I hope there is some one who know solutons of none easy problem in SAP

[kristjan]

I appreciate all of you for previous answers but i think is not help, so I must repeat question again. And i hope there is some one who know solutons of none easy problem.
Well I last time asked how we can model wall together with other frame element in software sap 2000, and i wonder is it posible in this software. The realy problem is automatic transfer load with wall on beam in sap2000.

Now i ask across a this egzample:
I have a concrete-beam who simply-supported and a concrete-wall above her.
Property of concrete- wall is:
height hw=3.0m
thickness ht=0.25m
length lw=6.0m
weight per unit volume of concrete W=25.0 kN/m3

Property of concrete-beam is:
height hg= in this cause we slack for better insight this problem.
width wg= -//-
length lg=6.0m.
Note : in sap 2000 we assigned a property ” other” material with W=0.0

Now i want to compute moment of beam in midspan ( loc=3 m ) and their reactions.And I do on two way. First by hand and second by sap2000.

First:
So if i slack a section of beam then i have a load of wall( kn/m ).
And I estimate as
g= W*ht*hw.( g=25*0.25*3 = 18.75 kn/m)
it follow
M(L=3.0m)=(1/8)*g*lg^2=(1/8)*18.75*6.0^2=84.375 knm
Ra=Rb=g*lg/2=18.75*6.0/2=56.25 kn

Second way
In Static load case box I put for type load a dead and for self weight multiplier =1.
I model a frame element as simply suported beam with all property as in first cause.
Then model,a shell element above their, with all property as in first cause.
Frame element divide into 8 by 1. Then mash shell into 8 by 1.
I suppose that wall and beam is connection so we have transfer load of wall(self weight-wall) at beam.
And when i run analysis i get a next value:
For frame:
(loc x= 3m) M33=My=0.46knm that is incorrect.
loc x=0;x=6) Ra=Rb= 56.250 kn that is correct

Notice: Reastion geted in both case is correct, but unfortunately moments is not.And we didn’t have a transfer load for wall on beam as allotment load(kn/m) already we did have transfer mass on end joints of frame.
If anyone know i hope that he answer a precisely and simply?!
Thanks.

 

[vonlueke]

KRISTJAN: Go back and review my first post. (Skip my second post for now.) I explain in my first post how you should add up the resultant moment at the midspan section cut. Notice, you must also include the wall shell elements in your section cut. You must not omit the x-direction element forces, because they create a giant force couple, which is the vast majority of your midspan moment.

See how small My is? My = 0.46 kN*m is probably correct, but you have omitted the big force couple created by all the axial forces in the frame and shell elements at loc = 3.0 m. Compute the resultant moment of all element x-direction forces about the centroid, then add your My, then it will equal 84.375 kN*m. Good luck.