Hi,
I tried two models with different Panel Calculation Model: one with Partial stiffening, and the other Without stiffening. I got two different diagrams for Mx, the former with uniform diagram and the latter with uniform varying diagram. Please find the attached pictures. I wonder why?
You have probably misunderstood my suggestion. I have suggested the following. Sorry for missing "each" word
"You have to check balance of nodal moments in all bars connected to each beam 9 ends."
Moments balance check (in global plane Zy) in node 4 for bars 9, 2, 20, 12 (moments balance should give about 0.5* 1.5 )
Moments balance check (in global plane ZY) in node 2 for bars 1, 9, 11, 19 (moments balance should give about another 0.5* 1.5 )
Moments have different direction
the same on other beam 9 end
2*0.76=1.52 which gives reasonable precision
Moments in beams (green ) are balanced by moment difference in columns (yellow)
I believe my verification is correct. If you think that I made mistake please present your proposition.
When balance in node is verified moments direction in respect to node (not to moment sign in bar) should be be considered. Positive bending moment can result in transferring moment to columns in opposite direction when we regard opposite end of the beam.
Example
When we consider node 4, we can presented moments direction in the following way.
For verification you should take into account difference in Mz in common node of bars 20 and 2 Depending on direction set as positive (in respect to verified node) it is
Mc= (8.53 -(-12.17))=20.7
or
Mc=(-12.17-8.53)=-20.7
Moments in remaining beams.
Mx in 9 and also My in 11 and 12, acts in opposite direction than Mc, so moments from those members should be added to Mc with opposite sign.
Verification from post 23 is correct.
Sign convention should be applied in respect to verified node not element.
I think you should think it over. I look at it in the following way.
I can simulate loads from slab as three bending moment, then I can simulate column as rigid support.
Then moment transferred by column can be presented as reaction in rigid support.
Slab is on the node right side - so all moments applied to column from loaded slab are clockwise
Reaction should balance applied forces so its direction (moment transferred by column) is counterclockwise
You missed correct direction of moment transferred to node from beam 12 in post 31
It is clockwise in respect to regarded node, since the following situation occurs
This load on beam gives the following moment distribution in the column
Forces distribution can be simulated in the following way
I appreciate your help.
You wrote:
You missed correct direction of moment transferred to node from beam 12 in post 31
It is clockwise in respect to regarded node, since the following situation occurs
Can you elaborate more on this? I'm not aware what you mean.
You wrote:
This load on beam gives the following moment distribution in the column
Forces distribution can be simulated in the following way
I think you are not clear on this.
Regarding your last questions.
ad1
In post 31 you have written that moment My from beam 12 acts counterclockwise. When we check moment balance in node 4, this moment (My from beam 12) is applied to node 4 in clockwise direction.
ad2
A. Moments applied to the node 4 from beam 12 (My), beam 9 (Mx), and also missing moment part 0.76, acting clockwise (in respect to node 4).
B. Since in statically stable structure sum of bending moments in each node, in each plane is 0, then moments from point A. are balanced by Mz moment difference in column part (difference between column part below node 4 and column part above node 4). This moment difference (equivalent moment in column) acts counterclockwise.
C. Sum of moments from point A (acting clockwise) and moments from point 2 (acting counterclockwise) gives 0 as expected for statically stable structure.
If anything remains not clear for you, or in your opinion, moments summation in node 4 should be different, please let me know. I would appreciate if you present your version of moments summation, then we can discuss it.
Let's do it step by step.
1- Is my sense of sign convention correct that if I pointed my right thumb in the direction of the local positive axis and curled my fingers, my fingers will be curled in the same direction as the positive moment about axis?
2- bar 9, case DL2, Mx
diagram for bar (not node)
Mx= -3.97
My=-15.97,
I understand that for moment summation in node 4, you set positive moment sign for moments acting in plane ZY as acting clockwise.
For that sign convention
MX from beam 6 ; M(6)= -3.97
My from beam 12; M(12)=-15.97
for that sign convention total moment from beam 6 and beam 12 applied to node 4
M(6)+M(12)= -19.97
