Team Flying Sheep Engineering Analysis Mark Berkobin John Nevin John Nott Christian Yaeger Michelle Rivero

System Review: Solid Edge Model

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 TorsionFBD 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 = 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 = ft Bending Stress = (Normal Stress)/(2) 0.5 Bending Stress = M/(W*a*(a+D)) = 5250/(14*0.3125*( )) = 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 = 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 = ft Bending Stress = (Normal Stress)/(2) 0.5 Bending Stress = M/(W*a*(a+D)) = 3450/(14*0.3125*( )) = 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 Supports Solid Edge Model

Horizontal Support Bending Analysis For the frame sliders side bar 600 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 = ((L outter ) 4 –(L inner ) 4 )/3 = ((1.5) 4 -(1.25) 4 )/3 =.873 in 4 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 beams 600 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 in 2 Maximum Allowable Shear Stress = 24.1 Kpsi Moment Max stress = P/A = 600/.5625 = 1,066 Safety Factor = Max Stress/Maximum Allowable Stress = 24,100/1,066 = 22.6

Vertical Support Solid Edge Model

Bike Interface Description Solid Edge Model Free Body Diagram

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 in 4 Max stress = (1,950*.75)/ /(.2*.5) = 7,675 psi Max allowable Stress = 36 kpsi Safety Factor = Max Stress/Maximum Allowable Stress = 36,000/7,675 = 4.69

Description of Ratcheting Mechanim Solid Edge Model Free Body Diagram

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

Description of Load Bearing Bolts Solid Edge Model In Tension In shear Free Body Diagram

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 kpsi150 kpsi130 kpsi

Bolt Analysis-Shear For Coarse ½” diameter bolt At= in 2 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

Questions?