Reply With QuoteWhy not wood?
The thickness increase of the Aluminum depends on the base steel strength. You should first figure out what the steel was exactly.
Hello all,
I am new to this forum, have some knowledge of metalworking, have onlyworked with aluminum minimally and am not an engineer so please barewith me !
Wehave these 6' tall “lifeguard style” chairs we need to re-build. The frame of the chair is basically shaped like an upside down“U”. The frames are made out 1.5” square steel .049” wall(18 gauge) and the remainder of the chair (other than the frame) ismade out of wood. Due to costal conditions and the salt air thesteel frames have rusted significantly and it is time to re-buildthese frames.
Iwould like to build the new frames out of aluminum. I want keep thesames OD on the tubing (1.5 “ square) but I know by using aluminumthat I will need to go to a larger wall thickness to maintain thesame strength as the .049” steel tube. I seems 6063 and 6061 maybe the most readily available square tubing, but that 6061 would bestronger. I would appreciates suggestions on which to use. Each tube has two 80 degree bends (5.9 inch centerline) as thelegs are slightly splayed. Also, the tubes werebent on a crush style bender on the existing steel tubes.
Somy question is what is the minimum wall thickness I could use on aluminum tubing and have it be equal in strength to the steel tube ?
Withthat information I think I can then check on available tubing optionsas well as speak with folks who do bending to see if they can makethe same radius bends in aluminum.
Thankyou very much for any help and guidance you can provide.
Chas
Why not wood?
The thickness increase of the Aluminum depends on the base steel strength. You should first figure out what the steel was exactly.
We don’t know some things about the problem but we know the .049” wall steel tube seems to have supported the load. We know the elastic modulus of steel is about 2.5 times that of aluminum. For discussion, let’s simplify the problem by considering one 1.5” x 1.5” tube as a simply loaded column.
Professor Rankine has suggested that we combine the load requires to crush the column with Euler’s buckling load to obtain the load that the column can withstand.
Buckling load = pi2 x E X I / (K X L2)
Where K is a factor based on the attachment of the ends.
If we consider the column failing at yield stress then the crush load is the area of the section multiplied by the cross sectional area.
Rankine Formulas: (1/Column Failure Load) = (1/crush load) + (1/buckling load)
A 72” length of 1.5 x 1.5 tubing is so slender that the buckling loads dominates the Rankine formula. This is particularly true if we consider the ends as being hinged or pivoting about a through bolt.
Because the Euler column buckling formula has only modulus and moment of inertia values which change in our problem, decreasing the modulus of elasticity by 2.5 times by changing to aluminum requires an increase in the moment of inertia (I) of the cross section by 2.5 times to remain equal in buckling load.
The original steel section had an inner dimension of 1.5 – (2 x .049) or 1.402”
The original moment of inertia is: (1.54 – 1.4024)/12 = .0999 in4
An equivalent aluminum section would have a moment of inertia approximately .25 in4..
This suggests a wall thickness a bit over .150”. The closest commercial wall size over this calculated equivalent is probably .1875” thick.
The radius of gyration of the 1.5” square with a 3/16” wall is:
.289 x (1.52 + 1.1252)1/2 = .5419 in.
The moment of inertia with a 3/16” wall = (1.54 – 1.1254)/12 = .288 in4
Dividing the 72” length by .5419” yields a “slenderness ration” of 133. While some new building codes have allowed columns with slenderness ratios up to 140, old school guys never like to exceed 120. (Of course you did just with the old design). Tying the legs of the chair together at some point above the ground might be considered.
Let’s see what the 3/16” wall 6061-T6 aluminum tube can withstand as a free column:
Crush Load = (1.52 – 1.1252) x 40,000 psi (yield strength) = 39,375 lbs
Buckling Load = 9.8696 x 12 x 106 x .288 / (1.0 x 722) = 6580 lbs
Failure Load as a single 72” vertical column = ((1/39375) + (1/6580))-1 = 5638 lbs
You have not revealed all of the details concerning the loading of the ‘chair’ and so the above analysis cannot be a definitive recommendation. It is offer it as a simplified method of answering your question. Someone with knowledge of the conditions that the ‘chair’ will encounter, the anchoring method, construction details and particularly the safety factors which are appropriate should be engaged to make design recommendations. The limited information that you presented does not permit an analysis of the safety of your original steel structure and it does not permit an evaluation of the safety of a design using aluminum.
Ref: http://www.engineersedge.com/column_...lumn_ideal.htm
Posting Permissions