# L beam - Strange result

I have the following system (at P and Q there is a ball joint; the length of L beam is l; the PB and PA forces are applied at l/2 and are known; Fz is not zero and is known):

The system has 0 degrees of freedom because each ball joint removes three degrees of freedom (6-2*3=0). I replaced each ball joint with three unknown forces.

The problem is that the second cardinal equation along x gives:

$$-F_z \, l = 0$$

How can I solve the problem?

Thank you very much for your time.

• For planar force system you can write 3 equilibrium equations and only 3 reactions can be found this way. Your example is statically indeterminate. One of your bearings has to have only one reaction component, the other can have two for your problem to become determinate. – Katarina May 23 '18 at 18:35
• Hello @Katarina, my system is in the space and thus has 6 degrees of freedom. I wrote 6 equations, 3 for translation equilibrium and 3 for rotation equilibrium. The only equation which gives me problem is the rotational equation along x. Maybe do I miss somithing? Thank you. – Gennaro Arguzzi May 23 '18 at 18:40
• I see, if this is 3D problem, you are missing PAz and PBz. – Katarina May 23 '18 at 18:49
• And you have - Fz x l+PAz x l/2 =0, or whichever direction of PAz you choose. – Katarina May 23 '18 at 18:53
• If points P and Q are ball joints and you have a nonzero force in the Z direction, what is preventing the model from rotating about the X axis through points P and Q? – JohnHoltz May 23 '18 at 20:12

The system is still statically indeterminate, but not due to lack of DOF. You have a singularity. Having correct DOFs only assures you have a chance of being statically determinate, but it does not guarantee this condition. In this case, the singularity is because the beam is allowed to rotate about the x axis unless $F_z$ is an unknown reaction.

As such, in this case, $F_zl = 0$ means $F_z = 0$. Here are your other equations, in no particular order:

\begin{alignat}{4} \sum& F_x &&= A + PB_x + D + PA_x &&= 0 \\ \sum& F_y &&= B + PB_y + E + PA_y &&= 0 \\ \sum& F_z &&= C + G + (F_z = 0) &&= 0 \\ \sum& M_{y@Q} &&= Cl &&= 0 \\ \sum& M_{x@Q} &&= F_zl &&= 0 \\ \sum& M_{z@Q} &&= -Bl + PB_yl/2 - PA_xl/2 &&= 0 \end{alignat}

We can immediately see $C = F_z = 0$. Because of $\sum F_z$, $G = 0$. We now have four unknowns and three equations. We can solve for $B$ immediately using:

$$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$

Rearranging, we can use this result for $B$ to solve for $E$ using:

$$\sum F_y = B + PB_y + E + PA_y = 0$$

But the final equation is singular:

$$\sum F_x = A + PB_x + D + PA_x = 0$$

This cannot be resolved using statics, but require static indeterminate methods.

• Hello @Mark, your answer is excellent. Thank you so much. – Gennaro Arguzzi May 23 '18 at 20:26