2nd session- dr.n.sivaprasad

Introduction to

Finite Element Analysis

by

Prof. N Siva Prasad

Department of Mechanical Engineering

Indian Institute of Technology Madras

2 Prof .N. Siva Prasad, Indian Institute of Technology Madras

Why Finite Element Modeling (FEM)?


Error

Why FEM? – Contd…

4

Error

Prof .N. Siva Prasad, Indian Institute of Technology Madras



Example

8

STRESSES AND EQULIBRIUM

Fig. 1 Three-dimensional body

A three-dimensional body occupying a volume V and having a surface S is

shown in Fig.1.

The deformation of a point is given by the three components of its

displacement: T=[ ]u,v,wu (1)

T( =[x,y,z] )x


9

The distributed force per unit volume, for example, the weight per unit

volume, is the vector f given by

T[ ]x yf , f , ff z (2)

The body force acting on the elemental volume dV is shown in Fig.1

The surface traction T may be given by

T]x y zT ,T ,TT =[ (3)

A load P acting at a point i is represented by its three components:

T

i i[ ]x y zP ,P ,PP (4)


10

The stresses acting on the elemental volume dV are shows Fig. 2.

Fig. 2 Equilibrium of elemental volume


11

The six independent components are

T

x y yz xz xy[ , , , , , ]z (5)

x y z, , yz xz xy, , , where are normal stresses and

are shear stresses. The equilibrium equations

0

0

0

xyx xzx

xy y yz

y

yzxz zz

fx y z

fx y z

fx y z

(6a)

(6b)

(6c)


12

STRAIN – DISPLACEMENT RELATIONS

T[ , , , , , ]x y z yz xz xy (7)

, ,x y z , ,yz xz

xy

where and are normal strains and and

are the engineering shear strains.


13

Fig. 3 Deformed elemental surface


14

The shear strain-displacement can be written as

xy

u v

y x

(8)

Considering the other faces y-z, and z-x,

T

, , , , ,u v w v w u w u v

x y z z y z x y x

(9)


15

yx zx

yx zy

yx zz

v vE E E

v vE E E

v vE E E

yz

yz

xzxz

xy

xy

G

G

G

(10)

Stress-strain relations

For linear elastic materials, the stress-strain relations come from the

generalized Hooke’s law. For isotropic materials, the two material

properties are Young’s modulus (or modulus of elasticity) E and

Poisson’s ratio . Considering an elemental cube inside the body,

Hooke’s law gives

v


The shear modulus (or modulus of rigidity), G is given by

2(1 )

EG

v

(11)

From Hooke’s law relationships (Eqn. 10), adding and LHS

(1 2 )( )x y z x y z

v

E

(12)

Substituting for and so on into Eq. 10, we get the inverse relations

D (13)

( )x z


17

D is the symmetric (6 X 6) material matrix given by

1 0 0 0

1 0 0 0

1 0 0 0

0 0 0 0.5 0 0(1 )(1 2 )

0 0 0 0 0.5 0

0 0 0 0 0 0.5

v v v

v v v

v v vE

vv v

v

v

D (14)


18

Fig. 4 (a) Plane stress

Plane Stress. A thin planar body subjected to in-plane loading on

its edge surface is said to be in plane stress. A ring press fitted on a

shaft, Fig. 4, is an example. Here stresses

are set as zero. The Hooke’s law relations (Eq. 10) then give us

, ,andz xz yz

2(1 )

( )

x

y

xy

z

yx vE E

yxvE E

vxyE

vx yE


19

(16)

Fig. 4(b) plane strain

1 0

1 01+ 1 2

0 0 0.5

v vx xE v vy yv v

vxy xy

(17)

D here is a (3x3) matrix.

Plane Strain.

If a long body of uniform cross section is subjected to transverse

loading along its length, are taken as zero. Stress

may not be zero in this case. The stress – strain relations can be

obtained

, ,andz xz yz

20

TEMPERATURE EFFECTS

The temperature strain is represented as an initial strain:

0

T

(18)


21

Strain Energy stored in the spring

U = ½ (force in the spring) (displacement )

= ½ (Su) u

= ½ s u2

Potential energy of the external load ‘p’

Wp = (load) (displacement from zero potential state)

= - p u

Total potential energy = total strain energy + Work potential

= U+ Wp

= ½ s u2 – p u

for minimum of , /u = 0

S u – p = 0

Su = p

Potential energy and equilibrium


P

u

P

22

Potential Energy,

The total potential energy of an elastic body, is defined as the

sum of total strain energy (U) and the work potential:

= Strain energy + Work potential

(U) (WP) (22)

For linear elastic materials, the strain energy per unit volume in the body

is . For an elastic body, the total strain energy is given by

12

V

1 T2

U dV

(23)

The work potential WP is given by


23

V S

T T TWP i ii

dV dS u f u T u P

(24)

The total potential for the general elastic body shown in Fig.1.1 is

T1 T T T2 i i

iV V S

dV dV dS u f u T u P (25)


24

Principle of Minimum Potential Energy

For conservative systems, of all the kinematically admissible

displacement field, those corresponding to equilibrium extremize the

total potential energy. If the extremum condition is a minimum, the

equilibrium state is stable.

Kinematically admissible displacements are those that satisfy the

single-valued nature of displacements and the boundary conditions.


25

Example 1

Fig. 5 shows a system of springs. The total potential energy is given by

Fig. 5 System of springs


Fixed

support

Fixed

support

F1

F3

26

2 2 2 2

1 2 3 4 1 1 3 31 2 3 41 1 1 12 2 2 2

k k k k F q F q (26)

where are extensions of the four springs. Since 1 2 3 4, , , and

1 1 2 2 2 3 3 2 4 3( ), , ( ),q q q q q and q substituting for i we can write as

22 2

1 1 2 2 3 3 2 4 1 1 3 32 3

21 1 1 12 2 2 2

k q q k q k q q k q F q F q

(27)

where q1,q2,andq3 are the displacements of nodes 1,2, and 3, respectively.


27

For equilibrium of this three degrees of freedom system, we need to

minimize with respect to q1, q2, and q3. The three equations are given

by

0 1,2,3iqi

(28)

which are

01 1 2 1

1

01 1 2 2 2 3 3 2

2

03 3 2 4 3 3

3

k q q Fq

k q q k q k q qq

k q q k q Fq

(29)


28

These equilibrium equations can be put in the form of matrix, Kq=F as

follows:

01 1 1 1

01 1 2 3 3 2

0 333 3 4

k k q F

k k k k k q

Fqk k k

(30)


29

One dimensional problems

In one dimensional problems, the stress, strain, displacement,

and loading depends only on the variable x

( )u xu ( )x ( )x ( )T xT ( )f xf

The stress-strain and strain-displacement relations are

E du

dx

The loading consists of three types

- body force f

-traction force T

-point load Pi

One dimensional problems


30

X

P2

P1

f

T

One dimensional bar loaded by traction, body and point loads

One dimensional problems contd..


31

Finite element modeling of the bar

.

.

X

.

X

1

4

3

2

1

2

5

3

4

.

.

.

.One dimensional problems contd..


.

32

Co-ordinates and shape functions

Consider a typical finite element in the local coordinate system (Fig a) we

define a natural co-ordinate system, denoted by

1

2 1

2( ) 1x x

x x

1

1 2

1

X1

X

X2

e 1 2

1

2

1( )

2

1( )

2

N

N

1 1

N1 N2

1 1 0 1 0


Fig.a Fig.b

e


33

Linear interpolation

e e

u1

u2 q1

q2

uunknown

ulinear

1 2 1 2

Linear displacement field within the element can be written in terms of the nodal

Displacement q1 and q2 as

1 1 2 2u N q N q

In matrix notation u Nq

The transformation from x to can be written in terms of N1 and N2 as

1 1 2 2x N x N x

Isoparametric formulation


1 2

1 2

,

,T

N N N

q q q

where


34

The general expression for the potential energy

1

2

T T T

i ii

Adx u fAdx u Tdx u P Since the continuum has been discretized into finite elements, the expression for

Potential energy becomes

1

2

T T T

i ie e e i

Adx u fAdx u Tdx QP The last term above assumes that point load Pi are applied at the nodes. This

assumption makes the present derivation simpler with respect to notation, and

is also a common modeling practice

T T

e i ie e e ie

U u fAdx u Tdx Q P where

1

2

T

eU Adx is the element strain energy



35

Element stiffness matrix

Consider the strain energy form 1

2

T

eU Adx

Substituting for and into the above yields EBq Bq

1[ ]

2

T T

e

e

U q B EBAdx q

In the finite element model, the cross sectional area of element e,denoted by A

Is constant. Also , B is a constant matrix, further the transformation from x to

yields

2eldx d

where 1 1 Le is the length of the element



36

The element strain energy Ue is now written as 1

1

1

2 2

T Tee e e

lU q A E B B d q

Where E is the young’s modulus of element and by using

1

1

2d

2

11 11 1

12

T

e e e e

e

U q A l E ql

Which results in

1 11

1 12

T e ee

e

A EU q q

l

The above equation is of the form

1

2

T e

eU q k q

Where the element stiffness matrix ke is given by

1 1

1 1e e

e

e

A Ek

l



37

The element body force term appearing in the total potential energy

is considered first. substituing u = N1q1 + N2q2 we have

T

e

u fAdx

1 1 2 2( )T

e

e

u fAdx A f N q N q dx

A and f are constant within the element and were consequently brought outside

The integral. the above equation can be written as

1

2

e

eT T

e e

e

A N dx

u fAdx qA f N dx

The integrals of the shape functions above can be readily evaluated

by making the substitution 2eldx d

1

1

1

1

2

1

1

2 2 2

1

2 2 2

e e

e

e e

e

l lN dx d

l lN dx d

The body force term in reduces to

1

12

e e eA l ff

Force vector



38

Traction force vector

The element traction force term appearing in the total potential energy

is now considered we have

1 1 2 2

T

e e

u Tdx N q N q Tdx

Since the traction force T is constant within the element, we have

1

2

eT T

e

e

T N dx

u Tdx qT N dx

Then the element traction force vector,Te is given by

1

12

e eTlT



39

The x coordinate system is mapped on to a coordinate system which is

given by the transformation

3

2 1

2( )x x

x x

1

2

3

1( ) (1 )

2

1( ) (1 )

2

( ) (1 )(1 )

N

N

N

1

X

1 2

1

0

1 2 3 3

Quadratic shape functions



40

Quadratic shape functions contd..

The displacement field within the element is written in terms is written in terms

Of the nodal displacement as

1 1 2 2 3 3u N q N q N q

or

u Nq

1 3 2

q1 q2 q3

u

1 1 1

N1 N2 N3

1 1 0 1 2

3 1

2 3 1 3 2



41

Two dimensional problems The displacement vector u is given as

T

u u v

Where u and v are the x and y components of u, respectively the stresses and

Strains are given by T

x y xy

T

x y xy

x

y

T

v

u

U=0

The fig. representing the two dimensional problem

in a general setting, the body force, traction vector,

And elemental volume are given by

T

x y

T

x y

f f f

T T T

and dv tdA

The strain displacement relations are given by

T

u u u v

x y y x

px

py

o


42

Constant strain triangle

Area coordinates

the shape functions can be physically represented

By area coordinates N1,N2,N3

1

2

3

(x,y)

A3

A2

A1

T

U=0

The independent shape functions are conveniently reperensented by the pair

1

2

3 1

N

N

N

1 2 3 1N N N

11

22

33

AN

A

AN

A

AN

A

Finite element Discretization


Two dimensional problems contd..

43

Isoparametric representation

The displacements inside the elements are now written using the shape function

And the nodal values of the unknown displacements

1 1 2 3 3 5

1 2 2 4 3 6

u N q N q N q

v N q N q N q

The relations can be expressed in matrix form by defining a shape function

1 2 3

1 2 3

0 0 0

0 0 0

N N NN

N N N

o or u Nq

For the triangular element, the coordinates x,y can be also be represented

by the nodal Coordinates by using the same shape functions

1 1 2 2 3 3

1 1 2 2 3 3

x N x N x N x

y N y N y N y



44

u u x u y

x y

u u x u y

x y

u x y u

x

uu x y

y

x y

x y

J

In evaluating strains ,the partial derivatives of u and v are to be taken with

respect to X and y coordinates, these derivatives can be expressed in terms

local coordinates by

Make use of chain rule

Where the 2 X 2 matrix is denoted as the Jacobian of transformation

13 13

23 23

x y

x y

J

Strain displacement relation



45

uu

x

u u

y

IJ

23 13

23 13

1

det

y y

x x

IJJ

x13 23 23 13det x y x y J

1det

2A J

Where J is the inverse of the Jacobian

Strain displacement relation contd..

and

The area of the triangle



46

Plane stress

t

0

0

0

z

yz

xz

0z

y

x

z

The constitutive relation

2

1 0

1 01

10 0

2

x x

y x

xy xy

E



47

0z

0

0

0

z

xz

yz


z

The constitutive relation

1 0

1 0(1 )(1 2 )

1 20 0

2

x x

y x

xy xy

E


Plane strain

y

x

48

Dynamic analysis

When loads are suddenly applied,

The mass and acceleration effects are come in to picture

Lagrangian

L T T Kinetic energy

Potential energy

Hamilton Princible

For an arbitrary time interval from t1 to t2,the state of motion of a body extremizes the

functional

2

1

t

t

I Ldt

If L can be expressed in terms of the generalized variables 1 2 3 4, , , ,..... ,nq q q q q

where ii

qq

t

Then the equations of motions are given by

0i i

d L L

dt q q

i=1 to n


49

1 1,x x

2 2,x xm2

m1

The kinetic energy and potentially energy are given by 2 2

1 1 2 2

2 2

1 1 2 2

1 1

2 2

1 1

2 2

T m x m x

k x k x

Using and L T 0i i

d L L

dt q q

i=1 to n

The equations of motions will be

1 1 1 1 2 2 1

11

2 2 2 2 1

22

( ) 0

( ) 0

d L Lm x k x k x x

dt xx

d L Lm x k x x

dt xx

In matrix form 1 1 1 2 2 1

2 2 2 22

0 ( )0

0

m x k k k x

m k k xx

Spring mass system

Stiffness matrix Mass matrix

Dynamic analysis contd..

50

Systems with Distributed mass

x

z

y

u

v

w

dv

= density

v The kinetic energy

The velocity vector of point at x with components

In the finite element method,we divide the body

into elements,and in each element

and

1

2

T

v

T u u dv

, ,u v w

u Nq u N q

The kinetic energy Te

1

2

1

2

T

v

T

v

T u u dv

T q N Ndv q

Mass matrix


Dynamic analysis contd..

51

Scalar field problems

• General Helmholtz equation given by

is the field variable that is to be determined

Q is the heat source or sink

( ) ( ) ( ) 0x y zk k k Qx x y y z z

( , , )x y z


52

Thermal problems

x y

T Tq k q k

x y

T=T(x,y) is a temperature field in the medium

qx and qy are the components of the heat flux ( W/m2 )

k is the thermal conductivity ( W/m˚ c )

,T T

x y

are the temperature gradients along x and y

Fourier’s law for two dimensional heat flow is given by


53

One dimensional steady state heat conduction

• Governing Equation

• Boundary conditions

The boundary conditions are mainly of three kinds

• Specified temperature

• Specified heat flux (or insulated)

• Specified convection

( ) 0d dT

k Qdx dx

QAdx

x dx

Left face

Right face

x


54

Ex.2

The inside surface of a wall is

insulated and outside is a convection

surface

0 0

( )

x

x L L

T T

q h T T


The boundary conditions for this problem are

(Specified)

L

,h T

LT

0TEx. 1

Consider the wall of a tank containing a hot liquid at temperature T0, with an air stream temperature T∞ passed on the outside, maintaining a wall temperature of TL at the boundary as shown in fig below

,h T

x

q=0(insulated)

LT

L

0 0xq ( )x L Lq h T T


55

Heat Dissipation

Hot gases

1 2 3 4

,h T

0T


Finite element model


56

Two dimensional steady state heat conduction

Governing equation

( ) ( ) 0

T Tk k Q

x x y y

ST: T=T0

Sq: qn=q0

Sc: qn=h(T-T∞)

Boundary conditions

Boundary conditions are of three types

specified temperature T=T0

specified heat flux qn=q0 on Sq

convection qn=h(T-T∞)on Sc


57


A A

Section A-A

Heat Transformation in chimney

Example 1


58

Example 2



59

0

0 T

0( )E

0 0 0 0

1 1( ) ( ) ( )

2 2

Tu E

0 0

1( ) ( )

2

T

LU E Adx

Thermal stresses If the change in temperature is ΔT(x), then the strain due to this temperature

Change is known as initial strain, , given as

T

is coefficient of thermal expansion

implies raise in temperature

The stress-strain relation in the presence of initial strain is given by

strain energy per unit volume

Total strain energy


60

1

0

12

e Tee e

lE A d

B

Thermal stress contd..

BqNoting that ,we get

0d

d

QFor minimization of the functional the necesssry condition

1 1

0

1 1

2

0

1( )

2 2 2

1

2 2

T T T Te ee e e e

e e

ee e

e

l lU E A d E A d

lE A

q B B q q B

Stiffness matrix Thermal load vector

1

0 0

1

1( ) ( )

2 2

Tee e

e

lU A E d

For a structure modeled using one-dimensional linear elements, the above

equation becomes

Constant term


61

After solving the finite element equations KQ=F for the displacements Q

The stress in each element

( )e e e

e

F f T P

( )E T Bq

The temperature load vector can be assembled along with the body force, traction

force, and point load vectors to yield the global load vector F, for the structure.

2 1

1

1

e e e eE A l T

x x

By substituting

2 1

11 1

x x

B

0 T and


Thermal stress contd..

62

Types of Elements

One dimensional elements

1. Beam (axial)

2. Beam (bending)

3. Pipe


63

Two dimensional elements

1. Triangular

inplane

bending

2. quadrilateral

inplane

bending


Types of Elements

64

Three dimensional elements

1. brick

2. Tetrahedral


Types of Elements

65

Modeling

• Element type must be consistent

• Finer mesh near the stress gradient

• Extremely fine mesh when forces to be applied near the stress

concentration areas such as fillets

• Uniform change in stress between adjacent elements

• Better aspect ratio


66

Modeling (cont’d)

• Gross element distortion should be avoided

• Adjoining elements must share common nodes and common

degrees of freedom

450

150


67

Debugging of FE models

• Geometry

• Material properties

• Applied forces

• Displacement constraints


68

Common symptoms and their possible causes

Symptoms

Causes

Excessive deflection, but

anticipate stress

Excessive deflection and

excessive stress

Internal discontinuity in stress and

deflection

Young’s modulus too low,

missing nodal constraints

Applied force too high,

nodal coordinates incorrect,

force applied at wrong nodes

Force applied at wrong nodes,

missing or double internal element


69

Symptoms

Causes

Discontinuity along boundary

Higher or lower frequency than

anticipated

-Static deflections, O.K.

-Static deflection not O.K.

Internal gap opening up in model

under load, stress discontinuity

Missing nodal constraint,

force applied at wrong node

Density incorrect

Young’s modulus incorrect

Improper nodal coupling


Common symptoms and their possible causes

70

Dynamic analysis

• Modal analysis - Natural frequency and mode shapes

• Harmonic analysis - Forced response of system to a sinusoidal

forcing

• Transient analysis - Forced response for non-harmonic loads

(impact, step or ramp forcing etc.)


71

Substructure

Rules for substructure

• A substructure may be generated from individual elements or other substructures

• Master nodes to be retained to be identified

• Nodal constraints will be retained in all subsequent uses of the substructure

• Along a substructure boundary that will be used for connection to the rest of the global model, all nodes must be retained as master nodes

• Cost effective


72

Guidelines for selection of dynamic degree of

freedom (DDOF)

• No. of the DOF must be 2 times highest mode of interest

• No. of reduced modes will be equal to the number of DDOF so that

only bottom half of the calculated modes should be considered

accurate

• DDOF should be placed in areas of large mass and rigidity

• DDOF should be distributed in such a way as to anticipate mode

shapes

• DDOF be selected at each point of dynamic force application

• DDOF must be used with gap elements

• For plate type elements emphasise DDOF in out of plane direction


Common Errors in Finite Element Analysis

Idealization error Discretization error

Idealization Error

•Posing the problem

•Establishing boundary conditions

•Stress-strain assumption

•Geometric simplification

•Specifying simplification

•Specifying material behaviour

•Loading assumptions

Prof .N. Siva Prasad, Indian Institute of Technology Madras Error - 1

Discretization Error

•Imposing boundary conditions

•Displacement assumption

•Poor strain approximation due to element distortion

•Feature representation

•Numerical integration

•Matrix ill-conditioning

•Degradation of accuracy during Gaussian elimination

•Lack of inter-element displacement compatibility

•Slope discontinuity between elements

Prof .N. Siva Prasad, Indian Institute of Technology Madras Error - 2

SOURCES OF NONLINEARITIES

Geometric

Material

Force Boundary Conditions

Displacement Boundary Conditions


GEOMETRIC NONLINEARITY

Physical source

Change in geometry as the structure deforms is taken into account

in setting up the strain displacement

and equilibrium equations.

Applications

1. Slender structures in aerospace, civil and mechanical

engineering applications.

2. Tensile structures such as cables and inflatable membranes.

3. Metal and plastic forming.

4. Stability analysis of all types.


GEOMETRIC NONLINEARITY CONTD..

Mathematical source Strain-displacement equations:

e = Du (2.1)

The operator D is nonlinear when finite strains (as opposed to infinitesimal strains) are expressed in terms of displacements.

Internal equilibrium equations:

b = −D∗σ (2.2)

In the classical linear theory of elasticity, D∗ = DT is the formal adjoint of D, but that is not necessarily true if geometric nonlinearities are considered.



Large strain

The strains themselves may be large, say over 5%.

Ex: rubber structures (tires, membranes)

Small strains

but finite displacements and/or rotations. Slender

structures undergoing finite displacements

Rotations

although the deformational strains may be treated

as infinitesimal.

Example: cables, springs


Linearized prebucking.

When both strains and displacements may be treated as

infinitesimal before loss of stability by buckling.These may

be viewed as initially stressed members.

Example:

Many civil engineering structures such as buildings and stiff

(non-suspended) bridges.



MATERIAL NONLINEARITY

Physical source

Material behavior depends on current deformation state and

possibly past history of the deformation.

Other constitutive variables (prestress, temperature, time,

moisture, electromagnetic fields, etc.) may be involved.

Applications

Structures undergoing

Nonlinear elasticity

Plasticity

Visco-elasticity

Creep, or inelastic rate effects.


MATERIAL NONLINEARITY CONTD..

Mathematical source

The constitutive equations that relate stresses and strains. For a linear elastic material

σ = Ee

where the matrix E contains elastic moduli.

Note:

If the material does not fit the elastic model, generalizations of this equation are necessary, and a whole branch of continuum mechanics is devoted to the formulation, study and validation of constitutive equations.



The engineering significance of material nonlinearities varies greatly

across disciplines.

Civil engineering deals with inherently nonlinear materials such as

concrete, soils and low-strength steel.

Mechanical engineering creep and plasticity are most important,

frequently occurring in combination with strain-rate and

thermal effects.

Aerospace engineering material nonlinearities are less important

and tend to be local in nature (for example, cracking and

“localization” failures of composite

materials).


Material nonlinearities may give rise to very complex

phenomena such as path dependence, hysteresis,

localization, shakedown, fatigue, progressive failure.

The detailed numerical simulation of these phenomena

in three dimensions is still beyond the capabilities of the

most powerful computers.



FORCE BC NONLINEARITY

Physical Source

Applied forces depend on deformation.

Applications

The most important engineering application concerns pressure loads

of fluids.

Ex:

1. Hydrostatic loads on submerged or container structures;

2. Aerodynamic and hydrodynamic loads caused by the motion of

aeriform and hydro form fluids (wind loads, wave loads, and drag

forces).


DISPLACEMENT BC NONLINEARITY Physical source

Displacement boundary conditions depend on the deformation of the structure.

Applications

The most important application is the contact problem,

in which no-interpenetration conditions are enforced on flexible

bodies while the extent of the contact area is unknown.

Non-structural applications of this problem pertain to the more

general class of free boundary problems,

example: ice melting, phase changes, flow in porous media.

The determination of the essential boundary conditions is a key

part of the solution process.


86

Some solution method For a time independent problem [K]{D}={F} For a linear analysis [K] and {R} are independent of [D]. For nonlinear analysis [K] and {R} are regarded as function of {D} Consider [K] is a function of {D} and can be computed for a given {D} Consider a nonlinear spring in Fig 1 Spring stiffness [K]=K0+KN

K0=constant term KN=depends on deformation (K0+KN)u=P Where u=displacement P=load And KN=f(u) and depends on [D] Note: 1.when KN is known in terms of u,P can be calculated in terms of u 2.Explicit solution for u is not available 3.u can be determined by iterative methods

K U

P

u

P Hardening KN>0

Softening KN<0

KN=0

Fig 1

87

10

APu

K

20 1( ( ))

APu

K f u

1 1 1

1 0 2 0 1 1 0, ( ) ,......, ( )A N A i Ni Au k p u k k p u k k p

Direct substitution Let KN<0 (softening spring PA is the load applied Assume KN=0 first iteration UA=displacement produced for the first iteration

Use u1 to compute the new stiffness. K0+KN1=K0+f(u)

Writing symbolically

This calculations are interpreted graphically in Fig 2.2

Note: 1.Approximate stiffness K0+KNi can be regarded as secants of the actual curve 2.After several iterations, the secant stiffness=K0+KN

3.stiffness=PA/UA u=uA is closely approximated

u

P

u1 u2 u3

1

2

3

a b c

K0

Slope=K0-KN1

FEA of Engine block for noise

and vibration

Case Studies (Automotives)

FE model of cylinder block

FE Model of oil sump

Main bearing forces

Second mode of vibration

FEA of single cylinder diesel

engine crankshaft

Crank shaft nomenclature

Solid model of crank shaft with flywheel

and balance weights

Meshed model

Element used: Tetrahedron

Alternate bending

stress

Alternate shear stress

Main bending stress Main shear stress

Tractor Rear Axle

Wheel Track

Flange to Flange

Wheel Stand out

Wheel To Fender clearance

Fender ROPS

Only Axle housing and Axle Shaft Modifications are considered in the project

Prof .N. Siva Prasad, Indian Institute of Technology Madras Tractor-3

Packaging Requirements

Flange to flange distance of 1524 mm (60”) Std wheel track of 1320 mm (52”) Fender to fender distance like existing model – To take care of the 3 point linkage fouling with the Fender Wheel to fender clearance of 60 mm Wheel stand out from fender should not be more than 40 mm Roll Over Protective Structures (ROPS) fitment suitability


Fender

and

ROPS

mtg

details

ROPS

fitment

Suitabilit

y

Fender

to

Fender

distance

Wheel

Standou

t

Fender

Wheel

Clearan

ce

3 point

Linkage

clearanc

e

Packaging with FENDER,ROPS And 3Pt Linkages

Existing Base plate of the ROPS has to be modified.

Fender to Fender Distance is important for 3 pt linkage clearance


Packaging

Fender and

ROPS mtg

details

ROPS

fitment

Suitability

Fender to

Fender

distance

Wheel

Standout

Fender

Wheel

Clearance

3 point

Linkage

clearance

Modifications carried out On axle housing to get all the Required features.

• Length of the housing modified • 3 or 4 slots tried out • Position of slots modified.

Tractor-8

Finalized Model of the Modified Housing


Constraints in Modified Housing Design Need for making use of the existing fixtures, gauges and machine tools, limits the modifications.

A New housing without all these limitations, but meeting all vehicle requirements were also designed.

It weighs just 55 kg against the original housing weight of 70.5kg

New Housing


FEA of the Axle housings

All three Axle housings, Existing Axle housing, Modified Axle housing and New Axle Housing were subjected for FEA as per Test Spec Loading.

If, the stress levels of the latter two housings are less than the Existing housing then the design is assumed to be safe

Since, there is no failure in the Existing axle housing so far.


Meshing Details

Parameter Existing

axle

housing

Modified

axle

housing

New axle

housing

Number of

nodes

172268 245479 295685

Number of

elements

94502 140515 168000

Weight, kg 70.5 62 55

Solver : ANSYS Meshing : ANSYS Element Type : Tetrahedron Element Name : SOLID 187

Existing Axle Housing Modified Axle Housing New Axle Housing

Tractor-13

Vertical Loading condition

Load Set 1 Load Set 2

Loading Condition

Hogging Loading condition

Tractor-14

Normal Stress plot of Existing axle housing with Load set 1


Normal Stress plot of Modified axle housing with Load set 2


Normal Stress plot of Modified axle housing with Hogging load


Loading

condition

Maximum stress

condition

Normal Stress in Axle housing, MPa

Existing Modified New

Load set 1 Tensile 321 143 170

Load set 1 Compressive 112 32 19

Load set 2 Tensile 632 281 335

Load set 2 Compressive 221 64 37

Comparison of normal stresses under vertical condition.

Loading

condition

Maximum stress

condition

Normal Stress in Axle housing, MPa

Existing Modified

Hogging Tensile 216 103

Hogging Compressive 32 16.5

Comparison of normal stresses under hogging.


FEA Results

In all conditions in Vertical loading, both modified and New axle housing stresses are far less than the Existing axle housing. Even in Hogging, Modified axle housing stresses are lesser than the Existing axle housing. Hence, it can be a safe design having same reliability like the existing housing.


(a) (b)

New Axle Housing (55 kg) Modified Axle Housing (62 kg)

Final design of Axle housings


FEA for Automotive application

CLUTCH

Prof .N. Siva Prasad, Indian Institute of Technology Madras Clutch - 1

Axial spring clutch system in a truck with

ceramic clutch disk

Face Plate

Clutch Plate

Pressure Plate

Flywheel

Bent Clutch Disc

Face Plate Back Plate Prof .N. Siva Prasad, Indian Institute of Technology Madras Clutch - 2

Problem Definition

• Due to extreme temperatures during clutch operation at interface, there is occurrence of pressure plate warpage. This leads to clutch failures

• Frictionally induced Thermo-mechanical instability causes breakdown of physical properties hence inducing warpage.

• The problem is a contact problem which is transient and non linear.


Material Properties

S.No Property Grey Cast Iron 97% Alumina

ceramic Units

1. Density 7.34 *10³ 3.69 *10³ kg/m³

2. Modulus of elasticity 124 300 GPa

3. Thermal expansion (20

ºC) 9.0*10-6 7.3*10-6 ºCˉ¹

4. Specific heat capacity 840 880 J/(kg*K)

5. Thermal conductivity 53.3 18 W/(m*K)

The two rotating discs were - Grey cast iron sliding against, alumina. The following properties were considered.


Analysis parameters S.No Parameter Definition

1. Element type Element with both structural and thermal capability

2. Analysis type Non-linear - Geometric Parameter is included.

3. No. of Sub-steps Multiple sub-step -Progressive time step Starting from 0

to 0.75 sec with sub-step of 0.005 sec

4. Pressure/ Angular Velocity Pressure and Angular velocity simultaneously varying

5. Coupling Rigid element coupling with master and slave nodes

6. Averaging method Multiple facets - A single element divided into sub

elements at the mid-nodes. Result is the average of the

sub elements.

Element with mid-side node Solid 226 Time varying Pressure / angular Velocity


FE Model / Boundary conditions

Master node defined at center with inner nodes as slave nodes. Rotational velocity given at the Master Node

Mesh using Solid 226

Contact Pair Creation at interface CONTA174 – upper nodes of lower disk TARGE170 – Lower nodes of upper disk

All DOF constrained for stationary disk at bottom end

Pressure applied at the rotating disk

surface


Comparison with Du & Fash Model and

simulation

Top surface Prof .N. Siva Prasad, Indian Institute of Technology Madras Clutch - 7

Bottom surface Anti-symmetric w.r.t top

surface


simulation


Top surface


simulation


Extension of Quarter disk to Annular Disk

– Focal hot-spots at Mid-plane


Crashworthiness of a Commercial Vehicle

using Finite Element Method


Need of crash analysis

Year Road Accidents

(In Thousands)

Persons Killed

(In Thousands)

Persons Injured

(In Thousands)

2004 429.8 92.5 464.6

2003 406.7 86.0 435.1

2002 407.5 84.7 408.7

2001 405.6 80.9 405.2

2000 391.4 79.8 399.3

1999 386.4 82.0 375.0

1998 385.0 79.9 390.7

1997 373.7 77.0 378.4

1996 371.2 74.6 369.5

1995 348.9 70.6 323.2

1994 320.4 64.0 311.5

(Source: Department of Road Transport and Highways, Ministry of Shipping, Road Transport and Highways, Government of India.)

Auto Accident Statistics

Prof .N. Siva Prasad, Indian Institute of Technology Madras Crash - 1

Objectives

• To evaluate the occupant response and the

structural damage according to the Federal

Motor Vehicle Safety Standard (FMVSS)

frontal impact regulation and ECE-R29

regulation in accordance with the basic

principles of automotive crashworthiness.

• To check the deformation length and the

amount of energy absorbed at high crash

velocities (resulting in less fatal injuries).


Finite Element Methods for analysis

The following softwares are used to carryout the analysis:

• Modeling

Catia/ Pro - E

• Meshing and Boundary Conditions HyperMesh

• Solver LS-DYNA

• Post-processing and viewing the results HyperView


Truck Modeling

Mass density (kg/m3) 7890

Young's Modulus (GPa) 210

Yield Stress (MPa) 270

Poisson Ratio 0.3

Vehicle Material Properties


Finite Element Model of the Truck

Number of Parts 157

Number of Nodes 38955

Number of Elements 36539

Number of Quad Elements 33124

Number of Tria Elements 2012

Number of Discrete Elements 10

Ford Truck Vehicle Model Summary


Load Cases simulated according

to ECE-R29 Regulation

Load Cases Variations

Pendulum striking the truck without constraining the cabin of the truck.

( kJ)

a. 36

b. 54

Pendulum striking the truck with the back wall of the cabin held to a fixture

( kJ)

a. 36

b. 54

ECE-R29 frontal impact test configuration

(ECE – Economic Commission for Europe)


Simulation results for 36 kJ of energy

transferred from pendulum

Displacement contour plot for pendulum

transferring 36 kJ of energy.

% Reduction in occupant space = 2 %



transferred from pendulum

Displacement contour plot for

pendulum transferring 54 kJ of energy.

% Reduction in occupant space = 5 - 6%



transferred with cabin constrained

Displacement contour plot for 36 kJ

energy transferred, cabin constrained




transferred with cabin constrained

Displacement contour plot for 54 kJ

energy transferred, cabin constrained



Load Cases simulated according to

FMVSS -208 Regulation

Load Cases Variations

Truck hitting a rigid wall with full head on collision (km/hr) a. 36

b. 54

Truck hitting a rigid wall with partial overlap (km/hr) a. 36

b. 54

c. 72

Truck hitting the wall obliquely at 30o (km/hr) 36

Partial overlap test configuration Oblique Impact test configuration


Simulation results for head-on collision with

truck at a speed of 36 kmph

Displacement contour for head-on collision at a

speed of 36km/hr % Reduction in occupant space = 11 - 13%


Simulation results for head-on collision with

truck at a speed of 54 kmph

Displacement contour plot for head-on collision at

a speed of 54km/hr



Simulation results for partial overlap with

truck at speed of 36 kmph

Displacement contour plot for partial overlap at a

speed of 36 km/hr



Simulation results for partial overlap with



speed of 54 km/hr



Simulation results for partial ovelap with



speed of 72 km/hr



Simulation results for oblique collision at

an angle of 30 degrees at 36 kmph

Displacement contour plot for impact at 30

degree at a speed of 36 km/hr



Simulation Results - Acceleration response

Acceleration response of front axle for head-on collision at a speed of 54km/hr


144

Drum brakes are used to control the speed of vehicles or to stop

them by friction caused by set of pads

Externally contracting brakes are extensively used for

locomotives

When brakes are applied, high amount of kinetic energy is

transformed into thermal energy in short periods

Present work is intended at predicting the temperature rise during

the process of braking, deceleration and stop time

Coupled field finite element analysis of internally expanding and externally contracting brakes

Prof .N. Siva Prasad, Indian Institute of Technology Madras Brake - 1

145

Objectives

Estimation of the contact pressure between brake liner

and counter body (drum/wheel)

Estimation of temperature distribution using coupled field

(structural-thermal) finite element method

Comparison of Finite element analysis results with

experimental values for externally contracting brakes


146

Average contact

pressure

(MPa)

Movement of the floor Present

analysis Reference

Leftward 1.001911 0.941538

Rightward 8.598652 9.663158

0

4

8

12

16

20

0 6 12 18 24 30

Distance from left end (mm)C

on

tact

pre

ssu

re (

MP

a)

rightward motion of the floor

leftward motion of the floor

Door stop problem is analysed to predict interface pressure


147

To validate the methodology for curved surfaces

0

2

4

6

8

10

12

14

0 20 40 60 80 100 120

Lining angle (deg.)

Pre

ssu

re (

MP

a)

Present analysis

Present analysis with chamfer

Day et al. [1]


148

A drum brake of commercial truck [4] is analysed to validate the

methodology for the prediction of temperature distribution


149

In reference

Thermal analysis is carried out by considering the cross section of

the drum as shown in figure

Heat flux calculated from kinetic energy (KE) loss and applied at

the interface

It represents the assumption of uniform pressure distribution, but it

is not uniform

In present method

Heat flux generated at the interface nodes is implicitly calculated

from contact pressure distribution which is more realistic

compared to the assumption of uniform heat flux to every node on

the inner periphery of the drum considered by reference[2]


150

Temperature measurement

Schematic diagram of the dynamometer test facility is shown in

figure

DC Motor

WheelFly wheels

Front view Side view of the wheel


151

Complete braking is solved in 100 load steps

From the experiment time taken to apply full load is 3.6seconds

20 load steps are used to ramp the load from 0-4.2T

In the 1st load step angular displacement corresponding to initial

velocity and a load of F/20 is applied

Analysis is solved

Now the stop time with the average torque of 1st load step is

calculated

Then the deceleration is calculated

Velocity after the load step is calculated

Energy available after the load step is determined

In the 2nd load step, a force of F/20 is added to the 1st load step

load and an angular displacement corresponding to velocity after

the 1st load step is applied

Analysis is solved

This procedure is repeated for 20 load steps

After 20 load steps there is no change in the load

Hence there is no change in the deceleration

With the deceleration after 20 load steps is used in the

simulation of remaining 80 load steps


152

Pressure distribution at different times during braking is shown in

the graph

Pressure distribution at leading end varies little with the time and is

due to thermal distortion of liner

0

0.5

1

1.5

2

2.5

0 5 10 15 20 25 30 35 40

Lining angle (deg.)

Pressu

re (

MP

a)

contact pressure @ 3.60 sec




153

Temperature distribution at a point on the outer periphery of the

wheel is shown in the graph along with the experimental values


154

Comparison of the speed calculated by FEA with experimental values

0

50

100

150

200

250

0 3 6 9 12 15 18 21

Time (sec)

Sp

eed

(rp

m)

FEA

Expt

FEA Experiment

Stop time 18.67 18.7

Temperature at the end of the braking process (0K) 335.14 332

Comparison of FEA and experimental values


155

Rigid beam elements

created from centre to

Brake torque (Nm) after

20 load

steps

40 load

steps

60 load

steps

80 load

steps

100 load

steps

Outer periphery 3507.60 3507.75 3507.53 3507.52 3507.69

Inner periphery 3508.74 3508.82 3508.91 3508.88 3509.20

In another analysis rigid beam elements are created from centre to inner

periphery of the wheel to consider the deformation of wheel in the

analysis

There is no significant variation in the torque by making wheel to behave

as rigid but computational time is less

Hence the wheel can be considered to behave as rigid


156

An analysis is carried out to know the variation of pressure

distribution with the velocity

In the first load step shoes are pressed against the drum

In the second load step different angular displacements are applied

Contact pressure doesn’t vary with the velocity

0

0.5

1

1.5

2

2.5

0 10 20 30 40

Lining angle (deg.)

Pre

ssu

re (

MP

a)

Second load step 4 radians in 1 sec

Second load step 10 radians in 1 sec


2nd session- dr.n.sivaprasad

Documents

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faces yz

x y z yz xz xy

error prof

c prof

z xprof

volume v

tx y zt