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I'm a PhD student in maths, and attended my last physics class some 15 years ago, so you can imagine my competences in the field. My supervisor (also not a mechanist) cant tell me how to proceed either, and after having spent already way too much time on wikipedia to try to understand the elementary concepts, I am turning to you, here is my problem:

Given is a discrete curve in $\mathbb{R}^3$, i.e. an ordered set of points $x_1,...,x_n$ all in $\mathbb{R}^3$, representing the branch of a (botanical) tree. $x_1$ denotes the point where the branch in question branches off from the trunk, $x_n$ is the branches ending point. This discrete curve may be curved and twisted arbitrarily. At each point $x_i$, there is a mass $M_i$ concentrated. Moreover all the branch radii, $r_i$, at point $x_i$ are known (which - I think I at least got this right - is relevant to calculate the "second moment of area"). (If it helps, I would also have a continuous version of the problem, i.e. a continuous curve $s\to g(s)$ instead of the points etc...)

The discrete curve $[x_1,...,x_n]$ describes the branch without gravity being taken into account and the task is to find the new curve $[x_1'(=x_1),x_2'...,x_n']$ resulting from the original curve when taking gravity into account. (By the way the trunk itself is considered to be unaffected by the branches growth and weight, thus remains vertical).

Apparently the problem is trivial if the curve is not twisted, in the way that all $x_i$ lie in one and the same plane perpendicular to the ground (discussed on the second half of page 2 of "Bending of apricot tree branches under the weight of axillary growth" by Alméras et al., in: Trees. Structure and Function, 2002.). However, I have unsuccessfully searched for a generalization for twisted branches.

Could you point me in the right direction? Or, if applicable, tell me that the problem is much more complicated than I think. Thanks a lot in advance.

PS: If, as it seems to me in the article mentioned, there exists a very easy way to approximate the real solution to some extend, making possibly questionable assumptions (the "small deflection assumption" seems to be of that sort), that's fine by me. I do not require an exact solution, any (halfway justified) rough approximation works perfectly for me.

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Are you really thinking of a real tree? Have a look at this rd.springer.com/article/10.1007/s00468-007-0182-7 . The main difficulty with your branch that has twists is that you will be introducing rotational forces about the axis of the twist. In a real tree this would be balanced by a twist in a new branch. Trees are alive. There are old olive trees which have a trunk like a corkscrew. If you dig up the roots you will see that the roots are twisting in the opposite direction, all due to finding a rock underground around which they twist to get to good earth. The trunk compensates. –  anna v Dec 20 '12 at 16:58
    
Thanks for the article, and quite interesting what you say about the twist and compensating effects. However I believe what serves my needs must be of a simpler nature: Assume you have a real tree (subject to gravity) in front of you, and you fully know the course of the (1D) branch curve (i.e. the curve passing through the centers of the cross sectional disks), as well as the cross sectional disks radii. Now you attach some christmas tree ball ornaments to a couple of spots on the branch, how does that alter the branch curve? Is this really that difficult to determine/approximate?... –  robster Dec 20 '12 at 17:41
    
If those branches are not cross each others, it seems that this type of question and somehow solved by the variational method. –  hwlau Dec 20 '12 at 18:08
    
Sounds promising, can you give me a concrete reference for that? –  robster Dec 20 '12 at 18:12

1 Answer 1

The standard engineering approximation, which makes a lot of simplifications, you should be able to find in any book by the name of Mechanics of Materials or Strength of Materials. The general ideas would be as follows...

First, within these simplifications elasticity is linear, so you can consider each force independently, and calculate the total stress, deformation, deflection... as the sum of the individual ones.

Each force will produce, at each cross section of your branch, a reaction force and moment. The force will have a component perpendicular to the cross-section (tensile force) and another in its plane (shear force). The moment will also have a perpendicular component (torsion moment) and another in the same plane (bending moment). You will need to figure this out for every cross section of your branch.

For long, straight, constant circular cross-section beams, each of these produce a specific set of strains and stresses, which translate into a definite set of displacements. These are the kind of things you will find in the books I mentioned above, or in this document. If your beam is twisted, and does not have a constant, or circular cross section, the approximation will be less accurate.

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Thank you very much, will I find the entire set of techniques in oder to be able to handle the twisted case in your document? Then I'll look into that and try my luck. General mechanics books scare me off a bit, you see, including gravity accounts only for a small part in my concepts, and I still expect the problem to be solvable in a few minutes if you know what to do, while it would take me an unreasonable amount of time to understand mechanics from scratch. Just as you physicists occasionally resort to a mathematical theorem without necessarily looking into all the theoretical background :) –  robster Dec 20 '12 at 20:58

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