Solving a 3D force-moment balance using two force application points

[daniel7campbell]

Hi,

I'm attempting to solve a 6x6 system of equations, but am returning an uninvertible matrix.
The system has an arbitrary center point and two load application points defined by [x1,y1,z1] and [x2,y2,z2].
The forces and moments at the center point are [Fx Fy Fz] and [Mx My Mz] respectively.
Using two forces at the load application points,([Fx1 Fy1 Fz1] and [Fx2 Fy2 Fz2]) to zero out the force/moment balance, there are 6 unknowns and 6 variables, yielding:

A=
[-y1 x1 0 -y2 x2 0
z1 0 -x1 z2 0 -x2
0 -z1 y1 0 -z2 y2
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 0 1]
x=
[Fx1;Fy1;Fz1;Fx2;Fy2;Fz2]
B=
[Mz;My;Mx;Fx;Fy;Fz]

Using Ax=b, this matrix cannot be inverted and I am therefore confused.

Any thoughts are appreciated.

Thanks

 

[rb1957]

you might have better luck swapping Mx and My ... taking the zero off the diagonal.

i'm not sure about your problem ... is your body a two force member, with the origin being a general point in the body ? it looks like you're calculating the internal forces due to the two applied loads.

 

[daniel7campbell]

Thanks for the response.

I have tried swapping the equations around to no avail.
The system actually has a lot more loads on it, summing to the forces and moments described by [Fx Fy Fz] and [Mx My Mz]. There are no direct forces on that point.
I am trying to balance all of the other forces by applying the two additional loads.

 

[rb1957]

? "I am trying to balance all of the other forces by applying the two additional loads." ... ok it took me a little while to realise you're summing the applied forces to get their sum force and sum moments at the origin ... sounds ok, unnecessary but ok.

i think your matrix is correct ... summing forces and moments at a point due to two generalised forces. but i think you should have a -ve sign in there ...
Fx1+Fx2 = -Fx
Fy1+Fy2 = -Fy
Fz1+Fz2 = -Fz
('cause forces at 1 and 2 are reacting the sum applied forces)

i think what you're trying to do with it isn't ... i don't think you can invert the matrix to determine the generalised forces. i've always reduced the LHS set of equations using elementary row operations ... you have 6 equations and 6 unknowns.

Fx1 = -Fx -Fx2
substitute into the moment equations to remove Fx1 (and Fy1 and Fz1) so you have three equations and three unknowns (Fx2, Fy2, Fz2)

 

[daniel7campbell]

Ok, I agree that the forces should sum to -Fx to equal zero, my mistake, but that does not affect the ability to invert A.
I had already gone down the reduction path, getting:
A =
[-y1+y2 x1-x2 o
z1-z2 0 -x1+x2
0 -z1+z2 y1-y2];
x =
[-Fx2; -Fy2; -Fz2];
B =
[Mz-Fx*y1+Fy*x1;
My+Fx*z1-Fz*x1;
Mx-Fy*z1+Fz*y1];

Again, A does not invert.

 

[zekeman]

Your problem is, as you may know, that the determinant is zero.
And that means that the first 3 rows are not independent
There is an obvious physical reason for this.
You are asking an arbitrary force vector, (Fx,Fy,Mz) at point at 0,0 to produce unique forces at 2 locations to put the system in equilibrium. For example, take the arbitrary force vector at 0,0 to be (0,0,Mz) then it is clear to me that you can't get a unique Fx1,Fy1, Fx2, Fy2, but an infinity of couples that will produce the Mz. Similarly, if you have (Fx,0,Mz), you would get an infinity of force vectors at points 1 and 2.

So, if you grapple with the first 3 rows as they do to reduce the matrix , you will eventually get all the rows to be 0, thus proving mathematically what I attempted to show above. Add col2 to col1, add col5 to col4, add row3 to row2 and you get row 2 to become 0 0 -x1 0 0 -x2. now add col2 to col3 and add col 5 to col 6 and finally add row 1 to row 2 to get 0 0 0 0 0 0 .

BTW, what re you trying to do?

 

[rb1957]

but given pt1 and pt2 then there is only one solution for the 6 force components.

i'd've thought that since there is a real solution (maybe that's an assumption) that can be determined by "elementary row operations" on a set of simultaneous equations then other matix methods would've worked. maybe not, maybe the determinate is zero (maybe 'cause the 6 components are not independent, but rather they are dependent on one another).

I'm sure that given 6 components of load at a point (whether they correctly sum the applied loads or not, i'd thought that an error in this summing might have thrown the math off) then for 2 given points there is a unique solution of 6 reactions components. if there is an error in the summing then the reactions won't balance the applied, but there'll still be a solution.

not sure why you're finding the reactions by two steps, summing the applied then balancing, but i don't think it's Wrong (just Odd).

 

[daniel7campbell]

I'll try to explain a little further.
The system is a fairly large mechanical device used on most aircraft (I'm afraid I can't give much more information than that). It is mounted to the A/C using a number of rods/lugs. From FEM runs, I have the loads at all of these A/C interface points for a certain condition, as well as g-loading and drag loading. I then sum all the reaction forces to the center point to get a known force defined by [Fx Fy Fz] and [Mx My Mz].
For a structural test of the system, and to replicate the g and drag loads, I am applying a force at two known, fixed points. The two loads at these points, with their 6 components, should be able to balance out the force at the center point created by all of the other reaction forces. This method then allows me to replicate the load condition with multiple g's in a 1g, zero drag, static test.

Now, I know the determinant of both the 6x6 and reduced 3x3 matrices are zero, implying #vars ~= #eqns. I just don't understand how this is possible, since I still don't see how the equations are dependent.

Does what I'm asking make sense?

Thanks for the help

 

[rb1957]

we've gone full circle (in my mind at least) ... you want two forces that are equivalent to a generalised load at the origin.

but i think the logic still applies (only maybe the -ve sign disappears, you want Fx1+Fx2 = Fx, though it doesn't change the math.

but instead of inverting the matrix, solve the simultuaneous equations. i'm suprised that it doesn't want to solve with matrix methods but maybe it doesn't (maybe 'cause the determinate is zero).

6 equations with 6 unknowns has a real solution. you've reduced this to 3 equations with 3 unknowns using substitution. solve the 3 equations, if you've forgotten the "olde school" methods, google "elementary row operations", "solve simulataneous equations".

 

[daniel7campbell]

I guess we are beating this to death at this point, but I believe the definition of a system of equations with a determinant of zero means that the equations are not independent.

I just attempted to solve the remaining system by hand and ended up with a constant = constant (all variables cancelled out).

I guess I will just have to add a third load application point and iteratively solve since this method does not seem to work.

Thanks

 

[rb1957]

post your 6 loads and your two points

 

[daniel7campbell]

F = [-0.000305554 1081.047698 47.62631473]
M = [-1409.758905 -1839.618482 -7996.19718]
x1=-25.705; x2=-25.705;
y1=-5.097; y2=5.097;
z1=-3.276; z2=-3.276;

 

[rb1957]

i started thinking that you could apply one force through the CG, but i couldn't fit your loads. the best i could do was match Mx and Mz by appling Fx,Fy,Fz at a point (-7.4,0,1.3). being short on My i could then add a couple (either Fx or Fz) to fill out the short-fall. i think this explain (in lay-man's terms) why the math fails ... there are multiple ways to resolve moments as couples and the math can't choose for you, but you can.

with essentially no Fx, My = -Fz*x, and Mz = Fy*x and Mx = Fz*y-Fy*z

a problem with your points is there is no way to determine the distribution of Fy between the two points. that may not matter to you in the real world, but i can see the math world being quite upset.

so fy1+fy2 = 1081
Mx = -1081*-3.276 = 3542
Mz = 1081*-25.705 = -27788

let fz1 = 47.6
Mx = 3542-47.6*5.097 = 3299
My = 47.6*25.705 = 1224
Mz = -27788

now use couples to modify the moments to the desired values.

 

[rb1957]

the problem is you can balance out Mx and Mz using couples, but not My

 

[zekeman]

"Now, I know the determinant of both the 6x6 and reduced 3x3 matrices are zero, implying #vars ~= #eqns. I just don't understand how this is possible, since I still don't see how the equations are dependent.'

If you wrote the equations properly ( you did) ,and you get a singular matrix, how can you question that the equations are not independent.

The onus is on you to either accept that fact or, if you still don't believe it, then find the proof.


If you looked at my post , I offered the proof with standard reduction methods to reduce one row of the matrix to 0 0 0 0 0 0 , thus proving singularity.


I don't understand why you and rb1957 would continue on this fruitless inersion path when it is not possible.

 

[zekeman]

I looked at your posted problem. Since it has all kinds of symmetry, it is very easy to do directly, although I have a problem with Mx which I find to be -3483 vs your -1410.
But if you are looking for a general solution, that is a matter for another day.

 

[FeX32]

I concur.

A singular matrix in A*x=b in a statics problem indicates a fundamental flaw in force/ moment sums. The most common mistake is dependancy of 2 or more static equations. A singular matrix A could also indicate motion in the system (not one to one system, ie. given reactions go back to applied forces).

You can do deeper if you like. You could look into determining the rank of your A matrix. This can tell you just how much information each row gives you. Your rank for a 6by6 should be full, 6.
The dimension of the range of a linear transformation is equal to the rank of its corresponding matrix.


Fe (IronX32)

 

[rb1957]

zeke, why "I have a problem with Mx which I find to be -3483 vs your -1410." ?

F and M are calc'd from a bunch of forces we haven't seen (as i understand it).

i think the reason the math approach (using matrix methods, which i'm not advocating) fails is that there are multiple ways to resolve the moment imbalance for Mx and Mz and no way to resolve the My imbalance.

you can resolve Mx with z-force couple at the two points (and Mz using x-force couple), but you can't do anything about My (since the line between the points is parallel with the y axis). My is due to z-force at the two points, but z-force is fixed, so you're skrewed ...

 

[GregLocock]

Is the problem that the OP is trying to solve that of reacting an arbitrary 6dof set of forces and moments on a rigid body, by the application of 2 forces in any direction at two points on the body? If so are the points allowed to be moved (I think not)?





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[zekeman]

I made an error in my last post;I misread the coordinates.

Rb1957,

"F and M are calc'd from a bunch of forces we haven't seen (as i understand it)."

The way I see it,. it is the other way around, or that not what you said?

From the given vector force , F and M at 0,0 the OP wants you to find the vector forces at x1,y1,z1 and x2,y2,z2. His approach fails because the matrix is singular.
However for the special symmetrical case he presents, there would be a solution providing he was consistent with My which in this case is

My=(Fz1+Fz2)*x1 =-Fz*x1=47*25=1180 (since Fx=0)

His My at 0,0 is 1840.

I didn't bother with the signs , but one or both of these are clearly wrong

So YOU are not skrewed after all.The problem is wrongly characterized.