Multilayer Shell Model

[Mousapoor]

How to arrange layers in section LayeredShell in order to build Multilayer Shell element? Layers arranged from top to bottom or from bottom to top?

[EricsonEncinaZ]

The layers will be stacked up in the local Z direction, so we can say that they are arranged from bottom to top in the local Z.

[Mousapoor]

thankful
I have modeled a 3D flat slab with multilayer shell element.a incrimental point load applied to middle (from top to bottom) and supports is placed in the slab edges (wihout applying in-plane restraints to the slab).
the local Z direction is from bottom to top. The layers are arranged from bottom to top in the local Z in section LayeredShell. Increasing the slab tensile reinforcement ( bottom rebar) does not have an effect on the flextural strength of the slab and stiffness, but Increasing the top slabs reinforcement will increase the slab flextural resistance and stiffness.
It does not seem logical, what is the reason??
Can I use the multilayer shell element to model the slab??

[EricsonEncinaZ]

Could you show your script or the relevant parts if it is too long?
i.e. node definitions, layered shell section definition, shell element definition and load definition.

[Mousapoor]

# Units: N, mm, sec
layered shell section definition:
# FLOOR ******************************************
# UnConfined Concrete
# fc ft fcu epsc0 epscu epstu stc
nDMaterial PlaneStressUserMaterial 1 40 7 40.50 3.168 0.00 -0.002 -0.007 0.00137 0.3
# nDMaterial PlateFromPlaneStress $matTag $PlaneStressMatTag $OutOfPlaneShearModulus
nDMaterial PlateFromPlaneStress 2 1 13258.252

#steel
# tag fy E0 b R0 cR1 cR2
uniaxialMaterial Steel02 3 552.0 2.0e5 0.0080 18.5 0.925 0.15
uniaxialMaterial MinMax 4 3 -min -0.25 -max 0.25 ;

#angle=90 longitudinal reinforced steel
nDMaterial PlateRebar 5 4 90

#angle=0 transverse reinforced steel
nDMaterial PlateRebar 6 4 0

# section LayeredShell $sectionTag $nLayers $matTag1 $thickness1...$matTagn $thicknessn
section LayeredShell 701 28 2 10 2 10 2 10 2 10 6 0.523333 5 0.523333 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 2 10.4894 5 0.4024 6 0.4024 2 10 2 10 2 10 2 10 ;

Define Elements (e.g):
#Shell element
#element ShellNLDKGQ $eleTag $iNode $jNode $kNode $lNode $secTag
element ShellNLDKGQ 4 10 22 23 24 701

[EricsonEncinaZ]

You missed nodes definition and load definition.

[Mousapoor]

excuse me.
I show a part of the nodes definition.
#Define Nodes (e.g.):
# X Y Z
node 6 1.500E+003 1.500E+003 3.000E+002
...
node 10 0.000E+000 0.000E+000 3.000E+002
node 22 9.790E+001 0.000E+000 3.000E+002
node 23 9.790E+001 9.790E+001 3.000E+002
node 24 0.000E+000 9.790E+001 3.000E+002
...

#Push Pattern:
## Load Case = PUSH
pattern Plain 2 Linear {
load 6 0.000E+000 0.000E+000 -1.000E+000 0.000E+000 0.000E+000 0.000E+000
}

[EricsonEncinaZ]

According to your code the shell is defined on a plane in the global X-Y. Se below a schematic representation. It is worth noting that the global Z is perpendicular to the screen and pointing towards you.

^ Y
|
|24-23
|10-22 ------->X

So your definition represents a 97.9mm side square shell defined by nodes 10, 22, 23, 24 in an X-Y plane at globla Z = 300mm

Due to the definition of your shell in this way the local x, y and z of the shell actually match the global X, Y and Z of the domain. Given that the local z - and also the global Z - are pointing to you, the stacking of the layers is preform from the screen towards you. Since your load pattern is pointing in minus global Z, or in other words your load is pointing into the screen, the steel layer of 0.523mm is actually the reinforcement that is in the tension side of your shell. So the model response is according to what I wrote in the previous post.

Remember that in the local z the layer's stacking is like this:

^ Local z
|
| . . .
| second layer defined
| first layer defined

[Mousapoor]

Thank you for your attention.
My question is that Increasing the slab tensile reinforcement ( bottom rebar= the steel layer of 0.523mm) does not have an effect on the flextural strength of the slab and stiffness, in the other word the Load-Deflection curves Slightly changed, but Increasing the top reinforcement ( the steel layer of 0.402mm )will increase the slab flextural resistance and stiffness, in the other word the Load-Deflection curves Much changed. pin supports is placed in the slab edges.

It does not seem logical,what is the reason??

[EricsonEncinaZ]

I don't know the reason. I thought I was wrong and I did a small model to check the layered shell stacking and I'm right, also compare them with fibre section responses and it was ok as well.

I just can think of 2 very unlikely reasons:
1. Are you sure that when you modify the reinforcement area (thickness) you keep the same overall thickness of the section? the overall thickness is the sum of all the layers, so when you increase the thickness of a reinforcement layer the overall section thickness changes, but anyway I would say that a variation in reo thicknesses are negliglible when compared to the whole section.
2. Are you sure that all the local axes of your shell elements have the same orientation? I dont know how much this could affect the results.

I hope you can solve your problem

[duguyihan]

Dear,

I also obtain significant difference when setting variables value for bottom and top rebar in shell element. My question is, how to determine the local z axis if my shell element is in the global X-Z plane?

[EricsonEncinaZ]

x cross product y = z as usual for vectors
for x and y read my post on 09 Jan 2018