Well...
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!

For aluminum, the allowable stress for axial loading is different than the allowable stress for bending type loading.  They depend somewhat on the cross-section shape of the member, the alloy, the unsupported length, the end conditions, if the member is bent about its strong or weak axis and if there is welding being done.  There are also different allowable stresses for flanges and returns as well as shear stresses and methods of analyzing combined bending and axial stresses.

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.

thank you very much for your help, i now beter understand what forces i am dealing with and some of the ways of determining the load, but to get any real data on the various available extrusions i will need to contact a supplier.  

thanks for the input.