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Plates, shells and laminates
in
Mentat & MARC
Tutorial
Eindhoven University of TechnologyDepartment of Mechanical EngineeringPiet Schreurs September 30, 2015
Contents
1 Linear plate bending 3
1.1 Deformation and loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Example: bending of a strip . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.1 Strains and stresses in a shell element . . . . . . . . . . . . . . . . . . 81.3 Example: bending of a square isotropic plate . . . . . . . . . . . . . . . . . . 111.4 Example: bending of a circular isotropic plate . . . . . . . . . . . . . . . . . . 14
1.4.1 Plate with central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4.2 Solid plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.5 Orthotropic and transversaly isotropic material . . . . . . . . . . . . . . . . . 161.5.1 Orthotropic plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6 Example: bending of a square orthotropic plate . . . . . . . . . . . . . . . . . 18
2 Laminates 22
2.1 Example: 4-ply laminate plate . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Example: laminated tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1 Linear plate bending
Linear plate bending theory is explained in the document Linear plate bending. In Cartesianand cylindrical coordinate systems, plate bending equations are derived. For the circularplates analytical solutions are elaborated.
In the following sections we consider first the definition of deformation and load variablesin MSC.Marc/Mentat, which differ from the definitions in the above-mentioned document.We also look at the definition and names of orthotropic and transversal isotropic materialparameters in MSC.Marc/Mentat.
1.1 Deformation and loads
In the document Linear plate bending the next sign conventions have been used for membranestrains and curvatures.
x
yz
εxx0
εxy0
εxy0εyy0
κxy
κxx
κyyκxy
Fig. 1.1 : Strain and curvature definitions
In accordance with this, the next sign conventions have been chosen for applied cross-sectionalforces and moments.
x
y
zNxy
Dx
Dy
Myy
Mxx
Mxy
Nyy
Nxy
Nxx
Mxy
Fig. 1.2 : Cross-sectional forces and moments
These loads are edge loads per unit of length with units [N/m] for forces and [Nm/m] formoments.
In MSC.Marc/Mentat the sign for applied nodal forces are equivalent. Applied nodal
moments are defined about x- and y-axis and are positive according to the right-hand rule.This is shown in the next figure. Equivalent sign conventions and definitions hold for strainsand curvatures.
x
y
zNx
Mx
My
My
Ny
Nx
Ny
Mx
Fig. 1.3 : Cross-sectional forces and moments in MSC.Marc/Mentat
In MSC.Marc/Mentat the loads are applied as nodal forces and nodal moments. If edge loadsper unit of length have to be prescribed the nodal forces and moments must be calculatedsuch that the total load is the same. NB: Values in corner nodes must be reduced.
1.2 Example: bending of a strip
We start here with a simple example, where we model a rectangular strip of material, clampedat one end (x = 0) and loaded with forces and moments at the free end (x = L).
My
x
L
W
H
z
Fx
Fzy
Fig. 1.4 : Strip loaded at the end x = L
As this strip can be seen as a beam, we have analytical solutions for the end displacementsux and uz and the end rotation φy and also for the axial strain εxx and the axial stress σxx.With L the lengh of the beam, W its width, H its heigth and E the Young’s modulus of thematerial, these values are :
ux =L
EWHFx
uz =L3
3EIFz +
L2
2EIMy ; φy =
L2
2EIFz +
L
EIMy
σxx,max =1
WHFx +
1
I
H
2(My + FzL) ; εxx,max =
σxx,max
E
where I = 112 WH3 is the area moment of inertia.
The strip is modelled subsequently with beam elements (type 52), with shell elements (type75) and with three-dimensional elements (type 7). In axial direction the number of elementsis always 20. Using shell or 3D elements, the width is subdivided in 2 elements. Using 3Delements, 5 elements in thickness (= height) direction are used.
Modelling with beam and 3D elements is described in other tutorials and will not berepeated here. The use of shell elements is described in detail below. Model parameter valuesare given in the next table.
length L 1 mwidth W 0.1 mheigth H 0.05 mYoung’s modulus E 150 GPaPoisson’s ratio ν 0.3 -
axial end force Fx 100 Nlateral end force Fz 100 Nbending end moment My 100 Nm
Modelling the strip with shell elements starts by making a square 4-noded element (QUAD4) ofdimensions 1 [m] × 0.1 [m] in the xy-plane. This element is subdivided – use SUBDIVIDE – in 20× 2 elements of uniform dimensions, so without a BIAS. Use the options SWEEP ALL and CHECK
UPSIDE DOWN to complete the mesh. The thickness is specified in GEOMETRIC PROPERTIES.Material parameters have to be specified in the MATREIAL PROPERTIES submenu. When we
enter this menu we have to select MATERIAL PROPERTIES. Moduli have to be given in [Pa =N/m2] as the dimensions of the plate is modelled in meter.
MATERIAL PROPERTIES
NEW
(type) STANDARD
STRUCTURAL
ELASTO-PLASTIC ISOTROPIC
YOUNG’S MODULUS 150e9
POISSON’S RATIO 0.3
OK
(ELEMENTS) ADD
(ALL) EXISTING
The plate has to be fixed at one end. This is done by suppressing all degrees of freedom ofthe mid-edge node. The other two edge nodes are free to move in y-direction, which meansthat the lateral contraction is free.
(MAIN MENU) (PREPROCESSING)
BOUNDARY CONDITIONS
ANALYSIS CLASS STRUCTURAL
NEW
TYPE FIXED DISPLACEMENT
PROPERTIES
DISPLACEMENT X 0
DISPLACEMENT Y 0
DISPLACEMENT Z 0
ROTATION X 0
ROTATION Y 0
ROTATION Z 0
OK
(NODE) ADD Select central edge node
END LIST
NEW
TYPE FIXED DISPLACEMENT
PROPERTIES
DISPLACEMENT X 0
DISPLACEMENT Z 0
ROTATION X 0
ROTATION Y 0
ROTATION Z 0
OK
(NODE) ADD Select corner edge node
END LIST
The edge forces are applied in the nodes at the free end at x = 1 meter. Values in the cornernodes are divided by 2.
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X 100/2
FORCE Z 100/2
OK
(NODE) ADD Select central edge node
END LIST
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X 50/2
FORCE Z 50/2
OK
(NODE) ADD Select corner edge nodes
END LIST
The edge moments are applied in the nodes at x = 1 meter. Note that the moment aboutthe y-axis is -100 Nm which is a bending moment of 100 Nm as indicated in the figure andused in the formulas.
NEW
TYPE POINT LOAD
PROPERTIES
MOMENT Y -100/2
OK
(NODE) ADD Select central edge node
END LIST
NEW
TYPE POINT LOAD
PROPERTIES
MOMENT Y -50/2
OK
(NODE) ADD Select corner edge nodes
END LIST
The JOB can be defined in the usual way. We have to select ELEMENT TYPE 75 for all elements.Output parameters have to be selected in JOB RESULTS. Obviously we want to be informed
about strains and stresses over the thickness of the strip. Therefore from the (AVAILABLE
ELEMENT TENSORS) Stress and Total Strain are selected. In the next subsection we will refer tothe calculation of the stresses in more detail, but for now this will do. In the default settingthe plate has 5 layers over the thicckness, which are used for numerical integration, as isexplained in the document Linear plate bending. In JOB>PROPERTIES>JOB PARAMETERS thenumber of layers can be changed, before doing the analysis. Because we always want to havea layer in the mid-plane of the shell, this number has to be odd. In these layers the strainsand stresses are calculated and to see all these values in the output, we have to select (toggle)(LAYERS) ALL. We can also select Shell Curvature, Shell Moment, Shell Membrane Strain and Shell
Membrane Force. These values are calculated in the mid-plane only, so there is no need toselect (LAYERS) ALL.
After saving the input, the analysis can be done through RUN. The result file can be openedafter a successfull analysis (3004). Displacements, rotations, reaction forces and momentsand the layer strains and stresses can be observed as color plots or as numerical values.
The figure below shows the color(band) plots of stress values.
1.000e+02
1.100e+02
1.200e+02
1.300e+02
1.400e+02
1.500e+02
1.600e+02
1.700e+02
1.800e+02
1.900e+02
2.000e+02
job1
Beam Bending Moment Local Y
Inc: 0Time: 0.000e+00
X Y
Z
4
-4.941e+06
-4.692e+06
-4.442e+06
-4.192e+06
-3.942e+06
-3.692e+06
-3.442e+06
-3.192e+06
-2.943e+06
-2.693e+06
-2.443e+06
job1
Comp 11 of Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
-4.315e+06
-3.447e+06
-2.580e+06
-1.712e+06
-8.442e+05
2.343e+04
8.911e+05
1.759e+06
2.626e+06
3.494e+06
4.362e+06
job1
Comp 11 of Stress
Inc: 0Time: 0.000e+00
X Y
Z
4
Fig. 1.5 : Stress plot for beam, shell and 3D models
The results for all models are compared to the exact analytical solution.
exact beam shell 3D
axial end displacement ux 0.133 0.133 0.133 0.133 µmlateral end displacement uz 0.533 0.533 0.531 0.392 mmend rotation φy 0.00096 0.00096 0.00097 - degmaximum axial stress σ 4.82 - 4.94 4.35 MPa
The analysis with the 3D solid elements is not very accurate. The results will improvewhen more elements are used in thickness direction, as in that case the linear stress over thethickness is represented better by the lineair elements, which have almost constant stress.Analysis time is much higher for the 3D solid model than for the beam and shell model. Itis also rather cumbersome to prescribe the bending moment at the end of the beam, becausethis has to be done with nodal forces.
Let us now take a closer look at the stresses in the shell element.
1.2.1 Strains and stresses in a shell element
In the example of the strip we have selected as output variables Stress and Total Strain. Theirvalues were (almost) the same as the exact solution. Let us now change the mesh slightly.This is done by defining first 6 nodes and subsequently 2 elements, which are subdivided.
(NODES) ADD
0 -0.05 01 -0.05 00 0 01 0 00 0.05 01 0.05 0
(ELEMENTS) ADD
1 2 3 44 6 5 3
Both elements are subdevided such that we end up with 20 elements along the length ofthe strip and 2 elements in width direction, just as we had in the initial example. The onlydifference is that elements in the two rows have a different orientation of the edge between thelocal nodes 1 and 2 (= EDGE12). The model can be completed as before by adding materialproperties and boundary conditions. From the results it can be observed that nodal values –displacements and rotations – are the same as before. The element Stress is different for theelements in the two rows along the length. This shown in the next figures where the Stress-11and Stress-22 values are plotted as color bands.
-4.941e+06
-4.447e+06
-3.953e+06
-3.459e+06
-2.964e+06
-2.470e+06
-1.976e+06
-1.482e+06
-9.874e+05
-4.932e+05
1.067e+03
job1
Comp 11 of Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
-4.941e+06
-4.447e+06
-3.953e+06
-3.459e+06
-2.964e+06
-2.470e+06
-1.976e+06
-1.482e+06
-9.874e+05
-4.932e+05
1.067e+03
job1
Comp 22 of Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
Fig. 1.6 : Stress plots with different element orientation in two rows in the width of the strip
So the question arises how these stresses are calculated.
Stress calculation in a shell element
In the figure below a shell element is shown with its local isoparametric coordinate system{ξ1, ξ2}. In the origin of this coordinate system the tangent vectors are calculated, whichconstitute a tangent plane.
~t1 =∂~x
∂ξ1
/
∣
∣
∣
∣
∂~x
∂ξ1
∣
∣
∣
∣
; ~t2 =∂~x
∂ξ2
/
∣
∣
∣
∣
∂~x
∂ξ2
∣
∣
∣
∣
~ey~ex
~x
~t1
~t2
ξ2
ξ1
~ez
Fig. 1.7 : Tangent vectors at local coordinate axes
New vectors are calculated from ~t1 and ~t2. They are orthogonal and normalised to have unitlength.
~s1 =(
~t1 + ~t2)
/√
2 ~s2 =(
~t1 − ~t2)
/√
2
~ey~ex
~x
~t1
~t2~s1
~s2
ξ2
ξ1
~ez
Fig. 1.8 : Ortonormal vectors in tangent plane
The final local coordinate system is defined by the orthonormal base vectors {~v1, ~v2, ~v3}.
~v1 = (~s1 + ~s2) /√
2 ; ~v2 = (~s1 − ~s2) /√
2 ; ~v3 = ~v1 ∗ ~v2
~ey~ex
~x ~s2
~v3
~v1
~v2 ~s1
ξ2
ξ1
~ez
Fig. 1.9 : Orthonormal local element vector base
Tensor components are defined in this local coordinate system, e.g. for the stresses :
σ = σij~vi~vj
Obviously the component values depend on the orientation of the element, i.e. the directionof the local coordinate system {ξ1, ξ2}, i.e. the local numbering of the nodes.
Due to the calculation in the local coordinate system, it may become difficult to interpretthe stress values. However, it is possible to select in JOB RESULTS from the (AVAILABLE ELEMENT
TENSORS) the Global Stress. These represent the stress components in the global coordinatesystem {x, y, z}. The figure shows the result.
-4.941e+06
-4.692e+06
-4.442e+06
-4.192e+06
-3.942e+06
-3.692e+06
-3.442e+06
-3.192e+06
-2.943e+06
-2.693e+06
-2.443e+06
job1
Comp 11 of Global Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
Fig. 1.10 : Global stress plot
1.3 Example: bending of a square isotropic plate
A square isotropic plate is loaded with edge loads and the deformation and resulting stresseswill be calculated. The plate has dimensions 1 × 1 meter and a thickness of 4 millimeter.
x
y
z1 [m]
1 [m]
Fig. 1.11 : Isotropic plate
We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] × 1 [m] in thexy-plane. This element is subdivided – use SUBDIVIDE – in 20 × 20 elements of uniformdimensions, so without a BIAS. Use the options SWEEP ALL and CHECK UPSIDE DOWN to completethe mesh. The thickness is specified in GEOMETRIC PROPERTIES.
The material parameters are
E = 150 [GPa] ; ν = 0.3
These parameters have to be specified in the MATREIAL PROPERTIES submenu. When we enterthis menu we have to select MATERIAL PROPERTIES. Moduli have to be given in [Pa = N/m2]as the dimensions of the plate are 1 × 1 [m].
MATERIAL PROPERTIES
NEW
(type) STANDARD
STRUCTURAL
ELASTO-PLASTIC ISOTROPIC
YOUNG’S MODULUS 150e9
POISSON’S RATIO 0.3
OK
(ELEMENTS) ADD
(ALL) EXISTING
The plate has to be fixed and this is done by suppressing all degrees of freedom of the centralnode. Boundary loads are defined on left- and right edge of the plate as is indicated in thefigure.
Fx = 100 [N/m] and My = 100 [Nm/m]
z x
y
My
Fx
My
Fx
Fig. 1.12 : Isotropic plate with edge loads
Forces and moments are applied as point loads so we have to calculate the nodal forces andmoments from the number of edge nodes. The corner nodes must be loaded with half theload values of the non-corner nodes.
First we fix the central node.
(MAIN MENU) (PREPROCESSING)
BOUNDARY CONDITIONS
ANALYSIS CLASS STRUCTURAL
NEW
TYPE FIXED DISPLACEMENT
PROPERTIES
DISPLACEMENT X 0
DISPLACEMENT Y 0
DISPLACEMENT Z 0
ROTATION X 0
ROTATION Y 0
ROTATION Z 0
OK
(NODE) ADD Select central node
END LIST
The edge forces are applied in the nodes.
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X 100/20
OK
(NODE) ADD Select central edge nodes right
END LIST
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X 50/2
OK
(NODE) ADD Select corner edge nodes right
END LIST
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X -100/20
OK
(NODE) ADD Select central edge nodes left
END LIST
NEW
TYPE POINT LOAD
PROPERTIES
FORCE X -50/2
OK
(NODE) ADD Select corner edge nodes left
END LIST
The edge moments are applied in the same way, where we have to be aware of the positivedirection of the edge moments. So we have to apply in total -100 Nm on the right edge and100 Nm on the left edge.
The JOB can be defined in the usual way and the plate can be analyzed.
In the RESULTS menu, we can observe the deformation and look at the values of strainsand stresses in the four layers. Some results are listed in the next table.
curvature κx -1.250e-01curvature κy 3.750e-02membrane strain εx 1.667e-07membrane strain εy -5.000e-08
strain top layer εxx -2.498e-04strain top layer εyy 7.495e-05stress top layer σxx -3.748e+07 Pastress top layer σyy ± 0 Pastress mid layer σxx 2.500e+04 Pastress mid layer σyy ± 0 Pa
In Job RESULTS we can also select the Shell Curvatures, Shell Membrane Strains, Shell Mem-brane Forces and Shell Moments. The Forces and Moments are per unit of thickness.
1.4 Example: bending of a circular isotropic plate
The bending of circular plates is described in the document Linear plate bending. Theoreticalsolutions for deformation and section forces and moments are presented for certain loadingconditions.
The same examples are presented here – only the results are shown – as they are modelledand analyzed with MSC.Marc/Mentat. Using correct symmetry conditions, only a quarterof the plate is modelled and analyzed. Four-noded shell elements type 75 are used, 30 incircumferential direction and 20 in radial direction, without a Bias.
1.4.1 Plate with central hole
A plate with a central hole is loaded with a uniform force per unit of area q [N/m2]. In Mentatthis is a Global Load. Values of geometric and material parameters are listed in the table.
z
r
R
2R
q
inner radius R 0.1 mouter radius 2R 0.2 mthickness h 0.01 mYoung’s modulus E 100 GPaPoisson’s ratio ν 0.3 -global load q -100 Pa
The plots show the z-displacement, the rotation and the two bending moments as a functionof the radial distance r. Note that the Moments are per unit of thickness.
0 0.05 0.1 0.15 0.2−1
−0.8
−0.6
−0.4
−0.2
0x 10
−7
r
w
0 0.05 0.1 0.15 0.20
0.2
0.4
0.6
0.8
1
1.2
1.4x 10
−6
r
dw/d
r
0 0.05 0.1 0.15 0.2−15
−10
−5
0
5
10
r
Mt
0 0.05 0.1 0.15 0.2−5
0
5
10
15
20
25
30
35
r
Mr
Fig. 1.13 : Calculated values as a function of radial distance.
1.4.2 Solid plate
A solid plate is loaded with a uniform force per unit of area q [N/m2]. In Mentat this is aGlobal Load. Values of geometric and material parameters are listed in the table.
r
q
R
z
inner radius R 0 mouter radius 2R 0.2 mthickness h 0.01 mYoung’s modulus E 100 GPaPoisson’s ratio ν 0.3 -global load q -100 Pa
The plots show the z-displacement, the rotation and the two bending moments as a functionof the radial distance r. Note that the Moments are per unit of thickness.
0 0.05 0.1 0.15 0.2−3
−2.5
−2
−1.5
−1
−0.5
0x 10
−7
r
w
0 0.05 0.1 0.15 0.20
0.5
1
1.5
2
2.5x 10
−6
r
dw/d
r0 0.05 0.1 0.15 0.2
−35
−30
−25
−20
−15
−10
−5
0
5
10
15
r
Mt
0 0.05 0.1 0.15 0.2−40
−30
−20
−10
0
10
20
30
40
50
r
Mr
Fig. 1.14 : Calculated values as a function of radial distance.
1.5 Orthotropic and transversaly isotropic material
Linear elastic orthotropic material behaviour is described by the next compliance and stiffnessmatrices :
ε11
ε22
ε33
γ12
γ23
γ31
=
E−11 −ν21E
−12 −ν31E
−13 0 0 0
−ν12E−11 E−1
2 −ν32E−13 0 0 0
−ν13E−11 −ν23E
−12 E−1
3 0 0 0
0 0 0 G−112 0 0
0 0 0 0 G−123 0
0 0 0 0 0 G−131
σ11
σ22
σ33
σ12
σ23
σ31
withν12
E1=
ν21
E2;
ν23
E2=
ν32
E3;
ν31
E3=
ν13
E1(Maxwell relations)
σ11
σ22
σ33
σ12
σ23
σ31
=1
∆s
1−ν32ν23
E2E3
ν31ν23+ν21
E2E3
ν21ν32+ν31
E2E30 0 0
ν13ν32+ν12
E1E3
1−ν31ν13
E1E3
ν12ν31+ν32
E1E30 0 0
ν12ν23+ν13
E1E2
ν21ν13+ν23
E1E2
1−ν12ν21
E1E20 0 0
0 0 0 ∆sG12 0 00 0 0 0 ∆sG23 00 0 0 0 0 ∆sG31
ε11
ε22
ε33
γ12
γ23
γ31
with ∆s =1 − ν12ν21 − ν23ν32 − ν31ν13 − ν12ν23ν31 − ν21ν32ν13
E1E2E3
In MSC.Marc/Mentat the definition of the material constants is the same, with E11 = E1,E22 = E2 and E33 = E3. The orthonormal 1, 2 and 3-directions are specified in the ORI-ENTATION option as the so-called material coordinate directions. In the input table theparameters ν21, ν32 and ν13 don’t have to be provided, as they can be calculated according
to the Maxwell relations.Transversely orthotropic materials have one isotropic plane and a direction perpendicular
to it with different elastic properties. In this subsection it is assumed that the 12-plane isthe isotropic plane. Variables in this plane are indicated with the index p. Compliance andstiffness matrices can then be written as follows.
ε11
ε22
ε33
γ12
γ23
γ31
=
E−1p −νpE
−1p −ν3pE
−13 0 0 0
−νpE−1p E−1
p −ν3pE−13 0 0 0
−νp3E−1p −νp3E
−1p E−1
3 0 0 0
0 0 0 G−1p 0 0
0 0 0 0 G−1p3 0
0 0 0 0 0 G−13p
σ11
σ22
σ33
σ12
σ23
σ31
withνp3
Ep
=ν3p
E3
σ11
σ22
σ33
σ12
σ23
σ31
=1
∆s
1−ν3pνp3
EpE3
ν3pνp3+νp
EpE3
νpν3p+ν3p
EpE30 0 0
νp3ν3p+νp
EpE3
1−ν3pνp3
EpE3
νpν3p+ν3p
EpE30 0 0
νpνp3+νp3
EpEp
νpνp3+νp3
EpEp
1−νpνp
EpEp
0 0 0
0 0 0 ∆sGp 0 00 0 0 0 ∆sGp3 00 0 0 0 0 ∆sG3p
ε11
ε22
ε33
γ12
γ23
γ31
with ∆s =1 − νpνp − νp3ν3p − ν3pνp3 − νpνp3ν3p − νpν3pνp3
EpEpE3
Because composites with fibers in one direction perpendicular to the isotropic plane havetransversely isotropic properties, the fiber direction is often indicated as longitudinal (indexl) and the isotropic plane directions as transversal (index t).
1.5.1 Orthotropic plate
In the plate shown below, the longitudinal material direction is indicated as the 1-direction.In MSC.Marc/Mentat it is defined in the ORIENTATION option. The 3-direction is per-pendicular to the plate and the 2-direction is in the plate perpendicular to the 1-direction.Properties in the transverse 2- and 3-directions are the same. So in this plate the 23-plane isthe isotropic plane.
1 = ℓ
2 = t
x
y
z = 3 = t
α
Fig. 1.15 : Material coordinate system in an orthotropic plate
The table gives the relation between the parameters as they are used in MSC.Marc/Mentat forplates and shells. Parameters which have to be given as input are indicated in red. Parameterswhich have to be known for the material involved are indicated in blue. Some input valuescan be calculated.
MSC
E1 E11 El
E2 E22 Et
E3 E33 Et
G12 G12 Glt
G23 G23Et
2(1 + νtt)
G31 G31 Glt
ν12 ν12 νlt
ν23 ν23 νtt
ν31 ν31E3
E1ν13 =
Et
El
νlt
ν21 ν21
ν32 ν32
ν13 ν13
So we have to known the following parameters
El , Et , Glt , νlt and νtt
and calculate other input parameters as indicated in the table.
1.6 Example: bending of a square orthotropic plate
A square orthotropic plate is loaded with edge loads and the deformation and resulting stresseswill be calculated. The plate has dimensions 1 × 1 meter and a thickness of 4 millimater.The angle of the 1-direction is 30o w.r.t. the global x-direction.
1 = ℓ
2 = t
x
y
z = 3 = t1 [m]
1 [m]
30o
Fig. 1.16 : Orthotropic plate
We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] × 1 [m] in thexy-plane. This element is subdivided – use SUBDIVIDE – in 20 × 20 elements of uniformdimensions, so without a BIAS. Use the options SWEEP ALL and CHECK UPSIDE DOWN to completethe mesh. The thickness is specified in GEOMETRIC PROPERTIES.
The material parameters are given below.
α E1 E2 E3 ν12 ν23 ν31 G12 G23 G31
o GPa GPa GPa - - - GPa GPa GPa
30 150 30 30 0.3 0.3 0.330
15010
30
2 ∗ (1 + 0.3)10
First we define the material coordinate system in the option ORIENTATION which can be foundas a submenu of MATERIAL PROPERTIES.
(MAIN MENU) (PREPROCESSING)
MATERIAL PROPERTIES
ORIENTATION
NEW
EDGE12
ANGLE 30
(ELEMENTS) ADD
(ALL) EXIST.
RETURN
The orthotropic material parameters have to be defined in the MATREIAL PROPERTIES sub-menu. When we enter this menu we have to select MATERIAL PROPERTIES. Moduli have to begiven in [Pa = N/m2] as the dimensions of the plate are 1 × 1 [m].
MATERIAL PROPERTIES
NEW
(type) STANDARD
STRUCTURAL
ELASTO-PLASTIC ORTHOTROPIC
E1 150e9
E2 30e9
E3 30e9
N12 0.3
N23 0.3
N31 0.3*(30e9/150e9)
G12 10e9
G23 30e9/(2*(1+0.3))
G31 10e9
OK
(ELEMENTS) ADD
(ALL) EXISTING
The plate has to be fixed and this is done by suppressing all degrees of freedom of the centralnode. Boundary loads are defined on left- and right edge of the plate as is indicated in thefigure.
Fx = 100 [N/m] and My = 100 [Nm/m]
z x
y
My
Fx
My
Fx
Fig. 1.17 : Orthotropic plate with edge loads
They are applied as point loads so we have to calculate the nodal forces and moments fromthe number of edge nodes. The corner nodes must be loaded with half the load values of thenon-corner nodes. This is done in the same way as in the example of the isotropic rectangularplate. Again : be aware of the positive direction of the edge moments. So we have to applyin total -100 Nm on the right edge and 100 Nm on the left edge.
The JOB can be defined in the usual way. Apart from the Stress and Total Strain, we also selectin JOB RESULTS the Stress in Preferred Sys(tem) and Elastic Strain in Preferred Sys(tem). Thesewill give us the stresses and strains in the material directions, as defined in ORIENTATION.
In the RESULTS menu, we can observe the deformation and look at the values of strainsand stresses in the four layers. The figure shows the deformation × 3 and the Stress-11 inthe top layer.
-3.748e+07
-3.744e+07
-3.740e+07
-3.736e+07
-3.733e+07
-3.729e+07
-3.725e+07
-3.721e+07
-3.718e+07
-3.714e+07
-3.710e+07
job1
Comp 11 of Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
Fig. 1.18 : Stress-11 in top layer of square plate
Some results are listed in the next table.
curvature κxx -4.469e-01curvature κyy 2.344e-01curvature κxy 4.438e-01membrane strain εxx 5.985e-07membrane strain εyy -3.125e-07membrane strain εxy -5.918e-07
strain top layer εxx -8.932e-04strain top layer εyy 4.684e-04strain top layer εxy 8.871e-04stress top layer σxx -3.748e+07 Pastress top layer σyy ± 0 Pastress top layer σxy ± 0 Pastress mid layer σxx 2.500e+04 Pastress mid layer σyy ± 0 Pastress mid layer σxy ± 0 Pa
strain top layer ε11 -1.686e-04strain top layer σ22 -2.561e-04strain top layer σ12 1.623e-03stress top layer σ11 -2.811e+07 Pastress top layer σ22 -9.369e+06 Pastress top layer σ12 1.623e+07 Pastress mid layer σ11 1.875e+04 Pastress mid layer σ22 6.250e+03 Pastress mid layer σ12 -1.083e+04 Pa
2 Laminates
Laminate theory is explained in the document Laminates. Some examples are shown, wheresquare laminate plates are build with a Matlab program, which subsequently calculates thedeformation and ply-strains and -stresses for a given edge load.
The first of the above-mentioned examples, a 4-ply laminate, is modelled and analyzed inMSC.Marc/Mentat.
2.1 Example: 4-ply laminate plate
A square laminated plate is build from four plies. It is loaded with edge loads and the defor-mation and resulting stresses will be calculated. The plate has dimensions 1 × 1 meter anda thickness of 4 millimater.
We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] × 1 [m] inthe xy-plane. This element is subdivided – use SUBDIVIDE – in 20 × 20 elements of uniformdimensions, so without a BIAS. Use the options SWEEP ALL and CHECK UPSIDE DOWN to completethe mesh.
There is no need to define GEOMETRIC PROPERTIES, because the thickness will be specifiedin the laminate definition.
The laminate has 4 laminas or plys, which have different properties. Each ply is orthotropicand all the used orthotropic materials have to be defined in the MATREIAL PROPERTIES sub-menu. When we enter this menu we have to select MATERIAL PROPERTIES. In our example wehave four ORTHOTROPIC materials for which we have to specify the parameters. Moduli haveto be given in [Pa = N/m2] as the dimensions of the plate are 1 × 1 [m].
First we define the material coordinate system in the option ORIENTATION which can befound as a submenu of MATERIAL PROPERTIES.
(MAIN MENU) (PREPROCESSING)
MATERIAL PROPERTIES
ORIENTATION
NEW
EDGE12
ANGLE 0
(ELEMENTS) ADD
(ALL) EXIST.
RETURN
We are back in MATERIAL PROPERTIES and select again MATERIAL PROPERTIES. The ANALYSIS
CLASS is STRUCTURAL. The first material is defined.
MATERIAL PROPERTIES
NEW
(type) STANDARD
STRUCTURAL
ELASTO-PLASTIC ORTHOTROPIC
E1 150e9
E2 30e9
E3 30e9
N12 0.3
N23 0.3
N31 0.3*(30e9/150e9)
G12 10e9
G23 30e9/(2*(1+0.3))
G31 10e9
OK
This is repeated for three NEW materials, with properties:
E1=100e9 ; E2=25e9 ; N12=0.2 ; N23=0.2 ; G12=20e9
E1=110e9 ; E2=21e9 ; N12=0.3 ; N23=0.3 ; G12=15e9
E1=90e9 ; E2=17e9 ; N12=0.2 ; N23=0.2 ; G12=10e9
Now that we have defined the orthotropc materials, we can define the laminate stacking. Thisis done in the submenu COMPOSITE which can be chosen as another NEW material. Here we se-lect the option which says that the thickness of each ply is an absolute value (in [m] of course).
NEW
(type) COMPOSITE
(DATA CATAGORIES) GENERAL
ABSOLUTE THICKNESS Toggle !
APPEND
material 1
THICKNESS 0.001
ANGLE 90
APPEND
material 2
THICKNESS 0.001
ANGLE 45
APPEND
material 3
THICKNESS 0.001
ANGLE 0
APPEND
material 4
THICKNESS 0.001
ANGLE 30
OK
(ELEMENTS) ADD
(ALL) EXIST.
RETURN
The plate has to be fixed and this is done by suppressing all degrees of freedom of the centralnode. Boundary loads are defined on left- and right edge of the plate as is indicated in thefigure.
z x
y
My
Fx
My
Fx
Fig. 2.19 : Orthotropic plate with edge loads
Fx = 100 [N/m] and My = 100 [Nm/m]
They are applied as point loads in the same way as in the earlier example of an isotropicsquare plate.
The JOB can be defined in the usual way. Apart from the Stress and Total Strain, we alsoselect in JOB RESULTS the Stress in Preferred Sys(tem) and Elastic Strain in Preferred Sys(tem).These will give us the stresses and strains in the material directions, as defined in ORIENTATION.In the RESULTS menu, we can observe the deformation and look at the values of strains andstresses in the four layers. The figure shows the deformation × 3 and the Stress-11 in the toplayer.
-2.429e+07
-2.426e+07
-2.424e+07
-2.421e+07
-2.419e+07
-2.417e+07
-2.414e+07
-2.412e+07
-2.409e+07
-2.407e+07
-2.404e+07
job1
Comp 11 of Stress Layer 1
Inc: 0Time: 0.000e+00
X Y
Z
4
Fig. 2.20 : Stress-11 in top layer of square laminate
Some results are listed in the next table.
curvature κxx -4.779e-01curvature κyy 6.885e-02curvature κxy 2.635e-01membrane strain εxx -9.737e-05membrane strain εyy -3.892e-05membrane strain εxy -4.646e-05
strain top ply εxx -8.143e-04strain top ply εyy 6.435e-05strain top ply εxy 3.488e-04stress top ply σxx -2.429e+07 Pastress top ply σyy 2.366e+06 Pastress top ply σxy 3.488e+06 Pa
strain top ply ε11 6.435e-05strain top ply σ22 -8.143e-04strain top ply σ12 -3.488e-04stress top ply σ11 2.366e+06 Pastress top ply σ22 -2.429e+07 Pastress top ply σ12 -3.488e+06 Pa
If you compare them with values from the report Laminates, you will notice that results differ.This is because the stacking of the laminate in the report is not symmetric with respect tothe plane z = 0. When it is made symmetric w.r.t. this plane, the results are the same asthose from MSC.Marc/Mentat, listed in the table.
2.2 Example: laminated tube
A cylindrical tube with circular cross-section is fixed on one side and loaded at the free end,as is shown in the figure below. The length of the tube is L and the diameter is D.
x
y
zF
W
L
D
y
z~ex
~ey
Fig. 2.21 : Tube with circular cross section
The wall is made of a composite material, which consists of a foam embedded between or-thotropic 2-ply laminates. The foam is considered to be an isotropic ply. As in the elementsthe local perpendicular coordinate axis ~v3 points in the outward direction plys 1 and 2 con-stitute the outer wall laminas, ply 3 is the foam and plys 4 and 5 the inner wall laminas.
Geometric and material parameters are listed in the next table. The fiber angle in the plysis given with respect to the axial z-direction, which is defined in the ORIENTATION option.
Length L 10 mDiameter D 2 mWall thickness w 40 mm
foam density ρf 200 kg/m3
foam Young’s modulus Ef 10 MPafoam Poisson’s ratio νf 0.3 -foam thickness wf 32 mm
ply density ρp 500 kg/m3
ply longitudinal modulus Eℓ 150 GPaply transverse modulus Et 30 GPaply Poisson’s ratio νℓt 0.3 -ply Poisson’s ratio νtt 0.3 -ply shear modulus Gℓt 10 GPaply 1,5 thickness wp15 2 mmply 1,5 angle αp15 120 degply 2,4 thickness wp24 2 mmply 2,4 angle αp24 60 deg
The geometry is defined by two CURVES of type CIRCLE. Between these curves a SURFACEis defined, being the tube wall. This surface is CONVERTED to equally sized QUAD4elements: 50 in longitudinal and 72 in circumferential direction. The tube is loaded by itsweight and by a lateral force ~F = −10~ey at the end x = L.
