Plates, shells and laminates in Mentat & MARC. Tutorial

Transcription

1 Plates, shells and laminates in Mentat & MARC Tutorial Eindhoven University of Technology Department of Mechanical Engineering Piet Schreurs September 30, 2015

2 Contents 1 Linear plate bending Deformation and loads Example: bending of a strip Strains and stresses in a shell element Example: bending of a square isotropic plate Example: bending of a circular isotropic plate Plate with central hole Solid plate Orthotropic and transversaly isotropic material Orthotropic plate Example: bending of a square orthotropic plate Laminates Example: 4-ply laminate plate Example: laminated tube

3 1 Linear plate bending Linear plate bending theory is explained in the document Linear plate bending. In Cartesian and cylindrical coordinate systems, plate bending equations are derived. For the circular plates analytical solutions are elaborated. In the following sections we consider first the definition of deformation and load variables in 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 material parameters in MSC.Marc/Mentat. 1.1 Deformation and loads In the document Linear plate bending the next sign conventions have been used for membrane strains and curvatures. z y ε xx0 ε xy0 ε xy0 ε yy0 x κ xx κ yy κ xy κ xy Fig. 1.1 : Strain and curvature definitions In accordance with this, the next sign conventions have been chosen for applied cross-sectional forces and moments. z y N xx N xy N xy D y N yy x D x M yy M xy M xx M xy 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] for moments. In MSC.Marc/Mentat the sign for applied nodal forces are equivalent. Applied nodal

4 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 strains and curvatures. z y N x N y N x N y x M x M y M y M x 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 loads per unit of length have to be prescribed the nodal forces and moments must be calculated such 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, clamped at one end (x = 0) and loaded with forces and moments at the free end (x = L). z L H W y F z x F x M y 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 displacements u x and u z and the end rotation φ y and also for the axial strain ε xx and the axial stress σ xx. With L the lengh of the beam, W its width, H its heigth and E the Young s modulus of the material, these values are : u x = L EWH F x u z = L3 3EI F z + L2 2EI M y ; φ y = L2 2EI F z + L EI M y σ xx,max = 1 WH F x + 1 I H 2 (M y + F z L) ; ε xx,max = σ xx,max E

5 where I = 1 12 WH3 is the area moment of inertia. The strip is modelled subsequently with beam elements (type 52), with shell elements (type 75) and with three-dimensional elements (type 7). In axial direction the number of elements is always 20. Using shell or 3D elements, the width is subdivided in 2 elements. Using 3D elements, 5 elements in thickness (= height) direction are used. Modelling with beam and 3D elements is described in other tutorials and will not be repeated here. The use of shell elements is described in detail below. Model parameter values are given in the next table. length L 1 m width W 0.1 m heigth H 0.05 m Young s modulus E 150 GPa Poisson s ratio ν axial end force F x 100 N lateral end force F z 100 N bending end moment M y 100 Nm Modelling the strip with shell elements starts by making a square 4-noded element (QUAD4) of dimensions 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/m 2 ] as the dimensions of the plate is modelled in meter. MATERIAL PROPERTIES (type) STANDARD STRUCTURAL ELASTO-PLASTIC ISOTROPIC YOUNG S MODULUS 150e9 POISSON S RATIO 0.3 (ELEMENTS) ADD (ALL) EXISTING The plate has to be fixed at one end. This is done by suppressing all degrees of freedom of the mid-edge node. The other two edge nodes are free to move in y-direction, which means that the lateral contraction is free. (MAIN MENU) (PREPROCESSING) BOUNDARY CONDITIONS ANALYSIS CLASS STRUCTURAL TYPE FIXED DISPLACEMENT PROPERTIES

6 DISPLACEMENT X 0 DISPLACEMENT Y 0 DISPLACEMENT Z 0 ROTATION X 0 ROTATION Y 0 ROTATION Z 0 (NODE) ADD Select central edge node END LIST TYPE FIXED DISPLACEMENT PROPERTIES DISPLACEMENT X 0 DISPLACEMENT Z 0 ROTATION X 0 ROTATION Y 0 ROTATION Z 0 (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 corner nodes are divided by 2. TYPE POINT LOAD PROPERTIES FORCE X 100/2 FORCE Z 100/2 (NODE) ADD Select central edge node END LIST TYPE POINT LOAD PROPERTIES FORCE X 50/2 FORCE Z 50/2 (NODE) ADD Select corner edge nodes END LIST The edge moments are applied in the nodes at x = 1 meter. Note that the moment about the y-axis is -100 Nm which is a bending moment of 100 Nm as indicated in the figure and used in the formulas.

7 TYPE POINT LOAD PROPERTIES MOMENT Y -100/2 (NODE) ADD Select central edge node END LIST TYPE POINT LOAD PROPERTIES MOMENT Y -50/2 (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 to the calculation of the stresses in more detail, but for now this will do. In the default setting the plate has 5 layers over the thicckness, which are used for numerical integration, as is explained in the document Linear plate bending. In JOB>PROPERTIES>JOB PARAMETERS the number of layers can be changed, before doing the analysis. Because we always want to have a layer in the mid-plane of the shell, this number has to be odd. In these layers the strains and 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 to select (LAYERS) ALL. After saving the input, the analysis can be done through RUN. The result file can be opened after a successfull analysis (3004). Displacements, rotations, reaction forces and moments and 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. Inc: 0 Time: 0.000e+00 Inc: 0 Time: 0.000e e e e e e e e e e e e e e e e e e e e e e+02 Z e+06 Z job1 X Beam Bending Moment Local Y Y 4 job1 Comp 11 of Stress Layer 1 X Y 4

8 Inc: 0 Time: 0.000e e e e e e e e e e e e+06 Z job1 Comp 11 of Stress X Y 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 u x µm lateral end displacement u z mm end rotation φ y deg maximum axial stress σ MPa The analysis with the 3D solid elements is not very accurate. The results will improve when more elements are used in thickness direction, as in that case the linear stress over the thickness 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. It is also rather cumbersome to prescribe the bending moment at the end of the beam, because this has to be done with nodal forces. Let us now take a closer look at the stresses in the shell element Strains and stresses in a shell element In the example of the strip we have selected as output variables Stress and Total Strain. Their values 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 (ELEMENTS) ADD

9 Both elements are subdevided such that we end up with 20 elements along the length of the strip and 2 elements in width direction, just as we had in the initial example. The only difference is that elements in the two rows have a different orientation of the edge between the local nodes 1 and 2 (= EDGE12). The model can be completed as before by adding material properties 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 the elements in the two rows along the length. This shown in the next figures where the Stress-11 and Stress-22 values are plotted as color bands. Inc: 0 Time: 0.000e+00 Inc: 0 Time: 0.000e e e e e e e e e e e e e e e e e e e e e e+06 Z e+06 Z job1 Comp 11 of Stress Layer 1 X Y 4 job1 Comp 22 of Stress Layer 1 X Y 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 {ξ 1,ξ 2 }. In the origin of this coordinate system the tangent vectors are calculated, which constitute a tangent plane. t 1 = x / x ξ 1 ξ 1 ; t 2 = x / x ξ 2 ξ 2 ξ 2 t 2 t 1 x ξ 1 e z e x e y Fig. 1.7 : Tangent vectors at local coordinate axes

10 New vectors are calculated from t 1 and t 2. They are orthogonal and normalised to have unit length. s 1 = ( t 1 + t 2 ) / 2 s2 = ( t 1 t 2 ) / 2 ξ 2 t s 1 2 t 1 x s 2 ξ 1 e z e y e x Fig. 1.8 : Ortonormal vectors in tangent plane The final local coordinate system is defined by the orthonormal base vectors { v 1, v 2, v 3 }. v 1 = ( s 1 + s 2 )/ 2 ; v 2 = ( s 1 s 2 )/ 2 ; v 3 = v 1 v 2 ξ 2 v 3 v 2 s1 x v 1 s 2 ξ 1 e z e y e x Fig. 1.9 : Orthonormal local element vector base Tensor components are defined in this local coordinate system, e.g. for the stresses : σ = σ ij v i v j Obviously the component values depend on the orientation of the element, i.e. the direction of the local coordinate system {ξ 1,ξ 2 }, i.e. the local numbering of the nodes. Due to the calculation in the local coordinate system, it may become difficult to interpret the 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 coordinate system {x,y,z}. The figure shows the result.

11 Inc: 0 Time: 0.000e e e e e e e e e e e e+06 Z job1 X Comp 11 of Global Stress Layer 1 Y 4 Fig : 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 stresses will be calculated. The plate has dimensions 1 1 meter and a thickness of 4 millimeter. y 1 [m] z x 1 [m] Fig : Isotropic plate We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] 1 [m] in the xy-plane. This element is subdivided use SUBDIVIDE in 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. The material parameters are E = 150 [GPa] ; ν = 0.3 These 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/m 2 ] as the dimensions of the plate are 1 1 [m].

12 MATERIAL PROPERTIES (type) STANDARD STRUCTURAL ELASTO-PLASTIC ISOTROPIC YOUNG S MODULUS 150e9 POISSON S RATIO 0.3 (ELEMENTS) ADD (ALL) EXISTING The plate has to be fixed and this is done by suppressing all degrees of freedom of the central node. Boundary loads are defined on left- and right edge of the plate as is indicated in the figure. F x = 100 [N/m] and M y = 100 [Nm/m] y M y F x z F x M y x Fig : Isotropic plate with edge loads Forces and moments are applied as point loads so we have to calculate the nodal forces and moments from the number of edge nodes. The corner nodes must be loaded with half the load values of the non-corner nodes. First we fix the central node. (MAIN MENU) (PREPROCESSING) BOUNDARY CONDITIONS ANALYSIS CLASS STRUCTURAL TYPE FIXED DISPLACEMENT PROPERTIES DISPLACEMENT X 0 DISPLACEMENT Y 0 DISPLACEMENT Z 0

13 ROTATION X 0 ROTATION Y 0 ROTATION Z 0 (NODE) ADD Select central node END LIST The edge forces are applied in the nodes. TYPE POINT LOAD PROPERTIES FORCE X 100/20 (NODE) ADD Select central edge nodes right END LIST TYPE POINT LOAD PROPERTIES FORCE X 50/2 (NODE) ADD Select corner edge nodes right END LIST TYPE POINT LOAD PROPERTIES FORCE X -100/20 (NODE) ADD Select central edge nodes left END LIST TYPE POINT LOAD PROPERTIES FORCE X -50/2 (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 positive direction of the edge moments. So we have to apply in total -100 Nm on the right edge and 100 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 strains and stresses in the four layers. Some results are listed in the next table.

14 curvature κ x e-01 curvature κ y 3.750e-02 membrane strain ε x 1.667e-07 membrane strain ε y e-08 strain top layer ε xx e-04 strain top layer ε yy 7.495e-05 stress top layer σ xx e+07 Pa stress top layer σ yy ± 0 Pa stress mid layer σ xx 2.500e+04 Pa stress mid layer σ yy ± 0 Pa In Job RESULTS we can also select the Shell Curvatures, Shell Membrane Strains, Shell Membrane 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. Theoretical solutions for deformation and section forces and moments are presented for certain loading conditions. The same examples are presented here only the results are shown as they are modelled and analyzed with MSC.Marc/Mentat. Using correct symmetry conditions, only a quarter of the plate is modelled and analyzed. Four-noded shell elements type 75 are used, 30 in circumferential direction and 20 in radial direction, without a Bias Plate with central hole A plate with a central hole is loaded with a uniform force per unit of area q [N/m 2 ]. In Mentat this is a Global Load. Values of geometric and material parameters are listed in the table. q z r R 2R inner radius R 0.1 m outer radius 2R 0.2 m thickness h 0.01 m Young s modulus E 100 GPa Poisson s ratio ν global load q -100 Pa

15 The plots show the z-displacement, the rotation and the two bending moments as a function of the radial distance r. Note that the Moments are per unit of thickness. w x r 10 5 dw/dr 1.4 x r M t M r r r Fig : Calculated values as a function of radial distance Solid plate A solid plate is loaded with a uniform force per unit of area q [N/m 2 ]. In Mentat this is a Global Load. Values of geometric and material parameters are listed in the table. q z r R inner radius R 0 m outer radius 2R 0.2 m thickness h 0.01 m Young s modulus E 100 GPa Poisson s ratio ν global load q -100 Pa The plots show the z-displacement, the rotation and the two bending moments as a function of the radial distance r. Note that the Moments are per unit of thickness.

16 0 x x w dw/dr M t r r M r r r Fig : 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 stiffness matrices : σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 ε 11 ε 22 ε 33 γ 12 γ 23 γ 31 with E1 1 ν 21 E2 1 ν 31 E ν 12 E1 1 E2 1 ν 32 E = ν 13 E1 1 ν 23 E2 1 E G G G 1 31 ν 12 = ν 21 ν 23 ; = ν 32 ν 31 ; = ν 13 E 1 E 2 E 2 E 3 E 3 E 1 = 1 s 1 ν 32 ν 23 E 2 E 3 ν 13 ν 32 +ν 12 ν 31 ν 23 +ν 21 E 2 E 3 1 ν 31 ν 13 ν 21 ν 32 +ν 31 ν 12 ν 31 +ν 32 E 1 E 3 ν 12 ν 23 +ν 13 E 1 E 3 ν 21 ν 13 +ν 23 1 ν 12 ν 21 E 1 E 2 E 1 E 2 σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 (Maxwell relations) E 2 E E 1 E E 1 E s G s G s G 31 with s = 1 ν 12ν 21 ν 23 ν 32 ν 31 ν 13 ν 12 ν 23 ν 31 ν 21 ν 32 ν 13 E 1 E 2 E 3 In MSC.Marc/Mentat the definition of the material constants is the same, with E 11 = E 1, E 22 = E 2 and E 33 = E 3. 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 the parameters ν 21, ν 32 and ν 13 don t have to be provided, as they can be calculated according ε 11 ε 22 ε 33 γ 12 γ 23 γ 31

17 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 is the isotropic plane. Variables in this plane are indicated with the index p. Compliance and stiffness matrices can then be written as follows. σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 ε 11 ε 22 ε 33 γ 12 γ 23 γ 31 with = = 1 s ν p Ep 1 ν 3p E Ep 1 ν 3p E ν p3 Ep 1 E G 1 p G 1 p G 1 3p E 1 p ν p E 1 p ν p3 E 1 p ν p3 = ν 3p E p E 3 1 ν 3p ν p3 E pe 3 ν p3 ν 3p +ν p E pe 3 ν pν p3 +ν p3 E pe p ν 3p ν p3 +ν p E pe 3 1 ν 3p ν p3 E pe 3 ν pν p3 +ν p3 E pe p ν pν 3p +ν 3p E pe ν pν 3p +ν 3p E pe ν pν p E pe p s G p s G p s G 3p σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 ε 11 ε 22 ε 33 γ 12 γ 23 γ 31 with s = 1 ν pν p ν p3 ν 3p ν 3p ν p3 ν p ν p3 ν 3p ν p ν 3p ν p3 E p E p E 3 Because composites with fibers in one direction perpendicular to the isotropic plane have transversely isotropic properties, the fiber direction is often indicated as longitudinal (index l) and the isotropic plane directions as transversal (index t) 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 perpendicular 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 is the isotropic plane. 2 = t y 1 = l α z = 3 = t x

18 Fig : Material coordinate system in an orthotropic plate The table gives the relation between the parameters as they are used in MSC.Marc/Mentat for plates and shells. Parameters which have to be given as input are indicated in red. Parameters which have to be known for the material involved are indicated in blue. Some input values can be calculated. MSC E 1 E 11 E l E 2 E 22 E t E 3 E 33 E t G 12 G 12 G lt G 23 G 23 E t 2(1 + ν tt ) G 31 G 31 G lt ν 12 ν 12 ν lt ν 23 ν 23 ν tt E 3 ν 31 ν 31 ν 13 = E t ν lt E 1 E l ν 21 ν 21 ν 32 ν 32 ν 13 ν 13 So we have to known the following parameters E l, E t, G lt, ν lt and ν tt 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 stresses will be calculated. The plate has dimensions 1 1 meter and a thickness of 4 millimater. The angle of the 1-direction is 30 o w.r.t. the global x-direction. 2 = t y 1 [m] z = 3 = t 30 o 1 = l x 1 [m]

19 Fig : Orthotropic plate We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] 1 [m] in the xy-plane. This element is subdivided use SUBDIVIDE in 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. The material parameters are given below. α E 1 E 2 E 3 ν 12 ν 23 ν 31 G 12 G 23 G 31 o GPa GPa GPa GPa GPa GPa ( ) 10 First we define the material coordinate system in the option ORIENTATION which can be found as a submenu of MATERIAL PROPERTIES. (MAIN MENU) (PREPROCESSING) MATERIAL PROPERTIES ORIENTATION EDGE12 ANGLE 30 (ELEMENTS) ADD (ALL) EXIST. RETURN The orthotropic material parameters have to be defined in the MATREIAL PROPERTIES submenu. When we enter this menu we have to select MATERIAL PROPERTIES. Moduli have to be given in [Pa = N/m 2 ] as the dimensions of the plate are 1 1 [m]. MATERIAL PROPERTIES (type) STANDARD STRUCTURAL ELASTO-PLASTIC ORTHOTROPIC E1 150e9 E2 30e9 E3 30e9 N N N31 0.3*(30e9/150e9) G12 10e9

20 G23 30e9/(2*(1+0.3)) G31 10e9 (ELEMENTS) ADD (ALL) EXISTING The plate has to be fixed and this is done by suppressing all degrees of freedom of the central node. Boundary loads are defined on left- and right edge of the plate as is indicated in the figure. F x = 100 [N/m] and M y = 100 [Nm/m] y M y F x z F x M y x Fig : Orthotropic plate with edge loads They are applied as point loads so we have to calculate the nodal forces and moments from the number of edge nodes. The corner nodes must be loaded with half the load values of the non-corner nodes. This is done in the same way as in the example of the isotropic rectangular plate. Again : be aware of the positive direction of the edge moments. So we have to apply in 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 select 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 and stresses in the four layers. The figure shows the deformation 3 and the Stress-11 in the top layer.

21 Inc: 0 Time: 0.000e e e e e e e e e e e e+07 Z job1 Comp 11 of Stress Layer 1 Fig : Stress-11 in top layer of square plate X Y 4 Some results are listed in the next table. curvature κ xx e-01 curvature κ yy 2.344e-01 curvature κ xy 4.438e-01 membrane strain ε xx 5.985e-07 membrane strain ε yy e-07 membrane strain ε xy e-07 strain top layer ε xx e-04 strain top layer ε yy 4.684e-04 strain top layer ε xy 8.871e-04 stress top layer σ xx e+07 Pa stress top layer σ yy ± 0 Pa stress top layer σ xy ± 0 Pa stress mid layer σ xx 2.500e+04 Pa stress mid layer σ yy ± 0 Pa stress mid layer σ xy ± 0 Pa strain top layer ε e-04 strain top layer σ e-04 strain top layer σ e-03 stress top layer σ e+07 Pa stress top layer σ e+06 Pa stress top layer σ e+07 Pa stress mid layer σ e+04 Pa stress mid layer σ e+03 Pa stress mid layer σ e+04 Pa

22 2 Laminates Laminate theory is explained in the document Laminates. Some examples are shown, where square laminate plates are build with a Matlab program, which subsequently calculates the deformation 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 in MSC.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 deformation and resulting stresses will be calculated. The plate has dimensions 1 1 meter and a thickness of 4 millimater. We start by making a square 4-noded element (QUAD4) of dimensions 1 [m] 1 [m] in the xy-plane. This element is subdivided use SUBDIVIDE in elements of uniform dimensions, so without a BIAS. Use the options SWEEP ALL and CHECK UPSIDE DOWN to complete the mesh. There is no need to define GEOMETRIC PROPERTIES, because the thickness will be specified in the laminate definition. The laminate has 4 laminas or plys, which have different properties. Each ply is orthotropic and all the used orthotropic materials have to be defined in the MATREIAL PROPERTIES submenu. When we enter this menu we have to select MATERIAL PROPERTIES. In our example we have four ORTHOTROPIC materials for which we have to specify the parameters. Moduli have to be given in [Pa = N/m 2 ] as the dimensions of the plate are 1 1 [m]. First we define the material coordinate system in the option ORIENTATION which can be found as a submenu of MATERIAL PROPERTIES. (MAIN MENU) (PREPROCESSING) MATERIAL PROPERTIES ORIENTATION 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 (type) STANDARD STRUCTURAL ELASTO-PLASTIC ORTHOTROPIC E1 150e9 E2 30e9

23 E3 30e9 N N N31 0.3*(30e9/150e9) G12 10e9 G23 30e9/(2*(1+0.3)) G31 10e9 This is repeated for three 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. This is done in the submenu COMPOSITE which can be chosen as another material. Here we select the option which says that the thickness of each ply is an absolute value (in [m] of course). (type) COMPOSITE (DATA CATAGORIES) GENERAL ABSOLUTE THICKNESS Toggle! APPEND material 1 THICKNESS ANGLE 90 APPEND material 2 THICKNESS ANGLE 45 APPEND material 3 THICKNESS ANGLE 0 APPEND material 4 THICKNESS ANGLE 30 (ELEMENTS) ADD (ALL) EXIST. RETURN The plate has to be fixed and this is done by suppressing all degrees of freedom of the central node. Boundary loads are defined on left- and right edge of the plate as is indicated in the figure.

24 y M y F x F x z x M y Fig : Orthotropic plate with edge loads F x = 100 [N/m] and M y = 100 [Nm/m] They are applied as point loads in the same way as in the earlier example of an isotropic square plate. The JOB can be defined in the usual way. Apart from the Stress and Total Strain, we also select 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 and stresses in the four layers. The figure shows the deformation 3 and the Stress-11 in the top layer. Inc: 0 Time: 0.000e e e e e e e e e e e e+07 Z job1 Comp 11 of Stress Layer 1 X Y 4 Fig : Stress-11 in top layer of square laminate

25 Some results are listed in the next table. curvature κ xx e-01 curvature κ yy 6.885e-02 curvature κ xy 2.635e-01 membrane strain ε xx e-05 membrane strain ε yy e-05 membrane strain ε xy e-05 strain top ply ε xx e-04 strain top ply ε yy 6.435e-05 strain top ply ε xy 3.488e-04 stress top ply σ xx e+07 Pa stress top ply σ yy 2.366e+06 Pa stress top ply σ xy 3.488e+06 Pa strain top ply ε e-05 strain top ply σ e-04 strain top ply σ e-04 stress top ply σ e+06 Pa stress top ply σ e+07 Pa stress top ply σ e+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 to the plane z = 0. When it is made symmetric w.r.t. this plane, the results are the same as those 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. y L y e y e x D F z z x W Fig : Tube with circular cross section The wall is made of a composite material, which consists of a foam embedded between orthotropic 2-ply laminates. The foam is considered to be an isotropic ply. As in the elements the local perpendicular coordinate axis v 3 points in the outward direction plys 1 and 2 constitute the outer wall laminas, ply 3 is the foam and plys 4 and 5 the inner wall laminas.

26 Geometric and material parameters are listed in the next table. The fiber angle in the plys is given with respect to the axial z-direction, which is defined in the ORIENTATION option. Length L 10 m Diameter D 2 m Wall thickness w 40 mm foam density ρ f 200 kg/m 3 foam Young s modulus E f 10 MPa foam Poisson s ratio ν f foam thickness w f 32 mm ply density ρ p 500 kg/m 3 ply longitudinal modulus E l 150 GPa ply transverse modulus E t 30 GPa ply Poisson s ratio ν lt ply Poisson s ratio ν tt ply shear modulus G lt 10 GPa ply 1,5 thickness w p15 2 mm ply 1,5 angle α p deg ply 2,4 thickness w p24 2 mm ply 2,4 angle α p24 60 deg The geometry is defined by two CURVES of type CIRCLE. Between these curves a SURFACE is defined, being the tube wall. This surface is CONVERTED to equally sized QUAD4 elements: 50 in longitudinal and 72 in circumferential direction. The tube is loaded by its weight and by a lateral force F = 10 e y at the end x = L.