Reaction forces in a two segment beam

[yaro99]

Homework Statement

Homework Equations

∑Fx=0
∑Fy=0
∑M=0

The Attempt at a Solution

[/B]
I am just concerned with part a.

I got the correct answers doing this:

First, FBD of segment AB:

+cw ∑M_A=0: -136 - 6*By = 0
By = -22.7N

two ways to get Ay:

one:
+cw ∑M_B=0: -136 + 6*Ay = 0
Ay = 22.7N

two:
∑Fy=0: -22.7 + Ay = 0
Ay = 22.7N

FDB entire beam:

∑Fy=0: 22.7 - 22.7 + Cy = 0
Cy = 0

∑Fx=0: -440 + 220 + Cx = 0
Cx = 220N

Ok, so I got the correct answers. However, here is where I get confused. If I make a FBD on segment BC I get a different answer for Cy.

My attempt at a FBD:

this leads to either:
∑Fy=0: -22.7 + Cy = 0
Cy = 22.7N

or

+cw ∑M_B=0: -136 + 3*Cy = 0
Cy = 45.3N

Clearly my FBD is wrong for segment BC. I assume the correct FBD would have a zero reaction at point B; this would give the correct answer (0) for Cy. In order for Cy to be zero, there must be no forces or moment at point B in the FBD, but this doesn't make sense to me.

 

[PhanthomJay]

Your free body diagram is wrong because when looking at that segment between the support at B and C , you must include the internal moment in the beam at B. Far simpler to take your FBD between the pin (just to the right of B ) and C. In this manner, the applied couple does not enter into your diagram.

 

[yaro99]

Your free body diagram is wrong because when looking at that segment between the support at B and C , you must include the internal moment in the beam at B.

So my fbd is just missing an internal moment at B, and the resulting equation would be:
+cw ∑M_B=0: -136 + 3*Cy + M_B = 0
I can't solve directly for Cy since I don't have M_B.
Is this correct?

Far simpler to take your FBD between the pin (just to the right of B ) and C. In this manner, the applied couple does not enter into your diagram.

Would I not still have to include an internal moment? Wouldn't a complete diagram look something like this?

 

[PhanthomJay]

So my fbd is just missing an internal moment at B, and the resulting equation would be:
+cw ∑M_B=0: -136 + 3*Cy + M_B = 0
I can't solve directly for Cy since I don't have M_B.
Is this correct?

yes

Would I not still have to include an internal moment? Wouldn't a complete diagram look something like this?

unlike at an externally pinned support , there can be no internal moment in the beam at a pinned connection.