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Simply Supported Beams In Real Life

  • 30-12-2009 11:51AM
    #1
    Closed Accounts Posts: 882 ✭✭✭


    A friend asked me an interesting question yesterday and I'm unsure of the actual answer....

    When a simply supported beam is designed we say it's pinned on one side and on rollers on the other. In real life is there actually a roller or does the beam (i.e. a bridge) just simply "rest" on the support?

    Thanks


Comments

  • Registered Users, Registered Users 2, Paid Member Posts: 1,643 ✭✭✭Turbulent Bill


    I'm not a civil, but I think in the vast majority of cases the roller end condition is just an idealisation - it makes the boundary condition easier to evaluate. In reality the actual boundary condition could be something between pinned, rollered etc., so you just design for the most conservative case.

    If there was very large expansion/contraction then maybe a real roller would be used.


  • Registered Users, Registered Users 2 Posts: 13,186 ✭✭✭✭bnt


    It depends on the application e.g. in a bridge you'll expect to see an expansion joint to allow for thermal expansion or deflection under load. With a simple beam it could just sit there, with no actual rollers - as long as it doesn't transfer significant lateral load to its supports.

    It depends on the application e.g. in a bridge you'll expect to see an expansion joint to allow for thermal expansion or deflection under load. With a simple beam it could just sit there, with no actual rollers.

    For anyone who's wondering what's behind this: in structural design, there are conditions that must be satisfied before a structure is "statically determinate", meaning that we can use the "simple" calculations to determine the safe loads and deflections. You need to have the same number of equilibrium equations as you do unknown forces or moments. The "roller" is how you indicate that a support takes a vertical load from e.g. a beam, but not a horizontal load: if the beam is pushed sideways, it will move, and must not transfer much horizontal force to its supports.

    If a structure doesn't meet the conditions, i.e. if there are too few or too many degrees of freedom, it's called statically indeterminate: either it's going to be unstable, or you have to break out the Finite Element Models (and/or Calculus) to accurately calculate how it will behave under load.

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



  • Closed Accounts Posts: 163 ✭✭Ballyhaunis


    bnt wrote: »
    It depends on the application e.g. in a bridge you'll expect to see an expansion joint to allow for thermal expansion or deflection under load. With a simple beam it could just sit there, with no actual rollers - as long as it doesn't transfer significant lateral load to its supports.

    It depends on the application e.g. in a bridge you'll expect to see an expansion joint to allow for thermal expansion or deflection under load. With a simple beam it could just sit there, with no actual rollers.

    For anyone who's wondering what's behind this: in structural design, there are conditions that must be satisfied before a structure is "statically determinate", meaning that we can use the "simple" calculations to determine the safe loads and deflections. You need to have the same number of equilibrium equations as you do unknown forces or moments. The "roller" is how you indicate that a support takes a vertical load from e.g. a beam, but not a horizontal load: if the beam is pushed sideways, it will move, and must not transfer much horizontal force to its supports.

    If a structure doesn't meet the conditions, i.e. if there are too few or too many degrees of freedom, it's called statically indeterminate: either it's going to be unstable, or you have to break out the Finite Element Models (and/or Calculus) to accurately calculate how it will behave under load.



    This man knows what hes talking about ;)


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