nisa verification manual 24april

S P E E D • A C C U R A C Y • V A L U E • T R U S T

VPG/NISA Verification Manual

Disclaimer

Cranes Software, Inc. makes no warranty or representations in connection with NISA/HEAT,NISA/EMAG, NISA/ROTOR, NISA/CIVIL, NISA II, DISPLAY III/IV, NISA-COMPOSITE, ENDURE, NISAOPT, FEAP, NISA/3D-FLUID, DYMES, or any othersoftware program of Cranes Software, Inc. (hereinafter referred to as NISA/DISPLAYSOFTWARE), express or implied, including any implied warranty of merchantability orfitness for a particular purpose. Cranes Software, Inc. makes no warranty and assumes noliability for the NISA/DISPLAY SOFTWARE. Cranes Software, Inc. assumes noresponsibility for the use of the programs or for the accuracy or validity of any resultsobtained from the NISA/DISPLAY SOFTWARE. Cranes Software, Inc. shall not be liablefor loss of profit, loss of business, or other financial loss which may be caused directly orindirectly by the NISA/DISPLAY SOFTWARE, when used for any purpose or use, or due toany defect or deficiency therein.

Any questions relating to the use or interpretation of the SOFTWARE or their operationshould be directed to:

Cranes Software, Inc.1133 E Maple Road, Suite 103Troy, Michigan, 48083, USA

Tel: (248) 689-0077Fax: (248) 689-7479

Notice for U.S. government users only: restricted rights legend.

Use, duplication or disclosure by the Government is subject to restrictions as set forth insubparagraph (c) (1) (ii) of the Rights in Technical Data and Computer Software clause atDFARS 252.227-7013.

Copyright 2005-2008 by the Cranes Software, Inc. 1133 E Maple Road, Suite 103, Troy,Michigan 48083, U.S.A.

All rights reserved. No part of this manual may be reproduced, stored in a retrieval system,or transmitted in any way without the prior authorization of Cranes Software, Inc.

Conten t s

Preface xiii

1 Introduction 1.1

1.1 Highlights of VPG/NISA Capabilities ....................................................................................1.1

1.2 Purpose of this Manual ............................................................................................................1.3

1.3 Verification Problem Format ..................................................................................................1.4

1.4 Problem Classifications ...........................................................................................................1.6

2 Static Analysis Verification Problems 2.1

2.1 Cantilever Beam Subjected to Multiple LoadingConditions ...............................................................................................................................2.1

2.2 Cantilever Beam Under an Off-Centered Load ......................................................................2.6

2.3 A Frame Structure Under General Loading ..........................................................................2.10

2.4 A Large Square Plate with a Circular Hole under Inplane Loading .....................................2.16

2.5 Thick Cylinder under Internal Pressure Load .......................................................................2.21

2.6 Simply Supported Circular Plate ...........................................................................................2.29

2.7 Uniform Thickness Rotating Disc .........................................................................................2.34

2.8 Solid Disc Subjected to Radially Varying Thermal Load .....................................................2.39

2.9 Simply Supported Circular Plate with Annular Line Load and Body Forces .......................2.45

2.10 Hemispherical Shell Under Edge-Moment Loading ...........................................................2.50

2.11 Clamped Square Plate with Central Concentrated Load .....................................................2.55

2.12 Cylindrical Shell Roof under Self-Weight ..........................................................................2.60

2.13 Thin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder) ..........................2.64

2.14 A Clamped Square Orthotropic Plate Under Pressure Loading ..........................................2.68

2.15 Analysis of Nine Layer Cross-Ply Orthotropic Plate ..........................................................2.73

2.16 This page is intentionally blank ..........................................................................................2.79

2.17 Analysis of a Three-layer Sandwich Plate ..........................................................................2.80

2.18 This page is intentionally blank ..........................................................................................2.85

2.19 Propped Cantilever Beam with a Gap .................................................................................2.86

2.20 Beam on a Tensionless Elastic Foundation .........................................................................2.90

2.21 Truss on Misaligned Supports .............................................................................................2.95

iii

2.22 Hollow Circular Cylinder Subjected to Nonaxisymmetric Loading ............................................................................................... 2.100

2.23 Cantilever Elbow Subjected to Multiple Loading Conditions ......................................................................................................................... 2.107

2.24 Cantilever Curved Beam Subjected to Multiple Loading Conditions ........................................................................................................... 2.112

2.25 Demonstration of the Use of Restarts in the Static and Buckling Analysis of the L-shaped Frame ...................................................................................................... 2.117

2.26 Interlaminar Shear Stresses for Eight-layer Antisymmetric Angle-ply Laminated Square Plates ........................................................ 2.125

2.27 Interlaminar Shear Stresses for Cross-Ply Laminated Spherical Shells of Rectangular Planform. ...................................................................................................... 2.133

2.28 Analysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed Load ........................................................................ 2.144

2.29 Analysis of a Simply Supported Three-layer Sandwich Plate Subjected to A Uniform Transverse Load ............................................................................................. 2.148

2.30 Static Analysis of a Symmetric Cross-ply Laminated Simply Supported Plate Subjected to a UniformTransverse Pressure Load ................................................... 2.153

3 Buckling Analysis Verification Problems 3.1

3.1 Buckling of a Uniaxially Compressed Clamped Square Plate ............................................... 3.1

3.2 Buckling of a Beam-Spring Structure ..................................................................................... 3.5

3.3 Buckling of a Cantilever Beam under Compressive Load ..................................................... 3.9

3.4 Buckling Analysis of a Simply Supported Tapered Beam ................................................... 3.13

3.5 Buckling of a Fixed-Free Column under Self Weight and Applied Load ............................ 3.20

4 Eigenvalue Analysis Verification Problems 4.1

4.1 Natural frequencies of a simply supported beam .................................................................... 4.1

4.2 Fundamental frequency of a cantilever beam ......................................................................... 4.6

4.3 Fundamental frequency of a clamped circular plate ............................................................. 4.10

4.4 Natural frequencies of a cantilever beam ............................................................................. 4.14

4.5 Natural frequencies of a simply supported square plate ....................................................... 4.19

4.6 Fundamental frequency of a cantilever triangular plate ....................................................... 4.25

4.7 Natural Frequencies of an Earth Dam ................................................................................... 4.29

4.8 Natural Frequencies of a Square Cantilever Plate ................................................................ 4.36

4.9 Free Vibration Analysis of a Cantilever Beam Using Solid Elements ................................. 4.43

4.10 Natural Frequencies of a Cylindrical Shell with a Rigid Diaphragm ................................. 4.51

iv

4.11 Natural Frequencies of a Simple Torsional System ............................................................4.58

4.12 Natural Frequencies of a Composite Square Plate ..............................................................4.62

4.13 Free Vibration Analysis of a Three-Layer Orthotropic Sandwich Plate .............................4.67

4.14 Natural Frequencies of a Free-Free (Unconstrained) Beam ................................................4.74

4.15 Natural Frequencies of a Free Edged Square Plate .............................................................4.78

4.16 Natural Frequencies of an Eccentric Mass-Spring System .................................................4.83

4.17 Natural Frequencies of a Rotating Cantilever Beam ...........................................................4.87

4.18 Natural Frequencies of a Simply Supported Square Plate using Guyan Reduction .................................................................................................................4.92

v

Lis t o f F igures

1 Introduction 1.1


2.1 Geometry and the Finite Element Mesh for the cantilever beam under multiple loading Conditions .................................................................................................................2.2

2.2 Geometry and the Finite Element Mesh for the Cantilever under an Off-centered Loading .............................................................................................................2.7

2.3 Geometry, Loading and Finite Element Mesh for the Frame Structure ...............................2.122.4 Load Case Combinations used to Obtain Solution of the Frame Structure

under General Loading (a) left half (b) right half ................................................................2.132.5 Geometry of the Square Plate with a Hole ...........................................................................2.172.6 Finite Element Mesh used for the Analysis of the Square Plate with a Hole ......................2.182.7 Geometry of the Cylinder and Finite Element Model with Boundary

Conditions ............................................................................................................................2.222.8 Stress Distribution along the Radius of a Cylinder

(a) Radial Stress (b) Circumferential Stress ........................................................................2.232.9 Simply Supported Circular Plate ..........................................................................................2.302.10 Finite Element Model of the Plate .....................................................................................2.312.11 Transverse Deflection of the Plate .....................................................................................2.312.12 Rotating Disk - NISA .........................................................................................................2.352.13 Radial Deflection ...............................................................................................................2.352.14 Tangential Stress ................................................................................................................2.362.15 Radial Stress .......................................................................................................................2.362.16 Solid Disk with Radially Varying Thermal Load ..............................................................2.402.17 Finite Element Model .........................................................................................................2.402.18 Radial Displacement ..........................................................................................................2.422.19 Radial and Circumferential Stresses ..................................................................................2.422.20 Geometry and Finite Element Mesh for Simply Supported Circular Plate ........................2.462.21 Geometry and Finite Element Mesh for a Hemispherical Shell with Edge

Loading ..............................................................................................................................2.512.22 Horizontal Displacement ....................................................................................................2.522.23 Clamped Plate with Central Concentrated Load ................................................................2.562.24 Finite Element Model .........................................................................................................2.562.25 Vertical Deflection along X-axis .......................................................................................2.572.26 Bottom Surface Stresses along X-axis ...............................................................................2.572.27 Geometry and Finite Element Mesh for the Cylindrical Shell Roof ..................................2.61

vii

2.28 Geometry and Finite Element Mesh for Thin Cylindrical Shell under Concentrated Force ............................................................................................................ 2.65

2.29 Geometry and Finite Element Mesh of a Clamped Square Plate under Pressure Loading ............................................................................................................... 2.69

2.30 Geometry and Finite Element Mesh of a Nine Layer Cross-Ply Orthotropic Plate ................................................................................................................ 2.74

2.31 Geometry and Finite Element Mesh of a Three-layer Sandwich Plate .............................. 2.812.32 Contour Plots of Stress and Resultant for a Simply Supported

Sandwich Plate under Uniform Pressure ........................................................................... 2.822.33 Propped Cantilever Beam with a Gap ................................................................................ 2.872.34 Geometry, Loading and Finite Element Model for a Beam on Elastic

Foundation ......................................................................................................................... 2.912.35 Truss on Misaligned Supports ........................................................................................... 2.962.36 Geometry and FE Model of Shell with Nonaxisymmetric Loading ................................ 2.1012.37 Top View of Cylinder ...................................................................................................... 2.1012.38 3-D Model of the Cylinder ............................................................................................... 2.1032.39 Geometry and the Finite Element Mesh for the Cantilever Elbow under

Multiple Loading Conditions ........................................................................................... 2.1092.40 Geometry and the Finite Element Mesh for the Cantilever Curved Beam

under Multiple Loading Conditions ................................................................................. 2.1132.41 Geometry and Dimensions ............................................................................................... 2.1182.42 Finite Element Model ...................................................................................................... 2.1182.43 Load Case No. 200 ........................................................................................................... 2.1182.44 Load combination ID No. 300 ......................................................................................... 2.1182.45 Simply Supported Plate Model ........................................................................................ 2.1262.46 Transverse Shear Stress Variation through the Thickness ............................................... 2.1272.47 A Simply Supported Spherical Shell of Rectangular Planform ....................................... 2.1342.48 Geometry and Finite Element Mesh for an Anti-Symmetric Cross-Ply

Laminated Clamped Plate ................................................................................................ 2.1452.49 Geometry and Finite Element Mesh of a Three-layer Sandwich Square Plate ............... 2.1492.50 Geometry and Finite Element Mesh for a Quarter of a Cross-ply Laminated

Simply Supported Plate ................................................................................................... 2.154


3.1 Square Plate under Uniaxial Compression ............................................................................ 3.23.2 Quarter Plate Model and Boundary Conditions ..................................................................... 3.23.3 Beam-Spring Structure and its Finite Element Mesh ............................................................ 3.63.4 Cantilever Beam under Compression Load ........................................................................ 3.103.5 Geometry and Finite Element Mesh for Buckling of a Simply Supported

Tapered Beam ...................................................................................................................... 3.143.6 Fixed Free Column Under Self Weight and Applied Load ................................................. 3.22

viii


4.1 FE Model for Simply Supported Beam ..................................................................................4.24.2 Finite Element Model for Cantilever Beam ...........................................................................4.74.3 Clamped Circular Plate ........................................................................................................4.114.4 Cantilever Beam ...................................................................................................................4.154.5 Finite Element Mesh for a Simply Supported Square Plate .................................................4.204.6 Mode No.1: Frequency = 83.979 Hz ....................................................................................4.214.7 Mode Shapes and Natural Frequencies ................................................................................4.224.8 Cantilever Triangular Plate ..................................................................................................4.264.9 Finite Element Mesh for an Earth Dam ...............................................................................4.304.10 Mode No. 1: Frequency = 1.237 Hz ...................................................................................4.304.11 Mode No. 2: Frequency = 1.993 Hz ...................................................................................4.304.12 Mode No. 3: Frequency = 2.365 Hz ...................................................................................4.314.13 Mode No. 4: Frequency = 2.966 Hz ...................................................................................4.314.14 Mode No. 5: Frequency = 3.465 Hz ...................................................................................4.324.15 Finite Element Mesh for Model 1 ......................................................................................4.374.16 Finite Element Mesh for Model 2 ......................................................................................4.374.17 Cantilever Beam .................................................................................................................4.444.18 Finite Element Mesh for Model 1 to 3 ...............................................................................4.444.19 Cylindrical Shell with Rigid Diaphragm ............................................................................4.524.20 Finite Element Mesh for Models 1 and 2 ...........................................................................4.534.21 Mode Shapes and Natural Frequencies ..............................................................................4.544.22 2 DOF Torsional System ....................................................................................................4.594.23 Finite Element Mesh for a Composite Square Plate ..........................................................4.634.24 Mode Shapes and Natural Frequencies ..............................................................................4.644.25 Finite Element Mesh for a 3 Layer Orthotropic Sandwich Plate .......................................4.684.26 Mode No. 1: Frequency = 23.4 Hz .....................................................................................4.694.27 Mode No. 2: Frequency = 44.8 Hz .....................................................................................4.704.28 Mode No. 3: Frequency = 71.3 Hz .....................................................................................4.704.29 Mode No. 4: Frequency = 81.1 Hz .....................................................................................4.714.30 Free-free Beam ..................................................................................................................4.754.31 Finite Element Mesh for a Free Edged Square Plate .........................................................4.794.32 Mode Shapes and Natural Frequencies ..............................................................................4.804.33 Eccentric Mass-Spring System ..........................................................................................4.844.34 Finite Element Model of the Rotating Cantilever Beam ....................................................4.884.35 Finite Element Mesh for a Simply Supported Square Plate ...............................................4.93

ix

Lis t o f Tab les

1 Introduction 1.1

1.1 Classification of verification problems by element type .......................................................1.6


2.1 Comparison of the NISA and the theoretical results ..............................................................2.32.2 Comparison of the NISA and the theoretical results ..............................................................2.72.3 Comparison of the NISA and the theoretical results ............................................................2.132.4 Stress concentration factor for circular hole in a plate .........................................................2.182.5 Deflection in thick cylinder due to internal pressure loading ..............................................2.232.6 Transverse deflection (in) ....................................................................................................2.312.7 Radial deflection and stress ..................................................................................................2.362.8 Radial displacement .............................................................................................................2.412.9 Radial and circumferential stresses ......................................................................................2.412.10 Comparison of deflections along the radius of circular plate ............................................2.462.11 Comparison of stresses at the center of circular plate ........................................................2.472.12 Displacements ....................................................................................................................2.522.13 Deflection at the Middle of the Free Edge of Roof for Different Mesh Sizes ...................2.612.14 Radial Deflection at the Point of Loading for a Pinched Cylinder ....................................2.652.15 Comparison of results for the clamped square orthotropic plate under

pressure loading .................................................................................................................2.702.16 Comparison of results for the simply supported square orthotropic plate

under pressure loading .......................................................................................................2.752.17 Comparison of results for the simply supported sandwich plate under

pressure loading .................................................................................................................2.822.18 Comparison of results for beam on tensionless elastic foundation ....................................2.922.19 Nodal displacement in inches .............................................................................................2.972.20 Extent of the contact gap ...................................................................................................2.972.21 Intermediate support forces ................................................................................................2.972.22 Comparison of the two models ........................................................................................2.1032.23 Comparison of displacement at node 45 in global direction ...........................................2.1042.24 The von Mises equivalent stress at nodes (1, 2, 3, 4, 5) ...................................................2.1042.25 Local displacements at node 45 .......................................................................................2.104

xi

2.26 Comparison of the NISA and theoretical results ............................................................. 2.1092.27 Comparison of the NISA and theoretical results ............................................................. 2.1142.28 Comparison of displacements and reactions at selected points ....................................... 2.1202.29 Transverse shear stresses of a cross-ply laminated spherical shell with

R/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=2. ............................................. 2.1352.30 Transverse shear stresses of a cross-ply laminated spherical shell with

R/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=3 .............................................. 2.1352.31 Transverse shear stresses of a cross-ply laminated spherical shell with

R/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=11 ............................................ 2.1352.32 Comparison of central transverse displacement of an anti-symmetric

cross-ply laminated clamped plate subjected to a uniform distributed load .................... 2.1462.33 Comparison of results for the simply supported sandwich plate subjected

to transverse loading ........................................................................................................ 2.1492.34 Comparison of central transverse displacements for a symmetric

cross-ply laminated simply supported plate subjected to a uniform transverse pressure load with span-to-depth ratios 10 and 100 ....................................... 2.155


3.1 Cross-Section Properties for Tapered beam ........................................................................ 3.15


4.1 Natural frequencies of a simply supported beam (Hz) .......................................................... 4.34.2 Natural frequencies of a cantilever beam (Hz) .................................................................... 4.154.3 Natural frequencies of a simply supported plate (Hz) ......................................................... 4.214.4 Natural frequencies of an earth dam (Hz) ........................................................................... 4.314.5 Natural frequencies of a square cantilever plate (Hz) .......................................................... 4.384.6 Natural frequencies (Hz) ...................................................................................................... 4.454.7 Natural frequencies (Hz) ...................................................................................................... 4.534.8 Frequencies in cycle/sec ...................................................................................................... 4.634.9 Natural frequencies of a simply supported 3 layer orthotropic sandwich plate (Hz) .......... 4.694.10 Frequencies corresponding to the elastic modes (HZ) ....................................................... 4.754.11 Frequencies corresponding to the elastic modes (HZ) ....................................................... 4.794.12 Natural frequencies of an eccentric mass-spring system (Hz) ........................................... 4.844.13 Natural Frequencies of a Rotating Cantilever Beam ......................................................... 4.894.14 Natural Frequencies of simply supported Plate ................................................................. 4.94

xii

Preface

You have with you the latest version of NISA; a proven and robust Finite Element Analysis software that has enjoyed a long-standing presence in the arena of Engineering Analysis and Design. Generations of scientists, engineers and researchers have come to depend on NISA to solve their most complex engineering problems.

With the take over of NISA by Cranes Software, Inc. in 2005, came an induction of fresh talent and resources which has taken NISA to where it is today. Over the last two years CSI has put together an exceptional team of engineers and substantial funds and transformed NISA into one of the market leaders once again. Many past users are delighted with NISA's new capabilities and have come back to the software they loved and were at ease with.

xiii

Ver i f i ca t ion Prob lem

1Introduction

1.1 Highlights of VPG/NISA CapabilitiesAnalysis Types:

NISA (Numerically Integrated elements for System Analysis) is a general purpose finite elementprogram to analyze a wide spectrum of problems encountered in engineering mechanics.

Analysis types available in VPG/NISA include: linear static and eigenvalue.

Element Library:

An extensive library of isoparametric finite elements with higher order displacement functions as wellas conventional elements is available in VPG/NISA. Such elements are suited for modeling 2-D, 3-D,axisyrnmetric solid and shell structures. In addition, other elements such as 3-D composite solid andsandwich shell, spars, prismatic and tapered beams, springs and concentrated masses are available.

Material Models:

Large number of material models are available in VPG/NISA. These include: linear elastic materialmodels such as isotropic, orthotropic and laminated composite.

Input and Output:

Input specification has been designed to make it easy for the user to define the problem. All the inputdata is in free format. The finite element model created for the static analysis can be used for dynamicand heat transfer studies with minor modifications and vice versa. Various output options are offered tomeet individual requirement, with comprehensive output control for all analysis types.

Pre- and Post-processing:

VPG/NISA directly interfaces with the pre- and post-processing modules of the VPG program; VPG/PrePost provides a wide range of tools that can be used to efficiently create meshes on CAD data using

1-1

Introduction Highlights of VPG/NISA Capabili t ies

an advanced automeshing algorithm, which eliminates most manual mesh manipulation andpostprocessing capabilities for finite element model generation and graphical representation of results..

A detailed description of VPG/NISA capabilities, input data setup and output features may be found inChapter 3 of the VPG/NISA User’s Manual.

VPG/NISA Verif ication Manual 1-2

Introduction Purpose of this Manual

1.2 Purpose of this ManualThe purpose of this manual is to provide a wide range of verification problems to serve the followingobjectives:

Overall program verification:VPG/NISA results of all problems in this manual are compared to theoretical results availablein the literature and/or other available numerical results. Close agreement is always evidentbetween VPG/NISA results and the results available in the literature. This helps new and cur-rent users establish credibility and build confidence in the results obtained for practical engi-neering problems using VPG/NISA.Provide wide range of example problems:The problems selected in this manual are designed to help the user get acquainted with theextensive capabilities of the VPG/NISA program. It is not feasible to provide a verification oran example problem for each program capability. However, the user is provided with verifica-tion or example problems for the major capabilities as well as many problems comparing vari-ous program options.It should be noted, however, that the problems in this manual are primarily intended for verifi-cation purposes. Therefore, step-by-step input and output printout is not included in this man-ual. The user is advised to refer to VPG/NISA training manual for detailed step-by-stepinstruction for preparing input for VPG/NISA. Provide comparison of various program options:In many cases more than one option or procedure will be available for the solution of a particu-lar problem. The user will find numerous problems in this manual which compare such optionsor procedures and point out applicability and specific use of each option, whenever possible.


Introduction Verification Problem Format

1.3 Verification Problem FormatA unified format is adopted for the presentation of all verification problems in this manual. A outlineof this format is as follow:

Title:A descriptive title of the problemElement Type:Element type description, example, beam, 2-D, 3-D, etc., and the applicable NKTP andNORDR numbers.Analysis Highlights:Description of the analysis type and options, example, linear static, eigenvalue with conven-tional or accelerated subspace iteration, etc.

Problem:A brief description of the physical problem, the loading and boundary conditions and therequired results or objectives:Properties:- Material:Listing the values of all material constants and/or material parameters used in the constitutivematerial model.- Geometric:Listing geometrical parameters such as thicknesses, cross sectional properties, etc.Finite Element Model:Specifying number of elements and nodes used in the model, specific boundary and loadingconditions applied, etc.Solution Procedures:Describing details of the analysis type used and various options compared. Results and Comparison:Presenting the VPG/NISA numerical results and giving discussion and various comparisonswith theoretical and/or numerical results available in the literatures.References:List of references from which the theoretical and/or numerical results may be obtained.


Introduction Verification Problem Format

Each problem description is followed by all the tables and figures used for problem description, modelspecification, results comparison and presentation.

Finally a list of the input data for the problem is given. If more than one case is studied in the problem,only one input deck is presented for a selected case. Some comment cards are given in the input deckto facilitate the interpretation of the input.


Introduction Problem Classifications

1.4 Problem ClassificationsThe main classification of the verification problems is done by analysis types. Each chapter in thismanual represent one type of analysis as follows:

A problem number is composed of two numbers; the first is the chapter number and the second is aserial number, example, Section 2.4 is in Chapter 2 (A large square plate with a circular hole underinplane loading) and is the fourth verification problem in the chapter. All tables and figures areidentified by the problem number followed by a serial number.

Refer to the table of contents for a complete list of all the problems in each chapter.

A second classification is given by element types or NKTP number. Table 1.1 gives a break-down ofall verification problems by element NKTP number and also identifies analysis type:

Table 1.1: Classification of verification problems by element type

Chapter 2 Verification problems for static analysis (STATIC)

Chapter 3 Verification problems for buckling analysis (BUCKLING)

Chapter 4 Verification problems for eigenvalue analysis (EIGENVALUE)

NKTP Element Shape Analysis Type Problem Number

1

4-node quadrilateral(NORDR = 1, 12) STATIC Section 2.4

8-node quadrilateral(NORDR = 2) EIGENVALUE Section 4.2

2 8-node quadrilateral(NORDR = 2)

STATIC Section 2.5

EIGENVALUE Section 4.7

3

8-node quadrilateral(NORDR = 2)

STATIC Section 2.5, Section 2.6


6-node triangle(NORDR = 11) STATIC Section 2.7, Section 2.8



4

8-node solid(NORDR = 1, 12)

STATIC Section 2.14

EIGENVALUE Section 4.9, Section 4.17

20-node solid(NORDR = 2)

STATIC Section 2.5

BUCKLING Section 3.3

7 20-node composite solid (NORDR = 2) STATIC Section 2.28

11 2-node tapered beam(NORDR = 1) BUCKLING Section 3.4

12 2-node beam (3-D)(NORDR = 1)



13 2-node beam(NORDR = 1)

STATICSection 2.3, Section 2.19, Section 2.20, Section 2.25

BUCKLING Section 2.27, Section 3.2


15 2-node spar (2-D)(NORDR = 1) STATIC Section 2.21

20 8-node quadrilateral(NORDR = 2)

STATIC Section 2.11

EIGENVALUESection 4.5, Section 4.6, Section 4.8,Section 4.18

21 2-node torsionalspring (3-D) EIGENVALUE Section 4.11

30 3-D general point mass with rotary inertia


32

8-node composite shell(NORDAR=2)

STATIC Section 2.15


3-node composite shell(NORDAR=10) STATIC Section 2.28




33

8-node sandwich shell(NORDR = 2)

STATIC Section 2.17


6-node triangular shell element (NORDR = 11)

STATIC Section 2.29

34

8-node axisymmetric solid with non axisymmetric loading (NORDR = 2)

STATIC Section 2.22

36

3-node axisymmetric shell(NORDR = 2)

STATIC Section 2.10

4-node axisymmetric shell (NORDR = 3) STATIC Section 2.9

38 2-node general spring(3 D) EIGENVALUE Section 4.16

39 2-node general beam (3-D, NORDR = 1)



40 4-node thin shell(NORDR = 1)




42 2-node gap STATIC Section 2.19, Section 2.20, Section 2.21

47 2-node elbow element (3-D) STATIC Section 2.23, Section 2.24




2Static Analysis Verification Problems

2.1 Cantilever Beam Subjected to Multiple LoadingConditions

Title:

Cantilever beam subjected to multiple loading conditions

Element Type:

3-D beam element (NKTP = 12, NORDR = 1)

Problem:

A long cantilever beam with 3cm square cross-section and 100 cm length Figure 2.1 is subjected to (a)concentrated tip force, (b) concentrated tip moment, (c) distributed pressure load, and (d) thermalloading. Each loading is applied separately.

The deflection at the free end, the bending moment and the maximum axial stress at the fixed end ofthe beam are computed for each load case.

Properties:

Material:EX = 207 GPa (Modulus of elasticity)NUXY = 0.3 (Poisson's ratio)

ALPX = 1.0 10-5(1/oC) (Thermal expansion coefficient)

Cross-section:

A = 9.0 cm2 (Cross-sectional area)

IYY = 6.75 cm4 (Area moment of inertia about beam Y-axis)

IZZ = 6.75 cm4 (Area moment of inertia about beam Z-axis)

×

2-1

Static Analysis Verification ProblemsCantilever Beam Subjected to Mult iple Loading Conditions

Finite Element Model:

The problem is modeled using one beam element Figure 2.1. All the six degrees of freedom areconstrained at the fixed end. Four different load cases are analyzed.

Case 1. A concentrated force FY = 1000 N at the free endCase 2. A concentrated moment MZ = 1000 N-m at the free endCase 3. Uniformly distributed line load of 10 N/cm in the Y directionCase 4. Thermal loading due to a 100°C temperature difference between the top and the bottom sur-

faces of the beam.

The stresses are computed at all the four corners of the cross-section at the fixed end, using stressrecovery points.

Results and Comparison:

The deflection at the free end, bending moment and the maximum bending stress at the fixed end of thebeam are tabulated in Table 2.1. The table also shows the theoretical values for these quantitiesobtained from the beam theory [1].

Reference:1. R.J. Roark, Formula for Stress and Strain, 4th Edition, McGraw-Hill Book Co., New York,

1965.

Figure 2.1: Geometry and the Finite Element Mesh for the cantilever beam under multiple loading Conditions

2-2 NISA II Verif ication Manual

Static Analysis Verification Problems Cantilever Beam Subjected to Mult iple Loading Conditions

Table 2.1: Comparison of the NISA and the theoretical results

LoadingDeflection(cm)

Bending Moment (N-cm)

Maximum Bending Stress (N/cm2) (at y = -1.5, z = -1.5)

NISA Theory [1] NISA Theory [I] NISA Theory [I]

Case 1 2.38564 2.38564 -1.0 105 -1.0 105 2.2222 104 2.2222 104

Case 2 3.57846 3.57846 -1.0 105 -1.0 105 2.2222 104 2.2222 104

Case 3 0.8946 1 0.89461 -5.0 104 -5.0 104 l.l l l l 104 l.l l l l 104

Case 4 1.66667 1.66667 0.0 0.0 0.0 0.0

× × × ×

× × × ×

× × × ×

NISA II Veri fication Manual 2-3

Static Analysis Verification ProblemsCantilever Beam Subjected to Mult iple Loading Conditions

Input Data for Verification Problem No. 2.**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE CANTILEVER BEAM SUBJECTED TO DIFFERENT LOADINGS.*ELTYPE 1, 12, 1*RCTABLE 1, 43,1, 0**_DISP3_: NAME =OTHER 9.0000000E+00, 6.7500000E+00, 6.7500000E+00, 1.3500000E+01, 0.0000000E+00, 0.0000000E+00, 3.0000000E+00, 3.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 4.0000000E+00, 0.0000000E+00, 1.5000000E+00, 1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,-1.5000000E+00, 1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,-1.5000000E+00,-1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.5000000E+00,-1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2,000000,000000, 0,*MATERIALEX , 1, 0, 2.07000E+07,NUXY, 1, 0, 3.00000E-01,ALPX, 1, 0, 1.00000E-05,*LDCASE, ID= 1 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*LCTITLE CONCENTRATED END FORCE.*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0


Static Analysis Verification Problems Cantilever Beam Subjected to Mult iple Loading Conditions

1,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 2,FY, 1.00000E+03,,, 0, 0*PRINTCNTLELST,0SLFO,0*LDCASE, ID= 2 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*LCTITLE CONCENTRATED END MOMENT.*CFORCE**_DISP3_: CFORCE, SET = 2 2,MZ, 1.00000E+05,,, 0, 0*PRINTCNTLELST,0SLFO,0*LDCASE, ID= 3 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*LCTITLE DISTRIBUTED PRESSURE LOAD.*PRESSURE**_DISP3_: PRESSURE, SET = 3 1,,,1,1, 0,0.0, 0, 0-.100E+02,-.100E+02,*PRINTCNTLELST,0SLFO,0*LDCASE, ID= 4 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*LCTITLE THERMAL LOADING.*NDTEMPDIF**_DISP3_: NDTEMPDIF, SET = 4 1,TEMP, 1.00000E+02 2,TEMP, 1.00000E+02*PRINTCNTLELST,0SLFO,0*ENDDATA


Static Analysis Verification ProblemsCantilever Beam Under an Off-Centered Load

2.2 Cantilever Beam Under an Off-Centered LoadTitle:

Cantilever beam under an off-centered load

Element Type:


Problem:

A long cantilever beam with 3 cm square cross-section and 100 cm length is subjected to an off-centered axial force of 1000 N (Figure 2.2). The force is offset by 10 cm.

The deflection at the free end, the bending moment and the maximum axial stress at the fixed end ofthe beam are computed.

Properties:


The problem is modeled using one beam element Figure 2.2. All the six degrees of freedom areconstrained at the fixed end. The off-centered force is applied using a rigid link, which connects thepoint of load application to the free end of the beam. The same problem can be solved using the offsetcapability of the 3-D beam element.

The stresses are computed at all the four corners of the cross-section at the fixed end, using stressrecovery points.


The deflection at the free end, bending moment and the maximum bending stress at the fixed end of thebeam are tabulated in Table 2.2. The table also shows the theoretical values for these quantitiesobtained from beam theory [1].

Material:

EX = 207 GPa (Modulus of elasticity)

NUXY = 0.3 (Poisson's ratio)

Cross-section:

A = 9.0 cm2 (Cross-sectional area)

IYY = 6.75 cm4 (Area moment of inertia about beam Y-axis)

IZZ = 6.75 cm4 (Area moment of inertia about beam Z-axis)


Static Analysis Verification Problems Cantilever Beam Under an Off-Centered Load

Reference:1. R.J. Roark, Formulas for Stress and Strain, 4th Edition, McGraw-Hill Book Co., New York,

1965.

Figure 2.2: Geometry and the Finite Element Mesh for the Cantilever under an Off-centered Loading


Result Deflection (cm)

Bending Moment (N-cm)

Maximum BendingStress (N/cm2)

NISA Analysis 0.357846 1.0 104 2.33333 103

Theory[1] 0.357846 1.0 104 2.33333 103

× ×

× ×


Static Analysis Verification ProblemsCantilever Beam Under an Off-Centered Load

Input Data for Verification Problem No. 2.2

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE CANTILEVER BEAM SUBJECTED TO AN OFF-CENTERED END FORCE USING RIGID LINK.*ELTYPE 1, 12, 1*RCTABLE 1, 43,1, 0**_DISP3_: NAME =OTHER 9.0000000E+00, 6.7500000E+00, 6.7500000E+00, 1.3500000E+01, 0.0000000E+00, 0.0000000E+00, 3.0000000E+00, 3.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 4.0000000E+00, 0.0000000E+00, 1.5000000E+00, 1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,-1.5000000E+00, 1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,-1.5000000E+00,-1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.5000000E+00,-1.5000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.00000E+02, 1.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2,000000,000000, 0,*RIGLINK 1, 3, 1,1,0, 2, 0,*MATERIALEX , 1, 0, 2.07000E+07,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*LCTITLE CONCENTRATED OFF-CENTERED END FORCE.*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0


Static Analysis Verification Problems Cantilever Beam Under an Off-Centered Load

1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 3,FX, 1.00000E+03,,, 0, 0*PRINTCNTLELST,0SLFO,0*ENDDATA


Static Analysis Verification ProblemsA Frame Structure Under General Loading

2.3 A Frame Structure Under General LoadingTitle:

A frame structure under general loading

Element Type:

2-D planar beam element (NKTP = 13, NORDR = 1)

Problem:

A frame structure is subjected to a uniform pressure of 5 Kips/ft and a concentrated force of 1500 lb asshown in Figure 2.3. Displacement and the end moments for the middle member are computed.

Properties:


Note that though the structure is symmetric, the problem is not symmetric because of thenonsymmetric loading applied to the structure. Although the solution may be obtained in one load caseusing a model of the entire frame, different loading conditions and load combinations are used forillustrative purposes. Only the left half of the structure is modeled using two beam elements as shownin Figure 2.4. Symmetric and antisymmetric boundary conditions are used, and the solution to theproblem is obtained using three load cases, followed by two load combinations as follows

Case 1. The model is analyzed for the pressure loading alone. Since for this case the structure aswell as the loading is symmetric, symmetric boundary conditions are imposed at the line ofsymmetry.

Material:

EX = 30.0 106 psi (Modulus of elasticity)


Cross-section:

For Vertical Members:

A = 1.5 105 in2 (Area of cross-section)

IYY = 110.0 in4 (Area moment of inertia about beam Y-axis)

IZZ = 110.0 in4 (Area moment of inertia about beam Z-axis)

For Horizontal Members:

A = 2.0 105 in2 (Area of cross-section)

IYY = 220.0 in4 (Area moment of inertia about beam Y-axis)


×

×

×


Static Analysis Verification Problems A Frame Structure Under General Loading

Case 2. The model is analyzed for half of the total concentrated load. Symmetric boundary condi-tions are imposed at the line of symmetry.

Case 3. Problem is solved for half of the total concentrated load with antisyrnmetric boundary con-ditions at the line of symmetry.

Case 4. The solution for the left half of the frame under total loading is obtained by adding theresults from the first three load cases (Figure 2.4a). This is done by using load case combi-nation capabilities of NISA.

Case 5. The solution for the right half of the frame under total loading is obtained by adding theresults from the first two load cases, and then subtracting the results of the third load case,Figure 2.4b. Again, this is done by using load case combination capabilities of NISA.


The horizontal displacement as well as the end moments for the middle member are computed, andtabulated in Table 2.3. The table also contains the theoretical values for these quantities. Note that thetheoretical results do not account for the axial deformation of the frame members. A very high value ofthe cross-sectional area is used for the beam elements to achieve similar results from the finite elementanalysis.

Reference:1. R.L. Ketter, G.C. Lee and S.P. Prawel, Structural Analysis and Design, McGraw-Hill Book Co.,

New York, 1979.

NISA II Verif ication Manual 2-11


Figure 2.3: Geometry, Loading and Finite Element Mesh for the Frame Structure



Figure 2.4: Load Case Combinations used to Obtain Solution of the Frame Structure under General Loading (a) left half (b) right half


AnalysisHorizontal Displacement of the Middle Member (in)

Bending Moment at the Left Corner (Kips-in)

Bending Moment at the Right Corner (Kips-in)

NISA 0.55227 347.529 -617.529

Theory [1] 0.55227 347.14 -617.14



Input Data for Verification Problem No. 2.3**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE SINGLE SPAN PORTAL FRAME.*ELTYPE 1, 13, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 1.5000000E+05, 1.1000000E+02, 2, 2,1, 0**_DISP3_: NAME =OTHER 2.0000000E+05, 2.2000000E+02,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 1.80000E+02, 0.00000E+00, 0 3,,,, 9.00000E+01, 1.80000E+02, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 2, 0 2, 3,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 0, 1, 3, 0,-1, 1, 0, 0.000, 0.000*LCTITLE PRESSURE LOADING, SYMMETRIC BOUNDARY CONDITIONS*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 3,UX , 0.00000E+00,,,,,,,, 0 3,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 2,,,1,1, 0,0.0, 0, 00.417E+03,0.417E+03,*PRINTCNTLELST,0



SLFO,0*LDCASE, ID= 2 0, 1, 3, 0,-1, 1, 0, 0.000, 0.000*LCTITLE CONCENTRATED FORCE, SYMMETRIC BOUNDARY CONDITIONS*CFORCE**_DISP3_: CFORCE, SET = 2 2,FX, 7.50000E+02,,, 0, 0*PRINTCNTLELST,0SLFO,0*LDCASE, ID= 3 0, 1, 3, 0,-1, 1, 0, 0.000, 0.000*LCTITLE CONCENTRATED FORCE, ANTISYMMETRIC BOUNDARY CONDITIONS*SPDISP**_DISP3_: SPDISP, SET = 3 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 3 2,FX, 7.50000E+02,,, 0, 0*PRINTCNTLELST,0SLFO,0*LDCOMB, ID = 41,1,11,1.0,2,1.0,3,1.0,*LCTITLE COMBINATION OF LOADCASES 1,2 AND 3. SOLUTION FOR THE LEFT SIDE*PRINTCNTLELST,0*LDCOMB, ID = 51,1,11,1.0,2,1.0,3,-1.0,*LCTITLE COMBINATION OF LOADCASES 1,2 AND 3. SOLUTION FOR THE RIGHT SIDE*PRINTCNTLELST,0*ENDDATA


Static Analysis Verification ProblemsA Large Square Plate with a Circular Hole under Inplane Loading

2.4 A Large Square Plate with a Circular Hole under Inplane Loading

Title:

A large square plate with a circular hole under inplane loading

Element Type:

2-D four node plane stress element (NKTP = 1, NORDR = 1)

Problem:

A square plate of 42 in side length and 1 in thickness has a hole of 6 in diameter in its center. The plateis subjected to a uniform in plane pressure of 1 psi along its width Figure 2.5.

Stress concentration due to the circular hole is computed using finite element analysis, and is comparedwith the theoretical solution of an infinite plate subjected to uniform inplane pressure.

Properties:


Due to the symmetry, only one quarter of the plate is modeled. The model consists of 30 elements.Symmetric boundary conditions (UX = 0.0 at X = 0.0 and UY = 0.0 at Y = 0.0) are applied Figure 2.5and Figure 2.6.


The stress concentration factor at the circular hole is computed by dividing the nodal stress in thedirection of applied load (SXX) at the circular hole (X = 0.0, Y = 3.0) by the applied pressure. Table2.4 shows the stress concentration factors obtained from NISA as well as the theoretical analysis [1].

Reference:1. S. Timoshenko, Strength of Materials Part II - Advanced Theory and Problems, 3rd Edition, D.

Van Nostrand Company, Inc., 1956.

Material:

EX = 7000.0 psi (Modulus of elasticity)



Static Analysis Verification Problems A Large Square Plate with a Circular Hole under Inplane Loading

Figure 2.5: Geometry of the Square Plate with a Hole



Figure 2.6: Finite Element Mesh used for the Analysis of the Square Plate with a Hole

Table 2.4: Stress concentration factor for circular hole in a plate

Analysis Stress concentration factor (SXX/P at X = 0, Y = 3)

NISA 3.0006

Theory [1] 3.0

Percent error 0.02


Static Analysis Verification Problems A Large Square Plate with a Circular Hole under Inplane Loading

Input Data for Verification Problem No. 2.4**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE A LARGE SQUARE PLATE WITH A CIRCULAR HOLE.*ELTYPE 1, 1, 1*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1, 1,,, 3.00000E+00, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 3.00000E+00, 1.50000E+01, 0.00000E+00, 0 3, 1,,, 3.00000E+00, 3.00000E+01, 0.00000E+00, 0 4, 1,,, 3.00000E+00, 4.50000E+01, 0.00000E+00, 0 : : 39,,,, 2.10000E+01, 2.10000E+01, 0.00000E+00, 0 40,,,, 1.40000E+01, 2.10000E+01, 0.00000E+00, 0 41,,,, 7.00000E+00, 2.10000E+01, 0.00000E+00, 0 42,,,, 0.00000E+00, 2.10000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 8, 9, 2, 2, 1, 1, 1, 0 2, 9, 10, 3, : : 29, 1, 1, 1, 0 33, 40, 41, 34, 30, 1, 1, 1, 0 34, 41, 42, 35,*MATERIALEX , 1, 0, 7.00000E+03,NUXY, 1, 0, 2.00000E-01,*SETS 1,S, 6, 2,S, 7, *LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*LCTITLE UNIFORMLY DISTRIBUTED END PRESSURE.



*SPDISP**_DISP3_: SPDISP, SET = 1 1,UY , 0.00000E+00,,,,,,,, 0 7,UX , 0.00000E+00,,,,,,,, 0 8,UY , 0.00000E+00,,,,,,,, 0 14,UX , 0.00000E+00,,,,,,,, 0 15,UY , 0.00000E+00,,,,,,,, 0 21,UX , 0.00000E+00,,,,,,,, 0 22,UY , 0.00000E+00,,,,,,,, 0 28,UX , 0.00000E+00,,,,,,,, 0 29,UY , 0.00000E+00,,,,,,,, 0 35,UX , 0.00000E+00,,,,,,,, 0 36,UY , 0.00000E+00,,,,,,,, 0 42,UX , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 25,,,2,0, 0, -1.0, 0, 0


Static Analysis Verification Problems Thick Cylinder under Internal Pressure Load

2.5 Thick Cylinder under Internal Pressure LoadTitle:

Thick cylinder under internal pressure load

Element Type:Case 1. 2-D eight node plane strain element (NKTP = 2, NORDR = 2)

Case 2. 2-D Eight node axisymmetric solid element (NKTP = 3, NORDR = 2)

Case 3. 3-D twenty node solid element (NKTP = 4, NORDR = 2)

Problem:

A long and thick cylinder with an inner diameter of 0.2 m and an outer diameter of 0.4 m is subjectedto an internal pressure of 100 MPa Figure 2.7.

The deflections and the radial stress distribution in the cylinder are compared with the exact solution.

Properties:

Finite Element Model:Case 1. Ten 8-node plain strain elements are used to model a ten degree sector of the cylinder.

Boundary conditions of no circumferential deformation are imposed using local coordinatesystems. This allows for the expansion of the cylinder in radial direction only Figure 2.7.

Case 2. Ten 8-node axisymmetric elements are used to model 0.1 m length of the cylinder. Pre-scribed boundary conditions of UY = 0 on top and bottom surfaces allow for the expansionof the cylinder in radial direction only Figure 2.7.

Case 3. Ten 20-node solid elements are used to model a ten degree sector of the cylinder. Boundaryconditions of no vertical and circumferential deformations are imposed as in case 1 and 2.Again, this allows for the expansion of the cylinder in radial direction only Figure 2.7.


The radial deflections at the inner and outer walls of the cylinder are computed in each case. Table 2.5shows these values as well as the theoretical solution [1]. Close agreement between the results isevident.

The variation of the radial and the circumferential component of the stresses is studied in each case.These values are compared with the theoretical solution. Figure 2.8a shows the variation of the radialstress along the radius of the cylinder. Figure 2.8b shows the variation of the circumferential stressalong the radius of the cylinder. Again, these results are compared with the theoretical results.

Material:

EX = 210 GPa (Modulus of elasticity)



Static Analysis Verification ProblemsThick Cylinder under Internal Pressure Load

Figure 2.7: Geometry of the Cylinder and Finite Element Model with Boundary Conditions

Reference:1. R.J. Roarli, Formulas for Stress and Strain, 4th Edition, McGraw-Hill Book Co., New York,

1965.



Table 2.5: Deflection in thick cylinder due to internal pressure loading

Figure 2.8: Stress Distribution along the Radius of a Cylinder (a) Radial Stress (b) Circumferential Stress

Deflection at the InnerWall of the Cylinder (m)

Deflection at the OuterWall of the Cylinder (m)

Case 1 7.93656 10-5 6.34922 10-5

Case 2 7.93651 10-5 6.34921 10-5

Case 3 7.93651 10-5 6.34920 10-5

Theory [1] 7.93651 10-5 6.34921 10-5

× ×

× ×

× ×

× ×



Input Data for Verification Problem No. 2.5A**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE THICK CYLINDER UNDER INTERNAL PRESSURE. PLANE STRAIN MODEL.*ELTYPE 1, 2, 2*NODES 1, 1,,, 1.00000E-01, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.05000E-01, 0.00000E+00, 0.00000E+00, 0 3, 1,,, 1.10000E-01, 0.00000E+00, 0.00000E+00, 0 4, 1,,, 1.15000E-01, 0.00000E+00, 0.00000E+00, 0 : : 218, 1,,, 1.85000E-01, 1.00000E+01, 0.00000E+00, 0 219, 1,,, 1.90000E-01, 1.00000E+01, 0.00000E+00, 0 220, 1,,, 1.95000E-01, 1.00000E+01, 0.00000E+00, 0 221, 1,,, 2.00000E-01, 1.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 103, 203, 202, 201, 101, 2, 1, 1, 1, 0 3, 4, 5, 105, 205, 204, 203, 103, : : 9, 1, 1, 1, 0 17, 18, 19, 119, 219, 218, 217, 117, 10, 1, 1, 1, 0 19, 20, 21, 121, 221, 220, 219, 119,*G2 201,0,0, 0.0,0.0,10.0 221,1,0, 0.0,0.0,10.0 *MATERIALEX , 1, 0, 2.10000E+11,NUXY, 1, 0, 0.00000E+00,*SETS 101,R, 1, 21, 1, *LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*LCTITLE UNIFORM INTERNAL PRESSURE.*SPDISP



Input Data for Verification Problem No. 2.5B

**_DISP3_: SPDISP, SET = 1 1,UY , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 : : 218,UY , 0.00000E+00,,,,,,,, 0 219,UY , 0.00000E+00,,,,,,,, 0 220,UY , 0.00000E+00,,,,,,,, 0 221,UY , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,4,0, 0, 0.1E+09, 0, 0*PRINTCNTLAVND,101DISP,101SLFO,0*ENDDATA

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE THICK CYLINDER UNDER INTERNAL PRESSURE. AXISYMMETRIC MODEL.*ELTYPE 1, 3, 2*NODES 1,,,, 1.00000E-01, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.05000E-01, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.10000E-01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.15000E-01, 0.00000E+00, 0.00000E+00, 0 : : 218,,,, 1.85000E-01, 1.00000E-02, 0.00000E+00, 0 219,,,, 1.90000E-01, 1.00000E-02, 0.00000E+00, 0 220,,,, 1.95000E-01, 1.00000E-02, 0.00000E+00, 0 221,,,, 2.00000E-01, 1.00000E-02, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 103, 203, 202, 201, 101, 2, 1, 1, 1, 0 3, 4, 5, 105, 205, 204, 203, 103, :



Input Data for Verification Problem No. 2.5C

: 9, 1, 1, 1, 0 17, 18, 19, 119, 219, 218, 217, 117, 10, 1, 1, 1, 0 19, 20, 21, 121, 221, 220, 219, 119,*MATERIALEX , 1, 0, 2.10000E+11,NUXY, 1, 0, 0.00000E+00,*SETS 101,R, 1, 21, 1, *LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*LCTITLE UNIFORM INTERNAL PRESSURE.*SPDISP**_DISP3_: SPDISP, SET = 1 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 : : 220,UY , 0.00000E+00,,,,,,,, 0 220,UZ , 0.00000E+00,,,,,,,, 0 221,UY , 0.00000E+00,,,,,,,, 0 221,UZ , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,4,0, 0, 0.1E+09, 0, 0*PRINTCNTLAVND,101DISP,101SLFO,0*ENDDATA

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE THICK CYLINDER UNDER INTERNAL PRESSURE. SOLID MODEL.*ELTYPE 1, 4, 2*LCSYSTEM



4,0,0, 901, 902, 903*NODES 1,,,, 1.00000E-01, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.05000E-01, 0.00000E+00, 0.00000E+00, 0 3, 1,,, 1.10000E-01, 0.00000E+00, 0.00000E+00, 0 4, 1,,, 1.15000E-01, 0.00000E+00, 0.00000E+00, 0 : : 821, 1,,, 2.00000E-01, 1.00000E+01, 1.00000E-02, 4 901,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 902,,,, 9.84808E-01, 1.73648E-01, 0.00000E+00, 0 903,,,, 0.00000E+00, 1.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 301, 601, 602, 603, 303, 3, 2, 101, 701, 703, 103, 201, 501, 801, 802, 803, 503, 203, 202, 2, 1, 1, 0, 0 : : 719, 119, 217, 517, 817, 818, 819, 519, 219, 218, 10, 1, 1, 0, 0 19, 319, 619, 620, 621, 321, 21, 20, 119, 719, 721, 121, 219, 519, 819, 820, 821, 521, 221, 220,*MATERIALEX , 1, 0, 2.10000E+11,NUXY, 1, 0, 0.00000E+00,*SETS 101,R, 1, 21, 1, *LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*LCTITLE UNIFORM INTERNAL PRESSURE.*SPDISP**_DISP3_: SPDISP, SET = 1 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 : : 820,UY , 0.00000E+00,,,,,,,, 0 820,UZ , 0.00000E+00,,,,,,,, 0 821,UY , 0.00000E+00,,,,,,,, 0 821,UZ , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,3,0, 0, 0.1E+09, 0, 0*PRINTCNTL



AVND,101DISP,101SLFO,0*ENDDATA


Static Analysis Verification Problems Simply Supported Circular Plate

2.6 Simply Supported Circular PlateTitle:

Simply supported circular plate

Element Type:

2-D axisymmetric solid element (NKTP = 3, NORDR = 2)

Problem:

A simply supported circular plate of 50 inch radius and one inch thickness Figure 2.9 is analyzed foruniform transverse loading. Three load cases are analyzed:

(a) Uniformly distributed pressure of 1 psi is applied on the top face of the plate

(b) Uniformly distributed pressure of 1 psi is applied on the bottom face of the plate

(c) Uniform self weight is applied as a body force of magnitude 1 lb/in3

All the load cases simulate the same loading conditions and hence are equivalent. The transversedeflection of the plate is computed along the radial direction.

Properties:


As the geometry is axisymmetric, a radial section is modeled using eight axisymmetric elements, seeFigure 2.10. Simply supported boundary conditions are imposed along the outer boundary (X = 50.0in) by setting UY = 0.0. At the center of the plate (X = 0.0 in), due to symmetry, UX = 0.0 at all nodes.

The problem is solved using three load cases; which are equivalent.

Case 1. Pressure load of 1 psi on top faceCase 2. Pressure load of 1 psi on bottom face

Case 3. Body force of 1 lb/in3

Material:

EX = 1 10-7psi (Modulus of elasticity)


DENS = 1.0 lb-sec2/in4 (Mass density)

×


Static Analysis Verification ProblemsSimply Supported Circular Plate


The transverse deflection of the plate along the radius is compared with the analytical solutionobtained using thin plate theory (Ref. [1]). The results are shown in Figure 2.11 as well as in Table 2.6.The results are in excellent agreement.

Figure 2.9: Simply Supported Circular Plate

Reference:1. S.P. Timoshenko, Theory of Elasticity, 3rd Edition, McGraw-Hill Book Co., New York, 1970.



Table 2.6: Transverse deflection (in)

Figure 2.10: Finite Element Model of the Plate

Figure 2.11: Transverse Deflection of the Plate

Node Number Case 1 Case 2 Case 3 Theory

35 -0.435 -0.435 -0.435 -0.435

37 -0.409 -0.409 -0.409 -0.409

39 -0.351 -0.351 -0.351 -0.351

41 -0.276 -0.276 -0.276 -0.2754

43 -0.195 -0.195 -0.195 -0.195

45 -0.132 -0.132 -0.132 -0.132

47 -0.08 -0.08 -0.08 -0.08

49 -0.04 -0.04 -0.04 -0.04

51 0.0 0.0 0.0 0.0


Static Analysis Verification ProblemsSimply Supported Circular Plate

Input Data for Verification Problem No. 2.6**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLESIMPLY SUPPORTED CIRCULAR PLATE*ELTYPE 1, 3, 2*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 5.50000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.10000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.55000E+01, 0.00000E+00, 0.00000E+00, 0 : : 48,,,, 4.55000E+01, 1.00000E+00, 0.00000E+00, 0 49,,,, 4.70000E+01, 1.00000E+00, 0.00000E+00, 0 50,,,, 4.85000E+01, 1.00000E+00, 0.00000E+00, 0 51,,,, 5.00000E+01, 1.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 20, 37, 36, 35, 18, 2, 1, 1, 0, 0 3, 4, 5, 22, 39, 38, 37, 20, : : 7, 1, 1, 0, 0 13, 14, 15, 32, 49, 48, 47, 30, 8, 1, 1, 0, 0 15, 16, 17, 34, 51, 50, 49, 32,*MATERIALEX , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 1.00000E+00,*LDCASE, ID= 1 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 17,UY , 0.00000E+00,,,,,,,, 0 18,UX , 0.00000E+00,,,,,,,, 0 35,UX , 0.00000E+00,,,,,,,, 0*PRESSURE



**_DISP3_: PRESSURE, SET = 1 1,,,3,0, 0, 1.0, 0, 0 2,,,3,0, 0, 1.0, 0, 0 3,,,3,0, 0, 1.0, 0, 0 4,,,3,0, 0, 1.0, 0, 0 5,,,3,0, 0, 1.0, 0, 0 6,,,3,0, 0, 1.0, 0, 0 7,,,3,0, 0, 1.0, 0, 0 8,,,3,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*PRINTCNTLSLFO,0*LDCASE, ID= 2 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*PRESSURE**_DISP3_: PRESSURE, SET = 2 1,,,1,0, 0, -1.0, 0, 0 2,,,1,0, 0, -1.0, 0, 0 3,,,1,0, 0, -1.0, 0, 0 4,,,1,0, 0, -1.0, 0, 0 5,,,1,0, 0, -1.0, 0, 0 6,,,1,0, 0, -1.0, 0, 0 7,,,1,0, 0, -1.0, 0, 0 8,,,1,0, 0, -1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*PRINTCNTLSLFO,0*LDCASE, ID= 3 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*BODYFORCE0.0,0.0,0.0,0.0,-1.0,0.0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*PRINTCNTLSLFO,0*ENDDATA


Static Analysis Verification ProblemsUniform Thickness Rotating Disc

2.7 Uniform Thickness Rotating DiscTitle:

Uniform thickness rotating disc

Element Type:


Problem:

A circular disc of 12 inches diameter and 2 inches thick is rotating at an angular speed of 1000 rad/sec.The stresses and displacements in the disc are computed.

Properties:


Since the problem is axisymmetric, a radial section of the disc is modeled using one cubic element asshown in Figure 2.12. Body force due to the angular velocity (centrifugal force) is applied. Due tosymmetry, only one half of the radial section is modeled. Symmetric boundary conditions are appliedalong the X-axis by setting UY = 0.0 and along the Y-axis by setting UX = 0.0.


The radial displacements, circumferential stress and radial stress are shown in Table 2.7 and are alsoplotted in Figure 2.13, Figure 2.14, and Figure 2.15, respectively. A comparison with the analyticalsolution (Ref. [1]) shows an excellent agreement.

Material:

EX = 3 107 psi (Modulus of elasticity)


DENS = 0.000733085 lb-sec2 /in4 (Mass density)

×


Static Analysis Verification Problems Uniform Thickness Rotating Disc

Reference:1. S.P. Timoshenko, Theory of Elasticity, 3rd Edition, McGraw-Hill Book Co., New York, 1970.

Figure 2.12: Rotating Disk - NISA

Figure 2.13: Radial Deflection



Table 2.7: Radial deflection and stress

Figure 2.14: Tangential Stress

Figure 2.15: Radial Stress

Node Number Deflection(10-3 in)

Circumferential Stress (SZZ) (1000 psi)

Radial Stress (SXX)(1000 psi)

1 0.0 10.79 10.79

2 0.479 10.05 9.53

3 0.825 7.97 5.93

4 0.908 4.53 -0.026


Static Analysis Verification Problems Uniform Thickness Rotating Disc

Input Data for Verification Problem No. 2.7**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEROTATING DISC*ELTYPE 1, 3, 3*NODES 1,,,, 0.00000E+00, 1.00000E+00, 0.00000E+00, 0 2,,,, 2.00000E+00, 1.00000E+00, 0.00000E+00, 0 3,,,, 4.00000E+00, 1.00000E+00, 0.00000E+00, 0 4,,,, 6.00000E+00, 1.00000E+00, 0.00000E+00, 0 5,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 2.00000E+00, 0.00000E+00, 0.00000E+00, 0 7,,,, 4.00000E+00, 0.00000E+00, 0.00000E+00, 0 8,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 9,,,, 0.00000E+00, 6.66667E-01, 0.00000E+00, 0 10,,,, 0.00000E+00, 3.33333E-01, 0.00000E+00, 0 11,,,, 6.00000E+00, 6.66667E-01, 0.00000E+00, 0 12,,,, 6.00000E+00, 3.33333E-01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 9, 10, 5, 6, 7, 8, 12, 11, 4, 3, 2,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.33085E-04,*LDCASE, ID= 1 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 6,UY , 0.00000E+00,,,,,,,, 0 7,UY , 0.00000E+00,,,,,,,, 0 8,UY , 0.00000E+00,,,,,,,, 0 9,UX , 0.00000E+00,,,,,,,, 0 10,UX , 0.00000E+00,,,,,,,, 0*BODYFORCE0.0,1000.0,0.0,0.0,0.0,0.0



*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*PRINTCNTLSLFO,0*ENDDATA


Static Analysis Verification Problems Solid Disc Subjected to Radially Varying Thermal Load

2.8 Solid Disc Subjected to Radially Varying Thermal LoadTitle:

Solid disc subjected to radially varying thermal load

Element Type:


Problem:

A solid circular disc of 0.1 inch radius and 0.005 inch thickness is subjected to a radially varyingthermal loading. The temperature T is assumed to vary in the radial direction as T = 100(1-x2/b2)where b is the outer radius. The geometry and temperature distribution are shown in Figure 2.16. Thedisplacements and stresses are computed along a radial section.

Properties:


As the problem is axisymmetric, a radial section of the disc is modeled using four cubic elements asshown in Figure 2.17. Radially varying thermal loading is applied as nodal temperatures as per thegiven function. Symmetric boundary conditions are applied along Y-axis by setting UX = 0.0 andalong the X-axis by setting UY = 0.0.


Radial displacements and radial as well as circumferential stresses are shown in Figure 2.18 andFigure 2.19. Table 2.8 shows the radial displacement and Table 2.9 shows the stresses at variouspoints. The results show good agreement with theory [1].

Reference:1. C.T. Wang, Applied Elasticity, McGraw-Hill Book Co., New York, 1985.

Material:

EX = l.0 107psi (Modulus of elasticity)

NUX = 0.3 (Poisson's ratio)

ALPX = 1.0 10-7(1/oF) (Coefficient of thermal expansion in X-direction)

×

×


Static Analysis Verification ProblemsSolid Disc Subjected to Radially Varying Thermal Load

Figure 2.16: Solid Disk with Radially Varying Thermal Load

Figure 2.17: Finite Element Model



Table 2.8: Radial displacement

Table 2.9: Radial and circumferential stresses

Node Number x (in) Radial Displacement (NISA)(UX) 10-6 in

1 0.0 0.0

2 0.003333 0.02521

3 0.006666 0.050366

4 0.01 0.07541

5 0.016666 0.12491

6 0.2333 0.17328

7 0.03 0.22004

8 0.04 0.28618

9 0.05 0.34613

10 0.06 0.39836

11 0.07333 0.45333

12 0.086666 0.48812

13 0.1 0.49919

Node number

Radial Stress (SXX) l02 psi

Circumferential Stress (SZZ) l02 psi

1 -0.2508 -0.2508

7 -0.2284 -0.1834

10 -0.1607 +0.0191

13 -0.0001 +0.4992



Figure 2.18: Radial Displacement

Figure 2.19: Radial and Circumferential Stresses



Input Data for Verification Problem No. 2.8**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLESOLID DISC SUBJECTED TO RADIALLY VARYING THERMAL LOAD*ELTYPE 1, 3, 3*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 3.33333E-03, 0.00000E+00, 0.00000E+00, 0 3,,,, 6.66667E-03, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.00000E-02, 0.00000E+00, 0.00000E+00, 0 : : 49,,,, 6.00000E-02, 5.00000E-03, 0.00000E+00, 0 50,,,, 7.33333E-02, 5.00000E-03, 0.00000E+00, 0 51,,,, 8.66667E-02, 5.00000E-03, 0.00000E+00, 0 52,,,, 1.00000E-01, 5.00000E-03, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 4, 17, 30, 43, 42, 41, 40, 27, 14, 2, 1, 1, 0, 0 4, 5, 6, 7, 20, 33, 46, 45, 44, 43, 30, 17, 3, 1, 1, 0, 0 7, 8, 9, 10, 23, 36, 49, 48, 47, 46, 33, 20, 4, 1, 1, 0, 0 10, 11, 12, 13, 26, 39, 52, 51, 50, 49, 36, 23,*MATERIALEX , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.00000E-02,ALPX, 1, 0, 1.00000E-07,*LDCASE, ID= 1 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0



3,UY , 0.00000E+00,,,,,,,, 0 4,UY , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 6,UY , 0.00000E+00,,,,,,,, 0 7,UY , 0.00000E+00,,,,,,,, 0 8,UY , 0.00000E+00,,,,,,,, 0 9,UY , 0.00000E+00,,,,,,,, 0 10,UY , 0.00000E+00,,,,,,,, 0 11,UY , 0.00000E+00,,,,,,,, 0 12,UY , 0.00000E+00,,,,,,,, 0 13,UY , 0.00000E+00,,,,,,,, 0 14,UX , 0.00000E+00,,,,,,,, 0 27,UX , 0.00000E+00,,,,,,,, 0 40,UX , 0.00000E+00,,,,,,,, 0*NDTEMPER**_DISP3_: NDTEMPER, SET = 1 1,TEMP, 1.00000E+02 2,TEMP, 9.98899E+01 3,TEMP, 9.95556E+01 4,TEMP, 9.90000E+01 : : 49,TEMP, 6.40000E+01 50,TEMP, 4.62223E+01 51,TEMP, 2.48889E+01 52,TEMP, 0.00000E+00*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*ENDDATA


Static Analysis Verification Problems Simply Supported Circular Plate with Annular Line Load and Body Forces

2.9 Simply Supported Circular Plate with Annular Line Load and Body Forces

Title:

Simply supported circular plate with annular line load and body forces

Element Type:

Axisymmetric shell elements (NKTP = 36, NORDR = 3)

Problem:

A circular plate of 10 inch radius and 1.0 inch thickness is simply supported at the outer edge Figure2.20. The plate is subjected to an annular line load, gravity load and centrifugal load. Each load isapplied separately.

The deflection and stresses at the center of the plate are calculated for the three loads.

Properties:


As the geometry is axisymmetric, the plate is modeled by five 4-noded axisymmetric shell elements,see Figure 2.20. Simply supported boundary conditions are imposed along the outer boundary (x = 5.0in) by setting UY = 0.0. Due to axial symmetry, radial displacement UX and rotation along the Z-axis,ROTZ are constrained at center. Three load cases are analyzed.

Case 1. An annular line load is applied at a radius of 3 inches. An annular line load is applied as aconcentrated load of 10 lbs (FY = -10) on the node at 3 inch radius.

Case 2. A body force due to the gravity is considered in this load case. The acceleration due togravity is specified as 384.6 in/sec2 in the transverse direction.

Case 3. A body force due to the angular rotation of the disk is considered. The centrifugal load isspecified as a constant angular velocity of 9.5493 rad/sec (3000 rpm) about the Y axis.


The transverse displacement (UY), the radial stress (SXX) and the hoop stress (SZZ) at the center areshown in Table 2.10 and Table 2.11. For case 1, the UY deflection at the node of radius 3 inches is alsogiven in Table 2.10. The tables also show the theoretical (Ref. [1]) values for the displacements andstresses. An excellent agreement between NISA and theoretical solution is found.

Material:



DENS = 0.283 lb-sec2 /in4 (Mass density)

×


Static Analysis Verification ProblemsSimply Supported Circular Plate with Annular Line Load and Body Forces

Figure 2.20: Geometry and Finite Element Mesh for Simply Supported Circular Plate

Reference:1. S.P. Timoshenko and J.N. Goodier, Theory of Elasticity, McGraw-Hill Book Co., New York,

1959.

Table 2.10: Comparison of deflections along the radius of circular plate

Load Case Number Location Component

Displacement (in)

NISA Theory

1 center UY 0.1536 10-4 0.1516 10-4

1 r = 3.0 UY 0.1347 10-4 0.1326 10-4

2 center UY 0.0256 0.0254

2 r = 3.0 UY 0.0228 0.0223

3 r = 3.0 UX 0.7187 10-4 0.7187 10-4

× ×

× ×

× ×



Table 2.11: Comparison of stresses at the center of circular plate

Load Case Number Component

Stress Value (psi)

NISA Theory

1 Radial (SXX) 8.994 8.994

1 Hoop (SZZ) 8.994 8.994

2 Radial (SXX) 13,532.2 13532.2

2 Hoop (SZZ) 13,532.3 13532.2

3 Radial (SXX) 1064.52 1064.52

3 Hoop (SZZ) 1064.52 1064.52


Static Analysis Verification ProblemsSimply Supported Circular Plate with Annular Line Load and Body Forces

Input Data for Verification Problem No. 2.9**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE 2.9 : A CIRCULAR PLATE WITH LINE LOAD AND BODY FORCES.*ELTYPE 1, 36, 3*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.00000E+00, 0.00000E+00, 0.00000E+00, 0 5,,,, 3.58333E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 4.16666E+00, 0.00000E+00, 0.00000E+00, 0 7,,,, 4.75000E+00, 0.00000E+00, 0.00000E+00, 0 8,,,, 5.33333E+00, 0.00000E+00, 0.00000E+00, 0 9,,,, 5.91666E+00, 0.00000E+00, 0.00000E+00, 0 10,,,, 6.50000E+00, 0.00000E+00, 0.00000E+00, 0 11,,,, 7.08333E+00, 0.00000E+00, 0.00000E+00, 0 12,,,, 7.66667E+00, 0.00000E+00, 0.00000E+00, 0 13,,,, 8.25000E+00, 0.00000E+00, 0.00000E+00, 0 14,,,, 8.83333E+00, 0.00000E+00, 0.00000E+00, 0 15,,,, 9.41667E+00, 0.00000E+00, 0.00000E+00, 0 16,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 4, 2, 1, 1, 1, 0 4, 5, 6, 7, 3, 1, 1, 1, 0 7, 8, 9, 10, 4, 1, 1, 1, 0 10, 11, 12, 13, 5, 1, 1, 1, 0 13, 14, 15, 16,*MATERIALEX , 1, 0, 3.00000E+07,



NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 2.83000E-01,*SETS 111,S, 1, 4, 222,S, 1, *LDCASE, ID= 1 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 16,UY , 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 4,FY,-1.00000E+01,,, 0, 0*PRINTCNTLAVND,222DISP,111SLFO,0*LDCASE, ID= 2 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*BODYFORCE0.0,0.0,0.0,0.0,-386.4,0.0*PRINTCNTLAVND,222DISP,222SLFO,0*LDCASE, ID= 3 0, 1, 3, 0, 0, 2, 0, 0.000, 0.000*BODYFORCE0.0,9.5493,0.0,0.0,0.0,0.0*PRINTCNTLAVND,222DISP,222SLFO,0*ENDDATA


Static Analysis Verification ProblemsHemispherical Shell Under Edge-Moment Loading

2.10 Hemispherical Shell Under Edge-Moment LoadingTitle:

Hemispherical shell under edge-moment loading

Element Type:

2-D axisymmetric shell element (NKTP = 36, NORDR = 2)

Problem:

A hemispherical shell has 100 inch mean radius and is 1 inch thick. It has an opening at the crown andis subjected to a unit edge moment as shown in Figure 2.21. The base of the shell is clamped. Thehorizontal displacement and the top surface nodal stress (STY) at specific nodal points are computed.

Properties:


Due to axisymmetry, the shell is modeled using 12 quadratic elements as shown in Figure 2.21. A totaledge moment of 314.16 [(2 ). (1 lb.-in)] is applied as a nodal value. Clamped boundary condition atthe base is enforced by setting UX, UY and ROTZ to zero at node 1.


The horizontal displacement and moment are plotted in Figure 2.22 as a function of the angle(azimuthal). The displacements are shown in Table 2.12. Comparison with the theoretical results (Ref.[1]) shows very good agreement.

Reference:1. P.A. Grafton and D.R. Strome, Analysis of Axisymmetric Shell by the Direct Stiffness Method,

AIAAJ, 1, 2342-2347, 1963.

Material:

EX = 1.0 107psi (Modulus of elasticity)


×

πr


Static Analysis Verification Problems Hemispherical Shell Under Edge-Moment Loading

Figure 2.21: Geometry and Finite Element Mesh for a Hemispherical Shell with Edge Loading



Table 2.12: Displacements

Figure 2.22: Horizontal Displacement

Node Angle Displacement 10-5 in

19 45o -0.0473

21 40o -0.2778

23 35o -0.3058

24 32.5o +0.2485

25 30o +1.5955

×


Static Analysis Verification Problems Hemispherical Shell Under Edge-Moment Loading

Input Data for Verification Problems No. 2.10**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEHEMISPHERICAL SHELL UNDER EDGE-MOMENT LOADING*ELTYPE 1, 36, 2*RCTABLE 1, 3,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1, 1,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.00000E+02, 2.50000E+00, 0.00000E+00, 0 3, 1,,, 1.00000E+02, 5.00000E+00, 0.00000E+00, 0 4, 1,,, 1.00000E+02, 7.50000E+00, 0.00000E+00, 0 5, 1,,, 1.00000E+02, 1.00000E+01, 0.00000E+00, 0 6, 1,,, 1.00000E+02, 1.25000E+01, 0.00000E+00, 0 7, 1,,, 1.00000E+02, 1.50000E+01, 0.00000E+00, 0 8, 1,,, 1.00000E+02, 1.75000E+01, 0.00000E+00, 0 9, 1,,, 1.00000E+02, 2.00000E+01, 0.00000E+00, 0 10, 1,,, 1.00000E+02, 2.25000E+01, 0.00000E+00, 0 11, 1,,, 1.00000E+02, 2.50000E+01, 0.00000E+00, 0 12, 1,,, 1.00000E+02, 2.75000E+01, 0.00000E+00, 0 13, 1,,, 1.00000E+02, 3.00000E+01, 0.00000E+00, 0 14, 1,,, 1.00000E+02, 3.25000E+01, 0.00000E+00, 0 15, 1,,, 1.00000E+02, 3.50000E+01, 0.00000E+00, 0 16, 1,,, 1.00000E+02, 3.75000E+01, 0.00000E+00, 0 17, 1,,, 1.00000E+02, 4.00000E+01, 0.00000E+00, 0 18, 1,,, 1.00000E+02, 4.25000E+01, 0.00000E+00, 0 19, 1,,, 1.00000E+02, 4.50000E+01, 0.00000E+00, 0 20, 1,,, 1.00000E+02, 4.75000E+01, 0.00000E+00, 0 21, 1,,, 1.00000E+02, 5.00000E+01, 0.00000E+00, 0 22, 1,,, 1.00000E+02, 5.25000E+01, 0.00000E+00, 0 23, 1,,, 1.00000E+02, 5.50000E+01, 0.00000E+00, 0 24, 1,,, 1.00000E+02, 5.75000E+01, 0.00000E+00, 0 25, 1,,, 1.00000E+02, 6.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 2, 1, 1, 1, 0 3, 4, 5,



3, 1, 1, 1, 0 5, 6, 7, 4, 1, 1, 1, 0 7, 8, 9, 5, 1, 1, 1, 0 9, 10, 11, 6, 1, 1, 1, 0 11, 12, 13, 7, 1, 1, 1, 0 13, 14, 15, 8, 1, 1, 1, 0 15, 16, 17, 9, 1, 1, 1, 0 17, 18, 19, 10, 1, 1, 1, 0 19, 20, 21, 11, 1, 1, 1, 0 21, 22, 23, 12, 1, 1, 1, 0 23, 24, 25,*MATERIALEX , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.30000E-01,DENS, 1, 0, 1.00000E+00,*LDCASE, ID= 1 1, 1, 3, 0, 3, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 25,MZ,-3.14159E+02,,, 0, 0*ENDDATA


Static Analysis Verification Problems Clamped Square Plate with Central Concentrated Load

2.11 Clamped Square Plate with Central Concentrated LoadTitle:

Clamped square plate with central concentrated load

Element Type:

3-D general shell element (NKTP = 20, NORDR = 2)

Problem:

A clamped square plate shown in Figure 2.23 is subjected to a central concentrated load of 10 lbs. Theplate has a side length of 8 inches and thickness of 0.12 inch. The objective is to computedisplacements and stresses along the X-axis of the plate.

Properties:


Due to symmetry, only a quarter of the plate is modeled using 2 2 and 4 4 meshes. Figure 2.23shows the 4 4 mesh. Clamped boundary conditions at X = 4.0 in and Y = 4.0 in are imposed bysetting all displacements and rotations to zero. Symmetric boundary conditions on Y = 0.0 are imposedby setting UY = 0.0 and ROTX = 0.0 and along X = 0.0 by setting UX and ROTY = 0.0.

Results and Comparisons:

The central vertical deflection obtained from NISA, for the 4 4 mesh, is 0.002497 inch which is invery good comparison with the theoretical value of 0.002489 inch (Ref. [1]).

The vertical deflection of the plate along the X axis is shown in Figure 2.25 for the two differentmeshes along with the theoretical solutions [1]. Convergence towards the theoretical solution is clearlyobserved as the mesh is refined. Figure 2.26 shows the bottom surface stress (SXX) along the x-axis incomparison with the finite element solution [2] where good agreement is found.

References:1. S.P. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells, 2nd Edition,

McGraw-Hill Book Co., New York, 1959.2. O.C. Zienkiewicz, The Finite Element Method, McGraw-Hill Book Co., London,

249-251, 1977.

Material:



×

× ×

×

×


Static Analysis Verification ProblemsClamped Square Plate with Central Concentrated Load

Figure 2.23: Clamped Plate with Central Concentrated Load

Figure 2.24: Finite Element Model



Figure 2.25: Vertical Deflection along X-axis

Figure 2.26: Bottom Surface Stresses along X-axis


Static Analysis Verification ProblemsClamped Square Plate with Central Concentrated Load

Input Data for Verification Problem No. 2.11**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = STATICSAVE = 26,27*TITLECLAMPED SQUARE PLATE WITH CENTRAL CONCENTRATED LOAD*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.2000000E-01, 1.2000000E-01, 1.2000000E-01, 1.2000000E-01, 1.2000000E-01, 1.2000000E-01, 1.2000000E-01, 1.2000000E-01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 5.00000E-01, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.50000E+00, 0.00000E+00, 0.00000E+00, 0 : : 75,,,, 1.00000E+00, 3.50000E+00, 0.00000E+00, 0 77,,,, 2.00000E+00, 3.50000E+00, 0.00000E+00, 0 79,,,, 3.00000E+00, 3.50000E+00, 0.00000E+00, 0 81,,,, 4.00000E+00, 3.50000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 48, 12, 11, 10, 46, 2, 1, 1, 1, 0,,,,, 0 3, 4, 5, 50, 14, 13, 12, 48, : : 15, 1, 1, 1, 0,,,,, 0 32, 33, 34, 79, 43, 42, 41, 77, 16, 1, 1, 1, 0,,,,, 0 34, 35, 36, 81, 45, 44, 43, 79,*MATERIALEX , 1, 0, 9.10000E+06,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 1, 1, 3, 0, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0



1,UY , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 : : 81,UZ , 0.00000E+00,,,,,,,, 0 81,ROTX, 0.00000E+00,,,,,,,, 0 81,ROTY, 0.00000E+00,,,,,,,, 0 81,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 1,FZ, 2.50000E+00,,, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*ENDDATA


Static Analysis Verification ProblemsCylindrical Shell Roof under Self-Weight

2.12 Cylindrical Shell Roof under Self-WeightTitle:

Cylindrical shell roof under self-weight

Element Type:

3-D four node thin shell element (NKTP = 40, NORDR = 1)

Problem:

A cylindrical shell roof with a radius 25 ft, length 50 ft and thickness 0.25 ft is subjected to self weightload of 90 lb/ft (along its length). The roof is supported by diaphragms on both the ends Figure 2.27.

Deflection along the direction of load application (i.e. Y-direction displacement) at the middle of thefree edge of the roof is computed.

Properties:


Due to the symmetry, only one quarter of the roof is modeled. The following boundary conditions areapplied:

1. At Z = 0.0 ft, Diaphragm boundary conditions UX = UY = 0.02. At Z = 25.0 ft, Symmetric boundary conditions UZ = ROTX = ROTY = 0.03. At X = 0.0 ft, Symmetric boundary conditions UX = ROTY = ROTZ = 0.0

Acceleration due to gravity is taken to be 32.0 ft/s2.

The problem is solved using three different mesh sizes of 4 4, 8 8 and 14 14, respectively.Figure 2.27 shows the 4 4 mesh.


Table 2.13 shows the vertical deflection at the center of the free edge of the roof (Z = 25.0 ft, at 40degrees), obtained from the NISA analysis. Analytical solutions for this problem are based on certainassumptions, and no exact solution is available [1]. NISA results are compared with the finite elementanalysis results [1] with the comparable number of degrees of freedom meshes.

Material:

EX = 432.0 106 lb/ft2 (Modulus of elasticity)


DENS = 11.25 lb-sec2/ft4 (Density)

×

× × ×

×


Static Analysis Verification Problems Cylindrical Shell Roof under Self-Weight

Figure 2.27: Geometry and Finite Element Mesh for the Cylindrical Shell Roof

Reference:1. R.D. Cook, Concepts and Applications of Finite Elenzent Analysis, 2nd Edition, John Wiley &

Sons, Inc., 1981.

Table 2.13: Deflection at the Middle of the Free Edge of Roof for Different Mesh Sizes

NISA Results 4-node Element Ref [1] Results 8-node Element

Mesh Deflection (ft) Deflection (ft)

4 4 0.3027 0.2935

8 8 0.3003 0.3019

14 14 0.3003 0.3015

×

×

×


Static Analysis Verification ProblemsCylindrical Shell Roof under Self-Weight

Input Data for Verification Problem No. 2.12**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE CYLINDRICAL ROOF SUBJECTED TO DEAD WEIGHT OF 90.0, 4X4 QUAD4 MESH*ELTYPE 1, 40, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01,*NODES 1, 1,,, 2.50000E+01, 9.00000E+01, 0.00000E+00, 0 2, 1,,, 2.50000E+01, 8.00000E+01, 0.00000E+00, 0 3, 1,,, 2.50000E+01, 7.00000E+01, 0.00000E+00, 0 4, 1,,, 2.50000E+01, 6.00000E+01, 0.00000E+00, 0 : : 82, 1,,, 2.50000E+01, 8.00000E+01,-2.50000E+01, 0 83, 1,,, 2.50000E+01, 7.00000E+01,-2.50000E+01, 0 84, 1,,, 2.50000E+01, 6.00000E+01,-2.50000E+01, 0 85, 1,,, 2.50000E+01, 5.00000E+01,-2.50000E+01, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 22, 21, 2, 1, 1, 1, 0,,,,, 0 2, 3, 23, 22, : : 15, 1, 1, 1, 0,,,,, 0 63, 64, 84, 83, 16, 1, 1, 1, 0,,,,, 0 64, 65, 85, 84,*MATERIALEX , 1, 0, 4.32000E+08,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.12500E+01,*LDCASE, ID= 1 0, 1, 3, 0,-1, 4, 0, 0.000, 0.000*LCTITLE UNIFORM DEAD WEIGHT FORCE.


Static Analysis Verification Problems Cylindrical Shell Roof under Self-Weight

*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 : : 84,ROTY, 0.00000E+00,,,,,,,, 0 85,UZ , 0.00000E+00,,,,,,,, 0 85,ROTX, 0.00000E+00,,,,,,,, 0 85,ROTY, 0.00000E+00,,,,,,,, 0*BODYFORCE0.0,0.0,0.0,0.0,32.0,0.0*ENDDATA


Static Analysis Verification ProblemsThin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder)

2.13 Thin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder)

Title:

Thin cylindrical shell under a concentrated force (pinched cylinder)

Element type:

3-D four node thin shell element (NKTP = 40, NORDR = 1)

Problem:

A cylindrical shell with a 300 inch radius, 600 inch length, and 3.0 inch thickness is subjected to aconcentrated force of 1 lb. (Figure 2.28).

The radial deflection of the cylinder at the point of load application is compared with the theoreticalsolution.

Properties:


Due to the symmetry, only one eighth of the cylinder is modeled with a 12 12 mesh (Figure 2.28).Following boundary conditions are applied:

1. At Z = 300 in, Symmetric boundary conditions UZ = ROTX = ROTY = 0.02. At X = 0.0 in, Symmetric boundary conditions UX = ROTY = ROTZ = 0.03. At Y = 0.0 in, Symmetric boundary conditions UY = ROTX = ROTZ = 0.0Results and Comparison:

Table 2.14 shows the radial deflection at the point of loading obtained from the finite element analysisand the theoretical analysis [1]. A good agreement between the results is evident.

Reference:1. S. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells, 2nd Edition, McGraw-

Hill Book Co. Inc., New York, 1959.

Material:



×

×


Static Analysis Verification Problems Thin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder)

Figure 2.28: Geometry and Finite Element Mesh for Thin Cylindrical Shell under Concentrated Force

Table 2.14: Radial Deflection at the Point of Loading for a Pinched Cylinder

Analysis Radial Deflection at thePoint of Loading (in)

NISA 4.62516 10-4

Theory [1] 4.51970 10-4

Percent error 2.28

×

×


Static Analysis Verification ProblemsThin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder)

Input Data for Verification Problem No. 2.13**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE PINCHED CYLINDRICAL SHELL WITH FREE ENDS. 12X12 QUAD4 MESH*ELTYPE 1, 40, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 3.0000000E+00, 3.0000000E+00, 3.0000000E+00, 3.0000000E+00, 3.0000000E+00, 3.0000000E+00, 3.0000000E+00, 3.0000000E+00,*NODES 1, 1,,, 3.00000E+02, 9.00000E+01, 3.00000E+02, 0 2, 1,,, 3.00000E+02, 8.25000E+01, 3.00000E+02, 0 3, 1,,, 3.00000E+02, 7.50000E+01, 3.00000E+02, 0 4, 1,,, 3.00000E+02, 6.75000E+01, 3.00000E+02, 0 : : 250, 1,,, 3.00000E+02, 2.25000E+01, 0.00000E+00, 0 251, 1,,, 3.00000E+02, 1.50000E+01, 0.00000E+00, 0 252, 1,,, 3.00000E+02, 7.50000E+00, 0.00000E+00, 0 253, 1,,, 3.00000E+02, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 22, 21, 2, 1, 1, 1, 0,,,,, 0 2, 3, 23, 22, : : 143, 1, 1, 1, 0,,,,, 0 231, 232, 252, 251, 144, 1, 1, 1, 0,,,,, 0 232, 233, 253, 252,*MATERIALEX , 1, 0, 3.00000E+06,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 0, 1, 3, 0,-1, 4, 0, 0.000, 0.000*LCTITLE CONCENTRATED FORCE.*SPDISP


Static Analysis Verification Problems Thin Cylindrical Shell Under A Concentrated Force (Pinched Cylinder)

**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 13,UY , 0.00000E+00,,,,,,,, 0 : : 253,UZ , 0.00000E+00,,,,,,,, 0 253,ROTX, 0.00000E+00,,,,,,,, 0 253,ROTY, 0.00000E+00,,,,,,,, 0 253,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 241,FY, 2.50000E-01,,, 0, 0*ENDDATA


Static Analysis Verification ProblemsA Clamped Square Orthotropic Plate Under Pressure Loading

2.14 A Clamped Square Orthotropic Plate Under Pressure Loading

Title:

A clamped square orthotropic plate under pressure loading

Element Type:

3-D eight noded solid element (NKTP = 4, NORDR = 1)

Problem:

A square orthotropic plate has unit length and 0.01 inch thickness. The plate is subjected to a pressureload of 1 psi (Figure 2.29). Compute the strain energy and transverse displacement at the center of theplate. The problem is analyzed for two sets of rotation angles for the material axes. The first set ofrotation angles is (0°, 0°, 15o) and the second (0°, 0°, 45o).

Properties:


Mesh sizes of 8 8 1 and 16 16 1 are used to model the plate. Figure 2.29 shows the 8 8 1mesh. The material rotation angles are defined on an element by element basis. Clamped boundaryconditions (UX = UY = UZ = 0.0) are imposed on all four sides of the plate. A pressure loading of 1psi is applied on the top face of the plate.


The strain energy and central transverse displacement are shown in Table 2.15 along with the results ofRef. [1]. The results are in good agreement. The convergence of the solution with the refinement of themesh can also be observed from the table

Material:

EX = 40000.0 psi

EY = 1000.0 psi (Moduli of elasticity in principal directions)

EZ = 1000.0 psi

GXY = 500.0 psi

GXZ = 500.0 psi (Shear moduli in material principal directions)

GYZ = 416.7 psi

NUXY = 0.25

NUXZ = 0.0 (Poisson's ratios in the material principal directions)

NUYZ = 0.0

× × × × × ×


Static Analysis Verification Problems A Clamped Square Orthotropic Plate Under Pressure Loading

Figure 2.29: Geometry and Finite Element Mesh of a Clamped Square Plate under Pressure Loading

Reference:1. A.K. Noor and M.D. Mathers, Finite Element Analysis of Anisotropic Plates, International Jour-

nal for Numerical Methods in Engineering, 11, 289-307, 1977.



Table 2.15: Comparison of results for the clamped square orthotropic plate under pressure loading

SourceRotation angle ( ) Strain energy Central Transverse

Displacement (w)

Mesh: 15o 0.1468 0.9094

Mesh: 15o 0.1688 0.9521

Reference 1 15o 0.17460 -

Mesh: 45o 0.1799 1.2738

Mesh: 45o 0.2184 1.5805

Reference 1 45o 0.2262 1.6311

θz

8 8 1××

16 16 1××

8 8 1××

16 16 1××


Static Analysis Verification Problems A Clamped Square Orthotropic Plate Under Pressure Loading

Input Data for Verification Problem No. 2.14**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLEPRESSURE LOADING ON AN ORTHOTROPIC CLAMPED PLATE, THETA=15*ELTYPE 1, 4, 1*NODES 1,,,,-5.00000E-01,-5.00000E-01,-5.00000E-03, 0 2,,,,-3.75000E-01,-5.00000E-01,-5.00000E-03, 0 3,,,,-2.50000E-01,-5.00000E-01,-5.00000E-03, 0 4,,,,-1.25000E-01,-5.00000E-01,-5.00000E-03, 0 : : 178,,,, 1.25000E-01, 5.00000E-01, 5.00000E-03, 0 179,,,, 2.50000E-01, 5.00000E-01, 5.00000E-03, 0 180,,,, 3.75000E-01, 5.00000E-01, 5.00000E-03, 0 181,,,, 5.00000E-01, 5.00000E-01, 5.00000E-03, 0*ELEMENTS 1, 1, 1, 0, 1 1, 2, 11, 10, 101, 102, 111, 110, 2, 1, 1, 0, 1 2, 3, 12, 11, 102, 103, 112, 111, : : 63, 1, 1, 0, 1 70, 71, 80, 79, 170, 171, 180, 179, 64, 1, 1, 0, 1 71, 72, 81, 80, 171, 172, 181, 180,*MATDIR2 1, 0, 0, 0.00000E+00, 0.00000E+00, 0.00000E+00,0 64, 1, 0, 0.00000E+00, 0.00000E+00, 0.00000E+00,0*MATERIALEX , 1, 0, 4.00000E+01,EY , 1, 0, 1.00000E+00,EZ , 1, 0, 1.00000E+00,NUXY, 1, 0, 2.50000E-01,NUXZ, 1, 0, 0.00000E+00,NUYZ, 1, 0, 0.00000E+00,GXY , 1, 0, 5.00000E-01,GXZ , 1, 0, 5.00000E-01,GYZ , 1, 0, 4.16670E-01,



*SETS 101,R, 20, 60, 20, 102,R, 10, 120, 10, *LDCASE, ID= 1 1, 1, 2, 1, 2, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 : : 180,UZ , 0.00000E+00,,,,,,,, 0 181,UX , 0.00000E+00,,,,,,,, 0 181,UY , 0.00000E+00,,,,,,,, 0 181,UZ , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,1,0, 0, 1.0, 0, 0 2,,,1,0, 0, 1.0, 0, 0 3,,,1,0, 0, 1.0, 0, 0 4,,,1,0, 0, 1.0, 0, 0 : : 61,,,1,0, 0, 1.0, 0, 0 62,,,1,0, 0, 1.0, 0, 0 63,,,1,0, 0, 1.0, 0, 0 64,,,1,0, 0, 1.0, 0, 0*PRINTCNTLAVND,102DISP,102ELFO,101ELSE,101ELST,101SLFO,0*ENDDATA


Static Analysis Verification Problems Analysis of Nine Layer Cross-Ply Orthotropic Plate

2.15 Analysis of Nine Layer Cross-Ply Orthotropic PlateTitle:

Analysis of nine layer cross-ply orthotropic plate

Element Type:

3-D eight noded laminated composite shell element (NKTP = 32, NORDR = 2)

Problem:

A nine layers laminated square plate, ten inches wide, is simply supported and subjected to a uniformpressure of 10 psi (Figure 2.30). The lamination is alternating at 0o and 90o with respect to global x-axis. The total thickness of the five layers at 0o is same as the total thickness of the four layers at 90o.

To show a wide applicability of the element, three problems are solved. (a) a very thin plate (b) a thinplate and (c) a thick plate. The central transverse displacement, total strain energy and the centralmoment resultant (MXX) are computed in each case.

Properties:

In the material directions, the elastic properties and allowable stresses are:

Note: The allowable stresses are input as material properties for laminated composite elements.

Material:

EX = 40.0 106 psi(Moduli of elasticity)

EY = 1.0 106 psi

GXY = GXZ = 0.6 106 psi(Shear moduli)

GYZ = 0.5 106 psi


FXT = 150.0 103 psi

(Allowable Stresses)

FXC = 100.0 103 psi

FYT = 10.0 103 psi

FYC = 15.0 103 psi

FS = 20.0 103 psi

×

×

×

×

×

×

×

×

×


Static Analysis Verification ProblemsAnalysis of Nine Layer Cross-Ply Orthotropic Plate

Figure 2.30: Geometry and Finite Element Mesh of a Nine Layer Cross-Ply Orthotropic Plate


Since the plate is orthotropic in the global coordinates, using symmetry in the problem, only onequarter of the geometry is modeled using a 2 2 mesh (Figure 2.30). The following boundaryconditions are applied:

At X = 0.0 in, simply supported boundary conditions UY = UZ = ROTX = 0.0At Y = 0.0 in, simply supported boundary conditions UX = UZ = ROTY = 0.0At X = 5 in, symmetric boundary conditions UX = ROTY = 0.0

×



At Y = 5 in, symmetric boundary conditions UY = ROTX = 0.0

Three different cases are analyzed for central transverse deflection, total strain energy and the centralmoment resultant.

Case 1. A very thin plate of total thickness 0.010 inCase 2. A thin plate of total thickness 0.10 inCase 3. A thick plate of total thickness 1.0 in


Table 2.16 shows the comparisons of central transverse displacement, total strain energy and centralmoment resultant (MXX) with exact and approximate solutions [I]. Results are in good agreement inall the three cases.

Reference:1. A.K. Noor and M.D. Mathers, Shear Flexible Finite Element Models of Laminated Composite

Plates and Shells, NASA TN D-8044, 1975.

Table 2.16: Comparison of results for the simply supported square orthotropic plate under pressure loading

Thickness to Length Ratio (t/a)

Central Transverse Deflection Strain Energy Central Moment

(MXX) Resultant

NISA Ref. [1] NISA Ref. [1] NISA Ref. [1]

0.1 0.0005844 0.0005848 0.126 0.126 91.68 84.24

0.01 0.4474 0.4486 92.8 93.0 96.42 88.79

0.001 365.30 447.2 74952.0 92700.0 80.60 88.85



Input Data for Verification Problem No. 2.15**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = STATICSAVE = 26,27*TITLESTATIC ANALYSIS OF NINE LAYER CROSS-PLY ORTHOTROPIC PLATES*ELTYPE 1, 32, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 1.2500000E-02, 3, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 4, 8,1, 0**_DISP3_: NAME =OTHER 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 5, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 6, 8,1, 0**_DISP3_: NAME =OTHER 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,890.0,90.0,90.0,90.0,90.0,90.0,90.0,90.0*LAMSQ2**_DISP3_: 1,NAME = 1,9,1,11,2,1,2,1,2,1,2,1,12,1,2,1,2,1,2,1,1,1



1,1,1,1,1,1,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.25000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.50000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.75000E+00, 0.00000E+00, 0.00000E+00, 0 5,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 0.00000E+00, 1.25000E+00, 0.00000E+00, 0 8,,,, 2.50000E+00, 1.25000E+00, 0.00000E+00, 0 10,,,, 5.00000E+00, 1.25000E+00, 0.00000E+00, 0 11,,,, 0.00000E+00, 2.50000E+00, 0.00000E+00, 0 12,,,, 1.25000E+00, 2.50000E+00, 0.00000E+00, 0 13,,,, 2.50000E+00, 2.50000E+00, 0.00000E+00, 0 14,,,, 3.75000E+00, 2.50000E+00, 0.00000E+00, 0 15,,,, 5.00000E+00, 2.50000E+00, 0.00000E+00, 0 16,,,, 0.00000E+00, 3.75000E+00, 0.00000E+00, 0 18,,,, 2.50000E+00, 3.75000E+00, 0.00000E+00, 0 20,,,, 5.00000E+00, 3.75000E+00, 0.00000E+00, 0 21,,,, 0.00000E+00, 5.00000E+00, 0.00000E+00, 0 22,,,, 1.25000E+00, 5.00000E+00, 0.00000E+00, 0 23,,,, 2.50000E+00, 5.00000E+00, 0.00000E+00, 0 24,,,, 3.75000E+00, 5.00000E+00, 0.00000E+00, 0 25,,,, 5.00000E+00, 5.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0,,,,, 0 1, 2, 3, 8, 13, 12, 11, 6, 2, 1, 1, 0, 0,,,,, 0 3, 4, 5, 10, 15, 14, 13, 8, 3, 1, 1, 0, 0,,,,, 0 11, 12, 13, 18, 23, 22, 21, 16, 4, 1, 1, 0, 0,,,,, 0 13, 14, 15, 20, 25, 24, 23, 18,*MATERIALEX , 1, 0, 4.00000E+07,EY , 1, 0, 1.00000E+06,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 6.00000E+05,GXZ , 1, 0, 6.00000E+05,GYZ , 1, 0, 5.00000E+05,DENS, 1, 0, 1.00010E+00,FXC , 1, 0, 1.00000E+05,FXT , 1, 0, 1.50000E+05,FYC , 1, 0, 1.50000E+04,FYT , 1, 0, 1.00000E+04,FS , 1, 0, 2.00000E+04,*SETS 101,S, 1, 4, 102,R, 1, 25, 2,



*LDCASE, ID= 1 1, 1, 2, 1, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 : : 25,UX , 0.00000E+00,,,,,,,, 0 25,UY , 0.00000E+00,,,,,,,, 0 25,ROTX, 0.00000E+00,,,,,,,, 0 25,ROTY, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,1,0, 0, 10.0, 0, 0 2,,,1,0, 0, 10.0, 0, 0 3,,,1,0, 0, 10.0, 0, 0 4,,,1,0, 0, 10.0, 0, 0*SFDCOMP0.0,0,0*PRINTCNTLAVND,102DISP,102ELFO,101ELSE,101ELST,101REAC,102SLFO,0*ENDDATA


Static Analysis Verification Problems This page is intentionally blank

2.16 This page is intentionally blank


Static Analysis Verification ProblemsAnalysis of a Three-layer Sandwich Plate

2.17 Analysis of a Three-layer Sandwich PlateTitle:

Analysis of a three-layer sandwich plate

Element Type:

3-D eight noded sandwich shell element (NKTP = 33, NORDR = 2)

Problem:

A simply supported square sandwich plate, ten inches wide, is subjected to a uniform normal pressureof 100 psi. The face sheets and the core material are orthotropic in global XY plane (i.e. material axescoincide with global X and Y axes). The face sheets and the core material are 0.028 inch and 0.75 inchthick, respectively.

The center transverse deflections is computed. Also, contour plots of the inplane stresses are shown.

Properties:

Note: The allowable stresses are input as material properties for laminated composite elements. A zerovalue for EX indicates that material type 2 is a core material.

Material:

In the material principal directions, the following properties are used

(a) Face Sheets

EX = 10.0 106 psi(Elastic moduli)

EY = 4.0 106 psi

GXY = 1.875 106 psi (Inplane shear modulus)


FXT = FXC = 50.0 103 psi

FYT = 15.0 103 psi (Allowable stresses)

FYC = 20.0 103 psi

(b) Core Material

EX = 0.0 (Elastic modulus)

GXZ = 3.0 104 psi (Shear moduli)

GYZ = 1.2 104psi

FTS = 2.0 104psi (Allowable stresses)

×

×

×

×

×

×

×

×

×


Static Analysis Verification Problems Analysis of a Three-layer Sandwich Plate

Figure 2.31: Geometry and Finite Element Mesh of a Three-layer Sandwich Plate

Finite Element Model:Using symmetry in the model, only one quarter of the geometry is modelled using 2 2 mesh(Figure 2.33). Following boundary conditions are applied:

- At x = 0 in, simply supported boundary conditions UX = UZ = ROTY = 0.0- At y = 0 in, simply supported boundary conditions UY = UZ = ROTX = 0.0- At x = 5 in, symmetric boundary conditions UX = ROTY = 0.0- At y = 5 in, symmetric boundary conditions UY = ROTX = 0.0- Rotations about the normal to the plate are suppressed at all nodes to avoid spurious

rotations.Results and Comparison:Table 2.17 shows the comparison of center deflections obtained from NISA with the analytical andfinite element results [1].

Figure 2.32 is a stress survey plot. Examination of this reveals that the inplane shear stress componentis 93% of allowable for element 1, while Y-direction normal stress component is most critical inelements 2, 3, and 4, the maximum being 95% of the allowable for element 4.

Contour plots of the three inplane stress components (Sxx, Syy, Sxy)in the top layer are shown in Figure2.32. Since the plate is in pure bending, the bottom layer stresses are the opposite of the top layer

×



stresses. The transverse shear stress resultant (Qx) contour plots show the distribution of transverseshear stress in the core.

Reference:1. T.P. Khatua and Y.K. Chueng, Bending and Vibration of Multilayer Sandwich Beams and Plates,

International Journal for Numerical Methods in Engineering, 6, 11-24, 1973.

Table 2.17: Comparison of results for the simply supported sandwich plate under pressure loading

Figure 2.32: Contour Plots of Stress and Resultant for a Simply Supported Sandwich Plate under Uniform Pressure

Source Central Transverse Deflection (w)

NISA 0.1254

Analytical solution [1] 0.123

Finite element solution [l] 0.1213


Static Analysis Verification Problems Analysis of a Three-layer Sandwich Plate

Input Data for Verification Problem No. 2.17**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE 2.19 : PROPPED CANTILVER BEAM WITH A GAP*ELTYPE 1, 13, 1 2, 42, 1*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 3.0000000E+00, 1.0000000E+01, 0.0000000E+00, 0.0000000E+00, 2, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 1.0000000E-01, 1.0000000E-24,-9.0000000E+01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.00000E+01,-1.00000E-01, 0.00000E+00, 0 4,,,, 1.00000E+01,-1.01000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 3, 4, 3, 1, 2, 2, 0 2, 3,*MATERIALEX , 1, 0, 3.33333E+02,NUXY, 1, 0, 0.00000E+00,*LDCASE, ID= 1 0, 1, 0, 0, 0, 0, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 4,UX , 0.00000E+00,,,,,,,, 0 4,UY , 0.00000E+00,,,,,,,, 0 4,ROTZ, 0.00000E+00,,,,,,,, 0



*CFORCE**_DISP3_: CFORCE, SET = 1 2,FY,-2.00000E+00,,, 0, 0*ENDDATA


Static Analysis Verification Problems This page is intentionally blank

2.18 This page is intentionally blank


Static Analysis Verification ProblemsPropped Cantilever Beam with a Gap

2.19 Propped Cantilever Beam with a GapTitle:

Propped cantilever beam with a gap

Element Type:


2-D gap element (NKTP = 42)

Problem:

A 10 inch long cantilever beam is propped at the free end by a column of 10 inch height. The beam andcolumn joint has a 0.1 inch gap (Figure 2.33). The beam and the column have same cross-sectionalarea and moment of inertia. The cantilever is subjected to a tip load of 2 pounds.

The displacement under the load and the axial force in the column are calculated.

Properties:

Finite Element Model:The beam and the column are modeled using one 2-D beam element each. The gap is modeled using 2-D gap element. The gap is oriented in the Y direction. The initial gap value is given as 0.1 inch. Thetolerance for iterative convergence for gap closing is 10-8. The nodes at the left end of the beam and atthe base of the column are constrained (UX = UY = ROTZ = 0.0). A concentrated force (FY = -2.0 lbs)is applied on the node at the right end of the beam.

Material:



Cross-section:

IZZ = 10 in4 (Second area moment about beam Z-axis)

A = 3 in 2 (Cross-sectional area)

×


Static Analysis Verification Problems Propped Cantilever Beam with a Gap

Figure 2.33: Propped Cantilever Beam with a Gap

Results and Comparison:The displacement (UY) under the load is found to be 0.109091 inch. The force (FX) in column is0.909091 pounds. The theoretical values for the displacement and the force are given by the followingformulas. These formulas can be easily derived using the basic rules of mechanics of structure. Theformulas for calculating UY and FX are:

(2.1)

(2.2)

The NISA results exactly matches the values calculated using the above formulas

UY P∗L3

3∗IZZ A∗L2+( )---------------------------------------- A∗L2∗GW

E∗ 3∗IZZ A∗L2+( )-----------------------------------------------+=

FX P∗L2∗A3∗IZZ A∗L2+( )

---------------------------------------- 3∗A∗EX∗IZZ∗GWL∗ 3∗IZZ A∗L2+( )----------------------------------------------+=


Static Analysis Verification ProblemsPropped Cantilever Beam with a Gap

Input Data for Verification Problem No. 2.19**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE 2.21 : TRUSS WITH MISALIGNED SUPPORTS*ELTYPE 1, 15, 1 2, 18, 1 3, 42, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 1.0000000E+01, 1.0000000E+01, 2, 1,1, 0**_DISP3_: NAME =OTHER 1.0000000E+09, 3, 1,1, 0**_DISP3_: NAME =OTHER 1.0000000E+06, 4, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 8.0000001E-01, 9.9999999E-09,-9.0000000E+01, 5, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 8.9999998E-01, 9.9999999E-09,-9.0000000E+01, 6, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 1.2000000E+00, 9.9999999E-09,-9.0000000E+01, 7, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 5.0000000E-01, 9.9999999E-09,-9.0000000E+01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+02, 1.20000E+02, 0.00000E+00, 0 3,,,, 2.00000E+02, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.00000E+02, 1.20000E+02, 0.00000E+00, 0 :


Static Analysis Verification Problems Propped Cantilever Beam with a Gap

: 19,,,, 6.00000E+02,-1.20000E+00, 0.00000E+00, 0 20,,,, 8.00000E+02,-1.00000E+02, 0.00000E+00, 0 21,,,, 8.00000E+02,-5.00000E-01, 0.00000E+00, 0 22,,,, 1.00000E+03,-1.00000E+02, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 1, 3, : : 28, 0, 3, 6, 0 7, 19, 29, 0, 3, 7, 0 9, 21,*MATERIALEX , 1, 0, 3.00000E+07,*SETS 111,R, 3, 9, 2, 15, 21, 2, *LDCASE, ID= 1 0, 1, 0, 0, 0, 0, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 6,UX , 0.00000E+00,,,,,,,, 0 12,UX , 0.00000E+00,,,,,,,, 0 12,UY , 0.00000E+00,,,,,,,, 0 14,UX , 0.00000E+00,,,,,,,, 0 : : 20,UY , 0.00000E+00,,,,,,,, 0 21,UX , 0.00000E+00,,,,,,,, 0 22,UX , 0.00000E+00,,,,,,,, 0 22,UY , 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 3,FY,-4.00000E+04,,, 0, 0 5,FY,-4.00000E+04,,, 0, 0 7,FY,-4.00000E+04,,, 0, 0 9,FY,-4.00000E+04,,, 0, 0*PRINTCNTLDISP,111SLFO,0*ENDDATA


Static Analysis Verification ProblemsBeam on a Tensionless Elastic Foundation

2.20 Beam on a Tensionless Elastic FoundationTitle:

Beam on a tensionless elastic foundation

Element Type:


2-D spring element (NKTP = 18, NORDR = 1)


Problem:

A 30 inch long beam (3 inches wide, 1 inch deep) rests on an elastic soil foundation (Figure 2.34). Thesoil is active in compression only. A concentrated load of 8000 lbs. is applied at the midspan of thebeam. The foundation modulus is 5000 lb./in3.

The midspan displacement and the contact length of the beam are calculated.

Properties:

Finite Element Model:Due to symmetry, only one half of the structure is modeled using 2-D beam, 2-D spring and 2-D gap(or compression-only spring) elements (Figure 2.34). The beam is divided into eight equal length beamelements with two nodes per element. The tensionless foundation is simulated by nine discreteequidistant springs. The two springs (near the midspan) are assumed to be active in tension andcompression. These two springs support the beam initially. The remaining seven compression-onlysprings are modeled by gap elements with negative flexibility coefficients and zero gap width.

The spring constant (K) is taken as foundation modulus times beam-width times spacing between thesprings (K = 5000*3*1.875 = 28125 lb/in). The flexibility coefficient (F) for a typical compression-only spring is the reciprocal of the spring constant (F = -1/K).

Material:

EX = 30.0 106 psi (Modulus of elasticity of beam material)


Cross-Section:

IZZ = 0.25 in4 (Second area moment about beam Z-axis)

A = 3 in2 (Cross-sectional area)

×


Static Analysis Verification Problems Beam on a Tensionless Elastic Foundation

Figure 2.34: Geometry, Loading and Finite Element Model for a Beam on Elastic Foundation


Results for the displacement under the concentrated load and the extent of the contact are compared tothe exact solution (Ref. [1]) and to the approximate solution of Ref. [2] in Table 2.18. The contactlength (projected on to the X-axis) is approximately computed by linear interpolation of the verticaldisplacement (UY) over the beam segment in which the deflection changes sign. Also shown in thetable are the results of two additional NISA models (input data not provided) with different number ofspring elements.

References:1. M. Hetenyi, Beams on Elastic Foundation, University of Michigan Press, Ann Arbor, Michigan,

1967.2. F. Faten Mahmoud, N.J. Salamon and W.R. Marks, A Direct Automated Procedure for Friction-

less Contact Problem, International Journal for Numerical Methods in Engineering, 18, 245-257, 1982.



Table 2.18: Comparison of results for beam on tensionless elastic foundation

* Input data not provided

Source Displacemet under the Load (inches)

Half Length ofContact (inches)

NISA model A (6 springs)* 0.043468 10.401

NISA model B (9 springs) 0.043471 10.4125

NISA model C (1 1 springs)* 0.043478 10.546

Ref. [1] (Exact) 0.043480 10.505

Ref. [2] 0.042420 11.792


Static Analysis Verification Problems Beam on a Tensionless Elastic Foundation

Input Data for Verification Problem No. 2.20 **EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEHOLLOW CIRCULAR CYLINDER SUBJECTED TO NONAXISYMMETRIC LOADING*ELTYPE 1, 34, 2*NODES 1,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 5.50000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 6.50000E+00, 0.00000E+00, 0.00000E+00, 0 : : 42,,,, 5.50000E+00, 8.00000E+00, 0.00000E+00, 0 43,,,, 6.00000E+00, 8.00000E+00, 0.00000E+00, 0 44,,,, 6.50000E+00, 8.00000E+00, 0.00000E+00, 0 45,,,, 7.00000E+00, 8.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 8, 13, 12, 11, 6, 2, 1, 1, 0, 0 3, 4, 5, 10, 15, 14, 13, 8, : : 7, 1, 1, 0, 0 31, 32, 33, 38, 43, 42, 41, 36, 8, 1, 1, 0, 0 33, 34, 35, 40, 45, 44, 43, 38,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,*SETS 101,S, 1, 2, 102,R, 35, 45, 10, *LDCASE, ID= 1 1, 1, 2, 1, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0



2,UX , 0.00000E+00,,,,,,,, 0 : : 4,UZ , 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0*FRCNTL1,6,0,5*ANGSEC0.0,18.0,54.0,90.0,180.0*FRCOEFFRSET,1,1.0COS,1,1.887COS,2,1.577COS,3,1.15COS,4,0.709COS,5,0.34*PRESSURE**_DISP3_: PRESSURE, SET = 1 2,,,2,0, 1, 227.36, 0, 0 4,,,2,0, 1, 227.36, 0, 0 6,,,2,0, 1, 227.36, 0, 0 8,,,2,0, 1, 227.36, 0, 0*PRINTCNTLAVND,102DISP,102ELFO,101ELSE,101ELST,101REAC,102SLFO,0*ENDDATA


Static Analysis Verification Problems Truss on Misal igned Supports

2.21 Truss on Misaligned SupportsTitle:

Truss on misaligned supports

Element Type:

2-D spar element (NKTP = 15)

2-D spring element (NKTP = 18)


Problem:

A 1000 inch span (5 panels of 200 inch each) truss has six (flexible) supports Figure 2.35. The foursupports, located in between the end supports, are misaligned. The misalignments and the three point-loads on the truss are shown in Figure 2.35. The displacements at the joints (at the bottom chord of thetruss) and the final gap widths are calculated.

Properties:


Two-dimensional spar elements of constant cross section are used to model the truss. The flexiblesupports are represented by spring elements. The end supports are made stiffer than the intermediatesupports Figure 2.35. The misalignment of the truss is modeled with gap elements. The spring supportsare fixed (UX = UY = 0.0) to the ground. The joint at the top center of the truss is restrained frommoving in horizontal (UX = 0.0) direction. Only vertical movement is allowed at the center supports.Four concentrated forces 40,000 lbs. each are applied at the bottom chord of the truss as shown.


The calculated displacements at selected nodes and final gap width are shown in Table 2.19 and Table2.20 respectively. The support forces on the intermediate springs are listed in Table 2.21. The tablesalso show the values obtained from Ref. [1]. As indicated, NISA results and values from Ref. [1]match exactly.

Material:


Cross-section:

A1 = 10.0 in2 (Cross-sectional area at left end)

A2 = 10.0 in2 (Cross-sectional area at right end)

×


Static Analysis Verification ProblemsTruss on Misaligned Supports

Figure 2.35: Truss on Misaligned Supports

Reference:1. W.R. Marks, Solution of Frictionless Contact Problems by a Conjugate Gradient Technique,

M.Sc.Thesis, University of Wisconsin, Milwaukee, 1979.



Table 2.19: Nodal displacement in inches

Table 2.20: Extent of the contact gap

Table 2.21: Intermediate support forces

NodeNISA Reference

X Y X Y

3 -1.1801 10-1 -5.9902 10-1 -1.1801 10-1 -5.9902 10-1

5 -3.8220 10-2 -9.1924 10-1 -3.8220 10-2 -9.1924 10-1

7 -5.3594 10-2 -9.0721 10-1 5.3594 10-2 -9.0721 10-1

9 1.2368 10-1 -5.3625 10-1 1.2368 10-1 -5.362 10-1

15 0.0 0.0 0.0 0.0

17 0.0 -1.9241 10-2 1.5285 10-13 -1.9241 10-2

19 0.0 0.0 0.0 0.0

21 0.0 -3.6254 10-2 4.3663 10-13 -3.6254 10-2

Node Contact Pair

Initial Gap Width

Extent of GapContact

Reference [l]Final Gap Width

3-15 0.8 No Contact No Contact

5-17 0.9 -1.9241 10-2 -1.9241 10-2

7-19 1.2 No Contact No Contact

9-21 0.5 -3.6254 10-2 -3.6254 10-2

Spring NISA Ref. [1]

2 0.0 0.0

3 19240 19240

4 0.0 0.0

5 36254 36250.

× × × ×

× × × ×

× × × ×

× × × ×

× × ×

× × ×

× ×

× ×


Static Analysis Verification ProblemsTruss on Misaligned Supports

Input Data for Verification Problem No. 2.21**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE 2.21 : TRUSS WITH MISALIGNED SUPPORTS*ELTYPE 1, 15, 1 2, 18, 1 3, 42, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 1.0000000E+01, 1.0000000E+01, 2, 1,1, 0**_DISP3_: NAME =OTHER 1.0000000E+09, 3, 1,1, 0**_DISP3_: NAME =OTHER 1.0000000E+06, 4, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 8.0000001E-01, 9.9999999E-09,-9.0000000E+01, 5, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 8.9999998E-01, 9.9999999E-09,-9.0000000E+01, 6, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 1.2000000E+00, 9.9999999E-09,-9.0000000E+01, 7, 5,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 5.0000000E-01, 9.9999999E-09,-9.0000000E+01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+02, 1.20000E+02, 0.00000E+00, 0 3,,,, 2.00000E+02, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.00000E+02, 1.20000E+02, 0.00000E+00, 0 :



: 19,,,, 6.00000E+02,-1.20000E+00, 0.00000E+00, 0 20,,,, 8.00000E+02,-1.00000E+02, 0.00000E+00, 0 21,,,, 8.00000E+02,-5.00000E-01, 0.00000E+00, 0 22,,,, 1.00000E+03,-1.00000E+02, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 1, 3, : : 28, 0, 3, 6, 0 7, 19, 29, 0, 3, 7, 0 9, 21,*MATERIALEX , 1, 0, 3.00000E+07,*SETS 111,R, 3, 9, 2, 15, 21, 2, *LDCASE, ID= 1 0, 1, 0, 0, 0, 0, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 6,UX , 0.00000E+00,,,,,,,, 0 12,UX , 0.00000E+00,,,,,,,, 0 12,UY , 0.00000E+00,,,,,,,, 0 14,UX , 0.00000E+00,,,,,,,, 0 : : 20,UY , 0.00000E+00,,,,,,,, 0 21,UX , 0.00000E+00,,,,,,,, 0 22,UX , 0.00000E+00,,,,,,,, 0 22,UY , 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 3,FY,-4.00000E+04,,, 0, 0 5,FY,-4.00000E+04,,, 0, 0 7,FY,-4.00000E+04,,, 0, 0 9,FY,-4.00000E+04,,, 0, 0*PRINTCNTLDISP,111SLFO,0*ENDDATA


Static Analysis Verification ProblemsHollow Circular Cylinder Subjected to Nonaxisymmetric Loading

2.22 Hollow Circular Cylinder Subjected to Nonaxisymmetric Loading

Title:

Hollow circular cylinder subjected to nonaxisymmetric loading

Element Type:

Axisymmetric solid element with nonaxisymmetric loading (NKTP = 34, NORDR = 2)

Problem:

A hollow circular cylinder, fixed at one end, has an inside radius of 5.0 inches, and outside radius of7.0 inches and length of 8.0 inches. It is subjected to an external pressure, P( ) = (Pi/2)(l+cos(m ))per unit area (where m = 10/3 and ) on a portion of the circumference as shown inFigure 2.36. The total load over the whole cylinder is 10,000 lb/in. It is desired to compute thedisplacement at the outer radius of the free end and stresses at the fixed end.

Properties:


For comparison and verification of results, two models are used in solving this problem. The first is anaxisymmetric model with non-axisymmetric loading, NKTP = 34 and NORDR = 2, whereas thesecond model is a three dimensional one, NKTP = 4 and NORDR = 2.

In the first model, a radial section is modeled using 2 4 in mesh of axisymmetric elements

(Figure 2.36). The loading is symmetric about line which when represented by Fourier serieswill have only the cosine terms. These terms are given by:

(2.3)

where and m = 10/3

The Fourier coefficients are computed as follows:

Material:



θ θπ m θ π m⁄≤ ≤⁄–

×

×

θ 0o=

ao1

2π------

Pi2----- 1 mθcos+( ) θd

π–

π∫=

an1π---

Pi2----- 1 mθcos+( ) nθ θdcos

π–

π∫=

π m⁄ 54o=

2-100 NISA II Veri fication Manual

Static Analysis Verification Problems Hollow Circular Cylinder Subjected to Nonaxisymmetric Loading

The value of Pi in the pressure distribution is calculated from the given total load ;

Figure 2.36: Geometry and FE Model of Shell with Nonaxisymmetric Loading

Figure 2.37: Top View of Cylinder

(2.4)

which gives:

(lb/in2)

Ptot

PtotPi2----- 1 mθcos+( )R θd

π–m------

πm----

∫= 10000 lb in⁄=

PimπR------- Ptot=



The value of the Fourier coefficients a, can be calculated by the integration. By choosing the first sixterms, we obtain:

where

In the second model, 3-D solid elements (NKTP = 4, NORDR = 2) are used to model 180' segment ofthe cylinder. A total of 80 elements and 577 nodes are used (see Figure 2.38).

A similar pressure load is applied on the element surface along the circumferential direction accordingto the function.

(2.5)

where and m=10/3

and the value of as obtained above is lb/in


Table 2.23 and Table 2.24 show the comparison of global displacements and von Mises stress atvarious angular positions. Good agreement between the two solutions is found. Table 2.25 also showsthe local displacements at node 45 at various angles as output from NISA. The global values may becomputed from these local values by suitable coordinate transformation.

a0 1.0Ptot=

a2 1.557Ptot=

a4 0.709Ptot=

a1 1.88Ptot=

a3 1.150Ptot=

a5 0.340Ptot=

PtotPtot2πR----------=

P θ( ) Pi 2⁄( ) 1 mθcos+( )=

π m⁄– θ π m⁄≤ ≤

Pi Pi103

------PtotπR--------- 1515.8= =



Figure 2.38: 3-D Model of the Cylinder

Table 2.22: Comparison of the two models

Model8 node Axisymmetric Elements with Nonaxisymmetric Loading

20-node Three Dimensional Solid Elements

Number of elements 8 80

Number of nodes 37 577

Total DOF 111 1731



Table 2.23: Comparison of displacement at node 45 in global direction

Table 2.24: The von Mises equivalent stress at nodes (1, 2, 3, 4, 5)

Table 2.25: Local displacements at node 45

Angle(deg.)

Axisymmetric Model Global Dis-placement Component 10-3 inch

3-D Model Global Displacement Component 10-3 inch

UX UY UZ UX UY UZ

0o -3.047 0.8282 0.0 -3.02 0.822 0.0

18o -2.51 0.6915 -0.394 -2.49 0.687 -0.397

54o -0.671 0.3218 0.545 -0.671 0.0346 0.529

90o -0.629 -0.2688 1.111 -0.627 -0.267 1.10

180o 0.1826 0.05329 0.0 0.179 0.0528 0.0

Angle Element TypeNodal Points Stress (psi)

Node-1 Node-2 Node-3 Node-4 Node-5

0o Axisymmetric3-D solid

7196.77309.0

4814.95232.0

4180.955052.0

8160.07970.0

12486.611948.0

18o Axisymmetric 3-D solid

6530.06499.0

4404.64624.0

3612.154243.0

6630.56454.0

10157.09688.0


2004.72009.0

1777.91828.0

1985.052032.0

2836.82729.0

3863.63722.0


111.0125.2

90.03101.7

94.5498.25

125.6123.4

168.9176.0

AngleUx: Radial

10-3 inchUy: Axial

10-3 inch Uz: Circumferential

10-3 inch

0o - 3.047 0.8282 0.0

18o -2.509 0.6915 0.4004

54o 0.04679 0.03218 0.8629

90o 1.111 -0.2688 0.6290

180o -0.1826 0.05329 0.0

× ×

× × ×



Input Data for Verification Problem No. 2.22**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEHOLLOW CIRCULAR CYLINDER SUBJECTED TO NONAXISYMMETRIC LOADING*ELTYPE 1, 34, 2*NODES 1,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 5.50000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 6.50000E+00, 0.00000E+00, 0.00000E+00, 0 : : 42,,,, 5.50000E+00, 8.00000E+00, 0.00000E+00, 0 43,,,, 6.00000E+00, 8.00000E+00, 0.00000E+00, 0 44,,,, 6.50000E+00, 8.00000E+00, 0.00000E+00, 0 45,,,, 7.00000E+00, 8.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 8, 13, 12, 11, 6, 2, 1, 1, 0, 0 3, 4, 5, 10, 15, 14, 13, 8, : : 7, 1, 1, 0, 0 31, 32, 33, 38, 43, 42, 41, 36, 8, 1, 1, 0, 0 33, 34, 35, 40, 45, 44, 43, 38,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,*SETS 101,S, 1, 2, 102,R, 35, 45, 10, *LDCASE, ID= 1 1, 1, 2, 1, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0



2,UX , 0.00000E+00,,,,,,,, 0 : : 4,UZ , 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0*FRCNTL1,6,0,5*ANGSEC0.0,18.0,54.0,90.0,180.0*FRCOEFFRSET,1,1.0COS,1,1.887COS,2,1.577COS,3,1.15COS,4,0.709COS,5,0.34*PRESSURE**_DISP3_: PRESSURE, SET = 1 2,,,2,0, 1, 227.36, 0, 0 4,,,2,0, 1, 227.36, 0, 0 6,,,2,0, 1, 227.36, 0, 0 8,,,2,0, 1, 227.36, 0, 0*PRINTCNTLAVND,102DISP,102ELFO,101ELSE,101ELST,101REAC,102SLFO,0*ENDDATA


Static Analysis Verification Problems Cantilever Elbow Subjected to Mult iple Loading Conditions

2.23 Cantilever Elbow Subjected to Multiple Loading Conditions

Title:

Cantilever elbow subjected to multiple loading conditions

Element Type:

3-D elbow element (NKTP = 47, NORDR = 1)

Problem:

A quarter bend cantilever elbow with outer diameter = 10 in, thickness = 0.25 in and radius ofcurvature = 100 in Figure 2.39 is subjected to (a) concentrated tip force FZ, (b) concentrated tipmoment MX, and (c) concentrated tip moment MY.

Properties:

Material:



Cross-section:

D = 10 in (Outer diameter)

t = 0.25 in (Pipe wall thickness)

R = 100 in (Radius of curvature)

Deformation Factors:

SHY = 1.887 (Default) (Shear correction factor)

FLX = (ASME) (Flexibility factor)

×

1.65r2

tR----------------dd


Static Analysis Verification ProblemsCantilever Elbow Subjected to Multiple Loading Conditions


The problem is modeled using one elbow element (Figure 2.39). All six degrees of freedom areconstrained at the fixed end. Three different load cases are analyzed.

Case 1. A concentrated force FZ = 100 lb at the free endCase 2. A concentrated moment MX = 100 lb-in at the free endCase 3. A concentrated moment MY = 100 lb-in at the free end


The out-of-plane deflection and rotations at the free end are compared with the theoretical valuesobtained from Ref. 1 in Table 2.26.

Reference:1. I. Vigness, Elastic Properties of Curved Tubes, Transactions of the ASME, February 1943.



Table 2.26: Comparison of the NISA and theoretical results

Figure 2.39: Geometry and the Finite Element Mesh for the Cantilever Elbow underMultiple Loading Conditions

ResultsCase 1 Case 2 Case 3

NISA Theory[1] NISA Theory[1] NISA Theory[1]

UZ1.87157

10-11.86168

10-1-1.57518

10-3-1.57518

10-3-1.04656

10-3-1.04654

10-3

X-1.57518

10-3-1.57518

10-32.47429

10-52.47427

10-51.47457

10-61.47456

10-6

Y-1.04656

10-3-1.04654

10-31.47457

10-61.47456

10-62.47429

10-52.47427

10-5

× × × × × ×

θ × × × × × ×

θ × × × × × ×


Static Analysis Verification ProblemsCantilever Elbow Subjected to Multiple Loading Conditions

Input Data for Verification Problem No. 2.23**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLECANTILEVER QUARTER BEND ELBOW SUBJECTED TO DIFFERENT LOADINGS*ELTYPE 1, 47, 1*RCTABLE 1, 12,1, 0**_DISP3_: NAME =OTHER 1.0000000E+01, 2.5000000E-01, 1.0000000E+02,-2.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1, 1,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.00000E+02, 9.00000E+01, 0.00000E+00, 0 10,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 10, 0,000000,000000,*MATERIALEX , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*LCTITLECONCENTRATED END FORCE FZ*SPDISP**_DISP3_: SPDISP, SET = 1 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 2,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 1,FZ, 1.00000E+02,,, 0, 0*LDCASE, ID= 2 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*LCTITLECONCENTRATED END FORCE MX



*CFORCE**_DISP3_: CFORCE, SET = 2 1,MX, 1.00000E+02,,, 0, 0*LDCASE, ID= 3 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*CFORCE**_DISP3_: CFORCE, SET = 3 1,MY, 1.00000E+02,,, 0, 0*ENDDATA


Static Analysis Verification ProblemsCantilever Curved Beam Subjected to Mult iple Loading Conditions

2.24 Cantilever Curved Beam Subjected to Multiple Loading Conditions

Title:

Cantilever curved beam subjected to multiple loading conditions

Element Type:

3-D elbow element (NKTP = 47, NORDR = 1)

Problem:

A quarter bend cantilever beam with a circular cross section of diameter = 10 in and a radius ofcurvature = 100 in Figure 2.40 is subjected to (a) concentrated tip force FZ, (b) concentrated tipmoment FY, and (c) concentrated tip moment MZ.

Properties:


The problem is modeled using one elbow element (Figure 2.40). All six degrees of freedom areconstrained at the fixed end. Three different load cases are analyzed.

Case 1. A concentrated force FX = 100 lb at the free endCase 2. A concentrated force FY = 100 lb at the free endCase 3. A concentrated moment MZ = 100 lb-in at the free end


The in-plane displacements and rotation at the free end are compared with the theoretical values inTable 2.27. The theoretical values are obtained by Castigliano's first theorem as given by Ref. [1] andthe deflections due to axial force which the Ref. [1] ignores.

Material:



Cross-section:

D = 10 in (Section diameter)

R = 100 in (Radius of curvature)

Deformation Factors:

ALPHY = 0.0 in (no shear deformation) (Shear correction factor)

FLX = 1.0 in (Flexibility Factor)

×


Static Analysis Verification Problems Cantilever Curved Beam Subjected to Mult iple Loading Conditions

Reference:1. R.J. Roark, Formula for Stress and Strain, 4th Edition, McGraw-Hill Book Co., New York,

1965.

Figure 2.40: Geometry and the Finite Element Mesh for the Cantilever Curved Beam under Multiple Loading Conditions



Table 2.27: Comparison of the NISA and theoretical results

Results UZ θX θY

Case 1NISA 1.60100 10-2 1.01796 10-2 2.03718 10-4

Theory[1] 1.60100 10-2 1.01796 10-2 2.03718 10-4

Case 2NISA 1.01796 10-2 7.26633 10-3 1.16282 10-4

Theory[1] 1.01796 10-2 7.26633 10-3 1.16281 10-4

Case 3NISA 2.03718 10-4 1.16282 10-4 3.20000 10-6

Theory[1] 2.03718 10-4 1.16281 10-4 3.20000 10-6

× × ×

× × ×

× × ×

× × ×

× × ×

× × ×


Static Analysis Verification Problems Cantilever Curved Beam Subjected to Mult iple Loading Conditions

Input Data for Verification Problem No. 2.24**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLECANTILEVER CURVED BEAM SUBJECTED TO DIFFERENT LOADINGS*ELTYPE 1, 47, 1*RCTABLE 1, 12,1, 0**_DISP3_: NAME =OTHER 1.0000000E+01, 5.0000000E+00, 1.0000000E+02, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1, 1,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.00000E+02, 9.00000E+01, 0.00000E+00, 0 10,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 10, 0,000000,000000,*MATERIALEX , 1, 0, 1.00000E+07,DENS, 1, 0, 1.00000E-03,ALPX, 1, 0, 1.00000E-03,*LDCASE, ID= 1 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*LCTITLECONCENTRATED END FORCE FX*SPDISP**_DISP3_: SPDISP, SET = 1 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 2,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 1,FX, 1.00000E+02,,, 0, 0*LDCASE, ID= 2 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*LCTITLE



CONCENTRATED END FORCE FY*CFORCE**_DISP3_: CFORCE, SET = 2 1,FY, 1.00000E+02,,, 0, 0*LDCASE, ID= 3 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*CFORCE**_DISP3_: CFORCE, SET = 3 1,MZ, 1.00000E+02,,, 0, 0*ENDDATA


Static Analysis Verification Problems Demonstration of the Use of Restarts in the Static and Buckling Analysis of the L-shaped Frame

2.25 Demonstration of the Use of Restarts in the Static and Buckling Analysis of the L-shaped Frame

Title:

Demonstration of the use of Restarts in the static and buckling analysis of the L-shaped frame shownin Figure 2.41.

Element Type:

2-D planar Beam element (NKTP= 13, NORDR= 1)

Problem:

Restarts are to be used to analyze the frame for the three loading situations shown in Figure 2.42through Figure 2.44. Also, the concentrated vertical load at B which will cause the frame to buckle isto be determined.

Properties:


The frame is modeled using 2-D planar beam elements. As shown in Figure 2.42, the portion AB of theframe has five elements and the portion BC one element. For purposes of static analysis, it suffices tohave just one element for AB as well. But a buckling analysis also needs to be done and the use of onlyone element to model the buckling of a column is insufficient in that it does not accurately model itsbuckled shape. In the present problem, the buckling of the frame due to a concentrated vertical load atB is likely to involve buckling of the member AB as a column; hence the use of 5 elements to modelAB.

The in-plane translational degrees of freedom are arrested at the hinged end A and all in-plane degreesof freedom are constrained at the built-in end C.

Material:

EX = 30.0 106 lbs/in2 (Modulus of elasticity)


Cross-section:

Square of side 4 in

A = 16 in2 (Area of cross-section)


×


Static Analysis Verification ProblemsDemonstration of the Use of Restarts in the Static and Buckling Analysis of the L-shaped Frame

Analysis Procedure and Results:

Once any finite element model is made, it is good practice to perform a model check by doing a checkrun (using the executive command EXEC=CHECK). It may be noted that File 26 is saved here so thata Restart 1 may be initiated later without having to input the model data all over again in the restart.

After the successful completion of the check run, a Static Restart 1 is done in order to find thedisplacement of the structure for load case ID no. 100. It may be seen that the model data block isabsent since it is accessed from File 26 of the previous run. It is for this reason that it is essential thatthe FILENAME prefix be the same in the initial and restart input files. Files 24, 27, 28 and 30 areexplicitly saved in this run since it is expected that further restarts may be done later. File 26 is savedautomatically at the end of this restart run because any old file saved in a preceding run is always savedin its new state if modified, and in its old state if unaltered. It may also be observed from *LDCASE

Figure 2.41: Geometry and Dimensions Figure 2.42: Finite Element Model

Figure 2.43: Load Case No. 200 Figure 2.44: Load combination ID No. 300



that only the displacement solution is computed for load case ID 100. The vertical displacement ofpoint B for this load case is shown in Table 2.28.

The next step in the procedure is to analyze the structure for a different loading situation, namely, loadcase ID 200. A Static Restart 3 is done now instead of starting from scratch, the advantage being thatinformation from previous runs (for example, model data, element stiffness matrices, decomposedstiffness matrix) can be readily utilized. The boundary conditions need not be specified in this restartrun because they are unchanged from the earlier run. Again from *LDCASE, it can be seen that onlydisplacements are computed for load case ID 200 at this time. The vertical displacement of point B forthis load case is also listed in Table 2.28.

The displacement of the frame when subjected to the loads in Figure 2.44 is easily computed by doinga Restart 2 now and combining load case ID 100 and load case ID 200 (*LDCOMB) with factors of0.5 and 1.0, respectively. Assume also that the end force/moments for load case ID 100 (not computedin the Restart 1 run) are now desired and that so also are the reactions for each of the two load cases.These can be obtained in this restart run by appropriate *LDCASE entries. The displacements of thestructure evaluated in earlier runs are used to compute these quantities. The end moment at B for loadcase ID 100 and the moment reactions at C for load case ID 100 and load case ID 200 are tabulated inTable 2.28.

The values of the various quantities determined by the above restarts can be verified either byperforming a simple indeterminate frame analysis as in Ref. [I], or determining the same quantities bycarrying out a scratch NISA run without employing restarts. The values obtained from the scratch runsare shown in the last column of Table 2.28. It can be seen that the two columns are identical.

Finally, a buckling analysis is performed to evaluate the concentrated vertical load at B which willcause the frame to buckle. This can be done now by a Buckling Restart 4 from the last static restartrun. Note that the *LDCASE data group refers to load case ID 100. If this data group had beenomitted, the latest *LDCASE data group specified in the earlier runs (i.e., load case ID 200) wouldhave been used for the buckling analysis. The first buckling load factor is found to be 1.1597 whichgives a buckling load of -1159.7 Kips, which is in close agreement with the theoretical value of -1158.3 Kips obtained from Ref [2].

References:1. R.L. Ketter, G.C. Lee and S.P. Prawel, Jr., Structural Analysis and Design, McGraw-Hill Book

Co., New York, 1979.2. A. Chajes, Principles of Structural Stability Theory, Civil Engineering and Engineering Mechan-

ics Series, N.M. Newmark and W.J. Hall, Editors, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1974.



Table 2.28: Comparison of displacements and reactions at selected points

Quantity Restart Run Scratch Run

Vertical displacement of point B for load case IDno. 100 -1.87289 10-1 (in) -1.87289 10-1 (in)

Vertical displacement of point B for load case IDno. 200 -6.01978 10-3 (in) -6.01978 10-3 (in)

Vertical displacement of point B for load combination ID no. 300 -9.96641 10-2 (in) -9.96641 10-2 (in)

Moment at B for load case ID no. 100 3.80416 101 (Kip-in) 3.80416 101(Kip-in)

Reaction moment at C for load case ID no. 100 -6.34151 101 (Kip-in) -6.34151 101(Kip-in)

Reaction moment at C for load case ID no. 200 -7.25286 102(Kip-in) -7.25286 102(Kip-in)

× ×

× ×

× ×

× ×

× ×

× ×



Input Data for Verification Problem No. 2.25**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONEXECUTION = CHECKFILE = S25RESEQUENCE = OFFSAVE = 26,27*TITLE DEMONSTRATION OF RESTARTS IN STATIC ANALYSIS: CHECK RUN*ELTYPE 1, 13, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 1.6000000E+01, 2.1333334E+01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 1.80000E+01, 0.00000E+00, 0 3,,,, 0.00000E+00, 3.60000E+01, 0.00000E+00, 0 4,,,, 0.00000E+00, 5.40000E+01, 0.00000E+00, 0 5,,,, 0.00000E+00, 7.20000E+01, 0.00000E+00, 0 6,,,, 0.00000E+00, 9.00000E+01, 0.00000E+00, 0 7,,,, 9.00000E+01, 9.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 2, 3, 3, 1, 1, 1, 0 3, 4, 4, 1, 1, 1, 0 4, 5, 5, 1, 1, 1, 0 5, 6, 6, 1, 1, 1, 0 6, 7,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 0.00000E+00,*LDCASE, ID= 100 0, 0, 0, 0, 0, 0, 0, 0.000, 0.000*LCTITLE A CONCENTRATED VERTICAL FORCE OF -1000 KIPS AT NODE 6*SPDISP**_DISP3_: SPDISP, SET = 100



Input File for Static Restart 1


1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 7,UX , 0.00000E+00,,,,,,,, 0 7,UY , 0.00000E+00,,,,,,,, 0 7,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 100 6,FY,-1.00000E+06,,, 0, 0*ENDDATA

PROBLEM=S25R1.NIS ANALYSIS=STATIC RESTART=1 RESEQ=OFF**** FILES 24,27,28 AND 30 ARE SAVED HERE SO THAT A STATIC** RESTART 2 OR 3 MAY BE DONE LATER. FILE 26 IS AUTOMATICALLY** SAVED SINCE IT IS ACCESSED FROM PREVIOUS RUN (RESTART 0). FILE=S25 SAVE=24,27,28,30***TITLE DEMONSTRATION OF RESTARTS IN STATIC ANALYSIS: RESTART 1****** ONLY DISPLACEMENTS ARE COMPUTED; ELEMENT STRAIN ENERGY, ELEMENT** INTERNAL FORCES, REACTIONS AND ELEMENT STRESSES/STRAINS ARE NOT** COMPUTED IN THIS RESTART RUN FOR LOAD CASE 100.*LDCASE, ID=100 0,0,0,0*****LCTITLE A CONCENTRATED VERTICAL FORCE OF -1000 KIPS AT NODE 6*CFORCE 6,FY,-1000000.*SPDISP 1,UX,0.0,0,0,UY 7,UX,0.0,0,0,UY,ROTZ*ENDDATA

PROBLEM=S25R2.NIS ANALYSIS=STATIC RESTART=2 RESEQ=OFF




FILE=S25*TITLE DEMONSTRATION OF RESTARTS IN STATIC ANALYSIS: RESTART 2****** IN THIS RESTART, REACTIONS AND ELEMENT STRESSES ARE COMPUTED** FOR LOAD CASE 100; ALSO, THE ELEMENT STRESSES ARE PRINTED IN** OUTPUT FILE*LDCASE, ID=100 0,1,2,0*PRINTCNTL ELST,0****** IN THIS RESTART, REACTIONS ARE COMPUTED FOR LOAD CASE 200*LDCASE, ID=200 0,1,0,0****** COMBINATION OF LOAD CASES 100 AND 200 IS DONE TO DETERMINE** DISPLACEMENTS OF STRUCTURE TO A NEW SET OF LOADS.*LDCOMB, ID=300 1,0 100,0.5,200,1.0*ENDDATA

PROBLEM=S25R3.NIS ANALYSIS=STATIC RESTART=3 RESEQ=OFF FILE=S25*TITLE DEMONSTRATION OF RESTARTS IN STATIC ANALYSIS: RESTART 3****** ONLY DISPLACEMENTS ARE COMPUTED; ELEMENT STRAIN ENERGY, ELEMENT** INTERNAL FORCES, REACTIONS AND ELEMENT STRESSES/STRAINS ARE NOT** COMPUTED IN THIS RESTART RUN FOR LOAD CASE 200.*LDCASE, ID=200 0,0,0,0,*****LCTITLE A UNIFORM VERTICAL LOAD OF -10 KIPS/FT ON ELEMENT 6*PRESSURE 6,,,1,0,0,833.333333,**



Input File for Buckling Restart 4

** NO SPDISP DATA GROUP IS SPECIFIED HERE; THUS THE SAME** SPDISP DATA AS IN LOAD CASE ID=100 IS USED.***ENDDATA

PROBLEM=S25R4.NIS ANALYSIS=BUCKLING RESTART=4 EIGEN=SUBSPACE,CONVENTIONAL RESEQ=OFF FILE=S25*TITLE DEMONSTRATION OF RESTARTS: BUCKLING RESTART 4 FROM A STATIC RESTART**** THE BUCKLING LOAD FACTORS ARE DETERMINED FOR LOAD CASE 100*LDCASE, ID=100 0,0,0,0,***EIGCNTL 2,0** NO SPDISP DATA GROUP IS SPECIFIED HERE; THUS THE SAME** SPDISP DATA AS IN LOAD CASE ID=100 IS USED.*ENDDATA


Static Analysis Verification Problems Interlaminar Shear Stresses for Eight-layer Antisymmetric Angle-ply Laminated Square Plates

2.26 Interlaminar Shear Stresses for Eight-layer Antisymmetric Angle-ply Laminated Square Plates

Title:

Interlaminar shear stresses for eight-layer antisymmetric angle-ply laminated square plates.

Element Type:1. Eight noded 3-D laminated composite shell element (NKTP=32,NORDR=2)2. Twelve noded 3-D laminated composite shell element (NKTP=32,NORDR=3)3. Six noded 3-D laminated composite shell element (NKTP=32,NORDR=ll)Problem:

Figure 2.45 shows a square plate of size 32 32 in. and a total thickness, h equal to 3.2 in., subjectedto a uniform distributed pressure, q equal to 1.0 ksi. Global axes, x and y, are placed at the middlesurface (reference surface) of the plate. Axis z is a normal to the reference surface. Nine layers, 0/45/45/45/45/-45/-45/-45/-45, are selected to represent eight-layer anti-symmetric angle-ply plate. Toplayer, with angle 0 (measured with respect to x-axis) and thickness 1.0 10-15, is considered toexpress interlaminar shear stresses with respect to the global axes.

Numerical results for interlaminar shear stresses obtained from NISA II are compared with that due toChaudhuri and Seide [1] as well as the CLT (classical lamination theory) solution.

Properties:

The following material properties are used:


The plate is modeled, using 16 16 mesh for NORDR = 2 and 3; 12 12 mesh for NORDR = 11, forthe following boundary conditions:

At X = 0.0 and a, UX=UY=UZ=ROTX=0.0

At Y = 0.0 and b, UX=UY=UZ=ROTY=0.0

Rotations (ROTZ) about the normal to the plate are suppressed at all nodes to avoid spurious rotations.

EX = 25 l06 psi

EY = 106 psi

GXY=GXZ = 5.0 l05 psi

GYZ = 2.0 l05 psi

NUXY = 0.25

×

×

×

×

×

× ×


Static Analysis Verification ProblemsInterlaminar Shear Stresses for Eight-layer Antisymmetric Angle-ply Laminated Square Plates


Figure 2.46 shows the comparison of interlaminar shear stresses, , obtained from NISA with theCLT and LCST (layerwise constant shear angle theory) results [1].

Figure 2.45: Simply Supported Plate Model

Reference:1. R.A. Chaudhuri and P. Seide, An Approximate Semi-Analytical Method for Prediction of Inter-

laminar Shear Stresses in an Arbitrarily Laminated Thick Plate, Computers & Structures, 25(4),627-636, 1987.

σxz



Figure 2.46: Transverse Shear Stress Variation through the Thickness



Input Data for Verification Problem No. 2.26A**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = S26ARESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE INTERLAMINER SHEAR STRESSES FOR EIGHT-LAYER ANTISYMMETRIC ANGLE-PLY SQUARE PLATES (SS4)*ELTYPE 1, 32, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,8-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.03,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*LAMSQ2**_DISP3_: 1,NAME = 1,9,1,12,1,1,1,1,1,1,1,1,31,1,1,1,2,2,2,2,1,11,1,1,1,1,1,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.00000E+00, 0.00000E+00, 0.00000E+00, 0 : : 1086,,,, 2.90000E+01, 3.20000E+01, 0.00000E+00, 0 1087,,,, 3.00000E+01, 3.20000E+01, 0.00000E+00, 0 1088,,,, 3.10000E+01, 3.20000E+01, 0.00000E+00, 0



1089,,,, 3.20000E+01, 3.20000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0,,,,, 0 1, 2, 3, 36, 69, 68, 67, 34, 2, 1, 1, 0, 0,,,,, 0 3, 4, 5, 38, 71, 70, 69, 36, : : 253, 1, 1, 0, 0,,,,, 0 1015, 1016, 1017, 1050, 1083, 1082, 1081, 1048, 254, 1, 1, 0, 0,,,,, 0 1017, 1018, 1019, 1052, 1085, 1084, 1083, 1050, 255, 1, 1, 0, 0,,,,, 0 1019, 1020, 1021, 1054, 1087, 1086, 1085, 1052, 256, 1, 1, 0, 0,,,,, 0 1021, 1022, 1023, 1056, 1089, 1088, 1087, 1054,*MATERIALEX , 1, 0, 2.50000E+04,EY , 1, 0, 1.00000E+03,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 5.00000E+02,GXZ , 1, 0, 2.00000E+02,GYZ , 1, 0, 5.00000E+02,*SETS 101,S, 1, 128, 144, 102,R, 1, 100, 5, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 1089,UZ , 0.00000E+00,,,,,,,, 0 1089,ROTX, 0.00000E+00,,,,,,,, 0 1089,ROTY, 0.00000E+00,,,,,,,, 0 1089,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 1.0, 0, 0 2,,,2,0, 0, 1.0, 0, 0 3,,,2,0, 0, 1.0, 0, 0 4,,,2,0, 0, 1.0, 0, 0 : :



Input Data for Verification Problem No. 2.26B

253,,,2,0, 0, 1.0, 0, 0 254,,,2,0, 0, 1.0, 0, 0 255,,,2,0, 0, 1.0, 0, 0 256,,,2,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.08,3,1*PRINTCNTLAVND,102DISP,102ELFO,101ELSE,101ELST,101REAC,102SLFO,0*ENDDATA

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE INTERLAMINER SHEAR STRESSES FOR EIGHT-LAYER ANTISYMMETRIC ANGLE-PLY SQUARE PLATES (SS4)*ELTYPE 1, 32, 11*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15,*LAMANGLE1,845.0,45.0,45.0,45.0,45.0,45.0,45.0,45.02,8-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.03,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0



*LAMSQ2**_DISP3_: 1,NAME = 1,9,1,12,1,1,1,1,1,1,1,1,31,1,1,1,2,2,2,2,1,11,1,1,1,1,1,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 2.66667E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 5.33333E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 8.00000E+00, 0.00000E+00, 0.00000E+00, 0 : : 166,,,, 2.40000E+01, 3.20004E+01, 0.00000E+00, 0 167,,,, 2.66667E+01, 3.20004E+01, 0.00000E+00, 0 168,,,, 2.93333E+01, 3.20004E+01, 0.00000E+00, 0 169,,,, 3.20000E+01, 3.20004E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 15, 29, 28, 27, 14, 2, 1, 1, 1, 0,,,,, 0 1, 2, 3, 16, 29, 15, : : 70, 1, 1, 1, 0,,,,, 0 139, 140, 141, 154, 167, 153, 71, 1, 1, 1, 0,,,,, 0 141, 155, 169, 168, 167, 154, 72, 1, 1, 1, 0,,,,, 0 141, 142, 143, 156, 169, 155,*MATERIALEX , 1, 0, 2.50000E+04,EY , 1, 0, 1.00000E+03,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 5.00000E+02,GXZ , 1, 0, 5.00000E+02,GYZ , 1, 0, 2.00000E+02,*SETS 101,S, 1, 35, 36, 102,R, 1, 100, 5, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0



: : 169,UZ , 0.00000E+00,,,,,,,, 0 169,ROTX, 0.00000E+00,,,,,,,, 0 169,ROTY, 0.00000E+00,,,,,,,, 0 169,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 1.0, 0, 0 2,,,2,0, 0, 1.0, 0, 0 3,,,2,0, 0, 1.0, 0, 0 4,,,2,0, 0, 1.0, 0, 0 : : 69,,,2,0, 0, 1.0, 0, 0 70,,,2,0, 0, 1.0, 0, 0 71,,,2,0, 0, 1.0, 0, 0 72,,,2,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.07,1,52*PRINTCNTLAVND,102DISP,102ELSE,101ELST,101SLFO,0*ENDDATA


Static Analysis Verification Problems Interlaminar Shear Stresses for Cross-Ply Laminated Spherical Shells of Rectangular Planform.

2.27 Interlaminar Shear Stresses for Cross-Ply Laminated Spherical Shells of Rectangular Planform.

Title:

Interlaminar shear stresses for cross-ply laminated spherical shells of rectangular planform.

Element Type:1. Eight noded 3-D laminated composite shell element (NKTP=32, NORDR=2)2. Twelve noded 3-D laminated composite shell element (NKTP=32,NORDR=3)3. Six noded 3-D laminated composite shell element (NKTP=32, NORDR=11)Problem:

Figure 2.47 shows geometry of a spherical shell of radius of curvature, R equal to 640 in. and totalthickness, h equal to 3.2 in., subjected to a uniform distributed pressure, q equal to 1.0 ksi. Local axes,x and y, are placed at middle surface (reference surface) of the shell thickness. Axis z is normal to thereference surface.

Numerical results for interlaminar shear stresses obtained from NISA II are compared with the exactsolution presented by Reddy [1].

Properties:

The following material properties are used:


The shell is modeled, using 16 16 mesh for NORDR = 2 and 3; and 16 8 mesh for NORDR=11,for the following boundary conditions:

At x = 0.0 and a, UX=UZ=ROTY=0.0

At y = 0.0 and b, UY=UZ=ROTX=0.0

Rotations (ROTZ) about the normal to the shell are suppressed at all nodes to avoid spurious rotations.

Material:

EX = 25 l06 psi

EY = 106 psi

GXY = GXZ = 5.0 105psi

GYZ = 2.0 l05 psi

NUXY = 0.25

×

×

×

× ×


Static Analysis Verification ProblemsInterlaminar Shear Stresses for Cross-Ply Laminated Spherical Shells of Rectangular Planform.


Table 2.29-Table 2.31 show the comparison of interlaminar shear stresses, and obtained fromNISA II with the exact solution [1].

Figure 2.47: A Simply Supported Spherical Shell of Rectangular Planform

Reference:1. J.N. Reddy, Exact Solutions of Moderately Thick Laminated Shells, ASCE Journal of Engineer-

ing Mechanics, 110(5), 794-809, 1984.

σxz σyz



Table 2.29: Transverse shear stresses of a cross-ply laminated spherical shell with R/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=2.

Table 2.30: Transverse shear stresses of a cross-ply laminated spherical shell withR/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=3

Table 2.31: Transverse shear stresses of a cross-ply laminated spherical shell with R/a = R/b = 20 and a/h = 10 for NKTP=32 & NORDR=11

* Stresses are constant within the element.

Laminations

, at X=-15.57, Y=14.42, ;

Gauss Point No. 3 ofElement No. 113

at X=-0.4243, Y=0.4225. ;


NISA EXACT[l] NISA EXACT[l]

-90/0 -687 (z=3h/4) -6.98 (z=3h/4) -6.76 (z=h/4) -6.98 (z=h/4)

-90/0/0/-90 -5.88 (z=h/2) -5.94 (z=h/2) -4.52 (z=h/2) -4.68 (z=h/2)

Laminations

, at X=-15.77, Y=15.77, ;


at X=-0.2270, Y=0.2253. ;



-90/0 -7.12 (z=3h/4) -7.26 (z=3h/4) -7.02 (z=h/4) -7.16 (z=h/4)

-90/0/0/-90 -6.05 (z=h/2) -6.11 (z=h/2) -4.80 (z=h/2) -4.97(z=h/2)

Laminations

, at X=-14.66, Y=13.66,

; Element No. 49*

at X=-2.666, Y=1.333. ; Element No. 7*


-90/0 -6.36 (z=3h/4) -6.61 (z=3h/4) -6.11 (z=h/4) -6.51 (z=h/4)

-90/0/0/-90 -5.45 (z=h/2) -5.72 (z=h/2) -3.98 (z=h/2) -4.28(z=h/2)

σxz

Z 640≈

σyz

Z 640≈

σxz

Z 640≈

σyz

Z 640≈

σxz

Z 640≈

σxz

Z 640≈



Input Data for Verification Problem No 2.27A**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEINTERLAMINAR SHEAR STRESSES FOR CROSS-PLY LAMINATED SPHERICAL SHELLS OF RECTANGULAR PLANFORM (R/A=20; A/H=10; 0/90/90/0; SS3)*ELTYPE 1, 32, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15,*LAMANGLE1,8-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.02,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.03,845.0,45.0,45.0,45.0,45.0,45.0,45.0,45.04,8-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.0,-45.0*LAMSQ2**_DISP3_: 1,NAME = 1,4,5,11,1,1,1,1,2,2,1,1,11,1*NODES 1,,,,-1.60000E+01, 0.00000E+00, 6.39800E+02, 0 2,,,,-1.50003E+01, 0.00000E+00, 6.39824E+02, 0 3,,,,-1.40006E+01, 0.00000E+00, 6.39847E+02, 0 4,,,,-1.30008E+01, 0.00000E+00, 6.39868E+02, 0 : : 830,,,, 1.29974E+01, 3.20000E+01, 6.39868E+02, 0 831,,,, 1.39972E+01, 3.20000E+01, 6.39846E+02, 0 832,,,, 1.49969E+01, 3.20000E+01, 6.39824E+02, 0



833,,,, 1.59966E+01, 3.20000E+01, 6.39800E+02, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 35, 53, 52, 51, 34, 2, 1, 1, 1, 0,,,,, 0 3, 4, 5, 36, 55, 54, 53, 35, : : 254, 1, 1, 1, 0,,,,, 0 777, 778, 779, 798, 829, 828, 827, 797, 255, 1, 1, 1, 0,,,,, 0 779, 780, 781, 799, 831, 830, 829, 798, 256, 1, 1, 1, 0,,,,, 0 781, 782, 783, 800, 833, 832, 831, 799,*G2 1,,, 1.432394, -1.432394 , 0.0 33,1,,1.432394, 1.432394 , 0.0 34,,, 1.34287, -1.432394 , 0.0 50,1,, 1.34287, 1.432394 , 0.0 : : 784,,, -1.342860, -1.432394 , 0.0 800,1,, -1.342860, 1.432394 , 0.0 801,,, -1.432384, -1.432394 , 0.0 833,1,, -1.432384, 1.432394 , 0.0 *MATERIALEX , 1, 0, 2.50000E+04,EY , 1, 0, 1.00000E+03,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 5.00000E+02,GXZ , 1, 0, 5.00000E+02,GYZ , 1, 0, 2.00000E+02,*SETS 101,S, 1, 8, 113, 102,R, 1, 100, 5, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 833,UZ , 0.00000E+00,,,,,,,, 0 833,ROTX, 0.00000E+00,,,,,,,, 0 833,ROTY, 0.00000E+00,,,,,,,, 0



Input Data for Verification Problem No 2.27B

833,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 1.0, 0, 0 2,,,2,0, 0, 1.0, 0, 0 3,,,2,0, 0, 1.0, 0, 0 4,,,2,0, 0, 1.0, 0, 0 : : 253,,,2,0, 0, 1.0, 0, 0 254,,,2,0, 0, 1.0, 0, 0 255,,,2,0, 0, 1.0, 0, 0 256,,,2,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.07,1,52*PRINTCNTLAVND,102DISP,102ELSE,101ELST,101SLFO,0*ENDDATA

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEINTERLAMINAR SHEAR STRESSES FOR CROSS-PLY LAMINATED SPHERICAL SHELLS OF RECTANGULAR PLANFORM (R/A=20; A/H=10; 0/90/90/0; SS3)*ELTYPE 1, 32, 3*RCTABLE 1, 12,1, 0**_DISP3_: NAME =OTHER 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01,*LAMANGLE1,12



-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.0-90.0,-90.0,-90.0,-90.02,120.0,0.0,0.0,0.0,0.0,0.0,0.0,0.00.0,0.0,0.0,0.0*LAMSQ2**_DISP3_: 1,NAME = 1,4,5,11,1,1,1,1,2,2,1,1,11,1*NODES 1,,,,-1.60000E+01, 0.00000E+00, 6.39800E+02, 0 2,,,,-1.53335E+01, 0.00000E+00, 6.39816E+02, 0 3,,,,-1.46671E+01, 0.00000E+00, 6.39832E+02, 0 4,,,,-1.40006E+01, 0.00000E+00, 6.39847E+02, 0 : : 1374,,,, 1.39971E+01, 3.20000E+01, 6.39846E+02, 0 1375,,,, 1.46636E+01, 3.20000E+01, 6.39832E+02, 0 1376,,,, 1.53301E+01, 3.20000E+01, 6.39816E+02, 0 1377,,,, 1.59966E+01, 3.20000E+01, 6.39800E+02, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 4, 51, 68, 87, 86, 85, 84, 67, 50, 2, 1, 1, 1, 0,,,,, 0 : : 255, 1, 1, 1, 0,,,,, 0 1288, 1289, 1290, 1291, 1310, 1327, 1374, 1373, 1372, 1371, 1326, 1309, 256, 1, 1, 1, 0,,,,, 0 1291, 1292, 1293, 1294, 1311, 1328, 1377, 1376, 1375, 1374, 1327, 1310,*G2 1,,, 1.432394, -1.432394 , 0.0 49,1,,1.432394, 1.432394 , 0.0 50,,, 1.3727110, -1.432394 , 0.0 66,1,, 1.3727110, 1.432394 , 0.0 : :980,,, -0.656515, -1.432394 , 0.0 996,1,, -0.656515, 1.432394 , 0.0 914,,, -0.596832, -1.432394 , 0.0 962,1,, -0.596832, 1.432394 , 0.0 *MATERIALEX , 1, 0, 2.50000E+04,EY , 1, 0, 1.00000E+03,



NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 5.00000E+02,GXZ , 1, 0, 5.00000E+02,GYZ , 1, 0, 2.00000E+02,*SETS 101,S, 1, 8, 18, 113, 120, 102,R, 1, 100, 5, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 1377,UZ , 0.00000E+00,,,,,,,, 0 1377,ROTX, 0.00000E+00,,,,,,,, 0 1377,ROTY, 0.00000E+00,,,,,,,, 0 1377,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 1.0, 0, 0 2,,,2,0, 0, 1.0, 0, 0 3,,,2,0, 0, 1.0, 0, 0 4,,,2,0, 0, 1.0, 0, 0 : : 253,,,2,0, 0, 1.0, 0, 0 254,,,2,0, 0, 1.0, 0, 0 255,,,2,0, 0, 1.0, 0, 0 256,,,2,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.08,3,1*PRINTCNTLAVND,102DISP,102ELSE,101ELST,101SLFO,0*ENDDATA



Input Data for Verification Problem No 2.27C**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLEINTERLAMINAR SHEAR STRESSES FOR CROSS-PLY LAMINATED SPHERICAL SHELLS OF RECTANGULAR PLANFORM (R/A=20; A/H=10; 0/90/90/0; SS3)*ELTYPE 1, 32, 11*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 8.0000001E-01, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15, 1.0000000E-15,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,8-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.0,-90.03,845.0,45.0,45.0,45.0,45.0,45.0,45.0,45.0*LAMSQ2**_DISP3_: 1,NAME = 1,4,5,11,1,1,1,2,1,1,2,1,11,1*NODES 1,,,,-1.60000E+01, 0.00000E+00, 6.39800E+02, 0 2,,,,-1.40006E+01, 0.00000E+00, 6.39847E+02, 0 3,,,,-1.20010E+01, 0.00000E+00, 6.39887E+02, 0 4,,,,-1.00013E+01, 0.00000E+00, 6.39921E+02, 0 : : 286,,,, 9.99786E+00, 3.20000E+01, 6.39922E+02, 0 287,,,, 1.19976E+01, 3.20000E+01, 6.39887E+02, 0 288,,,, 1.39971E+01, 3.20000E+01, 6.39846E+02, 0 289,,,, 1.59966E+01, 3.20000E+01, 6.39800E+02, 0*ELEMENTS



1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 19, 35, 18, 2, 1, 1, 1, 0,,,,, 0 3, 20, 37, 36, 35, 19, : : 126, 1, 1, 1, 0,,,,, 0 253, 270, 287, 286, 285, 269, 127, 1, 1, 1, 0,,,,, 0 253, 254, 255, 271, 287, 270, 128, 1, 1, 1, 0,,,,, 0 255, 272, 289, 288, 287, 271,*G2 1,,, 1.432394, -1.432394 , 0.0 17,1,,1.432394, 1.432394 , 0.0 18,,, 1.253346, -1.432394 , 0.0 34,1,, 1.253346, 1.432394 , 0.0 : : 171,,, -0.358106, -1.432394 , 0.0 187,1,, -0.358106, 1.432394 , 0.0 154,,, -0.179040, -1.432394 , 0.0 170,1,, -0.179040, 1.432394 , 0.0 *MATERIALEX , 1, 0, 2.50000E+04,EY , 1, 0, 1.00000E+03,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 5.00000E+02,GXZ , 1, 0, 5.00000E+02,GYZ , 1, 0, 2.00000E+02,*SETS 101,S, 1, 49, 7, 102,R, 1, 100, 5, 103,S, 1, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 289,UZ , 0.00000E+00,,,,,,,, 0 289,ROTX, 0.00000E+00,,,,,,,, 0 289,ROTY, 0.00000E+00,,,,,,,, 0 289,ROTZ, 0.00000E+00,,,,,,,, 0



*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 1.0, 0, 0 2,,,2,0, 0, 1.0, 0, 0 3,,,2,0, 0, 1.0, 0, 0 4,,,2,0, 0, 1.0, 0, 0 : : 125,,,2,0, 0, 1.0, 0, 0 126,,,2,0, 0, 1.0, 0, 0 127,,,2,0, 0, 1.0, 0, 0 128,,,2,0, 0, 1.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.8,3,1*PRINTCNTLAVND,102DISP,102ELSE,101ELST,101,103SLFO,0*ENDDATA


Static Analysis Verification ProblemsAnalysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed Load

2.28 Analysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed Load

Title:

Analysis of an anti-symmetric cross-ply laminated clamped plate subjected to a uniform distributedload.

Element Type:

3-D three noded laminated composite shell element (NKTP=32, NORDR=10)

Problem:

Figure 2.48 shows a 00/900 laminated square clamped plate of total thickness h = 0.64 in and span a =32 in, subjected to a uniform distributed load q = 10 psi. The numerical result of central transversedisplacement (UZ) is compared with that of Ref. [1].

Properties:

The following material properties are used for an orthotropic lamina:


Due to symmetry, a quarter of the clamped plate is modeled using 64 elements as shown in Figure 2.48.

The following boundary conditions are applied:

At X = 0.0 in, UX = UY = UZ = ROTX = ROTY = 0.0At Y = 0.0 in,UX=UY=UZ=ROTX=ROTY=0.0At X = 16.0 in, UX = ROTY = 0.0At Y = 16.0 in, UY = ROTX = 0.0

Rotations (ROTZ) about the normal to the plate are suppressed at all nodes to avoid spurious rotations.


Material:

EX = 40.0 106 psi

EY = 1.0 106 psi

GXY = GYZ = GXZ = 0.5 106 psi

NUXY = 0.25

×

×

×


Static Analysis Verification Problems Analysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed

Load

Table 2.32 shows the comparison of central transverse displacement with respect to the exact solutionas in Ref. [1]. The result of Ref. [1], based on the classical lamination theory, is in good agreementwith NISA.

Reference:1. J.M. Whitney, The Effect of Boundary Conditions on the Response of Laminated Composites,

Journal of Composite Materials, 4, 192-203, 1970.

Figure 2.48: Geometry and Finite Element Mesh for an Anti-Symmetric Cross-PlyLaminated Clamped Plate


Static Analysis Verification ProblemsAnalysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed Load

Table 2.32: Comparison of central transverse displacement of an anti-symmetric cross-ply laminated clamped plate subjected to a uniform distributed load

(2.6)

Central Transverse Displacement NISA Ref.[1] Error (%)

UZ 0.112 in 0.114 in 1.95

Normalized , UZ* 2.805 2.861 1.95†

†UZ∗ 103 UZ EY h3

qa4----------------------------------------=


Static Analysis Verification Problems Analysis of an Anti-symmetric Cross-ply Laminated Clamped Plate Subjected to A Uniform Distributed

Load

Input Data for Verification Problem No 2.28**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = PostSAVE = 26,27*ENDDATA


Static Analysis Verification ProblemsAnalysis of a Simply Supported Three-layer Sandwich Plate Subjected to A Uniform Transverse Load

2.29 Analysis of a Simply Supported Three-layer Sandwich Plate Subjected to A Uniform Transverse Load

Title:

Analysis of a simply supported three-layer sandwich plate subjected to a uniform transverse load.

Element Type:

3-D six noded triangular sandwich shell element (NKTP=33, NORDR=11)

Problem:

A simply supported square three-layer sandwich plate of side 10 inches and a total thickness of 0.806inch is shown in Figure 2.49. The plate is subjected to a uniform transverse load of 100 psi. ACartesian coordinate system (X,Y,Z) is placed at the mid-depth of the plate thickness, with axis Zperpendicular to the plate surface (X-Y plane).

The material properties of the face sheets and core are orthotorpic in X-Y plane. The central transversedisplacement is computed and compared with the available analytical and finite element solutions.

Properties:

The following are the material properties for face sheets and core:


Due to symmetry in the plate geometry and boundary conditions, only one quarter of the plate ismodeled using 8 elements as shown in Figure 2.49. The following boundary conditions are imposed:

At X = 0.0 in, UY = UZ = ROTX = 0.0At Y = 0.0 in, UX = UZ = ROTY = 0.0At X = 5.0 in, UX = ROTY = 0.0At Y = 5.0 in, UY = ROTX = 0.0

Rotations about the normal to the plate are suppressed at all nodes to avoid spurious rotations.

Face sheets

EX = 10.0 106 psi (Young's modulus along X-direction)

EY = 4.0 106 psi (Young's modulus along Y-direction)

GXY = 1.875 106 psi (Inplane shear modulus)

NUXY = 0.30 (Inplane Poisson's ratio)Core Material

GXZ = 3.00 104 psi (Shear modulus in X-Z plane)

GYZ = 1.20 104 psi (Shear modulus in Y-Z plane)

×

×

×

×

×


Static Analysis Verification Problems Analysis of a Simply Supported Three-layer Sandwich Plate Subjected to A Uniform Transverse Load

Figure 2.49: Geometry and Finite Element Mesh of a Three-layer Sandwich Square Plate


Table 2.33 shows a comparison of central transverse displacement obtained from NISA with theanalytical and finite element results [1]. They show a close agreement.

Reference:1. T.P. Khatua and Y.K. Chueng, Bending and Vibration of Multilayer Sandwich Beams and Plates,


Table 2.33: Comparison of results for the simply supported sandwich plate subjected to trans-verse loading

Source Central Transverse Displacement (UZ)

NISA 0.12504

Analytical solution [1] 0.123

Finite element solution [1] 0.1213



Input Data for Verification Problem No 2.29**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFF*TITLE ANALYSIS OF A SIMPLY SUPPORTED THREE-LAYER SANDWITCH PLATE SUBJECTED TO A UNIFORM DISTRIBUTED LOAD*ELTYPE 1, 33, 11*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2.8000001E-02, 2, 8,1, 0**_DISP3_: NAME =OTHER 7.5000000E-01, 7.5000000E-01, 7.5000000E-01, 7.5000000E-01, 7.5000000E-01, 7.5000000E-01, 7.5000000E-01, 7.5000000E-01,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.03,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*LAMSQ2**_DISP3_: 1,NAME = 1,3,1,11,2,1,1,1,1,1,2,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.25000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.50000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.75000E+00, 0.00000E+00, 0.00000E+00, 0 : : 22,,,, 1.25000E+00, 5.00000E+00, 0.00000E+00, 0 23,,,, 2.50000E+00, 5.00000E+00, 0.00000E+00, 0 24,,,, 3.75000E+00, 5.00000E+00, 0.00000E+00, 0 25,,,, 5.00000E+00, 5.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0


Static Analysis Verification Problems Analysis of a Simply Supported Three-layer Sandwich Plate Subjected to A Uniform Transverse Load

1, 7, 13, 12, 11, 6, 2, 1, 1, 1, 0,,,,, 0 1, 2, 3, 8, 13, 7, : : 7, 1, 1, 1, 0,,,,, 0 13, 19, 25, 24, 23, 18, 8, 1, 1, 1, 0,,,,, 0 13, 14, 15, 20, 25, 19,*MATERIALEX , 1, 0, 1.00000E+07,EY , 1, 0, 4.00000E+06,NUXY, 1, 0, 3.00000E-01,GXY , 1, 0, 1.87500E+06,GXZ , 1, 0, 0.00000E+00,GYZ , 1, 0, 0.00000E+00,EX , 2, 0, 0.00000E+00,EY , 2, 0, 0.00000E+00,NUXY, 2, 0, 2.50000E-01,GXY , 2, 0, 0.00000E+00,GXZ , 2, 0, 3.00000E+04,GYZ , 2, 0, 1.20000E+04,*SETS 101,S, 1, 35, 36, 102,R, 1, 100, 5, *LDCASE, ID= 1 1, 1, 3, 1, 0, 2, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 25,UY , 0.00000E+00,,,,,,,, 0 25,ROTX, 0.00000E+00,,,,,,,, 0 25,ROTY, 0.00000E+00,,,,,,,, 0 25,ROTZ, 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 1,,,2,0, 0, 100.0, 0, 0 2,,,2,0, 0, 100.0, 0, 0 3,,,2,0, 0, 100.0, 0, 0 4,,,2,0, 0, 100.0, 0, 0 5,,,2,0, 0, 100.0, 0, 0 6,,,2,0, 0, 100.0, 0, 0 7,,,2,0, 0, 100.0, 0, 0



8,,,2,0, 0, 100.0, 0, 0*STRSFILTER0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*SFDCOMP0.08,3,1*PRINTCNTLAVND,102DISP,102ELSE,101ELST,101SLFO,0*ENDDATA


Static Analysis Verification Problems Static Analysis of a Symmetric Cross-ply Laminated Simply Supported Plate Subjected to a Uniform

Transverse Pressure Load

2.30 Static Analysis of a Symmetric Cross-ply Laminated Simply Supported Plate Subjected to a UniformTransverse Pressure Load

Title:

Static analysis of a symmetric cross-ply laminated simply supported plate subjected to a uniformtransverse pressure load.

Element Type:

3-D twenty noded solid composite element (NKTP = 7, NORDR = 2)

Problem:

A symmetric cross-ply laminated simply supported square plate of dimension, a = b=10 in., subjectedto a uniform transverse pressure load is considered Figure 2.50. The lamination sequences andthickness of plies considered are shown in Figure 2.50. Numerical results, obtained for moderately-thick and thin plates, are compared with an exact solution based on ReissnerMindlin plate theory.

Properties:

In the material directions, the elastic properties are


Due to symmetries in lamination sequences, geometry, and boundary conditions, a quarter of the plateis modeled Figure 2.50 with 4 4 2 elements, where five layers are included in top 16 elements, andfour layers are added to the rest of the elements. The following boundary conditions are applied at:

Material:

EX = 40.0 106 psi

EY = 1.0 106 psi

GXY = GXZ = 0.6 106 psi

GYZ = 0.5 106 psi

NUXY = 0.25

NUXZ = 0.25

NUYZ = 0.25

×

×

×

×

× ×


Static Analysis Verification ProblemsStatic Analysis of a Symmetric Cross-ply Laminated Simply Supported Plate Subjected to a Uniform Transverse Pressure Load

The numerical results are obtained for span-to-depth ratios 10 and 100 to represent moderately-thick

and thin situations. A uniform transverse pressure load of 10 psi is applied.

A typical cross-section through the plate thickness.

Figure 2.50: Geometry and Finite Element Mesh for a Quarter of a Cross-ply Laminated Simply Supported Plate


Table 2.34 shows a comparison of central transverse displacement with respect to the exact solution[1]. The NISA result shows a better performance in moderately-thick range, while the result at thinsituation agrees reasonably well with reference [1].

X = 0.0 in UX = UZ = ROTX = 0.0

Y = 0.0 in UY = UZ = ROTY = 0.0

X = 5.0 in UX = ROTY = 0.0

Y = 5.0 in UY = ROTX = 0.0




Reference:1. A.K. Noor and M.D. Mathers, Shear Flexible Finite Element Models of Laminated Plates and

Shells, NASA TN D-8044, 1975.

Table 2.34: Comparison of central transverse displacements for a symmetric cross-ply laminated simply supported plate subjected to a uniform transverse pressure load with span-to-depth ratios 10 and 100

Span-to-depth RatioCentral Transverse Displacement (UZ)

NISA Ref. [l]

10 0.0006927 0.0005848

100 0.4466 0.4486



Input Data for Verification Problem No. 2.30**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27*TITLESTATIC ANALYSIS OF A SYMMETRIC CROSS-PLY LAMINATED SIMPLY SUPPORTED PLATESUBJECTED TO A UNIFORM TRANSVERSE PRESSURE LOAD.*ELTYPE 1, 7, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 1.0000000E-01, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01, 1.2500000E-01,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,890.0,90.0,90.0,90.0,90.0,90.0,90.0,90.0*LAMSQ2**_DISP3_: 1,NAME = 1,4,8,12,1,2,1,2,1,2,1,1,11,1**_DISP3_: 2,NAME = 2,5,8,11,2,1,2,1,1,2,1,2,11,1,1,1,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.25000E-01, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.25000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.87500E+00, 0.00000E+00, 0.00000E+00, 0 : : 478,,,, 3.12500E+00, 5.00000E+00, 1.00000E+00, 0 479,,,, 3.75000E+00, 5.00000E+00, 1.00000E+00, 0 480,,,, 4.37500E+00, 5.00000E+00, 1.00000E+00, 0 481,,,, 5.00000E+00, 5.00000E+00, 1.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0




1, 2, 3, 12, 21, 20, 19, 10, 101, 103, 121, 119, 201, 202, 203, 212, 221, 220, 219, 210, 2, 1, 1, 1, 0 : : 379, 377, 459, 460, 461, 470, 479, 478, 477, 468, 32, 2, 1, 1, 0 261, 262, 263, 272, 281, 280, 279, 270, 361, 363, 381, 379, 461, 462, 463, 472, 481, 480, 479, 470,*MATERIALEX , 1, 0, 4.00000E+07,EY , 1, 0, 1.00000E+06,EZ , 1, 0, 1.00000E+06,NUXY, 1, 0, 2.50000E-01,NUXZ, 1, 0, 2.50000E-01,NUYZ, 1, 0, 2.50000E-01,GXY , 1, 0, 6.00000E+05,GXZ , 1, 0, 6.00000E+05,GYZ , 1, 0, 5.00000E+05,FXC , 1, 0, 1.00000E+05,FXT , 1, 0, 1.54000E+05,FYC , 1, 0, 1.50000E+04,FYT , 1, 0, 1.00000E+04,FS , 1, 0, 2.00000E+04,*SETS 1001,R, 1, 4, 1, 1002,R, 1, 481, 1, *LDCASE, ID= 1 1, 1, 2, 1, 0, 1, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 : : 446,UX , 0.00000E+00,,,,,,,, 0 455,UX , 0.00000E+00,,,,,,,, 0 464,UX , 0.00000E+00,,,,,,,, 0 473,UX , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 17,,,2,0, 0, 10.0, 0, 0 18,,,2,0, 0, 10.0, 0, 0 19,,,2,0, 0, 10.0, 0, 0 20,,,2,0, 0, 10.0, 0, 0 : :



29,,,2,0, 0, 10.0, 0, 0 30,,,2,0, 0, 10.0, 0, 0 31,,,2,0, 0, 10.0, 0, 0 32,,,2,0, 0, 10.0, 0, 0*SFDCOMP0.0,0,0*PRINTCNTLAVND,1002DISP,1002ELFO,1001ELSE,1001ELST,1001SLFO,0*POSTCNTLNDSTRS,0NDSTRN,0GPSTRS,0*ENDDATA



3Buckling Analysis Verification Problems

3.1 Buckling of a Uniaxially Compressed Clamped Square PlateTitle:

Buckling of a uniaxially compressed clamped square plate

Element Type:

3-D thin shell element (NKTP = 40, NORDR = 1)

Problem:

A square plate of dimensions shown in Figure 3.1 is clamped along all four edges and compressed by auniformly distributed force of 1.0 lb acting along two opposite edges. The buckling load factor is to bedetermined.

Properties:


Due to symmetry only one quarter of the plate is modeled with a 4 4 mesh using 3-D thin shellelements (NKTP = 40, NORDR = 1). Symmetric boundary conditions are applied along the centurions(Figure 3.2). The plate is prevented from moving in the z-direction or rotating at the boundaries. Theedges of the plate are, however, free to move in the xy-plane. The rotation about normal to the plate issuppressed at all the nodes (refer to Figure 3.2 for details of the boundary conditions).

Material:



×

×

3-1

Buckling Analysis Verificat ion Problems Buckling of a Uniaxial ly Compressed Clamped Square Plate

Solution Procedure:

Conventional subspace iteration technique is used for the eigenvalue extraction.


The buckling load factor for the first buckling mode obtained from NISA is 101.2, which is in goodcomparison with the theoretical value of 100.7 (Ref. [1]).

Reference:1. A. Chajes, Principles of Structural Stability Theory, Prentice-Hall Inc., Englewood Cliffs, New

Jersey, 1974.

Figure 3.1: Square Plate under Uniaxial Compression

Figure 3.2: Quarter Plate Model and Boundary Conditions


Buckling Analysis Verification Problems Buckling of a Uniaxially Compressed Clamped Square Plate

Input Data for Verification Problem No.3.1**EXECUTIVE data deckANALYSIS = BUCK MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE BUCKLING OF A CLAMPED PLATE: 4X4 MESH*ELTYPE 1, 40, 1*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 9.9999998E-03, 9.9999998E-03, 9.9999998E-03, 9.9999998E-03,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.25000E-01, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.50000E-01, 0.00000E+00, 0.00000E+00, 0 4,,,, 3.75000E-01, 0.00000E+00, 0.00000E+00, 0 : : 22,,,, 1.25000E-01, 5.00000E-01, 0.00000E+00, 0 23,,,, 2.50000E-01, 5.00000E-01, 0.00000E+00, 0 24,,,, 3.75000E-01, 5.00000E-01, 0.00000E+00, 0 25,,,, 5.00000E-01, 5.00000E-01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 7, 6, 2, 1, 1, 1, 0,,,,, 0 2, 3, 8, 7, : : 15, 1, 1, 1, 0,,,,, 0 18, 19, 24, 23, 16, 1, 1, 1, 0,,,,, 0 19, 20, 25, 24,*MATERIALEX , 1, 0, 1.10640E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 2.58799E-04,


Buckling Analysis Verificat ion Problems Buckling of a Uniaxial ly Compressed Clamped Square Plate

*LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*EIGCNTL2,0,80,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 : : 25,UZ , 0.00000E+00,,,,,,,, 0 25,ROTX, 0.00000E+00,,,,,,,, 0 25,ROTY, 0.00000E+00,,,,,,,, 0 25,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 5,FX,-6.25000E-02,,, 0, 0 10,FX,-1.25000E-01,,, 0, 0 15,FX,-1.25000E-01,,, 0, 0 20,FX,-1.25000E-01,,, 0, 0 25,FX,-6.25000E-02,,, 0, 0*PRINTCNTLELST,0SLFO,0*ENDDATA


Buckling Analysis Verification Problems Buckling of a Beam-Spring Structure

3.2 Buckling of a Beam-Spring StructureTitle:

Buckling of a beam-spring structure

Element Type:

Problem:

A beam of constant cross section, of length 10 inches, is subjected to an axial thrust at its upper end.The base of the beam is fixed for both rotation and translation. At the upper end, the member is free torotate, but is constrained against translation by a spring of stiffness 18 lb/in as shown in Figure 3.3.The buckling load factor is to be determined.

Properties:


The structure is modeled with five beam elements (NKTP = 13, NORDR = 1) and one spring element(NKTP = 18, NORDR = 1). All the three degrees-of-freedom (UX, UY, ROTZ) of the node at the baseof the beam is constrained. The spring is attached to the upper end of the beam and is constrained at theother end.


2-D translational spring (NKTP = 18, NORDR = 1)

Material:

EX = 100.0 psi (Modulus of elasticity)

Cross-Section:

A = 1.0 in2 (Cross-sectional area)

IZZ = 10.0 in4 (Moment of inertia)


Buckling Analysis Verificat ion Problems Buckling of a Beam-Spring Structure

Solution Procedure:

Inverse iteration technique is used to extract the eigenvalues within the range of 0.0 and 5000.0.


The buckling load factor for the first buckling mode obtained from NISA is 143.9 which is in goodcomparison with the theoretical value of 145.0 (Ref. [1]).

Reference:1. R.L. Ketter, G.C. Lee and S.P. Prawel, Structural Analysis And Design, McGraw-Hill Book Co.,

New York, 1970.

Figure 3.3: Beam-Spring Structure and its Finite Element Mesh


Buckling Analysis Verification Problems Buckling of a Beam-Spring Structure

Input Data for Verification Problem No.3.2**EXECUTIVE data deckANALYSIS = BUCK MEMORY = 10000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27*TITLE BUCKLING OF A FIXED-SPRING SUPPORTED COLUMN*ELTYPE 1, 13, 1 2, 18, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+01, 2, 1,1, 0**_DISP3_: NAME =OTHER 1.8000000E+01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 2.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 4.00000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 5,,,, 8.00000E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0 7,,,, 1.00000E+01, 1.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 2, 3, 3, 1, 1, 1, 0 3, 4, 4, 1, 1, 1, 0 4, 5, 5, 1, 1, 1, 0 5, 6, 6, 1, 2, 2, 0 6, 7,*MATERIAL


Buckling Analysis Verificat ion Problems Buckling of a Beam-Spring Structure

EX , 1, 0, 1.00000E+02,*LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*EIGCNTL3,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 7,UX , 0.00000E+00,,,,,,,, 0 7,UY , 0.00000E+00,,,,,,,, 0 7,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 6,FX,-1.00000E+00,,, 0, 0*ENDDATA


Buckling Analysis Verification Problems Buckling of a Cantilever Beam under Compressive Load

3.3 Buckling of a Cantilever Beam under Compressive LoadTitle:

Buckling of a cantilever beam under compressive load

Element Type:

3-D solid element (NKTP = 4, NORDR = 2)

Problem:

A cantilever beam of dimensions shown in Figure 3.4 is subjected to a compressive end force of 1 lbThe critical load is to be determined.

Properties:


The beam is modeled with twenty, 3-D solid elements with 20 nodes per element (NKTP = 4, NORDR= 2). All nodal degrees-of-freedom at the fixed end are constrained. The compressive end force isobtained by applying a pressure of 400.0 psi at the free end.

Solution Procedure:

Conventional subspace iteration technique is used for the eigenvalue extraction.


The buckling load factor for the first buckling mode obtained from NISA is 1.005 which is in goodcomparison with the theoretical value of 1.0 (Ref. [1]).

Material:



×


Buckling Analysis Verificat ion Problems Buckling of a Cantilever Beam under Compressive Load

Reference:1. A. Chajes, Principles of Structural Stabilig Theory, Prentice-Hall Inc., Englewood Cliffs, New

Jersey, 1974.

Figure 3.4: Cantilever Beam under Compression Load


Buckling Analysis Verification Problems Buckling of a Cantilever Beam under Compressive Load

Input Data for Verification Problem No.3.3**EXECUTIVE data deckANALYSIS = BUCK MEMORY = 1000000SOLV = FRONFILE = BUCK3RESEQUENCE = OFFSAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE BUCKLING OF A CANTILEVER BEAM*ELTYPE 1, 4, 2*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 2.50000E-02, 0.00000E+00, 0.00000E+00, 0 3,,,, 5.00000E-02, 0.00000E+00, 0.00000E+00, 0 4,,,, 7.50000E-02, 0.00000E+00, 0.00000E+00, 0 : : 1120,,,, 9.25000E-01, 5.00000E-02, 5.00000E-02, 0 1121,,,, 9.50000E-01, 5.00000E-02, 5.00000E-02, 0 1122,,,, 9.75000E-01, 5.00000E-02, 5.00000E-02, 0 1123,,,, 1.00000E+00, 5.00000E-02, 5.00000E-02, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 44, 85, 84, 83, 42, 501, 503, 585, 583, 1001, 1002, 1003, 1044, 1085, 1084, 1083, 1042, 2, 1, 1, 1, 0 : : 621, 619, 1037, 1038, 1039, 1080, 1121, 1120, 1119, 1078, 20, 1, 1, 1, 0 39, 40, 41, 82, 123, 122, 121, 80, 539, 541, 623, 621, 1039, 1040, 1041, 1082, 1123, 1122, 1121, 1080,*MATERIALEX , 1, 0, 7.78147E+05,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 2.58799E-04,*LDCASE, ID= 1 0, 1, 3, 0, 0, 1, 0, 0.000, 0.000*EIGCNTL


Buckling Analysis Verificat ion Problems Buckling of a Cantilever Beam under Compressive Load

5,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 42,UX , 0.00000E+00,,,,,,,, 0 : : 1042,UZ , 0.00000E+00,,,,,,,, 0 1083,UX , 0.00000E+00,,,,,,,, 0 1083,UY , 0.00000E+00,,,,,,,, 0 1083,UZ , 0.00000E+00,,,,,,,, 0*PRESSURE**_DISP3_: PRESSURE, SET = 1 20,,,4,0, 0, 400.0, 0, 0*ENDDATA


Buckling Analysis Verification Problems Buckling Analysis of a Simply Supported Tapered Beam

3.4 Buckling Analysis of a Simply Supported Tapered BeamTitle:

Buckling analysis of a simply supported tapered beam

Element Type:

3-D tapered beam element (NKTP = 11, NORDR = 1)

Problem:

A linearly tapered I beam, of length 144 inch, is subjected to axial thrust along its centrodial axis. Bothends are simply supported and the deflection in the weak direction (y direction in Figure 3.5) isconstrained. The buckling load factor is to be determined for the beam.

Properties:

Material:EX = 3.0 107 psi (Modulus of elasticity)NUXY = 0.3 (Poisson’s ratio)Geometric:

A = (Cross-sectional area, see Table 3.1)

IZZ = (Area moment of inertia about beam Z-axis, see Table 3.1)

b = 4.0 in (width of the flange)t = 0.25 in (thickness of the flange)W = 0.1 in (thickness of the web)

= 6.0 in (height of the smallest cross-section)

= (height of the cross-section at a distance x)

×

2tb dx 2t–( )w+

w12------ dx 2t–( )3 bt

2----- dx t–( )2+

do

dx do 1 3xL------+⎝ ⎠

⎛ ⎞


Buckling Analysis Verificat ion Problems Buckling Analysis of a Simply Supported Tapered Beam


The beam is modeled by ten 3-D tapered beam elements (NKTP=11, NORDR=l). All displacements inY direction and rotations about Z axis are constrained. Displacements in Z and rotations about X axisare constrained at both ends, and no axial displacement is allowed at the left end. A concentrated axialforce is applied at the centroid of the right end of the beam.

Solution Procedure:

Subspace iteration technique is used to extract the lowest eigenvalue for the first buckling mode.

Figure 3.5: Geometry and Finite Element Mesh for Buckling of a Simply Supported Tapered Beam




The buckling load factor for the first buckling mode obtained from NISA is 1366538.0 which is veryclose to the theoretical value 1368723.8 (Ref. [1]).

Reference:1. R.L. Ketter, G.C. Lee and S.P. Prawel, Structural Analysis and Design, McGraw-Hill Book Co.,

New York, 1970.

Table 3.1: Cross-Section Properties for Tapered beam

A IYY

0 6.0 2.55 17.918

0.1 7.8 2.73 31.743

0.2 9.6 2.91 49.991

0.3 11.4 3.09 72.953

0.4 13.2 3.27 100.920

0.5 15.0 3.45 134.186

0.6 16.8 3.63 173.040

0.7 18.6 3.81 217.775

0.8 20.4 3.99 268.680

0.9 22.2 4.17 326.054

1.0 24 4.35 390.180

where,

x : location of the cross-section in x axis

L : length of the beam

A : cross-section area at distance x

IYY : moment of inertia at distance x

xL--- dx



Input Data for Verification Problem No. 3.4**EXECUTIVE data deckANALYSIS = BUCK MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLEBUCKLING ANALYSIS OF A SIMPLY SUPPORTED TAPER BEAM*ELTYPE 1, 11, 1*RCTABLE 1, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 2.5500000E+00, 1.7917700E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 2.7300000E+00, 3.1743000E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 2, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 2.7300000E+00, 3.1743000E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 2.9100001E+00, 4.9991001E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 3, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 2.9100001E+00, 4.9991001E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 3.0899999E+00, 7.2953003E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 4, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.0899999E+00, 7.2953003E+01, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,



1.0000000E+00, 3.2700000E+00, 1.0092000E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 5, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.2700000E+00, 1.0092000E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 3.4500000E+00, 1.3418600E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 6, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.4500000E+00, 1.3418600E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 3.6300001E+00, 1.7303999E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 7, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.6300001E+00, 1.7303999E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 3.8099999E+00, 2.1777499E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 8, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.8099999E+00, 2.1777499E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 3.9900000E+00, 2.6867999E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 9, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 3.9900000E+00, 2.6867999E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 4.1700001E+00, 3.2605399E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,



0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 10, 24,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 4.1700001E+00, 3.2605399E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 4.3499999E+00, 3.9017999E+02, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.44000E+01, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.88000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 4.32000E+01, 0.00000E+00, 0.00000E+00, 0 5,,,, 5.76000E+01, 0.00000E+00, 0.00000E+00, 0 6,,,, 7.20000E+01, 0.00000E+00, 0.00000E+00, 0 7,,,, 8.64000E+01, 0.00000E+00, 0.00000E+00, 0 8,,,, 1.00800E+02, 0.00000E+00, 0.00000E+00, 0 9,,,, 1.15200E+02, 0.00000E+00, 0.00000E+00, 0 10,,,, 1.29600E+02, 0.00000E+00, 0.00000E+00, 0 11,,,, 1.44000E+02, 0.00000E+00, 0.00000E+00, 0 100,,,, 0.00000E+00, 1.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2,000000,000000, 100 2, 1, 1, 2, 0 2, 3,000000,000000, 100 : : 9, 1, 1, 9, 0 9, 10,000000,000000, 100 10, 1, 1, 10, 0 10, 11,000000,000000, 100*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,*LDCASE, ID= 1 0, 0, 3, 0, 0, 0, 0, 0.000, 0.000*EIGCNTL2,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP



**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 11,UY , 0.00000E+00,,,,,,,, 0 11,UZ , 0.00000E+00,,,,,,,, 0 11,ROTX, 0.00000E+00,,,,,,,, 0 11,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 11,FX,-1.00000E+00,,, 0, 0*ENDDATA


Buckling Analysis Verificat ion Problems Buckling of a Fixed-Free Column under Self Weight and Applied Load

3.5 Buckling of a Fixed-Free Column under Self Weight and Applied Load

Title:

Buckling of a fixed-free column under self weight and applied load

Element Type:

3-D General Beam Element (NKTP = 39, NORDR = 1)

Analysis Highlights:

Buckling analysis of a stiffened structure.

Problem:

A fixed-free column of square cross section (1 1in) and of length 50 in shown in Figure 3.6 issubjected to a concentrated force P at the free end in addition to its self weight W. The critical value ofP that will cause buckling of the column is to be determined.

Properties:


The cantilever beam is modeled with ten 3-D General Beam Elements and all degrees of freedom areconstrained at the clamped end.

Material:

EX = 75.0 106 lb/in2

NLTXY = 0.0

DENS = 0.125lb-sec2/in4

g = 386.4 in/sec2

×

×


Buckling Analysis Verification Problems Buckling of a Fixed-Free Column under Self Weight and Applied Load

Solution Procedure:

First a linear static (STATIC) analysis is performed for a unit concentrated load at the free end of thecolumn. Files 26 and 27 are saved from this run. Next a nonlinear static (NLSTATIC) analysis isperformed considering geometric nonlinearity only, with 5 load increments for the self weight of thestructure using *BODYFORCE data. Total Lagrangian formulation and full Newton-Raphson iterationprocedure are used in the analysis. File 24 containing the element stiffness matrices under stiffeningfrom the self weight is saved from this run. Finally, a buckling analysis (BUCKLING) RESTART=4 isdone to obtain two eigenvalues using conventional subspace iteration. Files saved from the first tworuns are accessed in the restart by providing the same file name prefix. Gauss point stresses needed forthe buckling analysis were not saved in the static analysis. However, the program automaticallycomputes them in the buckling restart run. The buckling load factors obtained from this run are thoseto be applied to the concentrated force with the self weight held constant.


The results from NISA are compared to those obtained from Ref. [1]. Using the above data we obtain avalue of n = 0.9994. From Table 2.7 in Ref. [1], we get m = 1.72 for n = 1.0. Thus, the critical load canbe computed to be 4300 lbs, which is close to the value of 4293.4 obtained from NISA.

Reference:1. S.P. Timoshenko and J.M. Gere, Theory of Elastic Stability, 2nd Edition, McGraw-Hill Book

Co., New York, 1961.



Figure 3.6: Fixed Free Column Under Self Weight and Applied Load



Nonlinear Static Input Data for Verification Problem No. 3.5**EXECUTIVE data deckANALYSIS = NLSTAT MEMORY = 1000000SOLV = FRONFILE = BU05RESEQUENCE = OFFSAVE = 24,26,27NLTYPE = GEOMETRY*TITLE A FIXED-FREE COLUMN UNDER SELF WEIGHT - NONLINEAR STATIC RUN*ELTYPE 1, 39, 1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 5.00000E+00, 0.00000E+00, 0 3,,,, 0.00000E+00, 1.00000E+01, 0.00000E+00, 0 4,,,, 0.00000E+00, 1.50000E+01, 0.00000E+00, 0 5,,,, 0.00000E+00, 2.00000E+01, 0.00000E+00, 0 6,,,, 0.00000E+00, 2.50000E+01, 0.00000E+00, 0 7,,,, 0.00000E+00, 3.00000E+01, 0.00000E+00, 0 8,,,, 0.00000E+00, 3.50000E+01, 0.00000E+00, 0 9,,,, 0.00000E+00, 4.00000E+01, 0.00000E+00, 0 10,,,, 0.00000E+00, 4.50000E+01, 0.00000E+00, 0 11,,,, 0.00000E+00, 5.00000E+01, 0.00000E+00, 0 12,,,, 1.00000E+00, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 2, 3, 3, 1, 1, 1, 0 3, 4, 4, 1, 1, 1, 0 4, 5, 5, 1, 1, 1, 0 5, 6, 6, 1, 1, 1, 0 6, 7, 7, 1, 1, 1, 0 7, 8, 8, 1, 1, 1, 0



Static Input Data for Verification Problem No. 3.5

8, 9, 9, 1, 1, 1, 0 9, 10, 10, 1, 1, 1, 0 10, 11,*MATERIALEX , 1, 0, 7.50000E+07,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.25000E-01,*BMDATA1,12,0,0,0,0,01,0.00.0,0.0,0.0,0.0,0.0,0.0*BMSECT RECT,1,1,0,0,00.0,0.0*EVENT, ID = 1INCREMENTS = EQUAL, 5TIMEATEND = 1.0MAXITERATIONS = 100TOLERANCES = 0.01,0.01,0.01*SPDISP, TCRV = 0**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0*BODYFORCE, TCRV = 0,0,00.0,0.0,0.0,0.0,-986.4,0.0*ENDDATA

**EXECUTIVE data deckANALYSIS = STATIC MEMORY = 1000000SOLV = FRONFILE = BU05RESEQUENCE = OFFSAVE = 26,27*TITLE



A FIXED-FREE COLUMN UNDER APPLIED AXIAL LOAD - STATIC RUN*ELTYPE 1, 39, 1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 5.00000E+00, 0.00000E+00, 0 3,,,, 0.00000E+00, 1.00000E+01, 0.00000E+00, 0 4,,,, 0.00000E+00, 1.50000E+01, 0.00000E+00, 0 5,,,, 0.00000E+00, 2.00000E+01, 0.00000E+00, 0 6,,,, 0.00000E+00, 2.50000E+01, 0.00000E+00, 0 7,,,, 0.00000E+00, 3.00000E+01, 0.00000E+00, 0 8,,,, 0.00000E+00, 3.50000E+01, 0.00000E+00, 0 9,,,, 0.00000E+00, 4.00000E+01, 0.00000E+00, 0 10,,,, 0.00000E+00, 4.50000E+01, 0.00000E+00, 0 11,,,, 0.00000E+00, 5.00000E+01, 0.00000E+00, 0 12,,,, 1.00000E+00, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 2, 3, 3, 1, 1, 1, 0 3, 4, 4, 1, 1, 1, 0 4, 5, 5, 1, 1, 1, 0 5, 6, 6, 1, 1, 1, 0 6, 7, 7, 1, 1, 1, 0 7, 8, 8, 1, 1, 1, 0 8, 9, 9, 1, 1, 1, 0 9, 10, 10, 1, 1, 1, 0 10, 11,*MATERIALEX , 1, 0, 7.50000E+07,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.25000E-01,*BMDATA



Buckling Input Data for Verification Problem No. 3.5

1,12,0,0,0,0,01,0.00.0,0.0,0.0,0.0,0.0,0.0*BMSECT RECT,1,1,0,0,00.0,0.0*LDCASE, ID= 1 1, 1, 3, 0, 0, 0, 0, 0.000, 0.000*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0*CFORCE**_DISP3_: CFORCE, SET = 1 11,FY,-1.00000E+00,,, 0, 0*ENDDATA

PROB=BU05R4.nis ANALYSIS=BUCKLING EIGEN=SUBSPACE,CONV RESTART=4 AUTO_CONSTRAINT=OFF FILE=BU05*TITLEBUCKLING OF A FIXED-FREE COLUMN UNDER SELF WEIGHT AND AXIAL LOAD - RESTART 4*LDCASE,ID=10*EIGCNTL2,0*ENDDATA



4Eigenvalue Analysis Verification Problems

4.1 Natural frequencies of a simply supported beamTitle:

Natural frequencies of a simply supported beam

Element Type:


Problem:

The natural frequencies of a simply supported beam, with dimensions as shown in Figure 4.1, are to becomputed.

Properties:

Material:


NUXY = 0.3 (Poisson’s ratio)

DENS = 7.28 10-4 lb-sec2/in4 (Mass density)

Beam cross-section:


IZZ = 1.3333 in4 (Moment of inertia about Z-axis)

×

×

4-1

Eigenvalue Analysis Verificat ion Problems Natural frequencies of a simply supported beam


The beam is modeled using four 2-D beam elements (NKTP = 13, NORDR = 1). Translations UX andUY are constrained at both ends of the beam.

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used for eigenvalueextraction.


The first three natural frequencies of the simply supported beam obtained from NISA are compared tothe theoretical Ref. [1] values in Table 4.1.

Reference:1. R.W. Clough and J. Penzien, Dynamics of Structures, McGraw-Hill Book Co. Inc., New York,

1975.

Figure 4.1: FE Model for Simply Supported Beam


Eigenvalue Analysis Verification Problems Natural frequencies of a simply supported beam

Table 4.1: Natural frequencies of a simply supported beam (Hz)

Mode Theoretical (Ref.[1]) NISA

1 28.766 28.766

2 115.063 115.398

3 258.891 263.012


Eigenvalue Analysis Verificat ion Problems Natural frequencies of a simply supported beam

Input Data for Verification Problem No.4.1**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV01SAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A SIMPLY SUPPORTED BEAM*ELTYPE 1, 13, 1*RCTABLE 1, 2,1, 0**_DISP3_: NAME =OTHER 4.0000000E+00, 1.3333330E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 2.00000E+01, 0.00000E+00, 0.00000E+00, 0 3,,,, 4.00000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 6.00000E+01, 0.00000E+00, 0.00000E+00, 0 5,,,, 8.00000E+01, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 2, 1, 1, 1, 0 2, 3, 3, 1, 1, 1, 0 3, 4, 4, 1, 1, 1, 0 4, 5,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.28000E-04,*EIGCNTL3,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0


Eigenvalue Analysis Verification Problems Natural frequencies of a simply supported beam

5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a cantilever beam

4.2 Fundamental frequency of a cantilever beamTitle:

Fundamental frequency of a cantilever beam

Element Type:

2-D plane stress element (NKTP = 1, NORDR = 2)

Problem:

The fundamental frequency of a cantilever beam, with dimensions as shown in Figure 4.2, is to becomputed.

Properties:


The beam is modeled using three 2-D plane stress elements (NKTP = 1, NORDR = 2). TranslationsUX and UY are constrained at the fixed end of the beam.

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used for the eigenvalueextraction.


The fundamental frequency of the cantilever beam obtained from NISA is 51.6675 Hz which is ingood comparison with the theoretical value of 51.1059 Hz (Ref. [1]).

Material:



DENS = 9.989 10-4 lb-sec2/in4 (Mass Density)

×

×


Eigenvalue Analysis Verification Problems Fundamental frequency of a cantilever beam


1975.

Figure 4.2: Finite Element Model for Cantilever Beam


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a cantilever beam

Input Data for Verification Problem No.4.2**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV02SAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE FUNDAMENTAL FREQUENCY OF A CANTILEVER BEAM*ELTYPE 1, 1, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01, 4.0000001E-01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.66667E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 3.33333E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 5,,,, 6.66667E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 8.33333E+00, 0.00000E+00, 0.00000E+00, 0 7,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0 8,,,, 0.00000E+00, 5.00000E-01, 0.00000E+00, 0 9,,,, 1.66667E+00, 5.00000E-01, 0.00000E+00, 0 10,,,, 3.33333E+00, 5.00000E-01, 0.00000E+00, 0 11,,,, 5.00000E+00, 5.00000E-01, 0.00000E+00, 0 12,,,, 6.66667E+00, 5.00000E-01, 0.00000E+00, 0 13,,,, 8.33333E+00, 5.00000E-01, 0.00000E+00, 0 14,,,, 1.00000E+01, 5.00000E-01, 0.00000E+00, 0 15,,,, 0.00000E+00, 1.00000E+00, 0.00000E+00, 0 16,,,, 1.66667E+00, 1.00000E+00, 0.00000E+00, 0 17,,,, 3.33333E+00, 1.00000E+00, 0.00000E+00, 0 18,,,, 5.00000E+00, 1.00000E+00, 0.00000E+00, 0 19,,,, 6.66667E+00, 1.00000E+00, 0.00000E+00, 0 20,,,, 8.33333E+00, 1.00000E+00, 0.00000E+00, 0 21,,,, 1.00000E+01, 1.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2, 3, 10, 17, 16, 15, 8,


Eigenvalue Analysis Verification Problems Fundamental frequency of a cantilever beam

2, 1, 1, 1, 0 3, 4, 5, 12, 19, 18, 17, 10, 3, 1, 1, 1, 0 5, 6, 7, 14, 21, 20, 19, 12,*MATERIALEX , 1, 0, 1.00000E+06,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 9.98900E-04,*EIGCNTL1,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 8,UX , 0.00000E+00,,,,,,,, 0 8,UY , 0.00000E+00,,,,,,,, 0 15,UX , 0.00000E+00,,,,,,,, 0 15,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a clamped circular plate

4.3 Fundamental frequency of a clamped circular plateTitle:

Fundamental frequency of a clamped circular plate

Element Type:


Problem:

The fundamental frequency of a clamped circular plate, of 16in radius and 1in thickness, is to becomputed.

Properties:


A cross-section of the plate is modeled using six 2-D axisymmetric solid elements (NKTP = 3,NORDR = 2) as shown in Figure 4.3. The nodes on the outer edge are constrained in X and Ydirections.

Solution Procedure:

Conventional subspace iteration and lumped mass matrix formulation are used to extract thefundamental frequency of the system.


The fundamental frequency of the system obtained from NISA is 192.258 Hz which is in goodcomparison with the theoretical value of 192.09 Hz (Ref. [1]).

Reference:1. S. Timoshenko, D.H. Young and W. Weaver, Jr., Vibration Problems in Engineering, 4th Edition,

John Wiley & Sons, New York, 1974.

Material:




×


Eigenvalue Analysis Verification Problems Fundamental frequency of a clamped circular plate

Figure 4.3: Clamped Circular Plate


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a clamped circular plate

Input Data for Verification Problem No.4.3**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV03SAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONALMASS FORMULATION = LUMPED*TITLE FUNDAMENTAL FREQUENCY OF A CLAMPED CIRCULAR PLATE*ELTYPE 1, 3, 2*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 1.33333E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 2.66667E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 4.00000E+00, 0.00000E+00, 0.00000E+00, 0 : : 36,,,, 1.20000E+01, 1.00000E+00, 0.00000E+00, 0 37,,,, 1.33333E+01, 1.00000E+00, 0.00000E+00, 0 38,,,, 1.46667E+01, 1.00000E+00, 0.00000E+00, 0 39,,,, 1.60000E+01, 1.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 16, 29, 28, 27, 14, 2, 1, 1, 0, 0 3, 4, 5, 18, 31, 30, 29, 16, 3, 1, 1, 0, 0 5, 6, 7, 20, 33, 32, 31, 18, 4, 1, 1, 0, 0 7, 8, 9, 22, 35, 34, 33, 20, 5, 1, 1, 0, 0 9, 10, 11, 24, 37, 36, 35, 22, 6, 1, 1, 0, 0 11, 12, 13, 26, 39, 38, 37, 24,*MATERIALEX , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 1.00000E-03,*EIGCNTL


Eigenvalue Analysis Verification Problems Fundamental frequency of a clamped circular plate

1,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 13,UX , 0.00000E+00,,,,,,,, 0 13,UY , 0.00000E+00,,,,,,,, 0 26,UX , 0.00000E+00,,,,,,,, 0 26,UY , 0.00000E+00,,,,,,,, 0 39,UX , 0.00000E+00,,,,,,,, 0 39,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural frequencies of a cantilever beam

4.4 Natural frequencies of a cantilever beamTitle:

Natural frequencies of a cantilever beam

Element Type:


Problem:

The natural frequencies of a cantilever beam, with dimensions as shown in Figure 4.4, are to becomputed.

Properties:


The structure is modeled using ten 3-D beam elements (NKTP = 12, NORDR = 1). The problem isreduced to a 2-D problem, to eliminate twin frequencies, by constraining the out-of-plane translation(UZ) and the rotations about X and Y axis (ROTX and ROTY). All the six degrees of freedom areconstrained for the node on the cantilever end.

Material:




Beam cross-section:


IYY = 1.3333 in4 (Moment of inertia about Y-axis)


×

×


Eigenvalue Analysis Verification Problems Natural frequencies of a cantilever beam

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used to extract the firstthree eigenvalues of the system.


The first three natural frequencies of the cantilever beam calculated by NISA are compared to thetheoretical (Ref. [1]) values in Table 4.2, and they show very good agreement.


1975.

Figure 4.4: Cantilever Beam

Table 4.2: Natural frequencies of a cantilever beam (Hz)


1 10.246 10.246

2 64.218 64.168

3 179.830 179.503



Input Data for Verification Problem No.4.4**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV04SAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE EIGENVALUES OF A CANTILEVER BEAM*ELTYPE 1, 12, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 4.0000000E+00, 1.3333000E+00, 1.3333000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 8.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.60000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 2.40000E+01, 0.00000E+00, 0.00000E+00, 0 5,,,, 3.20000E+01, 0.00000E+00, 0.00000E+00, 0 6,,,, 4.00000E+01, 0.00000E+00, 0.00000E+00, 0 7,,,, 4.80000E+01, 0.00000E+00, 0.00000E+00, 0 8,,,, 5.60000E+01, 0.00000E+00, 0.00000E+00, 0 9,,,, 6.40000E+01, 0.00000E+00, 0.00000E+00, 0 10,,,, 7.20000E+01, 0.00000E+00, 0.00000E+00, 0 11,,,, 8.00000E+01, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2,000000,000000, 0, 2, 1, 1, 1, 0 2, 3,000000,000000, 0, 3, 1, 1, 1, 0 3, 4,000000,000000, 0, 4, 1, 1, 1, 0 4, 5,000000,000000, 0, 5, 1, 1, 1, 0 5, 6,000000,000000, 0, 6, 1, 1, 1, 0


Eigenvalue Analysis Verification Problems Natural frequencies of a cantilever beam

6, 7,000000,000000, 0, 7, 1, 1, 1, 0 7, 8,000000,000000, 0, 8, 1, 1, 1, 0 8, 9,000000,000000, 0, 9, 1, 1, 1, 0 9, 10,000000,000000, 0, 10, 1, 1, 1, 0 10, 11,000000,000000, 0,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.28000E-04,*EIGCNTL3,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0 3,ROTX, 0.00000E+00,,,,,,,, 0 3,ROTY, 0.00000E+00,,,,,,,, 0 4,UZ , 0.00000E+00,,,,,,,, 0 4,ROTX, 0.00000E+00,,,,,,,, 0 4,ROTY, 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0 5,ROTX, 0.00000E+00,,,,,,,, 0 5,ROTY, 0.00000E+00,,,,,,,, 0 6,UZ , 0.00000E+00,,,,,,,, 0 6,ROTX, 0.00000E+00,,,,,,,, 0 6,ROTY, 0.00000E+00,,,,,,,, 0 7,UZ , 0.00000E+00,,,,,,,, 0 7,ROTX, 0.00000E+00,,,,,,,, 0 7,ROTY, 0.00000E+00,,,,,,,, 0 8,UZ , 0.00000E+00,,,,,,,, 0



8,ROTX, 0.00000E+00,,,,,,,, 0 8,ROTY, 0.00000E+00,,,,,,,, 0 9,UZ , 0.00000E+00,,,,,,,, 0 9,ROTX, 0.00000E+00,,,,,,,, 0 9,ROTY, 0.00000E+00,,,,,,,, 0 10,UZ , 0.00000E+00,,,,,,,, 0 10,ROTX, 0.00000E+00,,,,,,,, 0 10,ROTY, 0.00000E+00,,,,,,,, 0 11,UZ , 0.00000E+00,,,,,,,, 0 11,ROTX, 0.00000E+00,,,,,,,, 0 11,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Natural frequencies of a simply supported square plate

4.5 Natural frequencies of a simply supported square plateTitle:

Natural frequencies of a simply supported square plate

Element Type:


Problem:

The first five natural frequencies of a simply supported square plate, with dimensions as shown inFigure 4.5, are to be computed.

Properties:


The plate is modeled using sixteen 3-D general shell elements (NKTP = 20, NORDR = 2). The nodeson the periphery of the plate are constrained in the normal (UZ) direction. In-plane translations (UXand UY) for all the nodes are constrained to obtain only the bending modes.

Solution Procedure:

Conventional subspace iteration and lumped mass matrix formulation are used to extract theeigenvalues.


The first five natural frequencies of the simply supported plate obtained from NISA are compared tothe theoretical (Ref. [1]) values in Table 4.3. The mode shape plots are shown in Figure 4.6 and Figure4.7

Material:




Geometry:

t = 1.0 in (thickness)

×

×


Eigenvalue Analysis Verificat ion Problems Natural frequencies of a simply supported square plate



Figure 4.5: Finite Element Mesh for a Simply Supported Square Plate



Figure 4.6: Mode No.1: Frequency = 83.979 Hz

Table 4.3: Natural frequencies of a simply supported plate (Hz)


1 84.095 83.979

2 210.240 210.451

3 210.240 210.451

4 336.380 358.286

5 420.471 422.376



Figure 4.7: Mode Shapes and Natural Frequencies



Input Data for Verification Problem No.4.5**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = STATICSAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONALMASS FORMULATION = LUMPED*TITLE NATURAL FREQUENCIES OF A SIMPLY SUPPORTED SQUARE PLATE*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.20000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.80000E+01, 0.00000E+00, 0.00000E+00, 0 : : 78,,,, 3.00000E+01, 4.80000E+01, 0.00000E+00, 0 79,,,, 3.60000E+01, 4.80000E+01, 0.00000E+00, 0 80,,,, 4.20000E+01, 4.80000E+01, 0.00000E+00, 0 81,,,, 4.80000E+01, 4.80000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 12, 21, 20, 19, 10, 2, 1, 1, 1, 0,,,,, 0 3, 4, 5, 14, 23, 22, 21, 12, : : 15, 1, 1, 1, 0,,,,, 0 59, 60, 61, 70, 79, 78, 77, 68, 16, 1, 1, 1, 0,,,,, 0 61, 62, 63, 72, 81, 80, 79, 70,*MATERIAL



EX , 1, 0, 1.05000E+07,NUXY, 1, 0, 3.33330E-01,DENS, 1, 0, 2.58799E-04,*EIGCNTL5,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 81,UY , 0.00000E+00,,,,,,,, 0 81,UZ , 0.00000E+00,,,,,,,, 0 81,ROTX, 0.00000E+00,,,,,,,, 0 81,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Fundamental frequency of a canti lever t riangular plate

4.6 Fundamental frequency of a cantilever triangular plateTitle:

Fundamental frequency of a cantilever triangular plate

Element Type:


Problem:

The fundamental frequency of a cantilever triangular plate, with dimensions as shown in Figure 4.8, isto be computed.

Properties:


The plate is modeled using three 3-D general shell elements (NKTP = 20, NORDR = 2). All the sixdegrees of freedom of the nodes along X = 0 are constrained.

Solution Procedure:

Conventional subspace iteration and lumped mass matrix formulation are used for the eigenvalueextraction.


The fundamental frequency of the plate obtained from NISA is 254.45 Hz which is in good agreementto the theoretical value of 259.16 (Ref. [1]).

Reference:1. S. Timoshenko, D.H. Young and W. Weaver, Jr., Vibration Problems in Engineering, 3rd Edi-

tion, John Wiley & Sons, New York, 1974.

Material:




×

×


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a cantilever tr iangular plate

Figure 4.8: Cantilever Triangular Plate


Eigenvalue Analysis Verification Problems Fundamental frequency of a canti lever t riangular plate

Input Data for Verification Problem No.4.6**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV06SAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONALMASS FORMULATION = LUMPED*TITLE FUNDAMENTAL FREQUENCY OF A TRIANGULAR PLATE*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 1.00000E+00, 0.00000E+00, 0 3,,,, 0.00000E+00, 2.00000E+00, 0.00000E+00, 0 4,,,, 0.00000E+00, 3.00000E+00, 0.00000E+00, 0 5,,,, 0.00000E+00, 4.00000E+00, 0.00000E+00, 0 6,,,, 4.00000E+00, 5.00000E-01, 0.00000E+00, 0 8,,,, 3.50000E+00, 2.00000E+00, 0.00000E+00, 0 10,,,, 4.00000E+00, 3.50000E+00, 0.00000E+00, 0 11,,,, 8.00000E+00, 1.00000E+00, 0.00000E+00, 0 12,,,, 7.50000E+00, 1.50000E+00, 0.00000E+00, 0 13,,,, 7.00000E+00, 2.00000E+00, 0.00000E+00, 0 14,,,, 7.50000E+00, 2.50000E+00, 0.00000E+00, 0 15,,,, 8.00000E+00, 3.00000E+00, 0.00000E+00, 0 16,,,, 1.20000E+01, 1.50000E+00, 0.00000E+00, 0 20,,,, 1.20000E+01, 2.50000E+00, 0.00000E+00, 0 21,,,, 1.60000E+01, 2.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 6, 11, 12, 13, 8, 3, 2, 2, 1, 1, 1, 0,,,,, 0 3, 8, 13, 14, 15, 10, 5, 4, 3, 1, 1, 1, 0,,,,, 0 11, 16, 21, 20, 15, 14, 13, 12,


Eigenvalue Analysis Verificat ion Problems Fundamental frequency of a cantilever tr iangular plate

*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.28000E-04,*EIGCNTL1,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 3,UX , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0 3,ROTX, 0.00000E+00,,,,,,,, 0 3,ROTY, 0.00000E+00,,,,,,,, 0 4,UX , 0.00000E+00,,,,,,,, 0 4,UY , 0.00000E+00,,,,,,,, 0 4,UZ , 0.00000E+00,,,,,,,, 0 4,ROTX, 0.00000E+00,,,,,,,, 0 4,ROTY, 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0 5,ROTX, 0.00000E+00,,,,,,,, 0 5,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Natural Frequencies of an Earth Dam

4.7 Natural Frequencies of an Earth DamTitle:

Natural frequencies of an earth dam

Element Type:

2-D plane strain element (NKTP = 2, NORDR = 2)

Problem:

The natural frequencies of an earth dam, with dimensions as shown in Figure 4.9, are to be computed.

Properties:


The earth dam is modeled using seven 2-D plane strain elements (NKTP = 2, NORDR = 2). The nodesalong the base of dam are constrained in both X and Y directions.

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used to evaluate theeigenvalues.


The first five natural frequencies of the earth dam obtained from NISA, using parabolic isoparametricplane strain elements, are compared to those obtained from Reference 1, using constant straintriangular elements, in Table 4.4. The plots for the mode shapes are shown in Figure 4.10. throughFigure 4.14.

Reference:1. R.W. Clough and A.K. Chopra, Earthquake Stress Analysis in Earth Dams, Proc. ASCE, J. EM

Dn., 92, 1966.

Material:




×


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of an Earth Dam

Figure 4.9: Finite Element Mesh for an Earth Dam

Figure 4.10: Mode No. 1: Frequency = 1.237 Hz




Table 4.4: Natural frequencies of an earth dam (Hz)




1 1.242 1.237

2 2.039 1.993

3 2.039 2.365

4 3.239 2.966

5 3.436 3.465



Input Data for Verification Problem No.4.7**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV07SAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF AN EARTH DAM*ELTYPE 1, 2, 2*NODES 1,,,,-4.50000E+02, 0.00000E+00, 0.00000E+00, 0 2,,,,-3.00000E+02, 0.00000E+00, 0.00000E+00, 0 3,,,,-1.50000E+02, 0.00000E+00, 0.00000E+00, 0 4,,,,-7.50000E+01, 0.00000E+00, 0.00000E+00, 0 5,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 6,,,, 7.50000E+01, 0.00000E+00, 0.00000E+00, 0 7,,,, 1.50000E+02, 0.00000E+00, 0.00000E+00, 0 8,,,, 3.00000E+02, 0.00000E+00, 0.00000E+00, 0 9,,,, 4.50000E+02, 0.00000E+00, 0.00000E+00, 0 10,,,,-3.75000E+02, 5.00000E+01, 0.00000E+00, 0 12,,,,-1.50000E+02, 5.00000E+01, 0.00000E+00, 0 14,,,, 0.00000E+00, 7.50000E+01, 0.00000E+00, 0 16,,,, 1.50000E+02, 5.00000E+01, 0.00000E+00, 0 18,,,, 3.75000E+02, 5.00000E+01, 0.00000E+00, 0 19,,,,-3.00000E+02, 1.00000E+02, 0.00000E+00, 0 20,,,,-2.25000E+02, 1.00000E+02, 0.00000E+00, 0 21,,,,-1.50000E+02, 1.00000E+02, 0.00000E+00, 0 22,,,,-7.50000E+01, 1.25000E+02, 0.00000E+00, 0 23,,,, 0.00000E+00, 1.50000E+02, 0.00000E+00, 0 24,,,, 7.50000E+01, 1.25000E+02, 0.00000E+00, 0 25,,,, 1.50000E+02, 1.00000E+02, 0.00000E+00, 0 26,,,, 2.25000E+02, 1.00000E+02, 0.00000E+00, 0 27,,,, 3.00000E+02, 1.00000E+02, 0.00000E+00, 0 28,,,,-2.06250E+02, 1.62500E+02, 0.00000E+00, 0 30,,,,-5.62500E+01, 1.87500E+02, 0.00000E+00, 0 32,,,, 5.62500E+01, 1.87500E+02, 0.00000E+00, 0 34,,,, 2.06250E+02, 1.62500E+02, 0.00000E+00, 0 36,,,,-1.12500E+02, 2.25000E+02, 0.00000E+00, 0



37,,,,-5.62500E+01, 2.62500E+02, 0.00000E+00, 0 38,,,, 0.00000E+00, 3.00000E+02, 0.00000E+00, 0 39,,,, 5.62500E+01, 2.62500E+02, 0.00000E+00, 0 40,,,, 1.12500E+02, 2.25000E+02, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 12, 21, 20, 19, 10, 2, 1, 1, 0, 0 3, 4, 5, 14, 23, 22, 21, 12, 3, 1, 1, 0, 0 5, 6, 7, 16, 25, 24, 23, 14, 4, 1, 1, 0, 0 7, 8, 9, 18, 27, 26, 25, 16, 5, 1, 1, 0, 0 19, 20, 21, 22, 23, 30, 36, 28, 6, 1, 1, 0, 0 23, 24, 25, 26, 27, 34, 40, 32, 7, 1, 1, 0, 0 36, 30, 23, 32, 40, 39, 38, 37,*MATERIALEX , 1, 0, 1.17070E+07,NUXY, 1, 0, 4.50000E-01,DENS, 1, 0, 4.04150E+00,*EIGCNTL5,-1,0,0,0.1,0.0,0.0,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 3,UX , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 4,UX , 0.00000E+00,,,,,,,, 0 4,UY , 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 6,UX , 0.00000E+00,,,,,,,, 0 6,UY , 0.00000E+00,,,,,,,, 0 7,UX , 0.00000E+00,,,,,,,, 0 7,UY , 0.00000E+00,,,,,,,, 0 8,UX , 0.00000E+00,,,,,,,, 0



8,UY , 0.00000E+00,,,,,,,, 0 9,UX , 0.00000E+00,,,,,,,, 0 9,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Square Cantilever Plate

4.8 Natural Frequencies of a Square Cantilever PlateTitle:

Natural frequencies of a square cantilever plate

Element Type:


Problem:

The natural frequencies of a square cantilever plate, (80 80 0.25 inch) are to be computed.

Properties:


Two different mesh sizes are used to model the plate using 3-D general shell elements (NKTP = 20,NORDR = 2). The two models are shown in Figure 4.15 and Figure 4.16. All six degrees of freedom ofthe nodes along X = 0 are constrained. The input data is presented only for Model 1.

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used to evaluate thenatural frequencies.


The first five natural frequencies of the cantilever plate obtained from NISA are compared totheoretical (Ref. [1]) values in Table 4.5.

Reference:1. S. Timoshenko, D.M. Young and W. Weaver, Jr., Vibration Problems in Engineering, 4th Edi-

tion, John Wiley & Sons, New York, 1974.

Material:




× ×

×

×


Eigenvalue Analysis Verification Problems Natural Frequencies of a Square Canti lever Plate

Figure 4.15: Finite Element Mesh for Model 1

Figure 4.16: Finite Element Mesh for Model 2



Table 4.5: Natural frequencies of a square cantilever plate (Hz)

Mode Theoretical (Ref.[1])

NISA (Model 1)

NISA (Model 2)

1 3.010 3.001 2.996

2 7.371 7.389 7.325

3 18.502 19.344 18.426

4 23.649 25.042 23.641

5 26.887 28.470 26.748



Input Data for Verification Problem No.4.8**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = STATICSAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A SQUARE CANTILEVER PLATE - MODEL 1*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 5.00000E+00, 0.00000E+00, 0 3,,,, 0.00000E+00, 1.00000E+01, 0.00000E+00, 0 4,,,, 0.00000E+00, 1.50000E+01, 0.00000E+00, 0 5,,,, 0.00000E+00, 2.00000E+01, 0.00000E+00, 0 11,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 13,,,, 5.00000E+00, 1.00000E+01, 0.00000E+00, 0 15,,,, 5.00000E+00, 2.00000E+01, 0.00000E+00, 0 21,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0 22,,,, 1.00000E+01, 5.00000E+00, 0.00000E+00, 0 23,,,, 1.00000E+01, 1.00000E+01, 0.00000E+00, 0 24,,,, 1.00000E+01, 1.50000E+01, 0.00000E+00, 0 25,,,, 1.00000E+01, 2.00000E+01, 0.00000E+00, 0 31,,,, 1.50000E+01, 0.00000E+00, 0.00000E+00, 0 33,,,, 1.50000E+01, 1.00000E+01, 0.00000E+00, 0 35,,,, 1.50000E+01, 2.00000E+01, 0.00000E+00, 0 41,,,, 2.00000E+01, 0.00000E+00, 0.00000E+00, 0 42,,,, 2.00000E+01, 5.00000E+00, 0.00000E+00, 0 43,,,, 2.00000E+01, 1.00000E+01, 0.00000E+00, 0 44,,,, 2.00000E+01, 1.50000E+01, 0.00000E+00, 0 45,,,, 2.00000E+01, 2.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 11, 21, 22, 23, 13, 3, 2,



2, 1, 1, 1, 0,,,,, 0 3, 13, 23, 24, 25, 15, 5, 4, 3, 1, 1, 1, 0,,,,, 0 21, 31, 41, 42, 43, 33, 23, 22, 4, 1, 1, 1, 0,,,,, 0 23, 33, 43, 44, 45, 35, 25, 24,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 3.66300E-02,*EIGCNTL5,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 3,UX , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0 3,ROTX, 0.00000E+00,,,,,,,, 0 3,ROTY, 0.00000E+00,,,,,,,, 0 4,UX , 0.00000E+00,,,,,,,, 0 4,UY , 0.00000E+00,,,,,,,, 0 4,UZ , 0.00000E+00,,,,,,,, 0 4,ROTX, 0.00000E+00,,,,,,,, 0 4,ROTY, 0.00000E+00,,,,,,,, 0 5,UX , 0.00000E+00,,,,,,,, 0 5,UY , 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0 5,ROTX, 0.00000E+00,,,,,,,, 0 5,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA



Input Data for Verification Problem No.4.8A**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000IDNUM = REGULARSOLV = FRONFILE = EV08SAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A SQUARE CANTILEVER PLATE - MODEL 2*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 2.50000E+00, 0.00000E+00, 0 3,,,, 0.00000E+00, 5.00000E+00, 0.00000E+00, 0 4,,,, 0.00000E+00, 7.50000E+00, 0.00000E+00, 0 : : 86,,,, 2.00000E+01, 1.25000E+01, 0.00000E+00, 0 87,,,, 2.00000E+01, 1.50000E+01, 0.00000E+00, 0 88,,,, 2.00000E+01, 1.75000E+01, 0.00000E+00, 0 89,,,, 2.00000E+01, 2.00000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 11, 21, 22, 23, 13, 3, 2, 2, 1, 1, 1, 0,,,,, 0 3, 13, 23, 24, 25, 15, 5, 4, : : 15, 1, 1, 1, 0,,,,, 0 65, 75, 85, 86, 87, 77, 67, 66, 16, 1, 1, 1, 0,,,,, 0 67, 77, 87, 88, 89, 79, 69, 68,*MATERIALEX , 1, 0, 3.00000E+07,



NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 3.66300E-02,*EIGCNTL5,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 9,UY , 0.00000E+00,,,,,,,, 0 9,UZ , 0.00000E+00,,,,,,,, 0 9,ROTX, 0.00000E+00,,,,,,,, 0 9,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Free Vibrat ion Analysis of a Canti lever Beam Using Solid Elements

4.9 Free Vibration Analysis of a Cantilever Beam Using Solid Elements

Title:

Free vibration analysis of a cantilever beam using solid elements

Element Type:

3-D solid element (NKTP = 4, NORDR = 1)

Problem:

The first eight natural frequencies of a cantilever beam shown in Figure 4.17 are to be computed usingsolid elements.

Properties:


Three models, having 8, 16 and 32 elements respectively, are created using 3-D solid elements (NKTP= 4, NORDR = 1) as shown in Figure 4.18. All translational degrees of freedom (UX,UY and UZ) areconstrained at the clamped end. All the nodes have been fixed in Y-direction to eliminate the bendingmodes in that direction. The input data is presented only for Model 1.

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used to extract theeigenvalues.

Material:




×


Eigenvalue Analysis Verificat ion Problems Free Vibration Analysis of a Canti lever Beam Using Solid Elements


The first eight natural frequencies from NISA for the three models are compared to the theoreticalresults obtained using simple beam theory (Ref. [1]) in Table 4.6. The axial mode (mode 8) frequencyfrom NISA are close to the theoretical value even for the crude mesh. For the bending modes, NISAresults approach theoretical frequencies as the mesh becomes more refined.

Reference:1. S. Timoshenko, D.H. Young and W. Weaver, Jr., Vibration Problems in Engineering, John Wiley

& Sons, Inc., 4th Edition, 1974.

Figure 4.17: Cantilever Beam

Figure 4.18: Finite Element Mesh for Model 1 to 3



Table 4.6: Natural frequencies (Hz)

Mode Theoretical (Ref.[1])

NISA

Model 1 Model 2 Model 3

Bending 1 0.5595 0.5606 0.5598 0.5596

Bending 2 3.5067 3.6428 3.5393 3.5139

Bending 3 9.8200 10.9689 10.0845 9.8776

Bending 4 19.2437 24.1866 20.3121 19.4761

Bending 5 31.8087 47.3817 34.8603 32.4651

Bending 6 47.5167 90.0299 54.6393 49.0114

Bending 7 66.3662 182.1130 80.9776 69.3334

Axial 1 86.6026 86.7417 86.6373 86.6113



Input Data for Verification Problem No.4.9**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = ON,LISTSAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE FREE VIBRATION ANALYSIS OF A CANTILEVER BEAM USING SOLID ELEMENTS - MODEL 1*ELTYPE 1, 4, 1*NODES 1,,,, 0.00000E+00,-2.50000E+00,-5.00000E-01, 0 2,,,, 0.00000E+00, 2.50000E+00,-5.00000E-01, 0 3,,,, 0.00000E+00, 2.50000E+00, 5.00000E-01, 0 4,,,, 0.00000E+00,-2.50000E+00, 5.00000E-01, 0 : : 33,,,, 1.00000E+02,-2.50000E+00,-5.00000E-01, 0 34,,,, 1.00000E+02, 2.50000E+00,-5.00000E-01, 0 35,,,, 1.00000E+02, 2.50000E+00, 5.00000E-01, 0 36,,,, 1.00000E+02,-2.50000E+00, 5.00000E-01, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 1, 0, 0 5, 6, 7, 8, 9, 10, 11, 12, : : 7, 1, 1, 0, 0 25, 26, 27, 28, 29, 30, 31, 32, 8, 1, 1, 0, 0 29, 30, 31, 32, 33, 34, 35, 36,*MATERIALEX , 1, 0, 1.20000E+09,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.00000E+00,*EIGCNTL8,0,20,0,0.0,0.0,0.1E-04,0.0



Input Data for Verification Problem No.4.9A

*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 : : 33,UY , 0.00000E+00,,,,,,,, 0 34,UY , 0.00000E+00,,,,,,,, 0 35,UY , 0.00000E+00,,,,,,,, 0 36,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA

**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = STATICRESEQUENCE = ON,LISTSAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE FREE VIBRATION ANALYSIS OF A CANTILEVER BEAM USING SOLID ELEMENTS - MODEL 2*ELTYPE 1, 4, 1*NODES 1,,,, 0.00000E+00,-2.50000E+00,-5.00000E-01, 0 2,,,, 0.00000E+00, 2.50000E+00,-5.00000E-01, 0 3,,,, 0.00000E+00, 2.50000E+00, 5.00000E-01, 0 4,,,, 0.00000E+00,-2.50000E+00, 5.00000E-01, 0 : : 65,,,, 1.00000E+02,-2.50000E+00,-5.00000E-01, 0 66,,,, 1.00000E+02, 2.50000E+00,-5.00000E-01, 0 67,,,, 1.00000E+02, 2.50000E+00, 5.00000E-01, 0



Input Data for Verification Problem No.4.9B

68,,,, 1.00000E+02,-2.50000E+00, 5.00000E-01, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 1, 0, 0 5, 6, 7, 8, 9, 10, 11, 12, : : 15, 1, 1, 0, 0 57, 58, 59, 60, 61, 62, 63, 64, 16, 1, 1, 0, 0 61, 62, 63, 64, 65, 66, 67, 68,*MATERIALEX , 1, 0, 1.20000E+09,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.00000E+00,*EIGCNTL8,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 : : 65,UY , 0.00000E+00,,,,,,,, 0 66,UY , 0.00000E+00,,,,,,,, 0 67,UY , 0.00000E+00,,,,,,,, 0 68,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA

**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV09



RESEQUENCE = ON,LISTSAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE FREE VIBRATION ANALYSIS OF A CANTILEVER BEAM USING SOLID ELEMENTS - MODEL 3*ELTYPE 1, 4, 1*NODES 1,,,, 0.00000E+00,-2.50000E+00,-5.00000E-01, 0 2,,,, 0.00000E+00, 2.50000E+00,-5.00000E-01, 0 3,,,, 0.00000E+00, 2.50000E+00, 5.00000E-01, 0 4,,,, 0.00000E+00,-2.50000E+00, 5.00000E-01, 0 : : 129,,,, 1.00000E+02,-2.50000E+00,-5.00000E-01, 0 130,,,, 1.00000E+02, 2.50000E+00,-5.00000E-01, 0 131,,,, 1.00000E+02, 2.50000E+00, 5.00000E-01, 0 132,,,, 1.00000E+02,-2.50000E+00, 5.00000E-01, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 1, 0, 0 5, 6, 7, 8, 9, 10, 11, 12, : : 31, 1, 1, 0, 0 121, 122, 123, 124, 125, 126, 127, 128, 32, 1, 1, 0, 0 125, 126, 127, 128, 129, 130, 131, 132,*MATERIALEX , 1, 0, 1.20000E+09,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.00000E+00,*EIGCNTL8,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0



: : 129,UY , 0.00000E+00,,,,,,,, 0 130,UY , 0.00000E+00,,,,,,,, 0 131,UY , 0.00000E+00,,,,,,,, 0 132,UY , 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Natural Frequencies of a Cylindrical Shell with a Rigid Diaphragm

4.10 Natural Frequencies of a Cylindrical Shell with a Rigid Dia-phragm

Title:

Natural frequencies of a cylindrical shell with a rigid diaphragm

Element Type:


Problem:

The first five natural frequencies and the corresponding mode shapes of a cylindrical shell with a rigiddiaphragm as shown in Figure 4.7 are to be evaluated.

Properties:


One quarter of the cylindrical shell is modeled taking advantage of symmetry. Two models, one with a10 5 mesh and another with a 20 5 mesh are created using 3-D thin shell elements (NKTP = 40,NORDR = 1) as shown in Figure 4.20. Symmetry boundary conditions (UZ = 0.0) are applied at midsection (Z = 100.0). Twin frequencies are eliminated by applying symmetry boundary condition (UY =0.0 and ROTZ = 0.0) along the edges (X = 100.0 and X = -100.0). The rotational degrees of freedomnormal to the shell are suppressed using AUTO = ON in the executive commands. The input data ispresented only for Model 1.

Solution Procedure:

Accelerated subspace iteration with both consistent and lumped mass matrix formulation is used toextract the eigenvalues.

Material:

EX = 20.6 107 N/mm2 (Modulus of elasticity)


DENS = 7.85 10-9 Ns2/mm4 (Mass density)

Geometry:

t = 2.0 mm (thickness)

×

×

× ×


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Cylindrical Shell with a Rigid Diaphragm


The first five natural frequencies obtained from NISA for both Model 1 and Model 2 using consistentand lumped mass matrix formulation are compared to the theoretical results obtained from (Ref.1) inTable 4.7. The frequencies obtained using lumped mass matrix formulation are closer to the theoreticalresults. Plots of the five modes obtained for Model 2, using consistent mass formulation, are shown inFigure 4.21.

Figure 4.19: Cylindrical Shell with Rigid Diaphragm



Figure 4.20: Finite Element Mesh for Models 1 and 2

Reference:1. W. Sidel, Vibrations of Shells and Plates, Marcel Dekker Inc., 1981.

Table 4.7: Natural frequencies (Hz)

Theoretical (Ref.[ 1])

NISA

Consistent Mass Lumped Mass

Model 1 Model 2 Model 1 Model 2

1 1342.5 1363.7 1345.9 1363.7 1347.0

2 1464.8 1499.6 1468.2 1479.9 1468.8

3 1725.9 1740.9 1730.3 1750.3 1731.3

4 1892.4 1961.2 1898.2 1884.9 1896.7

5 2493.1 2636.7 2504.2 2413.7 2496.6



Input Data for Verification Problem No.4.10**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27*TITLE NATURAL FREQUENCIES OF CYLINDRICAL SHELL WITH A RIGID DIAPHRAGM - MODEL 1*ELTYPE 1, 40, 1*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 2.0000000E+00, 2.0000000E+00, 2.0000000E+00,*NODES 1, 1,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.00000E+02, 1.80000E+01, 0.00000E+00, 0 3, 1,,, 1.00000E+02, 3.60000E+01, 0.00000E+00, 0 4, 1,,, 1.00000E+02, 5.40000E+01, 0.00000E+00, 0 : : 63, 1,,, 1.00000E+02, 1.26000E+02, 1.00000E+02, 0 64, 1,,, 1.00000E+02, 1.44000E+02, 1.00000E+02, 0 65, 1,,, 1.00000E+02, 1.62000E+02, 1.00000E+02, 0 66, 1,,, 1.00000E+02, 1.80000E+02, 1.00000E+02, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 13, 12, 2, 1, 1, 1, 0,,,,, 0 2, 3, 14, 13, : : 49, 1, 1, 1, 0,,,,, 0 53, 54, 65, 64, 50, 1, 1, 1, 0,,,,, 0 54, 55, 66, 65,*MATERIALEX , 1, 0, 2.06000E+05,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.85000E-09,*EIGCNTL



Input Data for Verification Problem No.4.10A

5,0,20,0,0.0,0.0,0.0,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 : : 66,UZ , 0.00000E+00,,,,,,,, 0 66,ROTX, 0.00000E+00,,,,,,,, 0 66,ROTY, 0.00000E+00,,,,,,,, 0 66,ROTZ, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA

**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27*TITLE NATURAL FREQUENCIES OF CYLINDRICAL SHELL WITH A RIGID DIAPHRAGM - MODEL 2*ELTYPE 1, 40, 1*RCTABLE 1, 4,1, 0**_DISP3_: NAME =OTHER 2.0000000E+00, 2.0000000E+00, 2.0000000E+00, 2.0000000E+00,*NODES 1, 1,,, 1.00000E+02, 0.00000E+00, 0.00000E+00, 0 2, 1,,, 1.00000E+02, 9.00000E+00, 0.00000E+00, 0 3, 1,,, 1.00000E+02, 1.80000E+01, 0.00000E+00, 0 4, 1,,, 1.00000E+02, 2.70000E+01, 0.00000E+00, 0 : : 123, 1,,, 1.00000E+02, 1.53000E+02, 1.00000E+02, 0



124, 1,,, 1.00000E+02, 1.62000E+02, 1.00000E+02, 0 125, 1,,, 1.00000E+02, 1.71000E+02, 1.00000E+02, 0 126, 1,,, 1.00000E+02, 1.80000E+02, 1.00000E+02, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 23, 22, 2, 1, 1, 1, 0,,,,, 0 2, 3, 24, 23, : : 99, 1, 1, 1, 0,,,,, 0 103, 104, 125, 124, 100, 1, 1, 1, 0,,,,, 0 104, 105, 126, 125,*MATERIALEX , 1, 0, 2.06000E+05,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.85000E-09,*EIGCNTL5,0,20,0,0.0,0.0,0.0,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 : : 126,UZ , 0.00000E+00,,,,,,,, 0 126,ROTX, 0.00000E+00,,,,,,,, 0 126,ROTY, 0.00000E+00,,,,,,,, 0 126,ROTZ, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Simple Torsional System

4.11 Natural Frequencies of a Simple Torsional SystemTitle:

Natural frequencies of a simple torsional system

Element Type:

Problem:

A simple 2-DOF torsional system shown in Figure 4.22 is to be analysed for natural frequencies andmode shapes.

Properties:


The system is modeled with two 3-D torsional spring elements (NKTP = 21) and two 3-D point masselements (NKTP = 30) as shown in Figure 4.22. Node 1 is fully constrained. All degrees of freedomexcept rotation about global x-direction are constrained at nodes 2 and 3.

Solution Procedure:

Conventional subspace iteration is used to extract both the eigenvalues of the system.


The frequencies obtained from NISA for the first and second modes are 6.18 and 16.18 rad/sec,respectively. These correspond exactly to the theoretical values reported in Ref. [1].

Reference:1. W.T. Thomson, Theory of Vibration with Applications, Prentice-Hall Inc., Englewood Cliffs,

New Jersey, 2nd Edition, 1981.

3-D torsional spring element (NKTP = 21)

3-D point mass element with rotary inertia (NKTP = 30)

Material:

K = 100.0 lb-in/rad (Torsional spring constant)

J = 1.0 lb-in-sec2 (Rotary inertia)


Eigenvalue Analysis Verification Problems Natural Frequencies of a Simple Torsional System

Figure 4.22: 2 DOF Torsional System


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Simple Torsional System

Input Data for Verification Problem No.4.11**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV11SAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A SIMPLE TORSIONAL SYSTEM*ELTYPE 1, 21, 1 2, 30, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+02, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 2, 8,1, 0**_DISP3_: NAME =OTHER 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 1.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 5.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.00000E+01, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 0, 1, 1, 0 1, 2, 2, 0, 1, 1, 0 2, 3, 3, 0, 2, 2, 0 2, 4, 0, 2, 2, 0 3,*EIGCNTL2,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0


Eigenvalue Analysis Verification Problems Natural Frequencies of a Simple Torsional System

1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 1,ROTZ, 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 2,UY , 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 2,ROTZ, 0.00000E+00,,,,,,,, 0 3,UX , 0.00000E+00,,,,,,,, 0 3,UY , 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0 3,ROTY, 0.00000E+00,,,,,,,, 0 3,ROTZ, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Composite Square Plate

4.12 Natural Frequencies of a Composite Square PlateTitle:

Natural frequencies of a composite square plate

Element Type:

3-D laminated composite general shell element (NKTP = 32, NORDR = 2)

Problem:

A simply supported square plate with 1 inch side has nine layers alternating at 0o and 90o with respectto the global Z-axis. Thicknesses of the layers at 0o and 90o are 0.001 inch and 0.00125 inch,respectively. The natural frequencies of the plate are to be computed.

Properties:


Taking advantage of symmetry, only one quarter of the plate is modeled using sixteen 3-D laminatedcomposite general shell elements (NKTP = 32, NORDR = 2) as shown in Figure 4.23. Symmetryboundary conditions UX = 0.0, ROTY = 0.0, ROTZ = 0.0 and UY = 0.0, ROTX = 0.0, ROTZ = 0.0 areapplied along X = 0.5 and Y = 0.5, respectively. Translations normal to the plate (UZ) are constrainedfor the nodes on the edges of the plate. In-plane translations (UX, UY) and rotations about normal tothe plate (ROTZ) are suppressed for all the nodes.

Material: In material principal directions,


EY = 1.0 106 psi (Modulus of elasticity)


GXY = 0.6 106 psi (Shear modulus)

GXZ = 0.6 106 psi (Shear modulus)

GYZ = 0.5 106 psi (Shear modulus)


×

×

×

×

×


Eigenvalue Analysis Verification Problems Natural Frequencies of a Composite Square Plate

Solution Procedure:

Conventional subspace iteration and consistent mass matrix formulation are used for the eigenvalueextraction.

Figure 4.23: Finite Element Mesh for a Composite Square Plate


The first four natural frequencies of the plate obtained from NISA are compared to those obtainedfrom Ref. [1] in Table 4.8 and they show good agreement. Mode shape plots for the above modes arepresented in Figure 4.24.

Reference:1. J.E. Ashton and J.M. Whitney, Theory of Laminated Plates, Progress in Materials Science

Series, Technomic Publishing Co., Inc., Stanford, Connecticut, 4, 70, 1970.

Table 4.8: Frequencies in cycle/sec

Mode Ref. [1] NISA

1 30.064 30.015

2 167.113 166.082

3 205.867 203.431

4 270.579 268.712


Eigenvalue Analysis Verification Problems Natural Frequencies of a Composite Square Plate

Input Data for Verification Problem No.4.12**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000IDNUM = REGULARSOLV = FRONEXECUTION = CGOFILE = EV12RESEQUENCE = OFFSAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A COMPOSITE SQUARE PLATE*ELTYPE 1, 32, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 1.0000000E-03, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03, 1.2500000E-03,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.02,890.0,90.0,90.0,90.0,90.0,90.0,90.0,90.0*LAMSQ2**_DISP3_: 1,NAME = 1,9,0,11,2,1,2,1,2,1,2,1,12,1,2,1,2,1,2,1,1,11,1,1,1,1,1,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.25000E-02, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.25000E-01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.87500E-01, 0.00000E+00, 0.00000E+00, 0 :



: 78,,,, 3.12500E-01, 5.00000E-01, 0.00000E+00, 0 79,,,, 3.75000E-01, 5.00000E-01, 0.00000E+00, 0 80,,,, 4.37500E-01, 5.00000E-01, 0.00000E+00, 0 81,,,, 5.00000E-01, 5.00000E-01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 2,,,,, 0 1, 2, 3, 12, 21, 20, 19, 10, 2, 1, 1, 1, 2,,,,, 0 3, 4, 5, 14, 23, 22, 21, 12, : : 15, 1, 1, 1, 2,,,,, 0 59, 60, 61, 70, 79, 78, 77, 68, 16, 1, 1, 1, 2,,,,, 0 61, 62, 63, 72, 81, 80, 79, 70,*MATERIALEX , 1, 0, 4.00000E+07,EY , 1, 0, 1.00000E+06,NUXY, 1, 0, 2.50000E-01,GXY , 1, 0, 6.00000E+05,GXZ , 1, 0, 6.00000E+05,GYZ , 1, 0, 5.00000E+05,DENS, 1, 0, 1.00000E+00,*EIGCNTL4,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 81,UY , 0.00000E+00,,,,,,,, 0 81,ROTX, 0.00000E+00,,,,,,,, 0 81,ROTY, 0.00000E+00,,,,,,,, 0 81,ROTZ, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Free Vibration Analysis of a Three-Layer Orthotropic Sandwich Plate

4.13 Free Vibration Analysis of a Three-Layer Orthotropic Sandwich Plate

Title:

Free vibration analysis of a three- layer orthotropic sandwich plate

Element Type:

3-D sandwich general shell element (NKTP = 33, NORDR = 2)

Problem:

The first four natural frequencies and the corresponding mode shapes for a simply supported 3 layerorthotropic sandwich plate shown in Figure 4.25 are to be computed.

Properties:

Material: In material principal directions,

Face sheets:

EX=EY = 10 106 psi (Modulus of elasticity)




Thickness = 0.016 in

Orientation angle = 0

Core:


GYZ = 0.75 104 psi (Shear modulus)


Thickness = 0.025 in

Orientation angle = 0

×

×

×

°

×

×

×

°


Eigenvalue Analysis Verificat ion Problems Free Vibration Analysis of a Three-Layer Orthotropic Sandwich Plate


A 6 4 grid of finite elements is used to model the entire plate as shown in Figure 4.25. Simplysupported boundary condition is applied at all the four sides. In addition, in-plane translations (UX andUY) and rotation (ROTZ) are constrained at all nodes.

Solution Procedure:

Consistent mass matrix formulation and accelerated subspace iteration are used for the eigenvalueextraction.

Figure 4.25: Finite Element Mesh for a 3 Layer Orthotropic Sandwich Plate


The first four natural frequencies obtained from NISA are compared with experimental results fromRef. [1] and finite element results using a 25 25 mesh from Reference 2 in Table 4.9 and they showgood agreement. The mode shape plots for the above modes are shown in Figure 4.26 through Figure4.29.

×

×



References:1. M.E. Raville and C.E.S. Ueng, Determination of Natural Frequencies of Vibration of a Sand-

wich Plate, Experimental Mechanics, 490-493, 1967.2. T.P. Khatua and Y.K. Cheng, Bending and Vibration of Multilayer Sandwich Beams and Plates,


Table 4.9: Natural frequencies of a simply supported 3 layer orthotropic sandwich plate (Hz)


Mode Ref. [1] Ref. [2] NISA

1 -- 23.0 23.4

2 45.0 45.0 44.8

3 69.0 71.0 71.3

4 78.0 82.0 81.1



Input Data for Verification Problem No.4.13**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV13SAVE = 26,27AUTO CONSTRAINT = OFF*TITLE FREE VIBRATION ANALYSIS OF A 3 LAYER ORTHOTROPIC SANDWICH PLATE*ELTYPE 1, 33, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 1.6000001E-02, 2, 8,1, 0**_DISP3_: NAME =OTHER 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01, 2.5000000E-01,*LAMANGLE1,80.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0*LAMSQ2**_DISP3_: 1,NAME = 1,3,4,11,2,1,1,1,1,1,2,1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.20000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.80000E+01, 0.00000E+00, 0.00000E+00, 0 : : 114,,,, 5.40000E+01, 4.80000E+01, 0.00000E+00, 0 115,,,, 6.00000E+01, 4.80000E+01, 0.00000E+00, 0 116,,,, 6.60000E+01, 4.80000E+01, 0.00000E+00, 0 117,,,, 7.20000E+01, 4.80000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 2,,,,, 0 1, 2, 3, 16, 29, 28, 27, 14,



2, 1, 1, 0, 2,,,,, 0 3, 4, 5, 18, 31, 30, 29, 16, : : 23, 1, 1, 0, 2,,,,, 0 87, 88, 89, 102, 115, 114, 113, 100, 24, 1, 1, 0, 2,,,,, 0 89, 90, 91, 104, 117, 116, 115, 102,*MATERIALEX , 1, 0, 1.00000E+07,EY , 1, 0, 1.00000E+07,NUXY, 1, 0, 3.30000E-01,GXY , 1, 0, 3.75900E+06,DENS, 1, 0, 2.59000E-04,EX , 2, 0, 0.00000E+00,GXZ , 2, 0, 1.95000E+04,GYZ , 2, 0, 7.50000E+03,DENS, 2, 0, 1.14000E-05,*EIGCNTL4,0,20,0,0.0,0.0,0.1E-04,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 : : 117,UZ , 0.00000E+00,,,,,,,, 0 117,ROTX, 0.00000E+00,,,,,,,, 0 117,ROTY, 0.00000E+00,,,,,,,, 0 117,ROTZ, 0.00000E+00,,,,,,,, 0*MODEOUT1,1,1,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Free-Free (Unconstrained) Beam

4.14 Natural Frequencies of a Free-Free (Unconstrained) BeamTitle:

Natural frequencies of a free-free (unconstrained) beam

Element Type:


Problem:

The natural frequencies corresponding to the first three elastic modes of a 2-D free-free beam shown inFigure 4.30 are to be evaluated.

Properties:


The beam is modeled with four 3-D beam elements (NKTP = 12, NORDR = 1) as shown in Figure4.10. The out-of-plane translation (UZ) and rotations (ROTX and ROTY) are constrained reducing theproblem to a planar problem.

Solution Procedure:

Conventional subspace iteration with lumped mass matrix formulation is used to evaluate the first sixeigenvalues including those corresponding to the three rigid body modes. A negative value for theinitial shift is used to avoid singularity during decomposition.

Material:




Cross Section:

A = 4.0 in2 (Area)

IYY = 1.3333 in4 (Moment of inertia about Y-axis)


×

×


Eigenvalue Analysis Verification Problems Natural Frequencies of a Free-Free (Unconstrained) Beam


The frequencies corresponding to the first three elastic modes of the beam obtained from NISA arecompared to the theoretical values (Ref. [1]) in Table 4.10 and they show good agreement.

Figure 4.30: Free-free Beam

Reference:1. W.T. Thomson, Theory of Vibration with Applications, Prentice-Hall Inc., Englewood Cliffs,

New Jersey, 2nd Edition, 1981.

Table 4.10: Frequencies corresponding to the elastic modes (HZ)

Mode Theoretical(Ref.[1]) NISA

1 65.29 65.20

2 179.83 180.35

3 352.66 353.46


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Free-Free (Unconstrained) Beam

Input Data for Verification Problem No.4.14**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = EV14SAVE = 26,27AUTO CONSTRAINT = OFFEIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF A FREE-FREE (UNCONSTRAINED) BEAM*ELTYPE 1, 12, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 4.0000000E+00, 1.3333330E+00, 1.3333330E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 2.00000E+01, 0.00000E+00, 0.00000E+00, 0 3,,,, 4.00000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 6.00000E+01, 0.00000E+00, 0.00000E+00, 0 5,,,, 8.00000E+01, 0.00000E+00, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0 1, 2,000000,000000, 0, 2, 1, 1, 1, 0 2, 3,000000,000000, 0, 3, 1, 1, 1, 0 3, 4,000000,000000, 0, 4, 1, 1, 1, 0 4, 5,000000,000000, 0,*MATERIALEX , 1, 0, 3.00000E+07,NUXY, 1, 0, 3.00000E-01,DENS, 1, 0, 7.28000E-04,*EIGCNTL6,0,20,0,0.0,0.0,0.1E-04,-1.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UZ , 0.00000E+00,,,,,,,, 0


Eigenvalue Analysis Verification Problems Natural Frequencies of a Free-Free (Unconstrained) Beam

1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0 3,UZ , 0.00000E+00,,,,,,,, 0 3,ROTX, 0.00000E+00,,,,,,,, 0 3,ROTY, 0.00000E+00,,,,,,,, 0 4,UZ , 0.00000E+00,,,,,,,, 0 4,ROTX, 0.00000E+00,,,,,,,, 0 4,ROTY, 0.00000E+00,,,,,,,, 0 5,UZ , 0.00000E+00,,,,,,,, 0 5,ROTX, 0.00000E+00,,,,,,,, 0 5,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT3,0,0,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Free Edged Square Plate

4.15 Natural Frequencies of a Free Edged Square PlateTitle:

Natural frequencies of a free edged square plate

Element Type:


Problem:

The natural frequencies of the square plate with free edges shown in Figure 4.31 corresponding to thefirst three elastic modes are to be evaluated.

Properties:


The plate is modeled using 64 3-D thin shell elements (NKTP = 40, NORDR = 1) as shown in Figure4.31. The rotational degrees of freedom normal to the plate are suppressed using AUTO = ON in theexecutive commands.

Solution Procedure:

Accelerated subspace iteration with lumped mass matrix formulation is used. Nine eigenvaluesincluding those corresponding to the six rigid body modes are extracted. A negative value for theinitial shift is used to avoid singularity during decomposition.


The frequencies corresponding to the first three elastic modes of the plate obtained from NISA arecompared to the theoretical values (Ref. [1]) in Table 4.11 and they show good agreement. The modeshape plots for the above modes are presented in Figure 4.32.

Material:




Geometry:

t = 1.0 in (thickness)

×

×


Eigenvalue Analysis Verification Problems Natural Frequencies of a Free Edged Square Plate

Reference:1. S. Timoshenko, D.H. Young and W. Weaver, Jr., Vibration Problems in Engineering, John Wiley

& Sons, Inc., 4th Edition, 1974.

Figure 4.31: Finite Element Mesh for a Free Edged Square Plate

Table 4.11: Frequencies corresponding to the elastic modes (HZ)


1 55.18 56.83

2 80.46 79.83

3 93.58 94.17


Eigenvalue Analysis Verification Problems Natural Frequencies of a Free Edged Square Plate

Input Data for Verification Problem No.4.15**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 1000000SOLV = FRONFILE = STATICSAVE = 26,27MASS FORMULATION = LUMPED*TITLE NATURAL FREQUENCIES OF A FREE EDGED SQUARE PLATE*ELTYPE 1, 40, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.20000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.80000E+01, 0.00000E+00, 0.00000E+00, 0 : : 78,,,, 3.00000E+01, 4.80000E+01, 0.00000E+00, 0 79,,,, 3.60000E+01, 4.80000E+01, 0.00000E+00, 0 80,,,, 4.20000E+01, 4.80000E+01, 0.00000E+00, 0 81,,,, 4.80000E+01, 4.80000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 11, 10, 2, 1, 1, 1, 0,,,,, 0 2, 3, 12, 11, : : 63, 1, 1, 1, 0,,,,, 0 70, 71, 80, 79, 64, 1, 1, 1, 0,,,,, 0 71, 72, 81, 80,*MATERIALEX , 1, 0, 1.05000E+07,NUXY, 1, 0, 2.25000E-01,



DENS, 1, 0, 2.58799E-04,*EIGCNTL9,0,20,0,0.0,0.0,0.1E-04,-1.0*MODEOUT7,0,0,1*EIGOUT, ID = 11,0,3,1,0,1,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Natural Frequencies of an Eccentric Mass-Spring System

4.16 Natural Frequencies of an Eccentric Mass-Spring System

Title:

Natural frequencies of an eccentric mass-spring system

Element Type:

Problem:

A mass of 100 lb-sec2/in with a rotary inertia of 10,000 lb-in-sec2 is connected at an eccentricity of 5inches along the x-direction and 10 inches along the y-direction to a system of translational androtational springs as shown in Figure 4.33. The natural frequencies of the system are to be computed.

Properties:


The structure is modeled with a 3-D general spring element (NKTP = 38) and a 3-D point masselement (NKTP = 30). They are connected by a rigid element, thus inducing the eccentricity. Out-of-plane displacements (UZ, ROTX and ROTZ) at all the nodes are constrained reducing the problem to aplanar problem.

Solution Procedure:

Conventional subspace iteration is used to extract all the three eigenvalues of the system.

Results and Comparisons:

Eliminating the degrees of freedom at the slave node 3 and writing the stiffness and mass matrices forthe master degrees of freedom at node 2.

3-D general spring element (NKTP = 38)

3-D point mass element with rotary inertia (NKTP = 30)

= 1.0 104 lb/in Translational spring constants

= 1.0 106 lb-in Rotational spring constant

= 100 lb-sec2/in Translational mass

= 104 lb-in-sec2 Rotational mass

Kx Ky, ×

Kθ ×

Mx My,

Mθ


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of an Eccentric Mass-Spring System

Figure 4.33: Eccentric Mass-Spring System

The frequencies of the system solved by hand computation are compared to those obtained from NISAin Table 4.12 and they show very good agreement.

Table 4.12: Natural frequencies of an eccentric mass-spring system (Hz)

Mode Hand computation NISA

1 0.9336 0.9336

2 1.5915 1.5915

3 2.7131 2.7130

K102 0 103–

0 102 5 102×

103– 5 102× 2.25 104×

=

M104 0 0

0 104 0

0 0 106

=


Eigenvalue Analysis Verification Problems Natural Frequencies of an Eccentric Mass-Spring System

Input Data for Verification Problem No. 4.16**EXECUTIVE data deckANALYSIS = EIGEN MEMORY = 10000000SOLV = FRONFILE = EV16SAVE = 26,27EIGEN EXTRACTION = SUBSPACE,CONVENTIONAL*TITLE NATURAL FREQUENCIES OF AN ECCENTRIC MASS-SPRING SYSTEM*ELTYPE 1, 38, 1 2, 30, 1*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+04, 1.0000000E+04, 1.0000000E+04, 1.0000000E+06, 1.0000000E+06, 1.0000000E+06, 0.0000000E+00, 0.0000000E+00, 2, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+02, 1.0000000E+02, 1.0000000E+02, 1.0000000E+04, 1.0000000E+04, 1.0000000E+04, 0.0000000E+00, 0.0000000E+00,*NODES 1,,,,-1.00000E+01, 0.00000E+00, 0.00000E+00, 0 2,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 5.00000E+00, 1.00000E+01, 0.00000E+00, 0*ELEMENTS 101, 0, 1, 1, 0 1, 2, 3, 102, 0, 2, 2, 0 3,*RIGLINK 201, 2, 1,1,0, 3, 0,*EIGCNTL3,0,20,0,0.0,0.0,0.0,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 1,ROTX, 0.00000E+00,,,,,,,, 0 1,ROTY, 0.00000E+00,,,,,,,, 0


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of an Eccentric Mass-Spring System

1,ROTZ, 0.00000E+00,,,,,,,, 0 2,UZ , 0.00000E+00,,,,,,,, 0 2,ROTX, 0.00000E+00,,,,,,,, 0 2,ROTY, 0.00000E+00,,,,,,,, 0*MODEOUT1,3,1,1*EIGOUT, ID = 11,0,2,0,0,0,0,0*ENDDATA


Eigenvalue Analysis Verification Problems Natural Frequencies of a Rotating Cantilever Beam

4.17 Natural Frequencies of a Rotating Cantilever BeamTitle:

Natural frequencies of a rotating cantilever beam

Element Type:

3-D Solid Element (NKTP=4, NORDR=1)

Analysis Highlights:

Eigenvalue analysis of a stress stiffened structure

Problem:

The first four natural frequencies of the cantilever beam shown in Figure 4.34 are to be determined forrotor speeds of 2, 5, 10, 20 and 50 rad/sec. about the global Z-direction.

Properties:


The cantilever beam is modeled with sixteen 3-D solid elements as shown in Figure 4.34. The inputdata is presented for a rotational speed, , of 50 rad/sec. All translational degrees of freedom areconstrained at the clamped end.

Solution Procedure:

First a nonlinear static analysis is performed considering geometric nonlinearity only, with 5 loadincrements. Total Lagrangian formulation and full Newton-Raphson iterative procedure are used in theanalysis. Files 24, 26, and 27 are saved from this run. Next an eigenvalue restart analysis(RESTART=4) is performed using conventional subspace iteration and consistent mass formulation.Files saved from the first run are accessed in the restart by providing the same file name prefix.

Material:




×

Ωz


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Rotat ing Canti lever Beam


The results from NISA are compared to those obtained from Refs. [1], [2] and [3] and those obtainedfrom Southwell Coefficient method (Refs. [2, 3]) in Table 4.13. The frequencies corresponding tovibration of the cantilever in the plane of rotation (X-Y plane) are known as the lead-lag frequenciesand those corresponding to the out of plane vibration (X-Z plane) are known as flap frequencies. Theseare indicated by wy,j and wz,j in the table. Note that Refs. [1] and [2] give identical results for both lead-lag and flap frequencies. Also note that the results of Ref. [1] are based on a Rayleigh-Ritz formulationwhereas the results of Refs. [2] and [3] are obtained from a finite element analysis with beam and plateelements, respectively, and finally the Southwell coefficient method is an approximate one forcalculating the natural frequencies.

References:1. M. El-Essawi, A Model for Nonlinear Structural Dynamics of Rotating Cantilevers, Ph.D. Dis-

sertation, Duke University, Durham, North Carolina, March 1982.2. S. Putter and H. Manor, Natural Frequencies of Rotating Radial Beams, Journal of Sound and

Vibration, 56(2), 175-185, 1978.3. M.A. Dokainish and S. Rawtani, Vibration Analysis of Rotating Cantilever Plates, International

Journal for Numerical Methods in Engineering, 3(2), 233-248, 1971.

Figure 4.34: Finite Element Model of the Rotating Cantilever Beam



Table 4.13: Natural Frequencies of a Rotating Cantilever Beam

* Results seem to be beyond range of adapted formula.

AngularVelocity Mode

NISANKTP=4 Ref. [1] Ref. [2] Ref. [3]

SouthwellCoeff.

Method

rad/sec rad/sec rad/sec rad/sec rad/sec rad/sec

2 wy,1 3.6223 3.6218 3.6218 3.6191 3.6131

wz,1 4.1376 - - 4.1318 4.1306

wy,2 22.7004 22.5263 22.5263 22.5280 22.5175

wz,2 22.7882 - - 22.6168 22.6168

5 wy,1 4.0734 4.0739 4.0739 4.0840 4.0850

wz,1 6.4486 - - 6.4559 6.4566

wy,2 25.1160 24.9501 24.9500 24.9539 24.9002

wz,2 25.6078 - - 25.4562 24.4562

10 wy,1 5.0473 5.0491 5.0490 5.0620 5.4463

wz,1 11.2004 - - 11.2082 11.3957

wy,2 32.2794 32.1199 32.1197 32.1764 31.9925

wz,2 33.7896 - - 33.6975 33.6975

20 wy,1 6.7725 6.7773 6.7757 1.1584 9.0312

wz,1 21.1134 - - 20.0334 21.9445

wy,2 51.5681 51.3547 51.3531 51.8142 51.3568

wz,2 55.3021 - - 55.5475 55.5475

50 wy,1 10.4718 10.5939 10.4806 * 21.0918

wz,1 51.0800 - - * 54.2666

wy,2 117.1007 116.9850 116.1996 119.2917 118.0488

wz,2 127.3042 - - 129.3658 129.3658


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Rotat ing Canti lever Beam

Nonlinear Static Input Data for Verification Problem No.4.17**EXECUTIVE data deckANALYSIS = NLSTAT MEMORY = 1000000SOLV = FRONFILE = EV17SAVE = 24,26,27NLTYPE = GEOMETRY*TITLENATURAL FREQUENCIES OF A ROTATING CANTILEVER BEAM - NONLINEAR STATIC RUN*ELTYPE 1, 4, 1*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 3.12500E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 6.25000E+00, 0.00000E+00, 0.00000E+00, 0 4,,,, 9.37500E+00, 0.00000E+00, 0.00000E+00, 0 : : 65,,,, 4.06250E+01, 1.00000E+00, 1.00000E+00, 0 66,,,, 4.37500E+01, 1.00000E+00, 1.00000E+00, 0 67,,,, 4.68750E+01, 1.00000E+00, 1.00000E+00, 0 68,,,, 5.00000E+01, 1.00000E+00, 1.00000E+00, 0*ELEMENTS 1, 1, 1, 0, 0 1, 2, 19, 18, 35, 36, 53, 52, 2, 1, 1, 0, 0 2, 3, 20, 19, 36, 37, 54, 53, : : 15, 1, 1, 0, 0 15, 16, 33, 32, 49, 50, 67, 66, 16, 1, 1, 0, 0 16, 17, 34, 33, 50, 51, 68, 67,*MATERIALEX , 1, 0, 7.50000E+07,NUXY, 1, 0, 0.00000E+00,DENS, 1, 0, 1.00000E+00,*TIMEAMP1,2,00.00000000,0.00000000,1.00000000,1.00000000*EVENT, ID = 1INCREMENTS = EQUAL, 5



Eigenvalue Input Data for Verification Problem No. 4.17

TIMEATEND = 1.0MAXITERATIONS = 100TOLERANCES = -.01, 0.01,0.01*SPDISP, TCRV = 0**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 18,UX , 0.00000E+00,,,,,,,, 0 18,UY , 0.00000E+00,,,,,,,, 0 18,UZ , 0.00000E+00,,,,,,,, 0 35,UX , 0.00000E+00,,,,,,,, 0 35,UY , 0.00000E+00,,,,,,,, 0 35,UZ , 0.00000E+00,,,,,,,, 0 52,UX , 0.00000E+00,,,,,,,, 0 52,UY , 0.00000E+00,,,,,,,, 0 52,UZ , 0.00000E+00,,,,,,,, 0*BODYFORCE, TCRV = 0,0,00.0,0.0,50.0,0.0,0.0,0.00.0,0.0,0.0,0.0,0.5,0.0*NLOUT1,3,0,1,0,1,0,0,0,0*PRINTCNTLELST,0LDVE,0SLFO,0*ENDDATA

PROB=EV17R4.NISANALYSIS=EIGENVALUERESTART=4FILE=EV17EIGENEXTRACTION=SUBSPACE,CONVENTIONALMASS=CONSISTENT*TITLENATURAL FREQUENCIES OF A ROTATING CANTILEVER BEAM - EIGENVALUE RESTART 4*EIGCNTL6,0,0*SPDISP1,UX,0.0,52,17,UY,UZ*ENDDATA


Eigenvalue Analysis Verificat ion Problems Natural Frequencies of a Simply Supported Square Plate using Guyan Reduction

4.18 Natural Frequencies of a Simply Supported Square Plate using Guyan Reduction

Title:

Natural Frequencies of a simply supported square plate using Guyan reduction

Element Type:


Problem:

The effect of Guyan reduction on the natural frequencies of the simply supported square plate inVerification Problem 4.5 is to be evaluated.

Solution Procedure:

Consistent mass formulation and Guyan reduction are used.


The first five natural frequencies are evaluated for 8 cases of Guyan masters as given in Table 4.14.The first 5 cases involve manual selection of masters and the last 3 cases involve automatic selectionby the program. The first four include UZ degrees of freedom at selected nodes (Figure 4.35). The fifthcase involves all degrees of freedom with non-zero mass as masters and hence represents the exactresult. As it can be seen, when UZ displacements of nodes evenly distributed over the plate are chosenas masters, the results are close to the exact values. The drawback of the automatic selection of mastersis evident in case 6 where the master degrees of freedom are unevenly distributed over the plate.




Eigenvalue Analysis Verification Problems Natural Frequencies of a Simply Supported Square Plate using Guyan Reduction

Figure 4.35: Finite Element Mesh for a Simply Supported Square Plate



Table 4.14: Natural Frequencies of simply supported Plate

CASE MASTERS Mode 1 Mode 2 Mode 3 Mode 4 Mode 5

1 1 UZ @ nodes 87.64 --- --- --- ---

2 9 UZ @ nodes 82.87 213.79 213.79 373.96 483.07

3 24 UZ @ nodes 211.11 211.11 211.11 353.48 438.98

4 33 UZ @ nodes 82.80 211.08 211.08 353.42 438.99

5 163 All dynamic degrees of freedom (exact) 82.80 211.08 211.08 353.41 438.99

6 9 Automatic 86.31 237.14 280.37 447.61 577.27

7 24 Automatic 82.80 211.09 211.09 353.48 438.99

8 33 Automatic 82.80 211.08 211.08 353.42 438.99


Eigenvalue Analysis Verification Problems Natural Frequencies of a Simply Supported Square Plate using Guyan Reduction

Input Data for Verification Problem No. 4.18 **EXECUTIVE data deckANALYSIS = EIGEN IDNUM = REGULARSOLV = FRONFILE = STATICRESEQUENCE = OFFSAVE = 26,27EIGEN EXTRACTION = GUYAN*TITLENATURAL FREQUENCIES OF A SS SQUARE PLATE - SAME AS PROBLEM 5.5 BUT USING GUYAN AND CONSISTENT MASS MATRIX*ELTYPE 1, 20, 2*RCTABLE 1, 8,1, 0**_DISP3_: NAME =OTHER 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00, 1.0000000E+00,*NODES 1,,,, 0.00000E+00, 0.00000E+00, 0.00000E+00, 0 2,,,, 6.00000E+00, 0.00000E+00, 0.00000E+00, 0 3,,,, 1.20000E+01, 0.00000E+00, 0.00000E+00, 0 4,,,, 1.80000E+01, 0.00000E+00, 0.00000E+00, 0 : : 78,,,, 3.00000E+01, 4.80000E+01, 0.00000E+00, 0 79,,,, 3.60000E+01, 4.80000E+01, 0.00000E+00, 0 80,,,, 4.20000E+01, 4.80000E+01, 0.00000E+00, 0 81,,,, 4.80000E+01, 4.80000E+01, 0.00000E+00, 0*ELEMENTS 1, 1, 1, 1, 0,,,,, 0 1, 2, 3, 12, 21, 20, 19, 10, 2, 1, 1, 1, 0,,,,, 0 3, 4, 5, 14, 23, 22, 21, 12, : : 15, 1, 1, 1, 0,,,,, 0 59, 60, 61, 70, 79, 78, 77, 68, 16, 1, 1, 1, 0,,,,, 0 61, 62, 63, 72, 81, 80, 79, 70,*MATERIAL



EX , 1, 0, 1.05000E+07,NUXY, 1, 0, 3.33330E-01,DENS, 1, 0, 2.58799E-04,*EIGCNTL5,0,0,0,0.0,0.0,0.1E-03,0.0*SPDISP**_DISP3_: SPDISP, SET = 1 1,UX , 0.00000E+00,,,,,,,, 0 1,UY , 0.00000E+00,,,,,,,, 0 1,UZ , 0.00000E+00,,,,,,,, 0 2,UX , 0.00000E+00,,,,,,,, 0 : : 80,UZ , 0.00000E+00,,,,,,,, 0 81,UX , 0.00000E+00,,,,,,,, 0 81,UY , 0.00000E+00,,,,,,,, 0 81,UZ , 0.00000E+00,,,,,,,, 0*MASTER21,UZ, 25,239,UZ, 43,257,UZ, 61,2*ENDDATA


MAIN OFFICES

USA

Cranes Software, Inc.1133 E Maple Road, Suite 103Troy, Michigan 48083 USAP.O. Box 696Troy, MI, 48099, USAPhone : (248) 689-0077Fax : (248) 689-7479 E-mail : [email protected]

INDIA

Cranes Software International Ltd., Cranes Engineering, #52, Bannerghatta Road, B T M Layout, 1st Stage, 1st Phase, Bangalore – 560029, India. Phone : +91-80 4128 1111 Fax : +91-80 4128 0203 E-mail : [email protected]

www.cranessoftware.com

www.nisasoftware.com

Prasanna.Prabhu

Distance Measurement

3.08 in

nisa verification manual 24april

Documents

files25 title

accelerated

conventional

john wiley

shear correction

multilayer

vpgnisa verification

layer antisymmetric