Engineering question
#1
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Engineering question
many of you know where i used to work, and the engineers i used to work with. i learned a lot about manufacturing, materials, fabrication, and design from them (eventhough my background is pure IT). when i asked every single one of them about design (when it comes to things like the chassis braces, strut bars, etc.), every single one of them said use hollow material when possible as opposed to solid pieces. supposedly you have more (engineers help me out here please) strength with a tube than with a solid bar stock in those types of applications - i don't quite remember the argument behind that. can anyone explain? RT? Chris? UL?
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Originally posted by mingster
many of you know where i used to work, and the engineers i used to work with. i learned a lot about manufacturing, materials, fabrication, and design from them (eventhough my background is pure IT). when i asked every single one of them about design (when it comes to things like the chassis braces, strut bars, etc.), every single one of them said use hollow material when possible as opposed to solid pieces. supposedly you have more (engineers help me out here please) strength with a tube than with a solid bar stock in those types of applications - i don't quite remember the argument behind that. can anyone explain? RT? Chris? UL?
many of you know where i used to work, and the engineers i used to work with. i learned a lot about manufacturing, materials, fabrication, and design from them (eventhough my background is pure IT). when i asked every single one of them about design (when it comes to things like the chassis braces, strut bars, etc.), every single one of them said use hollow material when possible as opposed to solid pieces. supposedly you have more (engineers help me out here please) strength with a tube than with a solid bar stock in those types of applications - i don't quite remember the argument behind that. can anyone explain? RT? Chris? UL?
In a member subject to bending, there is a line called the neutral axis, where the material is neither in tension or compression. Material in compression or tension provides resistance to bending and the further away from the neutral axis, the better. This is a geometric relationship.
A good analogy is comparison to a lever; the further away from the fulcrum, you receive more mechanical advantage.
By removing material close to the neutral axis (the center of a tube) you remove material that wasn't contributing to the member's strength with the benefit of weight reduction.
This is also why I-beams are shaped like they are.
Gowgom
#4
Originally posted by Fed
Damnit, I know I should have listened more in my Dynamics and Materials classes!!!!!!
Damnit, I know I should have listened more in my Dynamics and Materials classes!!!!!!
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beat me to it gowgom. It all has to do with the geometric configuration of the beam and it's resulting "I" value (rotational inertia). Want a quick look-see at the equations?
MIT PDF file
- Honer
Btw Mingster, at least your questions are getting a bit more complicated - beams are a little more complex than the circumference of a circle.
MIT PDF file
- Honer
Btw Mingster, at least your questions are getting a bit more complicated - beams are a little more complex than the circumference of a circle.
#7
Ming, a quick and dirty example:
1" x 2" solid bar vs 1" x 2" x 0.25" wall tube
Bar:
Ix = 0.667 in^4
Tube
Ix = 0.526 in^4
Ix is the second moment of area, about the X axis. Ix is used if you will be bending it along the 2" direction, there is a value for Iy if you bend it the other way.
Ix is then used (for a simple beam in flexure) in the equation:
stress = M * c / Ix, where M is your bending moment at a given point, c is the distance from the X (center axis), and Ix is aforementioned moment of area.
The mass of the tube/bar can be determined by the cross-sectional area, in my example 2 sq in for the bar and 1.25 sq in for the tube, multiplied by the length and density.
Summary:
Using the tube, you've gotten 79% of the strength, while weighing 63% as much. Strength to weight baby!
1" x 2" solid bar vs 1" x 2" x 0.25" wall tube
Bar:
Ix = 0.667 in^4
Tube
Ix = 0.526 in^4
Ix is the second moment of area, about the X axis. Ix is used if you will be bending it along the 2" direction, there is a value for Iy if you bend it the other way.
Ix is then used (for a simple beam in flexure) in the equation:
stress = M * c / Ix, where M is your bending moment at a given point, c is the distance from the X (center axis), and Ix is aforementioned moment of area.
The mass of the tube/bar can be determined by the cross-sectional area, in my example 2 sq in for the bar and 1.25 sq in for the tube, multiplied by the length and density.
Summary:
Using the tube, you've gotten 79% of the strength, while weighing 63% as much. Strength to weight baby!
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#8
As an addendum, a beam in torsion (being twisted) is similar - tubes have more strength for comparable weight versus solid bars. The equations are different but the results are similar.
A beam in pure tension or compression is affected only be the cross sectional area of the material, thus a tube has no advantage over a solid bar in this load case. This is why poured concrete columns are used for bridges, etc.
Your suspension examples all fall into the flexure category, so tubing or a channel (C-channel or I-beam) would be better than a solid.
<end brain fart>
A beam in pure tension or compression is affected only be the cross sectional area of the material, thus a tube has no advantage over a solid bar in this load case. This is why poured concrete columns are used for bridges, etc.
Your suspension examples all fall into the flexure category, so tubing or a channel (C-channel or I-beam) would be better than a solid.
<end brain fart>