem

Forming Gears

o

Methods

Sand casting

Shell molding

Investment casting

Permanent mold casting

Die casting

Centrifugal casting

Powder metallurgy

Extrusion

Form cutters

Cold forming

Milling

Shaping

Hobbing

Designing

o

Determine necessary power, speeds
and transmission ratio

Gear relation: contact enforced so
/2/2
A A B B
d d
 

for gears A and B

A A B
B B A
d
d
 
 
 

A A
B B
d
d

from energy conservation

o

Choose type of gear

Shaft alignment

Efficiency

Size

o

Size out gear

Estimate

module

(1, 1.5, 2 mm)

length/tooth = m*
π

Choose min N

usually >12 for good fit

Set diameter

pitch
d mN

Tooth avg thickness h

2
m
h

16

Tooth length L

2.2
L m

length
roughly 2.2 times module

Tooth width b

Make wide enough to keep stress down

Assume tooth is cantilever, max stress at base

max
2
6
y
My FL
SF I bh

  

b<d sets upper limit on width, should be less than diameter

Calculations

o

Forces come in at the power angle, only
tangential

part of interest to power
transfer, but normal part exists

Spur and Helical Gears

o

Basic calculation:

o

Fixed
-
free cantilever
, stress at base to yield

o

M = F*l

o

2
6
Fl
bh

where F is tangential load on gear tip, l is height of gear, b with
thickness of whole gear and h is width of tooth (in direction of force)

o

Set this to yield

Bevel and Worm Gears

o

Straight bevel
:
Like spur gear, up to about 1000 ft/mim (5 m/s) otherwise no
ise
too high

o

Spiral bevel
:
For higher speeds

o

Zerol bevel
:

o

Hypoid gear
:

For non
-
intersecting shafts

o

Spiroid gear
:
Halfway between spiral bevel and worm gear

Clutches, Brakes, Couplings, Flywheels

Clutch
-

both sides rotate

Brake
-

only one side rotating, often self
-
locking

Self acting if will increase normal force as load is applied
-

holds better

Internal expanding rim clutches and brakes

o

If no spring, torque transmitted goes with speed^2

External contracting rim clutches and brak
es

o

Solenoids

o

Levers

o

o

Hydraulic

Capstan effect

o

1
2
f
P
e
P

, P1 is load, P2 is holding effort, f is coefficient of friction and phi is
angle of contact

17

Cone clutch
-

uses conical inner surface, slides equivalent surface
over, makes near
radially normal surface contact with only axial motion

o

-
15deg angle best

Must dissipate heat
-

kinetic to thermal

Flexible Mechanical Elements

Belts

o

Long separation

o

Have creep

o

Idler/tension pulley removes slack

o

Like gears, enforce
displacement equality on surface of pulleys

o

Efficiency of around 70
-
98%

If belt crossed on itself, reversing

Can use clutch to shift belt between wheels

Flat belts are made of urethane, reinforced with wire

o

Quiet

o

Efficient

at high speed

o

Large amounts of
power

V belts are smooth, made of fabric

o

Less efficient than flat

o

Can be used together for multiple drive
-

more compact

Timing belts have teeth

o

Cannot slip

o

No speed problems

o

High torque

All must have initial tension, or will slip (except timing)

o

Belt
weight

o

Spring

o

Weighted pulley in belt return path

Flat metal belts

o

Has minimum pulley diameter to avoid breaking

o

Diameter of pulley over belt thickness related to lifetime

Roller chains are like timing belts

o

Want to minimize angle made when going around pu
lley

o

At least 1/3 teeth should be in contact

o

Do not provide constant angular speed
-

angle of component leaving wheel is not
always horizontal

o

High strength

o

High maintenance

o

60
-
80%

Wire rope

o

Regular lay
-

Strands and rope twisted in opposit
e directions

o

Lang lay

same direction, better for abrasion

o

Because of wrapping, elasticity of the rope is much higher than single wire

Pulley efficiencies can easily be around 95%

18

M
otors

Model

o

Torque
2
T rL r
 
  

where τ is shear stress, r is radius, and L is axial length
of motor, so shear stress times area times lever arm

o

Shear stress for dif technologies:

5
-
10 kPa

induction

30
-
40 kPa

high performance perm mag

130
-
140 kPa

large power turbine

o

Power
2
2 2
surf
P T r L rL r A v
       
         

is a product
of torque and rotational velocity w, can be rearranged to give shear stress*rotor
surface area*surface speed, or shear stress*rotor volume*rot speed

Sizing:

Optimal power transfer when impedance of

Should oversize
power rate
-
5 times

account for efficiency

DC motor

synchronous

o

Linear relation from torque to speed

o

No torque max speed

o

No speed max torque

o

Modeled as resistor, inductor and back
emf

in series

o

Brushed motors
-

coils on inside
, permanent magnet on outside

75
-
80% efficient

Low

cost, simple speed control

o

Brushless are vice versa

85
-
95% efficient

Highest performance, long lifespan, low maintenance

o

Axial pole (current runs radially through disk)

o

Brushes break off, add dust to system

o

Burn out possible failure, or over torque to
gears, shaft/bearing failure, wiring
failures

Induction Motor

o

A
sychronous

Rotating magnetic field in stator

Low cost, high power, not exact speed (changes with torque)

Induces current in center which drives conductor out of field

Slip
(due to magnetic fre
q dif from rotor freq
-

asynchronous)
frequency
determines the amount of torque, for small deviations is proportional

Beyond given slip, torque drops off

o

Synchronous

Brushless DC, but driven with sinusoidal current rather than on/off

Synchronous reluctance

o

Many reluctance / solenoid type torque actuators in series, enable semi continuous
torque

Motor Sizing Chart

Horsepower

Size

Cost

AC 1/100

3.3” dia
=
␳A
=
䅃‱A㔰
=
3.3” dia
=
␴A
=
䅃‱A㈵
=
3.3” dia
=
␵A
=
䅃‱A㄰
=
4.4” dia
=
␸A
=
䅃‱AQ
=
5 5/8”
=
␱㄰
=
䅃‱A2
=
5 5/8”
=
␱㌰
=
䅃‱
=
5 5/8”
=
␱㔰
=
䑃‱
=
Q
-
6” dia x1’
=
␶〰
=

19

o

Hard to start

o

Linear version called sawyer motor

Hysteresis

o

Induction in solid metal core
-

magnetic drag

o

Low torque/inertia ratio

o

No ripple

S
tepper

o

Reluctance with rotation sensor

o

Precision positioning

Type

Applications

AC Induction

Che
a
p, High power,
long life

Rotation Slips from
Frequency

Fans and
Appliances

AC Synchronous

Rotation in
-
sync
with frequency,

Long life

Expensive

Clocks, turntables,
tape drives

DC Stepper

Precision
positioning,

High holding torque

Slow speed,

Requires a
controller

Positioning in
printers and floppy
drives

DC Brushless motor

Long lifespan,

low maintenance,

High efficiency

High initial co
st,

Requires a
controller

Hard drives,

CD/DVD players,

electric vehicles

DC Brushed Motor

Low initial cost,

Simple speed
control

High maintenance,

Low lifespan, Slow
speed

exercisers,

automotive starters

Transmissions

o

power from a motor to an actuator or mechanism

o

optimal transmission ratio most efficiently distributes power to the
motor/drivetrain and the load, which are connected and must accelerate together

o

For a motor to create a rotational velocity
motor
TR
J
J
n

o

For friction or belt drive
motor
Pulley
M
J
r

o

motor
M
J
J

20

Pump
s

Displacement

o

Constant flow machines
-

indep of pressure

Must therefor have safety if
exit blocked

o

Rotary dra
w fluid through motion of piston

Gear
(lobe)
pumps
-

two gears meshed

Screw pumps
-

two scre
w
s with opposing thread, moving in opposite
rotations, parallel to one another
, progressive cavity pump

Moving vane
-

cylinder in housing with
little gap, rotation draws fluid

o

Reciprocating

Expanding cavity on suction side, decreasing cavity on discharge side

Piston (plunger) pulls fluid in one step, pushes it out other valve in second
step

Dynamics waste a bit of energy

D
iaphragm

Peristaltic
-

m
edical type, squeeze tube to push fluid through tube

Hydraulic ram
-

passive uses some fluid from input to hammer small
amount of the fluid up a gravitational gradient
-

siphons energy from flow
and condenses it into small amount. Water hammer effect drives

it up
tube with one
-
way valve.

Dynamic

Displacement Pumps (Piston, Gear)

21

o

Pressure based, can withstand blockages

Lower efficiencies, but less
maintenance

o

Centrifugal pump
-

spings the water out, pressurize

Enters near rotating axis, spins out radia
l

If just flows along

axis, ‘axial’, like propeller

>80% of all pumps are centrifugal

o

Screw centrigual impeller
-

single blade, axially extended at inlet, like a corkscrew

Efficiency is around 50
-
80%, power imparted to fluid vs. power into pump

o

Efficiency reaches peak at midway

through operating range of flow rate

o

Power to fluid is

Power PQ gH Q

 

output power Po,
pressure P,
density
ρ
, height gained H, gravity g, flow rate Q (m
3
/s)

Design considerations

o

Cost

o

Flow rate

o

Contamination

o

Efficiency

o

Viscosity

o

Duty
-
cycle

o

Ene
rgy source (AC, DC, hydraulic, air, gas or diesel engine, steam, etc.)

o

Size

o

o

Maintenance

Actuators

electromagnetic rotary motor

iBNlr
lr
B
i
2

electromagnetic linear motor

iBNl
F
l
B
i
F

electromagnetic reluctance:
2 2
2
2
0
N I
μ
F A
h

Electrostatic
2
2
1
r
q
q
k
F
e

chemical (internal combustion, rocket)

piezoelectric

hydraulic linear
in
in
out
out
F
A
A
F

pneumatic (linear & rotary)

thermal actuator

conduction polymers

shape memory alloy

screw

friction drive

rack and pinion

Centrifugal pump

22

Concepts:

Saint
-
Venant’s Principal

Several characteristic dimensions away from an effect the
effect is essentially dissipated

Design Process

o

Functional Requirements

o

Design Parameters

o

Analysis

o

References

o

Risks

o

Countermeasures

/Ballscrews

-
8
0%

Ballscrew efficiency can be up to
80
-
95%

Must have some clearance to avoid binding
-

results in backlash

o

At high frequency can vibrate

Can stack

ElectroMagnetic
Equations

d
d N
dt

 

E l

d Ni
 

H l

Ampere’s Law

0
d
 

B s

Gauss’s Law

N
 

Li

cross domain inductance

d
V
dt

2
N
L
R

2
1
2
E LI

dE
F
dx

Fluidic Equations

2
C L
P RQ Q
A

 
  
 
 

Fluidic resistance R relates pressure drop ΔP to volumetric flow
rate Q through length of pipe L, viscosity μ and pipe cross secti
onal area A

o

8
C

for round tubes

o

35
C

for triangular tubes

23

2
2
v
F C A

Air drag force, scales with density ρ, cross sectional area A, relative
velocity v and with coefficient corresponding to aerodynamic shaping

o

1
C

for cylinder

o

1
C

for aerodynamic shapes

o

1
C

for flat plates

Error Calculation with HTM

R
r n
v v

coordinates in reference
(r)
as outputs, local frame
(n)
between frames to local frame

rTipActual rTipIdeal
e = v v

error calculated by taking final operating

point (tip) calculation and
subtracting the no
-
error prediction (ideal) from the full prediction

1 0 0
0 1 0
0 0 1
0 0 0 1
x
y
z
 
 
 

 
 
 
x/y/z
R

linear translation, these numbers (x,y,z) describe the
location of
the origin of the n
-
frame with respect to the origin of the
reference (r
-
frame)

1 0 0 0
0 cos sin 0
0 sin cos 0
0 0 0 1
x x
x x
 
 
 
 

 

 
 
 
x
R

positive rotations θ around the x
-
axis.

cos 0 sin 0
0 0 0 0
sin 0 cos 0
0 0 0 1
y y
y y
 
 
 
 
 

 

 
 
y
R

positive rotations θ around the y
-
axis.

cos sin 0 0
sin cos 0 0
0 0 1 0
0 0 0 1
z z
z z
 
 

 
 
 

 
 
 
z
R

positive rotations θ around the z
-
axis.

Sensors

Capacitance

Hall effect se
nsor

Inclinometer

Inductive sensor

LVDT

Magnetic Scale

Magnetostricti
ve

Sensor

Mechanical switch

Piezoelectric

24

Potentiometer

Synchro (measures rotor location, like ac motor)

Ultrasonic

Velocity

o

LVT

o

Tachometer

Optical

o

Interferometer

o

O
ptical encoder

o

Laser
triangulation

o

photoelectric

Engines

Internal Combustion

o

Description:

chemical
-
mechanical transformer

oxidization of fuel occurs in confined space (combustion chamber)

exothermic, high pressure, temp

chamber expands, generating motion

-
75 Hp/liter

o

Four stroke

Intake
-

Combustible mixtures are emplaced in the combustion chamber

Compression
-

The mixtures are placed under pressure

Combustion
-

The mixture is burnt, the hot mixture is expanded, pressing
on and moving parts of the
engine and performing useful work. Ignition
can either be a spark ignition or a compression ignition where ignition
relies solely on heat and pressure created by the engine in its compression
process.

Exhaust
-

The cooled combustion products are exhausted

o

Two stroke

Intake and compression in down stroke

Combustion and exhaust in up stroke

Simpler, smaller, lighter for given power output

Less efficient, more polluting

Gas Turbine

o

Description

Rotary, extracts energy from continuous flow of combustion gas

effi
-
50%

high power/weight ratio

continuous flow, easy to scale up power generation

much more complex and expensive than IC

o

Brayton cycle:

isentropically compressed air

combustion at constant pressure

expansion over turbine isentropically back
to ambient pressure

25

Jet Engine

o

Expansion of gas used to generate thrust

o

Similar layout to gas turbine but exhaust forced through nozzle

Human Factors and Conversions

Person

o

Generate about 0.25 to 0.75 Hp or 200 to 600 Watts

o

1
-
2 kW for few seconds, peak
athletes

o

o

-
30% of body mass

o

Walk at 3
-
5 mph

o

Max speed about 5 Hz for pedaling, other motions

Low force at this and higher

Max power at around 1
-
2 hz

Useful facts

o

Energy in gasoline, 45 MJ/kg

o

Internal resistance of battery start
ohm for car batteries

Standard to metric conversion

o

3ft

=
1 m

o

2
mph =
1
m/s

o

2.2lbs = 1kg

o

1HP

= 550 ft lbs/s
= 33000 ft lbs/min
= 746W

o

1lbf = 4.45N

o

1in = 0.0254m

o

1psi =

7000Pa

o

1mile = 1.61km

o

1gal = 4L

o

10 rpm =