1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constructing a Deck with Beam supports

  1. Jun 20, 2011 #1
    Hello all,

    I am currently trying to figure out a cheap way to build a "bridge" that will go in my back yard. It will primarily be used to drive my 2000lb lawn mower over a small creek. I want to support it with 2 to beams at a 30 ft span as I will rest 2x8 or 2X10 pieces of wood on it. It will be flat and just have a small incline on each side to get on and off it. I need to figure out what beam size to use (preferably I-beam). Any formulas or any advice that applies to my situation helps a lot. Thank you in advance. I have also attached a sketch so you can visualize what I am discussing.

    Attached Files:

    • jhjk.jpg
      File size:
      18.4 KB
  2. jcsd
  3. Jun 22, 2011 #2
    Have you google Beam Theory? You should get quite a few hits (just ignore the ones about Bernoulli).

    Then, look for the formulas for the cases of a beam supported on both ends.

    Then, pick your beam (moment of innertia)

    Because you are going to operate on the elastic part, you can apply superposition and so, you need to apply the formula twice for two different loads and add the deflections up ...keep increasing your I-beam until an acceptable deflection in the middle...1 inch? 2 inch? I don't know ...I took some classes over 20 years ago...don't remember much..

    So, first, you need to apply a distributed load due to all the pieces of wood that you are planning to put on the bridge....there is a formula for this

    Second, you need to apply a concentrated load due to the lawn mower and place it in the worst place...in the middle of the bridge...there should be a formula for this already , too.

    If you are planning on using at least two I-beams, don't forget to divide the forces (weights) by 2 when applying to the formulas.

    Hope this helps.
  4. Jun 22, 2011 #3
    The way you've drawn the picture, it appears you are thinking of having the 2x boards span the 30 foot gap across the stream. Laying 2x boards flat, spanning from I-beam to I-beam will not provide a useful beam/bridge. 2x boards laid flat will not support much of a load.

    Consider making the bridge with 2 beams spanning the stream, with 2x boards spanning between the beams to form the roadway. The beams might be fabricated from 2x boards on edge, with several laminated together to form strong beams. Those beams would then rest on footers on each side of the stream. The footers could be the I-beams, or more economically they could be concrete blocks, setting on a pad to distribute the loading into the ground. You would not need to have the blocks run from beam to beam.

    I've not run numbers for these beams, but think of each beam being formed by laminating 4 2x10's. You would need to splice about 3 2x10 to get the length. The splice joints could be made by butting them end to end and sandwiching/tying them to the adjacent boards with long screws or bolts. Splice joints would need to be staggered such they are some distance apart.

    The roadway could then be laid either on top of the beams, or set on a ledger board fastened to the side of the beams. If the bridge is fairly narrow (4 ft?) you may be able to just use a single layer of 2x. However, I suspect you will need to at least double them up, and tie (screw) the layers together. Or you might consider fastening a 2x4 to the underside of each roadway piece to form a 'T' beam.

    Some calc's do need to be run, as suggested by Gsal.
  5. Jun 22, 2011 #4
    Oh, my, I just read DickL's message and made me look at the picture....

    I just noticed that my description of the problem does not correspond to the way you have drawn the bridge...instead, I went under the assumption that you were going to use the I-beams to span the width of the stream....just like DickL suggests....that's the way to go.

    So, please read my post previous post with that in mind and it should make much more sense.

    So, think of your bridge like an upside down railroad track...instead of the I-beams on top of the sleepers, you want to put 2 I-beams that span the stream (one end of the I-beam is on one side of the stream and the other one on the other side) and on top of the I-beams you want to put sleepers across so that you can drive over them...
  6. Jun 22, 2011 #5
    Thanks. And yes, i realized my first idea didn't make very much sense. I do plan to go with what you and DickL are talking about. I now believe I have to tackle two problems:

    1. determine the size of planks required to carry a 2000lb dynamic load across a 6' span since that is how far apart I plan to place the two I-Beams
    2. determine the size of a 30' I beam required to carry the weight of the planks plus a 2000# dynamic load mid-span

    Any suggestions, formulas, calculations....anything at all is much appreciated
  7. Jun 22, 2011 #6
    Here is a nice tutorial


    By the way, if you set your two I-beams as much apart as the tires of your lawn mower, it is almost as if your 2x8's or 3x6's (a bit thicker) wouldn't even have to come much into play since they would simply have I-beam support right underneath of where the load is being applied on top of them...even if you are a bit off, I wouldn't worry much about that...we would be talking about a 500 lb per point.

    Also, another refinement when applying the point load on the I-beam, you need to apply two 500 lb point loads as much apart as the distance between the front and rear axis of the lawn mower to one I-beam and make sure it can withstand...
  8. Jun 22, 2011 #7
    For maximum stength your I beams should be placed at the one third points of the cross span, not at the edges.

    How are you going to attach the cross timbers? You need to be aware that the inner flange surfaces of I beams are not parallel they taper.
    You can get special clamps google 'lindapter' for this.

    You must bolt down one end only of the I beams, leaving the other free to expand. If you bolt down both ends they will buckle.
  9. Jun 23, 2011 #8


    User Avatar
    Science Advisor
    Homework Helper

    That appears to be incorrect. For maximum strength, the I-beams probably would be placed at the one-sixth points of the joists (or deck beams), measured from the joist ends. (However, it might depend on the vehicle wheel base.)
  10. Jun 23, 2011 #9
    Hello, nvn I suppose it depends upon the way you usually look at these things.
    I normally consider the half section for symmetrical loading.

    Anyway an overhang of one third of the half or one sixth overall is not quite theoretically perfect but quite close.

    Since the vehicle is likely to have wide tyres and or rollers I have modelled a distributed load for the cross section.

    You can see the theoretical best place for a simple support is 0.414 times the hald width in from the end.

    Attached Files:

  11. Jun 23, 2011 #10


    User Avatar
    Science Advisor
    Homework Helper

    srw2104: You must establish not only a spacing between the two I-beam centerlines, but also the length of your deck lumber, and the maximum amount a wheel centerpoint can deviate from directly above an I-beam centerline. Assuming your lawn mower wheel can deviate from directly above an I-beam by up to 450 mm outboard, then 38 x 184 mm (or 38 x 235 mm) lumber would be inadequate. It appears 89 x 89 mm southern pine lumber, grade number 2, might be adequate, assuming the tensile side of the lumber cross section (directly over an I-beam) is not drilled nor damaged.
  12. Jun 23, 2011 #11
    Some quick calculations run as follows:

    Wt of 2 inch thick 6 foot wide timber decking 6 foot per I beam

    \frac{1}{2}*6*\frac{1}{6}*30 = 15cuft \\
    15cuft\,@\,40lbs/cuft = 600lbs. \\

    Trial Wt of I beam @ 100 lbs per foot = 100 * 30 =3000lbs.

    This design will be moment controlled with max pointload of say 1500 lbs in centre

    Max moment is

    [tex]M = \left\{ {\frac{{3600*30}}{8} + \frac{{1500*30}}{4}} \right\}lbs - ft[/tex]

    =297,000 lbs-ins

    The required section modulus at working stress of 20,000 psi is therefore

    [tex]\frac{{297000}}{{20000}}in{s^3} = 14.85in{s^3}[/tex]

    I do not have many american standard tables but what I have suggests that a
    20 x 7 x 95lbs/ft beam would just suffice (S=14) since I have deliberately overestimated for the trial.
    A 24 x 7.875 x 105.9 lbs per ft would be safer at S=20

    You should get these calcs professionally checked, they can probably be refined further.

    You should also note the approx weight of a suitable I beam at nearly 1.5 tons.
    Last edited: Jun 23, 2011
  13. Jun 24, 2011 #12


    User Avatar
    Science Advisor
    Homework Helper

    Studiot: If you get a chance, would you be able to state the I-beam size listed in post 11 as a Euronorm standard HE size, so we may know what size you are referring to?
  14. Jun 24, 2011 #13
    I was rather rushing last night and looked in the wrong column.
    When I look in the correct one I see that a much smaller I beam will suffice
    An 8 x 4 x 23lb/ft looks better.
    This reduces the beam weight to 700 lbs
    Thus the max moment is now 166500
    and the required section modulus greater than 8.3

    The 8 x 4 has a section modulus of 16 (correct this time).
    The spare capacity would leave room for bolts, braces and what have you.
    If you need a thicker deck then things may change again.

    Further refinement is possible.


    I have worked this in imperial units because I think the OP is in America, where they still use proper units.

    I do not have any information about Eurocodes, but can tell you that the 8 x 4 is roughly equivalent to a BS5950 or BS5400 203 x 102 I section.
  15. Jun 25, 2011 #14


    User Avatar
    Science Advisor
    Homework Helper

    Studiot: It appears you are referring to an imperial I-beam called S8 x 23. Your maximum moment is slightly incorrect. You might want to check your numbers.

    More importantly, your I-beam is overstressed. The laterally-unbraced length is 9.14 m; but it appears you ignored lateral-torsional buckling, which is a leading cause of steel structure failures.

    The I-beam would instead need to be, at least, a European standard HE 200 AA, or an HE 180 A, or an imperial W8 x 24.
  16. Jun 25, 2011 #15
    Yes you are right its 193500, making the required modulus 9.675.

    The American 8 x 4 I section will easily accomodate this.

    If this were a real design it would all be taken care of in the next cycle, note the allowance above.

    Attached Files:

    Last edited: Jun 25, 2011
  17. Jun 26, 2011 #16


    User Avatar
    Science Advisor
    Homework Helper

    No, as I mentioned in post 14, the S8 x 23 is overstressed. There is nothing really stable to brace to, in this particular structure, if you use an unstable I-beam. Regardless, for approximately the same mass per unit length, one can instead specify a stable I-beam (listed in the last paragraph of post 14), instead of an unstable I-beam.
  18. Jun 26, 2011 #17
    Conventional web stiffeners?
  19. Jun 27, 2011 #18


    User Avatar
    Science Advisor
    Homework Helper

    No, just buy the correct I-beam size (or larger), to start with. And then you do not need to waste time and money trying (unsuccessfully) to jury-rig an inadequate design. Even if srw2104 gets an I-beam 2 to 7 kg/m larger than the minimum required size, no problem.
  20. Jun 28, 2011 #19
    This was all very useful information...thank you for all the help. It is much appreciated
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook