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Bending strength of timber and modulus of elasticity.

  1. Nov 30, 2015 #1
    I am constructing a glutamine beam as part of a project and need to estimate the load at which the beam fails in bending and the load that will give deflection of length/200.

    I have been allowed to do a none destructive test on each piece of timber to be used in my beam. This has given me a range of natural frequencies from 1335Hz to 1532Hz, and have been given to equations to get modulus of elasticity (v=2*f*l, where f=frequency and l=length of sample, and E=v^2*d where v=velocity and d=density), I have looked into these and they don't seem completely accurate as the moe equation is missing the poissons ratio parts. I have learned that if I use cm for the units my moe gives me dynes/cm^2 which I have got 16204.9kN/m^2 for the specimen with 1523Hz that has l=180.8cm t=2.01cm and b=9.32cm (l=length, t=thickness and b=breadth) and weight of 1.8kg.

    I was wondering if there is a link between modulus of elasticity and bending strength of timber, as the equations for the failure of a beam in bending in eurocode 5 uses the bending strength characteristic, is this conversion possible as I guess this is how timber grading machines work but can't seem to find anything on it.

    v=2*f*l
    v=2*1523*180.8
    v=550716.8m/s

    E=v^2*d
    E=550716.8^2*0.000558(kg/cm^3)
    E=169,235,258.5dynes/cm^2
    Therfore,
    E=169,235,258.5*0.0001
    E=16923.5kN/m^2

    Can I get to bending strength with what I already have.
     
  2. jcsd
  3. Nov 30, 2015 #2
    Or if I could find ultimate tensile stress from what I have I could get a good approximation by calculating the load that would cause my piece to fail under tension, then Med=Ft*z followed by Mmax=Pa for a beam with 2 equal point loads symmetrically spaced.
    I have used,
    Deflection=((Pa)/24EI)(3(l^2)-4(a^2))
    Rearranging this I get
    P=(deflection*24EI)/(3(l^2)-4(a^2))a.
     
  4. Dec 1, 2015 #3

    PhanthomJay

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    I am not familiar with your formula for determining the elastic modulus , but the elastic modulus is more a function of species and moisture content than strength, and for the laminated wood glued timbers I use on occasion, the modulus is on the order 12 GPa, not the 17 MPa you are calculating, so your result is way off by a large multiple. Laminated wood ultimate breaking strengths are on the order of 50 MPa or so. Note that deflections are a function of the load , E modulus, moment of inertia, length, and are independent of the material strength. I had one tested recently for failure I forget the details I think it was Douglas fir laminated wood with a published breaking strength of 7000 psi ( that's about the 50 MPa in SI units I think) and it failed at just about that stress by splitting near the point of max moment at the outermost fiber glue layer.
     
  5. Dec 1, 2015 #4

    SteamKing

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    The units of v here are cm/s, rather than m/s

    Not sure I agree with this density. I calculate 0.0005314 kg/cc (0.5314 g/cc) based on the gross dimensions of the beam given above.
    I get E = 531.4 kg / m3 ⋅ 55072 m2/s2 = 16.12 GPa (Note: 1 Pa = 1 N / m2 = 1 kg / m-s2)

    I find there's less confusion if you work with only one metric system at the time.

    By comparison, steel has an E of approximately 200 GPa. E = 16 GPa seems roughly in the range for the MOE of common hardwoods.

    The bending strength is independent of the modulus of elasticity.
     
  6. Dec 1, 2015 #5

    SteamKing

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    If the material is stressed out of the elastic range, I'm not sure that the usual Bernoulli beam theory is still applicable. Certainly, metal beams loaded beyond the elastic range require special treatment; I sure that beams made out of other materials would also require careful consideration beyond just fiddling with an equation from a handbook.

    The only certain way that I know to find the ultimate tensile stress of a material is to test a sample in a tension test until it, you know, breaks. The sample is tested in pure tension, so that an accurate result can be obtained, since you measure the applied load directly and the area of the cross section can be obtained from measurements of the sample while it is loaded.
     
  7. Dec 2, 2015 #6
    Thank you for your assistance.
     
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