In the process of editing the question to put better pictures of my dimensions in, I realized that how I calculated my angles was incorrect, and that is why my answers were yielding inconsistent answers
I am writing a computer program to help me solve for stresses in motorcycle suspension and chassis members. I have a girder front fork that I have simplified into a truss, shown below.
The "Too long, didn't read" version is that when I use the method of joints to solve the axial force in the members starting at point A, then again at point B, then again at point C, I get different answers for the axial force in basically all of the members. If I do the analysis again starting at point D, then again at point B and C, I get different answers as well.
What am I doing wrong? Is there something that conceptually, I am not understanding?
The reaction at joint C is a pin joint reaction, and the reaction at point D resists X-forces but not Y-forces.
The dimensions are shown below.
First, I made sure that the structure statically determinant internally from the equation m + r = 2j -> (5 members + 3 reactions) = 2*(4 joints) ; 8 = 8. I also checked to make sure that the structure was stable using a program called itruss. With that in mind, I assumed there shouldn't be any reason why method of joints would not work so I continued my analysis.
Reaction Forces Next, I solved for the reaction forces. I used matrix trick x = A^(-1) * B throughout this analysis. The 3 systems of equations that I got were:
Fx: 534.857 + Rx.d + Rx.c = 0
Fy: 594.286 + Ry.c = 0
Moments positive clockwise about joint C: Rx.d*11.446 - 534.857*24.869 + 594.286*9.259 = 0
Solving the system of equations gave me: Rx.c = -1216.2 lbf; Ry.c = -594.29 lbf; Rx.d = 681.362 lbf (note that there probably will be some rounding error here compared to if somebody did it by hand with the values I gave because the dimensions in the diagram were rounded)
Method of Joints - Joint AThen I used the method of joints on Joint A to solve the axial force in tube a (F.a) and tube b (F.b). Assuming that all tubes are in tension got me the following system of equations:
Fx: 534.857 + F.aCOS(80.44) + F.bCOS(69.58) = 0
Fy: 594.286 + F.aSIN(80.44) + F.bSIN(69.58) = 0
The solution that I got for this system of equations was F.a = 1559.9 lbf; F.b = -2275.5 lbf
Method of Joints - Joint B Next, I used the method of joints to solve for the axial force in tube e and tube c. Assuming that all tubes were in tension got me the following system of equations:
Fx: F.cCOS(57.18) + F.eCOS(9.102) - F.aCOS(80.44) = 0 F.y: F.cSIN(57.18) - F.eSIN(9.102) - F.aSIN(80.44) = 0
The solution for this system of equations was: F.c = 1703.76 lbf and F.e = -672.84 lbf
Method of Joints - Joint C This was where I realized that I wasn't getting consistent answers. I solved the system of equations for joint C as though I did not already know the axial force in tube E and I got a different answer for the axial force in tube E. My system of equations:
Fx: Rx.c + F.dCOS(80.404) - F.eCOS(9.102) - F.b*COS(69.58) = 0
Fy: Ry.c + F.dSIN(80.404) + F.eSIN(9.102) - F.b*SIN(69.58) = 0
The solution to this was F.d = -1452.1193 lbf; F.e = -672.832 lbf
Originally, the calculated values for F.e did not match but now they do after I fixed how I calculated my angles. Sorry about that.