
#1
Oct2909, 08:11 AM

P: 71

ok, i dunno why, but this question has been baffling me since morning
we have two supports both with reactions in x and ydirections (in 2D), and one external load. now i have tried this with two sets of variables, but the problem i am facing is that in both cases i am getting different answers for different sets of equations. The problem is something like this: I have a load W acting at pt C (W can be assumed at say 330N) in negative y direction and i have supports at pts A and B, which are: A_{x}, A_{y},B_{x} and B_{y}, all are not known. assuming all reactions are in positive x and y directions Now i can get these equations: A_{x} = B_{x} as no other horizontal forces are there also A_{y} + B_{y} = W from equilibrium of forces and then obviously three equations from the equilibrium of moments at point A, B and C respectively. (all distances are known) since there are four variables, i am feeding these to an equation solver software however, the software rightly says that the problem is overspecified with 5 equations and 4 variables but, when i try to run it with any four of the the five equations, each time i get a different answer...which is the problem because i think, it should give the same answer from any randomly chosen four equations...no? what am i doing wrong here? 



#2
Oct2909, 07:26 PM

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#3
Oct3009, 12:32 AM

P: 71

hey Jay, thanks for your reply
I have made a gif of this problem...hope that can help Now the equations i formulated are: 1. [tex]\sum Fx [/tex] 2. [tex]\sum Fy [/tex] 3. [tex]\sum MA [/tex] 4. [tex]\sum MB [/tex] 5. [tex]\sum MC [/tex] As you say, and i also think too, that one equation here is redundant, but the point is that use of any four should give the same result as the use of any other four, no? Also, i don't think the horizontal reactions will be zero as you can see in the picture and btw, this is part of no homework or assignment, i wrote to the guy who put it in here, but no reply thanks 



#4
Oct3009, 04:05 PM

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'Simple' statics
Oh, this is different than I was visualizing. Is this a cable or a bent beam? If a cable, which can take tension only, then Ax and Ay are related by the fact that their resultant must be equal and opposite to the tension in the cable AC, and likewise for Bx and By, the same approach applies. And in which case, you don't worry about the moment equations. Please clarify. If a solid beam, there are more issues to deal with, because you need another equation to solve for 4 unknowns with only 3 available static equilibrium equations.




#5
Oct3109, 12:59 AM

P: 71

Actually there are no direct links between A,B and C as i have shown. I just simplified it. It is actually supposed to be fibreglass shell harnessed to a frame at A and B. Point C is the centre of mass of the shell




#6
Nov309, 11:52 AM

P: 71

Jay, you there mate?
anyone? 



#7
Nov309, 03:27 PM

P: 63

As I see the picture there will be horizontal reactions in supports, so its pretty more math in there.




#8
Nov309, 08:59 PM

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Again, you have 4 unknowns but only 3 static equilibrium equations, not 4. If you take the sum of M_A = 0 as one of your static equilibrium equations, then taking the sum of M_B =0 or sum of M_C =0, buys you nothing. It's sort of like saying sum of F_x =0, sum of F_y =0, both correct and necessary, but then trying to use sum of forces about an axis 45 degrees to the x axis =0, also correct, but it doesn't give you the extra equation you need. You need the extra equation from other info, such as B_x or B_y =0 (roller support). Otherwise, you'd have to remove one of the components of the pin support, say B_x, then solve the reactions A_x and A_y and B_y with the 3 equilibrium equations, then add back the B_x term and note the reactions under B_x only, with the stipulation that the total horizontal deflection of the support at B under all loads is 0. You say you have a software program to run this problem, does it allow you to enter that the support deflections at A and B in both the horizontal and vertical directions are 0? I would think that the program needs the end conditions anyway.
As an approximate solution, with the load aplied at C, you might want to consider this as a truss, the difference being that you have a continuous , not a pinned 'link' joint, at C, but certain tests have shown that this does not appreciably affect member forces and support reactions (many trusses are built with multiple bolted joint connections with gusset plates, tending to fix the joint), thus, you may wish to consider AC and BC as taking axial forces only, in which case A_x and A_y are related by A_y/A_x = tan theta, and likewise for B_x and B_y. 



#9
Nov409, 05:31 AM

P: 71





#10
Nov709, 04:56 PM

P: 692

If A and B are both pinned supports then they cannot move relative to each other, as they would be able to do if one of them were a roller. So the problem cannot be solved using equilibrium equations alone. You also need strain energy equations. Alternatively, if C were an internal hinge, you can take moments about C for BC alone (M=0). That unlocks it. Also moments about C for AC alone acts as a check (again M=0). This is a practical solution because with the configuration illustrated, axial loads rather than bending dominates the behaviour of this structure. However, the real problem is that the single arrow at C is not the distributed load that is being modelled, and so you really do need to use strain energy equations in addition to the equilibrium ones.



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