## sizing help for aluminum tubing

## sizing help for aluminum tubing

(OP)

o.k. maybe i don't belong here being that i'm not a degreed engineer, but i am hoping someone will come forward to help the calculus challenged.

i am trying to build a device that will be set at an angle similar to a ladder that needs to carry a load of up to 1,000 pounds, it will be at about a 70 degree angle from horizontal and have two rails 8 feet long, again similar to a ladder,

my question is how do i calculate or verify the required aluminum square tubing? weight and size are very important, i don't want overkill. would 1/8 1"x2" be adequate for example. i guess what i really need is a formula for my scenario that i can contact a supplier and plug in the tensile or deflection of a particular extrusion to see if it would work.

i really appreciate any help, i don't know were else to go.

## RE: sizing help for aluminum tubing

Some questions first....

Where on the beam is the centroid (center of gravity) of the load located? is it applied at the very top? what is the vertical support doing? how rigidly is the bottom fixed? the top?

An example:

If the load is located exactly 4ft along the angled beam(s) then it would look like this:

|\ (some force would be here in two directions)

| \

| \

| ^(force down direction)

| \

| \

|______\ (two directions of force here)

We'll use X as the horizontal direction Y is the vertical

t-top fixture, b-bottom fixture, L-load, e-beam direction

O-theta (angle beam makes with vertical), F-force, T-torque

l-length (of load center) from top mount, ll-overall length of beam

Remember SOH/CAH/TOA?

The force along the beam is given by Fe=L*cos(O)(L=1000lbs;O=20deg)

the force at the top in the x direction is:

Ftx=Fe*cos(90-O)

In the y direction is:

Fty=Fe*cos(O)

At the bottom:

Since the geometry is the same then the resultant forces are the same IE: Ftx=Fbx and Fty=Fby

Now there will be some torque acting at the top mounting point:

Tt=l*[L*cos(90-O)] (in the clockwise direction)

at the bottom:

Tb=(ll-l)*[L*cos(90-O)]

Hmm...

Well now I've exausted my knowledge of statics,.

but someone must know how to from the above and assuming ridgid mounting calculate the resultant stresses on the beam, then with the stresses and using the geometry (of your aluminium beam) you should be able to figure the stress on the material.

Nick

I love materials science!

## RE: sizing help for aluminum tubing

Sorry it is not simpler.

However, one place to start is to assume:

alloy 6061-T6

no welding

static loading

worst case position of the load

use 19,000 psi for allowable axial compression

use 21,000 psi for allowable compression from bending

Make sure that the actuals divided by the allowables sum to less than 1.00

Determine moment of inertia needed for the bending load, pick a shape with that value or greater. Then see if the compressive and combined stresses are OK. Then see if the assumptions for 19,000 and 21,000 are valid and review the shear, bearing and component stresses.

The Aluminum Association publishes (for a cost) guidelines on engineering structures, complete with factors of safety and allowable stresses.

## RE: sizing help for aluminum tubing

thanks for the input.