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Thread: Real World Beam Failure Analysis

  1. #1
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    Real World Beam Failure Analysis

    On a recent theatrical show, during the "load-out",
    it was noted that a "beam" failed sometime during the show.

    Later that week I was shown the "beam", which is actually a section of schedule 40 pipe, 4 feet long.
    The pipe was spanning a grid "well" (see drawing), a distance of 10 inches.

    In attempting to determine the nominal stress to show the strain as indicated,
    I'm thinking that this formula applies:

    http://www.engineersedge.com/beam_be...m_bending2.htm

    even tho the "beam" is pipe.

    Am I thinking correctly?


    Pipe failure analysis.jpg

  2. #2
    Engineer PierArg's Avatar
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    Hi dalecyr,
    Very interesting question!

    I think you are right. I would have used the same formulas but this doesn't mean that it is right

    I'm curious to know what the other members think.

    What is different between a pipe and a "classic beam" is the internal stress.
    As you see, the pipe is subject at the following stress:
    - bending: the moment of inertia of a pipe is different from a beam
    - transverse shear: the shear factor for a pipe is different from a beam.
    Last edited by PierArg; 03-07-2013 at 02:30 AM.

  3. #3
    Lead Engineer Cake of Doom's Avatar
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    You are right to use that calculator. Being in the temporary works game, specifically scaffold design, I work with "tube" quite a bit. PierArg is correct in noting that the section properties between "beam" and "pipe" are quite different. In bending, tube is rubbish. Once the FOS has been applied, we can only put a maximum of 1.12 kN.m on a piece of 48.3 mm (outside dia.) tube (UK). Even without the FOS, it's not much better. In addition, there will be different results between a suspended load as opposed to a directly applied load. It may not have been a bending failure but a buckling failure.

    Do you have a link to the show?

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    Engineer PierArg's Avatar
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    Hi Cake,
    what do you mean for FOS? Maybe "force"?
    (i'm sorry but i don't understand the abbreviation)

    I'd like to point out that the "pipe" is not recommended for that application (IMHO).
    You need a beam shape with a high moment of inertia when bended...such as a rectangular shape with the long side oriented in the vertical direction.
    Or just a beam with the "I" section.
    In this way the bending is limited.

  5. #5
    Lead Engineer Cake of Doom's Avatar
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    Factor of safety. Sorry. I'm so used to using this abbreviations, I sometimes forget that not everybody knows what they mean. Damn you, international internet!

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    Engineer PierArg's Avatar
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    Thanks Cake!

    What do you think about my opinion?
    Should have you used a pipe for that application?

  7. #7
    Lead Engineer Cake of Doom's Avatar
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    Over a shorter span to decrease bending and increase shear (pipe can do quite well in shear), it would have been fine. A beam or just a larger diameter pipe would have been better.

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    "On a recent theatrical show, during the "load-out",
    it was noted that a "beam" failed sometime during the show."

    Please describe the "failure"

  9. #9
    Technical Fellow jboggs's Avatar
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    In my understanding, the term "beam" has to do with the loading, not the shape. If the loading is along the axis, its a column. If the loading is perpendicular to the axis, its a beam. Think of the columns and beams in a building. Material of any cross-sectional shape can be subjected to those loads. The moment of inertia factor in the formulas accounts for the effects of the different shapes and how they respond to the applied loading.

    And yes, you are using the right formula.

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    Quote Originally Posted by zeke View Post
    "On a recent theatrical show, during the "load-out",
    it was noted that a "beam" failed sometime during the show."

    Please describe the "failure"
    Hi Zeke.

    By "failure" I specifically mean that the pipe was permanently deformed.
    The 4' long pipe was "bent" about 3/4" at the center point,
    resulting in the ends lifting off the grid, as shown in the right hand half of the drawing.

    Doom:
    This method of rigging is standard practice in the U.S. in this industry.
    However, a 4" schedule 40 pipe is normally used in that location in this house.
    I don't know why the 1.5" pipe was used this time.
    (I was not in the building at the time,
    but I have been talking to the other riggers who installed / deinstalled that point.)

    The load is a self-climbing chain motor, with a load attached under that.
    The chain motor capacity is nominally 1 ton (2000 pounds),
    with a brake capacity of (typically) 2-3 times that.

    The show maintained that the static load was less than 1200 pounds.
    Motor startup / stopping typically increases the static load by 2-3 times.

    Industry standards for FOS for this configuration is 5 times the dynamic load minimum.


    So, I have a question and two concerns.
    The question is 'how much force does it take to permanently deform that pipe by that much'.
    One concern is that we are much closer to the failure point than we realized.
    The other concern is that the show may not know the load amount accurately.

    Joe:
    That is my understanding also, so I'm off to find (actually, re-check) the E and I for 1.5" schd 40 pipe.
    The last answer I got didn't make sense...

  11. #11
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    Quote Originally Posted by zeke View Post
    "On a recent theatrical show, during the "load-out",
    it was noted that a "beam" failed sometime during the show."

    Please describe the "failure"
    Hi Zeke.

    By "failure" I specifically mean that the pipe was permanently deformed.
    The 4' long pipe was "bent" about 3/4" at the center point,
    resulting in the ends lifting off the grid, as shown in the right hand half of the drawing.

    Doom:
    This method of rigging is standard practice in the U.S. in this industry.
    However, a 4" schedule 40 pipe is normally used in that location in this house.
    I don't know why the 1.5" pipe was used this time.
    (I was not in the building at the time,
    but I have been talking to the other riggers who installed / deinstalled that point.)

    The load is a self-climbing chain motor, with a load attached under that.
    The chain motor capacity is nominally 1 ton (2000 pounds),
    with a brake capacity of (typically) 2-3 times that.

    The show maintained that the static load was less than 1200 pounds.
    Motor startup / stopping typically increases the static load by 2-3 times.

    Industry standards for FOS for this configuration is 5 times the dynamic load minimum.


    So, I have a question and two concerns.
    The question is 'how much force does it take to permanently deform that pipe by that much'.
    One concern is that we are much closer to the failure point than we realized.
    The other concern is that the show may not know the load amount accurately.

    Joe:
    That is my understanding also, so I'm off to find (actually, re-check) the E and I for 1.5" schd 40 pipe.
    The last answer I got didn't make sense...

  12. #12
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    "The load is a self-climbing chain motor, with a load attached under that.
    The chain motor capacity is nominally 1 ton (2000 pounds),
    with a brake capacity of (typically) 2-3 times that."

    The calculation is not for a static load but the fact that it is braking makes it dynamic and energy methods have to be employed to get the correct pipe in addition to considering the startup load which is derived from 5x nominal torque. It is NOT 5x the weight.

  13. #13
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    Quote Originally Posted by zeke View Post

    The calculation is not for a static load but the fact that it is braking makes it dynamic and energy methods have to be employed to get the correct pipe in addition to considering the startup load which is derived from 5x nominal torque. It is NOT 5x the weight.
    Ok, fair enough.
    But I'm showing that it would take well over 200,000 pounds of force to bend a schedule 40 pipe 3/4 of an inch over a span of 10 inches.

    (I=.3)

    Does that sound right?

  14. #14
    Lead Engineer Cake of Doom's Avatar
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    Just did a rough sums and this what I came up with:-

    I=5.98cm^4
    I/D = 3.07cm
    O/D = 3.81cm
    F = 90kN (9 metric ton based on a rough conversion)
    Max BM = 5.63kN.m
    Max Shear = 45kN
    Def = ~2.37mm

    Look about right? Our nearest equivalent section can only take 1.3kN.m so that would be a big no no.

  15. #15
    Engineer PierArg's Avatar
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    Please forgive me Cake!
    And the torsional stress? You said there's a motor...the part of the beam between the motor and the rope is subject to torque when the motor runs.

    Or not?

  16. #16
    Lead Engineer Cake of Doom's Avatar
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    I'll have to look in to that a bit later. This is a sneaky post from work and I've got a job on at the minute that is being a bit of a female dog.

  17. #17
    Engineer PierArg's Avatar
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    Quote Originally Posted by Cake of Doom View Post
    This is a sneaky post from work and I've got a job on at the minute that is being a bit of a female dog.
    what does it mean?

    As I told you I speak just an elementary english

  18. #18
    Lead Engineer Cake of Doom's Avatar
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    Mea culpa. Posted from my work computer whilst nobody else is looking.

  19. #19
    Lead Engineer Cake of Doom's Avatar
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    As far as I can tell the beam isn't fixed at any end so I don't think torsional stress should be a factor.

  20. #20
    Engineer PierArg's Avatar
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    I agree, there's no torsional stress when the ends are not fixed...but how does it turn?
    Where is applied the motor?I suppose there is a motor joined at the end (right or left side)

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