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I want to be sure that the way I like to think about the displacement method of analysis is correct.

Suppose that a beam as given by the figure below is to be solved. enter image description here

Under this conditions, the only unknown is the rotation of joint $B$, $\alpha_B$. As I see it, the steps taken to compute $\alpha_B$ are:

i) apply a moment $M_B^F$ at point $B$ of the beam such that no rotation of $B$ takes place. Such a moment is clockwise. Due to this, node $B$ is under the action of an equal and opposite moment of anticlockwise direction and, therefore, according to the standard conventions, of positive value $PL/8$, and, as such, not in equilibrium;

ii) for node B to be in equilibrium, as it effectively is in the original structure, apply at beam $B$ a moment whose value will depend on $\alpha$; Node $B$ will be subject to an equal and opposite moment $M(\alpha_B)$ such that $$M(\alpha_B)+ PL/8=0$$

Is this a correct interpretation of this method of analysis?

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  • $\begingroup$ Have you searched before posting? A similar question was answered recently - if you find it before me that’s fine... $\endgroup$
    – Solar Mike
    Jan 2 '18 at 21:48
  • $\begingroup$ Yes, I did. But I didn't find anything satisfying. $\endgroup$
    – muimerp
    Jan 2 '18 at 21:56
  • $\begingroup$ This seems relevant : engineering.stackexchange.com/q/13444/10902 $\endgroup$
    – Solar Mike
    Jan 2 '18 at 22:05
  • $\begingroup$ Thank you for the link, but it doesn't help me that much. But let me ask you: would you accept what I've written as acceptable? $\endgroup$
    – muimerp
    Jan 2 '18 at 22:36
  • $\begingroup$ How did you get PL/8? $\endgroup$
    – Jem Eripol
    Jan 3 '18 at 0:12
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You should get same final result as the table you gave in comment, this kind of tables are formulas obtained by solving indeterminate beam. There is procedure for solving statically indeterminate beams as you can see in the following link slope deflection. Those solutions should guide you to solve your problem.

Keep in mind basics for real beam from theory of conjugate beam conjugate beam

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  • $\begingroup$ I actually haven't solved the problem, so I don't know of which solution you're talking about. $\endgroup$
    – muimerp
    Jan 4 '18 at 18:22
  • $\begingroup$ I have mistaken your last formula for solution M at B as it is obviously 0. Sorry this simbols mix on my phone from time to time. Though author of document in my link provides solution of similar problem in example 12.2. $\endgroup$
    – Katarina
    Jan 4 '18 at 18:48
  • $\begingroup$ It's OK. I already knew about the tables provided by you, but hank you, anyway. $\endgroup$
    – muimerp
    Jan 4 '18 at 19:16
  • $\begingroup$ At B beam is free to rotate so MB equals 0 that is one condition and you have to determine moment at A. Maybe revise your question? $\endgroup$
    – Katarina
    Jan 4 '18 at 19:54
  • $\begingroup$ No, the point is to compute the rotation of point A. And I'm trying to be sure that I understand the logic of the displacement method, not the computations that go along with it. $\endgroup$
    – muimerp
    Jan 4 '18 at 20:04

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