Finite Element Analysis

IW-CAAD 2004

Understanding and Using

Finite Element Analysis

July 19-21, 2004

Understanding and Using

Finite Element Analysis

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The Objective

• To understand the fundamentals of the

Finite Element Method and the Finite Element Analysis

• To apply the Finite Element Analysis Tools for Modeling and Analysis of Structures

• Use SAP2000 as Tool for Finite Element Modeling and Analysis of Structures

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The Program

• What is FEM and Why it is needed

• Fundamental concepts in FEM and FEA • Concept of Stiffness

• Finite Elements and their Usage

• Constructing Finite Element Models • Applying Loads to FE Models

• Interpreting FE Results

• Modeling Different Types of Structures using FE • Intro to Non-linear and Dynamic Analysis

What is Finite Element Analysis

and Why do We Need It!

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The Structural System

EXCITATION Loads Vibrations Settlements Thermal Changes RESPONSES Displacements Strains Stresses Stress Resultants

STRUCTURE

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The Need For Analysis

We need to determine the

Response of the Structure to

Excitations

so that:

We can ensure that the structure

can sustain the excitation with an

acceptable level of response

Analysis

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Analysis of Structures

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

xx yy zz vx

x



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Real Structure is governed by “Partial Differential Equations” of various order

Direct solution is only possible for:

• Simple geometry • Simple Boundary • Simple Loading.

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The Need for Structural Model

Structural Model

EXCITATION

Loads Vibrations Settlements Thermal Changes

RESPONSES

Displacements Strains Stress Stress Resultants

STRUCTURE

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The Need for Modeling

A - Real Structure cannot be Analyzed:

It can only be “Load Tested” to determine

response

B - We can only analyze a

“Model” of the Structure

C - We therefore need tools to Model the

Structure and to Analyze the Model

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Finite Element Method and FEA

• Finite Element Analysis (FEA)

“A discretized solution to a continuum

problem using FEM”

• Finite Element Method (FEM)

“A numerical procedure for solving (partial)

differential equations associated with field problems, with an accuracy acceptable to engineers”

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From Classical to FEM

      xx yy zz vx x  y  z  p 0  t v t s t v dV p u dV p u ds _ _ _     Assumptions Equilibrium Compatibility Stress-Strain Law

(Principle of Virtual Work) “Partial Differential Equations” Classical Actual Structure

Kr



R

“Algebraic Equations” K = Stiffness r = Response R = Loads FEM Structural Model

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Simplified Structural System

Loads (F) Deformations (u)

F_v

F = K u

F

K

(Stiffness)

u

Equilibrium Equation

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The Total Structural System

EXCITATION RESPONSES

STRUCTURE

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• Static

• Dynamic

• Elastic

• Inelastic

• Linear

• Nonlinear

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The Main Equilibrium Equations

1. Linear-Static Elastic

2. Linear-Dynamic Elastic

3. Nonlinear - Static Elastic OR Inelastic

4. Nonlinear-Dynamic Elastic OR Inelastic

F Ku  ) ( ) ( ) ( ) (t Cu t Ku t F t u M            F F Ku _NL 

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The Basic Analysis Types

Excitation Structure Response Basic Analysis Type

Static Elastic Linear Linear-Elastic-Static Analysis

Static Elastic Nonlinear Nonlinear-Elastic-Static Analysis

Static Inelastic Linear Linear-Inelastic-Static Analysis

Static Inelastic Nonlinear Nonlinear-Inelastic-Static Analysis

Dynamic Elastic Linear Linear-Elastic-Dynamic Analysis

Dynamic Elastic Nonlinear Nonlinear-Elastic-Dynamic Analysis

Dynamic Inelastic Linear Linear-Inelastic-Dynamic Analysis

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Special Analysis Types

• Non-linear Analysis – P-Delta Analysis

– Buckling Analysis

– Static Pushover Analysis

– Fast Non-Linear Analysis (FNA) – Large Displacement Analysis • Dynamic Analysis

– Free Vibration and Modal Analysis – Response Spectrum Analysis

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The Finite Element Analysis Process

Evaluate Real Structure Create Structural Model

Discretize Model in FE Solve FE Model

Interpret FEA Results

Physical significance of Results Engineer

Software

The Fundamentals

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From Continuum to Structure

From Structure To Structural Model

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Solid – Structure - Model

Simplification (geometric) Discretization 3D SOLIDS CONTINUOUS MODEL OF STRUCTURE DISCRETE MODEL OF STRUCTURE 3D-CONTINUM MODEL

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      xx yy zz vx x  y  z  p 0  t v t s t v dV p u dV p u ds _ _ _     Assumptions Equilibrium Compatibility Stress-Strain Law

(Principle of Virtual Work) “Partial Differential Equations” Continuum Actual Structure

Kr



R

“Algebraic Equations” K = Stiffness r = Response R = Loads Structure Structural Model

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Continuum Vs Structure

• A continuum extends in all direction, has infinite particles, with continuous variation of material

properties, deformation characteristics and stress state

• A Structure is of finite size and is made up of an assemblage of substructures, components and members

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Physical Categorization of Structures

• Structures can be categorized in many ways.

• For modeling and analysis purposes, the overall physical behavior can be used as basis of

categorization

– Cable or Tension Structures – Skeletal or Framed Structures – Surface or Spatial Structures – Solid Structures

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Structure, Member, Element

• Structure can be considered as an assemblage of “Physical Components” called Members

– Slabs, Beams, Columns, Footings, etc.

• Physical Members can be modeled by using one or more “Conceptual Components” called

Elements

– 1D elements, 2D element, 3D elements

– Frame element, plate element, shell element, solid

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Structural Members

Dimensional Hierarchy of Structural Members Continuum Regular Solid (3D) Beam (1D) b h L>>(b,h)  b h t z Plate/Shell (2D) x z t<<(x,z)  x z y x L

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The Reference System

• To convert continuum to structures, the first step is to define a finite number of reference

dimensions

• The Four Dimensional Reference System:

– Three Space Dimensions, x, y, z – One Time Dimension, t

• The Entire Structural System is a function of Space and Time

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Global Axis and Local Axis

• Global Axis used to reference the overall structure and to locate its components:

Also called the Structure Axis

• Local Axis used to reference the

quantities on part of a structure or a member or an element:

Also called the Member Axis or Element Axis

X

Y Z

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The General Global Coordinate System

• The global coordinate system is a

three-dimensional, right-handed, rectangular coordinate system.

• The three axes, denoted X, Y, and Z, are mutually perpendicular and satisfy the right-hand rule.

• The location and orientation of the global system are arbitrary. The Z direction is normally upward, but this is not required.

• All other coordinates systems are converted or mapped back and forth to General Coordinate System

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Polar Coordinate Systems

• Polar coordinates include

– Cylindrical CR-CA-CZ

coordinates

– Spherical SB-SA-SR

coordinates.

• Polar coordinate systems

are always defined with

respect to a rectangular

X-Y-Z system.

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Local Coordinate Systems

• Each part (joint, element, or constraint) of the structural model has its own local co-ordinate system used to define the properties, loads, and response for that part.

• In general, the local co-ordinate systems may vary from joint to joint, element to element, and constraint to constraint

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Local Axis and Natural Axis

• The elements and

variation of fields can often be described best in terms “Natural

Coordinates”

• Natural coordinates may be linear or curvilinear • Shape functions can are

used to associate the

local system and natural system

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Primary Relationships

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The Basic Structural Quantities

• Loads • Actions • Deformations • Strains • Stresses • Stress Resultants

The main focus of

Structural Mechanics is to develop relationships

between these quantities The main focus of FEM is solve these relationships numerically

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Mechanics Relationships

Load Action Deformation Strain Stress Stress Resultant

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Primary Relationships

• Load – Action Relationship

• Action – Deformation Relationship • Deformation – Strain Relationship • Strain – Stress Relationship

• Stress – Stress Resultant Relationship • Stress Resultant – Action Relationship

• Most of these relationships can defined

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Action - Deformation Relationship

• This involves two types of relationships

– Deformations produced due to given

Actions

• Example:

– Actions needed to produce or restrain

certain Deformation

• Example:

• Moment-Curvatures, Load-Deflection Curves are samples of this relationship

• The represents to “Element Stiffness” 

P d P d M  f M M E A L P    L A E P 

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Simplified Examples of Action-Deformation

P V M P  M  V V v M P V M P V M P  P  M  V M  V V v M V v M E A L P         _  L M V EI L v 2 3 6 3       _  V L M EI L 2 2 2 

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Deformation – Strain Relationship

• In general, strain is the first derivative of deformation

• Basic Deformation and Corresponding Strains are:

– Shortening Axial Strain

– Curvature Axial Strain

– Shearing Shear Strain

– Twisting Shear Strain + Axial Strain

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Strain – Stress Relationship

• The resistance of the material to strain, derived from the stiffness of the material particles

• For a general Isotropic Material

• For 2D, Isotropic Material, V=0                                                                               zx yz xy z y x zx yz xy z y x v v v v v v v v v v v v v v E             2 2 1 0 0 0 0 0 0 2 2 1 0 0 0 0 0 0 2 2 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 2 1 1 x xx

E

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  kf_c   f_y xy xy

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The Stress Strain Components

• The Hook's law is simplified form of Stress-Strain

relationship

• Ultimately the six stress and strain components can be represented by 3 principal summations xx  yy  zz  xy  zx  yx  zy  xz  yz  x y z

At any point in a continuum, or solid, the stress state can be completely defined in terms of six stress components and six corresponding

xx  yy  zz  xy  zx  yx  zy  xz  yz  x y z

At any point in a continuum, or solid, the stress state can be completely defined in terms of six stress components and six corresponding

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Secondary Relationships

• Global Axis - Local Axis

– Geometric Transformations Matrices

• Local Axis - Natural Axis

– Shape Functions – Jacobian Matrix

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What are Shape Functions

• Shape Functions or Interpolation Functions provide a

means of computing value of any quantity (field) at some point based on the value specified at specific locations

• Shape Functions are used in FEM to relate the values ate Nodes to those within the Element

– Nodal Displacements to Element Deformation – Nodal Stresses to Stresses within the Element

• Shape Functions can be in 1D, 2D or in 3D • Shape Functions can be Liner or Polynomials

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One Dimensional Shape Functions

) 1 ( 5 . 0 ) ( 1 s s s N    ) 1 )( 1 ( ) ( 2 s s s N    ) 1 ( 5 . 0 ) ( 3 s s s N  



    3 3 3 2 2 1 1( ) ( ) ( ) ) (s N s w N s w N s w w S=0 S =-1 S =+1 S=1 S=0 ) 1 ( ) ( 1 s s N  

s

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N

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(

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The Jacobian Matrix

• The Strain is Derivative of Displacement

• Displacements are specified on nodes, in Element Local Axis

• For computing K. strains are needed in element in “Natural Coordinates”

• Shape Functions relate Nodal Displacements with Element Displacements

• Jacobian Matrix relates the derivative of Nodal Displacement, directly with Element Strains

w N w N w N w J             3 ₃ 2 2 1 1

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The Concept of DOF

• In a continuum, each point can move in infinite ways

• In Structure, movement of each point is

represented or resolved in limited number of ways, called Degrees Of Freedom (DOF)

• The DOF of range from 1 to 7 depending on type and level of structural model and the element

being considered

• Global and Local DOF have different meaning and significance

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The Basic Six DOF

• Three Translations along the reference axis

– Dx, Dy, Dz

• Three Rotations about the reference axis

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The Seven Degrees of Freedom

• The General Beam Element may have 7 degrees of freedom • The seventh degree

is Warping

• Warping is out-of plane distortion of the beam cross-section z y x x u y u z u x r y r z r z w

Each section on a beam member can have seven Degrees Of Freedom (DOF) with respect to its local axis. z y x x u y u z u x r y r z r z w

Each section on a beam member can have seven Degrees Of Freedom (DOF) with respect to its local axis.

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The Complete DOF Picture

§ u_z  Axial deformation  Axial strain  Axial stress

§ u_x  Shear deformation  Shear strain  Shear stress

§ _uy  Shear deformation  Shear strain  Shear stress

§ r_z  Torsion  Shear strain  Shear stress

§ r_y  Curvature  Axial strain  Axial stress

§ r_x  Curvature  Axial strain  Axial stress

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Global Structural DOF

• Only 3 DOF are really needed at Global Level

• The deformation of the structure can be defined completely in terms of 3 translations of points with respect to Global Axis

• Rotations may be defined arbitrarily at various locations for convenience of modeling and

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Local DOF and Natural DOF

• DOF can be defined for local movements of joints and elements in 3 Orthogonal reference system

• Natural DOF can be defined in terms of Natural Coordinates System of the element which may be orthogonal or curvilinear

• Relationship between Global, Local and Natural DOF is established through Transformation

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Types of DOF in SAP2000

• Active

– the displacement is computed during the analysis

• Restrained

– the displacement is specified, and the corresponding

reaction is computed during the analysis

• Constrained

– the displacement is determined from the

displacements at other degrees of freedom

• Null

– the displacement does not affect the structure and is

ignored by the analysis

• Unavailable

– The displacement has been explicitly excluded from

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Constraints and Restraints

• Restraints:

– Direct limits on the DOF

– External Boundary Conditions

– Fixed Support , Support Settlement

• Constraints

– Linked or dependent limits on DOF

– Internal linkages within the structure, in addition to

or in place of normal connections

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Body Constraints

• A Body Constraint causes all of its constrained joints to move together as a three-dimensional rigid body.

• All constrained joints are connected to each other by rigid links and cannot displace relative to each other.

• This Constraint can be used to:

– Model rigid connections, such as where several beams

and/or columns frame together

– Connect together different parts of the structural model

that were defined using separate meshes

– Connect Frame elements that are acting as eccentric

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Constraints in SAP2000

• A constraint is a set of two or more constrained joints.

• The displacements of each pair of joints in the constraint are related by constraint equations. • The types of behavior that can be enforced by

constraints are:

– Rigid-body behavior

– Equal-displacement behavior

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Constraints in SAP2000

• Rigid-body behavior

– Rigid Body: fully rigid for all displacements

– Rigid Diaphragm: rigid for membrane behavior in a

plane

– Rigid Plate: rigid for plate bending in a plane – Rigid Rod: rigid for extension along an axis

The Concept of

Stiffness

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What is Stiffness ?

• In structural terms, stiffness may be defined as

“Resistance to Deformation” • So for each type of

deformation, there is a corresponding stiffness

• Stiffness can be considered or evaluated at various levels • Stiffness is also the

“constant” in the Action-Deformation Relationship

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F

K

F

Ku

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The Structure Stiffness

Material Stiffness Section Stiffness Member Stiffness Structure Stiffness Material Stiffness Cross-section Geometry Member Geometry Structure Geometry Stress/Strain EA, EI EA/L

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Structure Stiffness

• The overall resistance of the structures to over all loads, called the

Global Structure Stiffness.

• Derived from the sum of stiffness of its

members, their

connectivity and the boundary or the

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Member and Element Stiffness

• The resistance of each Element to local actions

called the Element Stiffness This is derived from the

cross-section stiffness and the geometry of the

Element.

• In FEM, the Member

Stiffness can be derived from stiffness of Elements used to model the Member

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Beam Element Cross-section Stiffness

• The resistance of the cross-section to unit strains. This is derived from the cross-section geometry and the stiffness of the

materials from which it is made.

• For each of degree of freedom, there is a corresponding stiffness, and a corresponding cross-section property § u_z  Cross-section area, A_x

§ u_x  Shear Area along x, SA_x

§ u_y  Shear Area along y, SA_y

§ _rz  Torsional Constant, J

§ r_x  Moment of Inertia, I_xx

§ r_y  Moment of Inertia, I_yy

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Computing Element Stiffness

• Assume Nodal Displacements (Deformations)

• Determine Deformations within the element using “Shape Functions”

• Determine the Strains within the element using Strain-Displacement Relationship

• Determine Stress within the element using Stress-Strain Relationship

• Use the principle of Virtual Work and integrate the product of stress and strain over the volume of

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Internal Work

Deriving the Basic Stiffness Equation

W I W E.  . F W E. 





 V T dv W I.  





 D                 







V T T V T T V T dv B D B W I dv B D B W I dv D W I . . .



 V dv W I.       K F dv B D B F dv B D B F V T V T T T                   





External Work Equilibrium





 B Stress-Strain Strain-Disp.

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Stiffness Equation: An Example

F     L E 1   AL L E K dv L E K dv L E L K V V   





2 2 1 1 L B E D 1       B D  



 V T dv B D B K L  EA

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The Matrices in FEM

Element Nodal Deformations

Deformation in Element Space

Strain In Element Space

Stress in Element Space Global Nodal Deformations

T-Matrix Global-Local Cords. N-Matrix Shape Functions B-Matrix Strain-Deforrmation D-Matrix Stress-Strain

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What is Stiffness Matrix

• The actions and deformations of different DOF in an element are not independent

– One action may produce more than one

deformations

– One Deformation may be caused by more than one

Action

• A Stiffness Matrix relates various Deformation and actions within an Element

• A Stiffness Matrix is generalized expression of overall element stiffness

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Element Stiffness Matrix

R 1 K11 K12 K13 K14 K15 K16 r1 R 2 K21 K22 K23 K24 K25 K26 r2 R 3 K31 K32 K33 K34 K35 K36 r3 R 4 K41 K42 K43 K44 K45 K46 r4 R 5 K51 K52 K53 K54 K55 K56 r5 R 6 K61 K62 K63 K64 K65 K66 r6

=

Node1 _Node2 r1 r2 r3 r4 r5 _r6

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A 2D Frame Element Stiffness

Node1 _Node2 U1 U2 U3 U1 U2 U3 E ,A ,I ,L

(P1)₁ EA/L 0 0 -EA/L 0 0 (U1)₁

(P2)₁ 0 12EI/L3 _6EI/L2 _{0 -12EI/L}3 _6EI/L2 _(U2)

1

(P3)₁ 0 6EI/L2 _{4EI/L 0 -6EI/L}2 _{2EI/L (U3)}

1

(P1)₂ -EA/L 0 0 EA/L 0 0 (U1)₂

(P2)₂ 0 -12EI/L3 _-6EI/L2 _{0 12EI/L}3 _-6EI/L2 _(U2)

2

(P3)₂ 0 6EI/L2 _{2EI/L 0 -6EI/L}2 _{4EI/L (U3)}

2

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Direct Stiffness Method and FEM

• Basically there is no conceptual difference

between DSM and FEM. DSM is a special case of the general FEM

• Direct Stiffness Method (DSM)

– The terms of the element stiffness matrix are defined

explicitly and in close form (formulae)

– It is mostly applicable to 1D Elements (beam, truss)

• Finite Element Method

– The element stiffness matrix terms are computed by

numerical integration of the general stiffness equation

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Introduction

• In real world, the problem domains are such that they have no proper shape

• It is difficult to find the exact solution of the real problems

• Isoparametric elements are used to discretize a complex shape problem domain into a number of geometrical shapes

• Analysis is carried out on the simple discretized shapes and then the result is integrated over the actual problem domain to get the approximate numerical solution

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1D Isoparametric Shape

• Consider the example of a bar element • For simplification, let the bar lie in x-axis

• First, relate the Global coordinate X to natural coordinate system with variable r,

Z Y X, U x₁ x₂ U₁ U₂ r r = +1 r = -1

1

1







r

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1D Isoparametric Shape

2 1 (1 ) 2 1 ) 1 ( 2 1 X r X r X     ) 1 ( 2 1 ) 1 ( 2 1 2 1 r and h r h    

Transformation is given by:

are interpolation of shape functions



 2 h_iU_i U

The bar global displacements are shown by:

h1 _h2 Z Y X, U x₁ x₂ U₁ U₂ r r = +1 r = -1

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1D Isoparametric Shape

2 2 2 1 2 1 2 L X X dr dX and U U dr dU dX dr dr dU        

Element Strains can be calculated by:

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1D Isoparametric Shape

L

U

U

₂



₁





Therefore, we have



1

1



1





L

B

u

B ˆ







So, Strain displacement transformation matrix can be shown as:

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1D Isoparametric Shape

dV

EB

B

K

v T









Jdr L AE K 1 1 1 1 1 1 2        





The Stiffness Matrix is given by:

Where E is the Elasticity constant Therefore, we have

Where,

• A = area of the bar • J = Jacobian

relating an element length in the global coordinate system to an element length in the natural coordinate system

dr

J

dX



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1D Isoparametric Shape

         1 1 1 1 L AE K 2 1 (1 ) 2 1 ) 1 ( 2 1 X r X r X    



  2 1 i i iU h U Therefore, K is evaluated as Substituting the value of r from

And put in





2 / 2 / ) 2 1 ( L X X X r    To get

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Example 01

Derive • Interpolation Matrix H • Strain Displacement Interpolation Matrix B • Jacobian Operator J

for the three-node element as shown in figure X, U r = 0 r = +1 r = -1 L/2 L/2 x₁ 1 3 2

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Example 01

Finding the interpolation functions of the given

element r = -1 r = 0 r = +1 +1 ) 1 ( 2 1 r r h    +1 r = -1 r = 0 r = +1 2 3

1 r

h





r = -1 r = 0 r = +1 +1 ) 1 ( 2 2 r r h  

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Example 01

So,

The strain displacement matrix B is obtained by     _{   }     r r r J B dr dH J B 2 ) 2 1 ( ) 2 1 ( 1 1



h₁ h₂ h₃



H 

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Example 01

For Jacobian Operator

2 det ; 2 2 2 2 ) 2 )( 1 ( ) )( 1 ( 2 ) 1 ( 2 1 1 1 2 1 1 3 3 2 2 1 1 L J L J L J dr dx J r L L x x L x r L x r r x r r x x h x h x h x                            

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2D Isoparametric Element

• Linear and quadratic two-dimensional

isoparametric finite elements use the same shape function for specification of the element shape

and interpolation of the displacement field

2 1 3 4   4 1 2 3 5   4 6 5 7 

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2D Isoparametric Element

• Shape functions Ni are defined in local

coordinates

• The same shape

functions are used for interpolations of displacements of coordinates

)

1

,

1

(

,























    i i i i i i i i y N y x N x v N v u N u ; ;

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2D Isoparametric Element

• Shape functions for linear quadratic

two-dimensional isoparametric elements are shown here

• Linear Elements 4-node: _{(1 )(1 )}

4 1

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2D Isoparametric Element

Quadratic Elements 8-nodes

where 8 , 4 ) 1 )( 1 ( 2 1 6 , 2 ) 1 )( 1 ( 2 1 7 , 5 , 3 , 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 N 2 2 2 2 i                   i N i N i o i o i o o o o           i o i o      ; 

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Example 02

• Derive the expressions

needed for the calculation of Stiffness Matrix of the isoparametric 4-node

finite element shown in the figure. Assume plane stress or plane strain

conditions y, v 1 2 3 4 r or  s or  y₄ x_{4 x, u}

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Example 02

• The four interpolation functions for the linear quadratic isoparametric element are

y, v 1 2 3 4 r or  s or  y₄ x_{4 x, u} ) 1 )( 1 ( 4 1 h ) 1 )( 1 ( 4 1 h ) 1 )( 1 ( 4 1 h ) 1 )( 1 ( 4 1 h 4 3 2 1 s r s r s r s r            

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Example 02

The coordinate interpolations for the element is given by

Using the interpolation functions, the coordinate interpolations for this element are





    4 1 4 1 ; i i i i i ix y h y h x 4 3 2 1 4 3 2 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 y ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 x y s r y s r y s r y s r x s r x s r x s r x s r                        

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Example 02

The displacement interpolations for the element is given by

Using the interpolation functions, the coordinate interpolations for this element are





    4 1 1 4 1 ; _{i i} i i iu v hv h u 4 3 2 1 4 3 2 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 v ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 ) 1 )( 1 ( 4 1 u v s r v s r v s r v s r u s r u s r u s r u s r                        

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Example 02

The element strains are given by

To evaluate the displacement derivatives, we need to evaluate





x v y u y v x u xy yy xx xy yy xx T                     ; ; x J r or x y x r y r x r                                          

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Example 02

where 4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 s y ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 r y ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 s x ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 r x y r y r y r y r y s y s y s y s x s x r x r x r x s x s x s x s                                        

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Example 02

For any value of r and s

We can form the Jacobian matrix. Assuming we evaluate J at

1

1

1

1

















r

and

s

s r J y x                                  1 j i

and

s

s

r

r





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Example 02

To evaluate the element strains, we use

4 3 2 1 4 3 2 1 4 3 2 1 4 3 2 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 s v ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 r v ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 s u ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 ) 1 ( 4 1 r u v r v r v r v r v s v s v s v s u s u r u r u r u s u s u s u s                                        

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Example 02

Simplifying the above relations, we get

Where u s r r r s s s s J and u s r r r s s s s J ˆ ) 1 ( 0 ) 1 ( 0 1 0 1 0 1 0 ) 1 ( 0 ) 1 ( 0 1 0 4 1 y v x v ˆ 0 ) 1 ( 0 ) 1 ( 0 1 0 1 0 1 0 ) 1 ( 0 ) 1 ( 0 1 4 1 y u x u 1 1                                                                        





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Example 02

Strain-displacement transformation is given by

So, we can get

u

B

_ij ij



ˆ



                                   s r s r s r s r r r r r s s s s B_ij 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( 1 1 1 ) 1 ( 0 ) 1 ( 0 1 0 1 0 0 1 0 ) 1 ( 0 ) 1 ( 0 1 4 1

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Example 02

Stiffness Matrix K is given by

In the above expressions, C is the material property

matrix, t is the thickness of the element at the sampling point (r,s) ij ij T ij ij j i ij ij ij J CB B F where F t K det ,  





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Example 03

• Calculate the deflection uA of the structural model shown Section AA 0.5 cm2 _each 0.1cm 0.1cm U1 U2 U3 U4 U5 U6 U7= u_A U8 Y Bar with x-sectional area = 1cm2 6 cm 6 cm 8 cm E= 30 x 106 _N/cm2 3 . 0   Z A A

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Example 03

By symmetry and boundary conditions, we only need to

evaluate the stiffness coefficient

corresponding to u_A We know that U1 U2 U3 U4 U5 U6 U7= u_A U8 Y Bar with x-sectional area = 1cm2 6 cm 6 cm E= 30 x 106 _N/cm2 3 . 0   Z A A                      s y s x r y r x J

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Example 03

So, we have Now, calculating B        3 0 0 4 J               ) 1 ( 4 ... 0 ... ) 1 ( 3 48 1 r s B

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Example 03

Stiffness K for an Area is,

The stiffness of the truss is AE/L, or





cm N K ds dr r s s r s E K ds dr J t EB B K T / 34 . 1336996 ) 12 )( 1 . 0 ( ) 1 )( 1 ( 2 ) 1 ( 3 ) 1 ( 3 ) 1 ( 4 0 ) 1 ( 3 1 48 1 det 2 1 1 2 1 1                            





 

    _   cm N X k 3750000 / 8 ) 10 30 )( 1 ( 6 _ 

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Example 03

Hence, K_total = 6.424 x 106 _N/cm Now, since P = Ku Therefore, u = P/K cm X X u ₆ 9.34 10 4 10 424 . 6 6000 _  

cm

X

u



9

.

34

10

4

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Shell Element

• A Shell element is used to model shell,

membrane, and plate behavior in planar and three-dimensional structures

• The membrane behavior uses an isoparametric formulation that includes translational in-plane stiffness components and a rotational stiffness component in the direction normal to the plane of the element.

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Shell Element

Axis 1 Axis 2 Axis 3 J1 J2 J3 J4 Face 1 Face 2 Face 3 Face 4 Face5 Bottom Face6 Top

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Shell Elements

• A simple quadrilateral Shell Element

• Two dimensional plate bending and membrane elements are combined to form a four-node shell element + = x u y u z  x y z x  y  z  ux y u z  x  y  z u

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Shell Elements

• A simple quadrilateral Shell Element

• A thin-plate (Kirchhoff) formulation is normally used that neglects transverse shearing

deformation

• A thick plate (Mindlin/Reissner) formulation can also be chosen which includes the effects of

What are

The Finite Elements

(in SAP2000)

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Nodes and Finite Elements

• The Finite Elements are discretized

representation of the continuous structure • Generally they correspond to the physical

structural components but sometimes dummy or idealized elements my also be used

• Elements behavior is completely defined within its boundaries and is not directly related to other

elements

• Nodes are imaginary points used describe arbitrary quantities and serve to provide connectivity across element boundaries

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Basic Categories of Finite Elements

• 1 D Elements (Beam type)

– Only one dimension is actually modeled as a line, other

two dimensions are represented by stiffness properties

– Can be used in 1D, 2D and 2D

• 2 D Elements (Plate type)

– Only two dimensions are actually modeled as a

surface, third dimension is represented by stiffness properties

– Can be used in 2D and 3D Model

• 3 D Elements (Brick type)

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Basic Properties of Joints

• All elements are connected to the structure at the joints • The structure is supported at the joints using Restraints

and/or Springs

• Rigid-body behavior and symmetry conditions can be specified using Constraints that apply to the joints

• Concentrated loads may be applied at the joints

• Lumped masses and rotational inertia may be placed at the joints

• Loads and masses applied to the elements are transferred to the joints

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Joint Local Coordinates

• By default, the joint local 1-2-3 coordinate system is identical to the global X-Y-Z coordinate system

• It may be necessary to use different local coordinate systems at some or all joints in the following cases:

– Skewed Restraints (supports) are present

– Constraints are used to impose rotational symmetry

– Constraints are used to impose symmetry about a plane

that is not parallel to a global coordinate plane

– The principal axes for the joint mass (translational or

rotational) are not aligned with the global axes

– Joint displacement and force output is desired in

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Spring Restraints on Joints

• Any of the six degrees of freedom at any of the

joints in the structure can have translation or

rotational spring support conditions.

• Springs elastically connect the joint to the ground.

• The spring forces that act on a joint are related to

the displacements of that joint by a 6x6

symmetric matrix of spring stiffness coefficients.

– Simple Springs – Coupled Springs

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Simple Spring Restraints

• Independent spring stiffness in each DOF

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Coupled Spring Restraints

• General Spring Connection • Global and skewed springs • Coupled 6x6 user-defined

spring stiffness option (for foundation modeling)

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Stiffness Matrix for Spring Element

where u1 ,u2 ,u3 ,r1 ,r2 and r3 are the joint displacements and rotations, and the terms u1, u1u2, u2, ... are the specified spring stiffness

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Some Sample Finite Elements

Truss and Beam Elemen ts (1D,2D,3D)

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