How to calculate max load for a VLV beam?

  1. Jul 8, 2009 #1

    I'm looking for a formula to determine the max load on a piece of Laminated Veneer Lumber (LVL) supported at each end. Although I've calculated the load on each component of bridges a thousand times way back in my Statics class, nobody ever showed us how to find the max allowable load given a specific material. I have the MOE and MOR values. Does anybody know any useful formulas?

  2. jcsd
  3. Jul 9, 2009 #2
    If you're looking for max allowable loads, you're probably looking for some kind of a published table from the manufacturer. Otherwise the best you can do is guess... laminated beams are composite and their strength depends not only on the strength of the individual materials but also on the interface between the materials (between each layer and the glue used to stick them together).

  4. Jul 9, 2009 #3
    This question is hard to understand. Can you post some picture with value you have no idea how to calculate? In example, easy formula like R>N/A, you can calculate N<R/A [N-max load in kN, A-section area m^2, R- max elements strenght in compresion in kN/m^2]. If you know R and A get the value needed-N.
  5. Jul 19, 2009 #4
    Sorry it took so long to get back...been kinda hectic here.

    Anyways, I'm simply looking for the maximum load a beam can take before failure. The beam is horizontal and supported on each end, like a bridge. Let's call it a bridge and say I'm trying to figure out the heaviest truck that could go over this beam.

    Here is a PDF with values for the material I'm looking at:

    Can I just convert the compression values to metric and solve for N or am I looking for the Max Moment, or even something else?

    Thanks again.
  6. Jul 19, 2009 #5
    I assume that the beam is loaded with point load wich could be placed anywhere on the beam (truck).
    1) Max design moment M will be when the truck will be in the midle of the beam. M=qL/4
    M<R*W - > Your max load for moment is q<(Rb*W)/(4/L)
    q- design load kN
    W - design section modulus [w=bh^2)/6] h-section hight, b-width
    Rb - design beam resistance in bending. (2700PSI from link)
    2) Max shear force Q will be when the tuck will be near the support. So Q=q
    Your max load for the shear force is q< Rv*(I*b)/S
    Rv - design beam resistance in ''chopping''. (320PSI from link)
    I- design moment of inertia [I=bh^3/12]
    S- staticmoment [(b*h/2)*h/4 if i remeber right]
    3) Deformations... do not remember the formul for point load

    correct me if i am somewhere wrong
  7. Jul 20, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper

    (1) The equation in item 1 of post 5 is incorrect, and should instead be q < 0.6667*Rb*b*(h^2)/L, where q = applied midspan point load (N), h = cross-sectional height (mm), b = cross-sectional width (mm), L = beam length (mm), and Rb = allowable bending stress (MPa). From the link, Rb = 20.00 MPa.

    (2) Item 2 of post 5 is correct, and simplifies to q < 0.6667*Rv*b*h, where q = applied point load (N), h = cross-sectional height (mm), b = cross-sectional width (mm), and Rv = allowable shear stress (MPa). From the link, Rv = 2.206 MPa.
  8. Jul 20, 2009 #7
    ''The equation in item 1 of post 5 is incorrect, and should instead be q < 0.6667*Rb*b*(h^2)/L''
    Agree. Should be q<Rb*W*4/L
  9. Jul 20, 2009 #8
    Hmmm....with equation 1 I get that a 30 ft beam (8" X 12'') can hold about 6200 lbs. I know LVL is strong, but that seems impossible with a span of 30 feet.
  10. Jul 21, 2009 #9
    So im metric, for me it will be L=9.15m, h=30cm b=20cm, load N=28kN. With these numbers i have stress sig=21.3MPa more then Rb=20MPa. Doesnt hold for me maybe becouse of the round up. Nevertheless we dont know what is your construction, what doas it carry, nor i dont know your building codes to say it will hold or not. 6200lbs is design load not factual load you can put on. design load=factual load * safty factor. In example, by eurocode safty factor for living loads is 1.5. +If i was you i would worry about the deformations in such beam length too.
  11. Jul 21, 2009 #10
    No doubt about the safety factor. There normally are flaws in the beams and one needs to take into consideration any points of higher stress such as bolts and whatnot. The beam itself is actually supported at four points making the longest section 12 ft, but I wanted to see what the beam could hold if it were not supported anywhere other than the ends to try to determine if the numbers where accurate. Also a truck is not really a single point load due to wheelbase anyways. What I'm saying is I just wanted an idea of the strength of LVL, and now I can compare it to regular wood and aluminum.
    Last edited: Jul 21, 2009
  12. Jul 21, 2009 #11


    User Avatar
    Science Advisor
    Homework Helper

    giant016: No, the safety factor is already included in that allowable bending stress value, Rb. Your math in post 8 is correct. If the cross-sectional dimensions you listed are accurate, the beam would hold the value you computed, if you provide one lateral brace at midspan. (Let us know if you are not familiar with what lateral support to the compression flange is.) If you do not want to provide the midspan lateral support, and you still want to use a span length of 9144 mm, then you must use Rb = 18.70 MPa, instead of 20.00. (Note that this reduced Rb value is not a general rule and only applies to your given cross-sectional dimensions and span length.)

    The rounding errors in numbers posted by archis in post 9 happen to be all additive, giving a 6.8 % cumulative error in his answer.
  13. Jul 23, 2009 #12
    It would probably be easier for me to post an actual pic. I know that the main beams across are not necessarily the weakest points, but I just wanted to see what the max load would be if I did not have any supports. Here are two screenshots:


    The main beam is split into three sections with a total of 320 inches. The middle is 120 and the end sections are 100'' each. The support beams make a right angle. The design needs to be updated since I will be using 3 4X12 beams for the main support and also the right angle support beams will be shorter.

    Thanks for all the help so far.
  14. Jul 23, 2009 #13


    User Avatar
    Science Advisor
    Homework Helper

    Just to clarify, in post 11, I was referring to the hypothetical, 9144 mm beam you commented on in posts 8 and 10.
  15. Jul 24, 2009 #14
    Nice bridge design.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?