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Finding the displacement of a point

  1. Jun 13, 2006 #1
    The attachment below shows a structure ABC where AB=3 m, BC=2 m. Both these parts have the same value for E and I which are 210 000 MPa and 2*10^6 mm4. A force of 5 kN acts as shown. Do not consider axial deformation. I am to find both the vertical and horizontal diplacement of point C. My first idea here was to flip the structure 90 degrees to the right and then consider AB as a cantilever beam which would now be acted on by a torque created by the force. A formula for calculating the downward displacement for such a beam is:

    \delta = \frac{{M \cdot l^2 }}{{2EI}}

    With inserted values I get 107 mm, which sound realistic to me at least but correct me if it isn't. But now I'm completely stuck when it comes to finding the vertical displacement. What should I now do?

    Attached Files:

  2. jcsd
  3. Jun 13, 2006 #2


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    It depends on what have you learnt??

    If it was up to me i will solve it with an energetic method such as Virtual Work.

    What methods do you know? energetic and/or geometric?

    Btw, your horizontal displacement is correct.
    Last edited: Jun 14, 2006
  4. Jun 14, 2006 #3
    Virtual Work is totally unknown to me. I haven't really learned anything beyond normal statics and some basic things about deformation of materials.
  5. Jun 15, 2006 #4


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    Ok, i see. Are you taking intro to Mechanics of Materials?

    I recommend James M. Gere's book.

    Anyway, you should note that for the horizontal displacement all that was provoking it was the bending of member AB, so it was easy to just use the deflection on a cantilever beam by a couple or moment. In the case of the vertical displacement is different. The bending on member AB and the bending on member BC affects the total vertical displacement of point C.

    I did some calculations with virtual work and the equation you are looking is:

    [tex] \delta_{c} = \frac{P(BC)^{3}}{3EI} + \frac{P(BC)^{2} (AB)}{EI} \downarrow [/tex]

    There's a way to get this result through the method of superposition. You can get the same result by adding the deflection of member BC like a cantilever beam with a load on its endpoint and another combo which eludes me right now. I'll see if i can recall.
    Last edited: Jun 15, 2006
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