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Calculate the maximum beam deflection

  1. Mar 30, 2007 #1
    Hello-

    This isn't a homework problem, but a real life problem. I'm trying to size a beam for a set of stairs that I'm trying to design. I have some material picked out, and it looks strong enough, but when I try to run some numbers to confirm, they don't appear to match what I've experienced in the real world.

    1. The problem statement, all variables and given/known data

    Calculate the maximum deflection of a simply supported beam given the following:

    Total length: 27 feet
    Material: Steel (E=30x10^6 lb/in^2
    Cross section: See image
    Weight: 0.283 lb/in^3


    2. Relevant equations
    Maximum deflection of uniformly loaded simply supported beam (from Schaum's Outline Series, Strength of Materials 2/ed, P.161)

    y=5/384 * wL^4/EI


    3. The attempt at a solution

    Using the given above, along with the following:

    w = 104.107 lb/ft
    I = 23.05 in^4

    Using the maximum deflection equation from above solving for y yields:

    y=259.23 inches

    This means that the middle of the beam will deflect almost 260 inches due to it's own weight?? I must be doing something wrong.....any help would be appreciated.

    Thanks.

    [​IMG]
     
    Last edited: Mar 30, 2007
  2. jcsd
  3. Mar 30, 2007 #2

    AlephZero

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    I can't see your attachment for some reason but your numbers look inconsistent.

    Density of steel = 0.28 lb/in^3.

    A self weight of 104 lb/ft is a volume of 104/0.28 = 371 in^3 per foot length.

    So the area of beam is 371/12 = 31 in ^2.

    Assuming it was a solid section about 5.5in square (big beam, huh???) that would have I = a^4/12 = 31^2/12 = 80 in^4.

    If it was an open section (I beam, hollow tube, etc) then I would be a lot bigger than 80 in^4. Your 23.05 in^4 doesn't seem consistent with the self weight.
     
    Last edited: Mar 30, 2007
  4. Mar 31, 2007 #3
    AlephZero-

    Thanks for the reply. The 104 that I came up with was from figuring the cross sectional area, multiplying by the density:

    V = lwh
    V = lA
    V = (12)(2.46)
    V = 29.5 in^3

    If I take the volume and divide by the density, I get the weight per unit length:

    29.5 in^3 / (0.283 lb/in^3) = 104.1 lb / ft

    (Schaum's outline series gave me this method...)

    The box beam is a hollow rectangular channel....thin walled (0.090 in wall thickness)

    Thanks for the help....
     
  5. Mar 31, 2007 #4

    AlephZero

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    Oops, the mass of 29.5 in^3 = 29.5 TIMES 0.283 lb/in^3 = 8.35 lb/ft

    Check the units: in^3 x lb/in^3 = lb!

    Also, are you taking the area of a solid beam not a hollow one? The area of a thin walled rectangle is (near enough for practical purposes) (2w+2h)t where w is the width h the height and t the wall thickness, not wh.
     
  6. Mar 31, 2007 #5
    AlephZero-

    You're right....it should be 29.5 * 0.283...my bad...

    The area that I'm looking at is a hollow beam. I took the area of the larger rectangle and subtracted the area of the smaller rectangles, which should leave the area of the thin walled section. Is that incorrect?
     
  7. Mar 31, 2007 #6
    Okay...now that AlephZero showed me my first mistake...I'll try again...

    Using the corrected value for w:

    w=8.34 lb/ft (or 100.08 lb/in) for a uniformly distributed load
    I = 23.05 in^4 (same as above)

    Using the equation for deflection from above, and solving for y yields:

    y=0.00053

    This would indicate that a beam of that length would not deflect due to it's own weight....that seems reasonable...

    Does that look right?
     
  8. Apr 1, 2007 #7

    AlephZero

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    Er...

    Length = 27ft = 324 in
    E = 30x10^6 psi
    I = 23.05 in^4
    w = 8.34 lb/ft = 0.695 lb/in (not 100.08)

    So deflection = (5/384) x (0.695 x 324^4)/(30 x 10^6 x 23.05)
    = 0.144 in according to my calculator

    FWIW I'm still having problems with your I and A seeming inconsistent with each other. What are the actual dimensions of the beam?
     
  9. Apr 1, 2007 #8
    Wrong Formula?

    OK. This is what I get using formulae from Machinery's Handbook 27th edition.

    First, the variables.

    Steel rectangular hollow beam with external dimensions - 2" wide by 5" high. Internal dimensions (subtracting 0.09" beam wall thickness) - 1.82" wide by 4.82" high.

    Beam length = L = 27 feet, (or 12inches/foot x 27ft) = 324 inches

    Steel density ~ 0.284 lb/in^3

    Steel modulus of elasticity = E ~ 29 000 000 lb/in^2

    Beam volume = (beam external area - beam internal area) x beam length
    = (2 x 5 - 1.82 x 4.82) x 324
    = 397.74 in^3

    Beam weight = W = beam volume x density
    = 397.74 x 0.284
    = 112.96 lb

    Second moment of area (aka moment of inertia) = I = (EW x EH^3 - IW x IH^3)/12, where EW - external width, EH - external height, IW - internal width, IH - internal height.
    I = (2 x 5^3 - 1.82 x 4.82^3)/12 = 3.85 in^4

    Beam displacement (under own weight) at mid point = D = 5/384 x W/E x L^3/I. Please note that beam length L is raised to the third power!
    D = 5/384 x 112.96/29000000 x 324^3/3.85 = 0.448"
     
  10. Apr 1, 2007 #9
    Hey-

    I tried to reply to this earlier with my cell, but I guess that it didn't work. The I and A seem inconsistent with each other...interesting....

    The actual dimensions of the beam are as follows. I'm planning on using a 5" tall x 2" wide steel channel welded together such that the overall height will be 10" and the overall width of the cross section will be 2".

    You're right...I screwed up when I calculated the value for w. I also noted that I had an error in my Excel formula, and have since corrected it. The value that I get for the max deflection is:

    y = 0.144 in

    You and your calculator are correct.

    Thanks.
     
    Last edited: Apr 1, 2007
  11. Apr 1, 2007 #10
    Hello meeshu-

    Thanks for your reply....here are my comments....

    I got the same for the variables that I was using...with the exception of the density of the steel (I was using 0.283 lb/in^3) and the modulus of elasticity for steel (I used 30 x 10^6 lb/in^2)

    The figures that I got were different. I think that you were calculating for one channel. The cross section that I'm working with is two channels welded together. I got the following:

    volume = 795.48 in^3
    weight = 225.92 lb

    I got a different value for I, namely 23.04 in^4. The beam in the diagram is made up of two 5" x 2" channels welded together. I took this to be a new channel section with the following parameters:

    width = 2"
    height = 10"

    It looks like you calculated I for a single channel section. I used the parallel axis theorem to figure out I, but I might have screwed that up.

    I guess that I'm puzzled by this. I checked another reference book, An Introduction To The Mechanics Of Solids, Second Edition With SI Units, and it lists the formula for maximum deflection of a uniformly loaded beam as the same equation that I was using, with the L term raised to the 4th power. I'm attaching a scan of the page:

    [​IMG]

    Thanks for your reply. I finally got smart, and put all of this into an Excel spreadsheet. If I use the figures that you had for the density and E, I get the following for the max deflection:

    y = 0.1497

    I am not sure that I have the value for the distributed load correct. I figured that as follows:

    weight of beam / length of beam (lb / in)
    225.92 lb / 324 in = 0.697 lb/in

    Does that look correct?
     
    Last edited: Apr 1, 2007
  12. Apr 1, 2007 #11
    Sorry, I didn't realize that you were proposing TWO beams welded together!? My calculation was just for one beam. I'll have another look at your proposal for two beams shortly.

    The formula you show (5/384 x W/E x L^4/I) for calculating displacement, I believe to be incorrect (a printing error?). If you do a dimensional analysis to verify (final) units, you should find that the above formula will result in length^2.

    5 and 384 are dimensionless. W is in lb. E is in lb/in^2. L is in inches. I is in in^4. Put units all together, we have -

    5/384 x lb/lb/in^2 x in^4/in^4
    = 5/384 x lb.in^2/lb x in^4/in^4 (by algebraic manipulation)
    = 5/384 x in^2. (lb and in^4 cancel, they divide by themselves)

    This is clearly incorrect, as you can't have diplacement is inches^2!

    Machinery's Handbook formula (5/384 x W/E x L^3/I).

    Units analysis produces - 5/384 x lb/lb/in^2 x in^3/in^4
    = 5/384 x in^2 x in^3/in^4 (lb cancels out)
    = 5/394 x in^5/in^4
    = 5/384 x in.

    Therefore, this formula appears to produce the correct displacement units (inch, instead of inch^2).

    So, the other sources which indicate length^4, I believe to be incorrect, as the units do not work out.

    I'll get back on the other details shortly.

    EDIT:

    Total weight of two beams is 2 x 112.96lb = 225.92lb, and is 8.367lb/ft, or 0.697lb/ft

    Using top of upper beam as reference for calculating moments.

    Moment of upper beam is as calculated previously, = 3.85in^4

    Moment of lower beam is moment of beam calculated with reference to top of lower beam, plus adjustment of calculated moment with reference to top of upper beam using parallel axis theorem.

    Moment of lower beam (with reference to its top) is the same as upper beam moment, = 3.85in^4.
    Adjusting lower beam moment with reference to top of upper beam -
    I lower beam = initial calculated moment + area lower beam x difference in moment reference points^2
    =3.85in^4 + (5in x 2in - 4.82in x 1.82in) x 5^2in
    =3.85in^4 + 1.2276in^2 x 25in^2
    =34.54in^4

    Total moment of beams = 3.85in^4 + 34.54in^4 = 38.39in^4

    Beams displacement = 5/384 x 225.92lb/29000000lb/in^2 x 324^3in/38.39in^4 = 0.090in
     
    Last edited: Apr 2, 2007
  13. Apr 2, 2007 #12

    AlephZero

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    If W is the load per unit length, the correct formula has L^4.
    If W is the total load, the correct formula has L^3.

    That is not surprising, since the total load = load / unit length times the length.... but you are not the first people who have taken a formula from a handbook and misinterpreted the notation!

    I don't think meeshu's M.I calculation is right. Remember the neutral axis of the combined beam is in the centre not at the neutral axis of the lower beam.

    I get 2 * 3.85 (twice the single beam M.I)
    + 2 * 2.5^2 * 1.23 (parallel axis theorem for both beams)
    = 23.08

    which is the same as the OP.
     
  14. Apr 2, 2007 #13
    meeshu-

    Thanks for the reply. In looking at the formula that I was using, w is the distributed load, not the weight of the beam. The units are lb/in and when I checked the units, they seem to work out to me. I'll have to check when I get home later today.

    I will have to take a look at the calculations for I. I was assuming the neutral axis for the combined beam was along the interface point between them.

    Thanks again for your help.

    EDIT

    I just read Alephzero's reply. I must have been replying while he was as well. Thanks for your help.
     
    Last edited: Apr 2, 2007
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