UNIVERSITY OF HAWAIIble, triangular, Reissner/Mindlin shear-deformable plate elements, MIN-N, is...

UNIVERSITY OF HAWAIICOLLEGE OF ENGINEERING

D

EPARTMENT OF

C

IVIL AND

E

NVIRONMENTAL

E

NGINEERING

riggs

DEVELOPMENT OF THE MIN-N FAMILY OF TRIANGULAR ANISOPARAMETRIC MINDLIN PLATE ELEMENTS

riggs

Yan Jane Liu and H. R. Riggs

riggs

Research Report UHM/CE/02-02 July 2002

riggs

ii

ACKNOWLEDGMENTS

This report consists of the dissertation by Ms. Yan Jane Liu, submitted in partial ful-

fillment of the requirements for the degree of Doctor of Philosophy in Civil Engineering.

Prof. H.Ronald Riggs was her research advisor. Other committee members were Prof.

Harold S. Hamada, Prof. Craig M. Newtson, Prof. Ian Robertson, and Prof. Ronald. H.

Knapp.

The authors appreciate the financial support for this work from NASA Langley

Research Center under grant NAG-1-1850. Dr. Alexander Tessler was the NASA Techni-

cal Officer. The first author gratefully acknowledges the financial support provided in the

form of graduate teaching assistantship from the Department of Civil Engineering of Uni-

versity of Hawaii at Manoa.

iii

ABSTRACT

A general formulation for a family of N-node, higher-order, displacement-compati-

ble, triangular, Reissner/Mindlin shear-deformable plate elements, MIN-

N,

is presented in

this work. The development of MIN-

N

has been motivated primarily by the success of the

3-node, 9 degree-of-freedom, low-order, constant moment, anisoparametric triangular

plate element, MIN3. This element avoids shear locking by using so-called anisoparamet-

ric interpolation, which is to use interpolation functions one degree higher for transverse

displacement than for bending rotations. The methodology to derive members of the MIN-

N

family based on the anisoparametric strategy is presented. The family of MIN-

N

ele-

ments possesses complete, fully compatible kinematic fields. In the thin limit, the element

must satisfy the Kirchhoff constraints of zero transverse shear strains. General formulas

for these constraints are developed.

As an example of a higher-order member, the 6-node, 18 degree-of-freedom, trian-

gular element MIN6 is developed. MIN6 has a cubic variation of transverse displacement

and quadratic variation of rotational displacements. The element, with its straightforward,

pure, displacement-based formulation, is implemented in a finite element program and

tested extensively. Numerical results for both isotropic and orthotropic materials show that

MIN6 exhibits good performance for both static and dynamic analyses in the linear, elastic

regime. Explicit formulas for Kirchhoff constraints in the thin limit, in terms of element

degrees-of-freedom, are developed. The results illustrate that the fully-integrated MIN6

element neither locks nor is excessively stiff in the thin limit, even for coarse meshes.

iv

TABLE OF CONTENTS

ACKNOWLEDGMENTS .................................................................................................. ii

ABSTRACT....................................................................................................................... iii

TABLE OF CONTENTS................................................................................................... iv

LIST OF TABLES............................................................................................................. vi

LIST OF FIGURES .......................................................................................................... vii

CHAPTER 1: INTRODUCTION ...................................................................................... 11.1 Overview ................................................................................................................11.2 Scope of Work .......................................................................................................3

CHAPTER 2: REVIEW OF PLATE THEORY AND PLATE ELEMENTS ................... 52.1 General Comments ................................................................................................52.2 Plate Theories ........................................................................................................5

2.2.1 Kirchhoff Plate Theory ............................................................................... 72.2.2 Reissner/Mindlin Plate Theory ................................................................... 92.2.3 Potential Energy Functional...................................................................... 112.2.4 Kirchhoff Constraints................................................................................ 12

2.3 Mindlin Plate Element Modeling .........................................................................122.3.1 Finite Element Model and Shear Locking Phenomenon .......................... 122.3.2 Concept of Anisoparametric Interpolations .............................................. 13

2.4 Review of Plate Elements ....................................................................................152.4.1 Overview................................................................................................... 152.4.2 Existing Low-Order Anisoparametric Element, MIN3 ............................ 17

CHAPTER 3: DERIVATION OF THE MIN-N FAMILY.............................................. 233.1 Introduction ..........................................................................................................233.2 Basic Concepts of MIN-N Element Family .........................................................233.3 Virgin Elements ...................................................................................................263.4 Shear Constraints .................................................................................................283.5 General Derivation of MIN-N .............................................................................29

3.5.1 Constraint equations.................................................................................. 293.5.2 Examples................................................................................................... 313.5.3 Constrained Interpolations ........................................................................ 32

3.6 Element Matrices and Stress Resultants of MIN-N .............................................363.7 Kirchhoff Constraints of MIN-N .........................................................................383.8 Application: Element MIN3 ................................................................................40

v

3.8.1 Virgin Interpolations of MIN3.................................................................. 413.8.2 Constrained Interpolations of MIN3......................................................... 423.8.3 Kirchhoff Constraints of MIN3 ................................................................ 433.8.4 Kirchhoff Edge Constraints ...................................................................... 44

CHAPTER 4: FORMULATION OF MIN6..................................................................... 474.1 Introduction ..........................................................................................................474.2 Shear Constraints .................................................................................................474.3 Interpolation Functions of Virgin Element ..........................................................484.4 Constrained Interpolation Functions of MIN6 ....................................................484.5 Kirchhoff Constraints of MIN6 ...........................................................................50

4.5.1 Kirchhoff Element Constaints................................................................... 504.5.2 Kirchhoff ‘Edge’ Constraints.................................................................... 51

CHAPTER 5: NUMERICAL RESULTS FOR MIN6..................................................... 585.1 General Comments ..............................................................................................585.2 Shear Relaxation Coefficient ...............................................................................585.3 Performance for Thin Plates ................................................................................615.4 Test Problems ......................................................................................................69

5.4.1 Overview................................................................................................... 695.4.2 Patch Test for Thin Plate........................................................................... 695.4.3 Meshing Strategy ...................................................................................... 715.4.4 Isotropic Thin and Moderately Thick Square Plates (

L

/t =1000 and 10) . 725.4.5 Thin Circular Plates (2R/t = 100) ............................................................. 775.4.6 Twisted Ribbon Tests ............................................................................... 805.4.7 Cantilever Plates (L/t = 100; L/t = 5)........................................................ 845.4.8 Orthotropic Square Plate (L/t = 30) .......................................................... 865.4.9 Free Vibration of Thin and Moderately-Thick Plates............................... 90

5.5 Remarks on MIN6 ...............................................................................................91

CHAPTER 6: CONCLUSIONS AND RECOMMENDATIONS ................................... 926.1 Conclusions ..........................................................................................................926.2 Recommendations ................................................................................................93

APPENDIX I:

MIN6 Element User Guide ...............................................................................................94

APPENDIX II:

MIN6 Element Input Data for Test Problems ...................................................................96

REFERENCES ................................................................................................................123

vi

LIST OF TABLES

Page

Table 3.1 Displacement interpolation for virgin and constrained elements..................... 25

Table 5.1 Location of interior nodes ................................................................................ 70

Table 5.2 Exact solutions for center deflection and bending moment of a square plate.. 73

Table 5.3 Exact solutions for center deflection and bending moment of a circular plate 77

Table 5.4 Cantilever plate results using MIN6 elements ................................................. 85

Table 5.5 Non-dimensional frequencies for simply-supported square plates ................. 90

Table 5.6 Percent error (%) in frequencies for simply-supported square plates ............. 91

vii

LIST OF FIGURES

Page

Figure 2.1 Plate variables and sign convention [3] ............................................................6

Figure 2.2 Kirchhoff kinematics (x-z plane)......................................................................7

Figure 2.3 Kinematics of Reissner/Mindlin plate (x-z plane)............................................9

Figure 2.4 Nodal configurations for initial and constrained displacement field..............17

Figure 3.1 Virgin and constrained elements’ nodal configuration. (a) N = 3, Linear. (b) N = 6, Quadratic. (c) N = 10, Cubic. (d) N = 15, Quartic. .................................24

Figure 3.2 Notation for MIN-N element ..........................................................................25

Figure 3.3 Correlation between A

mn

and polynomial terms............................................40

Figure 3.4 MIN3 triangular plate element........................................................................41

Figure 3.5 Nodal configuration for MIN3’s edge constraints..........................................44

Figure 4.1 MIN6 triangular plate element........................................................................47

Figure 4.2 Nodal configuration for MIN6’s edge constaints ...........................................53

Figure 4.3 Nodal configuration for MIN6’s ‘interior’ Kirchhoff constraints ..................55

Figure 4.4 MIN6 cross-diagonal pattern ..........................................................................57

Figure 5.1 Meshes of one quadrant of a thin, symmetric, square plate with a concentrated center load.......................................................................................................60

Figure 5.2

Cs

for MIN6....................................................................................................61

Figure 5.3 Performance of MIN6 (

Cs

= 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (simply supported and center load).................................................................63


Cs

= 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (clamped and center load)...............................................................................64

viii


Cs

= 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (simply supported and uniform load) .............................................................65


Cs

= 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (clamped and uniform load)............................................................................66


Cs

= 0) and ISOMIN6 with varying L/t ratios and three meshes (simply supported) .............................................................................67


Cs

= 0) and ISOMIN6 with varying L/t ratios and three meshes (clamped) ...........................................................................................68

Figure 5.9 Patch test for plate:

a

= 0.12;

b

= 0.24;

t

= 0.001;

E

= 1.0x10

6

; and

ν =0.25.70

Figure 5.10 The cross-diagonal meshes of MIN6 and MIN3 for plate problems ..............73

Figure 5.11 Meshes for one quadrant of doubly-symmetric square plates (dashed lines in-dicate the MIN3 elements)..............................................................................73

Figure 5.12 Convergence of center deflection for a thin square plate with center load ....74

Figure 5.13 Convergence of center deflection for thin square plate with uniform load ....75

Figure 5.14 Convergence of center deflection for moderately thick square plate with uni-form load.........................................................................................................76

Figure 5.15 MIN6 and SHELL93 meshes for 1/4 thin circular plate.................................78

Figure 5.16 MIN3 meshes for 1/4 thin circular plate.........................................................78

Figure 5.17 Convergence of center deflection for thin circular plate (2

R/t

=100).............79

Figure 5.18 Twisted ribbon examples with corner forces, and moments ..........................80

Figure 5.19 Mesh I for twisted ribbon examples ...............................................................81

Figure 5.20 Mesh II for twisted ribbon examples ..............................................................81

Figure 5.21 Twisted ribbon tests for tip deflection (

E

= 10

7

,

ν =

0.25,

t

= 0.05)for Mesh I ....................................................................................................................82

Figure 5.22 Twisted ribbon tests for tip deflection (

E

= 10

7

,

ν =

0.25,

t

= 0.05) for Mesh II ....................................................................................................................83

ix

Figure 5.23 Cantilever plate: load case 1 and 2; L/t =100; L/t = 5 ....................................84

Figure 5.24 Cantilever plate: numerical load models for case 1 and 2 ..............................85

Figure 5.25 Mesh of cantilever plate with MIN6 elements ...............................................85

Figure 5.26 Convergence for center deflection of orthotropic square plate (

L/t

= 30) ......87

Figure 5.27 Convergence of moment

Mxy

of orthotropic square plate (

L/t

= 30) .............87

Figure 5.28 Convergence of moment

Mx

and

My


L/t

= 30) ..88

Figure 5.29 Convergence of shear force

Qx

and

Qy


L/t

= 30) ..................................................................................................................89

1

CHAPTER 1

INTRODUCTION

1.1 Overview

Application of the finite element method to the bending of plates dates to the early

1960s. Nevertheless, the subject remains an active area of research because of the impor-

tance of plate structures and the difficulty to develop accurate and robust plate finite ele-

ments. A very large number of plate bending elements have been developed based on the

most commonly used plate theories, those of Kirchhoff and Reissner/Mindlin [1]. The

former theory ignores transverse shear deformation, while the latter includes it.

Interest in the shear deformable Reissner/Mindlin plate theory has increased signifi-

cantly during the past two decades primarily because of the increased use of laminated

composite materials. These types of structures, such as adhesively bonded laminated

plates, usually exhibit much lower strength in the transverse direction and between the

layer interfaces. Therefore, transverse shearing behavior can be significant in many cases.

Consequently, the finite element method based on Reissner/Mindlin theory can be a very

effective tool to provide more accurate modelling of moderately thick and anisotropic

(composite) plates with significant shear deformation. Also, in the finite element model,

Reissner/ Mindlin plate theory allows that rotations and the displacement derivatives are

not directly coupled. Consequently, rotation fields and the displacement field are indepen-

dently introduced into the functional of total transverse strain energy. Hence, only C

0

dis-

2

placement and rotation continuity is required for the finite element formulations. This low

continuity requirement makes the implementation much easier than elements based on

Kirchhoff plate theory, which in theory requires C

1

continuity of displacement.

Many displacement-based finite elements have been proposed which are able to

model shear deformable behavior in plates [2]. However, many low-order Reissner/Mind-

lin plate elements become very stiff when used to model thin plates. This phenomenon is

called shear-locking. One of the more successful low-order Mindlin plate elements in the

literature is Tessler and Hughes’ MIN3 element [3], which is a 3-node, 9 degree-of-free-

dom (DOF) triangle. It avoids shear locking by using so-called anisoparametric interpola-

tion. The anisoparametric interpolation strategy is to use interpolation functions one

degree higher for transverse displacements than for bending rotations. This strategy avoids

shear locking and ill-conditioning in the thin limit for even low-order Reissner/Mindlin

plate elements. Therefore, a straightforward, pure displacement-based formulation with

full integration of the stiffness matrix is possible. In addition to these favorable character-

istics, MIN3 produces accurate numerical results. A more detailed review of the MIN3

element will be given in Chapter 2.

The interpolation functions for the MIN3 element have also been used to develop a

‘smoothing element analysis’ for improved stress recovery in finite element analysis [4-6].

The characteristics of the interpolation functions, when the Kirchhoff zero shear con-

straints are enforced, leads to a C

1

continuous recovered stress field.

The success of the low-order, constant moment, anisoparametric triangular plate ele-

3

ment, MIN3, has motivated the development of a family (MIN-

N

) of higher-order triangu-

lar anisoparametric Reissner/Mindlin plate elements. A secondary motivation is the

potential to use these higher-order interpolation functions to develop higher-order smooth-

ing elements. A general methodology to derive members of the MIN-

N

family is devel-

oped herein. The transverse displacement is interpolated by a polynomial one order higher

than the rotational displacements. The transverse displacement is coupled with the bend-

ing rotations by enforcing continuous shear constraints in the element. MIN3, with a qua-

dratic variation of transverse displacement, is the lowest order member of the MIN-

N

family. As an example of a higher-order member, the 6-node MIN6, with a cubic variation

of transverse displacement, is developed herein. MIN6 is a higher-order, displacement-

compatible, complete polynomial, fully-integrated, triangular Reissner/Mindlin plate ele-

ment, with neither shear locking nor excessive stiffness in the thin limit. The element is

implemented in a finite element program and tested extensively.

1.2 Scope of Work

The primary objective of this study is to develop a general formulation for a family

of higher-order, pure displacement-based, triangular Mindlin plate elements; use the gen-

eral formulation to derive a 6-node element, MIN6, of the MIN-

N

family; and evaluate the

numerical performance of the element for linear elastic problems with isotropic and ortho-

tropic material behavior.

The dissertation is organized as follows. In Chapter 2, the two most common plate

theories, Kirchhoff and Reissner/Mindlin theories, are reviewed briefly. Because the MIN-

4

N

element is based on Reissner/Mindlin theory, the general review of previous plate ele-

ments focuses on the different type of Mindlin plate elements. The low-order anisopara-

metric Mindlin beam and plate elements [3, 7, 8] and some discrete Kirchhoff plate

elements [9, 10] are emphasized in the review section. The original formulation of the

three-node, triangular, anisoparametric plate element MIN3 will be reviewed briefly, as

well as the well-known discrete Kirchhoff triangular plate element, DKT. Chapter 3 gives

the derivation of the general formulation of the family of higher order,

N

-node triangular

anisoparametric Mindlin plate elements, in which

N

is also the number of terms in a com-

plete 2-D polynomial (i.e., 3, 6, 10, etc.). As an application of the general MIN-

N

formula-

tion, Tessler’s 3-node MIN3 element [3], which was originally derived with a different

strategy, has been verified as a member (

N

= 3) of the MIN-

N

element family. In Chapter

4, the formulation of the 6-node, 18 degree-of-freedom (DOF), triangular, anisoparametric

plate element (MIN6) is presented in detail. Numerical results with MIN6 for thin and

moderately thick plates in the linear, elastic regime for both static and dynamic analyses

are presented in Chapter 5, and these results are compared with those for MIN3 and

SHELL93, which is an ANSYS 6 to 8-node shell element with transverse shear deforma-

tion. Finally, conclusions and recommendations are given in Chapter 6.

5

CHAPTER 2

REVIEW OF PLATE THEORY AND PLATE

ELEMENTS

2.1 General Comments

This chapter begins with a brief review of two widely used linear plate bending theo-

ries, the Kirchhoff (classical) and Reissner/Mindlin plate theories and their finite element

formulations. The basic concept of shear locking for Reissner/Mindlin plate elements and

different approaches to avoid shear locking in the literature are then reviewed. The finite

element formulation for the discrete Kirchhoff triangular plate element, DKT, and

Tessler’s anisoparametric Reissner/Mindlin plate element, MIN3, are reviewed in detail.

The review provides a background for the present work.

2.2 Plate Theories

Consider a plate in the

x-y

plane with thickness

t

and mid-plane area

A

. Let

and be the plate in-plane displacements of a point and

be the transverse displacement of a point on the mid-surface of the plate.

Let and be the rotations of the normal to the midplane about the

x

and

y

-

axes, respectively. See Figure 2.1 for the sign convention.

u x y z, ,( ) v x y z, ,( ) x y z, ,( )

w x y,( ) x y,( )

θx x y,( ) θy x y,( )

6

Figure 2.1 Plate variables and sign convention [3]

7

2.2.1 Kirchhoff Plate Theory

The kinematics of Kirchhoff (classical) plate theory are based on the assumption that

plane sections remain plane and normal to the deformed midplane, as depicted in Figure

2.2. Therefore,

, (2.1)

and

, (2.2)

A comma is used to denote partial differentiation. From equation (2.2), the in-plane strains

are

(2.3a)

Figure 2.2 Kirchhoff kinematics (

x

-

z plane)

θy w,x–= θx w,y–=

u zw,x–= v zw,y–=

εεεε

εx

εy

γ xy

z

w,xx–

w,yy–

2w,xy–

==

−θy

γxz = 0

w,xz,w

x, u

w,x

u(x, y, z) w(x,y)midplane

8

and the transverse shear strains are

(2.3b)

The plate bending deformations, i.e., curvatures, are

(2.3c)

Given the above strain definitions, the moment curvature relations for an isotropic material

are

(2.4)

in which is the bending rigidity; E is the modulus of elasticity; and ν is

Poisson’s ratio. Because of the assumption of zero shear strain in Kirchhoff plate theory,

the transverse shear forces and cannot be obtained from the stress strain relations

and , where is the shear modulus. Instead, they must be

obtained based on equilibrium:

(2.5)

Refer to Figure 2.1 for the positive definitions of the plate bending variables.

γγγγγ xz

γ yz w,x θy+

w,y θx+

0= = =

κκκκ

κxx

κyy

κxy w,xx–

w,yy–

w,xy–

= =

M

Mx

My

Mxy

D

w,xx νw,yy+

w,yy νw,xx+

1 ν–( )w,xy

–= =

D Et3

12 1 ν2–( )-------------------------=

Qx Qy

τxz Gγ xz= τyz Gγ yz= G

QQx

Qy Mx,x Mxy,y+

My,y Mxy,x+

= =

9

2.2.2 Reissner/Mindlin Plate Theory

The Reissner/Mindlin plate theory uses a generalization of the Kirchhoff hypothesis,

i.e., plane sections remain plane but not necessarily normal to the deformed midplane.

Based on this assumption we have

(2.6)

in which and are the rotations of the line that was initially normal to the unde-

formed midsurface.

The displacements are therefore

, and (2.7)

as depicted in Figure 2.3. Based on equation (2.7), the in-plane strains are

Figure 2.3 Kinematics of Reissner/Mindlin plate (x-z plane)

γ xz

γ yz w,x θy+

w,y θx+

=

θx θy

u x y z, ,( ) zθy x y,( )–= v x y z, ,( ) zθx x y,( )–= w x y, z,( ) w x y,( )=

−θy

γxz w,x

z,w

x, u

w,x

u(x, y, z) w(x,y)midplane

10

(2.8a)

and the transverse shear strains are

(2.8b)

The plate bending curvatures are

(2.8c)

Given the strain definitions in equations (2.8a) and (2.8c), the moment curvature relations

for an isotropic material are

(2.9)

and based on equation (2.8b), the shear forces are

(2.10)

εεεε

εx

εy

γ xy

= z

θy x,

θx y,

θy y, θx x,+

z

κxx

κyy

κxy

= =

γγγγγ xz

γ yz w,x θy+

w,y θx+ x∂

∂1 0

y∂∂

0 1

w

θy

θx

= = =

κκκκ

κxx

κyy

κxy θy x,

θx y,

θy y, θx x,+ x∂

∂0

0y∂

∂

y∂∂

x∂∂

θy

θx

= = =

M

Mx

My

Mxy

D

1 ν 0

ν 1 0

0 01 ν–

2------------

κxx

κyy

κxy

= =

QQx

Qy

Gk2t 1 0

0 1

γ xz

γ yz

–= =

11

in which is the shear correction factor for non-uniform shear distribution (usually with

the value 5/6).

2.2.3 Potential Energy Functional

The total strain energy of shear deformable plates can be decomposed into mem-

brane energy, , bending energy, , and transverse shearing energy, :

(2.11)

in which the membrane energy is considered as zero (i.e., assume the in-plane displace-

ments of the midplane are zero and linear kinematics). For a linear elastic, isotropic mate-

rial, the total transverse strain energy functional, including only the bending and shearing

energies, may be written as

(2.12)

in which .

From equation (2.12) the total transverse strain energy functional for Kirchhoff plate

can be reduced to:

(2.13)

It is clear from equation (2.12) that the rotation fields , and the transverse dis-

placement w are considered as independent variables in the formulation and the highest

derivatives of displacements and rotations are of first order, which means that only C0-

k2

Um Ub Us

U Um Ub Us+ +=

U12---D θy x,

2 2νθy x, θx y, θx y,2 1

2--- 1 ν–( ) θx x, θy y,+( )2+ + + A

λA--- w,x θy+( )2 w,y( θx )2+ + ][ ] Ad∫∫+

d∫∫

=

λ 6k2 1 ν–( ) At2----=

U12---D w,xx( )2 2νw,xxw,yy w,yy( )2 2 1 ν–( ) w,xy( )2+ + +[ ] Ad∫∫=

θx θy

12

continuity is required in the finite element formulation. This is a great simplification com-

pared to equation (2.13), the Kirchhoff formulation, in which C1-continuity is required for

the displacement field w.

2.2.4 Kirchhoff Constraints

In equation (2.12), the second integral is referred to as a penalty constraint func-

tional. represents a penalty parameter which tends to infinity as the plate thickness, t,

approaches zero. In the thin plate regime, this penalty parameter enforces the Kirchhoff

constraints:

and (2.14a)

or

and (2.14b)

The total strain energy given by equation (2.12) reduces to that of the classical (Kirchhoff)

plate theory in equation (2.13) when and , i.e., when the

transverse shear strains are assumed to be zero.

2.3 Mindlin Plate Element Modeling

2.3.1 Finite Element Model and Shear Locking Phenomenon

The general form of the finite element model can be obtained by expressing the dis-

placements and the rotations in terms of interpolation functions and nodal variables and

then minimizing the total transverse strain energy, resulting in

(Kb + Ks) d = F (2.15)

λ

γ xz 0= γ yz 0=

w,x θy+ 0= w,y θx+ 0=

w,x θy+ 0= w,y θx+ 0=

λ

13

Kb is the bending stiffness matrix, which results from the first integral in equation (2.12);

Ks is the shear stiffness matrix, which results from the second integral in equation (2.12);

d is the vector of nodal displacements; and F is the vector of equivalent nodal loads. As the

plate thickness approaches zero, (i.e., as and ), Us = dTKsd must

approach zero. However, as the thickness t goes to zero, approaches infinity. For the

strain energy Us to remain bounded, Ksd therefore must approach zero. This means either

d must approach zero, which results in shear locking, or Ks must be a singular matrix. The

latter is achieved in many elements by using selective-reduced integration. Unfortunately,

this solution may result in the introduction of spurious zero energy modes, which require

an additional stabilization procedure.

2.3.2 Concept of Anisoparametric Interpolations

To relieve the problems associated with shear locking, different methods have been

developed. Relatively successful solutions have involved discrete penalty constraints,

reduced-integration procedures, improved penalty-strain interpolations, and penalty-

parameter modifications [3]. The selective, reduced integration approach is one of the

most popular methods. However, this can result in a rank-deficient stiffness matrix.

Another approach is to use different degree of the interpolation functions for displacement

and rotation, the so-called anisoparametric interpolation functions [3]. This approach has

proven to be effective in avoiding shear locking.

To understand why the anisoparametric approach is effective in avoiding locking, let

us use a simple numerical model to demonstrate the mathematics of shear locking. Con-

γ xz 0→ γ yz 0→ 12---λ

λ

14

sider a 3-node isoparametric triangular Reissner/Mindlin element. The transverse dis-

placement, w, and rotational displacements, and , with linear variations are given by

the following equations:

(2.16a)

and , i = 1, 2, 3 (2.16b)

in which the are both area-parametric coordinates and linear interpolation functions,

and , and are nodal DOFs. The interpolation functions in terms of Cartesian

coordinates are

(2.16c)

A is the area of the triangle, and the coefficients , , and are given by

(2.16d)

with a cyclic permutation of the indices (i = 1, 2, 3; j = 2, 3, 1; k = 3, 1, 2). and are

the nodal coordinates of node i.

The transverse shear strain field is

(2.17)

Substitution of equations (2.16a) and (2.16b) into (2.17) yields the shear strain in the Car-

tesian coordinates x, y as

(2.18a)

θx θy

w ξiwii 1=

3

∑=

θx ξiθxii 1=

3

∑= θy ξiθyii 1=

3

∑=

ξi

wi θxi θyi

ξi1

2A------- ci bix aiy+ +( )=

ai bi ci

ai xk x j–= bi y j yk–= ci x jyk xky j–=

xi yi

γ xz w= ,x θy+

γ xz1

2A------- biwi ciθyi+( ) biθyix aiθyiy+ +[ ]

i 1=

3

∑=

15

Equation (2.18a) may be rewritten as

(2.18b)

In the thin plate limit, is imposed to satisfy the Kirchhoff constraint. Thus, the

coefficients A0, A1 and A2 must go to zero, which gives three Kirchhoff constraint equa-

tions. A0 = 0 establishes a linear relation between the nodal rotations and the nodal trans-

verse displacements. However, from A1 = 0 and A2 = 0 we obtain

and (2.19)

These equations imply that the nodal rotations have to be either zero or constant val-

ues (note that and ). As a result, the bending energy in equation

(2.12) vanishes and the element locks.

The idea of the anisoparamatric approach is that the interpolation for w must be one

degree higher than that for and . Therefore, the deflection slopes and are

represented by the same complete polynomials as and (i.e. linear variation), and it

is therefore possible to represent .

2.4 Review of Plate Elements

2.4.1 Overview

Recent research on displacement-based thin plate elements has focused on relaxing

the assumptions of Kirchhoff thin-plate theory. The elements produced by relaxing these

assumptions are called discrete Kirchhoff elements. The most successful of these elements

is the DKT (Discrete Kirchhoff Triangle) plate element which was reexamined in 1980 by

γ xz A0 wi θyi,( ) A1 θyi( )x A2 θyi( )y+ +=

γ xz 0=

biθyii 1=

3

∑ 0= aiθyii 1=

3

∑ 0=

θyi

aii 1=

3

∑ 0= bii 1=

3

∑ 0=

θx θy w,x w,y

θx θy

γ xy w,x= θy+ 0=

16

Batoz et al. [9] based on the original elements QQ3 and KC published by Stricklin et al.

and Dhatt in 1969. An explicit formulation has been presented by Batoz [10] also. The

DKT element only enforces the Kirchhoff condition of zero shear at discrete points in the

element. This 3-node, 9 degree-of-freedom DKT element is one of the best triangular plate

elements known [11]. It combines the advantages of both theories: shear locking free anal-

ysis of thin plate (Kirchhoff) and C0-continuity requirements for independently interpo-

lated displacements and rotations (Reissner/Mindlin).

Reissner/Mindlin theory of moderately thick plates has C0-continuity requirements

for the displacement assumption, which is an important advantage compared to the Kirch-

hoff theory, as discussed in the previous sections. Many successful C0 elements have been

developed by Hughes et al. [12], Hughes and Tezduyar [13], Pugh et al. [14], MacNeal,

[15], Crisfield [16], Belytschko, [17], and Tessler and Hughes [3, 18]. Tessler’s MIN3 ele-

ment will be presented in detail subsequently.

Mixed elements can be derived for Kirchhoff and Reissner/Mindlin theories. In

mixed elements, both displacements and stresses are approximated. The highest degree of

derivatives of displacement and moments occurring in the functionals is 1, which means

that C0-continuous shape functions are sufficient. This is a great advantage, and many suc-

cessful elements are available based on this approach. However, nodal variables are dis-

placements and moments, which make these elements difficult to implement. The system

of equations has zeros on the main diagonal, and therefore special equation solvers are

required. So-called hybrid elements use special procedures to eliminate the nodal moment

components in the interior mixed field. Therefore, hybrid elements have only nodal dis-

17

placement components. Hybrid elements are successfully used in the practical analysis of

thin plates.

2.4.2 Existing Low-Order Anisoparametric Element, MIN3

In this section the basic concept and formulation of MIN3 are reviewed. More details

of the derivation can be found in [3]. As mentioned in [2], many types of shear deformable

plate elements have been developed because of their advantages over elements based on

the thin plate theory of Kirchhoff. However, numerical results indicate that many of the

elements become excessively stiff, and sometimes rigid, for very thin plates. This phenom-

enon is known as shear locking because the excessive stiffness results from the ‘zero’

transverse shear deformation constraint which occurs in the thin limit. MIN3 uses aniso-

parametric interpolation functions and a penalty parameter modification to avoid shear

locking and excessive stiffness problems, respectively.

Figure 2.4 Nodal configurations for initial and constrained displacement field

2

3

1

2

3

1

5

4

6

18

The initial interpolation functions for MIN3 are given in terms of area-parametric

coordinates for the 6-node triangle shown in Figure 2.4. The quadratic deflection is inter-

polated as

(2.20a)

where the interpolation functions and nodal DOFs are

, , i = 1, 2,..., 6 (2.20b)

with

, , i = 1, 2, 3; k = 2, 3, 1 (2.20c)

These are the standard Lagrange type interpolation functions for a 6-node triangle. The

linear rotation fields are

and (2.21a)

where the linear interpolation functions and nodal DOFs are

, and , i = 1, 2, 3 (2.21b)

Note that in this initial, or ‘virgin,’ element, 6 nodes have transverse displacement

DOFs and only the 3 vertex nodes have rotational DOFs. This in inconvenient, and to

eliminate the 3 transverse displacements at the mid-edge nodes, the shear strain along the

element edges are required to be constant, i.e.,

, i = 1, 2, 3 (2.22)

where s denotes the edge coordinate and θn is the tangential edge rotation (refer to Figure

2.1). Enforcing these constraints leads to a constrained deflection field in terms of vertex

w Nwv=

NT Ni{ }= wv wi{ }=

Ni ξi 2ξi 1–( )= Ni 3+ 4ξiξk =

θx ξξξξθθθθx= θy ξξξξθθθθy=

ξξξξ ξi{ }= θθθθx θxi{ }= θθθθy θyi{ }=

γ sz,s w,s θn+( ),s ξi 0=0≡=

19

DOFs. The displacements and rotations become

(2.23a)

The shape functions

i =1, 2, 3 (2.23b)

are given in terms of area-parametric coordinates as [3]:

(2.23c)

(2.23d)

(2.23e)

with a cyclic permutation of the indices (i = 1, 2, 3; j = 2, 3, 1; k = 3, 1, 2).

The element stiffness matrix, K, may be expressed in terms of its bending, Kb, and

transverse shear, Ks, components as

K = Kb + Ks = , (2.24a)

where

(2.24b)

w N1 N2 N3 d= θx N1θθθθx= θy N1θθθθy=

N1T ξi{ }= N2

T N2i{ }= N3T N3i{ }=

ξi1

2A------- ci bix aiy+ +( )= ξi

i 1=

3

∑ 1=

N2i12--- bkξiξ j b jξkξi–( )= N2i

i 1=

3

∑ 0=

N3i12--- a jξkξi akξiξ j–( )= N3i

i 1=

3

∑ 0=

BbT∫

A∫ DbBb A Bs

T∫A∫ GsBs Ad+d

Bb

0 0 N1,x

0 N1,y 0

0 N1,x N1,y

=

20

(2.24c)

The bending and transverse shear rigidity matrices are, respectively,

(2.24d)

where

(2.24e)

with Cij denoting the plane stress elastic moduli and the shear correction factors.

The consistent load vector due to the distributed normal load, q, as well as bending

moments, , , and transverse shear force, , prescribed on the portion of the

element boundary may be written as

(2.25)

The element bending moment and transverse shear stress resultants may be deter-

mined from the relations

(2.26)

BsN1,x N2,x N3,x N1+

N1,y N2,y N1+ N3,y

=

Db

D11 D12 D16

D22 D26

sym. D66

= GsG11 G12

sym. G22

=

Dij112------h

3Cij= Gij kij

2hC6 i– 6 j–,=

kij2

Mxx Myy Q Γσ

Γ

F q N1 N2 N3

TAd∫

A∫ 0 MyyN1 MxxN1

TΓd

Γσ

∫Q N1 N2 N3

TΓd

Γσ

∫

+

+

=

Mxx

Myy

Mxy

DbBbd=

21

(2.27)

It is noted that full integration is used to evaluate the element matrices in equations (2.24a)

and (2.25). The stiffness matrix therefore has full rank.

Tessler [3] defined the shear relaxation factors in equation (2.24e) to incorporate

both the classical shear correction factor (taken as 5/6), , and a finite element relax-

ation factor, . That is, = , where subscripts i and j are used to denote a gen-

eral anisotropic material case. The element stiffness matrix K may be written as

(2.28a)

where is the unrelaxed element shear stiffness that includes the classical shear correc-

tion factor. The introduction of the element-appropriate shear relaxation factor, , can be

interpreted as follows. In the thin regime, the numerical values in can be very large

relative to those in . Therefore a numerically ill-conditioned matrix and subsequent

loss of accuracy in finite precision arithmetic can result. Hence, is introduced to avoid

this problem by appropriately relaxing the shear stiffness. Tessler [6] defined as

(2.28b)

where

(2.28c)

Qx

Qy

GsBsd=

kij2

ki j∗2

φ2 kij2 φ2 ki j∗

2

K Kb Ks+ Kb φ2Ks∗+= =

Ks∗

φ2

Ks∗

Kb

φ2

φ2

φ2 11 Csα+-------------------=

α

ksii∗i 4=

9

∑

kbiii 4=

9

∑-------------------=

22

The parameter α is the ratio of the sum of the diagonal shear and bending stiffness coeffi-

cients associated with the rotational DOFs, and Cs is a constant which was determined

numerically. Note that the factor tends to zero as the plate thickness approaches zero,

and that it is closer to 1 for thick plates. Another advantage of the additional relaxation

introduced by is that the excessive stiffness caused by coarse meshes is reduced.

The success of MIN3 is due to the anisoparametric interpolation functions, which

eliminate shear locking, and the element-appropriate shear relaxation factor, which results

in a well-conditioned element stiffness matrix over the entire range of element thickness-

to-area ratios. Consequently, the element has no deficiencies in either the classical (thin)

or shear-deformable (thick) regimes and at the same time produces rapidly convergent

solutions. MIN3 is an excellent element, based upon numerical testing on linear, elasto-

static problems.

φ2

φ2

23

CHAPTER 3

DERIVATION OF THE MIN-N FAMILY

3.1 Introduction

As mentioned in Chapter 1, the success of the existing low-order anisoparametric tri-

angular element MIN3 has motivated the development of a family of arbitrary, triangular,

anisoparametric Mindlin plate elements, MIN-N. In this chapter, the basic concepts and

the derivation of MIN-N are presented in detail. More detailed information of the existing

low-order, displacement-compatible beam, plate and shell elements that are based on

anisoparametric interpolations can be found in other sources [3, 7, 8, 19-22].

3.2 Basic Concepts of MIN-N Element Family

Based on the anisoparametric strategy, the desired triangular N-node and 3N-DOF

elements, MIN-N, must have a complete polynomial interpolation for the transverse dis-

placement, w, which is one degree higher than the interpolation of the normal rotations,

and .

The first step in the derivation is to generate a family of ‘virgin’ elements that are

based on independent interpolations of the kinematic variables (transverse displacement

and normal rotations). Lagrange-type interpolations for the rotations are used with the

standard triangular nodal configuration, as shown in Figure 3.1. Hierarchical interpolation

functions are employed for the higher-order terms of the transverse displacement field.

From the virgin elements, a family of constrained elements, MIN-N, can be developed by

θx

θy

24

enforcing a continuous shear constraint along any line L on the N-node element (Figure

3.2). As a result of imposing these constraints, the hierarchical DOFs are eliminated and

the transverse displacement of the constrained element is then coupled with the bending

rotations. The constrained element then has the same nodal configuration, with the same

nodal DOF, as the virgin element. The deflection field for virgin and constrained elements

is illustrated in Table 3.1. Once the interpolation functions for constrained elements are

developed, formulating element stiffness, consistent mass matrices, and consistent load

vectors follows the straightforward, standard procedure for displacement-based finite ele-

ments. The derivation of the MIN-N interpolation functions is discussed in the following

sections.

Figure 3.1 Virgin and constrained elements’ nodal configuration. (a) N = 3, Linear. (b) N = 6, Quadratic. (c) N = 10, Cubic. (d) N = 15, Quartic.

1 1112 22

3 333

4 44 5 5

56

6

7

6 878 9

9

10

1011

1214

15

13

(c)(a) (b) (d)

25

Figure 3.2 Notation for MIN-N element

Table 3.1 Displacement interpolation for virgin and constrained elements

Element type Virgin element Constrained element

Degree of interpolation fields; ,

(p + 1); p (p + 1); p

Degree of shear strain p p - 1

Number of nodes N N

Number of nodal DOF 3N 3N

Number of hierarchical parameters p + 2 0

Number of constraint equations 0 p +2

x

y

n

s

z, w

θx

θy

θs

θn

α

L

w x y,( ) θx θy

γ sz w,s θn+=

N p 1+( ) p 2+( ) 2⁄=

26

3.3 Virgin Elements

Let w represent the transverse displacement field, and and the rotations of

midsurface-normals about the x and y axes, respectively (Figure 3.2). The nodal displace-

ments of the N-node element are , and , i = 1, 2... N, and

(3.1a)

For the virgin elements, independent interpolation is used for w, , and .

Lagrange-type interpolation is used for the rotations, such that

(3.1b)

and N is the 1 x N vector of interpolation functions:

i = 1, 2,... N (3.1c)

These are the standard Lagrange interpolation functions, used extensively for plane trian-

gular elements, that can be found in many finite element texts [11]. These interpolation

functions have degree , such that p = 1 when N = 3, p = 2 when N

= 6, p = 3 when N = 10, etc.

The transverse displacement is interpolated with functions of degree p+1. The terms

of degree up to p come from the Lagrange interpolation functions Ni; hierarchical func-

tions are used for the p+2 higher-order terms. The nodeless degrees-of-freedom associated

with the hierarchical terms are represented by

j = 1, 2...(p+2) (3.1d)

The displacement can then be expressed as

θxi θyi

wi θxi θyi

w wi{ }= θθθθx θxi{ }= θθθθy θyi{ }=

θx θy

θx Nθθθθx= θy Nθθθθy=

NT Ni{ }=

p 1 8N+ 3–( ) 2⁄=

a a j{ }=

27

(3.2a)

in which is the 1 x (p+2) vector of hierarchical interpolation functions

j = 1, 2...(p+2) (3.2b)

and

(3.2c)

Equations (3.1b), (3.2a) and (3.2c) can be combined to obtain

(3.3a)

in which

(3.3b)

and

(3.3c)

Note that the virgin elements have 3N nodal degrees-of-freedom and p+2 nodeless

degrees-of-freedom.

w N 0 0 Nadv=

Na

NaT Naj{ }=

dv

w

θθθθx

θθθθy

a

=

u Nvdv=

u

w

θx

θy

=

Nv

N 0 0 Na

0 N 0 00 0 N 0

=

28

3.4 Shear Constraints

The transverse shear along any arbitrary line L that makes an angle α with the x-

axis (Figure 3.2) is defined as

(3.4a)

in which is the midsurface-normal rotation as depicted. Hence and are defined

as

; (3.4b)

can be expressed in terms of and as follows. Based on the directional deriva-

tive,

(3.4c)

The relation is

(3.4d)

Substitution of equations (3.4c) and (3.4d) into (3.4a) results in

(3.4e)

Our desired element, MIN-N, is obtained from the virgin element by eliminating the

p+2 nodeless degrees-of-freedom, a. The MIN-N elements have therefore 3N nodal

degrees-of-freedom only (displacement and 2 rotations at each node). As a result of the

elimination of a, the transverse displacement will be coupled with the rotational degrees-

of-freedom. p+2 constraints are needed to eliminate a. The p+2 constraint equations are

obtained from

γ sz

γ sz w s, θn+=

θn γ xz γ yz

γ xz w x, θy+= γ yz w y, θx+=

γ sz γ xz γ yz

w s, w x, α w y, αsin+cos=

θn

θn θy α θx αsin+cos=

γ sz γ xz α γ yz αsin+cos=

29

(3.5)

Equation (3.5) says that the p-th partial derivative of the shear strain along any direc-

tion is required to be zero. For example, for a 3-node element, p = 1 and the first partial

derivative has to be zero; for a 6-node element, p = 2 and the second partial derivative has

to be zero; for a 10-node element, p = 3 and the third partial derivative has to be zero; and

for an N-node element, the p-th order partial derivative should be zero. Note that this con-

straint reduces the degree of completeness of the shear strain polynomial by one.

The explicit derivation of the p + 2 shear constraint equations is given in Section 3.5.

3.5 General Derivation of MIN-N

3.5.1 Constraint equations

From equations (3.5) and (3.4e), we have

(3.6a)

where

γ sz ss…s,p

0≡

γ sz

sp

p

∂

∂ γ sz 0≡

30

(3.6b)

in which

and , . To satisfy equation (3.6a), because angle α is arbitrary,

each term in equation (3.6b) must be zero:

(3.7)

The p+2 equations represented by equation (3.7) can be simplified as

m = 0, 1, 2,...(p-1) (3.8)

sp

p

∂

∂ γ sz

x∂∂ αsin

y∂∂ αcos+

p= γ sz

x∂∂ αsin

y∂∂ αcos+

pγ xz αcos γ yz αsin+( )=

C pm αcos( )p m– αsin( )m γ xz αcos γ yz αsin+( )p∂

xp m– ym∂∂--------------------------------------------------------

m 0=

p

∑=

C pm

m 0=

p

∑ αcos( )p m– 1+ αsin( )m γ xzp∂

xp m– ym∂∂-------------------------- αcos( )p m– αsin( )m 1+ γ yz

p∂xp m– ym∂∂

--------------------------+=

C pm 1+

γ xzp∂

xp m 1+( )– ym 1+∂∂------------------------------------------- C p

mγ yz


--------------------------+ αcos( )p m– αsin( )m 1+

m 1–=

p

∑=

C pm p!

m! p m–( )!---------------------------=

C p0 1= C p

1– C pp 1+ 0==

C pm 1+

γ xzp∂

xp m 1+( )– ym 1+∂∂------------------------------------------- C p

mγ yz


--------------------------+ 0 m 1– 0 1 …p, , ,= =

xp

p

∂

∂ γ xz 0=

p m–( )x

p m 1+( )–ym 1+∂

p

∂

∂ γ xz m 1+( )x

p m–ym∂

p

∂

∂ γ yz+ 0=

yp

p

∂

∂ γ yz 0=

31

The constraint equations (3.8) can be used to eliminate the p+2 hierarchical degrees-of-

freedom.

3.5.2 Examples

From the general equation (3.8) the constraint equations for different elements in the

MIN-N family are easily obtained.

1. MIN3 is the first element in the family. It is a 3-node triangle with p = 1. The element

has a quadratic displacement field and linear rotation fields. The virgin element has 9

degrees-of-freedom, with three nodeless degrees-of-freedom. The three constraint

equations are

(3.9a)

2. MIN6 is the second element in the family. It is a 6-node triangle with p = 2. The ele-

ment has a cubic displacement field and quadratic rotation fields. The virgin element

has 18 nodal degrees-of-freedom and four nodeless degrees-of-freedom. The four

constraint equations are

(3.9b)

3. MIN10 is the third element in the family. It is a 10-node triangle with p = 3. The ele-

ment has a quartic displacement field and cubic rotation fields. The virgin element

γ xz x, 0=

γ xz y, γ yz x,+ 0=

γ yz y, 0=

γ xz xx, 0=

2γ xz xy, γ yz yy,+ 0=

γ xz yy, 2γ yz xy,+ 0=

γ yz yy, 0=

32

has 30 nodal degrees-of-freedom and five nodeless degrees-of-freedom. The five


(3.9c)

4. MIN15 is the fourth element in the family. It is a 15-node triangle with p = 4. The

element has a quintic displacement field and quartic rotation fields. The virgin ele-

ment has 45 nodal degrees-of-freedom and six nodeless degrees-of-freedom. The six


(3.9d)

3.5.3 Constrained Interpolations

The definitions of and in terms of displacement can be substituted in the con-

straint equation (3.8) to obtain the constraints in terms of the displacement variables:

γ xz xxx, 0=

3γ xz xxy, γ yz xxx,+ 0=

2γ xz xyy, 2γ yz xxy,+ 0=

γ xz yyy, 3γ yz xyy,+ 0=

γ yz yyy, 0=

γ xz xxxx, 0=

4γ xz xxxy, γ yz xxxx,+ 0=

3γ xz xxyy, 2γ yz xxxy,+ 0=

2γ xz xyyy, 3γ yz xxyy,+ 0=

γ xz yyyy, 4γ yz xyyy,+ 0=

γ yz yyyy, 0=

γ xz γ yz

33

(3.10a)

where m = 0, 1, 2,...(p - 1). Let

(3.10b)

in which is a (p+2) x 3 matrix. Substitution of equation (3.3a) and (3.10b) into equa-

tion (3.10a) results in

(3.11a)

or

(3.11b)

Let

xp 1+

p 1+

∂∂

0x

p

p

∂∂

p 1+( )x

p m–y

m 1+∂

p 1+

∂∂

m 1+( )x

p m–y

m∂

p

∂∂

p m–( )x

p m 1+( )–y

m 1+∂

p 1+

∂∂

yp 1+

p 1+

∂∂

yp

p

∂∂

0

w

θx

θy

0=

∂∂∂∂[ ]

xp 1+

p 1+

∂∂

0x

p

p

∂∂

p 1+( )x

p m–y

m 1+∂

p 1+

∂∂

m 1+( )x

p m–y

m∂

p

∂∂

p m–( )x

p m 1+( )–y

m 1+∂

p 1+

∂∂

yp 1+

p 1+

∂∂

yp

p

∂∂

0

=

∂∂∂∂[ ]

∂∂∂∂[ ]Nvdv 0=

∂∂∂∂[ ]N 0 0 Na

0 N 0 00 0 N 0

w

θθθθx

θθθθy

a

0=

34

(3.12a)

so that equation (3.11b) can be written as

(3.12b)

Based on equation (3.12b), the nodeless degrees-of-freedom a can be written in terms of

the nodal degrees-of-freedom as

(3.13)

Then

(3.14a)

where I is a 3N x 3N identity matrix, the transformation matrix T is (3N + p + 2) x 3N,

(3.14b)

and

Bc ∂∂∂∂[ ]N 0 0 Na

0 N 0 00 0 N 0

0 BθxBθy

Ba

p 2+( ) 3N p 2+ +( )×= =

0 BθxBθy

Ba

w

θθθθx

θθθθy

a

0=

a Ba1–

– 0 BθxBθy

w

θθθθx

θθθθy

=

dv

I

Ba1–

– 0 BθxBθy

w

θθθθx

θθθθy

Td= =

TI

Ba1–

– 0 BθxBθy

=

35

(3.14c)

With , , we obtain the constrained interpolation functions

(3.15a)

in which

(3.15b)

such that

(3.15c)

Nc is the 3 x 3N matrix of constrained interpolation functions for MIN-N elements. Notice

that as a result of enforcing the continuous shear constraints along any line in the element,

the transverse displacements are now coupled to the bending rotations and can be

expressed as

(3.16a)

in which

i = 1, 2,...N (3.16b)

L and M are 1 x N vectors and have degree p+1, one order higher than N. That is, the

transverse displacement is no longer interpolated independently of the rotations, as it is for

the virgin elements.

d

w

θθθθx

θθθθy

=

dv Td= u NvTd=

Nc NvT=

Nc

N L M0 N 00 0 N

=

u Ncd=

w N L M d=

LT Li{ }= MT Mi{ }=

36

To guarantee that the rigid body motions are satisfied, the interpolation functions N,

L, and M for MIN-N elements must satisfy the conditions

, , (3.17)

The restriction on Ni is clearly satisfied because these are the standard Lagrange-type

interpolation functions. It will be shown later that the other two conditions are satisfied for

specific elements. One can argue, however, that they will be satisfied in general. The virgin

interpolation functions certainly can represent rigid body motions. The constrained inter-

polation functions are obtained from the virgin functions by restricting the variation of the

shear strains (equation (3.6a)). Because rigid body motions involve zero shear, the con-

straints do not affect the ability of the constrained functions to represent rigid body

motion, and hence equation (3.17) will be satisfied.

3.6 Element Matrices and Stress Resultants of MIN-N

Once the interpolation functions of MIN-N are developed, formulating the element

stiffness, consistent mass matrices and consistent load vectors follows the straightforward

procedure in the displacement-based finite element formulation. Stiffness, consistent mass

matrices and consistent load vectors are integrated with full integration.

Using the displacement-based finite element model based on Mindlin plate theory,

the element stiffness matrix, K, can be expressed as the sum of the bending component,

Kb, and transverse shear component, Ks:

Nii 1=

n

∑ 1≡ Lii 1=

n

∑ 0≡ Mii 1=

n

∑ 0≡

37

K = Kb + Ks = (3.18a)

in which

(3.18b)

(3.18c)

The bending and transverse shear rigidity matrices are, respectively,

(3.18d)

in which

(3.18e)

with Cij are the plane stress elastic moduli and are the shear relaxation factors (the

same as used for MIN3 [3]). The shear relaxation factors are introduced here in the MIN-

N formulas to verify their effects for the higher-order elements.

When considering inertia forces, the element’s consistent mass matrix is

M = (3.19a)

in which

BbT∫

A∫ DbBb A Bs

T∫A∫ GsBs Ad+d

Bb

0 0 N,x

0 N,y 0

0 N,x N,y

=

BsN,x L,x M,x N+

N,y L,y N+ M,y

=

Db

D11 D12 D16

D22 D26

sym. D66

= GsG11 G12

sym. G22

=

Dij112------h

3Cij= Gij kij

2hC6 i– 6 j–,=

kij2

NcT∫

A∫ mNc Ad

38

(3.19b)

where is the mass per unit area of the plate and is the rotary inertia.

The consistent load vector for the distributed normal load, q, as well as applied

bending moments, , , and transverse shear force, , prescribed on the portion

of the element boundary may be written as

(3.20)

The element bending moment and transverse shear stress resultants may be deter-

mined from the relations

(3.21)

(3.22)

3.7 Kirchhoff Constraints of MIN-N

The p+2 shear constraints in equation (3.8) for MIN-N limit the variation of the

shear strains. The shear strains and are in the form

m

m 0 0

mt

2

12-------- 0

sym. mt

2

12--------

=

m mt2

12⁄

Mxx Myy Q

Γσ Γ

F q N L MT

Ad∫A∫ 0 MyyN MxxN

TΓd

Γσ

∫Q N L M

TΓd

Γσ

∫

+

+

=

Mxx

Myy

Mxy

DbBbd=

Qx

Qy

GsBsd=

γ xz γ yz

39

(3.23a)

(3.23b)

to satisfy the first and last shear constraints in equation (3.8). Figure 3.3 gives the relations

between the polynomial terms and the corresponding coefficient Amn and Bmn in equation

(3.23a). The relations for Bmn are similar. To satisfy the other p constraint equations

m = 0, 1, 2,...(p-1) (3.24)

we obtain the p relations

m = 1, 2,...p (3.25)

In the thin plate regime as , the Kirchhoff constraints

and (3.26)

are enforced over the entire element domain. Therefore, with (3.23a), (3.23b) and (3.25)

the Kirchhoff constraints are

m = 0, 1,...n; n = 0,1,...p-1; l = 1, 2,...p. (3.27)

These are the p(p+2) Kirchhoff constraint equations for MIN-N in the thin plate limit.

γ xz Amnxn m–

ym

m 0=

n

∑

Ampxp m–

ym

m 1=

p

∑+n 0=

p 1–

∑=

γ yz Bmnxn m–

ym

m 0=

n

∑

Bmpxp m–

ym

m 0=

p 1–

∑+n 0=

p 1–

∑=

p m–( )x

p m 1+( )–ym 1+∂

p

∂

∂ γ xz m 1+( )x

p m–ym∂

p

∂

∂ γ yz+ 0=

Amp B p m–( ) p+ 0=

L t⁄ ∞→

γ xz 0→ γ yz 0→

Amn Alp B p l–( ) p–=( ) Bmn 0→, ,

40

Figure 3.3 Correlation between Amn and polynomial terms

3.8 Application: Element MIN3

As stated previously, the success of the 3-node, triangular, constant-moment MIN3

element presented by [3] has motivated this development of the MIN-N family, which

includes higher-order elements. It is verified in this section that Tessler’s MIN3 element is

the lowest order member in the MIN-N family, even though his derivation and the deriva-

tion of MIN-N are somewhat different.

The detailed derivation of the constrained interpolation functions for MIN3 is now

presented. The nodal configuration for MIN3 is shown in Figure 3.4.

1

x y

x2 xy y2

x3 x2y xy2 y3

xp-1 xp-2y xyp-2 yp-1

xyp-1 yp. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . xp xp-1y

A00

A01 A11

A02 A12 A22

A03 A13 A23 A33

A1(p-1) A(p-2)(p-1)

A1p App

A(p-1)(p-1)A0(p-1)

A(p-1)pA0p

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

41

Figure 3.4 MIN3 triangular plate element

3.8.1 Virgin Interpolations of MIN3

The interpolation functions for MIN3’s virgin element are given in terms of area-

parametric coordinates as:

, (3.28a)

where

, i = 1, 2, 3

and

, i = 1, 2, 3; k = 2, 3, 1. (3.28b)

The Ni are also used as shape functions, and therefore the linear relation between Carte-

sian and area coordinates is

(3.28c)

As before, A is the area of the triangle, and the coefficients , , and are

x

y

2

3

1

NT Ni{ }= NaT Nai{ }=

Ni ξi =

Nai ξiξk =

ξi 1

2A------- ci bix aiy+ +( )=

ai bi ci

42

(3.28d)

with a cyclic permutation of the indices (i = 1, 2, 3; j = 2, 3, 1; k = 3, 1, 2).

3.8.2 Constrained Interpolations of MIN3

For MIN3, with p = 1, equation (3.11b) becomes

(3.29a)

in which

i = 1, 2, 3 (3.29b)

The constrained interpolation functions Nc can be obtained from equations (3.11b) to

(3.15a). The interpolation functions are

i =1, 2, 3 (3.29c)

in which

(3.29d)

(3.29e)

(3.29f)

Equations (3.29d), (3.29e) and (3.29f) are identical with those for Tessler’s MIN3

ai xk x j–= bi y j yk–= ci x jyk xky j–=

x2

2

∂∂

0x∂

∂

2x y∂

2

∂∂

x∂∂

y∂∂

y2

2

∂∂

y∂∂

0

N 0 0 Na

0 N 0 00 0 N 0

w

θθθθx

θθθθy

a

0=

w wi{ }= θθθθx θxi{ }= θθθθy θyi{ }= a ai{ }=

NT Ni{ }= LT Li{ }= MT Mi{ }=

Ni ξi =

Li12--- bkξiξ j b jξkξi–( )=

Mi12--- a jξkξi akξiξ j–( )=

43

formulation [3]. Therefore, it has been verified that Tessler’s MIN3 is the lowest order

member in the MIN-N family.

Note that

, , (3.30)

As stated, these relations are required for the interpolation functions to represent rigid

body modes correctly.

3.8.3 Kirchhoff Constraints of MIN3

The 3 shear constraints in equation (3.9a) for MIN3 limit the variation of the shear

strains. The shear strains and are in the form

, (3.31a)

to satisfy the first and third shear constraints in (3.9a). To satisfy the second constraint

equation

(3.31b)

the relation is

(3.31c)

In the thin plate regime as , the Kirchhoff constraints

and (3.32a)

are enforced over the entire element domain. Therefore, with (3.31a) and (3.31c), the

Kirchhoff constraints are

Nii 1=

3

∑ 1≡ Lii 1=

3

∑ 0≡ Mii 1=

3

∑ 0≡

γ xz γ yz

γ xz A00 A11y+= γ yz B00 B01x+=

γ xz y, γ yz x, A11 B01+=+ 0=

A11 B01+ 0=

L t⁄ ∞→


44

(3.32b)

which agree with Tessler’s analysis [18].

3.8.4 Kirchhoff Edge Constraints

Figure 3.5 Nodal configuration for MIN3’s edge constraints

Consider the element depicted in Figure 3.5. Let w(s) represent the transverse dis-

placement, with quadratic interpolation, along the edge ij. Let represent the rota-

tional displacement, which is a linear function. The displacements can be expressed as

(3.33a)

(3.33b)

in which

, (3.33c)

and

A00 A11 B01–=( ) B00 0→, ,

x

y

j

k

sn

lij

lki

ljk

z i

wi

awj

θni

θnj

θn s( )

w s( ) Niwi N jw j Naa+ +=

θ s( ) Niθni N jθnj+=

Ni

s lij–( )lij

-----------------–= N jslij----=

45

(3.33d)

where Ni and Nj are the Lagrange interpolation functions; Na is a higher-order hierarchical

function. The nodeless degree-of-freedom associated with the hierarchical term is a.

The shear constraint for MIN3 is

(3.34)

Substitution of equations (3.33a) and (3.33b) into (3.34) results in

(3.35a)

or

(3.35b)

From equation (3.35b), we have

(3.36)

The displacement w(s) can then be expressed in terms of nodal DOFs as

(3.37)

Na s s lij–( )=

γ sz,s w,ss θy,s+ 0≡=

s2

2

∂

∂ Ni

s2

2

∂

∂ N j

s∂∂Ni

s∂∂N j

s2

2

∂

∂ Na

wi

w j

θni

θnj

a

0=

0 0 1lij----–

1lij---- 2

wi

w j

θni

θnj

a

0=

aθni θnj–

2lij--------------------=

w s( ) Niwi N jw j Na

θni θnj–

2lij--------------------+ +=

46

If we impose the Kirchhoff constraint

(3.38)

on each edge, we obtain the three Kirchhoff edge constraints:

i = 1, 2, 3; j = 2, 3, 1 (3.39)

Note that these Kirchhoff constraints are equivalent to the Kirchhoff constraints in equa-

tion (3.32b). This form is just perhaps a bit cleaner. They are referred to herein as edge

constraints because each constraint involves the nodal degrees-of-freedom associated with

one edge only.

γ sz w s, θn+ 0= =

w j wi–θni θnj+( )lij

2------------------------------=

47

CHAPTER 4

FORMULATION OF MIN6

4.1 Introduction

The 6-node, 18 degree-of-freedom (DOF), triangular, anisoparametric plate element

(MIN6) is presented in this chapter as an example of a higher-order member in the MIN-N

family. The detailed derivation of MIN6 is based on the general formulation of MIN-N

presented previously. The Kirchhoff constraints of MIN6 in the thin plate limit are given

explicitly. Figure 4.1 shows the nodal configuration for MIN6.

Figure 4.1 MIN6 triangular plate element

4.2 Shear Constraints

To generate MIN6, first restate the shear constraints from equation (3.9a):

x

y

2

3

1

56

4

48

(4.1)

These 4 constraints are used to eliminate the 4 hierarchical degrees-of-freedom from

MIN6’s virgin element interpolation functions.

4.3 Interpolation Functions of Virgin Element

The interpolation functions for MIN6’s virgin element are

, i =1, 2...6; j = 1, 2, 3, 4 (4.2a)

where

, (4.2b)

and

, i = 1, 2, 3; k = 2, 3, 1. (4.2c)

Note that equation (4.2b) involves the standard, quadratic Lagrange-type interpolation

functions for a 6-node triangle. The hierarchical functions in equation (4.2c) add higher-

order cubic terms.

4.4 Constrained Interpolation Functions of MIN6

The 4 shear constraints in equation (4.1) can be written in terms of the nodal

degrees-of-freedom based on the interpolation functions of MIN6’s virgin element. From

equation (3.11b) with p = 2,

γ xz xx, 0=



γ yz yy, 0=

NT Ni{ }= NaT Naj{ }=

Ni ξi 2ξi 1–( ) = Ni 3+ 4ξiξk =

Nai ξiξk 2ξi 1–( ) = Na4 ξ1ξ2ξ3 =

49

(4.3a)

in which

i = 1, 2,...6 (4.3b)

Note that the interpolation functions are defined in terms of the area coordinates rather

than in terms of x and y. Therefore, the derivatives needed in equation (4.3a) are not imme-

diately available. The Jacobian matrix that relates derivations in between area coordinates

and Cartesian coordinates is needed. If the element is distorted, the Jacobian matrix is not

a constant, and the resulting constrained interpolation functions would not be polynomials

because of the inverse in equation (3.13). In addition, the nodeless degrees-of-freedom a

would not be constant. Hence, an undistorted triangle is assumed in the derivation of the

interpolation functions. As a result, the linear relation between Cartesian and area coordi-

nates that results from an undistorted triangle is valid:

(4.4)

Because of this assumption, it is anticipated that MIN6’s performance may be sensitive to

distortion.

x3

3

∂∂

0x

2

2

∂∂

3x

2y∂

3

∂∂

x2

2

∂∂

2x y∂

2

∂∂

3x y

2∂

3

∂∂

2x y∂

2

∂∂

y2

2

∂∂

y3

3

∂∂

y2

2

∂∂

0

N 0 0 Na

0 N 0 00 0 N 0

w

θθθθx

θθθθy

a

0=

w wi{ }= θθθθx θxi{ }= θθθθy θyi{ }= a ai{ }=

ξi

ξi 1

2A------- ci bix aiy+ +( )=

50

The constrained interpolation functions in terms of the area-parametric coordinates,

Nc, can be obtained by using equations (3.11b) to (3.15a). The result is

i =1, 2,...6 (4.5a)

in which

(4.5b)

(4.5c)

(4.5d)

with i = 1, 2, 3; j = 2, 3, 1; k = 3, 1, 2

It is readily verified that the conditions for the rigid body motions

(4.6)

are satisfied.

4.5 Kirchhoff Constraints of MIN6

In the thin limit, the Kirchhoff constraint of zero shear strains constrains the nodal

degrees-of-freedom so that the displacement field is consistent with zero shear strain.

These constraints are derived explicitly in this section.

4.5.1 Kirchhoff Element Constaints

From equations (3.23a) and (3.23b) when p = 2, the shear strains and have

the form

NT Ni{ }= LT Li{ }= MT Mi{ }=

Ni ξi 2ξi 1–( ) = Ni 3+ 4ξiξk =

Li Ni13--- bkξ j b jξk–( )= Li 3+ Ni 3+

13--- biξk bk ξi

12---–

–=

Mi N– i13--- akξ j a jξk–( )= Mi 3+ N– i 3+

13--- aiξk ak ξi

12---–

–=

Nii 1=

6

∑ 1≡ Lii 1=

6

∑ 0≡ Mii 1=

3

∑ 0≡

γ xz γ yz

51

(4.7a)

From equation (3.24),

(4.7b)

the relations are

, (4.7c)

In the thin plate regime ( ), the Kirchhoff constraints

and (4.8a)

are enforced over the entire element domain. Therefore, the following 8 Kirchhoff con-

straints must be enforced

(4.8b)

Also, equation (4.8b) can be directly obtained from the general equation (3.27) when p =

2. Note that these constraints can be written in terms of nodal DOFs to obtain nodal con-

straints. However, a somewhat ‘cleaner’ formulation of these constraints is presented next.

4.5.2 Kirchhoff ‘Edge’ Constraints

Consider the element in Figure 4.2. Along edge ij, the transverse displacement, w(s),

is interpolated with a cubic function. Similarly, the normal rotation, , is a quadratic

function. Both are expressed with respect to the edge coordinate s. The ‘virgin’ displace-

ment fields can be written as

γ xz A00 A01x A11y A12xy A22y2+ + + +=

γ yz B0 B01x B11y B02x2 B12xy+ + + +=



A12 B12+ 0= A22 B02+ 0=

L t⁄ ∞→


A00 A01 A11 A12 B12–=( ) A22 B02–=( ) B00 B01 B11, ,, 0→, , , ,

θn s( )

52

(4.9a)

(4.9b)

in which

(4.10a)

(4.10b)

(4.10c)

(4.10d)

where Ni, Nj and Ni+3 are the Lagrange interpolation functions; and Na is a higher-order

hierarchical function. The nodeless degree-of-freedom associated with the hierarchical

term is a.

w s( ) Niwi N jw j Ni 3+ wi 3+ Naa+ + +=

θn s( ) Niθni N jθnj Ni 3+ θni 3++ +=

Ni

2 s lij 2⁄–( ) s lij–( )

lij2

----------------------------------------------=

N j

2 s lij 2⁄–( )s

lij2

------------------------------=

Ni 3+

4s s lij–( )

lij2

-----------------------–=

Na s s lij–( ) s lij 2⁄–( )=

53

Figure 4.2 Nodal configuration for MIN6’s edge constaints

The shear constraint along any line on the MIN6 element is

(4.11a)

or

(4.11b)

Substitution of equations (4.9a) and (4.9b) into (4.11b), results in

(4.11c)

x

y

j

k

sn

lij

lki

ljk

z i

wi

wj

θni

θnj

wi+3

θni+3

i+3

j+3k+3

γ sz,ss 0≡

γ sz,ss w,sss θn,ss+ 0= =

s3

3

∂

∂ Ni

s3

3

∂

∂ N j

s3

3

∂

∂ Ni 3+

s2

2

∂

∂ Ni

s2

2

∂

∂ N j

s2

2

∂

∂ Ni 3+

s3

3

∂

∂ Na

wi

w j

wi 3+

θni

θnj

θni 3+

a

0=

54

or

(4.11d)

in which lij is the length of edge ij. From equation (4.11d), a must be

(4.12)

The displacement w(s) can now be expressed in terms of nodal DOFs:

(4.13)

Imposing the Kirchhoff constraint

(4.14)

leads to two Kirchhoff edge constraints. When this is done for each edge, six Kirchhoff

edge constraints result:

(4.15a)

(4.15b)

where i = 1, 2, 3; j = 2, 3, 1

From equation (4.8b) there are 8 Kirchhoff constraints in the thin limit, so two more

0 0 04

lij2

---- 4

lij2

---- 8

lij2

----– 6

wi

w j

wi 3+

θni

θnj

θni 3+

a

0=

a2

3lij2

-------- θ– ni θnj– 2θni 3++( )=

w s( ) Niwi N jw j Ni 3+ wi 3+ Na2

3lij2

-------- θ– ni θnj– 2θni 3++( )+ + +=

γ sz w s, θn+ 0= =

wi 3+12--- wi w j+( ) 1

8---lij θni θnj–( )–=

θni 3+3

2lij-------- wi w j–( ) 1

4--- θni θnj+( )–=

55

equations are needed, which are referred to as ‘interior’ constraints. To find these two con-

straints, first consider Figure 4.3. In the figure, lij, ljk and lki are the lengths of the element

edges (as before) and c is the centroid of the triangle. The line i(j+3) is defined as the line

from i to j+3 with the direction of s; its normal direction is n; and its length is li(j+3). Along

this line there are two actual nodes and an ‘artificial’ node at the center. Each node has two

degrees-of-freedom. The situation is similar for lines j(k+3) and k(i+3).

Figure 4.3 Nodal configuration for MIN6’s ‘interior’ Kirchhoff constraints

As before, the transverse displacement and normal rotation along the interior line

can be written as

(4.16a)

(4.16b)

x

y

z

wi

θni

k+3 θni,j+3

wc

wj+3

θnj+3

s

j

i

i+3

j+3

k

c

lij

lki

ljk

n

s

n

n

w s( ) Niwi N j 3+ w j 3+ Ncwc Naa+ + +=

θn s( ) Niθni N j 3+ θnj 3+ Ncθni j 3+( )+ +=

56

in which

(4.17a)

(4.17b)

(4.17c)

(4.17d)

Substitution of equations (4.16a) and (4.16b) into (4.11b), results in

(4.17e)

As before, this equation can be solved for a and the displacement w(s) can then be

expressed in terms of nodal DOFs. Imposing the Kirchhoff constraint

(4.18)

results in two Kirchhoff constraints from the line i(j+3). These constraints include nodal

DOFs at the center, wc and . Similarly, along the line j(k+3) and the line k(i+3) we

have four more equations with nodal DOFs and . These four center

Ni3

li j 3+( )2

--------------- s13---li j 3+( )–

s li j 3+( )–( )=

N j 3+3

li j 3+( )2

--------------- s13---li j 3+( )–

s=

Nc9

2li j 3+( )2

------------------- s li j 3+( )–( )s=

Na s13---li j 3+( )–

s li j 3+( )–( )s=

s3

3

∂

∂ Ni

s3

3

∂

∂ N j 3+

s3

3

∂

∂ Nc

s2

2

∂

∂ Ni

s2

2

∂

∂ N j 3+

s2

2

∂

∂ Nc

s3

3

∂

∂ Na

wi

w j 3+

wc

θni

θnj 3+

θni j 3+( )

a

0=

γ sz w s, θn+ 0= =

θni j 3+( )

θnj k 3+( ) θnk i 3+( )

57

DOFs (wc, , and ) can be eliminated by the 6 equations, and

then the ‘interior’ Kirchhoff constraints can be obtained as

i = 1, 2, 3; j = 2, 3, 1; k = 3, 1, 2. (4.19)

Only two equations above are independent (any two can be used).

The 8 Kirchhoff constraints of MIN6 are given explicitly in equations (4.15a),

(4.15b) and (4.19).

Figure 4.4 MIN6 cross-diagonal pattern

When 4 MIN6 elements are placed in a cross-diagonal pattern, as depicted in Figure

4.4, one would expect to have 24 Kirchhoff constraints (2 per edge plus 2 per element).

However, it can be shown that there are only 23 independent constraints. This fact explains

the superior performance of MIN6 when used in a cross-diagonal meshing scheme, as

compared to the performance in other meshing schemes. Conceptually, one obtains an

additional ‘free’ degree-of-freedom in each pattern by the elimination of one constraint.

(The same is true for MIN3.) Hence, this cross-diagonal meshing strategy is used for all

test problems shown in Chapter 5.

θni j 3+( ) θnj k 3+( ) θnk i 3+( )

7 wi wk–( ) 10 w j 3+ wi 3+–( ) 2l j 3 i,+ θni θnj 3+–( ) 2li 3 k,+ θni 3+ θnk–( )++ + 0=

1

2

3

4

5

6

7

8

910

1112

13

58

CHAPTER 5

NUMERICAL RESULTS FOR MIN6

5.1 General Comments

In this chapter the numerical performance of MIN6 for a number of test problems is

discussed in detail. An important factor in the performance of MIN3, especially for coarse

meshes, is the element-appropriate shear relaxation. In this chapter we discuss first what

the optimal value of the constant Cs in equation (2.28b) should be for MIN6 and if the per-

formance of MIN6 depends on the element-appropriate shear relaxation. The subsequent

analyses will use this value in testing the performance of MIN6 on both isotropic and

orthotropic plates.

Subsequent to the evaluation of Cs, results for several numerical examples demon-

strate the element behavior of MIN6. The tests are carried out for various types of plates

and different meshes. The test problems include isotropic, thin and thick square plates, and

a moderately-thick circular plate. Also, a twisted ribbon test is used to evaluate the perfor-

mance for different aspect ratios. Various loadings and boundary conditions are consid-

ered. An example using a Gr/Ep unidirectional (orthotropic) square plate under a

distributed sine load is included. These examples are used to test the element for robust-

ness and accuracy. Results are compared with MIN3 and SHELL93 in ANSYS [23]

5.2 Shear Relaxation Coefficient

To determine the appropriate value of Cs, the same approach is used that was used

59

for MIN3 [3]. A thin square plate (L/t = 100) subjected to a center concentrated load is

analyzed using the three meshes in Figure 5.1. Both clamped and simply supported bound-

ary conditions are considered. The support conditions for one quadrant of the simply-sup-

ported plate are

(5.1a)

The support conditions for one quadrant of the clamped plate are

(5.1b)

The solution is determined for a large range of Cs values. For each case, the percent

error in the strain energy is determined. For a single concentrated load, the error is simply

the error in the center displacement

% (5.2)

where is the finite element solution, and is the exact displacement. For a thin

simply supported plate = , and for a thin clamped plate =

, where P is the center point load, , and E and ν are

Young’s modulus and Poisson’s ratio, respectively [8].

For each boundary condition, a linear least squares fit of the error in energy as a

function of Cs is determined. Values of Cs for the two boundary conditions are chosen cor-

responding to zero error in the energy according to the least squares fit. The results of this

procedure are shown in Figure 5.2. Based on these results, MIN6 with the value of Cs = 0,

1/10, and 1/9 was used for the test problems described in the next section. A comparison

w 0 y,( ) θx 0 y,( ) θy x 0,( ) w x 0,( ) 0= = = =

w 0 y,( ) θx 0 y,( ) θy 0 y,( ) 0w x 0,( ) θx x 0,( ) θy x 0,( ) 0

= = == = =

Errorw fe

wex-------- 1–

100×=

w fe wex

wex 0.011603PL

2

D--------- wex

0.005595PL

2

D--------- D

Et3

12 1 ν2–( )-------------------------=

60

of the results presented there shows that the performance of MIN6 with Cs = 0 (no shear

relaxation) is superior over the entire range of problems.

Figure 5.1 Meshes of one quadrant of a thin, symmetric, square plate with a concentrated center load

y

x 2x2 4x4 8x8

L/2

L/2

c

c

-6

-4

-2

0

2

4

6

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

2x24x48x8

Err

or in

Ene

rgy

(%)

Cs

Meshes

Least Square Fit

Cs = 0.18

(a) Simply Supported Plate w/ Center Load

61

Figure 5.2 Cs for MIN6

5.3 Performance for Thin Plates

As the span to thickness ratio, L/t, increases, shear deformation typically becomes

less significant. In the limit, as , solutions for Reissner/Mindlin theory should

coincide with solutions for Kirchhoff theory. Theoretically, MIN6 should perform ‘lock-

ing-free’ in the thin limit, because of the anisoparametric interpolation functions. There-

fore, it is interesting to examine the element’s behavior as increases. For comparison,

a 6-node Lagrange, displacement-based, isoparametric Reissner/Mindlin plate element

with full integration is implemented, named ISOMIN6, which has the same nodal configu-

ration as that of MIN6.

The analyses involved square, simply supported and clamped plates subjected to a

-6

-4

-2

0

2

4

6

0 0.05 0.1 0.15 0.2 0.25

2x24x48x8

Err

or in

Ene

rgy

(%)

Cs

Meshes

Least Square Fit

Cs = 0.11

(b) Clamped Plate w/ Center Load

L t⁄ ∞→

L t⁄

62

concentrated center load and a uniform distributed load. The L/t ratio was varied from 10

to 1000. For each case, the three meshes shown in Figure 5.1 were used. Nondimensional

center displacements are plotted for MIN6 and ISOMIN6 in Figures 5.3 to 5.8. The results

for MIN6 show that it neither locks nor is excessively stiff, even for thin plates and the

coarse 2x2 meshes. However, ISOMIN6 results with the 2x2 coarse mesh show clearly

that the element has substantially poorer performance in the thin plate limit.

In addition, Figures 5.3 to 5.8 show the results for MIN6 with Cs = 0, 1/10, 1/9.

These results indicate that the results are not sensitive to Cs. Specifically, the results show

that shear relaxation is not needed with the higher-order members of the MIN-N family,

and Cs should be zero. Hence, all subsequent analyses use Cs = 0.

Figures 5.7 and 5.8 repeat the results for three meshes and Cs = 0 for an easier com-

parison.

63

Figure 5.3 Performance of MIN6 (Cs = 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (simply supported and center load)

0.6

0.7

0.8

0.9

1

1.1

1.2

10 100 1000

KirchhoffMIN6(C_s=0)MIN6(C_s=1/10)MIN6(C_s=1/9)ISOMIN6

L/t

W/W

thin

(a) 2x2 mesh

0.6

0.7

0.8

0.9

1

1.1

1.2

10 100 1000


L/t

W/W

thin

(b) 4x4 mesh

0.6

0.7

0.8

0.9

1

1.1

1.2

10 100 1000


L/t

W/W

thin

(c) 8x8 mesh

64

Figure 5.4 Performance of MIN6 (Cs = 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (clamped and center load)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

10 100 1000


W/W

thin

L/t(a) 2x2 mesh

0

0.25

0.5

0.75

1

1.25

1.5

10 100 1000

KirchhoffMIN6(C_s=0)MIN6(C_S=1/10)MIN6(C_s=1/9)ISOMIN6

L/t

C C ( es )

W/W

thin

(b) 4x4 mesh

0

0.25

0.5

0.75

1

1.25

1.5

10 100 1000


L/t

W/W

thin

(c) 8x8 mesh

65

Figure 5.5 Performance of MIN6 (Cs = 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (simply supported and uniform load)

0.8

0.85

0.9

0.95

1

1.05

1.1

10 100 1000


L/t

W/W

thin

(a) 2x2 mesh

0.8

0.85

0.9

0.95

1

1.05

1.1

10 100 1000


L/t

W/W

thin

(b) 4x4 mesh

0.8

0.85

0.9

0.95

1

1.05

1.1

10 100 1000


L/t

W/W

thin

(c) 8x8 mesh

66

Figure 5.6 Performance of MIN6 (Cs = 0, 1/10, 1/9) and ISOMIN6 with varying L/t ratios (clamped and uniform load)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

10 100 1000


L/t(a) 2x2 mesh

W/W

thin

0

0.25

0.5

0.75

1

1.25

1.5

10 100 1000


L/t(b) 4x4 mesh

W/W

thin

0

0.25

0.5

0.75

1

1.25

1.5

10 100 1000


L/t

W/W

thin

(c) 8x8 mesh

67

Figure 5.7 Performance of MIN6 (Cs = 0) and ISOMIN6 with varying L/t ratios and three meshes (simply supported)

0.6

0.7

0.8

0.9

1

1.1

1.2

10 100 1000

KirchhoffMIN6(2x2)MIN6(4x4)MIN6(8x8)ISOMIN6(2x2)ISOMIN6(4x4)ISOMIN6(8x8)

W/W

thin

L/t(a) Center load

0.8

0.85

0.9

0.95

1

1.05

1.1

10 100 1000


L/t(b) Uniform load

W/W

thin

68

Figure 5.8 Performance of MIN6 (Cs = 0) and ISOMIN6 with varying L/t ratios and three meshes (clamped)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

10 100 1000


L/t(a) Center load

W/W

thin

0

0.2

0.4

0.6

0.8

1

1.2

1.4

10 100 1000


L/t

W/W

thin

(b) Uniform load

69

5.4 Test Problems

5.4.1 Overview

The problems used to test the accuracy and robustness of MIN6 are

1. A patch test for a thin plate

2. Thin and thick square plates, simply-supported and clamped, subjected to a uniform

load and a concentrated center load

3. Moderately-thick circular plate subjected to a uniform load

4. ‘Twisted ribbon’ problem for thin plates

5. Cantilever plate

6. Orthotropic square plate subjected to a sine load

7. Free vibration of thin and moderately-thick plates

Problems 2 through 4 are the same as those used by Tessler and Hughes [3] to test MIN3.

Problem 5 has been used previously by Batoz and Lardeur [24] to examine the DKT ele-

ment. Problems 6 and 7 are the same as those used for the MIN4T shell element [8].

5.4.2 Patch Test for Thin Plate

The minimum requirement for any plate element is to pass the constant moment

patch test; that is, it must produce the exact solution for a constant moment problem. The

thin plate bending patch test used herein is based on that proposed by MacNeal and Harder

[25]. The mesh for the 6-node element is shown in Figure 5.9; the coordinates of the inte-

rior corner nodes are given in Table 5.1; the other nodes are at the middle of the element

edges.

70

Figure 5.9 Patch test for plate: a = 0.12; b = 0.24; t = 0.001; E = 1.0x106; and ν =0.25

Table 5.1 Location of interior nodes

In the patch test, an exact displacement field is specified. The exact displacements

are imposed at the 8 boundary nodes, and the displacements at the 17 interior nodes are

calculated based on the element stiffness. Three displacement fields are considered.

The first field corresponds to rigid body motion:

Interior node x y

1 0.04 0.02

2 0.18 0.03

3 0.16 0.08

4 0.08 0.08

12

4

b

a

x

y

3

71

(5.3a)

The second field represents a constant moment condition [25]:

(5.3b)

The third field involves a cubic displacement and linear moment variation:

(5.3c)

This test checks the completeness of the cubic shape functions for the transverse displace-

ment. Note that this field represents a Kirchhoff plate bending solution, i.e., zero shear

strain. MIN6 cannot represent this exactly, because the shear force, and hence shear strain,

are not zero. However, because the plate is very thin, it should approximate the solution

very well.

MIN6 produces the exact displacements and moments for all three cases.

5.4.3 Meshing Strategy

As stated previously, the preferred meshing strategy for both MIN6 and MIN3 is a

cross diagonal pattern, as depicted in Figure 5.10. The figure shows a basic quadrilateral

meshed with the same nodal pattern for both MIN6 and MIN3. SHELL93 meshes are

w 10 3– x y 1+ +( )θx w,y 10 3–

θy w,x– 103–

=

= =

= =

w12---10 3– x2 xy y2+ +( )

θx w,y12---10 3– 2y x+( )

θy w,x– 12---– 10 3– 2x y+( )

=

= =

= =

w13---10 3– x3 x2y xy2 y3+ + +( )

θx w,y13---10 3– x2 2xy 3y2+ +( )

θy w,x– 13---– 10 3– 3x2 2xy y2+ +( )

=

= =

= =

72

identical to the MIN6 meshes (Note that this configuration may not be optimal for

SHELL93.) Meshes for all three elements therefore have the same number of nodes and

degrees-of-freedom. Note that the MIN3 meshes have four times the number of elements

as the MIN6 meshes.

5.4.4 Isotropic Thin and Moderately Thick Square Plates (L/t =1000 and 10)

Convergence studies are carried out for a thin square plate with L/t of 1000 and a

moderately thick square plate with L/t of 10. Simply supported and clamped boundary

conditions are used, and the plates are subjected to a uniform load and a center concen-

trated load. The meshes are shown in Figure 5.11. Poisson’s ratio is 0.3 in all cases. The

center deflection and center bending moment were chosen to evaluate the numerical per-

formance. The exact solutions, which are used to nondimensionalize the results, are given

in Table 5.2. Wthin and Mthin [8] are Kirchhoff theory solutions of the center deflection and

bending moment and Wmind are Mindlin theory solutions of the center deflection. P is the

center point load, q is the uniform load, and , where E and ν are Young’s

modulus and Poisson’s ratio, respectively. For comparison, the results of MIN3 and

SHELL93 are also presented in the figures.

Convergence plots are shown in Figures 5.12 to 5.14. The results show that MIN6 is

as accurate as MIN3 and in some cases more accurate. Both elements are more accurate

than SHELL93. All elements, however, illustrate convergent behavior.

DEt3

12 1 ν2–( )-------------------------=

73

Figure 5.10 The cross-diagonal meshes of MIN6 and MIN3 for plate problems

Figure 5.11 Meshes for one quadrant of doubly-symmetric square plates (dashed lines indicate the MIN3 elements)

Table 5.2 Exact solutions for center deflection and bending moment of a square plate

Boundary Condition and Loading

Square Plate

Wthin Wmind Mthin

Simply supported with center point load

0.011603PL2/D ---- ----

Clamped with center point load

0.005595PL2/D ---- ----

Simply supported with uniform load

0.004066qL4/D 0.004270qL4/D 0.00479qL2

Clamped with uniform load

0.001264qL4/D 0.001500qL4/D 0.00231qL2

13 Nodes

4 MIN6 Elements13 Nodes

16 MIN3 Elements

y

x13 Nodes 41 Nodes 145 Nodes

L/2

L/2

c

c

74

Figure 5.12 Convergence of center deflection for a thin square plate with center load

0.9

0.95

1

1.05

MIN6MIN3SHELL93

20 60 100

W/W

thin

Number of Nodes

(a) Simply supported

0

0.5

1

1.5

MIN6MIN3SHELL93

20 60 100

W/W

thin

Number of Nodes

(a) Clamped supported

75

Figure 5.13 Convergence of center deflection for thin square plate with uniform load

0.94

0.96

0.98

1

1.02

1.04

MIN6MIN3SHELL93

20 60 100

W/W

thin

Number of Nodes(a) Simply supported

0.2

0.4

0.6

0.8

1

1.2

MIN6MIN3SHELL93

20 60 100

W/W

thin

Number of Nodes

(b) Clamped plate

76

Figure 5.14 Convergence of center deflection for moderately thick square plate with uni-form load

0.98

0.99

1

1.01

1.02

1.03

MIN6MIN3SHELL93

20 60 100Number of Nodes

(a) Simply supported

W/W

min

d

0.98

0.99

1

1.01

1.02

1.03

1.04

MIN6MIN3SHELL93

20 60 100

W/W

min

d

Number of Nodes

(b) Clamped plate

77

5.4.5 Thin Circular Plates (2R/t = 100)

A simply-supported, thin circular plate (2R/t = 100) is used to demonstrate the per-

formance of MIN6 when nonrectangular 4-element assemblies are used. The three meshes

shown in Figures 5.15 and 5.16 are used. MIN6 and SHELL93 have the same meshes and

MIN3’s meshes are shown in Figure 5.16. As anticipated, MIN6 is sensitive to element

distortion, and its performance degrades substantially. Therefore, it is recommended to use

it only undistorted. Therefore, for MIN6 meshes, only the element’s vertex nodes are

placed on the circular boundary. The transverse displacement of the nodes on the circular

plate boundary are constrained. Poisson’s ratio is 0.3. Center deflection results, as a ratio

of the exact solutions in Table 5.3, are shown in Figure 5.17. MIN6’s performance is

excellent for deflection in the thin regime. The element competes well with MIN3 and

SHELL93.

Table 5.3 Exact solutions for center deflection and bending moment of a circular plate

Boundary Condition and Loading Circular Plate

Wthin Mthin

Simply supported with center point load 0.05051PR2/D ----

Simply supported with uniform load 0.06370qR4/D 0.20625qR2

78

Figure 5.15 MIN6 and SHELL93 meshes for 1/4 thin circular plate

Figure 5.16 MIN3 meshes for 1/4 thin circular plate

Rc9 nodes 31 nodes 109 nodes

c

10 nodes 31 nodes 109 nodes

R c

c

79

Figure 5.17 Convergence of center deflection for thin circular plate (2R/t =100)

0.95

1

1.05

1.1

1.15

MIN6MIN3SHELL93

10 100

W/W

thin

Number of Nodes(a) Center load

0.95

1

1.05

1.1

1.15

MIN6MIN3SHELL93

10 100

W/W

thin

Number of Nodes(b) Uniform load

80

5.4.6 Twisted Ribbon Tests

A cantilevered, thin rectangular plate subject to a twisting moment at the free end is

considered as a strict test for plate bending elements with large aspect ratios. The prob-

lems are illustrated in Figure 5.18. The two meshes for MIN3 and MIN6 are shown in Fig-

ures 5.19 and 5.20. The computed tip deflections are compared with the benchmark

solutions [26] in Figures 5.21 and 5.22. In the comparison, MIN6 does not appear to per-

form as well as MIN3 for mesh I. The main reason is that, as shown in Figure 5.19, the

maximum side of element 1 and 2 in the MIN6 mesh has the same length as the entire

plate and three elements are constrained at the fixed boundary. Therefore, the MIN6 mesh

is too constrained, and the displacement is under predicted. The error increases as the

aspect ratio increases. However, for mesh II MIN6 performs very well, as shown in Figure

5.22.

Figure 5.18 Twisted ribbon examples with corner forces, and moments

L

1

1

1L

1

0.5

0.5

Corner forces Corner moments

81

Figure 5.19 Mesh I for twisted ribbon examples

Figure 5.20 Mesh II for twisted ribbon examples

L

MIN6 mesh4 elements with 13 nodes

MIN3 mesh16 elements with 13 nodes

41

231

L

MIN6 mesh8 elements with 23nodes

MIN3 mesh32 elements with 23nodes

1

82

Figure 5.21 Twisted ribbon tests for tip deflection (E = 107, ν = 0.25, t = 0.05)for Mesh I

0

0.005

0.01

0.015

0.02

0.025

0.03

2 4 6 8 10

MIN6MIN3SHELL93Benchmark

Dis

plac

emen

t

L(a) Corner forces

0

0.005

0.01

0.015

0.02

0.025

0.03

2 4 6 8 10

MIN6MIN3SHELL93Benchmark

Dis

plac

emen

t

L(b) Corner moment

83

Figure 5.22 Twisted ribbon tests for tip deflection (E = 107, ν = 0.25, t = 0.05) for Mesh II

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

2 4 6 8 10

MIN6MIN3Benchmark

Dis

plac

emen

t

L(a) Corner forces

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

2 4 6 8 10

MIN6MIN3Benchmark

Dis

plac

emen

t

L(b) Corner moment

84

5.4.7 Cantilever Plates (L/t = 100; L/t = 5)

A long cantilevered rectangular plate with two aspect ratios, L/t = 100 and L/t = 5, is

subjected at the free end to both out-of-plane tip shear force (case 1) and bending moment

(case 2). The geometry and loading conditions are shown in Figure 5.23. The loads for the

finite element analysis is given in Figure 5.24. This problem was used by Batoz and Lar-

deur (Batoz 1989) to test the DKT element. The numerical results with coarse mesh calcu-

lated for the deflection at the free end are given in Table 5.4. The results are normalized

with respect to the exact solutions from beam theory including shear deformation:

Case 1:

, , (5.4a)

Case 2:

, , (5.4b)

Figure 5.23 Cantilever plate: load case 1 and 2; L/t =100; L/t = 5

wexPL

3

3D--------- 1 α

4---+

= α 2k--- t

L---

2= k

56---= υ 0=

wexML

2

2D-----------= D

Et3b

12-----------= υ 0=

Case 1: tip force

L =101

PM

Case 2: tip moment

t

L =101

85

Figure 5.24 Cantilever plate: numerical load models for case 1 and 2

Figure 5.25 Mesh of cantilever plate with MIN6 elements

Table 5.4 Cantilever plate results using MIN6 elements

Even with the coarse mesh shown in Figure 5.25, MIN6 provides excellent results

for the tip displacement and rotation, bending moments and shear forces, as shown in

Table 5.4. For load case 1, MIN6 gives a constant shear force Qx and a linear bending

moment Mx. The displacement and rotation at the free end are exact. Case 2 is a patch test

with constant Mx. It leads to constant moment and zero shear force, and linear rotation θy.

Load case L/t

1 100 1.000 1.000 1.000 1.000

1 5 1.000 1.000 1.000 1.000

2 100 1.000 1.000 1.000 ----

2 5 1.000 1.000 1.000 ----

Case 1: tip force Case 2: tip moment

2P/3

P/6P/6

2M/3 M/6M/6

L

41

231

W max W ex⁄ θymax θyex⁄ Mxmax Mxex⁄ Qxmax Qxex⁄

86

5.4.8 Orthotropic Square Plate (L/t = 30)

A simply supported, orthotropic thin square plate (L/t = 30) is used to test the con-

vergence and accuracy of MIN6 for non-isotropic problems. The meshes are shown in Fig-

ure 5.11. Material properties are Ex = 22.9 x 106; Ey = Ez = 1.39 x 106; Gxy = Gxz = 0.86 x

106; Gyz = 0.468 x 106; νxy = νxz = 0.32 and νyz = 0.49. (These properties correspond to a

Gr/Ep unidirectional composite.) A uniform sine loading, q = sin(πx/L)sin(πy/L), is

applied. The load on the finite element model varies linearly within each triangle for both

MIN6 and MIN3. Convergence of the center deflection, bending moments and shear

forces are shown in Figures 5.26 to 5.29. For comparison, the results of MIN3 and exact

solutions are presented in the figures. MIN6’s results are excellent. For MIN6 and MIN3,

the bending moments and shear forces for the orthotropic square plate problem are deter-

mined by taking the values of the Gauss points from the triangles closest to the corner or

edges.

87

Figure 5.26 Convergence for center deflection of orthotropic square plate (L/t = 30)

Figure 5.27 Convergence of moment Mxy of orthotropic square plate (L/t = 30)

0.9400

0.9500

0.9600

0.9700

0.9800

0.9900

1.000

1.010

MIN6MIN3Exact

13 41 145

W/W

exac

t

Number of Nodes

(a) Center

4.400

4.800

5.200

5.600

6.000

MIN6MIN3Exact

13 41 145

Mxy

at C

orne

r

Number of Nodes(a) M

xy at corner (x=0; y=0)

88

Figure 5.28 Convergence of moment Mx and My of orthotropic square plate (L/t = 30)

-80.00

-75.00

-70.00

-65.00

-60.00

-55.00

MIN6MIN3Exact

13 41 145

Mx a

t Cen

ter

Number of Nodes

(a) Center

-7.000

-6.500

-6.000

-5.500

-5.000

-4.500

-4.000

MIN6MIN3Exact

13 41 145

My a

t Cen

ter

Number of Nodes

(b) Center

89

Figure 5.29 Convergence of shear force Qx and Qy of orthotropic square plate (L/t = 30)

-9.000

-8.500

-8.000

-7.500

-7.000

-6.500

-6.000

MIN6MIN3Exact

13 41 145

Qx a

t Edg

e

Number of Nodes

(a) Edge (x = 0; y = L/2)

-1.250

-1.200

-1.150

-1.100

MIN6MIN3Exact

13 41 145

Qy a

t Cen

ter

of E

dge

Number of Nodes(b) Q

y at edge (x=L/2; y=0)

90

5.4.9 Free Vibration of Thin and Moderately-Thick Plates

The natural frequencies of simply-supported thin (L/t = 104) and moderately thick

(L/t = 10) isotropic square plates have been computed with the 64 element mesh (145

nodes) shown in Figure 5.11. Poisson’s ratio is 0.3. The nondimensional frequencies

(where is the natural frequency for mode m, n and is the

mass density) for MIN6 and MIN3 are compared to the exact, Mindlin solution in Table

5.5. The percent error of the frequencies are shown in Table 5.6 for MIN6 and MIN3. To

compare with the exact solution for the symmetric modes, symmetric boundary conditions

were used. The consistent mass matrix for MIN6 was used and is calculated with full inte-

gration. The results demonstrate the good performance of MIN6.

Table 5.5 Non-dimensional frequencies for simply-supported square plates

Mode Number

Thin Plate (L/t = 104) Thick Plate (L/t = 10)

n m MIN6 MIN3 Mindlin MIN6 MIN3 Mindlin

1 1 5.97490 5.96294 5.97337 5.77045 5.76056 5.76932

1 3 30.24870 29.9370 29.8668 25.90368 25.9352 25.7337

3 3 54.73293 53.1777 53.7602 42.96241 42.8235 42.3832

1 5 84.96310 79.0499 77.6514 59.14034 59.6609 56.8948

3 5 106.73014 101.3926 101.555 73.26188 73.1938 69.8843

λnm ωnmL2 ρ Et2( )⁄= ωnm ρ

λnm

λnm λnm

91

Table 5.6 Percent error (%) in frequencies for simply-supported square plates

5.5 Remarks on MIN6

Based on numerical testing, MIN6 is accurate and robust. For meshes with the same

number of degrees-of-freedom, it is comparable in accuracy to the MIN3 and SHELL93

elements for both isotropic and orthotropic materials.

MIN6 is implemented in the computer program MANOA, Matrix And Numerical-

Oriented Analysis. The user guide for MIN6 is given in Appendix I, and it is also available

‘on-line’ in MANOA. Input data for a representative set of the test problems reported

herein are given in Appendix II.

Mode Number

Thin Plate (L/t = 104) Thick Plate (L/t = 10)

% error in % error in

n m MIN6 MIN3 MIN6 MIN3

1 1 0.0256 -0.1746 0.0196 -0.1518

1 3 1.2787 0.2350 0.6605 0.7830

3 3 1.8094 -1.0835 1.3666 1.0389

1 5 9.4161 1.8010 3.9468 4.8618

3 5 5.0959 -0.1599 4.8331 4.7357

λnm

λnm λnm

92

CHAPTER 6

CONCLUSIONS AND RECOMMENDATIONS

6.1 Conclusions

A general derivation for the interpolation functions of a family of higher-order, tri-

angular anisoparametric plate elements based on Mindlin theory is developed. The trans-

verse displacement is interpolated by a polynomial one order higher than the interpolation

of the rotations. The transverse displacement is coupled with the bending rotations by

enforcing continuous shear constraints along any line in the element. An efficient, higher

order, compatible, fully-integrated, six-node triangular Mindlin plate element (for isotro-

pic and orthotropic material) with neither shear locking nor excessive stiffness in the thin

limit is generated as an example of this approach. The element is implemented in a finite

element program.

Based on this study, the following conclusions are drawn:

1. The family of MIN-N elements possess a fully compatible kinematic field through

the continuous shear constraints enforced along any line in the element including the

edges of the element.

2. As an example element of the MIN-N family, MIN6 displays good accuracy and

robustness.

3. MIN6 has the straightforward formulation and implementation characteristics of

pure, displacement-based elements.

93

4. The numerical studies for the element shear correction factors indicate that there is

no element-appropriate shear relaxation necessary for the stiffness matrix of MIN6.

5. The numerical results show that MIN6 is competitive with SHELL93 and MIN3 ele-

ments and exhibits excellent performance for both static and dynamic analyses in the

linear, elastic regime.

6. The success of the development of a family of higher-order, triangular plate elements

presented herein as MIN-N demonstrates the anisoparametric methodology to be

effective for higher-order Mindlin plate elements.

7. A sensitivity to the element with curved edges for MIN6 is shown in the numerical

experiments.

6.2 Recommendations

It is recommended that the following work be carried out:

1. Investigate methods to decrease the sensitivity of the element to geometric distor-

tion.

2. Extend the members of the MIN-N family for shells.

3. Develop a higher-order smoothing element based on the interpolation functions of

MIN6.

94

APPENDIX I

MIN6 Element User Guide

The following describes the command to use MIN6 in the command-driven com-

puter program MANOA.

6-node, linear, triangular Mindlin plate bending element.Implementation assumes element is in the X-Y plane.

Command Syntax

min6 m=? n=? [q=q_vec] [m=]mat# [e=]E1,E2,E3 [g=]G12,G23,G13 [nu=]nu12,nu23,nu13 & [t=]thickness [mass=]mass [local=]local [C_s=]C_s [n=]nel [nodes=]node1,node2,node3,node4,node5,node6 [mat=]mat & [ print=]print [inc=]inc1,inc2,inc3 [gen=]gen & [inc_2d=]inc1_2d,inc2_2d,inc3_2d & [gen_2d=]gen_2d [inc_el=]inc_el & [pat=]pat [q=]q1,q2,q3,q4,q5,q6 m is the number of different materials n is the number of elements q_vec is the name of a vector in the database for which the ith element is the normal pressure at node i mat# is the material number E1, E2, E3 are the moduli of elasticity G12, G23, G13 are the shear moduli nu12, nu23, nu13 are the Poisson ratios thickness is the element thickness mass is the mass per unit volume C_s is the shear relaxation factor (default = 0.0) nel is the element number node1 thru node6 are node numbers mat is the material number for the element print .ne. 0 -> element results not printed inc1, inc2 inc3 are node increments in a "linear sequence" gen is the number of elements to generate in a sequence inc1_2d, inc2_2d, inc3_2d are node increments between sequences gen_2d is the number of linear sequences to generate inc_el is the element increment between sequences pat is the load pattern number for the normal pressures q1,q2,q3,q4,q5,q6 are normal pressures for nodes (these values override those defined by q_vec, if any.) If the optional identifiers (e.g., e=) are used on an input record, all data on that record must have the correct identifier. Nodes 1 to 3 are the vertex nodes, and are specified counterclockwise. Nodes 4 to 6 are the midnodes.

95

A "linear sequence" of elements can be generated by specifying inc1, inc2, inc3, and gen. In a linear sequence, nodes 1, 2, and 4 are incremented by inc1; nodes 5, 6 are incremented by inc2; and nodes 3 is incremented by inc3. gen is the number of elements to generate, so a sequence will have gen+1 elements. To generate a 2D patch of elements, multiple sequences can be specified; inc1_2d, inc2_2d, and inc3_2d are used to increment the node numbers from one sequence to the next. gen_2d is the number of additional sequences. The element numbers in two successive sequences differ by inc_el (default = numgen+1).

End input with a blank line.

The global coordinate system is the same as local coordinate system for triangle.

For an orthotropic material, the parameters are specified in the global coordinate system.

The normal pressures act in the local z-direction.

On input, created arrays are: .min6_mp(m,14) -> Ei, Gij, nuij, thickness, mass, (unused), C_s, (unused) .min6_el(8,n) -> node1 - node6, material #, print code .min6_q(7,n) -> pat, q1, q2, q3, q4, q5, q6

For stiffness calculation, bending and shear stiffness are calculated with 4 and 7 Gauss points for full integration, respectively. If a nonzero mass density is specified, uniform (gravitational) body forces are applied if the vector gravity(4) has been defined. The 4 components of gravity are: load pattern number, gx, gy, and gz, where gi is the gravitational acceleration in the global i direction. For state calculation, global coordinates and bending stress resultants are stored in .min6_stb(n, 4*8) in the order x, y, z, Mx, My, Mxy, Qx, and Qy. These values are calculated at the 4 Gauss points of the triangle. The element does not calculate the equivalent nodal forces in equilibrium with its stress state, and therefore cannot be used in a nonlinear analysis. The response option has not been implemented. The error estimation option has not been implemented. For state output, the stress resultants in .min6_stb are printed. See Also min3s min5s min4t pmin6 pstate

96

APPENDIX II

MIN6 Element Input Data for Test Problems

• Patch Test for Thin Plate

date

/* Bending plate patch test/* Tests min6 element stiffness/* Tests Out of Plane Stresses/* Use a=0.12; b=0.24; h=0.001/* C_s=~C_s

nodes #=251 4.00000E-02 2.000000E-02 0.00000E+002 1.80000E-01 3.000000E-02 0.00000E+003 1.60000E-01 8.000000E-02 0.00000E+004 8.00000E-02 8.000000E-02 0.00000E+005 0.00000E+00 0.000000E+00 0.00000E+006 2.40000E-01 0.000000E+00 0.00000E+007 2.40000E-01 1.200000E-01 0.00000E+008 0.00000E+00 1.200000E-01 0.00000E+009 1.10000E-01 2.500000E-02 0.00000E+0010 1.70000E-01 5.500000E-02 0.00000E+0011 1.20000E-01 8.000000E-02 0.00000E+0012 6.00000E-02 5.000000E-02 0.00000E+0013 1.20000E-01 0.000000E+00 0.00000E+0014 2.40000E-01 6.000000E-02 0.00000E+0015 1.20000E-01 1.200000E-01 0.00000E+0016 0.00000E+00 6.000000E-02 0.00000E+0017 2.00000E-02 1.000000E-02 0.00000E+0018 2.10000E-01 1.500000E-02 0.00000E+0019 2.00000E-01 1.000000E-01 0.00000E+0020 4.00000E-02 1.000000E-01 0.00000E+0021 1.40000E-01 1.000000E-02 0.00000E+0022 2.10000E-01 7.500000E-02 0.00000E+0023 1.60000E-01 1.000000E-01 0.00000E+0024 2.00000E-02 7.000000E-02 0.00000E+0025 1.30000E-01 4.000000E-02 0.00000E+00

nodef p=1

min6 m=1 n=10 m=1 e=1.0e6,1.0e6,1.0e6 g=4.0e5,4.0e5,4.0e5 nu=.25,.25,.25 t=.001 C_s=0n=1 nodes=3,4,1,11,12,25 mat=1 n=2 nodes=1,2,3,9,10,25 mat=1 n=3 nodes=5,6,1,13,21,17 mat=1 n=4 nodes=1,6,2,21,18,9 mat=1 n=5 nodes=2,7,3,22,19,10 mat=1

97

n=6 nodes=6,7,2,14,22,18 mat=1 n=7 nodes=7,8,4,15,20,23 mat=1 n=8 nodes=4,3,7,11,19,23 mat=1 n=9 nodes=5,1,8,17,24,16 mat=1 n=10 nodes=1,4,8,12,20,24 mat=1

pnodesbcid r=1,1,0,0,0,1,1pmin6

filein graphgset symbol type=0

form_k

/* imposed displ. function/* w(x,y)=(10^-3)*(x+y+1)/* thetax=w,y=10^-3/* thetay=-w,x=-10^-3

imposed_displn=5 dof=3 disp=1.0000E-03 n=5 dof=4 disp=1.0000E-03 n=5 dof=5 disp=-1.0000E-03 n=6 dof=3 disp=1.2400E-03 n=6 dof=4 disp=1.0000E-03 n=6 dof=5 disp=-1.0000E-03 n=7 dof=3 disp=1.3600E-03 n=7 dof=4 disp=1.0000E-03 n=7 dof=5 disp=-1.0000E-03 n=8 dof=3 disp=1.1200E-03 n=8 dof=4 disp=1.0000E-03 n=8 dof=5 disp=-1.0000E-03 n=13 dof=3 disp=1.1200E-03 n=13 dof=4 disp=1.0000E-03 n=13 dof=5 disp=-1.0000E-03 n=14 dof=3 disp=1.3000E-03 n=14 dof=4 disp=1.0000E-03 n=14 dof=5 disp=-1.0000E-03 n=15 dof=3 disp=1.2400E-03 n=15 dof=4 disp=1.0000E-03 n=15 dof=5 disp=-1.0000E-03 n=16 dof=3 disp=1.0600E-03 n=16 dof=4 disp=1.0000E-03 n=16 dof=5 disp=-1.0000E-03

lsolvegclosepdispstatepstate

return

98

date



nodef p=1

min6 m=1 n=10 m=1 e=1.0e6,1.0e6,1.0e6 g=4.0e5,4.0e5,4.0e5 nu=.25,.25,.25 t=0.001 C_s=0n=1 nodes=3,4,1,11,12,25 mat=1 n=2 nodes=1,2,3,9,10,25 mat=1 n=3 nodes=5,6,1,13,21,17 mat=1 n=4 nodes=1,6,2,21,18,9 mat=1 n=5 nodes=2,7,3,22,19,10 mat=1n=6 nodes=6,7,2,14,22,18 mat=1 n=7 nodes=7,8,4,15,20,23 mat=1 n=8 nodes=4,3,7,11,19,23 mat=1 n=9 nodes=5,1,8,17,24,16 mat=1 n=10 nodes=1,4,8,12,20,24 mat=1

pnodes

99

bcid r=1,1,0,0,0,1,1

pmin6


form_k

/* imposed displ. function/* w(x,y)=(10^-3)*(x^2+x*y+y^2)/2/* thetax=w,y=(10^-3)*(y+x/2)/* thetay=-w,x=-(10^-3)*(x+y/2)

imposed_displn=5 dof=3 disp=0.0000E+00 n=5 dof=4 disp=0.0000E+00 n=5 dof=5 disp=0.0000E+00 n=6 dof=3 disp=2.8800E-05 n=6 dof=4 disp=1.2000E-04 n=6 dof=5 disp=-2.4000E-04 n=7 dof=3 disp=5.0400E-05 n=7 dof=4 disp=2.4000E-04 n=7 dof=5 disp=-3.0000E-04 n=8 dof=3 disp=7.2000E-06 n=8 dof=4 disp=1.2000E-04 n=8 dof=5 disp=-6.0000E-05 n=13 dof=3 disp=7.2000E-06 n=13 dof=4 disp=6.0000E-05 n=13 dof=5 disp=-1.2000E-04 n=14 dof=3 disp=3.7800E-05 n=14 dof=4 disp=1.8000E-04 n=14 dof=5 disp=-2.7000E-04 n=15 dof=3 disp=2.1600E-05 n=15 dof=4 disp=1.8000E-04 n=15 dof=5 disp=-1.8000E-04 n=16 dof=3 disp=1.8000E-06 n=16 dof=4 disp=6.0000E-05 n=16 dof=5 disp=-3.0000E-05

lsolvegclose

pdisp

statepstate

return

100

date



nodef p=1

min6 m=1 n=10 m=1 e=1.0e6,1.0e6,1.0e6 g=4.0e5,4.0e5,4.0e5 nu=.25,.25,.25 t=.001 C_s=0n=1 nodes=3,4,1,11,12,25 mat=1 n=2 nodes=1,2,3,9,10,25 mat=1 n=3 nodes=5,6,1,13,21,17 mat=1 n=4 nodes=1,6,2,21,18,9 mat=1 n=5 nodes=2,7,3,22,19,10 mat=1n=6 nodes=6,7,2,14,22,18 mat=1 n=7 nodes=7,8,4,15,20,23 mat=1 n=8 nodes=4,3,7,11,19,23 mat=1 n=9 nodes=5,1,8,17,24,16 mat=1 n=10 nodes=1,4,8,12,20,24 mat=1

pnodes

101

bcid r=1,1,0,0,0,1,1

pmin6


form_k

/* imposed displ. function/* w(x,y)=(10^-3)*(x^3+y*x^2+x*y^2+y^3)/3/* thetax=w,y=(10^-3)*(x^2+2*x*y+3*y^2)/3/* thetay=-w,x=-(10^-3)*(3*x^2+2*x*y+y^2)/3

imposed_displn=5 dof=3 disp=0.0000E+00 n=5 dof=4 disp=0.0000E+00 n=5 dof=5 disp=0.0000E+00 n=6 dof=3 disp=4.6080E-06 n=6 dof=4 disp=1.9200E-05 n=6 dof=5 disp=-5.7600E-05 n=7 dof=3 disp=8.6400E-06 n=7 dof=4 disp=5.2800E-05 n=7 dof=5 disp=-8.1600E-05 n=8 dof=3 disp=5.7600E-07 n=8 dof=4 disp=1.4400E-05 n=8 dof=5 disp=-4.8000E-06 n=13 dof=3 disp=5.7600E-07 n=13 dof=4 disp=4.8000E-06 n=13 dof=5 disp=-1.4400E-05 n=14 dof=3 disp=0.00000612 n=14 dof=4 disp=3.2400E-05 n=14 dof=5 disp=-6.8400E-05 n=15 dof=3 disp=2.3040000000E-06 n=15 dof=4 disp=2.8800000000E-05n=15 dof=5 disp=-2.8800000000E-05 n=16 dof=3 disp=0.000000072 n=16 dof=4 disp=3.6000E-06 n=16 dof=5 disp=-1.2000E-06

lsolvegclose

pdisp

statepstate

return

102

• Isotropic Square Plates

date/* Simply supported square plate with point load - 1/4 plate modeled/* Tests min6 element (16 element)/* Tests Displacement/* w(max)= c0*p*L^2/D; p=point load; a=plate width; /* D=E*h^3/12*(1-v^2); c0=0.011603/* Use L=8; p=1; t=0.008, mesh=4x4(midnode(2,2)) square,

nodes #=41 1 0.00000E+00 0.00000E+00 0.00000E+00 2 1.00000E+00 0.00000E+00 0.00000E+00 3 2.00000E+00 0.00000E+00 0.00000E+00 4 3.00000E+00 0.00000E+00 0.00000E+00 5 4.00000E+00 0.00000E+00 0.00000E+00 6 0.00000E+00 1.00000E+00 0.00000E+00 7 1.00000E+00 1.00000E+00 0.00000E+00 8 2.00000E+00 1.00000E+00 0.00000E+00 9 3.00000E+00 1.00000E+00 0.00000E+00 10 4.00000E+00 1.00000E+00 0.00000E+00 11 0.00000E+00 2.00000E+00 0.00000E+00 12 1.00000E+00 2.00000E+00 0.00000E+00 13 2.00000E+00 2.00000E+00 0.00000E+00 14 3.00000E+00 2.00000E+00 0.00000E+00 15 4.00000E+00 2.00000E+00 0.00000E+00 16 0.00000E+00 3.00000E+00 0.00000E+00 17 1.00000E+00 3.00000E+00 0.00000E+00 18 2.00000E+00 3.00000E+00 0.00000E+00 19 3.00000E+00 3.00000E+00 0.00000E+00 20 4.00000E+00 3.00000E+00 0.00000E+00 21 0.00000E+00 4.00000E+00 0.00000E+00 22 1.00000E+00 4.00000E+00 0.00000E+00 23 2.00000E+00 4.00000E+00 0.00000E+00 24 3.00000E+00 4.00000E+00 0.00000E+00 25 4.00000E+00 4.00000E+00 0.00000E+00 26 0.50000E+00 0.50000E+00 0.00000E+00 27 1.50000E+00 0.50000E+00 0.00000E+00 28 2.50000E+00 0.50000E+00 0.00000E+00 29 3.50000E+00 0.50000E+00 0.00000E+00 30 0.50000E+00 1.50000E+00 0.00000E+00 31 1.50000E+00 1.50000E+00 0.00000E+00 32 2.50000E+00 1.50000E+00 0.00000E+00 33 3.50000E+00 1.50000E+00 0.00000E+00 34 0.50000E+00 2.50000E+00 0.00000E+00 35 1.50000E+00 2.50000E+00 0.00000E+00 36 2.50000E+00 2.50000E+00 0.00000E+00 37 3.50000E+00 2.50000E+00 0.00000E+00 38 0.50000E+00 3.50000E+00 0.00000E+00 39 1.50000E+00 3.50000E+00 0.00000E+00 40 2.50000E+00 3.50000E+00 0.00000E+00 41 3.50000E+00 3.50000E+00 0.00000E+00

nodef p=1 25 1 0 0 0.25 0 0 0

103

min6 m=1 n=16 m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=~t C_s=~C_sn=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 0 1 1 13 1 1 1 0 1 1 14 1 1 1 0 1 1 15 1 1 1 0 1 1 16 1 1 1 1 0 1 111 1 1 1 1 0 1 116 1 1 1 1 0 1 121 1 1 1 1 0 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6


form_klsolve k=1gclosepeqnspdisp

/*state/*pstate

nodal_disp dis nodes=25,25cp dis u_calc r=1 c=3 m=1

p u_calcreturn

104

date

/* Simply supported square plate with uniform load - 1/4 plate modeled/* Tests min6 element (16 elements)/* Tests Displacement/* w(max)= c0*q*L^4/D; q=uniform load; a=plate width; /* D=E*h^3/12*(1-v^2); c0=0.004066/* Use L=8; q=1; t=0.008, mesh=2x2(midnode(2,2)) square

zero pressure r=41 v=1


min6 m=1 n=16 q=pressure

105

m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=~t C_s=~C_sn=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 0 1 1 13 1 1 1 0 1 1 14 1 1 1 0 1 1 15 1 1 1 0 1 1 16 1 1 1 1 0 1 111 1 1 1 1 0 1 116 1 1 1 1 0 1 121 1 1 1 1 0 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6



/*state/*pstate


p u_calcreturn

106

date

/* Clamped square plate with point load - 1/4 plate modeled/* Tests min6 element (16 elements)/* Tests Displacement/* w(max)= c0*p*L^2/D; p=point load; a=plate width; /* D=E*h^3/12*(1-v^2); c0=0.005595/* Use L=8; p=1; t=0.008, mesh=4x4(midnode(2,2)) square,


nodef p=1 25 1 0 0 0.25 0 0 0

107

min6 m=1 n=16 m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=~t C_s=~C_sn=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 1 1 1 13 1 1 1 1 1 1 14 1 1 1 1 1 1 15 1 1 1 1 1 1 16 1 1 1 1 1 1 111 1 1 1 1 1 1 116 1 1 1 1 1 1 121 1 1 1 1 1 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6



/*state/*pstate


p u_calcreturn

108

date

/* clamped thin square plate under uniform load - 1/4 plate modeled/* Tests min6 element (4 elements)/* Tests Displacement/* w(max)= c0*p*L^4/D; q=uniform load; L=plate width; /* D=E*h^3/12*(1-v^2); c0=0.001264/* Use L=8; q=1; t=0.008, mesh=2x2(midnode(2,2)) square



109

min6 m=1 n=16 q=pressurem=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=~t C_s=~C_sn=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 1 1 1 13 1 1 1 1 1 1 14 1 1 1 1 1 1 15 1 1 1 1 1 1 16 1 1 1 1 1 1 111 1 1 1 1 1 1 116 1 1 1 1 1 1 121 1 1 1 1 1 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6



/*state/*pstate


p u_calcreturn

110

• Thin Circular Plate

date

/* Simply supported circular plate with center point load - 1/4 plate modeled/* Tests min6 element (6 node)/* Tests Displacement at the center of the plate/* w(max)= c0*p*L^2/D; p=point load; R=plate radius; /* D=E*h^3/12*(1-v^2); c0=0.05051/* Use 2R=8; p=1; h=0.08, mesh N=2 tri; L/t=100

nodes #=31 1 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00 2 2.00000000000 0.000000000000E+00 0.000000000000E+00 3 1.00000000000 0.000000000000E+00 0.000000000000E+00 4 0.999999999624 1.00000000038 0.000000000000E+00 5 1.49999999981 0.500000000188 0.000000000000E+00 6 0.499999999812 0.500000000188 0.000000000000E+00 7 1.41421356184 1.41421356291 0.000000000000E+00 8 1.70710678092 0.707106781453 0.000000000000E+00 9 1.20710678073 1.20710678164 0.000000000000E+00 10 0.000000000000E+00 2.00000000000 0.000000000000E+00 11 0.707106780920 1.70710678145 0.000000000000E+00 12 0.499999999812 1.50000000019 0.000000000000E+00 13 0.000000000000E+00 1.00000000000 0.000000000000E+00 14 4.00000000000 0.000000000000E+00 0.000000000000E+00 15 3.00000000000 0.000000000000E+00 0.000000000000E+00 16 2.27614237456 0.942809041937 0.000000000000E+00 17 3.13807118728 0.471404520968 0.000000000000E+00 18 2.13807118728 0.471404520968 0.000000000000E+00 19 1.84517796820 1.17851130242 0.000000000000E+00 20 2.82842712368 2.82842712581 0.000000000000E+00 21 3.41421356184 1.41421356291 0.000000000000E+00 22 2.55228474912 1.88561808387 0.000000000000E+00 23 2.12132034276 2.12132034436 0.000000000000E+00 24 0.942809041227 2.27614237527 0.000000000000E+00 25 1.88561808245 2.55228475054 0.000000000000E+00 26 1.17851130153 1.84517796909 0.000000000000E+00 27 0.471404520614 2.13807118764 0.000000000000E+00 28 0.000000000000E+00 4.00000000000 0.000000000000E+00 29 1.41421356184 3.41421356291 0.000000000000E+00 30 0.471404520614 3.13807118764 0.000000000000E+00 31 0.000000000000E+00 3.00000000000 0.000000000000E+00

nodef p=1 1 1 0 0 0.25 0 0 0 pnodef min6 m=1 n=12 m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=.08 C_s=0.0 n=1 nodes=2,4,1,5,6,3 mat=1 pat=1n=2 nodes=7,4,2,9,5,8 mat=1 pat=1n=3 nodes=4,7,10,9,11,12 mat=1 pat=1

111

n=4 nodes=4,10,1,12,13,6 mat=1 pat=1n=5 nodes=16,2,14,18,15,17 mat=1 pat=1n=6 nodes=16,7,2,19,8,18 mat=1 pat=1n=7 nodes=20,16,14,22,17,21 mat=1 pat=1n=8 nodes=7,16,20,19,22,23 mat=1 pat=1n=9 nodes=24,7,20,26,23,25 mat=1 pat=1n=10 nodes=7,24,10,26,27,11 mat=1 pat=1 n=11 nodes=24,20,28,25,29,30 mat=1 pat=1n=12 nodes=10,24,28,27,30,31 mat=1 pat=1

bcid r=1,1,0,0,0,1,11 1 1 0 1 1 1 114 1 1 1 1 0 1 12 1 1 0 1 0 1 13 1 1 0 1 0 1 115 1 1 0 1 0 1 121 1 1 0 0 0 1 120 1 1 1 0 0 1 129 1 1 0 0 0 1 113 1 1 0 0 1 1 110 1 1 0 0 1 1 131 1 1 0 0 1 1 128 1 1 1 0 1 1 1

pmin6



statepstate

nodal_disp dis nodes=1,1cp dis u_calc r=1 c=3 m=1p u_calc

return

112

date

/* Simply supported circular plate with uniform load - 1/4 plate modeled/* Tests min6 element (6 node only)/* Tests Displacement/* w(max)= c0*q*L^4/D; q=uniform load; a=plate width; /* D=E*h^3/12*(1-v^2); c0=0.06370/* Use 2R=8; q=1; h=0.08, mesh N=1; L/t=100


nodes #=31 1 0.000000000000E+00 0.000000000000E+00 0.000000000000E+00 2 2.00000000000 0.000000000000E+00 0.000000000000E+00 3 1.00000000000 0.000000000000E+00 0.000000000000E+00 4 0.999999999624 1.00000000038 0.000000000000E+00 5 1.49999999981 0.500000000188 0.000000000000E+00 6 0.499999999812 0.500000000188 0.000000000000E+00 7 1.41421356184 1.41421356291 0.000000000000E+00 8 1.70710678092 0.707106781453 0.000000000000E+00 9 1.20710678073 1.20710678164 0.000000000000E+00 10 0.000000000000E+00 2.00000000000 0.000000000000E+00 11 0.707106780920 1.70710678145 0.000000000000E+00 12 0.499999999812 1.50000000019 0.000000000000E+00 13 0.000000000000E+00 1.00000000000 0.000000000000E+00 14 4.00000000000 0.000000000000E+00 0.000000000000E+00 15 3.00000000000 0.000000000000E+00 0.000000000000E+00 16 2.27614237456 0.942809041937 0.000000000000E+00 17 3.13807118728 0.471404520968 0.000000000000E+00 18 2.13807118728 0.471404520968 0.000000000000E+00 19 1.84517796820 1.17851130242 0.000000000000E+00 20 2.82842712368 2.82842712581 0.000000000000E+00 21 3.41421356184 1.41421356291 0.000000000000E+00 22 2.55228474912 1.88561808387 0.000000000000E+00 23 2.12132034276 2.12132034436 0.000000000000E+00 24 0.942809041227 2.27614237527 0.000000000000E+00 25 1.88561808245 2.55228475054 0.000000000000E+00 26 1.17851130153 1.84517796909 0.000000000000E+00 27 0.471404520614 2.13807118764 0.000000000000E+00 28 0.000000000000E+00 4.00000000000 0.000000000000E+00 29 1.41421356184 3.41421356291 0.000000000000E+00 30 0.471404520614 3.13807118764 0.000000000000E+00 31 0.000000000000E+00 3.00000000000 0.000000000000E+00

min6 m=1 n=12 q=pressure m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=.08 C_s=0.0 n=1 nodes=2,4,1,5,6,3 mat=1 pat=1n=2 nodes=7,4,2,9,5,8 mat=1 pat=1n=3 nodes=4,7,10,9,11,12 mat=1 pat=1n=4 nodes=4,10,1,12,13,6 mat=1 pat=1n=5 nodes=16,2,14,18,15,17 mat=1 pat=1n=6 nodes=16,7,2,19,8,18 mat=1 pat=1n=7 nodes=20,16,14,22,17,21 mat=1 pat=1n=8 nodes=7,16,20,19,22,23 mat=1 pat=1n=9 nodes=24,7,20,26,23,25 mat=1 pat=1

113

n=10 nodes=7,24,10,26,27,11 mat=1 pat=1 n=11 nodes=24,20,28,25,29,30 mat=1 pat=1n=12 nodes=10,24,28,27,30,31 mat=1 pat=1

bcid r=1,1,0,0,0,1,11 1 1 0 1 1 1 114 1 1 1 1 0 1 12 1 1 0 1 0 1 13 1 1 0 1 0 1 115 1 1 0 1 0 1 121 1 1 0 0 0 1 120 1 1 1 0 0 1 129 1 1 0 0 0 1 113 1 1 0 0 1 1 110 1 1 0 0 1 1 131 1 1 0 0 1 1 128 1 1 1 0 1 1 1

pmin6



statepstate

nodal_disp dis nodes=1,1cp dis u_calc r=1 c=3 m=1p u_calc

return

114

• Twisted Ribbon

date

/*Twisted ribbon test via transverse corner forces (E=10e6, nu=0.25, t=.05)

/* cantilever plate with L=1.0 to 10 with b=1.0 two Cross-Diagonal Mesh - 4 elements

nodes #=23 1 0.0000E+00 0.0000E+00 0.0000E+00 2 2.5000E-01 0.0000E+00 0.0000E+00 3 5.0000E-01 0.0000E+00 0.0000E+00 4 7.5000E-01 0.0000E+00 0.0000E+00 5 1.0000E+00 0.0000E+00 0.0000E+00 6 1.2500E-01 2.5000E-01 0.0000E+00 7 3.7500E-01 2.5000E-01 0.0000E+00 8 6.2500E-01 2.5000E-01 0.0000E+00 9 8.7500E-01 2.5000E-01 0.0000E+00 10 0.0000E+00 5.0000E-01 0.0000E+00 11 2.5000E-01 5.0000E-01 0.0000E+00 12 5.0000E-01 5.0000E-01 0.0000E+00 13 7.5000E-01 5.0000E-01 0.0000E+00 14 1.0000E+00 5.0000E-01 0.0000E+00 15 1.2500E-01 7.5000E-01 0.0000E+00 16 3.7500E-01 7.5000E-01 0.0000E+00 17 6.2500E-01 7.5000E-01 0.0000E+00 18 8.7500E-01 7.5000E-01 0.0000E+00 19 0.0000E+00 1.0000E+00 0.0000E+00 20 2.5000E-01 1.0000E+00 0.0000E+00 21 5.0000E-01 1.0000E+00 0.0000E+00 22 7.5000E-01 1.0000E+00 0.0000E+00 23 1.0000E+00 1.0000E+00 0.0000E+00

nodef p=1 5 1 0 0 -1 23 1 0 0 1 pnodefmin6 m=1 n=8 m=1 e=10e6,10e6,10e6 g=4e6,4e6,4e6 nu=.25,.25,.25 t=.05 C_s=~C_sn=1 nodes=1,3,11,2,7,6 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=3 nodes=21,19,11,20,15,16 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=5 nodes=19,1,11,10,6,15 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=7 nodes=3,21,11,12,16,7 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2 pnodesbcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 110 1 1 1 1 1 1 119 1 1 1 1 1 1 1 pmin6

gopen

115

ginit -mesh_plotgdraw

formkpeqnslsolve k=1pdispstatepstate

/*gclose

nodal_disp dis nodes=23,23cp dis u_calc r=1 n=1 c=3 m=1

return

date

/*Twisted ribbon test via transverse corner moments (E=10e6, nu=0.25, t=.05)

/* cantilever plate with L=1.0 to 10 with b=1.0 two Cross-Diagonal Mesh - 4 elements

nodes #=23 1 0.0000E+00 0.0000E+00 0.0000E+00 2 2.5000E-01 0.0000E+00 0.0000E+00 3 5.0000E-01 0.0000E+00 0.0000E+00 4 7.5000E-01 0.0000E+00 0.0000E+00 5 1.0000E+00 0.0000E+00 0.0000E+00 6 1.2500E-01 2.5000E-01 0.0000E+00 7 3.7500E-01 2.5000E-01 0.0000E+00 8 6.2500E-01 2.5000E-01 0.0000E+00 9 8.7500E-01 2.5000E-01 0.0000E+00 10 0.0000E+00 5.0000E-01 0.0000E+00 11 2.5000E-01 5.0000E-01 0.0000E+00 12 5.0000E-01 5.0000E-01 0.0000E+00 13 7.5000E-01 5.0000E-01 0.0000E+00 14 1.0000E+00 5.0000E-01 0.0000E+00 15 1.2500E-01 7.5000E-01 0.0000E+00 16 3.7500E-01 7.5000E-01 0.0000E+00 17 6.2500E-01 7.5000E-01 0.0000E+00 18 8.7500E-01 7.5000E-01 0.0000E+00 19 0.0000E+00 1.0000E+00 0.0000E+00 20 2.5000E-01 1.0000E+00 0.0000E+00 21 5.0000E-01 1.0000E+00 0.0000E+00 22 7.5000E-01 1.0000E+00 0.0000E+00 23 1.0000E+00 1.0000E+00 0.0000E+00

nodef p=1 5 1 0 0 0 .5 23 1 0 0 0 .5

116

pnodefmin6 m=1 n=8 m=1 e=10e6,10e6,10e6 g=4e6,4e6,4e6 nu=.25,.25,.25 t=.05 C_s=~C_sn=1 nodes=1,3,11,2,7,6 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=3 nodes=21,19,11,20,15,16 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=5 nodes=19,1,11,10,6,15 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2n=7 nodes=3,21,11,12,16,7 mat=1 pat=1 inc=2,2,2 gen=1 inc_el=2 pnodesbcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 110 1 1 1 1 1 1 119 1 1 1 1 1 1 1 pmin6

formkpeqnslsolve k=1pdispstatepstate

nodal_disp dis nodes=23,23cp dis u_calc r=1 n=1 c=3 m=1

return

• Orthotropic Square Plale

date

/* Simply supported square plate with sine load - 1/4 plate modeled/* Min6 element (6-node Mindlin plate element)/* W(max)= -3.5687e-03; q=-sin(pi*x/L) *sin(pi*y/L)/* E11,22,33 =22.9e6,1.39e6,1.39e6 g/* G12,23,31 =0.86e6,0.468e6,0.86e6 /* NU12,23,31=.32,.49,.32/* Use L_tol=30; t=1, mesh=2x2(midnode(7.5,7.5)) square; L/t=30


nodes #=41 1 0.0000E+00 0.0000E+00 0.0000E+00 2 3.7500E+00 0.0000E+00 0.0000E+00 3 7.5000E+00 0.0000E+00 0.0000E+00 4 1.1250E+01 0.0000E+00 0.0000E+00 5 1.5000E+01 0.0000E+00 0.0000E+00 6 0.0000E+00 3.7500E+00 0.0000E+00 7 3.7500E+00 3.7500E+00 0.0000E+00 8 7.5000E+00 3.7500E+00 0.0000E+00

117

9 1.1250E+01 3.7500E+00 0.0000E+00 10 1.5000E+01 3.7500E+00 0.0000E+00 11 0.0000E+00 7.5000E+00 0.0000E+00 12 3.7500E+00 7.5000E+00 0.0000E+00 13 7.5000E+00 7.5000E+00 0.0000E+00 14 1.1250E+01 7.5000E+00 0.0000E+00 15 1.5000E+01 7.5000E+00 0.0000E+00 16 0.0000E+00 1.1250E+01 0.0000E+00 17 3.7500E+00 1.1250E+01 0.0000E+00 18 7.5000E+00 1.1250E+01 0.0000E+00 19 1.1250E+01 1.1250E+01 0.0000E+00 20 1.5000E+01 1.1250E+01 0.0000E+00 21 0.0000E+00 1.5000E+01 0.0000E+00 22 3.7500E+00 1.5000E+01 0.0000E+00 23 7.5000E+00 1.5000E+01 0.0000E+00 24 1.1250E+01 1.5000E+01 0.0000E+00 25 1.5000E+01 1.5000E+01 0.0000E+00 26 1.8750E+00 1.8750E+00 0.0000E+00 27 5.6250E+00 1.8750E+00 0.0000E+00 28 9.3750E+00 1.8750E+00 0.0000E+00 29 1.3125E+01 1.8750E+00 0.0000E+00 30 1.8750E+00 5.6250E+00 0.0000E+00 31 5.6250E+00 5.6250E+00 0.0000E+00 32 9.3750E+00 5.6250E+00 0.0000E+00 33 1.3125E+01 5.6250E+00 0.0000E+00 34 1.8750E+00 9.3750E+00 0.0000E+00 35 5.6250E+00 9.3750E+00 0.0000E+00 36 9.3750E+00 9.3750E+00 0.0000E+00 37 1.3125E+01 9.3750E+00 0.0000E+00 38 1.8750E+00 1.3125E+01 0.0000E+00 39 5.6250E+00 1.3125E+01 0.0000E+00 40 9.3750E+00 1.3125E+01 0.0000E+00 41 1.3125E+01 1.3125E+01 0.0000E+00

/* Define nodal pressurestrans .xyz tmpcp tmp xcor m=1cp tmp ycor c=2 m=1rm tmppi PIscale xcor v=~PIscale xcor v=30 -invscale ycor v=~PIscale ycor v=-30 -invsin xcorsin ycormult_elem xcor ycor pressurerm xcor ycor

min6 m=1 n=16 q=pressurem=1 e=22.9e6,1.39e6,1.39e6 g=0.86e6,0.468e6,0.86e6 nu=.32,.49,.32 t=1 C_s=0 iso=1n=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

118

n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 0 1 1 13 1 1 1 0 1 1 14 1 1 1 0 1 1 15 1 1 1 0 1 1 16 1 1 1 1 0 1 111 1 1 1 1 0 1 116 1 1 1 1 0 1 121 1 1 1 1 0 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6

/*filein graph/*gset symbol type=0


statepstate

/*Displacements at center of plate, x=15, y=15nodal_disp dis nodes=25,25cp dis u_calc r=1 c=3 m=1p u_calc

/*Moments and shear forces at center of plate, x=15, y=15cp .min6_stb mx1 r=8 n=1 c=12 m=1cp .min6_stb mx2 r=8 n=1 c=12 m=1add mx1 mx2 mx1_calcscale mx1_calc v=2 -invp mx1_calc

cp .min6_stb my1 r=8 n=1 c=13 m=1cp .min6_stb my2 r=8 n=1 c=13 m=1add my1 my2 my1_calcscale my1_calc v=2 -invp my1_calc

cp .min6_stb mxy1 r=8 n=1 c=14 m=1cp .min6_stb mxy2 r=8 n=1 c=14 m=1

119

add mxy1 mxy2 mxy1_calcscale mxy1_calc v=2 -invp mxy1_calc

cp .min6_stb qx1 r=8 n=1 c=15 m=1cp .min6_stb qx2 r=8 n=1 c=15 m=1add qx1 qx2 qx1_calcscale qx1_calc v=2 -invp qx1_calc

cp .min6_stb qy1 r=8 n=1 c=16 m=1cp .min6_stb qy2 r=8 n=1 c=16 m=1add qy1 qy2 qy1_calcscale qy1_calc v=2 -invp qy1_calc

/*Moments and forces at center of left edge, x=0, y=15cp .min6_stb mx1 r=11 n=1 c=12 m=1cp .min6_stb mx2 r=11 n=1 c=12 m=1add mx1 mx2 mx2_calcscale mx2_calc v=2 -invp mx2_calc


cp .min6_stb mxy1 r=11 n=1 c=14 m=1cp .min6_stb mxy2 r=11 n=1 c=14 m=1add mxy1 mxy2 mxy2_calcscale mxy2_calc v=2 -invp mxy2_calc




cp .min6_stb my1 r=1 n=1 c=13 m=1

120

cp .min6_stb my2 r=1 n=1 c=13 m=1add my1 my2 my3_calcscale my3_calc v=2 -invp my3_calc









return

121

• Free Vibration

date

/* Simply supported square plate with uniform load - 1/4 plate modeled/* Tests min6 element (16 elements in 1/4 plate, total 64)/* Tests vibration problems/* Use L=8; t=0.0008, mesh=2x2(midnode(2,2)) square,L/t = 10^4



min6 m=1 n=16 q=pressure

122

m=1 e=10.92e6,10.92e6,10.92e6 g=4.2e6,4.2e6,4.2e6 nu=.3,.3,.3 t=0.0008 C_s=0 mass=1n=1 nodes=1,3,7,2,27,26 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=5 nodes=13,11,7,12,30,31 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=9 nodes=11,1,7,6,26,30 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2n=13 nodes=3,13,7,8,31,27 mat=1 pat=1 inc=2,2,2 gen=1 inc_2d=10,8,10 gen_2d=1 inc_el=2

bcid r=1,1,0,0,0,1,11 1 1 1 1 1 1 12 1 1 1 0 1 1 13 1 1 1 0 1 1 14 1 1 1 0 1 1 15 1 1 1 0 1 1 16 1 1 1 1 0 1 111 1 1 1 1 0 1 116 1 1 1 1 0 1 121 1 1 1 1 0 1 110 1 1 0 0 1 1 115 1 1 0 0 1 1 120 1 1 0 0 1 1 122 1 1 0 1 0 1 123 1 1 0 1 0 1 124 1 1 0 1 0 1 125 1 1 0 1 1 1 1

pmin6

form_kptoful .kstr K .kdiag_loc

peqns

form_mptoful .mstr M .kdiag_loc

/*jacobi K M phi lambda

eigval #=11 p .omega2sqrt .omega2 omegap omega

/* calculate nondimensional frequencies

setr L v=80000cp .min6_mp ex r=1 n=1 c=1 m=1sqrt ex ex-2scale omega v=~Lscale omega ex-2 e=1,1 -inv

pf omega row#=1

return

123

REFERENCES

1. R. D. Mindlin, ‘Influence of Rotatory Inertia and Shear on Flexural Motions of Isotro-

pic, Elastic Plates,’ Journal of Applied Mechanics, March, 31-38 (1951).

2. R. C. Averill and J. N. Reddy, ‘Behaviour of Plate Elements Based on the First-Order

Shear Deformation Theory,’ Engineering Computations, 7, 57-74 (1990).

3. A. Tessler and T. J. R. Hughes, ‘A Three-Node Mindlin Plate Element with Improved

Transverse Shear,’ Computer Methods in Applied Mechanics and Engineering, 50,

71-101 (1985).

4. H. R. Riggs, A. Tessler and H. Chu, ‘C1-Continuous Stress Recovery in Finite Ele-

ment Analysis,’ Computer Methods in Applied Mechanics and Engineering, 143(3/4),

299-316 (1997).

5. A. Tessler, H. R. Riggs and S. C. Macy, ‘Application of a Variational Method for

Computing Smooth Stresses, Stress Gradients, and Error Estimation in Finite Element

Analysis,’ in J. R. Whiteman (eds.), The Mathematics of Finite Elements and Applica-

tions, John Wiley & Sons, Ltd., 189-198 (1994).

6. A. Tessler, H. R. Riggs and S. C. Macy, ‘A Variational Method for Finite Element

Stress Recovery and Error Estimation,’ Computer Methods in Applied Mechanics and

Engineering, 111, 369-382 (1994).

124

7. A. Tessler and S. B. Dong, ‘On a Hierarchy of Conforming Timoshenko Beam Ele-

ments,’ Computers & Structures, 14(3-4), 335-344 (1981).

8. J. Liu, H. R. Riggs and A. Tessler, ‘A Four-Node, Shear-Deformable Shell Element

Developed via Explicit Kirchhoff Constraints,’ International Journal for Numerical

Methods in Engineering, 49(8), 1065-1086 (2000).

9. J. L. Batoz, K. J. Bathe and L. W. Ho, ‘A Study of Three-Node Triangular Plate Bend-

ing Elements,’ International Journal for Numerical Methods in Engineering, 15,

1771-1812 (1980).

10. J. L. Batoz, ‘An Explicit Formulation for an Efficient Triangular Plate-Bending Ele-

ment,’ International Journal for Numerical Methods in Engineering, 18, 1077-1089

(1982).

11. R. D. Cook, D. S. Malkus and M. E. Plesha, Concepts and Applications of Finite Ele-

ment Analysis, John Wiley & Sons, New York, 1989.

12. T. J. R. Hughes, R. Taylor and W. Kanolkulchai, ‘A simple and efficient finite element

for plate bending,’ International Journal for Numerical Methods in Engineering, 11,

1529-1543 (1977).

13. T. J. R. Hughes and T. E. Tezduyar, ‘Finite Elements Based Upon Mindlin Plate The-

ory with Particular Reference to the Four-Node Bilinear Isoparametric Element,’

Journal of Applied Mechanics, 48, 587-596 (1981).

125

14. E. D. Pugh, E. Hinton and O. C. Zienkiewicz, ‘A study of quadrilateral plate bending

elements with “reduced” integration,’ International Journal for Numerical Methods

in Engineering, 12, 1059-1078 (1978).

15. R. H. MacNeal, ‘A Simple Quadrilateral Shell Element,’ Computers & Structures, 8,

175-183 (1978).

16. M. A. Crisfield, ‘A four-noded thin-plate bending element using shear constraints - a

modified version of Lyons' element,’ Computer Methods in Applied Mechanics and

Engineering, 38, 93-120 (1983).

17. T. Belytschko, H. Stolarski and N. Carpenter, ‘A C0 Triangular Plate Element with

One-Point Quadrature,’ International Journal for Numerical Methods in Engineering,

20, 787-802 (1984).

18. A. Tessler, ‘A Priori Identification of Shear Locking and Stiffening in Triangular

Mindlin Elements,’ Computer Methods in Applied Mechanics and Engineering,

53(2), 183-200 (1985).

19. A. Tessler and T. J. R. Hughes, ‘An Improved Treatment of Transverse Shear in the

Mindlin-Type Four-Node Quadrilateral Element,’ Computer Methods in Applied

Mechanics and Engineering, 39, 311-335 (1983).

126

20. A. Tessler, ‘A Two-Node Beam Element Including Transverse Shear and Transverse

Normal Deformations,’ International Journal for Numerical Methods in Engineering,

32, 1027-1039 (1991).

21. S. Oral, ‘Anisoparametric Interpolation in Hybrid-Stress Timoshenko Beam Element,’

Journal of Structural Engineering, 117(4), 1070-1078 (1991).

22. S. R. Marur and G. Prathap, ‘Consistency and Correctness Evaluation of Shear

Deformable Anisoparametric Formulations,’ International Journal of Solids and

Structures, 37(5), 701 (2000).

23. ANSYS, ‘ANSYS User's Manual,’ Report v. 5.5, 1998.

24. J. L. Batoz and P. Lardeur, ‘A Discrete Shear Triangular Nine D.O.F. Element for the

Analysis of Thick to Very Thin Plates,’ International Journal for Numerical Methods

in Engineering, 28, 533-560 (1989).

25. R. H. MacNeal and R. L. Harder, ‘A Proposed Standard Set of Problems to Test Finite

Element Accuracy,’ Finite Elements in Analysis and Design, 1, 3-20 (1985).

26. J. Robinson and G. W. Haggenmacher, ‘LORA-An Accurate Four Node Stress Plate

Bending Element,’ International Journal for Numerical Methods in Engineering, 14,

296-306 (1979).

UNIVERSITY OF HAWAIIble, triangular, Reissner/Mindlin shear-deformable plate elements, MIN-N, is...

Documents

Transcript of UNIVERSITY OF HAWAIIble, triangular, Reissner/Mindlin shear-deformable plate elements, MIN-N, is...