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Thread: Absorption of force by beam

  1. #1
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    Absorption of force by beam

    I would like to start by saying hello to everyone and thanking you for this amazing forum.

    I am presented with an issue at work an am hoping I might get some help here.

    I have a 12' X 24' table placed inside of a negative pressure environment. During testing the pressure will reach -500PSF (pulling upward on the top table surface) during the most extreme circumstance. The table is constructed on five(5) 6W20 beams placed 6' OC going across the 12' span. They are secured on the ends to a fixed point on the table and in the center they are secured to one(1) 6W-20 beam spanning the 24' length. The load is evenly distributed across the entire 288ft˛ of the table surface.

    My question has two parts. The first part is, does the assembly (beam structure) absorb any of the 144000lbs of uplift force and if so, how would I calculate that. The second part is, what is the maximum uplift force that the 24' center beam would exert on anchors (holding it to the floor) from below (total number is acceptable).

    My initial thought was to spread the load over the 3 linear supports (two side and center) and divide by 3, giving me 48,000lbs/beam Then it occurred to me that the load being transferred to the anchors would most likely be decreased due to the beam "absorbing" the load. Is this idea completely out in left field?

    Thank you in advance for your assistance.

  2. #2
    Administrator Kelly Bramble's Avatar
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    A drawing (Free Body Diagram) would be helpful..
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  3. #3
    Lead Engineer Cake of Doom's Avatar
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    Beams don't absorb load; they transfer a portion of it from one point to another with the other portion travelling in the opposite direction. As Kelly points out: draw up a FBD and the load paths should become pretty obvious. It may also do you a favour to analyse the whole thing as a plate so that no single element is acting in isolation to the rest. That would be an easier way to justify the total load being divided by the number of holding down bolts.

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    Hi Kelly and Cake,

    Thank you for your replies. I think this is the drawing you are looking for. I hope there is enough detail.My question is what is the maximum load exerted on the threaded rods.

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  5. #5
    Lead Engineer Cake of Doom's Avatar
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    Now turn that picture in to a FBD, note all know forces. I might even be inclined to do two diagrams: one as a UDL over 4 equal spans and then use those reaction to assemble a diagram for beam/bars section. You'll be hard pushed to get anybody here to do it for you unless you make, at least, a minimal effort to solve it yourself.

  6. #6
    Lead Engineer Cake of Doom's Avatar
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    If you make the effort for the easy top section, I'll walk you through the more painful lower section. Deal?

  7. #7
    Administrator Kelly Bramble's Avatar
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    Not a simple question... You have a complex section with an unusual applied negative load (like vacuum) applied. Not sure what the application is other lifting while glued to a plate.

    FEA software would make quick work of this.

    I don't know a quick solution without a derivation... I'll think about it.
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    As a general beam loading issue, what is the reason for variable spacing of the tension rods along the cross beam?

    The uniform pressure (vacuum) loading on the floor is going to result in the five equally spaced perpendicular beams acting as five loading points on the base beam connected to the rods and the irregular locations of the tie rods relative to those beams is going to result in varying tension loads on the rods and bending stresses on the cross beam.

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    I will be more than happy to make the effort, I suppose that I should have started with the fact that I'm not an engineer. I can understand the math if I am pointed in the right direction, and am more than happy to do all the work necessary. I guess I am unclear as to what other forces are acting on the system.

    Also, I believe I was unclear with my question. What I meant to ask, how do I find the TOTAL load placed on the four rods collectively. The reason they are offset is that the beams move given the assembly being tested at any given time. Sometimes there are six beams, sometimes four, but usually five.

    If I am asking too much for the forum, please let me know and I will look elsewhere. Thank you again.

  10. #10
    Technical Fellow jboggs's Avatar
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    When engineers say "negative pressure" generally they are describing a closed volume where the pressure on the inside is just slightly less than the pressure on the outside, otherwise known as a vacuum. They do this usually to make sure that any leaks only go in one direction. Is the source of your pressure actually a vacuum? If so, there's no way you can achieve 500 psi since standard atmospheric pressure is 14.7 psia. On the other hand, if by "negative pressure" you mean an upward force created by some device, then there's no real purpose in including the word "negative". You just say "upward". Not trying to be snarky, just trying to help you be clear.

  11. #11
    Lead Engineer Cake of Doom's Avatar
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    You'd need to repeat the process then every time the beam centres are altered. What is it you're trying to test? Currently, you may end up with one of the rods doing the lions share of the work whilst others will load in the opposite direction as the primary beam deflects. The total load will be a lie and could lead to you under sizing members.

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    It is a vacuum and the pressures will reach negative 500 pounds/ft˛ (approx -3.5psi) differential pressure

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    Administrator Kelly Bramble's Avatar
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    Quote Originally Posted by jp_rukus View Post
    What I meant to ask, how do I find the TOTAL load placed on the four rods collectively.
    That would be:

    Width x Depth = Area^2

    Area = 12' X 24' = 288 ft^2

    Load applied=500 lbs/ft^2

    Therefore:

    288 ft^2 x
    500 lbs/ft^2 = 144,000 lbs total
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    I am trying to replace the rods with a dynamic system and am trying to make sure that it is designed properly. The rods add an unfavorable danger to securing the table (having someone crawl under the table to adjust them before every test). The only function they provide is to stop the beam from deflecting too far upward.

  15. #15
    Lead Engineer Cake of Doom's Avatar
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    This problem has got us all chatting, so that's a positive (+).

  16. #16
    Lead Engineer Cake of Doom's Avatar
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    Quote Originally Posted by jp_rukus View Post
    I am trying to replace the rods with a dynamic system and am trying to make sure that it is designed properly. The rods add an unfavorable danger to securing the table (having someone crawl under the table to adjust them before every test). The only function they provide is to stop the beam from deflecting too far upward.
    In that case, can you move the rods? If you can get them evenly spaced, you could use continuous beam theory to obtain the reactions as they will be proportional to the applied loading. A lot simpler than the current set up.

  17. #17
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    I definitely appreciate everybody's help.

    Kelly, am I correct that 144,000 would be divided between the center beam and the two side channels that are attached to the table (also anchored to the floor) holding the cross members down?

  18. #18
    Lead Engineer Cake of Doom's Avatar
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    What does this rig look like on plan?

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    Moving the anchors would be a lot easier in the new design. If I'm right it will be taken care of with a winch and block system and Dyneem due to the low stretch.
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    Technical Fellow jboggs's Avatar
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    My bad! Sloppy reading. I am not used to seeing pressures expressed in pounds per square foot. Assumed psi.

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