
#1
Nov1112, 02:14 AM

P: 210

So I'm looking for the reaction forces at the left end of the truss before I solve for the forces in the members of the truss in terms of F and theta. I summed forces in the x, y direction, took moments about point A,B, to get 4 equations in 4 unknowns (Ax, Ay, Bx, By). However, although Ax,Bx can be solved easily, I'm having trouble finding Ay,By because it appears as if my force and moment equations are not independent of one another. I've never had this problem before in my Statics class (this is for a numerical methods class), so I'm wondering if this problem is statically indeterminant. Btw, each member is H long. Here are my equations: Force summation in x: Ax + Bx + 2Fcostheta = 0 Force summation in y: Ay + By + 2Fsintheta = 0 Moment summation at B: AxH + 2FsinthetaH + FcosthetaH = 0 Moment summation at A: BxH + 2FsinthetaH + FcosthetaH = 0 I solved the two moment equations for Ax and Bx. The force summation in the x direction confirms my Ax, Bx reaction forces. Thus, I still have two unknowns (Ay, By) in one equation (force summation in y). How do I solve for them? I've also tried summing moments at the righthand corners, but they gave me equations which are also dependent. 



#2
Nov1112, 02:10 PM

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The problem is statically indeterminate and requires deformation analysis unless the equal length diagonals are also of the same cross section area and within allowable buckling limits, in which case you can take advantage of symmetry to find the vertical reactions. Did the problem mention anything about equally sized members?




#3
Nov1112, 03:43 PM

P: 210

Member buckling can be ignored. All members are made of the same material with the same density and crosssection. How will this help in determining the remaining reaction forces?




#4
Nov1112, 04:24 PM

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Weird Statics Truss Problem  Finding reactions
Well then, from symmetry of geometry and member sizes, is there any good reason to believe that one support will see more vertical load than the other??




#5
Nov1112, 05:11 PM

P: 210

So you're saying that Ay = By. I thought so at first, but Ax does not equal Bx, which I found was weird. Is there any way to mathematically prove that Ay = By, rather than using the symmetry argument?




#6
Nov1112, 09:37 PM

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#7
Nov1112, 11:33 PM

P: 210

Ah, okay. This was really a problem I had within a larger problem. Here's the original problem:
I now have N sets of these "squares" of truss. For N = 1, it's easy to find for the reactions and forces within the members. But I'm stuck if I have, in general, N of these. How do I solve for all of the forces in the members if N is known, not to mention the reaction forces will also change for a general N case? (F and theta is supposedly known as well) Attempt: I have solved for the forces in the members for N = 1, and I'm thinking to go ahead and do N = 2 in hopes of recognizing some sort of pattern in the forces, but it would be a major headache to do it and realize I'm going nowhere. Is this worth a try? P.S. Matlab is required, but I'm making sure I know what's going on before I start to program. 



#8
Nov1212, 06:39 AM

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When N increases, Ax and Bx increase, but not Ay and By, they stay the same. This might help you find a 'pattern' for how load changes in the vertical members of each square as you increase the number of squares, and it should tell you something about the diagonals also.




#9
Nov1212, 02:11 PM

P: 210

Ay and By stay the same even when there is a weight associated with each member? I did not include the weights above for simplicity, but each member supposedly has its own weight.




#10
Nov1212, 03:07 PM

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Usually the weights of the members are small in comparison to the applied forces, but if you must consider them, half of each weight of the individual member should be distributed to its end joints. Then Ay and By will still be equal, but of course be larger with increasing values of N.




#11
Nov1312, 06:40 PM

P: 210

Here's what I have for N = 1 (just the reaction forces): Ax = 0.162Hg  Fcostheta  2Fsintheta Bx = 0.162Hg  Fcostheta +2Fsintheta Ay = Fsintheta + 0.135Hg By = Fsintheta + 0.135Hg And here's what I have for N = 2: Ax = 0.594Hg  4Fsintheta  Fcostheta Bx = 0.594Hg + 4Fsintheta  Fcostheta Ay = 0.27Hg  Fsintheta By = 0.27Hg  Fsintheta The Hg terms in the equations are found using H as the length of one member, g represents acceleration due to gravity, and density rho = 0.0027 kg/cm^3 and cross sectional area 20 cm^2. I placed 1/2 of each member weight on each pin. 



#12
Nov1412, 04:48 AM

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It is dificult to see a pattern when you work with letter variables and include the numerical member weights also. I would throw in some numbers for the variables and eliminate the weight part, to understand things more clearly. Let F = 1000, H = 2, and theta = 45 degrees. H is the length of a vertical member and the distance between supports. Now solve for the end reactions and member forces when N = 1, N = 2, and N = 3. That will help you see how, for the negligible weight case, Ay and By are independent of N, how the diagonal member forces are independent of N, and how the reaction and vertical member forces change with increasing values of N.



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