Only broken one rod so far - knock on wood, assembling LE and didn't have outer rod seated fully, gust came up and cracked end of rod! ::My Bad Still have 4 out of 5 original rods from my Rev 1!! Tough little buggers!!!

I wonder if anyone can translate the following into something that relates to tubular rods <grins>

Felix

"Imagine we have a solid circular rod, and consider the forces in the

middle as we try to bend it. I'm going to draw some cross sectional

diagrams.

Seen from the side:

y coordinate

| axes plane of interest

|___x v

<=== ===> F

---------------------------------------------

C _ ^ stretched : : _

/ \ |2r : deviation, / \ C

\> | : exaggerated </

v squashed *

---------------------------------------------

Seen end-on:

+----> expansion

y e

| ,-------. | /

|___z / \ | /

| | |/ graph of

| | / expansion

| | /| vs y

\ / / |

`-------' / |

So we have a rod of radius r. We apply a bending couple C. This will

cause it to bend slightly, into (by symmetry) a section of a very

large radius hoop. We will neglect the 2nd-order effects of squeezing

of the rod in the y direction and expansion in the z direction.

>From the geometry, we can see that the expansion or contraction

distance is proportional to y. At the middle (y=0) the rod neither

stretches or compresses; at the top edge (y=r) it is maximally

extended. Ie e(y) = d * y where e is the proportion by which the

material is extended and d is a value indicates how tight a bend we

get.

While the tube is still in its elastic region (ie, it hasn't started

to bend permanently ie fail), the force is proportional to the

extension. Since the force varies across the cross-section of the

rod, we want the force at each point - ie, the pressure. The

geometrical symmetry means the force varies only according to y and

not x or z.

Ie, P(y) = E * e(y) Hooke's law, where E is a material

property (Young's modulus).

Consider the component of the couple at the plane of interest,

provided by the horizontal strip of the plane of interest at the

vertical position y:

dC(y) = dF(y) * y

where dC is the sliver of the couple and dF is the sliver of the

normal force. But force is pressure times area:

dF(y) = P(y) * w(y) * dy

where w is the horizontal width across (ie, in the z direction) the bar

at the height y. Since the bar is circular in cross-section,

w(y) = sqrt( r^2 - y^2 )

So substituting all that in,

dC(y) = ( E * ( d * y ) ) * y * sqrt(r^2 - y^2) * dy

= E * d * y^2 * sqrt(r^2 - y^2) * dy

(NB I have neglected to think too hard about the overall sign of

this couple.)

Ie to calculate C we have to integrate the contribution of the P at

the various y:

r

/\

C = | E * d * y^2 * sqrt(r^2 - y^2) * dy

\/

y = -r

I didn't fancy remembering how to integrate that so I cheated and used

Wolfram Mathematica online integrator[1] which tells me:

C =

- E * d * 1/8 *

[ -1 y ] +r

[ y * sqrt(r^2-y^2) * (2*y^2 - r^2) + r^4 tan ------------- ]

[ sqrt(r^2-y^2) ] y=-r

The LH term is zero at both limits because sqrt(r^2-y^2) is zero at

both limits. So we are left with the RH term. This is slightly

awkward the way Wolfram has integrated it (using the 1-argument atan

rather than the computer person's 2-argument atan) because the

denominator is zero, but atan is actually well-defined if we take the

appropriate limit. So:

[ -1 y ] +r

C = -1/8 * E * d * [ lim l^4 * tan ------------- ]

[ y=0..->l sqrt(r^2-y^2) ] l=-r

The sqrt() denominator approaches 0 from the positive side in each

case. l^4 is always positive. The numerator is positive at the upper

limit and negative on the lower, so:

-1 -1

C = -1/8 * E * d * r^4 * ( lim tan f - lim tan f )

f->inf f->-inf

= -1/8 * E * d * r^4 * ( pi/2 - -pi/2 )

= -pi/8 * E * d * r^4

The minus sign is simply due to carelessness about the overall sign

and can be disregarded. Recollect that E is a material parameter and

d is a geometrical meausre of how much we've bent the rod. Let K =

pi/8 * E, and dropping the minus sign, we have:

C = K * d * r^4

So for any given material, and amount of bending, the couple

(informally, force) required to bend the rod by that amount is

proportional to the 4th power of the rod's radius. The rod's

stiffness against bending goes as the 4th power of the radius.

This is the Luggage's 4th power.

But if we are interested in the _failure strength_ we want to know

when the rod will fail. Failure occurs when the expansion of the rod

at any point exceeds the material's elastic limit, ie where:

r

max e(y) >= e

y=-r lim

As soon as that is reached anywhere that bit of the rod will start to

yield, and the other parts will need to take up the strain and won't

be able to do so, and it will buckle (this is not true for all

structures but it is for this one). e_lim is a material property too.

Since e(y) = d * y obviously the maximum e is at r. (It will

probably fail at the top, in tension, but if it fails in compression

that's -r, which will make no difference to the analysis.)

So we have, at failure:

d * r = e

lim

What we want to know is how C_lim (the couple at which the rod fails)

depends on r.

Above we had:

C = K * d * r^4

Substituing in to eliminate d, we get:

e

lim

C = K * ---- * r^4

lim r

or

C = K * e * r^3

lim lim

So the rod's _strength_ against bending is only proportional to the

third power of its thickness, not the fourth.

The discrepancy between third power for strength and the fourth power

for stiffness is because a thicker rod bends less before it fails.

[1]

http://integrals.wol...expr=x^2 * Sqrt[r^2+-+x^2]"