Transcript ASME Design Competition

Group 11- ASME Design Competition
• Alicia Christie
• Desmond Bourgeois
• Toddrick Ruff
ASME Design Competition
Rule Update
•Rocks do need to touch
receiving area
•1 rock in each circle
Schematic of Test Course
2 – 1700 pt rocks
1 – 1600 pt
2 - 1500pt
2 – 1000pt
Score
S = Σ (R*t) +1000P - W - A - 1000T – 5s
R = designated rock score
t = target multiplier
W = Weight of the vehicle in grams
A = milliamp-hours available to the device according
to the battery labels
T = Times device touches border tape
s = seconds to complete task, maximum 240
P = bonus for parking vehicle at end of task (1 =
parked, 0 = not parked)
Decision from Last Presentation
3 – Wheel Design
Street Sweeper
Mechatronics, Ariel University Center
Design Concept
rotate
move
Must be in an equilateral
triange configuration for
symmetry
3.500
L
2*cos(30)[r+s/2]
r
4.000
s
r + 2*cos(30)*[r+s/2] = 3.75 in
Wheel set distance D = 4r + s : if r = 1 in, s = 1.18 in, D = 5.18 in
D*2 + 4 in (box size with guides in back) = 14.36 in
with wheels and box touching
Max length is 14.5 in
Design Concept
?
3.5-r
3.50
d
r
r
4.00
d
2
4  ( 3.5  r)
2
in
Final Design
•Wheels are offset
with front wheels
on inside
barrier
barrier
Idea for body expansion
• Switch blades
Approximate
Extension Length
Interim Prototype
3 Wheel Operating Principle
Exploded View
Assembled View
Gear Force Analysis
Assumptions
Radial
Tangential
Axial
•Const. load with uniform shock
•Const. mesh with no backlash
•Axial load is negligible
•Qv < 5, low pinion velocity < 13m/s
•99% reliability
•Full-depth teeth with tip loading
• Pressure angle 20º
•Operating temps < 200º
• Teeth form standard AGMA profile
Gear Force Analysis
Given
TP 
H
p
  20 deg
H  .121hp
.p  250 rpm
TP  3.447 N m
TP  2.542 ft  lbf
Resulting Forces
TP
wt
wr  wt tan   
wresultant 
wt  229.768 N
wr  83.629 N
wresultant  244.514 N
wt  51.654 lbf
wr  18.8 lbf
wresultant  54.969 lbf
wt 
dp
cos   
2
Rock Collector
Free Body Diagram
1lb
Position #1
• Assumptions
– Weight of box = 2lbs
– Weight of Front = 7lbs
– Tipping will not occur
on back wheel
∑M = ∑F*d
3lb
3lb
2lb
M  T ;WBox  d1  cos( )  WFront  d 2  cos( )  T
3.5in
1.5in
5.9in
10.225in
M = 34.04 lb-in, moment created by rock collector would have to
exceed this limit to tip over giving a negative moment
Cost Analysis
Cost Analysis
Current Total: $943.70
Future Plan
• FEM Analysis to determine ways
to remove material
•Amount of motors, gears,
other control components
•Exact gear placement to minimize
motor use
•Program control components
•Power input and velocity
calculations
•Exact placement of wheels and
expandable body to fit within size
requirement
•Placement of batteries and
motors
•Control configuration