Difference between revisions of "Suspension Forces"
FsaeIllumina (talk | contribs) |
FsaeIllumina (talk | contribs) |
||
| Line 29: | Line 29: | ||
Apply sum of forces equals zero (Fx, Fy, and Fz are 3 equations), and sum of moments (torques) equals zero (Mx, My, and Mz are 3 equations) to the wheel assembly. Break up all 6 suspension tube force vectors into their x, y, and z components multiplied by the unknown magnitude of the force in each arm. The 6 equations and 6 unknowns form a solvable 6x6 linear system. [[File:Free Body Diagram.png|right|middle|thumb|Free Body Diagram Showing 3 of 6 Suspension Arms]] | Apply sum of forces equals zero (Fx, Fy, and Fz are 3 equations), and sum of moments (torques) equals zero (Mx, My, and Mz are 3 equations) to the wheel assembly. Break up all 6 suspension tube force vectors into their x, y, and z components multiplied by the unknown magnitude of the force in each arm. The 6 equations and 6 unknowns form a solvable 6x6 linear system. [[File:Free Body Diagram.png|right|middle|thumb|Free Body Diagram Showing 3 of 6 Suspension Arms]] | ||
===Solving=== | ===Solving=== | ||
| − | [[File:Suspension Forces 1.png|right|middle|thumb|Breaking up Force Vectors into x, y, z components]] | + | [[File:Suspension Forces 1.png|right|middle|thumb|Breaking up Force Vectors into x, y, z components]] Breaking up Force Vectors into x, y, z components. First determine the x, y, z components of each of the 6 tubes. In vector form, this is the same as the unit vectors of each tube, multiplied by the magnitude of the force of each arm. The latter of which will be left as a variable because it is unknown. [[File:Equations With Highlights.png|right|middle|thumb|Force Equations with Unit Vectors Highlighted in Red and Unknowns Highlighted in Orange]]<br /> |
Revision as of 17:00, 31 December 2020
Contents
Forces in 6 Suspension Tubes Per Corner
In double wishbone suspension systems typically seen on FSAE cars, there are 6 tubes connecting the wheel assembly to the vehicle:
- Upper Wishbone, Fore
- Upper Wishbone, Aft
- Lower Wishbone, Fore
- Lower Wishbone, Aft
- Push/Pull Rod or Spring/Damper (direct suspension)
- Toe Rod or Steering Tie Rod
Applications
- Proper design and selection of suspension tubes
- Proper design of upright / knuckle
- Better understanding of forces during different loadcases
- Proper design of bellcrank
- Proper design of mounting brackets of suspension tubes onto chassis
- Reduce failures, while keeping weight low
- Compliance analysis
Static Free Body Diagram 6x6 Matrix Method
Assumptions
Because all of these have spherical bearings on both ends, they are two-force-members, so they will only see tension/compression forces. If the push/pull rod is mounted to a control arm, then it will introduce bending forces in that control arm. In order to calculate the axial forces in all 6 tubes, it will be assumed that they are all two-force members.
Acceleration of the wheel assembly can be ignored.
Theory
Apply sum of forces equals zero (Fx, Fy, and Fz are 3 equations), and sum of moments (torques) equals zero (Mx, My, and Mz are 3 equations) to the wheel assembly. Break up all 6 suspension tube force vectors into their x, y, and z components multiplied by the unknown magnitude of the force in each arm. The 6 equations and 6 unknowns form a solvable 6x6 linear system.
Solving
Breaking up Force Vectors into x, y, z components. First determine the x, y, z components of each of the 6 tubes. In vector form, this is the same as the unit vectors of each tube, multiplied by the magnitude of the force of each arm. The latter of which will be left as a variable because it is unknown.
