Moment distibution (stiffness factor modifications) for a frame

[fonseh]

Homework Statement

For the KAb and KCD , they are connected by the same method ... At B and C , we can see that they are connected by the same method , although i am not sure it's fixed or pinned .

Homework Equations

The Attempt at a Solution

I assume they are fixed at B and C , so i agree that KAB is 4EI / L for far end fixed case . So , i think KCD should be 4EI / L as well .

 

[haruspex]

far end fixed

Aren't the "far ends" A and D? One is fixed and one is pinned.

 

[fonseh]

Aren't the "far ends" A and D? One is fixed and one is pinned.

B and C are fixed ? How do we know that ? It's not stated in the question

 

[haruspex]

B and C are fixed ? How do we know that ? It's not stated in the question

No, not B and C; A and D. As I read it, the stiffness factor at one end of a beam depends on whether the far end (other end) is fixed or only pinned. That makes intuitive sense to me.
A, B and C are fixed, so for the interactions at B and C:
AB is 4 because A is fixed
BC is 4 at each end because the other end is fixed
CD is 3 because D is pinned
CE is 3 because E is pinned

 

[fonseh]

BC is 4 at each end because the other end is fixed

Which end is fixed ? B or C ?

 

[haruspex]

Which end is fixed ? B or C ?

Both are "fixed". (I think this just means it's a rigid joint, not a hinge.)
But hey, I'm only inferring all this from the text.

 

[fonseh]

Both are "fixed". (I think this just means it's a rigid joint, not a hinge.)
But hey, I'm only inferring all this from the text.

You just assume they are fixed ? So , for this type of question , we just assume they are always fixed ?? Since the author didnt show how's the connection at B and C

 

[haruspex]

You just assume they are fixed ? So , for this type of question , we just assume they are always fixed ?? Since the author didnt show how's the connection at B and C

I'm not sure how you were supposed to know. They certainly look rigid in the drawing, but then so do D and E.

 

[fonseh]

S

I'm not sure how you were supposed to know. They certainly look rigid in the drawing, but then so do D and E.

Since they look ' rigid' , so they are assumed to be fixed at B and C ?

 

[haruspex]

S

Since they look ' rigid' , so they are assumed to be fixed at B and C ?

I might have it...

The question of "far end fixed" is to do with sideways flexing at the "near end" being inhibited by the far end.
At the top of CD, sideways movement of C is not resisted by the joint D since that is a hinge.
At the top of AB, sideways movement of B is resisted by the joint at A.
In the beam BC, vertical movement at B is resisted by the combination of the rigidity of BCE and constraints at D and E. C cannot move up or down because of D. The rigidity of BCE then inhibits vertical movement at B.
Similarly, vertical movement at C is resisted by the combination of the rigidity of BCE and constraints at AB and E. B cannot move up or down because of A.

Hope that makes sense.