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I have a statics problem that I am running into some difficulties with. What is being asked here is to determine the support reaction at the points A and C. This would normally be no problem, but when I set up my equations (sum of the forces in the x-direction, sum of the forces in the y-direction, and total moment) I realize that there are too many unknowns for the problem to be solved.

Is there a way to assume something in order to eliminate one or two of the unknowns? (I am guessing it will be in the x-direction, but that's just a hypothesis.) Or should the problem be solved in a completely different way? Thanks for the help!

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The question has not defined any section properties so we assume all members are rigid. Then by inspection, we see this structure reduces to a 2 member frame AB and BC connected at a hinge at B and supported at pins on ends.

AB is at 45 degrees angle and BC is arctan(5/15) = arctan(1/3).

EDIT

After the OP correcting my arithmetic error.

$$ \Sigma M_a= -10kN*10m + C_x*15m - C_y*25m =0 \quad 15C_x -25C_y=-100kNm \quad C_x=3 C_y\ \text{as per geometry of the triangle BCCh} $$

$$ 20C_y=-100kN\ , \\ C_y= -5kN \\ C_x=15kN $$

$$ \Sigma Fx=0\ A_x= 15kN \ ,\\ \Sigma Fy=10=A_y-10-5=0, \ A_y= 15kN $$

check my numbers, please.

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  • $\begingroup$ Wouldn't Cx have a perpendicular distance of 15m and Cy have a perpendicular distance of 25m from point A? And since AB is a two-force member we can just ignore point B essentially? $\endgroup$ – ThatOneNerdyBoy Oct 11 at 1:43
  • $\begingroup$ No, remember the moment is about point A. All distances are from A, i am talking the point B. let me double check, will get back to you. $\endgroup$ – kamran Oct 11 at 2:08
  • $\begingroup$ @ThatOneNerdyBoy, I corrected my error in Cx and Cy after your remark. My apologies. Thank you. check it now, I had dinner with friends, distracted. $\endgroup$ – kamran Oct 11 at 3:58
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In this kind of questions, I believe that most of the attention should be focused on the general approach and the philosophy of attacking. Afterwards, each individual may proceed by writing its own equations and solving them in its prefer way.

In our particular case, I guess you examined the whole structure, realized that there are 4 reactions (Ax, Ay, Cx and Cy) and thought to yourself - OK, here we have a statically indeterminate system (3 equations and 4 unknowns), so we must apply a unique technique in order to solve it.

However, the trick here is hidden in the middle joint. The structure is divided into two parts (AB and BC) which are connected through a pin joint. This joint brings another two reactions to the table (Bx and By). You may first be tempted to think that "hi, these are internal forces, they cannot add any useful information. I could alternatively create an artificial cut in any other place along the structure and look for the internal reactions there".

However, this would be a wrong way of thinking. If we cut our structure in any other zone, we would have to deal with 3 reactions in the cross section (two forces and a bending moment). The pin joint actually eliminates the bending moment which in turn allows us the solve the problem. You may look at each part separately now. Part AB has 4 reactions and 3 equilibrium equations and the same is true for part BC. Since two of these reactions are the same (Bx and By) we end up with 6 equilibrium equations and 6 unknown (Ax, Ay, Bx, By, Cx and Cy). At this point we realize that we can solve the problem without applying sophisticated methods (using stiffness and calculating deflections and etc.)

@kamran showed you a good way of solving it. Just pay attention to the nice approach he used when writing the equations of part BC - Since there are no external forces exerted on it, the total reaction of joint C must be directed towards joint B. Otherwise, there would be a total moment applied on Joint B. The same is true, of course, to point C. By finding a relationship between Cy and Cx he was able to come up with the 4th equation you just look for.

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  • $\begingroup$ Thanks for this answer as well! I guess my method of solving should be slowed down and I should evaluate what I have and what I need more precisely! $\endgroup$ – ThatOneNerdyBoy Oct 11 at 11:07

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