Difference between revisions of "Suspension Forces"
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* Compliance analysis | * Compliance analysis | ||
| − | == | + | ==Analysis of Only the Push/Pull Rod (Incorrect Method)== |
| − | Some teams have assumed that the vertical force of the tire at the contact patch is exactly equal to the vertical component of the push/pull rod force, and used the component forces / similar triangles method to calculate the force in the push/pull rod, ignoring the additional forces of the other 5 suspension tubes. '''This is incorrect''', and can underestimate forces by a factor of 2 or more. The method is most inaccurate on pull-rod suspension. Here, the upper wishbone applies an additional vertical force to the wheel assembly, which increases the loads on the pull-rod. [[File:Pullrod Forces2.png|right|middle|thumb|Upper Wishbone Increasing Forces on Pull-rod]] | + | Some teams have assumed that the vertical force of the tire at the contact patch is exactly equal to the vertical component of the push/pull rod force, and used the component forces / similar triangles / Trigonometry method to calculate the force in the push/pull rod, ignoring the additional forces of the other 5 suspension tubes. '''This is incorrect''', and can underestimate forces by a factor of 2 or more. The method is most inaccurate on pull-rod suspension. Here, the upper wishbone applies an additional vertical force to the wheel assembly, which increases the loads on the pull-rod. [[File:Pullrod Forces2.png|right|middle|thumb|Upper Wishbone Increasing Forces on Pull-rod]] |
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==Static Free Body Diagram 6x6 Matrix Method== | ==Static Free Body Diagram 6x6 Matrix Method== | ||
| − | Numerous papers on this method exist. A google search for "Formula SAE Suspension Forces Matrix" shows a couple, or check the [[Suspension Forces#See Also|See Also]] | + | Numerous papers on this method exist. A google search for "Formula SAE Suspension Forces Matrix" shows a couple, or check the [[Suspension Forces#See Also|See Also]] Section. The rest of this section provides only a summary with a few pictures. The video provides a detailed explanation in a very easy to understand format. |
| − | + | A ready-to-fill-in excel file, which accompanies the video is available at [https://github.com/fsaeillumina/suspension-forces https://github.com/fsaeillumina/suspension-forces] {{#ev:youtube|https://youtu.be/rhqZQ9Lic2o|600|center|Formula SAE Suspension Forces Matrix Video|frame}} | |
| − | + | ===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 is ignored here for simplicity. | ||
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| + | ===Inboard vs outboard drive/brake=== | ||
| + | The forces through the suspension members also depend on the brake/drive setup. It is crucial to consider this in the calculation to properly represent the load paths. | ||
| − | + | For an inboard drive/brake, the driveshaft helps and reacts the moment that originates from the contact patch as the driveshaft itself is being stressed. For outboard drive/brake, the driveshaft is not stressed, and cannot transmit the torque like in the inboard scenario. Therefore, with an inboard layout, the driveshaft acts like a 7th link and typically reduces the loads going through the other suspension links. This can be compensated for in the calculation by the additional moment, M<sub>y</sub>, when analysing inboard drive/brakes. | |
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===Theory=== | ===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. [[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]] | ||
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===Solving=== | ===Solving=== | ||
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A summary of the solution is given here. It is a decently long process. It is recommended the reader follow through a paper, or watch a video. | A summary of the solution is given here. It is a decently long process. It is recommended the reader follow through a paper, or watch a video. | ||
| − | 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: | + | 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:Suspension Forces 1.png|right|middle|thumb|Breaking up Force Vectors into x, y, z components]] |
| − | + | [[File:Equations With Highlights.png|right|middle|thumb|Force Equations with Unit Vectors Highlighted in Red and Unknowns Highlighted in Orange]] | |
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| − | Calculating the moments from each tube is best done in vector form as the cross product of the moment arm vector and the force vector. This will result in a moment vector that has a Mx, My, and Mz components, which can be placed into their respective " | + | Calculating the moments from each tube is best done in vector form as the cross product of the moment arm vector and the force vector. This will result in a moment vector that has a Mx, My, and Mz components, which can be placed into their respective <math display="inline"> \sum M_x = 0, \sum M_y = 0, \sum M_z = 0,</math> equation. |
The 6 equations can be rearranged into a matrix equation of the form A*X=b, which can be solved easily with linear algebra. | The 6 equations can be rearranged into a matrix equation of the form A*X=b, which can be solved easily with linear algebra. | ||
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[http://scholarworks.csun.edu/bitstream/handle/10211.3/123383/Flickinger-Evan-thesis-2014.pdf;sequence=1 DESIGN AND ANALYSIS OF FORMULA SAECAR SUSPENSION MEMBERS] | [http://scholarworks.csun.edu/bitstream/handle/10211.3/123383/Flickinger-Evan-thesis-2014.pdf;sequence=1 DESIGN AND ANALYSIS OF FORMULA SAECAR SUSPENSION MEMBERS] | ||
| + | [http://www.iaeme.com/MasterAdmin/uploadfolder/IJMET_07_02_018/IJMET_07_02_018.pdf FORCE CALCULATION IN UPRIGHT OF A FSAE RACE CAR] | ||
| − | + | [[Category:Suspension]] | |
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Latest revision as of 11:40, 9 April 2024
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
Analysis of Only the Push/Pull Rod (Incorrect Method)
Some teams have assumed that the vertical force of the tire at the contact patch is exactly equal to the vertical component of the push/pull rod force, and used the component forces / similar triangles / Trigonometry method to calculate the force in the push/pull rod, ignoring the additional forces of the other 5 suspension tubes. This is incorrect, and can underestimate forces by a factor of 2 or more. The method is most inaccurate on pull-rod suspension. Here, the upper wishbone applies an additional vertical force to the wheel assembly, which increases the loads on the pull-rod.
Static Free Body Diagram 6x6 Matrix Method
Numerous papers on this method exist. A google search for "Formula SAE Suspension Forces Matrix" shows a couple, or check the See Also Section. The rest of this section provides only a summary with a few pictures. The video provides a detailed explanation in a very easy to understand format.
A ready-to-fill-in excel file, which accompanies the video is available at https://github.com/fsaeillumina/suspension-forces
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 is ignored here for simplicity.
Inboard vs outboard drive/brake
The forces through the suspension members also depend on the brake/drive setup. It is crucial to consider this in the calculation to properly represent the load paths.
For an inboard drive/brake, the driveshaft helps and reacts the moment that originates from the contact patch as the driveshaft itself is being stressed. For outboard drive/brake, the driveshaft is not stressed, and cannot transmit the torque like in the inboard scenario. Therefore, with an inboard layout, the driveshaft acts like a 7th link and typically reduces the loads going through the other suspension links. This can be compensated for in the calculation by the additional moment, My, when analysing inboard drive/brakes.
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
A summary of the solution is given here. It is a decently long process. It is recommended the reader follow through a paper, or watch a video.
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.
Calculating the moments from each tube is best done in vector form as the cross product of the moment arm vector and the force vector. This will result in a moment vector that has a Mx, My, and Mz components, which can be placed into their respective equation.
The 6 equations can be rearranged into a matrix equation of the form A*X=b, which can be solved easily with linear algebra.
