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Basic Bending Moment Help!

  1. Jun 14, 2012 #1

    Im relatively new and need some help on a simple problem. If i have a force in one corner of an object and want a resultant bending moment in the opposite corner, how do i find this? see attached image.

    Obviously i have two bending moments, 100Nx1.7m and 100Nx1.2m, how do I find a 'total' bending force in the resultant corner? - or can i take a diagonal length from the applied force to the reaction to find one bending moment?!


    Attached Files:

  2. jcsd
  3. Jun 14, 2012 #2


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    welcome to pf!

    hi brianslowsnai! welcome to pf! :smile:

    i'm not sure what you mean by "bending moment" :confused:

    if you mean the moment of the force (also called the torque) about the opposite corner,

    then there's only one moment (torque), and it's 100 N x 1.2 m …

    all that matters is the line of the force (the point on the line doesn't matter) …

    by how much does that line miss the corner? :wink:
  4. Jun 14, 2012 #3
    Hi tiny-tim!

    Thanks for your help!

    Yes, I meant moment of the force, not bending moment... I'm still a little confused. The main view shows two crosshairs, where the forces are being applied (opposite corners).

    I understand your moment calculation 100 x 1.2 using my previous drawing, but what I'm confused about is that I could draw the forces for the same example in a different way. (see new attachment). This shows the same forces I am trying to get to grips with.

    - with the new pic, would you now say the moment is 100N x 1.7m because of the way I have displayed it? - This is why I am asking whether there is a 'combination' or 'total' resultant moment?!

    ...sorry if i have got the completely wrong end of things! I'd really like to understand!



    Attached Files:

  5. Jun 14, 2012 #4


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    Hi Jamie!

    (is that brian from the magic roundabout?)
    you're not "drawing the forces for the same example in a different way",

    you're draw the forces for a different example :wink:

    a force pointing north is different from a force pointing east

    in each diagram, you have only one force

    i don't see two things for the moment to be a combination (or resultant) of :confused:
  6. Jun 14, 2012 #5
    Hi tiny-tim,

    Thanks for pursuing my question, I think I'm either confusing myself or I'm not explaining it correctly.

    In each of the attachments, I've drawn effectively a thin plate, in third angle projection. I've imagined the lower left corner being held (resultant) and a force being applied (into the screen) in the top right corner of the front view (1.2m x 1.7m plane). As the force is being applied into the screen I have drawn it (in third angle) in each of the other views (once in each attachment) - hence demonstrating the same force, but in a different way.

    I understand the moment is either 1.7x100, or 1.2x100 depending which direction of moment I want to consider. If i take the first attachment:

    moment = 100N x 1.2m = 120Nm

    However, my aim is to find a 'resultant force' of the lower left corner, and there must be an effect of the 1.7m since the force is being applied 1.7m?

    I want to know how do I combine, or what do I do to my 120Nm moment and 1.7m in order to find the overall effect of the force on the opposite corner?

    Thanks very much
  7. Jun 14, 2012 #6


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    ohh, you mean the force is 100 N perpendicular to the rectangle, and you want the moment about an axis also perpendicular to the rectangle, through the opposite corner?

    then the moment is 100 N times the perpendicular distance from the axis, which is the length of the diagonal

    (there's no way of splitting the moment into two moments, since you can't split the force into two forces …

    it's not like a diagonal force which can be split into "x" and "y" components, this is a "z" force)​
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