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approx strength of wood vs. aluminum

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  • approx strength of wood vs. aluminum

    Hi all,
    I have a project where I want to replace some plywood with some aluminum plate. I want to retain roughly the same strength, but it is not life & limb critical.

    The plywood to be replaced is actually 5/8" thick OSB (oriented strand board, a.k.a. "waferboard"). I want to use 6061-T6 aluminum plate to replace it, and I am wondering what thickness of aluminum plate would give me roughly the same strength as the original 5/8" OSB. My initial guess was to try half the thickness, which would be 5/16" 6061-T6 plate, but honestly I have no idea. I'd rather not waste unnecessary money if 3/16" or 1/4" would do the trick. I'm guessing 1/8" would be noticeably weaker than the 5/8" OSB.

    Does anyone have some advice, or an online resource that gives approximate equivalences of strength between OSB and aluminum?

    Thanks in advance for any help you can offer.
    -Eric

  • #2
    Originally posted by ridesideways View Post
    Hi all,
    I have a project where I want to replace some plywood with some aluminum plate. I want to retain roughly the same strength, but it is not life & limb critical.

    Thanks in advance for any help you can offer.
    -Eric
    Need a few more details . . . photos or dimensions. If you have room to add a few ribs, you can make a structure with 1/8 inch plate.

    Jim

    Comment


    • #3
      I'm assuming that when you are talking about having the same "strength" you really mean you want the same "stiffness."

      Stiffness is the product of the material modulus of elasticity and the section moment of inertia, or "EI." It determines deflection (sponginess), which is generally the controlling criteria anyway.

      For a rectangular cross section, the moment of inertia is (bh^3)/12, where b is width and h is depth.

      The modulus of elasticity of some materials is as follows:

      Steel: 30,000 ksi
      Aluminum: 10,000 ksi
      OSB: 1,000 ksi

      So the stiffness for a 12" wide strip of 5/8 nominal OSB (actual thickness 19/32") is:

      12*(.594^3)/12*1,000,000=209,300lb*in^2

      Now to get the same stiffness from aluminum, find X...

      12*(x^3)/12*10,000,000=209,300

      x=.276"

      So to match 5/8 OSB, you need .276" thick aluminum. If you use 1/4", you'll get:

      12*(.25^3)/12*10,000,000=156,200 lb in^2 which is only 75% as stiff as the OSB.

      Go with 5/16" aluminum and you'll get:

      12*(.25^3)/12*10,000,000=305,200 lb*in^2 which is 146% as stiff as the OSB.

      Note that the extra 1/16" thickness doubled the stiffness. Also note that any grade of aluminum will be about this stiff. T6 does not make it stiffer.

      Is this the answer you were looking for?

      Comment


      • #4
        I hope a few high school math teachers saw this thread. It make a great argument for why math matters to people in everyday life. Great work Bodybagger!

        Comment


        • #5
          And after you price that, then the wood will seem like a super bargain

          Comment


          • #6
            Originally posted by Bodybagger View Post
            I'm assuming that when you are talking about having the same "strength" you really mean you want the same "stiffness."

            Is this the answer you were looking for?
            Wow that is exactly what I was looking for! Thank you so much!

            Yes, when I say "same strength," I really mean same stiffness. Stiffness is what I care about here-- I really don't anticipate the thing actually *breaking* or anything.

            Incidentally, it is a platform on which my two race motorcycles sit. The platform deck is about 8' x 10'.

            The whole thing is currently made of 6061-T6 aluminum structural members, except for the deck, which is the 5/8" OSB, and therefore not very weather resistant. The structural supports are AL channel spaced every 16". 5/16" Al plate is obviously pricey, so it might make sense to weld in more supports, so that I can go with something thinner for the deck.

            It looks like some more calculations are in order on my part. Obviously the goal is figure out the least costly solution: more joists & thinner plate, or fewer joists and thicker plate. I think given BB's formulas, it should be straightforward to calculate.

            I am still a little confused as to why the 6061-T6 "grade" of aluminum has no impact on the stiffness. For example, if I take some 1/16" sheet of 6061-T6, and run my acetylene torch over it to anneal it, the AL sheet becomes noticeably easier to bend. Is my thinking wrong here?

            Comment


            • #7
              Originally posted by ridesideways View Post
              I am still a little confused as to why the 6061-T6 "grade" of aluminum has no impact on the stiffness. For example, if I take some 1/16" sheet of 6061-T6, and run my acetylene torch over it to anneal it, the AL sheet becomes noticeably easier to bend. Is my thinking wrong here?
              The following applies to 6061-T6 and 6061-T0 at room temperature
              .
              When you bend it, you are exceeding the yield point. Below the yield point, a material will exhibit elastic behavior and return to its original shape after being deformed... like a spring

              The heat treatment changes the yield point. At T6, the yield point is around 40,000psi. Annealed (T0), the yield point is around 8,000 psi.

              So, the modulus of elasticity (along with the section moment of inertia) determines how stiff the member is in the elastic range.

              Yield point determines when the material "gives way" and deforms.

              Heat treat only changes the yield point - not the modulus of elasticity. Both grades will exhibit the same stiffness right up to the point where they give way... and the T0 will get to that point WAY before the T6 does.

              The following applies to almost every metal when heated...

              When a metal is heated, its modulus of elasticity goes down. It becomes less stiff.

              When you heat your 6061-T6 with a torch to bend it, you are doing BOTH... you are removing the temper (lowering the yield point) AND lowering the modulus of elasticity. When it cools to room temperature, the modulus of elasticity will return to 10,000,000 psi. The yield point may only return to 8,000 psi. But the stiffness depends only on the modulus of elasticity and the section moment of inertia (EI), NOT the ultimate strength or the yield point.

              Does this explain, or was it too complicated?

              Comment


              • #8
                Complicated as it may sound... you still explained it in terms that could be understood.
                Thanks on my part

                Comment


                • #9
                  Originally posted by ridesideways View Post
                  The structural supports are AL channel spaced every 16". 5/16" Al plate is obviously pricey, so it might make sense to weld in more supports, so that I can go with something thinner for the deck.

                  It looks like some more calculations are in order on my part. Obviously the goal is figure out the least costly solution: more joists & thinner plate, or fewer joists and thicker plate. I think given BB's formulas, it should be straightforward to calculate.
                  Well, not exactly very straightforward...

                  OK, you are spanning 16" joists. To simplify things, we will analyze a 16" long simply supported 12" wide sheet of 3/16" aluminum.

                  E=10,000,000 psi
                  I=(12*(3/16)^3)/12=.00659 in^4
                  Live load=100psf->8.33lb/in
                  Dead load=2.7psf->.225lb/in
                  Ultimate bending moment=((1.6LL+1.2DL)*span^2)/8=[{1.6(8.33)+1.2(.225)*16^2]/8=435 in*lb
                  Ultimate shear= (1.6LL+1.2DL)*span/2=[{1.6(8.33)+1.2(.225)*16]/2=109lb
                  Max deflection=(5*(LL+DL)*span^4)/(384*E*I)=(5*(8.33+.225)*16^4)/(384*10,000,000*.00659)=0.108" which is L/148 (which exceeds the L/360 you'd use for floors)
                  Ultimate stress in bending = 6Mu/(bh^2)= 6*435/(12*(3/16)^2)=6,191psi
                  Ultimate stress in shear = Vu/bh=109/(12*(3/16))=48.3psi

                  As you can see, all the stresses are easily well below the 40,000psi yield point of 6061-T6. However, the deflection is already higher than ideal. Reasonable, but still more than you'd like.

                  You see, deflection almost always controls. I'd probably use 3/16 in one or two large sheets and rivet it to the joists pretty regularly.

                  Simple huh? BTW, you'd repeat the calculations for every thickness you want to try. This is the simple version that does not take into account the effects of spanning multiple joists.

                  Comment


                  • #10
                    Originally posted by Bodybagger View Post
                    Well, not exactly very straightforward...

                    OK, you are spanning 16" joists. To simplify things, we will analyze a 16" long simply supported 12" wide sheet of 3/16" aluminum.

                    E=10,000,000 psi
                    I=(12*(3/16)^3)/12=.00659 in^4
                    Live load=100psf->8.33lb/in
                    Dead load=2.7psf->.225lb/in
                    Ultimate bending moment=((1.6LL+1.2DL)*span^2)/8=[{1.6(8.33)+1.2(.225)*16^2]/8=435 in*lb
                    Ultimate shear= (1.6LL+1.2DL)*span/2=[{1.6(8.33)+1.2(.225)*16]/2=109lb
                    Max deflection=(5*(LL+DL)*span^4)/(384*E*I)=(5*(8.33+.225)*16^4)/(384*10,000,000*.00659)=0.108" which is L/148 (which exceeds the L/360 you'd use for floors)
                    Okay, I understand the calculations, but where did come up with the Live load, Dead load . . . and the factors 1.6 and 1.2?

                    Jim

                    Comment


                    • #11
                      100 psf seems like a good design load for this application. There isn't a category for "motorcycle stand," so you have to estimate what's going to take place. To compare to other design loads, residential floors are generally 40psf live and 10 psf dead, decks and balconies are 60 or 100psf depending on area, and 10psf dead load.

                      2.7psf is the actual dead weight of a 3/16" thick sheet of aluminum.

                      The 1.6 multiplier is a standard factor of safety for live load in LRFD design to account for impact effects and uncertainty in load magnitude... as is 1.2 for dead load (lower uncertainty).

                      Comment


                      • #12
                        Nice explanation. I bet you picked up some civil engineering knowledge somewhere.

                        Comment


                        • #13
                          ok, i defninitely follow the logic of your explanation about temper affecting the yield point (at which permanent deformation occurs), but temper not affecting temporary elastic deflection where the object will return to the original shape. but actually all along i have been referring to temporary elastic deflection.

                          in my example when i said "heat it up with my acetylene torch and it is easier to bend," i was referring to temporary elastic deflection. what i really should have said more precisely was the following: i temporarily heat up the 1/16" sheet with the torch, then let it gradually cool to anneal it. after it is cooled off, i "bend" it, as in flop around like a flat spring in my hands (i didn't mean bend as in fold it in half to permanently deform it).

                          i am imagining the following experiment: take two 2" x 36" strips of 1/16" AL sheet 6061-T6. anneal one sheet and leave the other at T6 temper. then take both sheets and lay them across a 30" wide span (not fixing the endpoints). they will obviously both sag somewhat in the middle. but i am thinking the annealed sheet will sag more. neither strip of sheet metal has been permanently deformed though, since they both return to the original shape when removed.

                          from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
                          Last edited by ridesideways; 02-20-2009, 12:41 PM.

                          Comment


                          • #14
                            Originally posted by ridesideways View Post
                            from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
                            Believe it . . . it's true.

                            The only thing I would add, if you heat your 6061-T6 to 400 F, the yield strength will drop to 15,000 psi and elongation increase from 17% to 28% . . . it will bend more easily, it will bend farther, but it will not be annealed. When cooled back to room temperature, it will return back to 40,000 psi yield strength and 17% elongation as long as the heat cycle is not prolonged . . . 30-60 minutes will not appreciably affect the properties. Heating it for longer periods will over-age it and reduce the yield strength, but not all the way to -O temper.

                            Jim

                            Comment


                            • #15
                              Originally posted by ridesideways View Post
                              i am imagining the following experiment: take two 2" x 36" strips of 1/16" AL sheet 6061-T6. anneal one sheet and leave the other at T6 temper. then take both sheets and lay them across a 30" wide span (not fixing the endpoints). they will obviously both sag somewhat in the middle. but i am thinking the annealed sheet will sag more. neither strip of sheet metal has been permanently deformed though, since they both return to the original shape when removed.

                              from what you are saying, both sheets will sag the same amount. is that really true? if so, i would never have guessed that.
                              The Euler-Bernoulli beam equation contains modulus of elasticity E, section moment of inertia I, and uniform load as the only variables affecting the slope of the beam at any given point.

                              They will sag the same amount provided you have not somehow altered E through metallurgical changes, I through deformation, and the span must remain the same.

                              They may not be exactly the same, but they will be darn close - within experimental error.

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