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3 Member force system-Could use some input

  1. Jun 10, 2006 #1
    I have a simple statics problem that I am trying to solve. The rigid body to be analyzed is an upside down L-shaped beam connected together at the corner by a pin. The object has a height of 4.5 meters vertically, by 3.5 meters in the horizontal. Additionally, the object is secured to the ground by means of a fixed support (cemented into the ground). In the corner of the upside down L connected diagonally is a structural element connected by pin joints at both ends. The lower end of the structural element is connected to the upside down L object 3 meters above the ground. The higher end is connected two meters from the corner of the L.

    One 3kn force (straight down) is applied at 1 meter from the corner of the upside-down L-shaped object on the horizontal member. On the same member at the very end (3.5m from the corner) is a 2kn force applied straight down. Lastly there is a given 4kn force applied applied horizontally at a point 3.5 meters above the ground on the vertical member of the upside down L.

    PS: I have tried to work it out on my own by combining the 3 and 2kn forces and then summing the moments about the fixed support. And summing the forces in the X and Y. I am not happy with my answers though and would like to know what someone else thinks about it. Additionally I have been reading everything I could on the internet since morning.

    PHEW! Here is my question. It seems to me that I have a 3 force member problem here. But I am not sure how to deal with the 4kn force. It seems to me that the 4kn force and the X componant of the force at the corner of the upside down L would make a force couple.

    As a three-force member is defined as a rigid body with no force couples, acted upon by a system of forces composed of or reducible to, three forces at three different locations I am not sure what I have here. How do I deal with the 4kn force?

    SOLVE....find the reactions at the fixed support (where the beam is cemented into the ground).
    Last edited: Jun 10, 2006
  2. jcsd
  3. Jun 10, 2006 #2


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    In order to solve a problem one needs to know what is to be solved for. You have not supplied this information...:wink:
  4. Jun 10, 2006 #3


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    A picture would be nice :smile:, but so far is an isostatic problem.
  5. Jun 10, 2006 #4
    There I edited my original post. So far as a picture.... I wish I could, but I don't know how. Any suggestions? I could find a similar picture and try to cut and paste it into a post.
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