NM Lecture 6 on 15th Sept 2015.pdf

7/25/2019 NM Lecture 6 on 15th Sept 2015.pdf

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NUMERICAL METHODS

Lecture 6

Dr. P V Ramana 1NM Dr P V RAMANA



Function (u) a+bx a+bx+cx2 a+bx+cx2+dx3

Function (u) a+bx+cy a+bx+cy+dx2+exy+fy2

Function u a+bx+c +dx a+bx+c +dx2+ex +f 2+ x2 2+ x2

Function (u) a+bx+cy+dz+exy+fzy+gxz+hxyz; a+bx+cy+dx2+exy+fy2+…

NM Dr P V RAMANA 2



Forces are acting in transverse

direction

3NM Dr P V

RAMANA



Characteristics (Properties) of global stiffness matrix K IJ

Symmetric Matrix (KIJ = KJI Linear relationship between load and displacement)

Diagonal terms are positive KII (KII= 0 or –ve unless structure is unstable) = . .

order less than the rank of K).

Becomes non-singular by imposing boundary conditions.

NM Dr P V RAMANA 4



•

1 DOF

define the displacement of every material point.

• Usually use low order polynomials• Here

u = a + b x

– u is axial displacement – a b are constants to be determined

– x is local coordinate along member

NM Dr P V RAMANA 5



uJ

F F

uI

I J

•

known nodal displacements ui, u j at nodes

i and j= +

u j = a + b x j

• ui, u j are nodal displacements

• i, j 6NM Dr P V RAMANA



=I

• e ng xi = , x j = , ne can ge

– a = ui

– b = (u j-ui)/L

•

}]{ N[1 d u x x

u i

– N = matrix of element sha e functions or

u j

interpolation functions

– d = nodal dis lacements

7NM Dr P V RAMANA



1 2 N N N

xN 1 L

N1=1

2N

Pro ertiesVariation of N1

N 1 at node i and zero at all other nodes

2=

i N 1 Variation of N

1 2i.e. at any point in the element N N 1 8NM Dr P V RAMANA



}]{ N[1 d u x x

u i

•

du d[N]dx dx

1

• where [B] is a matrix relating strain to nodalL

displacement (matrix of derivatives of shape

functiondV E

V ]B[]B[]k [

NM Dr P V RAMANA9 Adx E T ]B[]B[



dx AE dx AE T LT

]1,1-[1

]1,1-[1

]B[]B[]k [

• for 1D one can get

11]k[

L

EA

• -

and 3-D elements

10NM Dr P V RAMANA

P bl 1



For a system of springs

Problem 1

1k Fx

2k 0d1x

1xd2xd 3xd

,

as:

2211 k k k k

222

111 k k k k

11 0k k

221121 k k k k K K K

11

NM Dr P V RAMANA 22



32



3

3

2

210)( xa xa xaa xv 2 DOF

321 32)( xa xaa xv

NM Dr P V RAMANA 13



Use the following equations into above equation.

3 (0)d v

M M

V V

22323

3231

11

2,32 xL L x L x L

N L L x x L

N

3

2

2

(0)i

i

dx

F V d v EI

M M dx

3433 ,

L L 3

2

2

( )

j

j

v EI

M dx M

d v L

EI dx

d N v

d dxdx

vd d dxdx

vd d dxdx

dv ,,33

3

22

2

EI N d 22

2

Ldx,,,

3

NM Dr P V RAMANA 15 L L Ldx

6,12,6,1233



32 231

x x N

322 x x212

L L x

3

3

2

2

3

x x N

32

x x

224

L L

operties :Pr

x L N N N 342

31NM Dr P V RAMANA 16



The beam element equation relating nodal forces and nodal displacements is given as

ii v L LF 612612

ii

v L L

L L L L

L

EI

F

M 22

3 612612

2646

j j L L L L M 22 4626

d k f or

Where [k] is the element stiffness matrix for a beam element with neglected axial effects.

NM Dr P V RAMANA 17



NM Dr P V RAMANA 19

AE AE

3 DOF



3 2 3 2

AE AE0 0 0 0

L L

12EI 6EI 12EI 6EI0 0

L L L L

6EI 4EI 6EI 2EI

1 3

3 DOF

rame: combination of bar

nd beam

2 2

3 2 3 2

2 2

L L L L[k]

AE AE0 0 0 0

L L12EI 6EI 12EI 6EI

0 0L L L L

6EI 2EI 6EI 4EI0 0

2

E, A, I, LQ1 ,

v1

Q3 ,

v2P1 , P2 ,

Q2 ,

1

u1 Q4 ,

2

u2

0000

AE AE

1

23231

6120

6120 u

L

EI

L

EI

L

EI

L

EI L L

P

2

122

2

100

u AE AE

L L L L

P

M

2

2

23232

2 6120

6120

v

L

EI

L

EI

L

EI

L

EI L L

M

Q

22460260 L EI

L EI

L EI

L EI

20

NM Dr P V RAMANA



GJ GJ

L L

3 2 3 20 0

L L L L

3 30 0

L L L L

GJ GJ0 0 0 0

12EI 6EI 12EI 6EI0 0

6EI 2EI 6EI 4EI

2 2L L L L 21

NM Dr P V RAMANA

F d t l C t



Fundamental Concepts

1 2 3 4 5 6, a a a ; , a a a .u x y x y v x y x y

Required Properties of the Approximate Solution:

• Completeness: approximate solution must be able to representtwo special displacement states exactly –

1. Constant strain state: all normal strains and shearing strains have afixed value everywhere in the element.

.

displacement.

NM Dr P V

RAMANA

22

F d t l C t



Fundamental Concepts

Required Properties of the Approximate Solution:• e -pose ness: e num er o s ape unc ons use

in the approximate solution must equal the number of degrees

.

Degrees of freedom are the unknowns in the local problem;

.

E.g., Galerkin and Rayleigh’s method

NM Dr P V RAMANA 23

El t F l ti



Element Formulation

The Constant Strain Triangle element:

• 2D element used in plane

stress and plane strain

roblems

• Nominal thickness = h

(small; can be variable)

• Three corner nodes withcoordinates (xi, yi)

node: (ui, vi)

1 0

NM Dr P V

RAMANA

24 2

12

1 010 0 1

D

Element Formulation



Element Formulation

Three approaches for generating shape

“Interpolation approach”:-

• Works best for small numbers of d.o.f.

“Direct approach”:ore geome r c me o

Works best for higher-order elements

“Area based approach”:

NM Dr P V RAMANA 25

Large structures or nearly volumetric Works best for higher-order elements

Basics:



Basics:

6 d.o.f. total 6 sha e functions

Fundamental unknowns are horizontal dis lacement

u(x,y) and vertical displacement v(x,y)

Each dis lacement ex ected to use 3 sha efunctions

Simple shape functions = better shape functionseasier to inte rate, more widel a licable, …

The derivatives of order n in variational rinci le, itis best to choose the shape functions so that they

can form a com lete ol nomial of order n

NM Dr P V RAMANA 26

(Gives control over errors, faster convergence, …)

Element Formulation



Element Formulation

“Interpolation approach”:st , ,

1 2 3 4 5 6, a a a ; , a a a .u x y x y v x y x y

– At each node, require u(xi,yi) = ui and v(xi,yi) = vi :

1 1 2 1 3 1 1 4 5 1 6 1

2 1 2 2 3 2 2 4 5 2 6 2

.

a a a ; a a a .u x y v x y

3 1 2 3 3 3 3 4 5 3 6 3.

6 e uations for the 6 unknowns!

NM Dr P V

RAMANA

27

1 1 2 1 3 1 1 4 5 1 6 1a a a ; a a a .u x y v x y

Element Formulation



2 1 2 2 3 2 2 4 5 2 6 2a a a ; a a a .u x y v x y Element Formulation3 1 2 3 3 3 3 4 5 3 6 3a a a ; a a a .u x y v x y

“Interpolation approach”:

– Write this in matrix form:

1 1 1 1

1 1 1 2

a

0 0 0 1 a

u x y

v x y

2 2 2 3

2 2 2 4

1 0 0 0 a .0 0 0 1 a

u x yv x y

d C a

3 3 3 5

3 3 3 6

1 0 0 0 a

0 0 0 1 a

u x y

v x y

– Solution (in symbolic form) is 1

.

a C d

NM Dr P V RAMANA 28

Element Formulation



Element Formulation

“Interpolation approach”:1 2 3 4 5 6, a a a ; , a a a .u x y x y v x y x y

– ow, rewr e n erpo a on unc ons n ma r x vec or orm:

1a

2

3

a

, a1 0 0 0u x y x y

4

5

, a0 0 0 1a

v x y x y

–

6a

1

.

N x

NM Dr P V RAMANA 29

Element Formulation



Element Formulation

“Interpolation approach”: – For CST, can show that

1

1

u

v

21 3 5

22 4 6

, , , ,,

, 0 , 0 , 0 ,

uu x y x y x y x y

vv x y N x y N x y N x y

u

3

1 2 2 3 3 2 2 3 3 2 , , 2 ;

v

N x y N x y x y x y x y y y x x A

3 4 3 1 1 3 3 1 1 3 , , N x y N x y x y x y x y y y x x

5 6 1 2 2 1 1 2 2 1

2 ;

, , 2 ;

A

N x y N x y x y x y x y y y x x A

1 1

12 22

1

area of triangle = det 1 .

x y

A x y

3 3

x y

NM Dr P V RAMANA 30

Element Formulation



Element Formulation

Notes on “Interpolation approach”:

– s approac genera zes o eren s apes, eren no e

locations, and different numbers of d.o.f.

– However, the matrix [C] is not always invertible for general

choices of nodal locations.

– As number of d.o.f. increases, matrix inversion becomes more

difficult, and thus exact functions become harder to determine.

NM Dr P V

RAMANA

31

Element Formulation



Element Formulation

“Direct approach”: Need two “facts” about shape functions – – st, ,

1 2 3 1 1 2 2 3 3, a a a , , , ;u x y x y u N x y u N x y u N x y

4 5 6 1 4 2 5 3 6, , , , .v x y x y v x y v x y v x y Shape functions must be linear in both x and y.

– uppose one new e s ape unc ons a rea y:

1 1 2 2 3 3, , , , .u x y u N x y u N x y u N x y

1 1 2 2 3 3, , , ,i i i i i i i i iu u x y u N x y u N x y u N x y

1 2 3 , , .

1 if=

i jKronecker delta ro ert

NM Dr P V RAMANA32

, .0 if

j i i

i j

Element Formulation



Element Formulation

• Visually, this looks like:

NM Dr P V RAMANA 33



Shape functions construction




1 1

2 2 2 3 3 2 2 3 1 3 2 1

11 1 1

1 [( ) ( ) ( ) ]

2 2 2e

x y

x y x y x y y y x x x y P

3 3 x y

Area of triangle Moment matrix

Substitute a1, b1 and c1 back into N1 = a1 + b1x + c1y:

1 2 3 2 3 2 2

1[( )( ) ( )( )]

2 e

N y y x x x x y y

A

35NM Dr P V RAMANA





Similarly,

2 1 1

2 2 2

( , ) 0

( , ) 1

N x y

N x y

2 3 1 1 3 3 1 1 3

1[( ) ( ) ( ) ]

2 e

N x y x y y y x x x y A

2 3 3( , ) 0 N x y 3 1 3 1 3 3

1[( )( ) ( )( )]

2 e

y y x x x x y y A

3 1 1( , ) 0

0

N x y

N x

3 1 2 1 1 1 2 2 11 [( ) ( ) ( ) ]2 e

N x y x y y y x x x y A

3 3 3( , ) 1 N x y 1 2 1 2 1 1

1[( )( ) ( )( )]

2 e

y y x x x x y y A

1 2 3 2 3 2 2

1[( )( ) ( )( )]

2 N y y x x x x y y

NM Dr P V RAMANA 36





i i i i N a b x c y

1

where

2

1

i j k k j

e A

i= 1, 2, 3

21

i j k

e A

J, k determined from cyclic permutation

2i k j

e A

ii = 1, 2

Large polynomials to model bending stress

jk

j = 2, 3k = 3 1

along the other direction of load are poor

Spurious shear stress when bent, strains

NM Dr P V RAMANA 37

Using area coordinates



Using area coordinates

• Alternative method of constructing shape functions

1

1 1

x y

2-3-P:

1 2 2 2 3 3 2 2 3 3 2

3 3

2 21 x y

k , 3 y 11

e

A L

A1 Similarly, 3-1-P A2 22

A L

i, 1

P

1-2-P A3

e

33

A L

38

, 2

xe

NM Dr P V RAMANA

Using area coordinates1

1 1 x y



Using area coordinates 1 11 A x x x x x x

3 3

2 2

1 x y

1 2 3 1 L L L Partitions of unity: 11

e

L A

elta function property: e.g. L1 = 0 at if P at nodes 2 or 3

Therefore,

1 1 2 2 3 3, , N L N L N L

h

NM Dr P V RAMANA 39

, ,e

Strain matrix i i i i N a b x c y 1

( )2

i j k k ja x y x yA



Strain matrix 2e A

u

( )

21

i j k

e

b y y

A

xx x

v

x

2i k j

e A

yy

y

u v

y

y x

y x BdLNdLU

0

321 000 bbb

0 y

B LN N

321 000

2bcbcbc

ccc A

B

40 y x

(constant strain element)NM Dr P V RAMANA

Element matrices



Element matrices

Constant matrix

1 0 E

D

1

2

10 0 1

41NM Dr P V RAMANA



1 2 3 2 3 2 2

1[( )( ) ( )( )]

2 N y y x x x x y y

A



2e

A

2 3 1 1 3 3 1 1 3

1[( ) ( ) ( ) ]

21

e

N x y x y y y x x x y

A

3 1 3 1 3 32 e A

3 1 2 1 1 1 2 2 1

1[( ) ( ) ( ) ] N x y x y y y x x x y

1 2 1 2 1 1

1[( )( ) ( )( )]

2

e

e

y y x x x x y y A

ˆ0u

1 3 5 1 3 5

2 4 6

ˆ ˆ, , 0 , 0 , 0 *( ).

, 0 , 0 , 0 , 0 0

ˆ

u x y N x y N x y N x y u u N N N

v x y N x y N x y N x y

0

u

NM Dr P V RAMANA 43

1 3 5 2 3 3 2 3 1 1 3 1 2 2 1 , , , .



Constant Strain Triangle (CST) : Simplest 2D finite element



v3v1

u3

u1

v2

v(x1,y1)

(x3,y3)

y

u22

(x,y)

x

2, 2

• 3 nodes per element

• 2 dofs per node (each node can move in x- and y- directions)• Hence 6 dofs per element

NM Dr P V

RAMANA

45

The displacement approximation in terms of shape functions is



1 1 2 2 3 3u (x,y) u u u N N N

1

v

u1 1 2 2 3 3v x,y v v v

2321

v

u

000

0 N0 N0 N

xv

y)(x,uu

3

vu

166212 d Nu

321

N0 N0 N0

0 N0 N0 N N

NM Dr P V

RAMANA

46

Formula for the shape functions are

cxba



c xba

yc xba N

A2222

1

v

3

1

u11

(x3,y3)

yc xba N

A2

3333

y

u3v23

(x,y)

vu

where u22 (x2,y2)

11x11

y

x

33

22

x12

y

23132123321

x xcb x xa

x xc y yb y x y xa

12321312213 x xc y yb y x y xa NM Dr P V

RAMANA

47

Properties of the shape functions:



. 1, 2 3

and y inodeat ''1

N

N

nodesother at 0

1

N1

1 13

y 3

1

23

31

21

2

x

NM Dr P V

RAMANA

48

2. At every point in the domain



3

1i

i

3

i1i i

3

i

1i

i

NM Dr P V

RAMANA

49

3. Geometric interpretation of the shape functions

At an oint P x that the sha e functions are evaluated



At an oint P x that the sha e functions are evaluated

A1

1P (x,y)

y 3 A1 A3

A2 22

2

x3

NM Dr P V

RAMANA

50

Approximation of the strains

u



u

v x

y

x

B d

x y u v

y x

321 0y)(x, N

0y)(x, N

0y)(x, N

x x x yc xba 111

321 y)(x, N

0

y)(x, N

0

y)(x, N

0 y y y Bc xba 2

321 000 bbb

x y x y x y A22

332211

321 0002

bcbcbc

ccc A

A N

2

3333

NM Dr P V

RAMANA

51

Inside each element, all components of strain are constant: hence

the name Constant Strain Triangle



g

Element stresses (constant inside each element)

dBD

IMPORTANT NOTE:

1. The displacement field is continuous across element

boundaries. e s ra ns an s resses are con nuous across

element boundaries

NM Dr P V

RAMANA

52

Element stiffness matrix

321

321

000

000

2

1ccc

bbb

B



2

eV k dVBDBT

t 332211 ccc

1 0

Since B is constant

2

1

1 01

0 0 1

D

At k eV BDBdVBDBTT

t=thickness of the element A=surface area of the element

T T

S

eT

b

e

f

S S

f

V

NM Dr P V

RAMANA

53

Element nodal load vector due to body forces



ee

T T

bdA X t dV X f N N

dA X N t

e

e

A b

A

yb

xb

dA X N t f

f

11

1

f b3

f b1

y1

e A a

yb

xbb

dA X N t

dA X N t

f

f f 2

2

2f b3xxf b2y 3

XbXa

e

e

A a

A

b

xbdA X N t

f

f 3

3

3f b2x

2

,

e A b dA X N t 3x

NM Dr P V

RAMANA

54

LINEAR RECTANGULAR ELEMENTS



• Non-constant strain matrix

1 2 3 4

• More accurate representation of stress and strain

• Regular shape makes formulation easy

55NM Dr P V RAMANA



–

f y y z

x y x

x

x

Plane stress Plane strain

56NM Dr P V RAMANA

Rectangular Element:



each other through special points (“nodes”)

py

3v

3x4 3u3

v4 1u

Su

v emen e

1 4u

v2

2

1

u

v

yx

y

2 2v1

3

2

u

vd

x

u

xv

1u1

3

u

v

u 4vNM Dr P V

RAMANA

57

Element Formulation



Three approaches for generating shape


-• Works best for small numbers of d.o.f.

“Direct approach”:ore geome r c me o

Works best for higher-order elements

“Area based approach”:

NM Dr P V RAMANA 58Large structures or nearly volumetric Works best for higher-order elements

Basics:

8 d.o.f. total 8 sha e functions3

u3

v3

v



Fundamental unknowns are 4u

v4

v2

horizontal displacement u(x,y) and

vertical dis lacement v x,

y

2u2v1

Each displacement expected to

use 4 sha e functionsx

v

u

1u1

Sim le sha e functions = better sha e functions

(easier to integrate, more widely applicable, …)

The derivatives of order n in variational principle, it

is best to choose the sha e functions so that the

NM Dr P V RAMANA 59can form a complete polynomial of order n(Gives control over errors, faster convergence, …)

Element Formulation



“Interpolation approach”:st , ,

xya ya xaa y xv xya ya xaa y xu 87654321 ),(;),(

– At each node, require u(xi,yi) = ui and v(xi,yi) = vi :

xaa xaav xaa xaau

228272652224232212 ; y xa ya xaav y xa ya xaau

448474654444434214

338373653334333213

;

;

y xa ya xaav y xa ya xaau


NM Dr P V

RAMANA

60

Element Formulation; yxayaxaavyxayaxaau



228272652224232212

118171651114131211

;

;




– Write this in matrix form:448474654444434214

338373653334333213

; y xa ya xaav y xa ya xaau

– Solution (in symbolic form) is 1

.a C d

• Non-constant strain matrix

• More accurate representation of stress and strain

•

NM Dr P V RAMANA 61

Element Formulation

)()(



“Interpolation approach”: xya ya xaa y xv xya ya xaa y xu87654321

),(;),(

– ow, rewr e n erpo a on unc ons n ma r x vec or orm:

1

.

xya ya xaa y xv xya ya xaa y xu ),(;),( 87654321

aP y xu xu ),()(

–1

,

.

N x

NM Dr P V RAMANA 62



Consider a rectangular element

, v

3 ( x3, y3)4 ( x4, y4)

1 displacements at node 1u

v

u3, v3

sy

sx

4, 4

2b 2

2 displacements at node 2

e

u

u

d

1 ( x1, y1) 2 ( x2, y2)

2a3

3

4

displacements at node 3

dis lacements at node 4

u

u

, u

, ,4u

63NM Dr P V RAMANA



, v

3 ( x3, y3)4 ( x4, y4)

,

(u3, v3)

, +

(u4, v4)

3, 3

sy

sx

4, 4

2b

2b

1 ( x1, y1)(u1, v1)

2 ( x2, y2) (u2, v2)

1 (1, 1)

u v

2 (1, 1)

u2, v2

2a

, u

, b ya x

( , ) ( , )h

e x y x yU N d

31 2 4

31 2 4

00 0 0

00 0 0

N N N N

N N N N

N

where

NM Dr P V RAMANA 64

(Interpolation)

Summary: For each element

Rectangular Element:

Displacement approximation in terms of shape functions



Displacement approximation in terms of shape functions

d Nu

Strain approximation in terms of strain-displacement matrix

Stress approximation

dBε

dBD

e

V

k dVBDBT

Element nodal load vector

T T

S

eT

b

e

f

S

f

V

NM Dr P V

RAMANA

65

Lagrange family

Serendipity family

Second Approach



4-noded rectangular element with edges parallel to the coordinate axes:,

4, 4

4

uxxu N

(x,y)

vu 2b

1i

4

y

(x2,y2)1 22a

1

i,,

i

i

x

,

• 4 nodes per element

• 2 dofs per node (each node can move in x- and y- directions)

•8 dofs per element NM Dr P V RAMANA 66

Generation of N1:2 x x

1xx



y 21 x x

has the property

34l1(y)

0)(

1)(

21

11

xl

xl

2b Similarly

41

41 )(

y y

y y yl

x

1 2

2a

N1

has the property1)( 11 yl

l1(x)

10)( 41 yl

Hence choose the shape function at node 1 a

4242

111

1)()( y y x x

y y x x yl xl N

4121 a y y x x

NM Dr P V RAMANA 67



Second Approach

b ya xab

y y x xab

N 4

1

4

1421

by

a xyyxxN



Lagrange familySerendi it famil 11 b ya x

b y

ab

y y x x

ab

N

144

312

Lagrange family

1144243

ab y y

ab

b ya x

y y x x N

11

134

4-noded rectangle

y

In local coordinate system

a a

12

b b xaab

N

41

y

l1(

x

b

yb xa

ab

))((

42

2b

y

3 4

yb xa N

ab

))((

43

1 22a

N1 1

ab4NM Dr P V RAMANA 69l1(x)1

1 Polynomial completeness



1. Polynomial completeness

Convergencerate dis lacement

y x4 node; p=2

3223 y xy y x x

y xy x9 node; p=3

54322345

432234

x x x x x y xy y x y x x

La ran e sha e functions contain hi her order terms but miss out lowerorder terms

NM Dr P V RAMANA 70

Properties of the shape functions:



1. The shape functions N1, N2 , N3 and N4 are bilinearfunctions of x and

2. Kronecker delta property

nodesother at

inodeat y x

0

''1),( N i

3. Completeness

y

a a 12

4

1i

i 1 N x

x x i

4

1i

i N 3 4

y y i 1i

i N NM Dr P V RAMANA 71

3. Along lines parallel to the x- or y-axes, the shape functions

are linear. But along any other line they are nonlinear.



4. An element shape function related to a specific nodal point

point.

5. The displacement field is continuous across elements

. e s ra ns an s resses are no cons an w n an

element nor are they continuous across element boundaries.

y

a a12

x

b

NM Dr P V RAMANA 723 4

The strain-displacement relationship

4214

1 y y x x

ab N

34

2

l1(y)

x

3124

1 y y x x

ab N b

N

1



xy

y

2434

1

4

y y x xab

N

ab

x1 22

N1

1

1

1

4321

v

u

y)(x,y)(x, Ny)(x,y)(x, N

1344

1 y y x x

ab N a

1

1

3

2

2

4321

u

v

u

y)(x, N0

y)(x, N0

y)(x, N0

y)(x, N0

y y y y

x x x x

4

3

B

44332211

u

vy)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, N

x y x y x y x y

4

y y y y y y y y 1234 0000

y y x x y y x x y y x x y y x x

x x x x x x x xab

B

13243142

3412 00004

o ce a e s ra ns an ence s resses are cons an w n an

element NM Dr P V RAMANA 73

Computation of the terms in the stiffness matrix of 2D elements (recap)



The B-matrix (strain-displacement) corresponding to this element is34v4 v3

uu4

y

v

1 2 3 4 N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0

u1 v1u2 v2

u3 u4v3 v4

(x,y) u

1 2

2v1

u1u2

1 2 3 4

1 1 2 2 3 3 4 4

N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0

N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y)

y y y y

y x y x y x y x

Denote the columns of the B-matrix as

x

1 N (x,y)0

N x x

1 1

1

1

; ; an so on...

N (x,y) N (x,y)

u v y

y

x

NM Dr P V RAMANA 74

The stiffness matrix corresponding to this element is

eV

k dVBDB which has the following form



V

u1 v1u2 v2 u3

u4v3v4

1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8

2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8

k k k k k k k k

k k k k k k k k

u1

v1

3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8

4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8

k k k k k k k k

k k k k k k k k k

u2

v2

5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8

6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8k k k k k k k k k k k k k k k k

u3

u4

v3

8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8k k k k k k k k v4

1 1 1 1 1 2

T T T

11 12 13B D B dV; B D B dV; B D B dV,...e e eu u u v u u

V V V k k k

1 1 1 1

T T

21 21B D B dV; B D B dV;.....e ev u v v

V V

k k

NM Dr P V RAMANA 75



76NM Dr P V RAMANA

NM Lecture 6 on 15th Sept 2015.pdf

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Transcript of NM Lecture 6 on 15th Sept 2015.pdf