Multi-layer Shell Element Problem

[vahidbarzegar]

Hi,
I used Multi layer shell element to perform a pushover analysis on a 84x21 foot RC shear wall. The pushover diagram reaches a peak at a 0.01 drift and then drops and tends to a constant. But as it is a validation analysis, the original one goes up constantly until 2.5% drift achieved.
In my opinion there is something wrong with the material definition. I used expected values of material properties but the behavior was the same. Since the material definition is very straight forward in this model, I don' know what is the problem (first define a PlaneStressUserMaterial, then assign it to a PlateFormPlaneStress nDMaterial and finally a uniaxialMaterial Steel is defined).

nDMaterial PlaneStressUserMaterial 1 40 7 4000 560 -800 -0.003 -0.01 0.00028 1
nDMaterial PlateFromPlaneStress 4 1 1812971.7
#6:d=0.75in
uniaxialMaterial Steel02 7 68000 29000000 0.010 18.5 0.925 0.15
nDMaterial PlateRebar 8 7 90
section LayeredShell 1 12 4 1 4 1 8 0.03682 4 0.96318 4 1 4 1 4 1 / 4 1 4 0.96318 8 0.03682 4 1 4 1

[samsonama]

hei,
i have a similar problem. Can you explain me how do you define the asbolute thickness of the rebars in the section layeredShell?

thanks

[vahidbarzegar]

I used a shell with equal area with area of rebars.

[samsonama]

hei,
i am reading the article "http://www.luxinzheng.net/publication6/ ... D_2015.htm", where the model is described. I still don't get how to combine the thickness with the area of the rebars? is it the total area of the rebars or what?
#middle region
# material absolute thickness angle(steel) material tag
##cover 12.5 4
##d=6transverse 0.2403 0 11
##d=6longitudinal 0.2356 90 10
##core 49.5241 4
##core 49.5241 4
##d=6longitudinal 0.2356 90 10
##d=6transverse 0.2403 0 11
##cover 12.5 4

section LayeredShell 2 8 4 0.0125 11 0.0002403 10 0.0002356 4 0.0495241 4 0.0495241 10 0.0002356 11 0.0002403 4 0.0125

[vahidbarzegar]

total area of the rebars in a direction perpendicular to the rebars. for example longitudinal rebars for specimen sw1-1 is transformed as:
d=6mm==>As=28.27mm^2 ==> total Ast=5*As=141.37 ==> equivalent thickness=141.37/600=0.2356

[samsonama]

you made my day

[samsonama]

and for the transverse rebars?

[vahidbarzegar]

Ast=480.66
t=480.66/2000=0.2403
Please let me know if you understand anything about my problem

[samsonama]

is much easier if you post the script