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Module 5: FEM for Two and Three Dimensional Solids

Lecture 1: Constant Strain Triangle

The triangular elements with different numbers of nodes are used for solving two dimensional solid

members. The linear triangular element was the first type of element developed for the finite element

analysis of 2D solids. However, it is observed that the linear triangular element is less accurate

compared to linear quadrilateral elements. But the triangular element is still a very useful element for

its adaptivity to complex geometry. These are used if the geometry of the 2D model is complex in

nature. Constant strain triangle (CST) is the simplest element to develop mathematically. In CST,

strain inside the element has no variation (Ref. module 3, lecture 2) and hence element size should

be small enough to obtain accurate results. As indicated earlier, the displacement is expressed in two

orthogonal directions in case of 2D solid elements. Thus the displacement field can be written as

{ }u

d v

ì üï ïï ï= í ï ïï ïî

(5.1.1)

Here, u and v are the displacements parallel to x and y directions respectively.

5.1.1 Element Stiffness Matrix for CST

A typical triangular element assumed to represent a subdomain of a plane body under plane

stress/strain condition is represented in Fig. 5.1.1. The displacement (u, v) of any point P is

represented in terms of nodal displacements

1 1 2 2 3 3

1 1 2 2 3 3

u N u N u N u

v N v N v N v

(5.1.2)

Where, N1, N2, N3 are the shape functions as described in module 3, lecture 2.

Fig. 5.1.1 Linear triangular element for plane stress/strain

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The strain-displacement relationship for two dimensional plane stress/strain problem can be

simplified in the following form from three dimensional cases (eq.1.3.9 to1.3.14).

2 2

2 2

1

2

1

2

x

y

xy

u u v

x x x

v u v

y y y

v u u u v v

x y x y x y

é ùæ ö æ ö¶ ¶ ¶ê ú÷ ÷ç çe = + +÷ ÷ç çê ú÷ ÷ç çè ø è ø¶ ¶ ¶ê úë ûé ùæ ö æ ö¶ ¶ ¶ê ú÷ ÷ç çe = + ÷ + ÷ç çê ú÷ ÷ç ç÷ ÷ç ç¶ ¶ ¶è ø è øê úë û

é ù¶ ¶ ¶ ¶ ¶ ¶ê úg = + + +ê ú¶ ¶ ¶ ¶ ¶ ¶ë û

(5.1.3)

In case of small amplitude of displacement, one can ignore the nonlinear term of the above equation

and will reach the following expression.

x

y

xy

u x

v

y

v u

x y

(5.1.4)

Hence the element strain components can be represented as,

31 21 2 3

31 21 2 3

3 31 2 1 21 2 3 1 2 3

x

y

xy

N N N uu u u

x x x x

N N N vv v v

y y y y

N N N N N N u vu u u v v v

y x y y y x x x

Or,

131 2

2

331 2

1

23 31 2 1 2

3

0 0 0

0 0 0

x

y

xy

u N N N u x x x

u N N N v y y y

v N N N N N N

y y y x x x v

(5.1.5)

Or, B d (5.1.6)

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In the above equation [B] is called as strain displacement relationship matrix. The shape functions

for the 3 node triangular element in Cartesian coordinate is represented as,

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 3 3 2 2 3 3 2

1

2 3 1 1 3 3 1 1 3

3

1 2 2 1 1 2 2 1

1x y x y y y x x x y

2A N

1 N x y x y y y x x x y2A

N1

x y x y y y x x x y2A

ì üï ïé ùï ï- + - + -ï ïë ûï ïì ü ï ïï ïï ïï ï ï ïï ïï ï ï ïé ù= - + - + -í í ë ûï ï ï ïï ï ï ïï ï ï ïï ïî ï ïï ïé ù- + - + -ï ïë ûï ïï ïî

Or,

[ ]

[ ]

[ ]

1 1 1

1

2 2 2 2

3

3 3 3

1x y

2A N1

N x y

2A N1

x y2A

ì üï ïï ïa +b + gï ïï ïì ü ï ïï ï ï ïï ï ï ïï ïï ï ï ï= a + b + gí í

ï ï ï ïï ï ï ïï ï ï ïï ïî ï ïï ïa + b + gï ïï ïï ïî

(5.1.7)

Where,

( )1 2 3 3 2x y x y ,a = -

( )2 3 1 1 3x y x y ,a = - ( )3 1 2 2 1

x y x y ,a = -

( )1 2 3y y ,b = - ( )2 3 1y y ,b = - ( )3 1 2y y ,b = - (5.1.8)

( )1 3 2x x ,g = - ( )2 2 1x x ,g = - ( )3 2 1x x ,g = -

Hence the required partial derivatives of shape functions are,

1 1 ,2

N

x A

2 2 ,

2

N

x A

3 3 ,

2

N

x A

1 1 ,2

N

y A

2 2 ,

2

N

x A

3 3 ,

2

N

x A

Hence the value of [B] becomes:

31 2

31 2

3 31 2 1 2

0 0 0

0 0 0

N N N

x x x

N N N B

y y y N N N N N N

y y y x x x

Or, 1 2 3

1 2 3

1 2 3 1 2 3

0 0 01

0 0 02

B A

(5.1.9)

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According to Variational principle described in module 2, lecture 1, the stiffness matrix is

represented as,

T

k B D B d

(5.1.10)

Since, [B] and [D] are constant matrices; the above expression can be expressed as

T T

V

k B D B dV B D B V (5.1.11)

For a constant thickness (t ), the volume of the element will become A.t . Hence the above equation

becomes,

T

k B D B At (5.1.12)

For plane stress condition, [D] matrix will become:

2

1 01 0

11

0 02

E D

(5.1.13)

Therefore, for a plane stress problem, the element stiffness matrix becomes,

1 1

2 2

1 2 3

3 3

1 2 321 1

1 2 3 1 2 32 2

3 3

0

01 0 0 0 0

01 0 0 0 0

04 1

10 002

0

Et k

A

(5.1.14)

Or,

2 2

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

2 2

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

2 223 3 3 1 1 3 3 2 2 3 3 3

2 2

1 1 1 2 1 2 1 3 1 3

2 2

2 2 2 3

1

2

1

2

14 1

2

.

C C C C C

C C C C

Et k

C C C A

C C C

Sym C C

2 3

2 2

3 3C

(5.1.15)

Where, 1

2C

Similarly for plane strain condition, [D] matrix is equal to,

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1 0

1 01 1 2

1 20 0

2

E D

(5.1.16)

Hence the element stiffness matrix will become:

2 2

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

2 2

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

2 2

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

2 2

1 1 1 2 1 2 1 3 1 3

2 2

2 2 2 3 2 3

3

( 1)

1

1

2 1

.

M M M

M M

M C Et k

A M M M

Sym M M

M

2 2

3

(5.1.17)

Where 1 M

5.1.2 Nodal Load Vector for CST

From the principle of virtual work,

{ } { } { } { } { } { }T T T

d u F d u F d G W

W G W

d e s W = d G + d Wò ò ò (5.1.18)

Where, F, and F are the surface and body forces respectively. Using the relationship between

stress-stain and strain displacement, one can derive the following expressions:

and D B d , B d u N d (5.1.19)

Hence eq. (5.1.18) can be rewritten as,

{ } [ ] [ ][ ]{ } { } { } { } [ ] { }TT T T T TSd B D B d d d N F d d N F d G W

W G W

é ùd W= d G + d Wê úë ûò ò ò (5.1.20)

Or, [ ] [ ][ ]{ } { } [ ] { }TT TSB D B d d N F d N F d G W

W G W

é ùW = G + Wê úë ûò ò ò (5.1.21)

Here, [ N s] is the shape function along the boundary where forces are prescribed. Eq.(5.1.21) is

equivalent to[ ]{ } { }k d F= , and thus, the nodal load vector becomes

{ } [ ] { } [ ] { }T TS

F N F d N F d G W

G W

= G + Wò ò (5.1.22)

For a constant thickness of the triangular element eq.(5.1.22) can be rewritten as

{ } [ ] { } [ ] { }T TS

S A

F t N F ds t N F dAG W

= +ò ò (5.1.23)

For the a three node triangular two dimensional element, one can represent FW and FG as,

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{ } x

y

FF

F

WW

W

ì üï ïï ï=í ï ïï ïî

and { } x

y

FF

F

GG

G

ì üï ïï ï=í ï ïï ïî

For example, in case of gravity load on CST element, { } x

y

F 0F

F g

WW

W

ì ü ì üï ï ï ïï ï ï ï= =í í

ï ï ï ï-rï ïî ï ïî

For this case, the shape functions in terms of area coordinates are:

[ ] 1 2 3

1 2 3

L L L 0 0 0 N

0 0 0 L L L

é ùê ú= ê úë û

(5.1.24)

As a result, the force vector on the element considering only gravity load, will become,

{ }

1

2

3

1 1 1A A A

2 2 2

3 3 3

L 0 0 0

L 0 0 0

L 0 0 00F t dA t dA dA

0 L L g Lg

0 L L g L

0 L L g L

gt= = = -- r-r

- r

- r

é ù ì ü ì üï ï ï ïï ï ï ïê ú ï ï ï ïê ú ï ï ï ïï ï ï ïê ú ï ï ï ïï ï ï ïê úì üï ï ï ï ï ïï ï ï ï ï ïê ú rí í í ê úï ï ï ï ï ïï ïî ï ï ï ïê ú ï ï ï ïï ï ï ïê ú ï ï ï ïê ú ï ï ï ïï ï ï ïê ú ï ï ï ïï ï ï ïë û î î

ò ò ò (5.1.25)

The integration in terms of area coordinate is given by,

1 2 3 22

p q r

A

p ! q ! r ! L L L dA A

( p q r )!

(5.1.26)

Thus, the nodal load vector will finally become

{ }F

0 00 0

0 0

1!0!0! gAt2Agt (1 0 0 2)! 3

0!1!0! gAt2A

(0 1 0 2)! 3

gAt0!0!1!2A

3(0 0 1 2)!

= -

ì üï ï ì üï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïï ï ï ïrï ï ï ïï ï -ï ï ï ïr =+ + +í í ï ï ï ïï ï ï ïï ï ï ïrï ï ï ï-ï ï ï ïï ï ï ï+ + +ï ï ï ïï ï ï ïï ï ï rï ï ï-ï ï ïï ï ïïî + + +ï ïï ïî

0

0

0gAt

13

1

1

ì üï ïï ïï ïï ïï ïï ïï ïr ï ïï ï= - í ï ïï ïï ïï ïï ïï ïï ïï ïï ïî ïïïïï

(5.1.27)

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