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What's this steel bearing plate going to do under this steel I beam?

  1. Aug 23, 2012 #1
    I'm doing some work on my house at the moment, and have had a design produced by a structural engineer for a steel I-beam to span a newly made (4 meter) opening in a brick wall. They've given me a design (see the attached picture for my interpretation) for the bearing of the I-beam which I can't get my head around (I've phoned and checked that's what they definitely intended, and they've confirmed that it i...). I have no engineering, but I did study Physics and Maths+Mechanics to English "A-Level"...

    The steel I beam has a 12mm steel "spreader" plate underneath it (or rather under the 150mm piece of it which bears on the wall), which is not welded to the I beam. As far as I understand it, the job of the spreader plate is to distribute the load which is imposed by the steel beam over the underlying masonry so as not to put it under excessive compressive force, but I can't see how that spreader plate is going to do this. Moreover, won't the masonry pier itself end up with 47.3N acting on a line only 75mm from it's edge (maybe closer to the edge if the beam deflects a bit)?

    If both the wall and the I beam ran the other way (i.e. at right angles to the span of the I beam when viewed on plan), then I can see that the spreader plate would work just fine (and I've seen such designs both on paper and built). Similarly, if the I beam extended the extra 180mm over the top of the spreader plate, then I assume there would be no problem there either (but then the plate itself would be redundant)?

    I'm hoping that my doubts are unfounded, and the design is fine... Any comments?
     

    Attached Files:

  2. jcsd
  3. Aug 24, 2012 #2
    You are assuming that the "spreader" plate it is there in order to "distribute" the bearing load.

    Perhaps this is not the case?

    Instead, maybe the "spreader" plate will be bolted down and is placed there in order to resist the tension in the row of bricks just under the I-beam. The engineer doesn't want that edge brick to come off, for example. This is plain old unreinforced brick, right?

    Image search for corbel design and you see similar ideas.
     
  4. Aug 24, 2012 #3
    The design calls for no bolting or welding - from the bottom up there is, unreinforced on this a mortar bed, then the steel plate (just bedded on the mortar, no other connections to the masonry), then the I beam sits on that.

    The engineer refers to it as a "spreader plate", stating:

    "All padstones to be 330x100x12mm thick spreader plate. All beams to have 150mm end bearing".

    The sketch included in the design then shows the arrangement which I've drawn above (I've checked through this with her on the phone, but I haven't yet sent through the graphic which I did above).

    I did a check using some beam design software "eurobeam", and that software includes a portion which designs a concrete padstone or steel spreader plate to go under the end-bearing of the steel I beam. Using that module of the software - a 330x100 horizontal padstone or plate was sufficient, but 150x100 was insufficient to distribute load onto the masonry without exceeding the maximum allowable compression in the underlying masonry. This software takes into account brick/block + mortar strength and defects etc.

    So I'm reasonably sure that the intention behind the spreader plate is to "distribute" the load over a larger portion of the masonry, so-as not to overload it - but I don't see how it can do that if it's not welded or bolted to the I beam (I've checked, and there is no weld or bolting in the proposed design).
     
  5. Aug 24, 2012 #4
    Personally I doubt that the proposed steel bearing plate meets the codes for bearing.

    The bearing code bearing capacity of new ordinary brickwork is around 2 to 5 N/mm2. London building control requires less than 0.5 N/mm2 for old brickwork.

    Using your figures and some fag packet calculations I could not see proposed beam sizes so I have estimated my own.

    Calling your end load 50kN that is 100kN distributed load on a 4000mm beam ie 25kN/m

    Limiting the deflection to say 5mm (span/800) and entering table 16.4 of the Steel Designer Manual with

    α = 800 extracting K =4.7

    Thus Irequired = KWL2 = 4.7 *100*42 = say 7500 cm4

    Entering Universal Beam tables the nearest above is a 305 x 127 x 42kg/m beam Ixx = 8159 cm4

    This has a web thickness of 12mm and a flange thickness of 8mm.

    Assuming the 50kN load spreads at 45° a point 50kN load corresponds to a pressure of

    At the base of the web vertical shear stress = 50000/150*12 = 28 N/mm2

    At the base of an 8mm flange this reduces to 50000/150*(12+8+8) = 12 N/mm2,
    which is still too high.


    Adding a 12 mm thick pad further reduces this to 50000/150*(12+20+20) = 6.5 N/mm2


    This would be acceptable on high stength concrete blockwork or engineering bricks.
     
    Last edited: Aug 24, 2012
  6. Aug 24, 2012 #5
    OP, why do you keep mentioning connecting the spreader plate to the beam? How would that help?

    Studiot, I don't understand what stress demands you are calculating.. Without the spreader plate wouldn't the flange WIDTH be the only relevant dimension that you need once youve sized the beam? With the spreader plate, why do you need to know anything about the beam?
     
  7. Aug 24, 2012 #6
    The vertical shear force in the web appears as a shear stress equal to the force divided by the sectional area of the web over the support. I have taken this area to be 150mm x web thickness.

    This is therefore the stress at the root of the web ie the junction of the web and the flange.

    This stress does not magically spread over the full width of the flange immediately on entry.

    I have taken the conventional assumption of a 45° spread of this stress which allow calculation of the stress at any lower section eg the bottom of the flange.

    Making the flange effectively thicker continues that spread.

    You are correct there is no structural difference whether the plate and flange are connected or not.
     
    Last edited: Aug 24, 2012
  8. Aug 24, 2012 #7

    AlephZero

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    I think the main purpose of the plate is to provide a flat level surface for the beam to sit on, and the mortar bed under it ensures that the loiad is distributed over the whole area, rather than just the points where the flange of the beam might touch the uneven brickwork.

    You can ensure the plates at each end of the beam are properly aligned with each other and allow the mortar to set and fix them in position, without having to deal with the weight of the beam and whatever loads it will support.

    I'm not a CEng, so no comment on whether the area of the plates or brickwork is sufficient to carry the load. I would expect that meeting the building regulations in your country is more relevant than trying to do an "accurate" calculation.
     
  9. Aug 25, 2012 #8
    Why do you think a plate would achieve this, better than proper bedding?

    I could understand the plate as a bed to slide on if thermal considerations were important but say the temperature varies ±10°C the expansion will be less than 1 mm.
     
  10. Aug 25, 2012 #9
    I had a feeling that if it was unrestrained, then under load the plate might rotate about the end of the beam when a load was introduced (since the beam is bearing on less than 50% of the plate, but the reaction to the plate is over 100% of the length). I've since done an experiment which seems to suggest that this might actually be correct:

    http://youtu.be/Hoe_0PADGrs

    ... standing in for the I beam, I have a length of 18mm OSB/3, the spreader plate is being impersonated by a piece of hardboard (probably a bit harsh on the spreader plate, the hardboard has way too much flex - maybe I should try and find a bit of ply), and I have a piece of polyethylene foam to take the place of the masonry. Although the polythene deflects way more than the masonry, hopefully it shows where the load would be concentrated, as the masonry will deform slightly prior to cracking?

    I put the plate in two different arrangements:

    In the first, the beam bears onto the whole of the plate.

    In the second, the plate is positioned roughly as it would be in the proposed design (i.e. the beam bears down on a little under 50% of the plate).

    In both cases, the load was imposed vertically (as best I could manage), and was approximately the same magnitude (just enough to lift my knees off the ground).

    For the first case, the load distribution seems to be uniform - as expected.

    For the second case, it looks to me like the plate might be acting as a lever and that perhaps the load is actually being concentrated at the extreme edge of the masonry (i.e. the opposite of the desired effect). This seems like a bad idea to me, since isn't that exactly where the masonry will most prone to failure by shear cracking? When this shear cracking occurs, then the next bit of masonry will now have the greatest load, and may crack, and so-on.

    With the beam covering more than 50% of the bearing plate, this effect doesn't occur, and the compression seems to be uniform over the whole polythene block once again (ignoring any plate deflection effects - which presumably ARE correctly accounted for in the design).

    Any thoughts greatly appreciated!

    Tim.
     
    Last edited: Aug 25, 2012
  11. Aug 25, 2012 #10
    I share your doubts about the allowable bearing pressure onto the masonry wall (unless the stub wall is rebuilt) - and that information is very useful as the wall itself is about 60 miles from London in Brighton, and made from similarly naff old bricks with poor mortar.

    Setting that aside for the time being, I was suspicious about the entire design of the bearing plate orientation, since if that's not actually spreading the load in the location and orientation that has been specified (and the engineer has confirmed that this is their intended orientation + position) then it doesn't matter how big you make the plate - all bets are off!

    So, I wanted to get that established before progressing further! The fact the the engineer didn't make any comment about the original masonry bearing pressure, or give any calcs for the bearing plate made me more suspicious.

    So... is the bearing plate doing what the engineer thinks it is (uniform load distribution) or not?
     
  12. Aug 25, 2012 #11
    Well you were obviously not interested in mine before.

    I doubt your building inspector will be interested in yours.
     
  13. Aug 25, 2012 #12
    Sorry, I didn't wish to cause offence, I was working my way through the comments, and hadn't yet replied your second comment (although I did just get in there with a response to your first before you posted the one above)!
     
  14. Aug 25, 2012 #13
    Of course it is, however the thicker it is the the lower that pressure.

    Look more carefully at my earlier post and try to provide the information I had to guess at. Then a more accurate calculation can be substtituted.

    As a matter of interest if you cast a proper concrete bearing pad the allowable bearing pressure is 0.5 - 0.6 fcu
     
  15. Aug 25, 2012 #14
    I'd be extremely happy to do that, but I'd really like to try and satisfy my doubts about the bearing plate orientation first. In the video I made, the plate starts to rotate about the end of the beam when under load. I'm quite prepared to believe that this wouldn't be an issue in the building, but I cannot yet convince myself that is the case...

    It's this rotation and perhaps the potential for the load to be concentrated at the extreme edge of the stub wall which is unsettling me - what is the difference between my model and the actual proposed design which means that rotation cannot happen? What are the forces like at the extreme end of the I beam if that does happen? Ordinarily (or at least in all designs which I've seen), the centre of mass of the plate is underneath the beam (in fact it's underneath the beam web), but in this case it's 15mm horizontally beyond the end of the beam...
     
    Last edited: Aug 25, 2012
  16. Aug 25, 2012 #15
    OK, how about we consider some hypothetical extreme cases, which have some relation to the problem. If we assume:

    1. The steel beam and plate do not deflect.
    2. The masonry is slightly compressible (and uniformly compressible), and for the sake of argument only in the vertical direction.
    3. The steel beam rests on the plate, but is not connected to it by any other means.

    The length (in side view) of the bearing plate is "b1"
    The length (in side view) that the beam bears onto the plate is "L".
    The force "f" is imposed down by the beam onto the plate.

    and then try to ask the question:

    What is the pressure imposed on the masonry in each case, and how does it vary if:

    i. The beam bears on the plate over it's full length (L == b1)
    ii. The beam bears on a small portion of the plate at its edge only (L is much less than b1 e.g. L == (b1/30))
    iii. An intermediate case.

    See attached pictures.

    I think that:

    In case i. all of f ends up imposed as uniform pressure onto the masonry.
    In case ii. the pressure under the plate is greater at the edge of the masonry, and decreases as you move along the plate, away from the beam.

    I also think that in the general (case iii.) the pressure under the plate is uniform, if and only if:

    L > (b1/2)

    (which incidentally is not the case for the design which I've been given).


    I'm reasonably sure of those statements, but why won't it cause a problem in the real world if L == 150 and b1 == 330?

    Thanks!

    Tim.
     
    Last edited: Aug 25, 2012
  17. Aug 25, 2012 #16
    Hmm, illustrative files didn't attach properly the first time, lets try again...
     

    Attached Files:

  18. Aug 25, 2012 #17
    What does hypothetical have do do with it, I thought you were building an extension?

    All beams deflect and rotate at the ends, if not fully restrained. The question is how much.

    The effect of end rotation will be seen as a tipping upwards of the ends so the beam no longer rests squarely and evenly on the pads.

    Without the information I asked for 'how much' is all guesswork and pointless.

    Your engineer has designed the size of the plate to brickwork size for a single leaf. The only question is; is it thick enough?

    The answer to that again depends upon the information I asked for.

    Personally I wouldn't use a steel plate for the end rotation for reasons described above and because it would introduce practical construction difficulties that are difficult ( and uneccessary) to surmount.
     
    Last edited: Aug 25, 2012
  19. Aug 25, 2012 #18
    You might want to model this in BeamCalc or some other tool. The spreader plate will tend to distribute the load across the entire 330 mm span, modulo a decrease in shear as you go across. If this is in a seismic zone I would visualize the entire structure placed on a shaking table and secure it with bolts and rebar accordingly.
     
  20. Aug 25, 2012 #19
    The spreader plate will do little, if anything to distribute the load or reduce the bearing stress. As the beam is many times stiffer than the plate, the load will essentially all be carried by the brick which is directly under the beam. Bearing plates, or spreader plates, are used to distribute concentric loads, not eccentric loading like this.
     
  21. Aug 26, 2012 #20

    nvn

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    TimSmall: (1) The steel bearing plate ("spreader" plate) is necessary. The thickness of your current steel bearing plate is adequate.

    (2) Your I-beam web crippling stress is probably OK, but we cannot be sure unless you post your I-beam size.

    (3) Nominal bearing pressure, p, on brick with cement mortar must not exceed the brickwork allowable bearing stress, Sbra = 1.70 MPa. Your current nominal bearing pressure is, p = V/(b*L) = (47 300 N)/[(100 mm)(150 mm)] = 3.15 MPa, which exceeds Sbra, where V = applied load, b = bearing plate width, and L = bearing length. Therefore, the masonry currently appears overstressed.

    It appears your 150 mm bearing length instead needs to be, L = V/(b*Sbra) = (47 300 N)/[(100 mm)(1.70 MPa)] = ~280 mm. Therefore, if your given end reaction force (V) is correct, it currently appears you need to increase your 150 mm bearing length to 280 mm.

    (4) You might want to also see the last sentence in post 18, if you are in a seismic zone, or if you might experience high winds, tornadoes, or minor hurricanes.

    (5) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 12 mm, not 12mm. See the international standard for writing units (ISO 31-0).
     
    Last edited: Aug 26, 2012
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