Basics FEA Case Studies

8/13/2019 Basics FEA Case Studies

1/156

Introduction to

Finite Element Analysisby

Prof. N Siva Prasad

Department of Mechanical Engineering

Indian Institute of Technology Madras


2/156


3/156

3

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

Error

Why FEM?Contd


4/156

4

Error


Why FEM?Contd


5/156

5


Why FEM?Contd


6/156

6


Example


7/156

7Prof .N. Siva Prasad, Indian Institute of Technology Madras


8/156

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



9/156

9

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

volume, is the vector fgiven by

T

[ ]x yf , f , ffz (2)

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

The surface traction Tmay be given by

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

A load Pacting at a point iis represented by its three components:

T

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



10/156

10

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

Fig. 2 Equilibrium of elemental volume



11/156

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)



12/156

12

STRAINDISPLACEMENT 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.



13/156

13

Fig. 3 Deformed elemental surface



14/156

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)



15/156

15

yx zx

yx zy

yx zz

v vE E E

v v

E 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 Hookes law. For isotropic materials, the two material

properties are Youngs modulus (or modulus of elasticity) E and

Poissons ratio . Considering an elemental cube inside the body,

Hookeslaw gives

v



16/156

The shear modulus (or modulus of rigidity), Gis given by

2(1 )EG

v

(11)

From Hookes 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

16Prof .N. Siva Prasad, Indian Institute of Technology Madras


17/156

17

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

1 0 0 01 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 vv v v

v v vE

vv v

v

v

D (14)



18/156

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 Hookeslaw relations (Eq. 10) then give us

, , andz xz yz

2(1 )

( )

x

y

xy

z

yx vE E

yxv E E

vxyE

vx yE



19/156

19

(16)

Fig. 4(b) plane strain

1 0

1 01+ 1 2

0 0 0.5

v v

x xEv 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 beobtained

, , andz xz yz


20/156

20

TEMPERATURE EFFECTS

The temperature strain is represented as an initial strain:

0

T

(18)



21/156

21

Strain Energy stored in the spring

U = (force in the spring) (displacement )

= (Su) u

= 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

= s u2p u

for minimum of , /u = 0

S up = 0

Su = p

Potential energy and equilibrium


P

u

P


22/156

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 bodyis . For an elastic body, the total strain energy is given by1

2

V

1 T

2U dV

(23)

The work potential WP is given by



23/156

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 iiV V S

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



24/156

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, theequilibrium state is stable.

Kinematically admissible displacements are those that satisfy the

single-valued nature of displacementsandthe boundaryconditions.



25/156

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


26/156

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. Since1 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,andq3are the displacements of nodes 1,2, and 3, respectively.



27/156

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

0

1 1 2 2 2 3 3 22

03 3 2 4 3 3

3

k q q F q

k q q k q k q q

q

k q q k q F q

(29)



28/156


29/156

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 dudx

The loading consists of three types

- body force f

-traction force T

-point load Pi

One dimensional problems



30/156

30

X

P2

P1

f

T

One dimensional bar loaded by traction, body and point loads

One dimensional problems contd..



31/156

31

Finite element modeling of the bar

.

.

X

.

X

1

4

3

2

1

2

5

3

4

..

.

.



.


32/156

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

e1 2

1

2

1( )

2

1( )

2

N

N

1 1

N1 N2

1 1 0 1 0


Fig.a Fig.b

e



33/156

33

Linear interpolation

e e

u1u2

q1

q2

uunknown

ulinear

1 2 1 2

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

Displacement q1and q2as

1 1 2 2u N q N q

In matrix notationu Nq

The transformation from x to can be written in terms of N1and N2as

1 1 2 2x N x N x

Isoparametric formulation


1 2

1 2

,

, T

N N N

q q q

where



34/156

34

The general expression for the potential energy

1

2

T T T

i iiAdx 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 i

e e e i

Adx u fAdx u Tdx Q P The last term above assumes that point load Piare 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 i

e e e ie

U u fAdx u Tdx Q P where

1

2

T

eU Adx is the element strain energy




35/156

35

Element stiffness matrix

Consider the strain energy form1

2

T

eU Adx

Substituting for and into the above yieldsEBq 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

2

eldx d

where1 1 Le is the length of the element




36/156

36

The element strain energy Ue is now written as1

1

1

2 2

T Tee e e

lU q A E B B d q

Where E is the youngs modulus of element and by using

1

1

2d

211 1

1 112

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 keis given by

1 1

1 1

e ee

e

A Ek

l




37/156

37

The element body force term appearing in the total potential energy

is considered first. substituing u = N1q1 + N2q2we 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 q

A f N dx

The integrals of the shape functions above can be readily evaluated

by making the substitution2

el

dx d

1

1

1

1

2

1

1

2 2 21

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




38/156

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,Teis given by

112

e eTlT




39/156

O di i l bl d


40/156

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 Nq1 3 2

q1q2 q3

u

1 11

N1 N2N3

1 1 01 2

31

23 1 3 2



T di i l bl


41/156

41

Two dimensional problemsThe displacement vector uis given as

Tu u v

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

T

x y xy

T

x y xy

x

y

T

v

u

U=0

The fig. representing the two dimensional problemin 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


T di i l bl td


42/156

42

Constant strain triangle

Area coordinates

the shape functions can be physically representedBy area coordinates N1,N2,N3

1

2

3

(x,y)

A3

A2A1

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

AA

NA

Finite element Discretization


Two dimensional problems contd..


43/156

T di i l bl td


44/156

44

u u x u y

x y

u u x u y

x y

u x y u

x

uu x yy

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



T di i l bl td


45/156

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



T di i l bl td


46/156

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



T di i l bl td


47/156

47

0z

0

00

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

Dynamic analysis


48/156

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 extremize

functional

2

1

t

t

I Ldt

If L can be expressed in terms of the generalized variables1 2 3 4, , , ,..... ,nq q q q q

where

i

i

qq

t

Then the equations of motions are given by

0i i

d L L

dt q q

i=1 to n


Dynamic analysis contd


49/156

49

1 1,x x

2 2,x xm2

m1

The kinetic energy and potentially energy are given by2 2

1 1 2 2

2 2

1 1 2 2

1 1

2 21 1

2 2

T m x m x

k x k x

Using andL 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 matrixMass matrix

Dynamic analysis contd..

Dynamic analysis contd


50/156

50

Systems with Distributed mass

x

z

y

u

v

w

dv

= density

vThe 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

12

1

2

T

v

T

v

T u u dv

T q N Ndv q

Mass matrix


Dynamic analysis contd..

Scalar field problems


51/156

51

Scalar field problems

General Helmholtz equation given by

is the field variable that is to be determinedQ is the heat source or sink

( ) ( ) ( ) 0x y z

k k k Qx x y y z z

( , , )x y z


Thermal problems


52/156

52

Thermal problems

x yT Tq k q k x y

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

qxand qyare the components of the heat flux ( W/m2 )

k is the thermal conductivity ( W/mc )

,T T

x y

are the temperature gradients along x and y

Fouriers law for two dimensional heat flow is given by


One dimensional steady state heat conduction


53/156

53


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




54/156

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 temperatureT0, with an air stream temperature Tpassed on the outside, maintaining awall temperature of TLat theboundary as shown in fig below

,h T

x

q=0(insulated)

LT

L

0 0

xq ( )x L Lq h T T




55/156

55

Heat Dissipation

Hot gases

12 3 4

,h T

0T


Finite element model


Two dimensional steady state heat conduction


56/156

56


Governing equation

( ) ( ) 0T T

k k Qx 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




57/156

57


A A

Section A-A

Heat Transformation in chimney

Example 1




58/156

58

Example 2



Thermal stresses


59/156

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 stressesIf the change in temperature isT(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


Thermal stress contd


60/156

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 T e 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 matrixThermal load vector

1

0 01

1( ) ( )

2 2

Te

e ee

lU A E d

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

equation becomes

Constant term



61/156

Types of Elements


62/156

62

Types of Elements

One dimensional elements

1. Beam (axial)

2. Beam (bending)

3. Pipe


Types of Elements


63/156

63

Two dimensional elements

1. Triangularinplane

bending

2. quadrilateral

inplane

bending


Types of Elements

Types of Elements


64/156

64

Three dimensional elements

1. brick

2. Tetrahedral


Types of Elements

Modeling


65/156

65

Modeling

Element type must be consistent

Finer mesh near the stress gradient

Extremely fine mesh when forces to be applied near the stressconcentration areas such as fillets

Uniform change in stress between adjacent elements

Better aspect ratio


Modeling (contd)


66/156

66

Modeling (cont d)

Gross element distortion should be avoided

Adjoining elements must share common nodes and common

degrees of freedom

450

150


Debugging of FE models


67/156

67

Debugging of FE models

Geometry

Material properties

Applied forces

Displacement constraints


Common symptoms and their possible causes


68/156

68

y p p

Symptoms Causes

Excessive deflection, but

anticipate stress

Excessive deflection and

excessive stress

Internal discontinuity in stress and

deflection

Youngs 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


Common symptoms and their possible causes


69/156

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

Youngs modulus incorrect

Improper nodal coupling


y p p

Dynamic analysis


70/156

70

y y

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.)


Substructure


71/156

71

Rules for substructure

A substructure may be generated from individual elements or othersubstructures

Master nodes to be retained to be identified

Nodal constraints will be retained in all subsequent uses of thesubstructure

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

Cost effective


Guidelines for selection of dynamic degree of


72/156

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


73/156

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 MadrasError - 1

Discretization Error


74/156

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 MadrasError - 2

SOURCES OF NONLINEARITIES


75/156

SOURCES OF NONLINEARITIES

Geometric

Material

Force Boundary Conditions

Displacement Boundary Conditions


GEOMETRIC NONLINEARITY


76/156

Physical source

Change in geometry as the structure deforms is taken into accountin 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..


77/156

Mathematical source

Strain-displacement equations:e = Du (2.1)

The operator D is nonlinear when finite strains (asopposed to infinitesimal strains) are expressed in termsof displacements.

Internal equilibrium equations:

b = D (2.2)

In the classical linear theory of elasticity, D= DT is theformal adjoint of D, but that is not necessarily true ifgeometric nonlinearities are considered.




78/156

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



79/156

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


80/156

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..


81/156

Mathematical source

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

= 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 awhole branch of continuum mechanics is devoted to the

formulation, study and validation of constitutiveequations.




82/156

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 andlocalizationfailures of composite

materials).


MATERIAL NONLINEARITY CONTD


83/156

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


84/156

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 Ph i l


85/156

Physical source

Displacement boundary conditions depend on thedeformation of the structure.

ApplicationsThe 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.


Some solution method KU


86/156

86

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 regardedas 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 KNis 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

P

u

PHardening

KN>0Softening

KN


87/156

87

10

APuK

20 1( ( ))

APuK 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

N ( g p g

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+K

N1=K

0+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+KN3.stiffness=PA/UAu=uA is closely approximated

u

P

u1 u

2 u

3

1

2

3

a b c

K0

Slope=K0-KN1


88/156

FEA of Engine block for noiseand vibration

Case Studies (Automotives)

FE model of cylinder block


89/156

FE model of cylinder block

FE M d l f il


90/156

FE Model of oil sump

Main bearing forces


91/156

g

Second mode of vibration


92/156


93/156

FEA of single cylinder diesel

engine crankshaft

Crank shaft nomenclature


94/156

Crank shaft nomenclature

Solid model of crank shaft with flywheel


95/156

and balance weights

Meshed model


96/156

Element used: Tetrahedron


97/156


98/156

Main bending stressMain shear stress


99/156

Tractor Rear Axle Wheel Stand outFenderROPS


100/156

Wheel Track

Flange to Flange

Wheel

ToFenderclearance

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

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

Packaging Requirements


101/156

g g q

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


ROPS

fitment

Suitabilit

Wheel

Standou

t

Packaging with FENDER,ROPSAnd 3Pt Linkages


102/156

Fender

and

ROPSmtg

details

y

Fender

to

Fender

distance

t

Fender

Wheel

Clearan

ce

3 point

Linkage

clearanc

e

And 3Pt Linkages

Existing Base plate of the ROPS has to be modified.

Fender to Fender Distance is important for 3 pt linkage clearance



103/156

Finalized Model of the Modified Housing


104/156



105/156

FEA of the Axle housings


106/156

g

All three Axle housings, Existing Axle housing,

Modified Axle housing and New Axle Housing weresubjected 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 assumedto be safe

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

far.


Meshing DetailsExisting Axle Housing Modified Axle Housing New Axle Housing


107/156

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 : ANSYSMeshing : ANSYSElement Type : Tetrahedron

Element Name : SOLID 187

g g g g

Tractor-13

V ti l L di diti

Loading Condition


108/156

Vertical Loading condition

Load Set 1 Load Set 2

Hogging Loadingcondition

Tractor-14

Normal Stress plot of Existing axle housing with Load set


109/156


Normal Stress plot of Modified axle housing with Load set 2


110/156


Normal Stress plot of Modified axle housing with Hogging load


111/156


Loading Maximum stress Normal Stress in Axle housing MPa

Comparison of normal stresses under vertical condition.


112/156

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

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


113/156

In all conditions in Vertical loading, both modified and

New axle housing stresses are far less than theExisting 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.


Final design of Axle housings


114/156

(a) (b)

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



115/156

FEA for Automotive application

CLUTCH

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

Axial spring clutch system in a truck with

ceramic clutch disk


116/156

ceramic clutch disk

Face Plate

Clutch Plate

Pressure Plate

Flywheel

Bent Clutch Disc

Face Plate Back PlateProf .N. Siva Prasad, Indian Institute of Technology MadrasClutch - 2

Problem Definition


117/156

Problem Definition

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

Frictionally induced Thermo-mechanical instability causesbreakdown of physical properties hence inducing warpage.

The problem is a contact problem which is transient and

non linear.

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

Clutch - 3

Material Properties


118/156

p

S.No Property Grey Cast Iron97% Alumina

ceramicUnits

1. Density 7.34 *10 3.69 *10 kg/m

2. Modulus of elasticity 124 300 GPa

3.Thermal expansion (20

C)9.0*10-6 7.3*10-6 C

4. Specific heat capacity840 880 J/(kg*K)

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

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

Prof .N. Siva Prasad, Indian Institute of Technology Madras Clutch - 4

Analysis parametersS.No Parameter Definition


119/156

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 thesub elements.

Element with mid-side node

Solid 226 Time varying Pressure / angular Velocity


FE Model / Boundary conditionsh l d


120/156

Master node defined at center withinner 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 atthe rotating disk

surface


Comparison with Du & Fash Model and


121/156

simulation

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



122/156

Bottom surfaceAnti-symmetric w.r.t top

surface

simulation




123/156

Top surface

simulation


Extension of Quarter disk to Annular Disk

Focal hot-spots at Mid-plane


124/156

Focal hot-spots at Mid-plane



125/156

Crashworthiness of a Commercial Vehicleusing Finite Element Method


Need of crash analysisAuto Accident Statistics


126/156

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.)

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

Objectives


127/156

To evaluate the occupant response and the

structural damage according to the FederalMotor 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 crashvelocities (resulting in less fatal injuries).


Finite Element Methods for analysis


128/156

The following softwares are used to carryout the analysis:

ModelingCatia/ Pro -

E

Meshing and Boundary

ConditionsHyperMesh

SolverLS-DYNA

Post-processing andviewing the resultsHyperView


Truck Modeling


129/156

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


130/156

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



131/156


132/156

Simulation results for 54 kJ of energy

transferred from pendulum


133/156

transferred from pendulum

Displacement contour plot for

pendulum transferring 54 kJ of

energy.

% Reduction in occupant space = 5 - 6%



transferred with cabin constrained


134/156


Displacement contour plot for 36

kJ energy transferred, cabin

constrained% Reduction in occupant space = 7 - 8%





135/156


Displacement contour plot for 54

kJ energy transferred, cabin

constrained% Reduction in occupant space = 24 - 26%


Load Cases simulated according toFMVSS -208 Regulation


136/156

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 configuratio


Simulation results for head-on collision with

truck at a speed of 36 kmph


137/156

Displacement contour for head-on collision at

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


Simulation results for head-on collision with

truck at a speed of 54 kmph


138/156

Displacement contour plot for head-on

collision at a speed of 54km/hr




139/156

Simulation results for partial overlap with

truck at speed of 54 kmph


140/156

Displacement contour plot for partial overlapa speed of 54 km/hr



Simulation results for partial ovelap withtruck at speed of 72 kmph


141/156

Displacement contour plot for partial overlap

a speed of 72 km/hr



Simulation results for oblique collision atan angle of 30 degrees at 36 kmph


142/156

Displacement contour plot for impact at 30degree at a speed of 36 km/hr



Simulation Results - Acceleration response


143/156

Acceleration response of front axle for head-on collision at a speed of

54km/hr


Coupled field finite element analysis of internally

expanding and externally contracting brakes


144/156

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

p g y g

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


145/156

20

rightward motion of the floor

Door stop problem is analysed to predict interface pressure


146/156

146

Average contact

pressure(MPa)

Movement of the floorPresent

analysisReference

Leftward 1.001911 0.941538

Rightward 8.598652 9.663158

0

4

8

12

16

0 6 12 18 24 30

Distance from left end (mm)

Contactpressure(MPa)

leftward motion of the floor


To validate the methodology for curved surfaces


147/156

147

0

2

4

6

8

10

12

14

0 20 40 60 80 100 120

Lining angle (deg.)

Pressure(MPa

)

Present analysis

Present analysis with chamferDay et al. [1]


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

methodology for the prediction of temperature distribution


148/156

148Prof .N. Siva Prasad, Indian Institute of Technology Madras Brake - 5

In reference

Thermal analysis is carried out by considering the cross section of

the drum as shown in figure


149/156

149

the drum as shown in figure

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


Temperature measurement

Schematic diagram of the dynamometer test facility is shown in

figure


150/156

150

figure

DC Motor

WheelFly wheels

Front view Side view of the wheel


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 1stload step angular displacement corresponding to initial


151/156

151

velocity and a load of F/20 is applied

Analysis is solved

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

Then the deceleration is calculated

Velocity after the load step is calculated

Energy available after the load step is determined

In the 2ndload step, a force of F/20 is added to the 1stload step

load and an angular displacement corresponding to velocity after

the 1stload 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


Pressure distribution at different times during braking is shown in

the graph


152/156

152

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.)

Pressure(MPa)

contact pressure @ 3.60 sec




Temperature distribution at a point on the outer periphery of thewheel is shown in the graph along with the experimental values


153/156

153Prof .N. Siva Prasad, Indian Institute of Technology Madras Brake - 10

FEA Experiment

Stop time 18.67 18.7

Comparison of FEA and experimental values


154/156

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)

Speed(rpm)

FEA

Expt

p

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


In another analysis rigid beam elements are created from centre to innerperiphery of the wheel to consider the deformation of wheel in the

analysis


155/156

155

Rigid beam elements

created from centre to

Brake torque (Nm) after

20 loadsteps

40 loadsteps

60 loadsteps

80 loadsteps

100 loadsteps

Outer periphery 3507.60 3507.75 3507.53 3507.52 3507.69

Inner periphery 3508.74 3508.82 3508.91 3508.88 3509.20

There is no significant variation in the torque by making wheel to behaveas rigid but computational time is less

Hence the wheel can be considered to behave as rigid


An analysis is carried out to know the variation of pressuredistribution with the velocity

In the first load step shoes are pressed against the drum


156/156

p p g

In the second load step different angular displacements are applied

0

0.5

1

1.5

2

2.5

0 10 20 30 40Lining angle (deg.)

Pressur

e(MPa)

Second load step 4 radians in 1 sec

Second load step 10 radians in 1 sec

Basics FEA Case Studies

Documents

Transcript of Basics FEA Case Studies