djconnel wrote:Jason:
Thanks for replying!
No problem! This is really fun stuff for me.
djconnel wrote: The bar alone had 5% increased stiffness and the bar + stem combo had 7% increased stiffness with a change in the mid-bar diameter from 26.0 mm to 31.8 mm or a 22% increase. So it was sub-linear. But the center section is only a small fraction of the total bar. The width of the stem clamp is 39 mm: the bar is effectively constrained in this region. The total width of the raised section is 60 mm, so that's 21 mm outside that range. Then there's a 20 mm taper zone on each side where the diameter decreases to 23.8 mm, and the rest is 23.8 mm. So the total bar length is on order 80 cm, so if I allocate 1/4 the taper to the wide section (third power dependence), that's around 3 cm/80 cm = 3.8% of the bar being wider.
The cylinder is hollow with a fixed thickness so your 4th power dependence drops to a 3rd power dependence, I believe, but a 22% diameter increase becomes +82%, which multiplied by 3.8% of the bar length becomes 3.1%, not terribly far from the +5% result, especially considering the center should have a greater influence than the ends.
Yes indeed; that hand calc result is a pretty great approximation. And you're exactly right: stiffness goes to a third-power dependence in this case. Good catch.
djconnel wrote: On the shape of the elements: good question! 4-point tetrahedra are better for a linear model, while the 10-point tetrahedra are better for nonlinear integration. It's true 10-point are better for the same number of elements, but since linear integration is faster, I'm not sure if you allow the tradeoff of using more elements the nonlinear elements are still better. I'm pretty sure these are 4-point tetrahedra. Since the displacements are so small it's still well within a linear regime (the displacement versus force makes this clear), so I think that should work well. In any case I was rather aggressive on mesh density, because why not.... it was fast enough.
Yes, this is exactly the part that's non-obvious even to people who are sharp and technical but who haven't had a lot of experience with FEA. A couple of points, in ascending order of importance:
- FEA geeks (including me) are sloppy with their terminology. 4-node tets are called linear and ten-node tets are called quadratic (or second-order) for the reasons you mention, but they're both useful for analyses of linear structural response. In other words, while there are both linear and nonlinear analyses, those terms don't have much bearing on whether you choose linear or quadratic elements.
(1) 4-node tets record a single strain state
(2) within their volume, while 10-node tets can record a strain state that varies linearly through their volume. So you can approximate the strain in a linear analysis with many single-strain-value tets or fewer 10-node linearly-varying-strain-value tets. Put another way, for any given strain contour, you can capture it adequately with fewer 10-node tets than with 4-node tets.
- This is the big one that even trained FEA analysts overlook: a part under a bending load has linearly varying shear strain through its thickness. (If the top of the part is in compression, the bottom part is in tension, or vice versa). Furthermore, most thin-walled shells (like a frame tube or a handlebar) end up being meshed with a single element through their thicknesses. To capture the bending stress of a plate or shell with 4-node tets or 6-node bricks, you'd need to mesh it with many elements through the thickness...on the order of five or six. That's the real reason to use 10-node tets in your model: to capture the bending strains effectively.
- The caveat to the big point above is that because a handlebar is a hollow tube, the geometry itself does a fair job of capturing the linearly varying strain state through its thickness. In other words, if you're pulling up on the left side, the upper portion of the bar near the stem is in compression while the lower portion is in tension, and the shear is distributed linearly as you pass through the neutral plane. So the question of 4-node vs. 10-node tets really shouldn't matter, right? Well, it depends what you're trying to learn. If you're interested in stresses and strains near the bar clamp, where the handlebar wall is itself in bending (as opposed to the bar as a whole), then it matters a lot.
Swapping your 4-node tets for 10-node tets will likely not change your peak displacements much (though I'd expect they'd increase a smidge) but it should change the magnitudes and shapes of the stress contours near the stem.
djconnel wrote: I specifically did not attempt carbon fiber because it's a anisotropic material. I recently reviewed a paper for a graphene nanoelectromechanical device and I questioned the authors' use of a simple elastic anisotropic model, but they defended it (it's a bit outside my expertise). But with aluminum, no problem, at least sufficiently far from yield.
I think this is wise; carbon would only confuse the issue and, even if it didn't, you wouldn't learn anything more from an anisotropic model than you would from an isotropic one.
Graphene is mildly anisotropic when you compare transverse and shear directions...you were right to raise that question. Even if modeling it as isotropic is defensible, you certainly showed you were paying attention. I'd be more troubled, though, if that paper's authors were assuming that out-of-plane properties were also anisotropic. That's much less of a given.
Aluminum is isotropic whether we're talking about elastic deformations or plastic (beyond yield) deformations.
Fun trivia fact: shim stock, which is basically thin, cold-rolled stainless steel, is pretty highly anisotropic for a homogenous metal. It's something like 10% stiffer in the longitudinal direction than in the transverse direction. The rolling process aligns the crystal structure with the rolling direction, resulting in anisotropy. Repeat that one at your next get-together and you'll be the life of the party!
Cheers,
Jason
(1) FEA dorks about 15 years older than myself used to engage in holy wars about whether to use many linear elements or fewer quadratic elements. But Moore's Law pretty much rendered the argument moot; most people now use quadratic elements (10-node tets or 20-node hexes) by default and tailor their mesh density to capture the detail they need. If you're doing a nonlinear analysis, you're almost always using quadratic elements, but just that's a coincidence of terminology.
For this reason, I generally refer to elements not as linear or quadratic but as having or not having midside nodes. This covers any element shape and is also less prone to confusion.
(2) I'm talking about strain here rather than stress because strain is the direct result of an FEA solution. The basic equation is F=[K][u] where F is force, [K] is the stiffness matrix and [u] is the displacement (or strain) matrix. Stress is a derived quantity in FEA analyses.
