Finite Element Analysis

7/17/2019 Finite Element Analysis

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Asian Center for Engineering Computations and Software

Asian Institute of Technology, Thailand

IW-CAAD 2004

Understanding and Using

Finite Element Analysis

July 19-21, 2004

Moratuwa, Sri Lanka



Understanding and Using

Finite Element Analysis

Buddhi S. Shrama



F i n i t e E l e m e n t A n a l y

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ACECOMS, AIT

The Objective

• To understand the fundamentals of theFinite Element Method and the Finite Element

Analysis

• To apply the Finite Element Analysis Tools forModeling 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

VibrationsSettlements

Thermal Changes

RESPONSES

Displacements

StrainsStresses

Stress Resultants

STRUCTURE

pv




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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

Design




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Analysis of Structures

pv

xx yy zz

vx x y z p 0

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

VibrationsSettlements

Thermal Changes

RESPONSES Displacements

StrainsStress

Stress Resultants

STRUCTURE

pv




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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 procedu re for solvin g (part ial)

dif ferent ial equat ion s associated with f ield

prob lems, with an accuracy acceptable to

engineers”




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From Classical to FEM

xx yy zzvx 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 = Stiff ness

r = Response

R = Loads

FEM

Structural Model




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Simplified Structural System

Loads (F) Deformations (u)

Fv

F = K u

F

K (Stiffness)

u

Equilibrium Equation



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The Total Structural System

EXCITATION RESPONSES

STRUCTURE

pv

• Static

• Dynamic

• Elastic

• Inelastic

• Linear

• Nonlinear

Eight types of equilibrium equations are possible!



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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 F t Kut uC t u M

)()()()()( t F t F t Kut uC t u M NL

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

Dynamic Inelastic Nonlinear Nonlinear-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

– Steady State Dynamic 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

Engineer



The Fundamentals

In Finite Element Method




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From Continuum to Structure

From Structure To Structural Model




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Solid – Structure - Model

(Governed by partial

differential equations)

Simplification (geometric)

Discretization 3D SOLIDS

CONTINUOUS MODEL

OF STRUCTURE

(Governed by either

partial or total dif-

ferential equations)

DISCRETE MODEL

OF STRUCTURE

(Governed by algebraic

equations)

3D-CONTINUM

MODEL



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xx

yy zz

vx x y z p 0

t vt

st

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 = Stiff ness 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




ACECOMS, AIT

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

– Mixed 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

element, etc.




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Structural Members

Dimensional Hierarchy of Structural Members

Continuum

Regular Solid

(3D)

Beam (1D)

b h

L>>(b,h)

b

ht

z

Plate/Shell (2D)

x z

t<<(x,z)

xz

y

xL




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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

– S (x, y, z, t)




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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

G G C i S




ACECOMS, AIT

The General Global Coordinate System

• The global coordinate system is a three-

dimensional, right-handed, rectangular coordinatesystem.

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

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

• All other coordinates systems are converted ormapped back and forth to General Coordinate

System

P l C di S




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Polar Coordinate Systems

• Polar coordinates include

– Cylindrical CR-CA-CZcoordinates

– Spherical SB-SA-SR

coordinates.

• Polar coordinate systemsare always defined with

respect to a rectangular X-

Y-Z system.

L l C di t S t




ACECOMS, AIT

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, andconstraint to constraint

L l A i d N t l A i




ACECOMS, AIT

Local Axis and Natural Axis

• The elements and

variation of fields canoften be described bestin terms “NaturalCoordinates”

• Natural coordinates maybe linear or curvilinear

• Shape functions can areused to associate thelocal system and natural

system




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Primary Relationships

Th B i St t l Q titi




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The Basic Structural Quantities

• Loads

• Actions

• Deformations

• Strains

• Stresses• Stress Resultants

The main focus ofStructural Mechanics is to

develop relationships

between these quantities

The main focus of FEM issolve these relationships

numerically

Mechanics Relationships




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Mechanics Relationships

Load

Action Deformation

StrainStressStress Resultant




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• 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

mathematically, numerically and by testing

Action

Deformation Relationship



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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

Simplified Examples of Action

Deformation



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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 Lv 326

3

V L

M

EI

L 2

2

2

Deformation

Strain Relationship



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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

• Total Strain is summation of strains from differentdeformations

Strain

–

Stress Relationship



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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

vvv

vvv

vvv

vv

E

2

2100000

02

210000

002

21000

0001

0001

0001

211

x xx E

kf c

f y

xy xy G

The Stress Strain Components



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The Stress Strain Components

• The Hook's law is

simplified form ofStress-Strain

relationship

• Ultimately the six

stress and strain

components can berepresented 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 correspondingstrains.

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 correspondingstrains.

Secondary Relationships



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Secondary Relationships

• Global Axis - Local Axis

– Geometric Transformations Matrices

• Local Axis - Natural Axis

– Shape Functions

– Jacobian Matrix

What are Shape Functions



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What are Shape Functions

• Shape Functions or Interpolation Functions provide a

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

• Shape Functions are used in FEM to relate the values ateNodes 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

One Dimensional Shape Functions



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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

1

332211

)(

)()()()(

i

ii w N sw

w s N w s N w s N sw

S=0 S =-1 S =+1

S=1 S=0 )1()(1 s s N

s s N )(1

S is the “Natural Coordinate System”

The Jacobian Matrix



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The Jacobian Matrix

• The Strain is Derivative of Displacement

• Displacements are specified on nodes, in ElementLocal Axis

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

• Shape Functions relate Nodal Displacements withElement Displacements

• Jacobian Matrix relates the derivative of Nodal

Displacement, directly with Element Strains

i si w N J

w s

N

w s

N

w s

N

s

w

J

,

33

22

11



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The Concept of DOF

The Concept of DOF



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e Co cep o O

• 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 typeand level of structural model and the element

being considered

• Global and Local DOF have different meaning

and significance

The Basic Six DOF



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• Three Translations along the

reference axis

– Dx, Dy, Dz

• Three Rotations about the

reference axis

– Rx, Ry, Rz

The Seven Degrees of Freedom



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ACECOMS, AIT

g

• The General Beam

Element may have 7degrees of freedom

• The seventh degree

is Warping

• Warping is out-ofplane distortion of

the beam cross-

section

z

y

x

xu

y

u

z u

xr

yr

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

xu

y

u

z u

xr

yr

z r

z w

Each section on a beam

member can have seven

Degrees Of Freedom

(DOF) with respect to its

local axis.

Actions and DOF



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The Complete DOF Picture



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p

§ u z Axial deformation Axial strain Axial stress

§ u x Shear deformation Shear strain Shear stress

§ u y 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

§ w z

Warping Axial strain Axial stress

Global Structural DOF



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• Only 3 DOF are real ly n eeded 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

interpretation

Local DOF and Natural DOF



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ACECOMS, AIT

• 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

Matrices

Types of DOF in SAP2000



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• Active

– the displacement is computed during the analysis• Restrained

– the displacement is specified, and the correspondingreaction is computed during the analysis

• Constrained

– the displacement is determined from thedisplacements 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

the analysis

Constraints and Restraints



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ACECOMS, AIT

• 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 toor in place of normal connections

– Rigid Diaphragm, Master-Slave DOF

Body Constraints



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ACECOMS, AIT

• 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

stiffeners to Shell elements

Constraints in SAP2000



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• 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

– Symmetry and anti-symmetry conditions

Constraints in SAP2000



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• 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

– Rigid Beam: rigid for beam bending on an axis



The Concept of

Stiffness

What is Stiffness ?



F i n i t e E


ACECOMS, AIT

• In structural terms, stiffness

may be defined as“Resistance to Deformation”

• So for each type ofdeformation, there is acorresponding stiffness

• Stiffness can be consideredor evaluated at various levels

• Stiffness is also the“constant” in the Action-

Deformation Relationship

u

F K

F Ku

F u

For Linear Response

The Structure Stiffness



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Material Stiffness

Section Stiffness

Member Stiffness

Structure Stiffness

Material Stiffness

Cross-section Geometry

Member Geometry

Structure Geometry

Stress/Strain

EA, EI

EA/L

Structure Stiffness



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• The overall resistance

of the structures to overall loads, called theGlobal StructureStiffness.

• Derived from the sumof stiffness of itsmembers, theirconnectivity and the

boundary or therestraining conditions.

Member and Element Stiffness



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• The resistance of each

Element to local actionscalled the Element StiffnessThis is derived from thecross-section stiffness andthe geometry of theElement.

• In FEM, the MemberStiffness can be derivedfrom stiffness of Elements

used to model the Member

Beam Element Cross

-

section Stiffness



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• 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 § r z

Torsional Constant, J § r x Moment of Inertia, I xx § r y Moment of Inertia, I yy § w z Warping Constant, W zz or C w

Computing Element Stiffness



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• Assume Nodal Displacements (Deformations)

• Determine Deformations within the element using“Shape Functions”

• Determine the Strains within the element usingStrain-Displacement Relationship

• Determine Stress within the element usingStress-Strain Relationship

• Use the principle of Virtual Work and integrate theproduct of stress and strain over the volume ofthe element to obtain the Stiffness

Deriving the Basic Stiffness Equation



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Internal Work

W I W E ..

F W E .

V

T dvW I .

D

V

T T

V

T T

V

T

dv B D BW I

dv B D BW I

dv DW I

.

.

.

V

dvW 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.

Stiffness Equation: An Example



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F

L

E

1

L

EA K

AL L

E K

dv L

E K

dv

L

E

L

K

V

V

2

2

11

L B

E D

1

B

D

V

T dv B D B K

L

EA

The Matrices in FEM



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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-MatrixStrain-Deforrmation

D-Matrix

Stress-Strain

What is Stiffness Matrix



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• 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

Element Stiffness Matrix




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ACECOMS, AIT

R 1

K 11

K 12

K 13

K 14

K 15

K 16

r1

R 2 K 21 K 22 K 23 K 24 K 25 K 26 r2

R 3

K 31

K 32

K 33

K 34

K 35

K 36

r3

R 4

K 41

K 42

K 43

K 44

K 45

K 46

r4

R 5

K 51

K 52

K 53

K 54

K 55

K 56

r5

R 6

K 61

K 62

K 63

K 64

K 65

K 66

r6

=

Node1 Node2

r1

r2r3

r4

r5 r6

A 2D Frame Element Stiffness

U2




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Node1 Node2

U1

U2U3

U1

U2U3

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/L3 6EI/L2 (U2)1

(P3)1 0 6EI/L2 4EI/L 0 -6EI/L2 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/L3 -6EI/L2 (U2)2

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

( U1)1 (U2)1 (U3)1 (U1)2 (U2)2 (U3)3

=

Direct Stiffness Method and FEM




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ACECOMS, AIT

• Basically there is no conceptual difference

between DSM and FEM. DSM is a special caseof the general FEM

• Direct Stiffness Method (DSM)

– The terms of the element stiffness matrix are definedexplicitly and in close form (formulae)

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

• Finite Element Method

– The element stiffness matrix terms are computed bynumerical integration of the general stiffness

equation



Isoparametric Elements

Introduction




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• 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 ofgeometrical shapes

• Analysis is carried out on the simple discretized

shapes and then the result is integrated over the

actual problem domain to get the approximatenumerical solution

1D Isoparametric Shape




l y s i s

ACECOMS, AIT

• 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 = +1r = -1

11 r





l y s i s

ACECOMS, AIT

21 )1(21)1(

21 X r X r X

)1(2

1)1(

2

121 r hand r h

Transformation is given by:

are interpolation of

shape functions

2

1i

iiU hU

The bar global displacements are shown by:

h1 h2

Z

Y

X, U

x1

x2

U1 U2

r r = +1r = -1





l y s i s

ACECOMS, AIT

22

2

12

12

L X X

dr

dX

and

U U

dr

dU

dX dr

dr dU

Element Strains can be calculated by:

Where L is the length of the bar





l y s i s

ACECOMS, AIT

LU U 12

Therefore, we have

111

L

B

u B ˆ

So, Strain displacement transformation

matrix can be shown as:


Where




l y s i s

ACECOMS, AIT

dV EB B K v

T

Jdr L

AE K 11

1

11

1

2

The Stiffness Matrix is given by:

Where E is the Elasticity constant

Therefore, we have

Where,

• A = area of the bar

• J = Jacobianrelating an element

length in the global

coordinate system

to an element

length in thenatural coordinate

system

2

L J so

dr J dX


Therefore K is evaluated as




l y s i s

ACECOMS, AIT

11

11

L

AE K

21 )1(2

1)1(

2

1 X r X r X

2

1i

iiU hU

Therefore, K is evaluated as

Substituting the value of r from

And put in

2/

2/)21(

L

X X X r

To get

Example 01




ACECOMS, AIT

Derive

• Interpolation Matrix H• Strain Displacement

Interpolation Matrix B

• Jacobian Operator J

for the three-node element

as shown in figure

X, Ur = 0 r = +1r = -1

L/2L/2x1

1 3 2

Example 01




ACECOMS, AIT

Finding the interpolation

functions of the given

element r = -1 r = +1r = 0

+1

)1(2

1 r r

h

+1

r = -1 r = +1r = 0

2

3 1 r h

r = -1 r = +1r = 0

+1

)1(2

2 r r

h

Example 01




ACECOMS, AIT

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

321 hhh H

Example 01




ACECOMS, AIT

For Jacobian Operator

2

det;2

2

22

)2

)(1())(1(2

)1(2

1

1

12

11

332211

L J

L

J

L J

dr dx J

r L L

x x

L xr L xr

r xr

r x

xh xh xh x

2D Isoparametric Element




ACECOMS, AIT

• Linear and quadratic two-dimensional

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

and interpolation of the displacement field

2

1

3

4

4

1

2 3

5

1

23

4

65

7




F i n i t e

E l e m e n t A n a l y s i s

ACECOMS, AIT

• Shape functions Ni are

defined in localcoordinates

• The same shape

functions are used forinterpolations of

displacements of

coordinates

)1,1(,

iiii

iiii

y N y x N x

v N vu N u

;

;




F i n i t e


ACECOMS, AIT

• Shape functions for linear quadratic two-

dimensional isoparametric elements are shownhere

• Linear Elements 4-node: )1)(1(

4

1 Ni oo


Q d ti El t 8 d



F i n i t e


ACECOMS, AIT

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(41)1)(1(

41 N

2

2

2

2i

i N

i N

i

oi

oi

o

ooo

ioio ;

Example 02

D i th i



F i n i t e


ACECOMS, AIT

• Derive the expressions

needed for the calculationof 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

sor

y4

x4 x, u

Example 02

Th f i t l ti f ti f th li



F i n i t e


ACECOMS, AIT

• The four interpolation functions for the linear

quadratic isoparametric element are

y, v

1

2

3

4

r or

sor

y4

x4 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

sr

sr

sr

sr

Example 02

The coordinate interpolations for the element is given by



F i n i t e


ACECOMS, AIT

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

ii

i

ii yh y xh x

4321

4321

)1)(1(4

1)1)(1(

4

1)1)(1(

4

1)1)(1(

4

1y

)1)(1(41)1)(1(

41)1)(1(

41)1)(1(

41x

y sr y sr y sr y sr

x sr x sr x sr x sr

Example 02

The displacement interpolations for the element is given



F i n i t e


ACECOMS, AIT

The displacement interpolations for the element is given

by

Using the interpolation functions, the coordinate

interpolations for this element are

4

11

4

1

; ii

i

ii vhvuhu

4321

4321

)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

1u

v sr v sr v sr v sr

u sr u sr u sr u sr

Example 02

The element strains are given by



F i n i t e


ACECOMS, AIT

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

y

x

s

y

s

x

r

y

r

x

s

r

Example 02



F i n i t e


ACECOMS, AIT

where

4321

4321

4321

4321

)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

yr yr yr yr

y s y s y s y s

x s xr xr xr

x s x s x s x s

Example 02

For any value of r and s



F i n i t e


ACECOMS, AIT

For any value of r and s

We can form the Jacobian matrix. Assuming we evaluate

J at

1111 sand r

ji s sand r r at

s

r J

y

x

1

ji s sand r r

Example 02



F i n i t e


ACECOMS, AIT

To evaluate the element strains, we use

4321

4321

4321

4321

)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

vr vr vr vr

v sv sv sv s

u sur ur ur

u su su su s

Example 02

Simplifying the above relations we get



F i n i t e


ACECOMS, AIT

Simplifying the above relations, we get

Where

u sr r r

s s s s J

and

u sr r r

s s s s J

ˆ)1(0)1(01010

10)1(0)1(010

4

1

y

v

x

v

ˆ0)1(0)1(0101

010)1(0)1(01

4

1

y

u

xu

1

1

ji

T

s sand r r where

vuvuvuvuu

44332211ˆ

Example 02

Strain-displacement transformation is given by



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Strain displacement transformation is given by

So, we can get

u Bijij ˆ

sr sr sr sr

r r r r s s s s

Bij

1)1()1()1()1(111

)1(0)1(01010010)1(0)1(01

4

1

Example 02

Stiffness Matrix K is given by



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

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 samplingpoint (r,s)

ijij

T

ijij

ji

ijijij

J CB B F where

F t K

det

,

Example 03

• Calculate the



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Calculate the

deflection uAof 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 =

1cm26 cm

6 cm

8 cm

E= 30 x 106 N/cm2

3.0

Z

A

A

Example 03

By symmetry and boundary

diti l d tZ



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

conditions, we only need to

evaluate the stiffness coefficientcorresponding to u A

We know that

U1

U2

U3

U4

U5

U6

U7= u A

U8

Y

Bar wi th x-

sectional

area =

1cm26 cm

6 cm

8 cm

E= 30 x 106 N/cm

3.0

Z

A

A

s

y

s

x

r y

r x

J

Example 03

So, we have



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Now, calculating B

3004 J

)1(4

...0...

)1(3

48

1

r

s

B

Example 03

Stiffness K for an Area is,



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

The stiffness of the truss is AE/L, or

cm N K

dsdr

r

s

s

r s E

K

dsdr J t EB B K T

/34.1336996

)12)(1.0(

)1)(1(2

)1(3

)1(3

)1(40)1(3148

1

det

2

1

1

21

1

cm N X

k /37500008

)1030)(1( 6

Example 03

Hence,

K = 6 424 x 106 N/cm



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Ktotal = 6.424 x 106 N/cm

Now, since P = Ku

Therefore, u = P/K

cm X X

u 4

61034.9

10424.66000

cm X u 41034.9

Shell Element

• A Shell element is used to model shell,



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

membrane, and plate behavior in planar andthree-dimensional structures

• The membrane behavior uses an isoparametricformulation that includes translational in-plane

stiffness components and a rotational stiffnesscomponent in the direction normal to the plane ofthe element.

Shell Element

Axis 3



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Axis 1 Axis 2

J1

J2

J3

J4

Face 1

Face 2

Face 3

Face 4

Face5 Bottom

Face6 Top

Shell Elements

• A simple quadrilateral Shell Element



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

• Two dimensional plate bending and membraneelements are combined to form a four-node shell

element

+ =

xu

yu

z

xy

z

x

y

z xu

yu

z

x

y

z u

Plate Bending Element Membrane Element Shell Element

Shell Elements

• A simple quadrilateral Shell Element



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

• 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

transverse shearing deformation



What are

The Finite Elements(in SAP2000)

Nodes and Finite Elements

• The Finite Elements are discretizedt ti f th ti t t



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

representation of the continuous structure

• Generally they correspond to the physicalstructural components but sometimes dummy oridealized elements my also be used

• Elements behavior is completely defined within its

boundaries and is not directly related to otherelements

• Nodes are imaginary points used describearbitrary quantities and serve to provideconnectivity across element boundaries

Basic Categories of Finite Elements

• 1 D Elements (Beam type)

O l di i i t ll d l d li th



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

– 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 stiffnessproperties

– Can be used in 2D and 3D Model

• 3 D Elements (Brick type)

– All three dimensions are modeled as a solid

– Can be used in 3D Model

The Joint or Node



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

The Joint or Node

Basic Properties of Joints

• All elements are connected to the structure at the joints

Th t t i t d t th j i t i R t i t



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

• 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 atthe joints

• Loads and masses applied to the elements are transferred

to the joints

• Joints are the primary locations in the structure at which

the displacements are known (the supports) or are to bedetermined

Joint Local Coordinates

• By default, the joint local 1-2-3 coordinate system is

identical to the global X-Y-Z coordinate system



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

identical to the global X-Y-Z coordinate system

• It may be necessary to use different local coordinatesystems 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

another coordinate system

Joint Local Coordinates



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Spring Restraints on Joints

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



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

joints in the structure can have translation orrotational 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 6x6symmetric matrix of spring stiffness coefficients.

– Simple Springs

– Coupled Springs

Simple Spring Restraints

• Independent spring

stiffness in each DOF



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

stiffness in each DOF

Coupled Spring Restraints

• General Spring Connection

•

Global and skewed springs



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Global and skewed springs

• Coupled 6x6 user -defined

spring stiffness option (for

foundation modeling)

Stiffness Matrix for Spring Element



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

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

coefficients.

Some Sample Finite Elements



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

Truss and Beam Elements (1D,2D,3D)

Plane Stress, Plane Strain, Axisymmetric, Plate and Shell Elements (2D,3D)

Brick Elements

One Dimensional Elements



F i n i t e

E l e m e n t A n

a l y s i s

ACECOMS, AIT

DOF for 1D Elements

Dy

DxDz

DyDy

Rz



F i n i t e


ACECOMS, AIT

Dx

Dx

Dx

Dy

Rz

Dy

RxRz DxDz

Dy

Rx

Rz

Ry

2D Truss 2D Beam 3D Truss

2D Frame 2D Grid 3D Frame

Variation of 1D Elements

• Based on DOF

–

2D Truss

• Non

-

Linear Elements

–

NL Link




ACECOMS, AIT

– 3D Truss

– 2D Beam

– 3D Beam

– 2D Grid

• Based on Behavior – Thick Beam/ Thin Beam

– Liner/ Isoperimetric

– Gap Element

– Tension Only

– Compression Only

– Friction

– Cable – Damper

Usage of 1D Elements




ACECOMS, AIT

3D Frame

2D Grid

2D Frame

Nonlinear Link Element in SAP2000




ACECOMS, AIT

Two Dimensional Elements




ACECOMS, AIT

DOF for 2D Elements

DyDy

Ry ?

Dy

Ry ?




ACECOMS, AIT

Dx

Dy

RzRx

Dz Rx

Rz

Dx

Membrane Plate Shell

Membrane Element

General•

Total DOF per Node = 3 (or 2)




ACECOMS, AIT

• Total Displacements per Node = 2• Total Rotations per Node = 1 (or 0)

• Membranes are modeled for flat

surfaces

Application• For Modeling surface elements

carrying

in- plane loadsMembrane

U1

Node 1

R3U2

U1

Node 3

R3U2

U1

Node 4

R3

U2

U1

Node 2

U2

3 2

1

Variation of Membrane Elements

1 unit

Plain-Strain

Assumptions x




ACECOMS, AIT

1 unit

x1

x3

x2

3D Problem

2D Problem

x

Plane Stress ProblemPlane Strain Problem

Plate Element

General

• Total DOF per Node = 3




ACECOMS, AIT

• Total Displacements per Node = 1

• Total Rotations per Node = 2

• Plates are for flat surfaces

Application

• For Modeling surface

elements carrying

out of plane loads

R1

Node 1

U3R2

1

23

R1

Node 2

U3R2

R1

Node 3

U3R2

R1

Node 4

U3R2

Plate

Shell Element

General•

Total DOF per Node = 6 (or 5)




ACECOMS, AIT

• Total Displacements per Node = 3• Total Rotations per Node = 3

• Used for curved surfaces

Application• For Modeling surface elements

carrying general loads

1

23

U1, R1

Node 3

U3, R3

U2, R2

U1, R1

Node 1

U3, R3 U2, R2

U1, R1

Node 4

U3, R3

U2, R2

U1, R1

Node 2

U3, R3

U2, R2

Shell

Variations of Plate Elements

– Based on Behavior

–

2D Plane Stress

– Based on Number of Nodes

–

3 Node, 6 Node




ACECOMS, AIT

– 2D Plane Strain – Axisymetric Solid

– Plate

– Shell

– Based on Material Model

– Rubber

– Soil

– Laminates

– Isotropic/ Orthotropic

– 4 Node, 8 Node, (9 Node)

Shell Elements in SAP2000




ACECOMS, AIT

Shell Elements in SAP2000




ACECOMS, AIT

Local Cords for Shell Element

• Each Shell elementhas its own localcoordinate system




ACECOMS, AIT

used to defineMaterial properties,loads and output.

• The axes of this local

system are denoted 1,2 and 3. The first twoaxes lie in the plane ofthe element the thirdaxis is normal

Three Dimensional Elements




ACECOMS, AIT

DOF for 3D Elements

D

Dy




ACECOMS, AIT

DxDz

Solid/ Brick

Brick Element in SAP2000

• 8-Node Brick

• Bricks can be



F i n i t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

added by using

Text Generation in

V7. New version

8 will have

graphical interfacefor Bricks

Connecting Dissimilar Elements




n a l y s i s

ACECOMS, AIT

Connecting Different Types of Elements

Truss Frame Membrane Plate Shell Solid

TrussOK OK Dz OK OK OK



F i n i t

e E l e m e n t A

n a l y s i s

ACECOMS, AIT

FrameRx, Ry, Rz OK

Rx, Ry, Rz,

Dz

Rx ?

Dx, DyRx ? Rx, Ry, Rz

MembraneOK OK OK Dx, Dy OK OK

PlateRx, Rz OK Rx, Rz OK OK Rx, Rz

ShellRx, Ry, Rz OK

Rx, Ry, Rz,

DzDx, Dz OK Rx, Rz

SolidOK OK Dz Dx, Dz OK OK

0

Orphan Degrees Of Freedom:

1 2 3 4

Connecting Dissimilar Elements• When elements with different degree of freedom at

ends connect with each other, special measures

may need to be taken to provide proper connectivity

d di S ft C bilit



F i n i t

e E l e m e n t A

n a l y s i s

ACECOMS, AIT

depending on Software Capability

Beams to Plates Beam to Brick Plates to Brick

Connecting Dissimilar Elements

• When members with mesh of different size or

configuration need to be connected we may

have to:



F i n i t

e E l e m e n t A

n a l y s i s

ACECOMS, AIT

have to: – Use special connecting elements

– Use special Constraints

– Use mesh grading and subdivision

– Use in-compatible elements (Zipper Elements inETABS)

– Automatic “Node” detection and internal

meshing by the Software

Connecting Beams with Membrane



F i n i t

e E l e m e n t A

n a l y s i s

ACECOMS, AIT

Modeling Shear-Wallsusing Panels only

(No Moment continuity

with Beams and Columns unless

6 DOF Shell is used)

Modeling Shear-Walls using

Panels, Beams, Columns

(Full Moment continuity

with Beams and Columns is restored

by using additional beams)

Meshing Slabs and Walls



F i n i t

e E l e m e n t A

n a l y s i s

ACECOMS, AIT

In general the mesh in the slab

should match with mesh in the

wall to establish connection

Some software automatically

establishes connectivity by using

constraints or “Zipper” elements

“Zipper”



How to Apply Loads to

Finite Element Model

Loads To Design Actions

• Loads

L d C




n a l y s i s

ACECOMS, AIT

• Load Cases

• Load Combinations

• Design Envelopes

• Design Actions

Load Cases

• Load cases are defined by the user and used foranalysis purpose only

St ti L d C




n a l y s i s

ACECOMS, AIT

• Static Load Cases

– Dead Load

– Live Load

– Wind Load

• Earthquake Load Cases – Response Spectrum Load Cases

– Time History Load Cases

• Static Non-Linear Load Cases

Load Combinations

• The Load Combinations may be created by theprogram, user defined or a combination of both.

• Some Examples: [Created by the program]




n a l y s i s

ACECOMS, AIT

• Some Examples: [Created by the program] – 1.4ΣDL

– 1.4ΣDL + 1.7(ΣLL + ΣRLL)

– 0.75[1.4ΣDL + 1.7(ΣLL + ΣRLL) + 1.7WL]

– 0.75[1.4ΣDL + 1.7(ΣLL + ΣRLL) - 1.7WL]

– 0.9ΣDL + 1.3WL – 0.9ΣDL - 1.3WL

– 1.1 [1.2ΣDL + 0.5(ΣLL + ΣRLL) + 1.0E]

– 1.1 [1.2ΣDL + 0.5(ΣLL + ΣRLL) - 1.0E]

– 1.1 (0.9ΣDL + 1.0E)

– 1.1 (0.9ΣDL - 1.0E)

Applying Gravity Loads

• All gravity loads are basically “Volume Loads” generated

due to mass contained in a volume




n a l y s i s

ACECOMS, AIT

• Mechanism and path must be found to transfer these loads

to the “Supports” through a Medium

• All type of Gravity Loads can be represented as:

– Point Loads

– Line Loads

– Area Loads

– Volume Loads

Load Transfer Path

• The Load is transferred through a medium which may be:

– A Point

– A Line

An Area




n a l y s i s

ACECOMS, AIT

– An Area

– A Volume

– A system consisting of combination of several mediums

• The supports may be represented as:

– Point Supports

– Line Supports

– Area Supports

– Volume Supports

Graphic Object RepresentationObject

Point Load

Concentrated LoadNode

Point Support

Column SupportPoint

LoadGeometry

Medium

Support

Boundary




n a l y s i s

ACECOMS, AIT

Line

Area

Volume

Beam Load

Wall Load

Slab Load

Slab LoadWind Load

Seismic Load

Liquid Load

Beam / Truss

Connection Element

Spring Element

Plate Element

Shell ElementPanel/ Plane

Solid Element

Line Support

Wall Support

Beam Support

Soil Support

Soil Support

ETABS and SAP200 uses graphic object modeling concept

Load Transfer Path is difficult to Determine

• Complexity of Load Transfer

Mechanism depend on:

A

Vol.

Load




n a l y s i s

ACECOMS, AIT

– Complexity of Load

– Complexity of Medium

– Complexity of BoundaryPoint Line Area Volume

Line

Area

LineArea

Volume

Medium

Boundary

Load Transfer Path is difficult to Determine




n a l y s i s

ACECOMS, AIT

Transfer of a Point Load to Point Supports Through Various Mediums

Line AreaVolume

Simplified Load Transfer




n a l y s i s

ACECOMS, AIT

Transfer of Area Load

To Lines To Points To Lines and Points

Applying Wind Loads

• At least 3 basic Wind Load Cases should beconsidered

– Along X-Direction




n a l y s i s

ACECOMS, AIT

g – Along Y Direction

– Along Diagonal

• Each Basic Wind Load Case should be enteredseparately into load combinations twice, oncewith (+ve) and once with (-ve) sign

• Total of 6 Wind Load Cases should considered inCombinations, but only 3 Load Cases need to bedefined and analyzed

Applying Wind Loads

WxAt least 3 Basic Load

Case for Wind Load




n a l y s i s

ACECOMS, AIT

Wy

Wxy

should be considered

Diagonal wind load may

be critical for special

types and layouts of

buildings

Wind Load CombinationsComb1 Comb2 Comb3 Comb4 Comb5 Comb6

Wx +f -f 0 0 0 0

Wy 0 0 +f -f 0 0




n a l y s i s

ACECOMS, AIT

Wxy 0 0 0 0 +f -f

(f) Is the load factor specified for Wind in

the design codes

Six Additional Load Combinations are

required where ever “Wind” is

mentioned in the basic Load

Combinations

Example:

Comb = 0.75(1.4D + 1.7W) will need Six

Actual Combinations

Comb1= 0.75(1.4D + 1.7Wx)

Comb2 = 0.75(1.4D - 1.7Wx)

Comb3 = 0.75(1.4D + 1.7Wy)

Comb4 = 0.75(1.4D - 1.7Wy)

Comb5 = 0.75(1.4D + 1.7Wxy)

Comb6 = 0.75(1.4D - 1.7Wxy)

Nature of Dynamic Loads

• Free Vibration

• Forced Vibration

• Random Vibration




n a l y s i s

ACECOMS, AIT

• Seismic Excitation

• Response Spectrum

• Time History

• Steady-State Harmonic Load• Impact

• Blast



Getting and Interpreting

Finite Element Results

What Results Can We Get ?

(in SAP2000)




n a l y s i s

ACECOMS, AIT

At Joints

• Joint Displacements

• Spring Reactions

• Restrained Reactions




n a l y s i s

ACECOMS, AIT

• Constrained Forces

• Results Available For:

– For all Available DOF

– Given on the “Local Joint Coordinates” – Given for all Load Case, Mode

Shapes,Response Spectrums, Time Histories,Moving Loads, and Load Combinations

For Frame Elements

• The Actions Corresponding to Six DOF at Both

Ends, in Local Coordinate System

12 12




n a l y s i s

ACECOMS, AIT

1

3

3

2

+P+V2

+V3

+V3

+V2+P

3

3

2

+T+M2

+M3

+M3

+M2+T

For Shell Element

• The Shell element internal forces (also called stress

resultants) are the forces and moments that result from

integrating the stresses over the element thickness.• The results include the “Membrane Results” (in plane



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

• The results include the Membrane Results (in-plane

forces) and “Plate Bending Results”

• The results are given for Element Local Axis

• It is very important to note that these stress resultantsare forces and moments per unit of in-plane lengt h

Shell Stress Resultants



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

Membrane Results



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

Plate Bending Results



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

Obtaining Design Actions From Basic Results



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

Obtaining Envelop Results

Comb1 Comb2 Comb3 Comb N

Load Case -1

L d C 2



F i n i

t e E l e m e n t A

n a l y s i s

ACECOMS, AIT

TotalMax, PMin, P

Load Case - 2

Load Case - 3

Load Case - M

Envelop Results

P1 P2 P3 P N

Can Envelop Results be Used for Design ?

P• Actions Interact with each other, effecting the

stresses

• For Column Design: P, Mx, My




n a l y s i s

ACECOMS, AIT

Mx

My

• For Beam Design: Mx, Vy, Tz

• For Slabs: Mx, My, Mxy

– At least 3 Actions from each combination must be

considered together as set

• Therefore, Envelop Results Can Not be Used

• Every Load Combinations must be used for

design with complete “Action Set”

Design Actions For Static Loads

• For static loads, Design

Actions are obtained as

the cumulative result

from each load

bi ti t f

Combinations

a d C

a s e s




n a l y s i s

ACECOMS, AIT

combination, as set for

all interacting actions

• The final or criticalresults from design of

all load combinations

are adopted

L o a

Design Actions

Obtained as set

from all

Combinations

Static, Dynamic and Nonlinear ResultsFor a Single Action:

Static Load Case

1

+




n a l y s i s

ACECOMS, AIT

Response Spectrum Load Case

Time History Load Case

Static Non-linear Load Case

+

-

1 for each Time Step

OR 1 for envelop

1 for each Load Step

Load

Combination

Table

OR 1 for Envelop

Response Spectrum Case – All response spectrum cases are assumed to be

earthquake load cases

– The output from a response spectrum is all positive.




n a l y s i s

ACECOMS, AIT

– Design load combination that includes a responsespectrum load case is checked for all possiblecombinations of signs (+, -) on the response spectrum

values

– A 3D element will have eight possible combinations ofP, M2 and M3 and eight combinations for M3, V, T

Response Spectrum Results for Action SetDesign Actions needed for Columns:

+P, +Mx, +My




n a l y s i s

ACECOMS, AIT

+P, +Mx, -My

+P, -Mx, +My

+P, -Mx, -My

-P, +Mx, +My-P, +Mx, -My

-P, -Mx, +My

-P, -Mx, -My

Maximum Results obtained by:

SRSS, CQC, etc.

P, Mx, My>

L o a d C o m b

i n a t i o n T a b l e

Time History Analysis Results

Max Val

Option

–

2:

Design For All Values

(At each tim e step)




n a l y s i s

ACECOMS, AIT

Min Val

Response Curve for One Action

Option – 1:

Envelope DesignT (sec)

Time

-

History Results – The default design load combinations do not include any time

history results

– Define the load combination, to include time history forces in a

design load combination




n a l y s i s

ACECOMS, AIT

– Can perform design for each step of Time History or design forenvelops for those results

– For envelope design, the design is for the maximum of eachresponse quantity (axial load, moment, etc.) as if they occurredsimultaneously.

– Designing for each step of a time history gives correctcorrespondence between different response quantities

Time History Results

– The program gets a maximum and a minimum value foreach response quantity from the envelope results for a timehistory

– For a design load combination any load combination that




n a l y s i s

ACECOMS, AIT

g yincludes a time history load case in it is checked for allpossible combinations of maximum and minimum timehistory design values.

– If a single design load combination has more than one timehistory case in it, that design load combination is designedfor the envelopes of the time histories, regardless of what isspecified for the Time History Design item in the preferences.

s

Static Non Linear Results

– The default design load combinations do not

include any Static Nonlinear results

– Define the load combination, to include Static




n a l y s i s

ACECOMS, AIT

,

Nonlinear Results in a design load combination

– For a single static nonlinear load case the design isperformed for each step of the static nonlinear

analysis.

s

Obtaining Reinforcement From Actions




n a l y s i s

ACECOMS, AIT

s

Computing Rebars For Beam Elements• For Beam type elements (1D

elements) design actions like Axial

force, moments, and shear force are

output directly.•

These actions can be used directly for

design purposes

y

z

x

M z

M y

T x

N x

V y

V z




n a l y s i s

ACECOMS, AIT

design purposes

• Generally, design is carried out in two

parts

• Axial- Flexural: P, Mx, My

• Shear Torsion: T, Vx, Vy

• Beam Design: Mx, Vy, T

• Column Design: Mx, My, P

y

z

x

M z

M y

N x

Biaxial & Load

3D Beam Column

s

Computing Rebars For Beam Elements A sc + Al /4: To resist compression due to

moment Mx (doubly reinforced beams) and

tension due to Torsion

A st : To resist

tension due to My

A sc : To resist compressiondue to My (may not be needed)




ACECOMS, AIT

A st + Al /4 : To resist main tension due tomoment and tension due to Torsion

A sw + Al /4 : To resist secondary tension

in deep beams due to moment and due to

Torsion

A svt + A sv /2: To resist shear due to

Torsion. Must be closed hoops on sides of

the section

s

Computing Rebars For Plate Elements• Moment output for plate type elements in Finite Element

Analysis is reported in moment per unit width along the

local axis of the plate element. These need to be

converted to moments along x and y for design purposes.

• The following procedure can be used:




ACECOMS, AIT

g p

The portion of a plate element bounded by a crack is

shown in the Adjoining figure. The moment about an axisparallel to the crack may be given as:

sindymkdymcoskdymdymdsm xy y xy xc

xy y xc kmmk mdx

dym 2

2

2

dx =k dy

dy

ds

Crack

s

Computing Rebars For Plate Elements• The plate needs to be reinforced

with bars in the x and y direction

• The corresponding moment

capacity at the assumed crack is

m xy kdy

my kdy

m xy dy

m x dy ms ds

mry kdy2




ACECOMS, AIT

• Where mrc must equal or exceed

mc solving for the minimum we

get

ryrx m ,m

Positive moment

capacities per unit width

mry kdy

mrx dy mrc ds

ryrxrc mk mdx

dym

2

2

xy yry mk

mm1

s

Computing Rebars For Plate Elements• The reinforcement at

the bottom of the slab in

each direction is

designed to provide

resistance for the

positive moment

xy xrx

xy yry

mmm

mmm

mry and mrx are set to zero if they

yield a negative value



F i n i t e E l e m e n t A n a l y s i

ACECOMS, AIT

p

• The reinforcement at

the top of the slab in

each direction isdesigned to provide

resistance for the

negative moment

xy xrx

xy yry

mmm

mmm

mry and mrx are set to zero if they

yield a positive value

s

Computing Rebars For Brick Elements• For Brick elements the FEA results in the nodal stresses

and strains.

• The stresses on the brick elements need to be

integrated along x and y direction to obtain forces.

•

Stress variation in both the directions may be

considered and integrated.



F i n i t e E l e m e n t A n a l y s i

ACECOMS, AIT

g

• These forces are then used to find the moment aboutthe two orthogonal axes and the net axial force. Similarapproach is used to obtain shear forces in two directions

• After the axial forces, moments and shear forces areobtained then the section can be designed as arectangular beam

i s

Computing Rebars For Brick ElementsSample Calculations for P and M

Following equations are based on the

assumption that there is no stress

variation in the transverse direction



F i n

i t e E l e m e n t A n a l y s i

ACECOMS, AIT

C1

C2TCL

x1

x2

x3

n

i

i

i

M M

.......Tx xC xC M

1

32211

n

i

i

i

P P

........T C C P

1

21

Modeling



Structures

Using FEM

i s

Global Modeling or ”Macro Model” • A model of the Whole Structure

• Objective is to get Overall StructuralResponse

• Results in the form of member forces andstress patterns



F i n


ACECOMS, AIT

• Global Modeling is same for nearly allMaterials

• Material distinction is made by using specificmaterial properties

• Global Model may be a simple 2D beam/frame model or a sophisticated full 3D finiteelement model

• Generally adequate for design of usualstructures

i s

Local Model or “Micro Model” • Model of Single Member or part of a Member

• Model of the Cross-section, Opening, Joints,

connection

• Objective: To determine local stress



F i n


ACECOMS, AIT

concentration, cross-section behavior,

modeling of cracking, bond, anchorage etc.

• Needs finite element modeling, often usingvery fine mesh, advance element features,

non-linear analysis

• Mostly suitable for research, simulation,

experiment verification and theoretical studies

i s

Global Modeling of Structural Geometry

(a) Real Structure



F i n


ACECOMS, AIT

(b) Solid Model (c) 3D Plate-Frame (d) 3D Frame

(e) 2D Frame

Fig. 1 Various Ways to Model a Real Struture

(f) Grid-Plate

i s

The Basic Issues• Which Model to be used ?

– 3D or 2D

– Frame or Grid

– Plate, Membrane, Shell, Solid

• Which Elements to be used ?



F i n

i t e E l e m e n t A n a l y s

ACECOMS, AIT

– Beam, Plate, Brick

– Size and number of elements

• Which Solution to be used ?

– Linear or Nonlinear

– Static or Dynamic

– Linear static or Nonlinear dynamic

– Linear dynamic or Nonlinear static

s i s

Overall Procedure

–

Linear Static

• Setup the Units to be used

• Define Basic Material Properties

• Define Cross-sections to be used• Draw, generate Nodes and Elements



F i n


ACECOMS, AIT

• Assign XSections, Restraints, Constraints etc.

• Apply Loads to Nodes and Elements

• Run the Analysis

• Check Basic Equilibrium and Deformations

• Interpret and use the Results

What Type of Analysis



Should be Carried out

s i s

The type of Analysis to be carr ied out

depends on the Structural System



F i n


ACECOMS, AIT

– The Type of Excitation (Loads)

– The Type Structure (Material and Geometry) – The Type Response

s i s

• P-Delta Analysis

• Buckling Analysis

• Static Pushover Analysis• Response Spectrum Analysis



F i n


ACECOMS, AIT

• Fast Non-Linear Analysis (FNA)

• Steady State Dynamic Analysis

• Free Vibration and Modal Analysis

• Large Displacement Analysis

s i s

• Static Excitation

– When the Excitation (Load) does not vary rapidly with Time

– When the Load can be assumed to be applied “Slowly”• Dynamic Excitation

Wh th E it ti i idl ith Ti



F i n


ACECOMS, AIT

– When the Excitation varies rapidly with Time

– When the “Inertial Force” becomes significant

• Most Real Excitation are Dynamic but are considered

“Quasi Static”

• Most Dynamic Excitation can be converted to

“Equivalent Static Loads”

s i s

Excitation/ Load Static Dynamic

Self Load Normal Operation At lifting/ placement

Superimposed Dead

Load

Normal Operation At placement

Live Load Normal Operation Depends on type



F i n


ACECOMS, AIT

Highway Traffic Quasi Static Impact

Water/ Liquid Normal Operation Filling, Sloshing

Creep, Shrinkage Static No DynamicComponent

Wind Equivalent Static Random Vibration

Seismic Excitation Equivalent Static Response Spectrum,

Time HistoryVibratory Machines Equivalent Static Impulse At Startup

Stead State at

s i s

• Elastic Material

– Follows the same path during loading and unloading and

returns to initial state of deformation, stress, strain etc. after

removal of load/ excitation

• Inelastic Material



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

– Does not follow the same path during loading and unloading and

may not returns to initial state of deformation, stress, strain etc.

after removal of load/ excitation

• Most materials exhibit both, elastic and inelastic behavior

depending upon level of loading.

y s i s

Creating Finite Element Models



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

y s i s

Model Creation Tools

• Defining Individual Nodes and Elements

• Using Graphical Modeling Tools

• Using Numerical Generation• Using Mathematical Generation

U i C d R li ti



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

• Using Copy and Replication

• Using Subdivision and Meshing

• Using Geometric Extrusions

• Using Parametric Structures

•

y s i s

Graphic Object Modeling

• Use basic Geometric Entities to create FE Models

• Simple Graphic Objects

– Point Object Represents Node

– Line Object Represents 1D Elements

– Area Object Represents 2D Elements



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

– Brick Object Represents 3D Elements

• Graphic Objects can be used to represent

geometry, boundary and loads• SAP2000, ETABS and SAFE use the concept of

Graphic Objects

y s i s

Modeling Objects and Finite Elements• Structural Members are representation of actual

structural components

• Finite Elements are discretized representation ofStructural Members

• The concept of Graphic Objects can be used tot b th th St t l M b ll



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

represent both, the Structural Members as well asFinite Elements

• In ETABS, the Graphic Objects representing theStructural Members are automatically divided intoFinite Elements for analysis and then back tostructural members for result interpretation

y s i s

Unstable Structures



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

y s i s

When is Structure Unstable in FEM Solution

• When the Global Stiffness Matrix is Singular

– The determinant of matrix is zero

– Any diagonal element in the matrix is zero• When the Global Stiffness Matrix is Ill-

Conditioned



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

Conditioned

– The numerical values in various matrix cells are of

grossly different order – Numerical values are either too small or too large

y s i s

Why are the FEM Models Unstable

• Restraint Instability

– Not enough Boundary Restraints

• Geometric Instability

– Not enough Elements

– Not enough stiffness of Elements



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

– Elements not connected properly

– Presence of Orphan Degrees Of Freedom

• Material Instability – Not enough Material Stiffness, (E, G)

– Not enough Cross-section Stiffness (A, I, J, ..)

y s i s

Structure Types

• Cable Structures• Cable Nets

• Cable Stayed

• Bar Structures• 2D/3D Trusses

• 2D/3D Frames, Grids



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

2D/3D Frames, Grids

• Surface Structures• Plate, Shell

• In-Plane, Plane Stress

• Solid Structures

How to Model the



Foundations

y s i s

Soil

-

Structure Interaction

• Simple Supports

•

Fix, Pin, Roller etc.

•

Support Settlement

• Elastic Supports

•

Spring to represent soil



F i n

i t e E l e m e n t

A n a l y

ACECOMS, AIT

Spring to represent soil

• Using Modulus of Sub-grade reaction

• Full Structure-Soil Model

• Use 2D plane stress elements

• Use 3D Solid Elements

y s i s

Modeling of Foundations and Mats

Beam Plate BrickConstraint

Yes Yes Yes

Modeling of Mat



F i n

i t e E l e m e n t

A n a l

ACECOMS, AIT

SpringYes Yes Yes

Brick No Yes Yes

S o i l

l y s i s

Computing Soil Spring

• A = Spacing of

Springs in X

•

B = Spacing of

Springs in Y

•

Ks = Modulus of

B



F i n

i t e E l e m e n t

A n a l

ACECOMS, AIT

sub-grade reaction

(t/cu m etc.)

• K = Spring constant

(t/m etc) A

K= ks*A*B

A

B

l y s i s

Raft as Beam

-

Grid, Soil as Spring

• The raft is represented as a

grillage of beams

representing slab strips in

both directions• The soil is represented by

spring

Thi h i



F i n i t e E l e m e n t

A n a l

ACECOMS, AIT

• This approach is

approximate and does not

consider the Mxy or thetorsional rigidity of the mat

l y s i s

Raft as Plate, Soil as Spring

• The raft is modeled using

Plate (or Shell) elements

• At least 9-16 elements

should be used in onepanel

• Soil springs may be




A n a l

ACECOMS, AIT

located or every node or

at alternate nodes

• Not suitable fro very thickrafts like thick pile caps

etc

l y s i s

Raft as Brick, Soil as Spring

• The raft is represented by

brick elements, soil as

springs

• More than one layer ofbrick elements should be

used along thickness

(usually 3 5) unless higher




A n a l

ACECOMS, AIT

(usually 3-5) unless higher

order elements are used

• Suitable for very thickmats and pile caps etc.

• Difficult to determine

rebars from brick results

l y s i s

Raft as Plate, Soil as Brick

• The raft is represented by

plate elements, soil as

bricks

• Soil around the mat shouldalso be modeled (min 2

times width)




A n a

ACECOMS, AIT

a l y s i s

Raft as Brick, Soil as Brick

• The raft is represented bybrick elements, soil as bricksalso

• More than one layer of brick

elements should be usedalong thickness (usually 3-5)unless higher order elementsare used




A n a

ACECOMS, AIT

• Soil around the mat shouldalso be modeled (min 2 times

width)• Suitable for very thick mats

and pile caps etc.

• Difficult to determine rebarsfrom brick results

a l y s i s

Modeling of Cellular Mats

• The top slab, the walls

and the bottom slab

should be modeled using

plate elements• More than one plate

element layer should be

used in the walls




A n a

ACECOMS, AIT

used in the walls

• The soil may be

represented by springs orby bricks

a l y s i s

Modeling of Piles

• For analysis and designof individual Pile, it canbe modeled as beam

element and soilaround it as series oflateral and verticalsprings




A n a

ACECOMS, AIT

springs

• For analysis of superstructure, entire pilecan be represented bya single a set of springs

a l y s i s

Using Nonlinear Springs to Model Soil

• The springs used to represent may be either

linear or non linear

• The non-linear response of the soil can be

obtained from actual tests

• The non-linear response can then be used to

determine “K” for various levels of load or




A n a

ACECOMS, AIT

determine K for various levels of load or

deformation

• Nonlinear springs are especially useful for

vertical as well as lateral response of piles

and pile groups

a l y s i s

Modeling of Shear Walls




ACECOMS, AIT

a l y s i s

Modeling of Planner Walls

Using Truss




ACECOMS, AIT

Using Beam and Column Using Panels, Plates and Beams

a l y s i s

Frame Model for Planer Walls

• Specially Suitable when H/B

is more than 5

• The shear wall is

represented by a column ofsection “B x t”

• The beam up to the edge of

the wall is modeled asB

H

t




ACECOMS, AIT

Rigid Zones

the wall is modeled as

normal beam

• The “column” is connected to

beam by rigid zones or very

large cross-section

B

a l y s i s

Using Plates to Model Walls

Multiple elements greater accuracy in determination of stress

distribution and allow easy modeling of openings




ACECOMS, AIT

Using Plate Elements only

(No Moment continuity

with Beams and Columns unless

6 DOF Shell is used)

Using Plate Elements with

Beams, Columns

(Full Moment continuity

with Beams and Columns)

a l y s i s

Truss Model for Planner Walls

• For the purpose of analysis, assumethe main truss layout based on wallwidth and floor levels

• Initial member sizes can beestimated as t x 2t for main axial

members and t x t for diagonalmembers

• Use frame elements to model thetruss. It is not necessary to usetruss elements

t x t



F i n i t e E l e m e n t A n

ACECOMS, AIT

truss elements

• Generally single diagonal is sufficient

for modeling but double diagonalmay be used for easier interpretationof results

• The floor beams and slabs can beconnected directly to truss elements

C

tB

t x 2t

n a l y s i s

Modeling of Cellular Shear Walls




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Uniaxial Biaxial

n a l y s i s

Modeling Walls With Openings




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Plate-Shell Model Rigid Frame Model Truss Model

Introduction



To Dynamic Analysis

n a l y s i s

What is Seismic Analysis

Determination of Structural Response due to Seismic

Excitation

• The Seismic Excitation is Dynamic in nature• So the Response is governed by



F i

n i t e E l e m e n t A n

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“The Dynamic Equilibrium Equation”

• The question is how to solve this equation?

n a l y s i s The General



F i


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Dynamic nalysis

n a l y s i s

•

Capture the Realistic Behavior of Structures

•

No Conservative Approximations in Analysis

•

Puts Check on Structural Irregularities

•

Identifies Ductility Demands

Why Dynamic Analysis

–

In General



F i


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• Lower Base Shears• Required by Code

n a l y s i s

P(u,a)

K

Static Elastic Only:

Displacement (U)=Force (P) /Stiffness(K)

U = P/K or K u = P

M

Basic Dynamic Equilibrium

–

No Damping



F i


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Inertia Only :

Acceleration (a)=Force (P) / Mass(M)

a = P/M or Ma = P

BOTH :

Ma+Ku=P

n a l y s i s

F

I

+ F

D

+ F

S

= F

F(t)

I

+ F(t)

D

+ F(t)

S

= F(t)

M a(t) + C v(t) + K u(t) = F (t)

F = External Force

FS = I nternal Forces

FD = Energy Dissipation Forces

FI = Inertial Force (t) = Varies with time

u’’ = Acceleration (a)

u’ = Velocity (v)

F

Basic Dynamic Equilibrium

–

With Damping



F i


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M u’’(t) +C u’(t) + K u(t) = F(t)

(Second order di ff erential equation for

li near structural behavior)

u = D isplacement

M = Mass

C = Damping

K = Stif fness

n a l y s i s

Basics of Structure Dynamics

• Idealization for a Single

Floor

– Mass less Column, Entiremass is concentrated on the

roof

– Rigid roof, Rigid ground

Roof

Column

Ground



F i


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g , g g

– Column is flexible in lateral

direction but rigid in vertical

direction

n a l y s i s

What is Dynamic Response ?

• If the roof is displaced laterally by a

distance uo and then released the

structure will oscillate around its

equilibrium position.

Roof

Column

Ground

One Cycle



F i


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uo

1 2

-uo

3 4

uo

5

n a l y s i s

Dynamic Response•

The oscillation will continue

forever with the same

amplitude u

o

and the structure

will never come to rest.

• Actual structure will oscillate

with decreasing amplitude

and will eventually come to

rest.

1

2

3

4

5

time

u

o

-

u

o

Displacement

Amplitude




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uo

1 2

-uo

3 4

uo

5

A n a l y s i s

Damped Dynamic Response

• To incorporate damping or

dying out of dynamic

response feature into the

idealized structure, an energy

absorbing element should be

introduced.

Mass m

Stiffness K

Damping C




ACECOMS, AIT

Idealized One storeyBuilding

• Viscous damper is the most

commonly used energyabsorbing element in the

dynamic modeling of

structures

A n a l y s i s

Displacement, Velocity, Acceleration

• Displacement Change in Location

•

Velocity Rate of Change of Displacement wrt Time

•

Acceleration Rate of Change of Velocity wrt Time• Time Period

The time taken to complete one cycle

• Frequency

The no. of cycles per second

u



F i n i t e E l e m e n

t A

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2

2

dt

ud uva

dt

duuv

u

A n a l y s i s

Free Vibration nalysis




t A

ACECOMS, AIT

A n a l y s i s

• Definition

–

Natural vibration of a structure released from initial

condition and subjected to no external load or

damping

•

Main governing equation

-

Eigen Value Problem

t t

t t

P u K ucu M

Free Vibration Analysis




t A

ACECOMS, AIT

• Solution gives – Natural Frequencies

– Associated mode shapes

– An insight into the dynamic behavior and response

of the structure

A n a l y s i s

Free Vibration

M u’’(t) +C u’(t) + K u(t) = F(t)

M u’’(t) + K u(t) = 0

Which leads to eigenvalue

problem

• No external force is applied

•

No damping of the system

= natural f requencies

= M ode shape

0d

0

2

2

2

MK

M K

M K

nn

nnn

w

w

A mode shape is set of

relative (not absolute)

nodal displacement for a




t A

ACECOMS, AIT

0det 2 M K nw

Solution of above equation yields a

polynomial of order n for w , which in

turn gives n mode shapes

nodal displacement for a

particular mode of freevibration for a specific

natural frequency

A n a l y s i s

Modal Analysis

• Determination of

natural frequencies

and mode shapes.

• No external load or

excitation is appliedto the structure.

• Obtained from

eigenvalue analysis.

Th




t A

ACECOMS, AIT

• There are as many

modes as there areDOF in the system

A n a l y s i s nalysis for

Ground Motion




t A

ACECOMS, AIT

A n a l y s i s

g

g

umumumum

umkuucum

mcm

k

ummg um F

F kuucum

22

2;

ww

w w

Basic Dynamic for Ground Motion




t A

ACECOMS, AIT

• The unknown is displacement and its

derivatives ( velocity, acceleration)

• Variables are ground acceleration, damping

ratio and circular frequency

g

g

uuuu

umumumum

22

2

w w

w w

A n a l y s i s

g uuuu 22 w w

Ground Motion Input and Displacement Output




t A

ACECOMS, AIT

A n a l y s i s

Response History Analysis

• Determination the totaldynamic response ofstructure as the sum ofresponse of all modeshapes using the groundacceleration at each timestep

0 15

+ Damping Ratio for each mode




t A

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-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 5 10 15 20 25 30 35

Time (Second)

Acceleration

(a/

g)

A n a l y s i s

Modal Displacements for Ground Motion




t A

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A n a l y s i s Response

Spectrum




t A

ACECOMS, AIT

nalysis

A n a l y s i s

What are Response Spectra

• For a ground acceleration at particular time, for a giventime period and damping ratio, a single value ofdisplacement, velocity and acceleration can be obtained

• Output of the above (u, v, a) equation are the dynamicresponse to the ground motion for a structure consideredas a single DOF

g uuuu 22 w w




t A

ACECOMS, AIT

as a single DOF

• A plot of the “maximum” response for different groundmotion history, different time period and damping ratio givethe “Spectrum of Response”

A n a l y s i s

Response Spectrum Generation




t A

ACECOMS, AIT

A n a l y s i s

Response Spectrum Generation




t

ACECOMS, AIT

A n a l y s i s

• Spectral Displacement Sd

•

Pseudo Spectral Velocity Sv

•

Pseudo Spectral Acceleration Sa

2

2

dt

ud uva

dt duuv

u

d va

d v

S S S

S S

2w w

w

Spectral Parameters



F

i n i t e E l e m e n

t

ACECOMS, AIT

A n a l y s i s

Spectra For Different Soils



F


t

ACECOMS, AIT

A n a l y s i s

How to Use Response Spectra

• For each mode of free vibration, corresponding Time

Period is obtained.

• For each Time Period and specified damping ratio, the

specified Response Spectrum is read to obtain the

corresponding Acceleration

• For each Spectral Acceleration, corresponding velocity

and displacements response for the particular degree

of freedom is obtained



F


t

ACECOMS, AIT

• The displacement response is then used to obtain thecorresponding stress resultants

• The stress resultants for each mode are then added

using some combination rule to obtain the final

response envelop

t A n a l y s i s

Modal combination Rules

• ABS SUM Rule• Add the absolute maximum value from

each mode. Not so popular and notused in practice

• SRSS• Square Root of Sum of Squares of the

peak response from each mode.Suitable for well separated naturalfrequencies

N

n

no r r 1

0

N

n

no r r 1

2

0



F


t

ACECOMS, AIT

frequencies.

• CQC• Complete Quadric Combination is

applicable to large range of structuralresponse and gives better results than

SRSS.

N

i

N

n

niino r r r 1 1

00

t A n a l y s i s

Response Spectrum Analysis

• Determination ofpeak response of thestructure based on adesign or specifiedresponse spectrumand the specifiedmode shapes

• Uses modal 1

1.2

1.4

eartion

0%



F


t

ACECOMS, AIT

combination rules todetermine total peakresponse from allmodes

0

0.2

0.4

0.60.8

1

0 1 2 3 4 5

Time Period (Sec)

SpectralAccele

2%

5%

Introduction to

Non-linear Analysis



t A n a l y s i s

Basic Sources of Non

-

Linearity

• Geometric Non-Linearity

• Material Non-Linearity

•

• Compound Non-Linearity P

P

M

f

M M



F


t

ACECOMS, AIT

• Large Displacementsdd

t A n a l y s i s

Geometric Non Linearity

• The deformations change the

basic relationships in the

stiffness evaluation

• Example: Axial LoadChanges Bending Stiffness

• The deformation produce

additional actions, not present



F


t

ACECOMS, AIT

at initial conditions• Example: Axial load causes

additional moments

t A n a l y s i s

Material Non

-

Linearity

• The basic material

“constants” (E, G, v)

etc. change with level of

strain• Example: Stress-Strain

curve is non-linear

• The cross-section

kf c

b

Kd



F


t

ACECOMS, AIT

properties change withlevel of strain

• Example: Cracking in

reinforced concrete

reduces A, I etc

d

Kd

yt

N.A

A s

t A n a l y s i s

Material Non

-

linearity

Semi-confined,High Strength Concrete

Moment

–

Curvature

curve generated for a

rectangular column

with circular core. The

outer portion is

modeled by stress

- strain curve for low

strength unconfined

concrete where as the

core is modeled by



F


ACECOMS, AIT

Rectangular Whitney Curve

lightly confinedconcrete. Observe the

drop in moment

capacity as the outer

concrete fails.

n t A n a l y s i s

Types of Non

-

Linearity

• Smooth , Continuous

–

Softening

–

Hardening

•

Discontinuous

• Snap

-

through

•

Bifurcation



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ACECOMS, AIT

• Elastic Buckling• In-Elastic Buckling

• P-Delta

n t A n a l y s i s

Non

-

linear Analysis in SAP2000

• The non-linear analysis is “Always” carried out

together with Time History Dynamic analysis

• Non-linear behavior can be modeled by:

– NL Link Element – For Dynamic – Nonlinear

• Elastic Stiffness for Linear Analysis

• Gap, Hook, Damper, Isolator for Nonlinear

– Hinge Element – For Static Pushover



F


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• Material Non linearity

• Load-Deflection Curves

Introduction to

Push-over Analysis



n t A n a l y s i s

Why Pushover Analysis

• Buildings do not respond as l inearly elastic systems

during strong ground shaking

• Improve Understanding of Building Behavior

- More accurate prediction of global displacement

- More realistic prediction of earthquake demand on individual

components and elements

- More reliable identification of “bad actors”

• Reduce Impact and Cost of Seismic Retrofit



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ACECOMS, AIT

- Less conservative acceptance criteria

- Less extensive construction

• Advance the State of the Practice

n t A n a l y s i s

Performance Based Design

-

Basics

• Design is based not on Ultimate Strength butrather on Expected Performance

– Basic Ultimate Strength does not tell us what will beperformance of the structure at Ultimate Capacity

• Performance Based Design Levels

– Fully Operational

– Operational



F


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– Life Safe

– Near Collapse

– Collapse

n t A n a l y s i s

Pushover Spectrum



F


ACECOMS, AIT

n t A n a l y s i s

Pushover Demand Curves



F


ACECOMS, AIT

n t A n a l y s i s

Earthquake Push on Building



F


ACECOMS, AIT

n t A n a l y s i s

The Pushover Curve



F


ACECOMS, AIT

n t A n a l y s i s

Pushover Capacity Curves



F

i n i t e E l e m e

ACECOMS, AIT

n t A n a l y s i s

Demand Vs Capacity



F

i n i t e E l e m e

ACECOMS, AIT

n t A n a l y s i s

Non

-

linearity Considered in Pushover

• Material nonlinearity at discrete, user-defined hinges in

frame/line elements.

1. Material nonlinearity in the link elements.

• Gap (compression only), hook (tension only), uniaxial plasticity

base isolators (biaxial plasticity and biaxial friction/pendulum)..

2. Geometric nonlinearity in all elements.

• Only P-delta effects

• P-delta effects plus large displacements



F

i n i t e E l e m e

ACECOMS, AIT

3. Staged (sequential) construction.• Members can be added or removed in a sequence of stages

during each analysis case.

n t A n a l y s i s

Important Considerations

• Nonlinear analysis takes time and patience

• Each nonlinear problem is different

• Start simple and build up gradually.

• Run linear static loads and modal analysis first

• Add hinges gradually beginning with the areas

where you expect the most non-linearity.

• Perform initial analyses without geometric non-



F

i n i t e E l e m e

ACECOMS, AIT

linearity. Add P-delta effects, and large

deformations, much later.

n t A n a l y s i s

Important Considerations

• Mathematically, static nonlinear analysis does not

always guarantee a unique solution.

• Small changes in properties or loading can cause

large changes in nonlinear response.• It is Important to consider many different loading

cases, and sensitivity studies on the effect of

varying the properties of the structure

N li l i t k ti d ti



F

i n i t e E l e m e

ACECOMS, AIT

• Nonlinear analysis takes time and patience.Don’t Rush it or Push to Hard

n t A n a l y s i s

Procedure for Static Pushover Analysis

1. Create a model just like for any other analysis.

2. Define the static load cases, if any, needed for use in thestatic nonlinear analysis (Define > Static Load Cases).

3. Define any other static and dynamic analysis cases that

may be needed for steel or concrete design of frameelements.

4. Define hinge properties, if any (Define > Frame NonlinearHinge Properties).

5. Assign hinge properties, if any, to frame/line elements

(A i F /Li F N li Hi )



F

i n i t e E l e m e

ACECOMS, AIT

(Assign > Frame/Line > Frame Nonlinear Hinges).6. Define nonlinear link properties, if any (Define > Link

Properties).

n t A n a l y s i s


7. Assign link properties, if any, to frame/line elements(Assign > Frame/Line > Link Properties).

8. Run the basic linear and dynamic analyses (Analyze >Run).

9. Perform concrete design/steel design so that reinforcingsteel/ section is determined for concrete/steel hinge ifproperties are based on default values to be computed bythe program.

10. For staged construction, define groups that represent thevarious completed stages of construction.



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i n i t e E l e m e

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11. Define the static nonlinear load cases (Define > StaticNonlinear/Pushover Cases).

n t A n a l y s i s


12.Run the static nonlinear analysis (Analyze > Run

Static Nonlinear Analysis).

13.Review the static nonlinear results (Display >

Show Static Pushover Curve), (Display > ShowDeformed Shape), (Display > Show Member

Forces/Stress Diagram), and (File > Print Tables

> Analysis Output).

14 P f d i h k th t tili t ti



F

i n i t e E l e m e

ACECOMS, AIT

14.Perform any design checks that utilize staticnonlinear cases.

15.Revise the model as necessary and repeat.

Finite Element Analysis

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