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Thanks Wasabi for the Latex editing.
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Mark
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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.

InAs 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.

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, $F_zl = 0$ 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.

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.

added 67 characters in body
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Wasabi
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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, $F_zl = 0$ means $F_z = 0$. Here are your other equations, in no particular order:

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

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

$\Sigma M_{z@Q} = -B*l + PB_y*l/2 - PA_x*l/2 = 0$$$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$

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

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

But the final equation is singular:

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

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

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

$\Sigma F_x = A + PB_x + D + PA_x = 0; \Sigma F_y = B + PB_y + E + PA_y = 0; \Sigma F_z = C + G + (F_z = 0) = 0; \Sigma M_{y@Q} = Cl = 0; \Sigma M_{x@Q} = F_zl = 0; \Sigma M_{z@Q} = -B*l + PB_y*l/2 - PA_x*l/2 = 0$

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

$\Sigma M_{z@Q} = -B*l + PB_y*l/2 - PA_x*l/2 = 0$

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

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

But the final equation is singular:

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

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

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, $F_zl = 0$ 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.

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Mark
  • 5.3k
  • 15
  • 47

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

$\Sigma F_x = A + PB_x + D + PA_x = 0; \Sigma F_y = B + PB_y + E + PA_y = 0; \Sigma F_z = C + G + (F_z = 0) = 0; \Sigma M_{y@Q} = Cl = 0; \Sigma M_{x@Q} = F_zl = 0; \Sigma M_{z@Q} = -B*l + PB_y*l/2 - PA_x*l/2 = 0$

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

$\Sigma M_{z@Q} = -B*l + PB_y*l/2 - PA_x*l/2 = 0$

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

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

But the final equation is singular:

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

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