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Bending stress

  1. Jun 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Rafters are to be used on 24in centers for a roof span of 16ft. Live load is 20psf (without snow) and the dead load is 15psf, including the weight of the rafters. Find the rafter size required for Douglas fir-larch of No 1 grade and No 2 grade, based on bending stress

    2. Relevant equations



    3. The attempt at a solution

    S= M/Fb

    I dont know the Moment though??? so I dont know where to start on this problem
     
  2. jcsd
  3. Jun 16, 2012 #2
    It looks like the length of your beam is 16 ft, and the beam has a uniform load distribution along its length. Each beam supports a 2 ft width of roof. First find the weight supported per unit length of each beam (rafter). It probably would be reasonable to assume that each beam is simply supported at its ends. So determine the vertical shear force as a function of position along the beam (assuming that each end supports half the weight), and then integrate the bending moment subject to the boundary condition that the moments on the ends are both equal to zero. Calculate the displacement distribution next (based on the assumed cross sectional dimensions of the rafters, possibly using commercially available dimensions). This will allow you to get the local radius of curvature, and then the bending strain on the outside of the bend. This will give you the bending stress on the outside of the bend. Look up the tensile strength of Douglas fir larch, and compare this with your calculated tensile stress. If the tensile stress exceeds the tensile strength, choose a more robust commercial cross section.
     
  4. Jun 16, 2012 #3

    PhanthomJay

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    Alternatively, to save some steps, once you determine the max moment of a simply supported beam under a uniformly distributed load, use your formula to determine S and then cross sections of available timber that yield that minimum value.
     
  5. Jun 16, 2012 #4
    So the Live load is just Area(LL)=Area(DL), which equals 1120.
    So M=WL/8 and V=W/2
    M=2240
    V=560
    No 1 lumber Fb=1000
    No 2 lumber Fb=900

    S=M/Fb
    2240(12)/1000 = 26.88 in^3
    2240(12)/900 = 29.87 in^3

    Which the table says 2x12 in both lumber sizes will cover it
     
  6. Jun 17, 2012 #5
    Yes. This is an excellent point. Once you know the maximum bending moment, you don't need to solve for the downward displacement. You can get the maximum stress at the outside of the bend directly from the bending moment, knowing the cross section of the beam.
     
  7. Jun 17, 2012 #6

    PhanthomJay

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    And the 2 x 12's look good!
     
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