I am a student that knows little about stress loads, I have sort of a
man-made support frame on the back of a truck I want to assembly. I
wish to calculate the stress load on one member. I have a 2.0" X 2.0 X
.125" Alum tube top frame member,attached to a existing upper square
frame, attached are (2) lower brackets
supporting a 300 Lbs load box. What is the shear stress load on the
single alum 2x2x.125" tube member. Only a rough estimate only, I just
wanted to see the
approx results of this member.

thank you

Rob

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Clarification of Question by groundteam8-ga on 11 Feb 2006 16:07 PST

___________
            |           |
            |  300 lbS  |
            |           |
            |___________|
___________________________________..........Tube is 5 feet long supported by 
|__________2 X 2 X .125" SQ________|         same tube running Vertically.
| |                              | |

Tube top would be the 2x2x.125" square Alum, that the 300 Lbs box is
resting on. I drew this pic to give sample; basiclly I know little
about calculations,
I realize this is only a partial picture, What formula would i use to
calculate this? Thank you very Much

you had me at 'tube top'

Could you give a better description of the layout?  Or, if this truck
thing is a commercial product, can you provide a link to a picture?

The max shear stress could be found as value of stress force towards
to normal of tube axis divide by (area of normal section of this tube
multiple on 2).
the 2 is defined by symmetrical load scheme.
At this case it means 
shearstress=300 lbs / 2*((2*2)-(1.75*1.75))square inches=300 lbs
/2*0.9375 square inches = 160 lbs/square inches

Above mentioned appproximation is mean value of stress trough
section. If you need to have distribution of shear stress in section
you need to use next equations:

(tau(h))=(Qy*Sx(h))/(b*Jx),
where
tau - stress at height of section h - vertical distance from our
height counted at center of symmetry of section
Qy - 150 lbs as most worst at this case
Sx(h) - static mement of part of section above h
Sx(h)=((2x(1-h))*(h+(1-h)/2) at h(0.875..1.0)
Sx(h)=((2x(1-h))*(h+(1-h)/2)-(1.75*(0.875-h)*((h+(0.875-h)/2)) at h(0.0..0.875)
situation symmetrical throw line at center of bar
b = b(h) - thickness of section at h
b= 2 at h(0.875..1)
b= 0.25 at h(0.0..0.875)
situation symmetrical throw line at center of bar
Jx - Axial moment  of section inertia
Jx=1/12(2*(2)^3-1.75*(1.75)^3) at this case
therefore you achive the result
it be at maximum at center (h=0) 
tau = 358.7 lbs /(squared inch)
This is more better approximation.