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Simple beam

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Dense No 1 grade of Douglas Fir-Larch, 6x12 is to be used as a simple beam spanning 10 feet with a 2 feet overhang. The beam has the loading shown in the sketch I've attached. Determine the maximum value of the concentrated load "P" to maintain the structural integrity associated with bearing. The beam rest on 4x12 members at both supports.


    2. Relevant equations
    fv= VQ/Ib
    fb= Mc/I

    fc[itex]\bot[/itex] = bearing/common area


    3. The attempt at a solution

    I found these for the 4x12:
    S=73.83
    I= 415.3
    A=39.38
    Q=72
    c=6
    fc[itex]\bot[/itex] =625
    common area = 3.5(5.5)???
    bearing=12031.25???

    I'm not sure if this is right? So the max "P" would be 12031.25#?

    or Fc[itex]\bot[/itex] = 625 for 2-4in boards and Fc[itex]\bot[/itex] = 730 for 5" and bigger

    730= V/3.5x3.5 - V=8942.5
    625= V/3.5x3.5 - V=7656.25
     

    Attached Files:

    Last edited: Jun 19, 2012
  2. jcsd
  3. Jun 19, 2012 #2
    Wood tends to be quite strong in compression - think railroad ties - so the part of the structure that limits the load is not the 4X12 supports. You must determine the reactions at the supports, then determine the maximum moment in the beam. To determine the reactions, you should include the weight of the beam itself. Then adjust P so that the maximum allowable stress is attained.
     
  4. Jun 19, 2012 #3
    This is what I've came up with:

    By adding 16,075# 5.75ft from R2
    R1(10)+2000(1)=16075(5.75)+10000(5)
    R1=14043.125
    R2=14031.875

    Fc allowable is 730 for Dense No1

    729.51=14043.125/(3.5)(5.5) for R1
    728.93=14031.875/(3.5)(5.5) for R2

    So am I getting the correct answer?? Max allowable Load for P would be 16,075#
     
  5. Jun 20, 2012 #4

    PhanthomJay

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    I don't know where your 5.75 comes from, or your 16 075. But essentially, your method
    for determining P based on allowable bearing pressure at the supports is ok. However , as noted in post 2, bending stress in the beam may control the design.
     
  6. Jun 20, 2012 #5
    The 5.75 is the distance "P" Is from the R2 reaction or 4.25 from R1. 16075 is the load "P" in lbs plus the 1000#/1 on the beam that gives me my reactions on R1and R2.

    Bending stress is fb? Allowable is Fb? I think I maybe over thinking this problem
     
  7. Jun 20, 2012 #6

    PhanthomJay

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    Oh I didn't realize that the distance from P to the supports was given...your sketch didn't show that . So yes you could put that 16 k load there and your support bearing stress is ok. But although the problem didn't ask about bending stresses, they soould be checked by calculating the max moment in the beam and the max bending stress fb vs the allowable bending stress Fb (about 1500 psi ?) and that may end up controlling the value of P.
     
  8. Jun 20, 2012 #7
    The distance from P was not given, its what I calculated and solved for when I did my reactions.

    I did the bending stresses and calculations this is what I came up with:

    Fb=1550
    fb=M/S =417.92psi
    M=50652
    S=121.2

    So by my calculations it would hold
     
  9. Jun 20, 2012 #8

    PhanthomJay

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    There are a couple of issues here. Your value for the max moment is in foot pounds. This must first be converted to inch pounds. Beyond that, you seem to have chosen a location for P that yields equal left and right support reaction forces. Is the problem worded that way?
     
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