FEM Notes5

8/12/2019 FEM Notes5

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FINITE ELEMENT METHOD BASIC DEFINITIONS

1. DISCRETIZATION PROCESS:The various considerations to be taken in the discretization process are as given below:1.1 Type of Elements or Various Elements Shapes or Element Library: The discretization of thestructure or body into finite elements forms the first step in finite element analysis of a complicatedstructural system. These elements coincide with the geometry of the structure and represent the geometry

and mechanical properties in the regions. Therefore, the first decision the engineer must make is to selectthe shape of the basic elements to be used in the analysis. The selection of finite element in FEM isdepends on the following factors. They are:

. !imension of the problem.". #hape of the body or continuum.$. %ccuracy of the results re&uired.'. (ost and time for the analysis.Fig. . shows some typical elements used to discretize the continuum.

" $ "(a) 2-noded e e!en" (#) $-noded e e!en"

(a) One d%!en&%ona #a' e e!en"

(a) (#) ( ) (d)

(#) T o d%!en&%ona e e!en"&

( ) T*'ee d%!en&%ona e e!en"&

F%+. 1.1: T, % a e e!en"&

1D Elements: )hen the geometry, material properties and field variable of the problem is described in

terms of only one spatial coordinate *for e+ample + coordinate-, then line elements shown in Fig. *a- canbe used. % one dimensional element may be represented by a straight line whose ends are nodal pointsas shown in Fig. *a-. These nodal points, numbered and " are called external nodes because theyrepresent connecting points to the ad/acent elements. #ome applications re&uire additional nodal pointsuch as node $ as shown in Fig. a*b-. 0ecause no connection to other elements occurs at this node, it iscalled internal node.2D Elements: )hen the configuration and other parameters of the problems are described in terms of twospatial coordinates, then two dimensional elements are used. Triangular or &uadrilateral elements *Fig. b-are used for two dimensional problems such as plane stress or plain strain or plate bending problems.The corner nodes of these elements are called primary external nodes and additional nodes occur on thesides of the elements, like ', 1 and 2 in 2 nodded triangular element or 1, 2, 3 and 4 in 4 noddedcurved element are called secondary external nodes. #ome times internal node such as 3 as shown inFig. b *b- is also used for these elements.

3D Elements: )hen the geometry, material properties and other parameters of the problems aredescribed by three spatial coordinates, then three dimensional elements shown in Fig. *c- are used. For $ dimensional problems, tetrahedron or he+ahedron elements are used. % tetrahedron has ' primarye+ternal nodes and he+ahedron has 4 primary e+ternal nodes.

65

1 2

3

456

1

2

3

4

5

67

8


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1.2 Size of Elements: The size of elements influences the convergence of the solution directly and henceit has to be chosen with care. 5f the size of the elements is small, the final solution is e+pected to be moreaccurate. 6owever, the use of elements of smaller size will take more computational time. #ometimes, wemay have to use elements of different sizes in the same body. For e+ample, in the case of stress analysisof the plate in Figure ."*a-, the size of all the elements can be appro+imately the same, as shown inFigure ."*b-. 6owever, in the case of stress analysis of a plate with a hole shown in Figure ."*c-,elements of different sizes have to be used, as shown in Figure ."*d-. The size of elements has to be very

small near the hole *where stress concentration is e+pected- compared to far away places. 5n general,whenever steep gradients of the field variable are e+pected, we have to use a, finer mesh in those regions. %nother characteristic related to the size of elements that affects the finite element solution is the aspectratio of the elements. The aspect ratio describes the shape of the element in the assemblage of elements.For two dimensional elements, the aspect ratio is taken as the ratio of the largest dimension of the elementto the smallest dimension. Elements with an aspect ratio of nearly unity generally yield best results.

F%+. 1.2( ) O'%+%na S"' " 'e F%+. 1.2(d) F%n%"e E e!en" D%& 'e"%/a"%on

1.3 Location of No es: 5f the body has no abrupt changes in geometry, material properties, and e+ternalconditions *e.g., load and temperature-, the body can be divided into e&ual subdivisions and hence thespacing of the nodes can be uniform. 7n the other hand, if there are any discontinuities in the problem,nodes have to be introduced at these discontinuities, as shown in Fig .$

(a) D%& on"%n %", %n Load%n+

(#) D%& on"%n %", %n 0eo!e"',

( ) D%& on"%n %", %n Ma"e'%a P'o e'"%e&

66

F%+.1.2(#) F%n%"e E e!en" D%& 'e"%/a"%onF%+.1.2(a) O'%+%na S"' " 'e


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2. DISPLACEMENT OR INTERPOLATION F NCTION:

The basic principle of FEM is to sub divide the solution domain into number of finite elements andrepresenting the solution within each element by relatively simple function. This simple function is called8displacement function or interpolation function 9. The polynomial function is commonly used for thisdisplacement model in FEM due the following two reasons.

. The polynomials are easier for the mathematical operation i.e. the differentiation and integration ofpolynomials are easy as compared to trigonometric functions.". % polynomial of arbitrary order permits us to have a re&uired appro+imation i.e by selecting the order ofpolynomial we can vary the degree of appro+imation.The degree of appro+imation is mainly depends on the order of the polynomial selected for thedisplacement model as shown in Fig. ". 5n this Fig." an e+act solution for the displacement )( xu for asimple one dimensional problem is appro+imated by the following degree of polynomials of the form,

nn x x x xu 1

2321)( +++++= *a-

#imilarly, for two and three dimensional problems, the polynomial form of interpolation functions can bee+pressed as

nm y y xy x y x y xu +++++++= 26524321),( *b-

nm z z y x z y x z y xu +++++++= 2726254321),,( *c-

The greater the number of terms included in the polynomial, the solution is more close to the e+act solutionas shown in Fig. "*a- to "*c-. The coefficient of the polynomial, the 9s in E&n. *a- are called generalizedcoordinates .

(a) Con&"an" o ,no!%a (#) L%nea' o ,no!%a

( ) ad'a"% o ,no!%a

F%+. 2: Po ,no!%a a 'o3%!a"%on %n one d%!en&%on

5n most of the cases, the order of the polynomial in interpolation function is taken as one, two or three.Thus, the above e&uations reduce to te following form for various cases.!or Linear "o el #i.e. n$1%For ! problem: x xu 21)( += For "! problem: y x y xu 321),( ++=For $! problem: z y x z y xu 4321),,( +++=!or &ua ratic "o el #n$2%

For ! problem:2

321)( x x xu ++=For "! problem: 265

24321),( y xy x y x y xu +++++=

For $! problem: zx yz xy z y x z y x z y xu 10982

72

62

54321),,( +++++++++=67


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$. NODAL DE0REES OF FREEDOM:

F%+. 4: 5 nodded "'%an+ a' e e!en"

The nodal displacement, rotation and or strains necessary to specify completely the deformation of theelement are referred as degrees of freedom of the element. The nodal degrees of freedom are the nodaldisplacement, rotations and or strains corresponding to the e+ternal nodes of an element. 7n the otherhand, the internal degrees of freedom are corresponds to the internal node of an element.For e+ample, in a triangular element shown in Fig. ', the nodes , " and $ are the primary e+ternal nodesand nodes ', 1 and 2 are secondary e+ternal nodes. The node 3 is an internal node. The displacementscorresponding to these primary and secondary e+ternal nodes to 2 are called nodal degrees of freedomand the displacement corresponding to the internal node 3 is called internal degrees of freedom .

4 SELECTION OF THE ORDER OF THE INTERPOLATION POL6NOMIAL: )hile selecting the order ofthe polynomial in interpolation function, the following considerations have to be taken into account:

. The interpolation polynomial should satisfy, as far as possible, the convergence re&uirements.". The pattern of variation of field variable resulting from the polynomial model should be independent of

local coordinate system.$. The number of generalized coordinates * 9s- should be e&ual to the number of nodal degrees of

freedom of the element.

7. SIMPLE89 COMPLE8 AND M LTIPLE8 ELEMENTS: Finite elements can be classified into threecategories as simple+, comple+, and multiple+ elements depending on the geometry of the element andthe order of the polynomial used in the interpolation function.Simple' Elements: The simple+ elements are those for which the appro+imating polynomial consists ofconstant and linear terms. Thus, the following polynomials represent the simple+ functions for one , two ,

and three dimensional elements.For ! problem: x xu 21)( += *a-For "! problem: y x y xu 321),( ++= - *b-For "! problem: z y x z y xu 4321),,( +++= *c-The simple+ element in one dimension is a " noded bar element and for two dimensions is a triangle withthree nodes *corners-.(omple' Elements: The comple+ elements are those for which the appro+imating polynomial consists of&uadratic, cubic, and higher order terms, according to the need, in addition to the constant and linearterms. Thus, the following polynomials denote comple+ functions.For ! problem: 2321)( x x xu ++= *d-For "! problem: 265

24321),( y xy x y x y xu +++++= *e-

For $! problem: zx yz xy z y x z y x z y xu 10982726254321),,( +++++++++=The comple+ elements may have the same shape as the simple+ elements but they have additionale+ternal and internal nodes. For e+ample, e&. *e- represents the &uadratic interpolation function for a twodimensional element. #ince this e&uation has 2 9s, the corresponding element should have si+ nodes andsi+ noded triangular element forms a comple+ element.

(a) Fo' 1 D 'o# e! (#) Fo' 2 D 'o# e!F%+ (a) Co! e3 E e!en"&"ultiple' Elements: Multiple+ elements are those whose boundaries are parallel to the coordinate a+es toachieve interelement continuity as shown in Fig. *b- and whose polynomials contain higher order terms.

68

1 2

3

4

567

1 2

3

4

561 23

y


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F%+. (#) M "% e3 E e!en"

. H%+*e' O'de' E e!en"&:5n certain applications, additional degrees of freedom behind the minimum number may be included forany element. This can be achieved by adding secondary e+ternal nodes. Then, the elements with theseadditional degrees of freedom are called as higher order elements. Fig. 1 shows the higher order elementsused in ! and " ! problems.

(a) Fo' 1 D 'o# e! (#) Fo' 2 D 'o# e!

F%+. 7: H%+*e' o'de' e e!en"&

5. CON;ER0ENCE RE IREMENTS: 5n FEM se&uence of appro+imate solutions are obtained as theelement size is reduced successively. These solutions are converge to the e+act solutions if theinterpolation polynomial satisfies the following convergence re&uirements:

1. The displacement model must be continues within the element and the displacement must becompatible between ad acent elements! The condition can be satisfied by choosing polynomials which are

inherently continues functions for the displacement model. The second condition indicates that when theelements deforms there should not be any discontinuity between the elements *elements should notoverlap or separate-.

". The displacement model must be capable of representing rigid body displacement of the element! Thiscondition states that when the nodes are given such displacements corresponding to a rigid body motion,the element should not e+perience any strains and hence leads zero nodal forces. The constant termpresent in the polynomial functions satisfy this condition.

#. The displacement model must be capable of representing constant strain state within the element! )henthe elements are made smaller and smaller size, the strain in each elements approach constant value. 5nsuch case, the assumed displacement model should include the terms to represent the constant strainstate within the element. For one, two and three dimensional elasticity problems, the linear terms *i.e.

2 - present in the polynomial functions satisfy this re&uirement.5n order to satisfy the above mentioned three re&uirements, the displacement model must contain at leastconstant and linear terms in the polynomials.


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F%+. 5: Pa& a "'%an+ e =o' 2 D 'o# e!&

For the polynomial functions used in displacement model to be balanced one, it should notincludes any terms from one side of the a+is of the symmetry of the triangle without including itscounter part from the other side. For e+ample, if we wish to construct a cubic model with 4 terms,the following are the geometrically isotropic.

. %ll the constant, linear and &uadratic terms plus 3 x and 3 y .i.e. 38

37

265

24321),( y x y xy x y x y xu +++++++=

". %ll the constant, linear and &uadratic terms plus y x 2 and 2 xy .i.e. 28

27

265

24321),( xy y x y xy x y x y xu +++++++=

The displacement model formed by any of the above two balanced polynomials will have thegeometric isotropy.

>. Coo'd%na"e S,&"e!:

The following coordinate systems are used in FEM.. ;lobal coordinate system

".


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e&uations derived in terms of local coordinate are to be transformed to the common globalcoordinate system.

F%+.


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". They simplify the integration and differentiation, which are re&uired to obtain the elementstiffness matri+ and load vectors.

A'ea oo'd%na"e: 5f any point >*+,y- divides the total area of the triangular element % into % , %"and % $ as shown in Fig. ?, then areas % , %" and % $ are called area coordinates. These are alsonatural coordinates and these can be used to e+press the natural coordinates 321 &, L L L as

A

A L

A

A L

A

A L 33

22

11 and;; === F%+.>

1?. S*a e F n "%on o' In"e' o a"%on F n "%on:

5n FEM, it is necessary in many cases to deal with functions whose analytical form is totallyunknown. 5n such case, the given function is replaced by another function which can be moreeasily handled. This operation of replacing a given function by simple one is known as)nterpolation or shape function.#hape function is a function used to interpolate nodal variables to field variables at any pointinside the element. @sing these shape functions, the field variable can be e+pressed in terms of

nodal variables asnn u N u N u N u ++= 2211

where n is the number of nodes in a element.)roperties of Shape functions:

The shape functions, which are used to interpolate nodal variable to field variable, are having thefollowing properties

. The shape functions are having unit value at one nodal point and zero value at all other nodalpoints For e+ample, in a ! bar element having nodal points ),( ji , the above mentionedproperty of shape functions can be written in mathematical form as

%t node i , 1=i N 0= j N %t node j , 0=i N 1= j N

". The sum of the shape functions is e&ual to unity.i.e 1=+ ji N N

$. The derivatives of a shape function is constant

i.e. ==dx

dN

dxdN ji (onstant

11. De'%@a"%on o= E3 'e&&%on& =o' S*a e F n "%on&:

(a) Fo' a 2-noded #a' e e!en":

72

i j

i j

1+ 1+i N j N

1 2

3

),,( 321 L L L P 1 A2 A

3 A


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*i+ )n terms of natural coordinate!

Fig. shows a two nodded bar element. The values of natural coordinates corresponding to node and " are as shown in Fig.


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Fig. shows a two nodded bar element. The values of cartesian coordinates corresponding tonode and " are as shown in Fig.


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No"e: -tudents are ad$ised to practice the deri$ation of shape functions either in terms ofcartesian or natural coordinate system and to remember the expressions for shape functions interms of both coordinates.

(#) Fo' a $noded #a' e e!en" ( ad'a"% #a' e e!en"):

Fig. shows a ! three nodded bar element. The values of natural coordinates correspondingto node , " and $ are as shown in Fig.


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[ ]

=

3

2

12

121

21

021

21

100

1

u

u

u

u

( )

+

+=

3

2

1222 1

2

1

2

1

2

1

2

1

u

uu

u

[ ]=3

2

1

321

u

u

u

N N N u

where,

)1(2

1

2

1

2

1 21 =

+= N

)1(21

21

21 2

2 +=

+= N

( 23 1 = N are the shape functions for the three nodded bar element and are varies over the length ofelement as shown below.

( ) Fo' a $noded "'%an+ a' e e!en" (CST e e!en"):

*i+ )n terms of natural coordinates!

Fig. shows a $ noded triangular element. The values of natural coordinates corresponding tonode , " and $ are as shown in Fig.


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1u , 2u and 3u respectively. Then the field variable *i.e. displacement- u at any point insidethe element can be e+pressed in natural coordinate as

321 ++=u *a-

[ ]=3

2

1

1

u *b-

#ubstitute the values of and corresponding to node , " and $ into E&n. *a-, we get thenodal displacementsi.e. at node , 0;1 == and 1uu =

211 += u * -at node ", 1;0 == and 2uu =

312 += u *"-at node $, 0;0 == and 3uu =

13 = u

31 u= *i-#ubstitute into E&n. * -, we get

231 += uu312 uu = *ii-

#imilarly, substitute into E&n. *"-, we get323 uu = *iii-

E&ns. *i- to *iii- can be written in matri+ form as

=

3

2

1

3

2

1

110

101

100

u

u

u

#ubstitute into E&n. *b-, we get

[ ]=

3

2

1

110

101

100

1

u

u

u

u A[ ]3

2

1

)1(

u

u

u

[ ]=3

2

1

321

u

u

u

N N N u

where, == 21 ; N N and =13 N are the shape functions for a three nodded triangular*(#T- element.

*ii+ )n terms of cartesian coordinate!

77

),( y x P

),( 11 y x

2

3

),( 22 y x

),( 33 y x

x

y


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Fig. shows a $ noded triangular element. The values of natural coordinates corresponding tonode , " and $ are as shown in Fig.


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=3

2

1

321

321

321

3

2

1

2

1

u

u

u

ccc

bbb

aaa

Ae

*e-

substitute E&n. *e- into E&n. *b-, we get

[ ]=3

2

1

321

321

321

211

u

u

u

ccc

bbb

aaa

A y xu

e

++++++=3

2

1

333222111 )(21

)(2

1)(

21

u

u

u

yc xba A

yc xba A

yc xba A

ueee

[ ] 3322113

2

1

321 u N u N u N

u

u

u

N N N u ++==

where shape functions )(2

1 yc xba

A N iii

ei ++= 3,2,1( =i -

(d) Fo' 4-noded Te"'a*ed'on E e!en":

Fig. shows a ' noded tetrahedron element. The values of natural coordinates corresponding tonode , ",$ and ' are as shown in Fig.


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=

4

3

2

1

444

333

222

111

4

3

2

1

1

1

1

1

z y x

z y x

z y x

z y x

uu

u

u

or { } [ ]{ } Aq ={ } [ ] { }q A 1= *c-

The inversion of matri+ ( is given by

=

4321

4321

4321

4321

1

61

d d d d

cccc

bbbb

aaaa

V A

e*d-

where,

444

333

222

111

11

1

1

6

1

z y x z y x

z y x

z y x

V e = B Colume of element

5n general,

444

333

222

111 1

z y x

z y x

z y x

M a +== B44

33

22

121

1

1

1

1

z y

z y

z y

M b == B

44

33

22

131

1

1

1

1

z x

z x

z x

M c +== B44

33

22

141

1

1

1

1

y x

y x

y x

M d ==

where, 4,3,2,1 are in cyclic order.

#ubstitute E&n. *d- into *c-, we get

=

4

3

2

1

4321

4321

4321

4321

4

3

2

1

61

u

u

u

u

d d d d

cccc

bbbb

aaaa

V e

*e-

#ubstitute E&n. *e- into E&n. *b-, we get

[ ]=4

32

1

4321

43214321

4321

611

u

u

u

u

d d d d

cccc

bbbb

aaaa

V z y xu

e

[ ] 443322114

3

2

1

4321 u N u N u N u N

uu

u

u

N N N N u +++==

where shape functions )(61

z d yc xbaV N iiiiei +++= 4,3,2,1( =i -

(e) Fo' a 4-noded ad'% a"e'a e e!en":

80


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Fig. shows a " !, ' noded &uadrilateral element. The values of natural coordinatescorresponding to node , ", $ and ' are as shown in Fig.


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#ubstitute for 42 from E&n. *3-, we get( ) 212212234 422 uuuuuu +=+=

( )43212 41

uuuu ++= *ii-#ubstitute into E&n. *3-, we get

( ) 2143214 41

*22 uuuuuu +++= ( )43214 4

1uuuu += *iv-

E&ns. *i- to *iv- can be written in matri+ form as

=

4

3

2

1

4

3

2

1

1111

1111

1111

1111

4

1

u

u

u

u


[ ]

=

4

3

2

1

1111

1111

1111

1111

41

1

u

u

u

u

u

( ) ( ) ( ) ( )++++++=

4

3

2

1

141

141

141

141

u

u

u

u

u

[ ]=

4

3

2

1

4321

u

u

u

u

N N N N u

where, ( ) ( )( ) =+= 1141

141

1 N

( )( ) += 1141

2 N

( )( ) ++= 1141

3 N

( )( ) += 1141

4 N

Note!5n general, the above shape functions for a ' noded &uadrilateral element can be written as

( )( )iii N ++= 1141

B 4,3,2,1=i

where, i and i are the values of natural coordinates corresponding to node * 4,3,2,1=i -.For example at node1*i.e. 1=i + 1,1 11 == then shape function

82


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( )( ) ( )( ) =++= 1141

)1(1)1(141

1 N

-imilarly at node " *i.e. 2=i + 1,1 22 =+= then shape function( )( ) += 11

41

2 N so on.

For a "/noded bar element the general expression for shape functions can also be written as

( )ii N += 121 0 2,1=i

at node 1 11 = therefore ( ) = 121

1 N and at node " 12 += therefore ( ) += 121

2 N

as we obtained in section *a+. #f% !or a 2*no e beam element:

(onsider a typical " noded beam element as shown in Fig. %t each node, the unknowns aredisplacement w and slop and hence there are " d.o.f node and ' d.o.f element. Therefore,the deformation of beam element at any point can be e+pressed as

34

2321 x x xw +++= *a-

and slop is given by differentiating w w.r.t x2

432 32 x xdx

dw ++== *b-

E&n. *a- can also be written in matri+ form as

[ ]=4

3

2

1

321

x x xw *c-


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=

2323

221

1212

13230010

0001

eeee

eeee

l l l l

l l l l A

#ubstitution of 1 A into E&n.*d- yields

=

2

2

1

1

2323

22

4

3

2

1

1212

13230010

0001

w

w

l l l l

l l l l

eeee

eeee

#ubstitute above E&n. into E&n. *c-, we get

[ ]

=

2

2

1

1

2323

2232

1212

13230010

0001

1

w

w

l l l l

l l l l x x xw

eeee

eeee

[ ] 242312112

2

1

1

4321

N w N N w N w

w

N N N N w +++==

where,

3

3

2

2

123

1ee l

x

l

x N += B 2

32

22

ee l

x

l

x x N +=

3

3

2

2

323

ee l

x

l

x N = B 2

32

4ee l

x

l

x N +=

No"e: The 1

A matrix can be written from the shape function expression nowing that elementsin 1 st " nd #rd and 2 th row are respecti$ely the coefficients of 210 ,, x x x and 3 x terms of shapefunctions.

i.e.

=

2323

221

1212

13230010

0001

eeee

eeee

l l l l

l l l l A

84

C e!!icien"s ! "e#$ ! s%a&e !'nc"i ns





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12. La+'an+e In"e' o a"%on F n "%on o' La+'an+e o ,no!%a &:


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Fig. shows a $ noded bar element. The values of natural coordinates corresponding to node , "and $ are as shown in Fig. The shape functions 1 N , 2 N and 3 N for a $ node bar elementcan be derived using


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)1(21

)()( 23 +== L L )1(23 +==

For node and ', the


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)()()( 597 L L L == 597 == and

)()()( 152 L L L == 152 ==)()()( 896 L L L == 896 ==)()()( 473 L L L == 473 ==

Then shape functions are obtained as

)1)(1(41

)()( 111 == L L N B )1)(1(41

)()( 222 +== L L N

)1)(1(41

)()( 333 ++== L L N B )1)(1(41

)()( 444 +== L L N

)1)(1(21

)()( 2555 == L L N B )1)(1(21

)()( 2666 +== L L N

)1)(1(21

)()( 2777 +== L L N B )1)(1(21

)()( 2888 == L L N

)1)(1()()( 22999 == L L N

1$. La+'an+e and Se'end% %", e e!en"&:La+ran+e elements:

The shape functions of a ! bar element can be easily derived using


25/65

FEM Dr. T.N

14. De'%@a"%on o= S"'a%n-d%& a e!en" !a"'%3 and S"'e&& !a"'%3

*a+ For a "/noded bar element!

Fig. shows a two nodded bar element. The values of cartesian coordinates corresponding tonode and " are as shown in Fig.


26/65

FEM Dr. T.N

[ ] =ee

e

l l B

11

5t may be noted that, the strain displacement matri+ 6 is depends only on length of elementwhich is constant. 6ence, strain displacement matri+ 6 for a " noded bar element is constant.

*2+ -tress/strain relation! For a ! element, the stress strain relation can be e+pressed ase xe

e x E = .

#ubstitute for e x from E&n. *c-, we get

[ ] { }eeee x q B E = is the stress matri+.

*b+ For a #/noded triangular *,.-.T+ element!

Fig. shows a $ noded triangular element. The values of cartesian coordinates corresponding tonode , " and $ are as shown in Fig.


27/65

FEM Dr. T.N

*#+ -train/displacement relations! for a " ! problem, the strain displacement relation ise+pressed as

xu

x = B

xv

y = and

xv

yu

xy +

=

!ifferentiate E&ns. *b- w.r.t + and y and substitute into the above e&uation, we get

33

22

11 u

x

N u

x

N u

x

N

x

u x

+

+

=

=

33

22

11 v

y N

v y

N v

y N

yv

y +

+

=

= and

33

22

11

33

22

11 v

x N

v x

N v

x N

u y

N u

y N

u y

N xv

yu

xy +

+

+

+

+

=

+

=

The above e&uation can be written in matri+ form as

{ }

==

3

3

2

2

1

1

332211

321

321

000

000

v

uv

u

v

u

x

N

y

N

x

N

y

N

x

N

y

N y

N

y

N

y

N x

N

x

N

x

N

xy

y

x

{ } [ ] { }eee q B= is the strain matri+. *c-

)here, [ ]

=

x

N

y

N

x

N

y

N

x

N

y

N y

N

y

N

y

N x

N

x

N

x

N

B e

332211

321

321

000

000

is strain displacement matri+.

5n general differentiation of shape functions )3,2,1( =i N i w.r.t + and y yields

ie

i b A x

N 2

1=

and i

e

i c A y

N

21=

. 6ence, substitution of the above derivatives into the above

e+pression yields

[ ]=332211

321

321

000

000

2

1

bcbcbc

ccc

bbb

A

B

e

e is the strain displacement matri+ for $ noded triangular

element.5t may be noted that strain displacement matri+ for a $ noded triangular element is only dependson nodal coordinates which are constant for the given element and hence it is constant.Therefore, $ noded triangular element is called as ,onstant strain Triangular *,.-.T+ element.

*2+ -tress/strain relation! for " ! problem, the stress strain relation is e+pressed as{ } [ ] { }eee D = , where, [ ]e D constitutive relation which relate the strain with stress.#ubstitute for { }e from e&n. *c-, we get{ } [ ] [ ] { }eeee q B D= is the stress matri+

Note! For "/D plain stress and plain strain problems the constituti$e matrix [ ] D in the abo$ee3uation is gi$en by

91


28/65

FEM Dr. T.N

[ ]

=

21

00

01

01

)1( 2

E

D for plain stress problems

[ ]

+=

221

00

01

01

)21)(1(

E D for plain strain problems

*c+ For a #/noded bar element or 1/D 7uadratic element!

Fig. shows a $ noded bar element. The values of natural coordinates corresponding to node , "and $ are as shown in Fig. The strain displacement, strain and stress matrices can be derivedusing the following steps.*1+ -hape functions! The shape functions for a $ noded bar * ! &uadratic- element can bee+pressed in natural coordinate system as

)1(21

1 = N , )1(21

2 += N and )1( 21 = N *a-

where,el

x x )(2 3= *b-

*"+ Field $ariable! 5s e+pressed as

[ ]=++=3

2

1

321332211

u

u

u

N N N u N u N u N u *c-

*#+ -train/displacement relation! 5s e+pressed as

xu

x = . !ifferentiate E&n. *c- w.r.t + , we get

33

22

11 u

x

N u

x

N u

x

N

xu

x +

+=

=

[ ] { }eee x q Bu

u

u

x N

x N

x N =

=

3

2

1321

where, [ ]

=

x

N

x N

x N

B e 321 *d-

is strain displacement matri+. %s the shape functions )3,2,1( =i N i are functions of , their derivatives w.r.t + can be writtenas

x N

x N ii

=

.

6ence, diff. E&n. *b- w.r.t + we get

92

1 2

11 = 12 += 3

03 =

1u 3u 2u

el


29/65

FEM Dr. T.N

ee l l x2)01(2 ==

. Therefore, above e&uation becomes as

=

ie

i N l x

N 2)3,2,1( =i

!ifferentiate E&n.*a- w.r.t and substitute we get

== )12(

2122 11

ee l N

l x N

+=

=

)12(2

122 22 ee l

N

l x

N

( )

222 33 =

=

ee l

N

l x

N

substitute into e&n. *d-, we get

[ ] += 2)12(21

)12(212

e

e

l B

17. De'%@a"%on o= S"%==ne&& !a"'%3 and Load ;e "o'&

#a% !or a 2*no e bar element:

(onsider a " noded bar element as shown in Fig. The elemental stiffness matri+ and loadvectors can be derived from strain energy and work potential of total potential energy functional.For an the element, the total potential energy can be written as

= s

T

V

T

V

T dsT udV f udV 21 *a-

where, first part of the above e&uation gives strain energy and second and third parts gives workpotential. @sing strain energy and work potential, the elemental stiffness and load vectors can bederived for the given element as follows:

#a% Stiffness "atri': (onsider strain energy functional of total potential energy and is e+pressed as

eV

T e dV e

= 21 *b-0ut, the stress strain relation yields

D= *i-where, matri+ D is called constitutive matri+. The strain displacement relation yields

Bq= *ii-#ubstitution of into the above e&uation yields

DBq= D Bq T T T = * D D T = as D matri+ is symmetric-

#ubstitution of T and into E&n. *b- yields

( ) q ! qq DBdV BqdV DBq Bq eT V eT T V eT T e ee 21

21

21

=

==

where, e ! is elemental stiffness matrix and is e+pressed as

93

1 2

1 x x = 2 x x =

1u 2u

)( x P

el

e


30/65

FEM Dr. T.N

=eV

eT e DBdV B ! *c-

For ! bar element, e E D = and dx AdV ee = . #ubstitute into E&n.*c-, we get

==ee l

T eee

l

eT e dx B B E Adx A B E B !

00 matri+ B is constant matri+

where, B is strain displacement matri+ and for " noded bar element it is e+pressed as

=ee l l

B 11

=

e

eT

l

l B

1

1

6ence,

=

=

22

22

11

11

1

111

ee

ee

e

e

ee

T

l l

l l

l

l l l

B B

#ubstitute into e ! and integrate we get

=

=

11

1111

11

22

22

e

eee

ee

eeee

e

l

E Al

l l

l l E A ! is the elemental stiffness matri+ for a " noded bar

element.

#b% Loa ,ectors:


31/65

FEM Dr. T.N

dx N

N f Adx f N A f

ee l e

xe

l e

xT

ee ==

0 2

1

0

@sing the following integration rule for " noded bar element, the above e&uation can beintegrated.

e

l q # l

q #

q #dx N N

e

++=

0 21 )1(

== e

ee xe

l e

xee

l

l f Adx

N

N f A f

e

2121

0 2

1

=1

1

2

e xeee f l A f is the load vector due to body force for the case of " noded bar element.

*i+ 8oad $ector due to surface force! From E&n. *f-, the work potential due to surface force is

e+pressed aseT

l e

xT T

f s T qdxT N q"P e

=

=

0

where, eT is elemental load vector due to surface force and is e+pressed as

dx N

N T dxT N T

ee l e

x

l e

xT e ==

0 2

1

0

@sing the above mentioned integration rule for " noded bar element, the above e&uation can beintegrated as

== e

ee x

l e

xe

l

l T dx

N N T T

e

212

1

0 2

1

=11

2

e xee T l T is the load vector due to surface force for the case of " noded bar element.

#b% !or a 3*no e trian+ular #(.S.T% element:

= s

T

V

T

V


95

eee

l

eee

l

l l l dx N

l l l dx N

e

e

21

211

)11(1

21

211

)11(1

02

01

=

=+

=

==+=

),( y x P

),( 11 y x

2

3),( 33 y x

x

y

Fig. shows a $ noded triangular element.The values of cartesian coordinatescorresponding to node , " and $ are asshown in Fig. The elemental stiffness matri+and load vectors can be derived from strainenergy and work potential of total potentialenergy functional. For an the element, thetotal potential energy can be written as


32/65

FEM Dr. T.N

where, first part of the above e&uation gives strain energy and second and third parts gives workpotential. @sing strain energy and work potential, the elemental stiffness and load vectors can bederived for the given element as follows:

#a% Stiffness "atri': >roceed same as case *a- upto E&n. *c- and show that

=e

V e

T e DBdV B ! *c-

For " ! element, eee dAt dV = , where et is thickness of element. #ubstitute into E&n.*c-, we get

==e

e

Ae

T eee

l T e dA DB Bt dAt DB B !

0

DB Bt A ! T eee = is elemental stiffness matri+.

6ere, D is constitutive relation for two dimensional problem and is e+pressed as

=

2

100

01

01

)1( 2

E

D for plain stress problems

+=

221

00

01

01

)21)(1(

E

D for plain strain problems

#b% Loa ,ector:


33/65

FEM Dr. T.N

=eV

eT T

f b fdV N q"P *f-

6ere z y x f f f f = is body force vector and for a " ! element { }T y x f f f = and *i.e.the components of body force along + and y directions-. Then, the above e&uation becomes as

eT

V e

y

xT f b f qdV f

f

N

N

N

N

N

N

q"P ==

3

3

2

2

1

1

0

0

0

0

0

0

*g-

where, e f is elemental load vector due to body force and is e+pressed as

e A

y

x

y

x

y

x

e A

e y

xe

V e

y

xe dA

f N

f N

f N

f N

f N

f N

t dA f

f

N

N

N

N

N

N

t dV f

f

N

N

N

N

N

N

f eee

===

3

3

2

2

1

1

3

3

2

2

11

3

3

2

2

11

0

0

0

0

0

0

0

0

0

0

0

0

For " ! element, eee dAt dV = where, et is thickness of element.@sing the following integration rule for $ noded triangular element, the above e&uation can beintegrated.

e A

$ q # A$ q #$ q #dA N N N

e

2)2(321 +++=

i.e 32

3211

2)21(

113

12

11

eee

Ae

Ae

Ae

A A AdA N dA N dA N

eee

=

=+

===

=

y x

y

x

y

x

eee

f

f

f

f

f

f

t A f

3

is the load vector due to body force for the case of $ noded triangular element.

*ii+ 8oad $ector due to surface force! The surface force *traction- is a distributed load acts on theedge of an element connecting the boundary nodes.

97

x- and y- c $& nen"s a" n de-1



3 x

y


34/65

FEM Dr. T.N

(onsider a triangular element sub/ected to uniformly varying load *i.e. traction force, force perunit surface area- on side " as shown in Fig. The component of this traction force in + and ydirections are xT and yT respectively.


35/65

FEM Dr. T.N

#ubstituting T N and T into E&n.*f-, we get

eT

y

x

y

x

l e

T f % T qdl

T T T T

N N

N N

N N

N

N

t q"P ==

2

2

1

1

21

21

2

2

1

1

2100

00

00

0

0

where, eT is elemental load vector due to surface force and is e+pressed as

dl

T T T

T

N N N N

N

N N

N

t T

y

x

y

x

l e

e=

2

2

1

1

21

21

2

2

1

1

2100

00

0

0

0

0

dl

T T T T

N

N

N N

N N

N N

N N

N

N

t T

y

x

y

x

l e

e=

2

2

1

1

22

22

21

21

21

21

21

21

21

0

0

0

0

0

0

0

0

@sing the following integration rule for line element *because side " is considered as line-, theabove e&uation can be integrated.

212121

)1(

++= l

q #q #

dl N N l

q #

i.e 3)12(2 21

212

12

1

2121

=+

==

l l dl N dl N

l l and 6)111(

11 2121

12

11

21

=++

=

l l dl N N

l

99


36/65

FEM Dr. T.N

=

2

2

1

1

21

21

21

21

21

21

21

21

30

03

60

06

60

06

30

03

y

x

y

x

ee

T T T

T

l

l

l

l

l

l

l

l

t T

++++

=

21

21

21

21

21

22

22

6 y y

x x

y y

x x

ee

T T T T T T T T

l t T *g-

is the load vector due to surface force for the case of $ noded triangular element.

Note!1. )f traction force is u.d.l 21 x x T T = and 21 y y T T = then E3n. *g+ reduces to

=

2

2

1

1

21

2

y

x

y

x

ee

T T T T

l t T

". 5f side " $ of triangular element is sub/ected to traction force, then 21

l is replaced by 32

l and suffices " of xT and yT are replaced by " $ respectively in E&n. *g-.

*c+ For a #/noded bar or 1/D 7uadratic element!

(onsider a &uadratic element having $ nodes as shown in Fig. The values of natural coordinatesat nodal points are as shown in Fig. The elemental stiffness matri+ and load vectors can be

100

1 2

11 = 12 += 3

03 =

1u 3u 2u

el


37/65

FEM Dr. T.N

derived from strain energy and work potential of total potential energy functional. For an the element, the total potential energy can be written as

= s

T

V

T

V


@sing strain energy and work potential, the elemental stiffness and load vectors can be derivedfor the given element as follows:

#a% Stiffness "atri': >roceed same as case *a- and show that

q Bdx B E Aq el

T ee

T e

=

02

1

q ! q eT e 21=

where = el

T ee

e Bdx B E A ! 0

is a stiffness matri+

Note: -train/displacement matrix 6 for a #/noded bar element is not constant and hence it

cannot ta e out side the integration. Further matrix 6 is a function of and hence the abo$eintegration w.r.t x should be modified to integrate w.r.t .

)e have,el

x x )(2 3= el dx

d 2= or d l

dx e2

=

#ubstitute into e ! , we get

d B Bl

E A ! T eeee

+

=

1

12*b-

*limits el 0 in the above E&n. changes to to -

6ere, += 2)12(21

)12(212

el B

+

+

=

2)12(21

)12(21

2

)12(21

)12(21

42e

T

l B B

+

+++

+

=2

2

2

2

4)12(21

2)12(21

2

2)12(21

)12(41

)12)(12(41

2)12(21)12)(12(

41)12(

41

4

e

T

l B B

d %ymml d B B e

T

+

+

+++

+

=

1

1 2

22

222

2

1

14

)2()414(41

)2()14(41

)414(41

4

*c-

101


38/65

FEM Dr. T.N

5ntegration of the following terms yields,

+

+

=+==

1

1

1

1

32

32

)11(31

3

d B +

+

===

1

1

1

1

2

0)11(21

2

d and +

+ =+==

1

1

11 2)11( d

#ubstitute into E&n. *c-, we get

++

+

= +

32

4

)32

2()04232

4(41

)0322()2

324(

41)042

324(

41

42

1

1

%ymml

d B Be

T

= +

32161416214

3

11

12

%ymml

Bd Be

T


=32

1614

16214

3

12 2

%ymml

l E A !

e

eeee

=

=1688

871

817

316

87

817

3 e

ee

e

eee

l E A

%ymml E A

!

(#) Load ;e "o':


39/65

FEM Dr. T.N

d N

N

N f l A

d N f l A

f e

xeeT e

xeee +

+

==

1

13

2

11

1 22

For a $ node bar element, )(2

1)1(

2

1 21 +== N B )(

2

1)1(

2

1 22 +=+= N and

23 1 = N . #ubstitute into above e&uation we get

d f Ael

f e

xee +

+

=1

1 2

2

2

1

)(21

)(21

2*f-

5ntegration of following terms yields

+

+

=+==1

1

1

1

32

3

2)11(

3

1

3

d B

+

+

===1

1

1

1

2

0)11(

2

1

2

d and

+

+ =+==

1

1

11 2)11( d

#ubstitute into E&n. *f-, we get

+

=

3

22

)03

2(

2

1

)03

2(

2

1

2

e xee f Ael f . Finally, the load vector due to body force is e+pressed as

=

3

26

161

e xe

e f Ael f

*ii+ 8oad $ector due to surface force! From E&n. *e-, the work potential due to surface force ise+pressed as

eT l

T e

xeT f s T qd N

T l q"P

e

== 02where eT is elemental load vector due to surface force and is e+pressed as

d T l

d

N

N

N T l

d N T l

T e

xee

xeT e

xee +

+

+

+

===1

1 2

2

2

1

13

2

11

11

)(21

)(21

222

#ubstitution of integration terms discussed above yields

103

* ad a" n de-1

* ad a" n de-2

* ad a" n de-3


40/65

FEM Dr. T.N

=

3

22

32

21

32

21

2

e xee T l T . Finally, the load vector due to surface force is e+pressed as

=

3

26

16

1

e xe

e T l T

1 . A&&e!# , o= "*e 0 o#a &"%==ne&& !a"'%3 and Load @e "o'

The potential energy for an e th element can be e+pressed as

=eV e%

eT

eT

eV

eT e Td% u fdV udV

where, the st term in D6# of the above e&uation represents strain energy due to internalstresses and " nd and $ rd terms represent the work potential due to body force, and surface forcerespectively. %lso, Bq= B D Bq T T T = and T T T N qu = .#ubstitution of the above terms into the above e&uation yields

=

eee % e

T T

V e

T T

V e

T T e Td% N q fdV N qq DBdV Bq21

eT eT eT e

T q f qq ! q = 21

where, e ! A element stiffness matri+

=q element nodal displacement vector e f and eT A element load vectors due to body force and surface force

The total potential energy of the entire body, which is divided into en number of elements, isobtained by the summation of the potential energy of each element.

=

e % e

T T

e V e

T T

e V e

T T

eee

Td% N q fdV N qq DBdV Bq21

%fter performing assembly process over each element, the above e&uation can be e+pressed inshort form as

& ' !'' T T =21

where, ! A ;lobal stiffness matri+ ' A global nodal displacement vector & A ;lobal force vector that includes global load vectors due to body force, surface

force and point loads *if any-.The global stiffness matri+ 9 and load vector F are respectively obtained by the assembly ofelement stiffness matrices and element load vectors using nodal connectivity table.

&rob! Fig.*a+ shows a stepped bar. For each element % i ' the cross sectional area and length are ( i and 8 i respecti$ely. Each element is sub ected to a traction force T i per unit length and a bodyforce f i per unit $olume. :oung's modulus of the material is E and a concentrated load & " isapplied at node ". ;btain the global stiffness matrix and nodal load $ector.

104

* ad a" n de-1

* ad a" n de-2

* ad a" n de-3

11 , L A

22 , L A

33 , L A

1T

2T

3T

2 P

1

2

3

4

x

11 , L A

22 , L A

33 , L A

1T

2T

3T

2 P

1

2

3

4

x

1

2

3


41/65

FEM Dr. T.N

F%+. (a) F%+. (#)

#olution: The given stepped bars are discretized into $ elements as shown in Fig. *b-.

(a) To o#"a%n 0 o#a S"%==ne&& Ma"'%3:

5n general element stiffness matri+ is e+pressed as

=11

11

e

eee

l

E A ! .

Then, for elements ," and $ it is e+pressed as

2

1

1

1

1

1

1

1

1

1

1

=

l

A

l

Al

Al

A

E ! 3

2

2

2

2

2

2

2

2

2

2

=

l

A

l

Al

Al

A

E ! and4

3

3

3

3

3

3

3

3

3

3

=

l

A

l

Al

Al

A

E ! *a-

(onnectivity table:Element =os.


42/65

FEM Dr. T.N

4

3

2

1

00

0

0

00

3

3

3

3

33

33

22

22

2

2

2

2

1

1

1

1

1

1

1

1

+

+

=

l A

l A

l A

l A

l Al A

l A

l A

l A

l A

l A

l A

E ! is the global stiff. Matri+.

(#) To o#"a%n + o#a oad @e "o':

*i-


43/65

FEM Dr. T.N

4

3

2

1

2

22

22

2

333

333222

222111

111

(

++=

f l A

f l A f l A

f l A f l A

f l A

& f b

*ii-


44/65

FEM Dr. T.N

4

3

2

1

2

22

22

2

33

3322

2211

11

(

++=

T l

T l T l

T l T l

T l

& f s

*iii-


45/65


46/65

FEM Dr. T.N

Element body force $ectors! is e+pressed as =11

2

e xeee f l A f

For element : here 361 10676 mm N f x

=

==

241(0

241(0

1

1

2

10*6(76*120*5(52 61 f

For element ": 362 10676 mm N f x=

==

143(0143(0

11

210*6(76*120*5(37

62 f

) To a&&e!# e &"%==ne&& !a"'%3 and oad @e "o'&:-tiffness matrix! The assembly of element stiffness matrices can be done by using the followingconnectivity table:Element =os.


47/65

FEM Dr. T.N

=+=1720

4131002410

P & & & B =

;lobal system e&uation can be written as

=

172(0

413(100

241(0

10*625(010*625(00

10*625(0510(110*875(0

010*875(010*875(0

3

2

1

55

555

55

*%-

d) To &o @e "*e &,&"e! e a"%on: (pplication of boundary condition! %s the plate is at node , we have 011 == a . Therefore,using elimination method E&. *%- can be modified by eliminating st row and st column. Themodified system e&uation is written as

=

=

172(0

413(100

172(0

413(100

10*625(010*625(0

10*625(010*5(1

131

121

3

2

55

55

a !

a !

*#ince, 01 =a -

The above e&uation can be written in e+panded form as41310010*625010*51 3

52

5 = *i-172010*625010*6250 3

52

5 =+ *ii-

%dd *i- and *ii-, we get 58510010*8750 25 =

mm 32 10*141 = . #ubstitute A " into E&. *i- or *ii- and solve for A #, we get

mm 3

3 10*151

=Therefore, displacement vector is { } mm' T 33 10*15110*1410 e) To e@a a"e e e!en" &"'e&&e&: 5n general element stress is given by

eee

e ' B E = , where =ee

e

l l B

11 and e' is the nodal displacement vector corresponding

to e th element.

111


48/65

FEM Dr. T.N

For element : == 3

3

2

1

111

1

10*14(1

0

120

1

120

110*200

11

l l E

9110*141*120

1*10*200 33 == = mm "

For element ": ==

3

33

3

2

222

2

10*15(1

10*14(1

120

1

120

110*200

11

l l E

AG.G 3 = mm"

=) To de"e'!%ne & o'" 'ea "%on: 5n elimination method, the support reactions at node and $are obtained as follows:

%t support : 1131312111 ( & ! ! ! =++

7r it is simply obtained by adding 1 ( to the D6# of st e&uation of E&. *%- and is written as

125

15 241010*875010*8750 ( +=

7r 1009999241010*141*10*8750 351 == ( =

&rob."! For the following figure *bar+ find the nodal displacements. The cross sectional areadecreases linearly from 1>>> mm " to B>> mm " . Ase two elements. Ta e E = "x1> C Mpa

* TA an. ">>#+

#olution:750

250

375"an == a 2250

750

375*250mma ==

( s area at midpoint A GGG "H a A GGG "H "1 A 31G mm "

%verage ( s area of element , ( 1 A 28752

7501000mm=+

112

375 $$


49/65

FEM Dr. T.N

%verage ( s area of element ", ( " A 26252

500750mm=+

>roceed similar to the previous problem and determine nodal displacement.

&rob #! ( load & = C> N is applied as shown in Fig *a+. Determine the displacement field stressand support reactions in the body. Ta e E = ">x1> # N@mm" . Ase *a+ Elimination method and *b+&enalty method to handle the boundary conditions.#olution: !ivide the given bar into two elements such that the point load lies on a node point asshown in Fig. *b-.5n general element stiffness matri+ for an e th element is e+pressed as

= 1111

eeee

l

E A !

Therefore,For element :

=

=

55

5531

1033010330

103301033011

11

1502501020

!

For element ":

== 55

5512

1033010330

1033010330 ! ! since, ( 1=( " E 1=E " and l 1=l "

;lobal stiff. Matri+ is

+

=55

5555

55

10330103300

10330)1033010330(10330

01033010330

! = mm

113

1 2 3

(b)

$31 mm $31 mm

431 mm "2"1 mm "

GGG =2 3


50/65

FEM Dr. T.N

%s there are no body and traction forces, the total global load vector is only due to point load andis given by

N & =0

10*60

03

;lobal system e&uation is

=

0

10*60

0

10330103300

103301066010330

010330103303

3

2

1

55

555

55

*%-

%pplication of boundary conditions:*a- Elimination method: 6ere 0011 == a and mma 2133 ==6ence, using elimination method, E& *%- is modified by eliminating st row and st column and $ rd

row and $ rd column and is given by

{ } 21*10*3300*10*33010*60**2110*6010*660 553

3231

3

2

5

++== a ! a !

52

5 10*996010*660 = mm 512 =

Therefore, nodal displacement field is { } { } mm ' T T 21510321 ==E e!en"a &"'e&&e&: 5n general, the elemental stress is e+pressed as

eee

ee

e ' B E E == Then,For element :

51*150

1*10*20

51

0

150

1

150

110*20

11 33

2

1

11

11 ===

l l E

A "GG = mm " *tensile-For element ":

+=== 21*

150

151*

150

110*20

21

51

150

1

150

110*20

11 33

3

2

222

2

l l E

A 'G = mm " *compressive-

S o'" 'ea "%on&: 5n elimination method, the support reactions at node and $ are obtained asfollows:

%t support : 1131312111 ( & ! ! ! =++

7r it is simply obtained by adding 1 ( to the D6# of st e&uation of E&. *%- and is written as

125

15 010*33010*330 ( +=

7r 4950051*10*330 51 == ( = * - %t support ": 33331332131 ( & ! ! ! =++7r it is obtained by adding 3 ( to the D6# of $ rd e&uation of E&. *%-

335

25 0*10*330*10*330 ( +=+

7r 990010)21*33051*330( 53 =+= ( = * -

(#) Pena ", a 'oa *: 6ere 0011 == a and mma 2133 ==410ij ! Max1 =

From E&. *%-, 510*660=ij ! Max 945 10*66010*10*660 == 1 6ence, using penalty approach, E&. *%- is modified as

114


51/65

FEM Dr. T.N

+

+=

+

+

21*10*6600

10*60

0*10*6600

10*6601033010330 0

1033010660 10330

01033010*66010330

9

3

9

3

2

1

955

555

595

#olve the above e&uation for unknown A 9s, we getmm mm mm 2000151and5000451;10*57 32

51 ===

E e!en"a &"'e&&e&: 5n general, the elemental stress is e+pressed asee

ee

ee ' B E E ==

Then,For element :

==

5000451

10*57150

1150

110*20

11 53

2

1

111

1

l l E

A 0162005000451*150

110*57*

1501

10*20 53 =

+ = mm " *tensile-

For element ":

==2000151

5000451

150

1

150

110*20

11 3

3

2

222

2

l l E

A

+ 2000151*

1501

5000451*150

110*20 3 A 'G.GG' = mm " *compressive-

S o'" 'ea "%on&: 5n penalty approach, the support reactions at node and $ are obtained asfollows:

%t support : 49500)010*57(10*660)( 59111 === a 1 ( =

%t support ": 9900)212000151(10*660)( 9332 === a 1 ( =

>rob. ': !etermine the nodal displacements, element stresses and support reactions of thea+ially loaded bar shown in Fig. @se elimination method for handling the boundary conditions.Take E A "GG ;pa and load > A $GG k=. *CT@ Feb. "GG"-

#olution: !ivide the given system into two elements so such that point load lies on a node pointas shown below.

Elemental stiffness matrices!

115

1 2 3

1 3 2 4 E 2 344x54 6 N7mm3

P 2 644x54 6 N


52/65

FEM Dr. T.N

=11

11

e

eee

l

E A !

For element :

=

=

55

5531

1033310333

103331033311

11

15025010200

!

For element ":

== 55

5512

1033310333

1033310333 ! ! since, ( 1=( " E 1=E " and l 1=l "

For element $:

=

=

55

5531

1067210672

106721067211

11

30040010200

!

;lobal stiff. Matri+ is

+

+

=55

5555

5555

55

10*67210*67200

10*672)10*67210*333(10*3330

010*333)10*33310*333(10*333

0010*33310*333

! = mm

%s there are no body and traction forces, the total global load vector is only due to point load andis given by

N & =

00

10*300

03

;lobal system e&uation is

=

0

0

10*300

0

10*67210*67200

10*67210*610*3330

010*33310*66610*333

0010*33310*3333

4

3

2

1

55

555

555

55

*%-

%pplication of boundary conditions:*a- Elimination method: 6ere 0011 == a and 044 == a

6ence, using elimination method, E& *%- is modified by eliminating st

row and st

column and 'th

row and ' th column and is given by

==

0

10*300

0

10*300

10*610*33(3

10*33(310*66(6 5

434131

4241213

3

255

55

a ! a !

a ! a !

since, 041 == aa

33

52

5 10*30010*33310*666 = *i-116


53/65

FEM Dr. T.N

010*610*333 35

25 =+

7r 32 8021 = . #ubstitute into E&. *i-, we get3

35

35 10*30010*3338021*10*666 =

mm 34603 = and mm 62403460*80212 ==Thus, nodal displacements are { } mm' T 0346062400=E e!en" S"'e&&: eeee ' B E =

832624(0

0

150

1

150

110*200

11 3

2

1

111

1 ===

l l E = mm " *tensile-

67(370346(0

624(0

150

1

150

110*200

11 3

3

2

222

2 ===

l l E = mm " *compressive-

===0346(0

300

1

300

110*200

11 3

4

3

333

3

l l E "$G.23 = mm " *compressive-

S o'" Rea "%on&: %t support : %dd 1 to the D6# of st e&uation of E&. *%-, we get

2077926240*10*33310*33310*333 525151 === ( = *here A 1AG- %t support ': %dd 2 to the D6# of ' th e&uation of E&. *%-, we get

923823460*10*67210*67210*672 545

35

4 ==+= ( = *here A 2=G-

&rob. B! ( component shown in Fig. is sub ected to a load of B N. Determine the following.i+ Element stiffness matrices ii+ 6 G Matrices iii+ Displacements and strainsi$+ -tresses and reactions.

;btain the stiffness matrix and load $ector assuming two elements. * TA uly">>2+

117


54/65

FEM Dr. T.N

%) E e!en" &"%==ne&& !a"'% e&:For element :

=

= 33333

1

10*5010*50

10*5010*5011

11

100010*100*500

!

For element ":

= = 33333

1

10*6710610*6710610*6710610*67106

1111

75010*200*400 !

%%) B Ma"'% e&:5n general, the 0 Matri+ for " node bar element is e+pressed as

=ee

e

l l B

11

For element , =1000

11000

11 B and for element ", =750

1750

12 B

%%%) D%& a e!en"& and &"'a%n&:The displacements and strains are obtained by the assemblyprocess and solution of assembled system e&uation. The assembled or global stiffness matri+ is

obtained as

+

=

33

3333

33

10*6710610*671060

10*67106)10*6710610*50(10*50

010*5010*50

!


55/65

FEM Dr. T.N

=

3

3

233

33

10*5

0

10*67(10610*67(106

10*67(10610*67(156

010*6710610*67156 33

23 = *i-

33

32

3 10*510*6710610*67106 =+ *ii- %dd *i- and *ii-, we getB>H1> #A " = BH1> #. Therefore, A " A >.1 mm#ubstitute A " into E&. *i- and solve, we get A # A G. '3 mm.Therefore, displacements field is { } mm' T 1470100=To determine strains: we have eee ' B=

41 10*11(0

010001

10001 == and 42 10*627(0

147(01(0

7501

7501 ==

%@) S"'e&&e& and 'ea "%on&:#tresses are 1010*1*10*100 4311

1 === E = mm " and531210*6270*10*200 4322

2 === E = mm "Deaction is obtained by adding 1 to the D6# of st e&uation of E&. *%- as

510*10*50*10*50 323

1 === ( k=

&rob. C! ;btain the displacement field stresses and support reactions for the stepped bar shownin Fig. Ase elimination and penalty approach to handle the boundary conditions. erify theanswer.

119

in". Ob"ain "%e / ba sys"e$ e 'a"i nsi$i a# " "%e e i 's b e$s and s ese&a#a"e y 'sin/ e i$ina"i n $e"% d and

&ena "y a& ac% %e / ba ad ec" # in"%is case is

{ } N & T 010*100 3= , ne/a"i e si/n ! #"%e a&& ied ad is beca'se i" ac"s in ne/a"i ex-di#ec"i n


56/65

FEM Dr. T.N

=

= 55553

1

10*3610*36

10*3610*3611

11

20010*70*1800

!

=

= 55553

2

10*45910*459

10*45910*45911

11

20010*105*1800

!

(onnectivity tableElement =os.


57/65

FEM Dr. T.N

=

0

10*385

0

10*45(910*45(90

10*45(910*75(1510*3(6

010*3(610*3(63

3

2

1

55

555

55

0oundary conditions: 01 = and 03 = . Eliminate st row and st column and $ rd row and $ rdcolumn, we get

32

5 10*38510*7515 = mm 24402 = *=ode " moves towards ve + direction-

#tresses:

23

2

1

211

1 ,4(85244(0

0

200

1

200

110*70

11mm N

l l E ===

23

3***

2

3

222

2

,2(256200

244(0*10*105

244(00

2001

200110*10511

mm N

l l E

==

==

HHH =ote that the nodal displacements corresponding to the local node and " of element " are

A # and A " respectively.#upport reactions::N ( 721532440*10*36 51 +== and :N ( 582302440*10*459

53 ==

&rob. I! ( bar element carries a distributed load of 3 N@m which $aries from 3 1 at one end to 3 "at other end. ,alculate consistent nodal loads. * TA May ">>J+

121

q N7m

1q 2q

el 1 2

1u 2u


58/65

FEM Dr. T.N

(onsider a " noded bar element sub/ected to an uniformly varying traction force which variesfrom 1q to 2q at node and " respectively as shown in Fig. Then, load vector due to thistraction force can be obtained from work potential due to surface force.

==el

T

s

T & % qdxuTdsu"

0

*for bar element qT T e x == and dxds = +

[ ] u Nquu N N u N u N u ==+=2

1212211

T T u

T N qu = %gain, since traction force varies from one node another node, it can be interpolated as

[ ]=+=2

1

212211 q

q N N q N q N q

#ubstitute T u and q into 5 s.F , we get

[ ] ==el

eT u

T T u & % f qdxq

q N N N q"

0 2

121

where, e f is consistent nodal load vector and is e+pressed as

[ ] dxqq

N N N

N N N dxqq

N N N N

f

ee lle

== 21

02221

2121

2

1

021

2

1

@sing the integration rule for bar element, it can be shown that

eee

l l

l l l dx N dx N

ee

31

3*2*12*1

)12(2

0

22

0

21 ==+== and

=++

=el

ee l l dx N N 021 6

1)111(

11

122


59/65

FEM Dr. T.N

#ubstitute into above e&uation we get

+

+

==21

21

2

1

2

2

6

36

63

qq

qql qql l

l l

f eee

ee

e

1 . I&o a'a!e"'% E e!en"&:

For the analysis of elasticity problems which are having comple+ or curved shapes, simpletriangular or rectangular elements are no longer sufficient. This needs to develop an element ofarbitrary shape called )soparametric element.The concept of isoparametric element is based on the transformation of the parent element inlocal or natural coordinate system to an arbitrary shape in cartesian coordinate system as shownin Fig. G. This can be done by using shape functions and nodal values of cartesian coordinates.Thus the cartesian coordinate of any point in an element may be e+pressed as

(a) Pa'en" e e!en" %n na" 'a (#) I&o a'a!e"'% ad'% a"e'a oo'd%na"e& e e!en"

F%+. 1?: 4-noded I&o a'a!e"'% ad'% a"e'a e e!en"

44332211 x N x N x N x N x +++=

[ ]{ }m x N x = and similarly [ ]{ }m y N y = *a-Further, the variation of displacement at any point within an element can also be e+pressed as

44332211 u N u N u N u N u +++=[ ]{ }uq N u = and similarly [ ]{ }vq N v = *b-

5f the shape functions [ ] N in E&n. *a- are same as the shape functions [ ] N in E&n. *b-, thenthe element is called as )soparametric element.Thus, the isoparametric element is Ione in which safe functions used to define the geometry ofthe element and variation of displacements within the elements are sameJ.

S e'- a'a!e"'% e e!en": 5s one which is having higher order shape functions used to define

geometry than the shape functions used to define the variation of displacement. 5n case ofsuperparametric element the number of nodes used to define the geometry of the element aremore than the number of nodes used to define the variation of displacement as shown in Fig. .

123


60/65

FEM Dr. T.N

F%+. 11: S e'- a'a!e"'% e e!en"

S #- a'a!e"'% e e!en": 5s one, which having lower order shape functions to define geometryof the element than the shape functions used to define variation of displacement. 5n case of subparametric element the number of nodes used to define the geometry of the element are lessthan the number of nodes used to define the variation of displacement as shown in Fig. ".


61/65


62/65


63/65

FEM Dr. T.N

=

yu xu

y x

y x

u

u

[ ]

=

yu x

u

> u

u

*d-

where, matri+ > is called acobian matrix and is given by

[ ]

=

y x

y x

> *e-

!ifferentiate E&n.*b- w.r.t and , we get

44

33

22

11

x N

x

N

x N

x N x

+

+

+

=

44

33

22

11 y

N y

N y

N y

N y

+

+

+=

44

33

22

11 x

N x

N x

N x

N x

++

+

=

44

33

22

11 y

N y

N y

N y

N y

++

+

=

=ow, diff. E&n.*c- w.r.t and , we get

4)1(1

=

N

B4

)1(2

+=

N B

4)1(3

++=

N B

4)1(4

+=

N

4)1(1

=

N

B4

)1(2

+=

N

B4

)1(3

++= N

B4

)1(4

=

N

#ubstitute into the above e&uations, we get

4321 4)1(

4)1(

4)1(

4)1(

x x x x x

++++=

4321 4)1(

4)1(

4)1(

4)1(

y y y y y

++++=

4321 4)1(

4)1(

4)1(

4)1(

x x x x x

++++=

4321 4)1(

4)1(

4)1(

4)1(

y y y y y

+

++

+

=

#ubstitution into E&n.*e- yields Kacobian matri+ as

[ ] ( ) ( )( ) ( )++++++++

++++++++=

43214321

43214321

)1()1()1()1()1()1()1()1()1()1()1()1()1()1()1()1(

41

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

>

&rob.#! E$aluate the shape functions 321 ,, N N N at the interior point P whose coordinatesare )84,853( for the triangular element shown in Fig. *a+ below.-olution! The cartesian co ordinate at point p can be e+pressed in terms of nodal coordinates as

332211 x N x N x N x ++=321 4751853 N N N ++= *i-

332211 y N y N y N y ++=

321 753284 N N N ++= *ii-127


64/65

FEM Dr. T.N

also we have 1321 =++ N N N *iii-#olving E&ns. *i- *iii-, we get

50;20;30 321 === N N N

&rob.2! determine the acobian matrix and 6 matrix for the triangular element shown in Fig.*a+below.

* TA uly ">>2+

#a%

-olution! The Kacobian matri+ is

[ ]

=

y x

y x

>

The cartesian co ordinate at point p can be e+pressed in terms of nodal coordinates as332211 x N x N x N x ++=

)1(4751 ++= x332211 y N y N y N y ++=

)1(7532 ++= y52451 ==

x

0572 ==

y

0347 ==

x

53753 ==

y

6ence, [ ] =

5303

0552 >

For a triangular element, the strain displacement matri+ is

[ ]=332211

321

321000

000

21

bcbcbc

cccbbb

A B

e

e

where, j: : j x x y y == ii c ; b *suffi+ : ji ,, ,",$ are in cyclic order-.53753321 === y yb 374231 === x xc

527132 === y yb 52451312 === x xc51532213 === y yb 55517123 === x xc

( ))()()(121

1

1

1

21

2312312332

33

22

11

x x y y y x y x y x

y x

y x

y x

Ae +==

&rob.#! The nodal coordinates of a triangular element are shown in Fig. *b+. (t interior point &the x/ coordinate is #.# and N 1=>.#. Determine N " N # and y/ coordinate at point &.

128

1

2

3

x

y

:


65/65

FEM Dr. T.N

!i+. #b% !i+. #c%

&rob.2! For a point & located inside the triangular element shown in Fig.*c+ the shape functionsN 1 and N " are >.JB and >."B respecti$ely. Determine x and y coordinates of point &.

&rob.B! ,alculate the displacements at point & whose coordinates are gi$en by *" #+ if the nodaldisplacements of a triangle are gi$en by

mmvmmvmmvmmummummu

: ji

: ji

4;3;22;4;3

======

Nodes ji , and : of the triangle are gi$en by the coordinates *1 1+ *2 1+ and *1 B+respecti$ely.

321332211 243 N N N u N u N u N u ++=++= ---(a)321332211 432 N N N v N v N v N v ++=++= ///*b+

332211 x N x N x N x ++=321 1412 N N N ++= *i-

332211 y N y N y N y ++=

321 5113 N N N ++= ///*ii+1321 =++ N N N ///*iii+

-ol$e E3ns. *i+ G *iii+ to get 321 ,, N N N and substitute into E3ns. *a+ and *b+ to obtain vu ,

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FEM Notes5