Which equation(s)?

Re: Which equation(s)?


Use the gravity equation and the equation for acceleration: e=mc^2

a=0.5*g*t^2

In torque or as a simple bending moment?

In torque, the amount of twisting deflection is a function (among other things) of the length of the rod, but failure is a function of cumulative shear.

With bending moments, with ductile materials the failure can be a combination of two types, deformation and shear. Deformation is a function of the bending moment (usually at the middle) and shear (usually at the beam roots).

Now both of these failures are for static loading. Dynamic loading failure is a whole different bundle of kittens.

BTW I'd recommend looking at Marks, Escbach, or any good Civil Engineering Handbook. Simple static beam loading analysis has been tabulated.

You'll have to do is change the moment of inertia for round beams.

Rap

Thanks raprap. I'm not exacly looking to calculate torque, although that might be the best way to do it. My end goal is just to produce a simulation that *looks* moderately realistic.

I'm talking specifically about a tree branch (round). It's connected at one side at an angle and it can obviously hold more weight the more vertical the branch is.

I could fudge it by doing something like this:

w = weight
a = angle between beam and vertical
r = radius
b = bend factor
s = strength

w*a/(r*b) > s then snap
otherwise bend = w*a*b/r

(with some constant factor)

However, before I commit to fudging it, I want to see what the real equations involved are and see if it wouldn't be too much trouble to make it a bit more realistic.

Chances are I won't have all the necessary variables anyway, but looking at the real equations would allow me to make a more realistic fudge.

Why is dynamic loading failure so different, out of curiousity? It seems that if you had a static load, you could calculate the result, then add more load, and calculate another static result, etc.

A Tree branch is effectively a cantilevered beam. Your right, the more vertical the angle the stronger the beam. Two reasons; First the load will be concentrated closer to the roof (the attachment to the bole); Second is by your angle reasoning.

Be careful about cantilevered beams, they are much weaker than a simple beam, and the ultimate failure is a function (if memory serves me) as a function of the fourth power of the distance between the load concentration and the root.

Dynamic loading is weaker because of resonance and shock. Everything resonates--if the tea dancers had not been dancing in resonance on that elevated skywalk in Kansas City in 1989--some of them would be enjoying their grandchildren.

BTW here's a simple beam calculator menu web page

Rap

Ok, then I'm dealing with a cantilevered beam...the load is not applied at any particular point but rather is the weight of the branch itself, so I think I could compute this by considering intervals along the branch, and computing the deflection at each interval where the load is the weight beyond that interval.

This appears to be the equation:

"Deflection Between Support and Load" at
http://www.engineersedge.com/beam_bending/beam_bending10.htm

The only thing is, what do you think the distance "x" is representing? It is not very clearly marked in the diagram.

Also, these all deal with the beams that are coming out at 90 degrees, any idea where to look for info on other angles?

Here's a neat experiment where I could measure the modulus of elasticity:

http://ceaspub.eas.asu.edu/imtl/HTML/Manuals/MC101_Modulus_of_Elasticity.html

The moment of interta I = mr^2, so I assume in the context of this equation it would simply be

I = (W/g) l^2

Since l is the distance to the load, and the mass would just be the load divided by gravity..

Ok, it turns out that in this equation x is the point on the beam where we want to calculate the displacement.

But, this considers only a point load...we really have 2 situations here, one is a distributed load from the self-weight and then a point load from the self weight of all child segments.

So the actual calculation I will use is this:

1) Calculate distributed load(W) on segment from self-weight

D1 = WL^4 / 3EI

2) Now consider the weight of all successive branches(P) as a point load at the endpoint

D2 = PL^3 / 3EI

Total displacement = WL^4 / 3EI + PL^3 / 3EI