Difference between revisions of "Suspension Forces"
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* Compliance analysis | * Compliance analysis | ||
| + | ==Similar Triangles of Only the Push/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 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]] | ||
==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]] | ||
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| + | Section. | ||
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===Assumptions=== | ===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. | 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 | + | Acceleration of the wheel assembly is ignored ignored here for simplicity. |
===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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| + | <a class="image"><img alt="" src="/images/thumb/7/7d/Free_Body_Diagram.png/300px-Free_Body_Diagram.png" decoding="async" class="thumbimage" srcset="/images/thumb/7/7d/Free_Body_Diagram.png/450px-Free_Body_Diagram.png 1.5x, /images/thumb/7/7d/Free_Body_Diagram.png/600px-Free_Body_Diagram.png 2x" width="300" height="243"></a> <a class="internal" title="Enlarge"></a>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]] 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 /> | + | [[File:Suspension Forces 1.png|right|middle|thumb|Breaking up Force Vectors into x, y, z components]] |
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| + | <a class="image"><img alt="" src="/images/thumb/d/d5/Suspension_Forces_1.png/300px-Suspension_Forces_1.png" decoding="async" class="thumbimage" srcset="/images/thumb/d/d5/Suspension_Forces_1.png/450px-Suspension_Forces_1.png 1.5x, /images/thumb/d/d5/Suspension_Forces_1.png/600px-Suspension_Forces_1.png 2x" width="300" height="269"></a> <a class="internal" title="Enlarge"></a>Breaking up Force Vectors into x, y, z componentsA 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. | ||
| + | |||
| + | |||
| + | 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]] | ||
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| + | <a class="image"><img alt="" src="/images/thumb/b/ba/Equations_With_Highlights.png/300px-Equations_With_Highlights.png" decoding="async" class="thumbimage" srcset="/images/thumb/b/ba/Equations_With_Highlights.png/450px-Equations_With_Highlights.png 1.5x, /images/thumb/b/ba/Equations_With_Highlights.png/600px-Equations_With_Highlights.png 2x" width="300" height="80"></a> <a class="internal" title="Enlarge"></a>Force Equations with Unit Vectors Highlighted in Red and Unknowns Highlighted in Orange<br /> | ||
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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 "Sum of moments in x/y/z = 0" equation. | ||
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| + | The 6 equations can be rearranged into a matrix equation of the form A*X=b, which can be solved easily with linear algebra. | ||
| + | ==See Also== | ||
| + | [http://scholarworks.csun.edu/bitstream/handle/10211.3/123383/Flickinger-Evan-thesis-2014.pdf;sequence=1 DESIGN AND ANALYSIS OF FORMULA SAECAR SUSPENSION MEMBERS] | ||
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| + | [http://www.iaeme.com/MasterAdmin/uploadfolder/IJMET FORCE CALCULATION IN UPRIGHT OF A FSAE RACE CAR] | ||
Revision as of 17:37, 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
Similar Triangles of Only the Push/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 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.
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 ignored here for simplicity.
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.
<a class="image"><img alt="" src="/images/thumb/7/7d/Free_Body_Diagram.png/300px-Free_Body_Diagram.png" decoding="async" class="thumbimage" srcset="/images/thumb/7/7d/Free_Body_Diagram.png/450px-Free_Body_Diagram.png 1.5x, /images/thumb/7/7d/Free_Body_Diagram.png/600px-Free_Body_Diagram.png 2x" width="300" height="243"></a> <a class="internal" title="Enlarge"></a>Free Body Diagram Showing 3 of 6 Suspension Arms
Solving
<a class="image"><img alt="" src="/images/thumb/d/d5/Suspension_Forces_1.png/300px-Suspension_Forces_1.png" decoding="async" class="thumbimage" srcset="/images/thumb/d/d5/Suspension_Forces_1.png/450px-Suspension_Forces_1.png 1.5x, /images/thumb/d/d5/Suspension_Forces_1.png/600px-Suspension_Forces_1.png 2x" width="300" height="269"></a> <a class="internal" title="Enlarge"></a>Breaking up Force Vectors into x, y, z componentsA 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.
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.
<a class="image"><img alt="" src="/images/thumb/b/ba/Equations_With_Highlights.png/300px-Equations_With_Highlights.png" decoding="async" class="thumbimage" srcset="/images/thumb/b/ba/Equations_With_Highlights.png/450px-Equations_With_Highlights.png 1.5x, /images/thumb/b/ba/Equations_With_Highlights.png/600px-Equations_With_Highlights.png 2x" width="300" height="80"></a> <a class="internal" title="Enlarge"></a>Force Equations with Unit Vectors Highlighted in Red and Unknowns Highlighted in Orange
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 "Sum of moments in x/y/z = 0" 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.
See Also
DESIGN AND ANALYSIS OF FORMULA SAECAR SUSPENSION MEMBERS
