Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Structures Question Frames

  • 11-01-2010 7:53pm
    #1
    Registered Users, Registered Users 2 Posts: 155 ✭✭


    Preety basic question here but cant find anything on it and wont be in the library till tommorow

    In lectures we have only done frames with members that are perpindicular and all forces perpindicular.
    ____________________
    |xxxxxxxxxxxxxxxxxxxxxx|
    |xxxxxxxxxxxxxxxxxxxxxx|
    |xxxxxxxxxxxxxxxxxxxxxx|
    |xxxxxxxxxxxxxxxxxxxxxx|

    Now I am trying a frame thats members are sloped, I cant seem to figure out how to get/draw the shear and benging moment diagrams.

    _______________
    /xxxxxxxxxxxxxxxxx\
    Y /xxxxxxxxxxxxxxxxxxx\
    / xxxxxxxxxxxxxxxxxxxx Fxb Fzb
    Fxa Fza

    Sorry about the awful diagrams,(had to turn the blank spaces to x's cause boards deleted them)
    say a vertical force at Y

    Do I resolve all forces so that they are either perpendicular or parallel to the member?

    Thanks for any insight


Comments

  • Registered Users, Registered Users 2 Posts: 13,074 ✭✭✭✭bnt


    Well, as I understand it, to analyse any given member, you need to resolve the forces on it in to axial (in line with the member) or perpendicular to it (shear forces). This goes for self-weight too. The axial forces are compression/tension, and the perpendicular (shear) forces generate a bending moment in the member.

    Possibly oversimplified example: if you have a beam of length L planted in the ground at 30° to the vertical, with only its self-weight w (no applied forces), you treat it as a Cantilever (which is what it is, just not horizontal) when it comes to calculating its shear force and bending moments.
    • You'd have axial forces of w.L.cos(30°), and max shear forces at the base of w.L.sin(30°).
    • The bending moment would also be calculated in the usual way for a Cantilever, using the reduced shear force: the maximum would be at the base, ½ w.L².sin(30°)
    When calculating whether a particular member is up to the job, you have to make allowances for axial forces and bending moments. This is all covered under Limit State Design in the Eurocodes - I don't know whether you've looked at those yet. (If not, well, you will!)

    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



Advertisement