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Team Flying SheepEngineering Analysis
Mark BerkobinJohn NevinJohn Nott
Christian YaegerMichelle Rivero
Possible Points of Failure for Analysis Welds between horizontal supports and tilt
mechanism
Horizontal Supports in Bending and Shear
Vertical Supports in bending and shear
Bike Interface in bending and shear
Ratcheting Mechanism
Load Bearing Bolts
Not Analyzed Stability: System is Constrained by AISI 1018 HR Low
Carbon Steel parts, stability should not be an issue
Fatigue Analysis: It is not expected that the system will be used more than once every several hours. Fatigue should not be an issue
Human interface: Is built into the mechanism of a popular, proven jack system incorporated into the system
All failure points on the jack itself: The jack is a proven commercial product that we are incorporating into the system, we assume it will continue not to fail
Welds Between Horizontal Supports and the Tilt Mechanism on the Base
FBD For Welds in Torsion FBD for Welds in Shear
Weld Analysis in Torsion Maximum length of from point of force application to weld: 8.75 inches
Length of Weld: 14 inches = 1.17 ft
Height of welded part (D): 1.5 inches = 0.125 in
Maximum Force at point of application = 600 lb
Maximum possible moment: 600 lb * 8.75 = 5250 lb-in
Weld thickness (a value) = 5/16 inch = 0.0260 ft
Bending Stress = (Normal Stress)/(2)0.5
Bending Stress = M/(W*a*(a+D)) = 5250/(14*0.3125*(.3125+1.5)) = 662 psi
Assume: Welds are Equivalent Strength to Steel: Max Allowable Stress = 36 Kpsi
Safety Factor = 36,000 psi/662 psi = 54.4
Weld Analysis in Shear Maximum length of from point of force application to weld: 8.75
inches Length of Weld: 14 inches = 1.17 ft Height of welded part (D): 1.5 inches = 0.125 in Maximum Force at point of application = 600 lb Maximum possible moment: 600 lb * 5.75 = 3450 lb-in Weld thickness (a value) = 5/16 inch = 0.0260 ft Bending Stress = (Normal Stress)/(2)0.5
Bending Stress = M/(W*a*(a+D)) = 3450/(14*0.3125*(.3125+1.5)) = 435 psi
Assume: Welds are Equivalent Strength to Steel: Max Allowable Shear Stress = 24.1 kpsi
Safety Factor = 24,100 psi/435 psi = 55.4
Horizontal Support Bending Analysis
For the frame sliders side bar600 lbs
17 inches, from center point;
Assume system is symmetrical around the midpoint
Max stress = M*c/I
M = r X F =17 in x 600 lbs = 10,200 lb-in
c = 0.75
I = ((Loutter) 4 –(Linner)4)/3 = ((1.5)4-(1.25)4)/3
= .873 in4
Max stress = (10,200*.75)/.873 = 8,762 psi
Max allowable Stress = 36 kpsi
Safety Factor = Max Stress/Maximum Allowable Stress
= 36,000/8,762 = 4.11
Horizontal Support Shear Stress Analysis
At the support beams600 lbs
Cross Sectional Area
Side of square tube: 1.25 in
Thickness:1/8 inches
Area affected in shear:
A=(1.25)2-(1.25-2(.125*2))2
A=.5625 in2
Maximum Allowable Shear Stress = 24.1 KpsiMoment
Max stress = P/A = 600/.5625 = 1,066
Safety Factor = Max Stress/Maximum Allowable Stress = 24,100/1,066 = 22.6
Vertical Support Bending Analysis For purposes of this analysis the bike interface with the foot pegs (half tube) and the
frame sliders (whole tube) will be treated as the same in terms of failure
Max stress = M*c/I
M = r X F = 3.25 in x 600 lbs = 1,950 lb-in
c = 0.75 I = ((Loutter) 4 –(Linner)4)/3 = ((1.5)4-(1.25)4)/3
= .873 in4
Max stress = (1,950*.75)/.873 +600/(.2*.5) = 7,675 psi
Max allowable Stress = 36 kpsi
Safety Factor = Max Stress/Maximum Allowable Stress
= 36,000/7,675 = 4.69
Ratcheting Analysis
Design Characteristics
Tooth height is .2 inches due to the need for adjustability
Tooth width is variable to meet requirements, currently planned at 0.5 inches
Mode of failure
Pressure of bike will be straight down on the tooth, tooth likely to fail in shear
Worst Case scenario: all of the bike’s weight resting on one “tooth”
Ratcheting Analysis-Continued Maximum Shear Stress: = F/A =
600/(.2*.5) = 6000 psi = 6 kpsi
Maximum Allowable Shear Stress for AISI 1018 HR Low Carbon Steel: 32 kpsi
Safety Factor: (32 kpsi)/(6kpsi) = 5.34
Bolt Analysis Bolts Used throughout this project have been
chosen as ½” diameter SAE Grade 8 bolts
Grade 8 bolts have the following properties:
Minimum Proof Strength
Minimum Tensile Strength
Minimum Yield Strength
120 kpsi 150 kpsi 130 kpsi
Bolt Analysis-Shear For Coarse ½” diameter bolt
At=0.1419 in2
Proof Load
Fp=AtSp
Fp=17028 lbs
Maximum expected Fp = 600 lb (weight of the motorcycle)Safety factor: 17028/600 = 23.4
Bolt Analysis-Tension
T=FL/HA
L= 1.90 inches H = 3.99 inches
T = (600*1.90)/(3.99*0.1419) = 2,013 psi
Safety Factor = 150/2.013 = 74.5