Finite Element Analysis of the Finite Element Analysis of the Steel Horse Rear SuspensionSteel Horse Rear Suspension

Presented by:Presented by:

Erika RamirezErika Ramirez

April 29, 2003April 29, 2003

Problem StatementProblem Statement

The rear suspension of The rear suspension of any mini baja car must be any mini baja car must be able to withstand the able to withstand the rough terrains associated rough terrains associated with the competition’s with the competition’s motor cross track. motor cross track.

The objective of this The objective of this project is to successfully project is to successfully analyze the rear analyze the rear suspension with an suspension with an impact load at the rear of impact load at the rear of

the tire.the tire.

Problem FormulationProblem Formulation

The rear suspension The rear suspension to be analyzed is a to be analyzed is a single a-arm. It is single a-arm. It is made out of 4130 made out of 4130 Chromoly tubing with Chromoly tubing with an outer diameter of an outer diameter of 1” and thickness 1” and thickness of .058”.of .058”.

Impact Force Impact Force Impact Force Impact Force Another car traveling at 30 mph hits our car and comes to rest at one Another car traveling at 30 mph hits our car and comes to rest at one

second or the car hits an object traveling at 30MPH form the side:second or the car hits an object traveling at 30MPH form the side:

Vi=30 MPH Vi=30 MPH Vf=0 MPH Vf=0 MPH

t=1s t=1s m=600lb m=600lb

Vf=Vi+atVf=Vi+at

a(impact)=Vf-Vi/ta(impact)=Vf-Vi/t

F=m*a(imp)F=m*a(imp)

F = 820 lbfF = 820 lbf

Worst Case ScenarioWorst Case Scenario

The car lands in one tire.The car lands in one tire.Weight of the vehicle (600 lbs)Weight of the vehicle (600 lbs)

The tire hits a rock, bump, etc.The tire hits a rock, bump, etc. Use the 820 lbf loadUse the 820 lbf load

Use an Actuator to model the motor – cross Use an Actuator to model the motor – cross tracktrack

Material PropertiesMaterial Properties

DensityDensity .284 lb/in^3.284 lb/in^3

Modulus of ElasticityModulus of Elasticity205 GPa205 GPa

Yield StressYield Stress8e8 Pa8e8 Pa

UTSUTS9.65e9 Pa9.65e9 Pa

Pro-E ModelPro-E Model

REVOLUTE JOINTS

ACTUATOR

IMPACT FORCE = 820 lb

SPRING/DAMPER (SHOCK) – 175 lb/in and C=10.2

Finite Element Modeling (Stresses)Finite Element Modeling (Stresses)

Concentration of Stresses located at the revolute joints. Concentration of Stresses located at the revolute joints.

DisplacementsDisplacements

Max Displacement occurs at the Hub tabs near the tire – also at one of the revolute joints.

Factor of SafetyFactor of Safety

Making the ends .01” thickerMaking the ends .01” thicker

Von Misses Stresses

Modified DisplacementsModified Displacements

Modified Factor of SafetyModified Factor of Safety

Comparing StressesComparing Stresses

The Maximum Principle Stresses are greatly reduced by making the revolute joints .01 thicker.

Comparing DisplacementsComparing Displacements

Displacements were increased a little from .75891 to .82149 in.

Von Misses StressesVon Misses Stresses

Von Misses stresses were also reduced.

Comparing Factor of SafetyComparing Factor of Safety

Minimum Factor of Safety was increased – and there are less areas of concern.

Comments / SummaryComments / Summary

H-Adaptivity was not used because of the H-Adaptivity was not used because of the complexity of the part.complexity of the part.

The Mesh size used was .02 Feet, this The Mesh size used was .02 Feet, this was done after converging on the same was done after converging on the same maximum value of stresses (no significant maximum value of stresses (no significant change for a smaller mesh size). change for a smaller mesh size).

A simple way to reduce the stress A simple way to reduce the stress concentrations would be to introduce more concentrations would be to introduce more material on the revolute ends. material on the revolute ends.

Questions?Questions?