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Simple static load problem giving me hell! (sketch provided)

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    This is not actually a homework problem but a design evaluation problem I have encountered.
    I am trying to solve for all the loads at the pins and have used WorkingModel to try and give me some quick answers. However, the solutions offered by the software do not make sense to me/can't verify with hand calcs.

    The image below shows the values of Fx, Fy and F as calculated by the software at each pin. I have added the co-ordinates of where the pin joints are and where the load acts (in mm). +X is to the right, +Y is to up. The load acting on this system is a 10kgF acting directly down as shown. Each grid square is 50mm wide. The mass of the elements is set to zero.

    [PLAIN]http://img833.imageshack.us/img833/7528/forces.jpg [Broken]

    The unknowns are the reaction forces (Fx, Fy) at each of fixed pins 11 and 3 and the force acting on the pin at 13.

    2. Relevant equations

    Ʃ forces in X direction = 0
    Ʃ forces in Y direction = 0
    Ʃ moments about any point = 0
    M=F.x
    cos(θ)=A/H
    sin(theta)=O/H
    tan(θ)=O/A

    3. The attempt at a solution

    Unless I have severely overlooked something, I immediately dismiss the solutions offered by the software because a simple check of the equation:

    Ʃ forces in X direction = 0
    ...reveals that this is NOT the case.

    Ʃ forces in X direction = Fx@pin11 + Fx@pin3 + Fx@LOAD
    = (-5.186) + (16.305) + (0)
    ≠ 0!!!!!

    ?????????????????????

    What is going on here? Am i going loco or is this solution just plain out wrong?

    Thanks
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 28, 2011 #2
    (hand calc removed because of error)
     
    Last edited: Oct 28, 2011
  4. Oct 30, 2011 #3
    I used a ruler on your image and took the torques about pin 11 and got,

    - 45mm*16.5 + 78mm*10 = 0 ?

    - 742.5 + 780 = 0 ? Close

    The load tends to rotate the arm about pin 11 clockwise and the force of pin 3 on the arm, via the link, tend to rotate the arm counterclockwise.

    Given the distortion of your image (doted squares measure on screen 30mmW X 22mmH) those numbers look close. What confuses me is why the force tables for pin 13 and 3 are not equal and opposite?
     
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