Bending Strength Of Aluminum Vs Wood?

Yes you guess! Obviously.

 

Maybe it is obvious for some but not for others

 

original poster

you can see here where I have glued 6mm American White oak to the alloy box sections, the span was actually your 9 feet and often 10 people standing there in the deck salon

 

wow - great job on the veneer, looks nice! What thickness decking or ply do you have over those? What spacing? (looks about 2' ?) What wall thickness on those box sections?

 

the boxes were 130 x50 x 3mm, , that hole is 32 inches, over the girders(fore aft) we had 10mm ply them 50mm foam then 12 ply, the foam was High Modulus(company) high density, all glued up, mainly cos it was over the engineAnyways it would support the engine, and ddi before we veneered the sole, a cummins about 509kg
i do not knowq what yo ua re trying to do exactly, but IF you sketched something it'd help

 

jdworld,

I spanned my 9 foot houseboat roof with 2 1/4"H X 1 5/8"W laminated

spruce beams. I made them on a jig, six at a time. They had an 8 inch

vertical curve (about an 18 foot radius). I glued and screwed (removing the

screws once the epoxy had set) three, 3/4" strips. The spring back was

about an inch when I removed them from the jig. I tested one by jumping

up and down on it repeatedly (I was 175 pounds then). It made some

cracking noise but never failed. I placed them on 18 inch centres with 1/4"

plywood, 2 inch foam and glassed, 1/4 plywood all screwed to the beam

with 3 1/2 inch #10 plated screws from the top. I painted them white as

they were exposed inside and then tacked a 5/16" mahogany strip along the

bottom that over hung 3/8" on each side (radiused). It feels solid to walk

on and held up 2 feet of wet snow in the '96 storm.

-Tom

 

60-63 is thew most common off shelf, and it is perfectly great for building this sort of structure inside. 6063 is however not suitable for immersion in salt water
With all my adult life, building such ,this guy apex 1 who knows nothing is , in fact used to consult me abt al building, is castigating me here. I have asked the moderater to stop this sort of drunken rubbish
we will see
Interesting Apex one can not see my gallery he is quite blind
It seems that he can call each and every person he does not like, an amatuer and laymenand a liar with impunity, this despite my having posted my qualifications, references, here
All the yachts I have built have been build under survey, that does not mean just To survey
Yet see his? all he does is well quite frankly nothing
He spends his whole bitter life here, thugging others
Sorry Pal the truth prevails, and at least I DO have the courage to put my face on here And you? you promised to ignore my posts, but you can not resist
FACT I made my life from building

 

It depends. Is the wood Douglas fir, or southern yellow pine? These are VERY different woods ITO strength and stiffness, the southern yellow being superior.

Furthermore, there are a wide variety of "off the shelf" rectangular aluminum alloy box sections, some hardenable, which means you then have to specify which hardness (absolutely critical for a stiffness calc, less so for UTS) . There's even a further question of the corner radius and fillet for the various types of aluminum box, the types with larger radius and fillet being able to be loaded more heavily and having longer fatigue life under cyclic loading.

To PAR's point:

There's a minimum of two obvious wood choices: Douglas fir or southern yellow pine, and probably no less than 8 aluminum alloy choices, all fitting the definition you established, which is "off the shelf", and a dimensional limit of 2" X 4", and .125" wall, and all these choices have WIDELY disparate specific strength characteristics! I count that as 16 possible combinations to compare! It would help a lot if you could be more specific!

The people posting so far are not trying to be evasive; but you are looking for a specific, precise answer to an engineering question while only providing the vaguest details about the materials to be used. How stiff; how much deflection under load 'X' is a very precise question, unlike the much more general question of whether or not the structure will fail under load "X".



Jimbo

 

nice Tom, yes there are many ways and experience tells us what to do,
thing us we HAVE to get it right, as boats are under warranty

 

That's just it, I'm not really needing a precisely engineered answer at this point - just an "in general" answer.

But ok, I know "in general" doesn't compute with engineer's brains so if you insist.....9' span, 300 lb point load midspan. Which deflects more (or fails first)?

2x4x8 DF

OR

2x4 x .25" 6063 Alum tube

 

300, thats simple spruce, , an other softer timbers will deflect a lot, oak hardly and t5 or 6 alloy in the sizes I gave , also hardly!!with this load neither the oak or the al al will fail cant speak for other timbers cos these are the only materials I have worked with But only a ship would use oak, its so heavy)

 

jdworld,

Okay, I give up. Why don't you try what you're proposing and see?

Let us know the result. Why would you put your trust in complete strangers

who have no vested interest in what you're doing just to save you the effort

of doing the test? Obviously my input was of no use to you as you didn't

even acknowledge it. Before I went to the effort of trying to help you I

should have simply asked: "Why do you want to know?"

Is this a bar bet you're trying to win or did you just wake up one morning

wondering which was more rigid or brittle?

-Tom

 

beam deflection flap

The OP asked a simple question that need not have generated a controversy. The answer is not difficult to ascertain. There a several variables that we need to agree upon. Basicly, the deal is this: place a beam across a pair of sawhorses, the support points are 9 feet apart, the beam is merely lying there, it is not rigidlgy attached to its' support. We hang a 300 pound weight in the exact center of the beam. The beam will deflect downward as sure as God made little green apples.

Let us look at the equation that establishes one component that describes the relative stiffness of a beam. It is the equation that calculates moment of inertia (I). Simple equation is I = bd^3/12... b is the breadth or width of the part in question, d is the depth of the part and that dimension is raised to the third power. The product of b times d to the third is now divided by the constant 12. Notice that increasing the width of the beam adds stiffness but only in a linear way. More importantly, increasing the depth of the part raises the stiffness factor in an exponential way. It is plain to see that depth is more important than width. Analyzing the solid wood part is easy enough. The hollow aluminum part needs a little more simple math. To analyze the aluminum box beam, find I for the outer dimensions (2x4) That'll come to about 10.667. Now do the math for the inner set of dimensions. Let us say that the wall thickness of the box beam is one eigth (0.125) We now have a space of 3.75 x 1.75 and I = 7.690 .....Subtract one from the other....result is roughly 2.976.

To be fair and compare apples with apples, let the wood beam have the same outer dimension as the aluminum part or 2.0 x 4.0 I for the wood beam is obviously the same as the outer I for the Aluminum part or 10.667 At this point it seems that the wood wins by a large margin.......But there is more to consider. Wood fiber is not as strong as aluminum fiber. That gets us to the next variable.

The relative fiber strength is modulous of elasticity (e). Wood modulous varies all over the lot. Wet wood is not as strong as dry wood, stiffness varies with respect to grain orientation,etc... Aluminum has an e in the region of 10 x 10^6. Despite a lot of ******** claims, common aluminum does not vary far from that figure. Woods vary from about 0.88 to 2.05 x 10^6. In general you can use a figure like 1.2 x 10^6 and be in the ball park. .............hang in there we are getting to the answers presently.

Stiffness is, in the simplistic sense, the product of I times e. You can see that in a comparison of elastic moduli, aluminum fiber is about eight times stronger than wood. We saw that I for the wood beam is about 4.8 times as large as I for the alum box beam. Taking the products of the two types we see that the aluminum wins the stiffness contest.

OK how much does the hypothetical beams supported by sawhorses and loaded with a 300 pound weight in the exact center deflect????

Formula is pretty simple for this scenario. Deflection = PL^3/192eI. P is the weight 300 pounds, L is the length between the supporting structures of the beam, 192 is merely a constant for this type of load situation and this one alone, then there is the e and the I... Aha!!! We see immediately that the length of the beam is the critical thing because length (L) is raised to the third power. Doing the math for deflection we find that the box beam will deflect signifigantly less than the wood beam. Had we used the common two by four that is not really a 2 x 4, the difference would be dramatic because of the exponential influence of the depth of beam. Corrolary: a little extra depth is more valuable than extra width and the length of the beam is a serious consideration, both on account of the exponents.

 

messabout, you're absolutely right, it's not difficult to ascertain at all.

You just stand on them like the OP said...

-Tom

 

Sub T; You are a most practical individual. I too would resort to the real world test if I had both beam test examples, a pair of sturdy sawhorses, and a 300 pound weight at hand. I'm not so sure about lifting that much weight however.

My essay was intended only to point out that the length and depth of a beam is the main determinate of whatever material and configuration one might select and how the strength of the material might play into our decision. But then we know that intuitively don,t we? Yes, and Par is correct in that an analysis of the suitability of a particular part amounts to considerably more than an exercise in simple arithmetic.