How does Force Effect https://forceeffect.autodesk.com do it? Given a 10m long bar angled as the hypotenuse of a 3, 4, 5 triangle with an xy pin constraint a (0,0), another xy constraint at (4,3), and a vertical force of 100N applied down at (10,6) Force Effect calculates the x and y reaction forces below at C and A. I am not as interested in this particular answer in how to get to it by a codeable algorithm.

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So the equations are:

Sum Moments about C = 0 => Y$_{ca}$(R$_{ax}$) + X$_{ca}$(R$_{ay}$) + X$_{cb}$(F$_{1}$) = 0 => 3 R$_{ax}$ - 4 R$_{ay}$ = -4(-100)

Sum Moments about A = 0 => Y$_{ac}$(R$_{cx}$) + X$_{ac}$(R$_{cy}$) + X$_{ab}$(F$_{1}$) = 0 => - 3 R$_{cx}$ + 4 R$_{cy}$ = -8(-100)

Sum Forces Y = 0 => R$_{ay}$ + R$_{cy}$ + F$_{1}$ = 0 => R$_{ay}$ + R$_{cy}$ = -(-100)

Sum Forces X = 0 => R$_{ax}$ + R$_{cx}$ = 0

Which could translate to a 4x4 matrix: Ax = b

|  3.   -4.    0.    0. |   | Rax |  =  | 400 |
|  0.    0.   -3.    4. |   | Ray |  =  | 800 |
|  0.    1.    0.    1. |   | Rcx |  =  | 100 |
|  1.    0.    1.    0. |   | Rcy |  =  |   0 |

And attempting to solve for x = inv(A)* b gives the following error in SciLab (or MatLab) -->inv(A) !--error 19 Problem is singular.

So the matrix is not invertible- whatever that means!

I’d be open to some sort of substitution method to solve it if that is any easier.

This is not homework. This is a simplified version of a real-world problem of a large free body being held in place by two fixed pins. It is not dynamic – nothing is moving.


your system of equations is underdetermined, which means that you don't have enough information to solve it. You can proof it by calculating the determinant. For a unique solution it has to be unequal zero. http://en.wikipedia.org/wiki/System_of_linear_equations

For static problems the determination can be assessed by counting the degrees of freedoms and the degrees of constraints.

The real world problem is solvable. You need more equations, which you can use to get enough information to describe a determined problem. You need the elastic modulus of the material which your beam consists of. Everything afterwards is rather a mathematical methodology http://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory#Statically_indeterminate_beams

Classical authors in the field of mechanics like Timoshenko cover the theme in more depths in their books.

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  • $\begingroup$ I assume Autodesk uses equal elastic modulus and cross section for all the elements. Thus the results will differ from the real world applications. $\endgroup$ – John Alexiou Jul 21 '15 at 20:56

The problem is relatively simple if the supports have infinite stiffness. The way to then solve it conceptually is to apply the principle of superposition to:

(a) system with a support at $C$, applied force at $B$, and no constraint at $A$ $\rightarrow$ get deflection $\delta_{A1}$ at $A$; (b) system with force $F_A$ at $A$, reaction force at $C$, and no constraint at $B$ $\rightarrow$ get deflection $\delta_{A2} = k{F_A}$ of $A$ as a function of $F_A$.

Then you must have: $\delta_{A1} + \delta_{A2} = 0$ for the superposition, so you can solve for $F_A$.

For real world problems with finite elasticity I concur with mark_r - consult Wikipedia and Timoshenko!

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