Week 10 MECH3361

Recap Finite Element Method (FEM) Bar Element Method 1 (Displacement Method) Method 2 (Energy Method)Displacement u u( x )=(1−x / L)ui+( x /L )u j u(ξ )=N i(ξ )ui+N j( ξ )u j=(1−ξ )u i+ξu j

ε=u j−ui

L= Δ

Lε= du

dx= d

dx( Nu )=( d

dxN )u=Bu

σ=Eε= EΔL

σ=Eε=E Bu

Force or W=U F=σA= EΔL

A=( EAL )Δ=kΔ U 1

2uT [∫V ( BT EB ) dV ]u=1

2uT f

Equilibrium eqn ku= f ku= f

Stiffness matrix k=[ k −k−k k ]=EA

L [ 1 −1−1 1 ] k=EA

L [ 1 −1−1 1 ]

Finite Element (FE) equilibrium equation: ku=f1D Spring element [ k −k

−k k ]{ui

u j}={f i

f j}

1D Bar Elements:

EAL [ 1 −1

−1 1 ]{ui

u j}={f i

f j}

2D Bar Element

EAL [ l2 lm −l2 −lm

lm m2 −lm −m2

−l2 −lm l2 lm−lm −m2 lm m2 ]{ui

'

v i'

u j'

v j' }={f i

'

0f j

'

0}

Beam Element

Distributed Load – Equivalent nodal loadSingle bar element

Multiple bar

elements

Beam element

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Week 10 MECH3361

8.5 Two-Dimensional Problems

General Formula for the Stiffness MatrixRecall the displacement in terms of “shape function” in bar element

u=[ N i N j ]{ui

u j}=Nu

We can extend it to 2D, where the displacements (u, v) in a plane element are interpolated from nodal displacements (ui, vi) using shape functions Ni as follows,

{u=u( x , y )=N1 ( x , y )u1+N2 ( x , y )u2+⋯¿ ¿¿¿

In the matrix form:

u=¿ {u¿ }¿{}=[N1 0 N2 0 ⋯0 N1 0 N2 ⋯]{

u1

v1u2v2¿⋮

}=Nd

where N is the shape function matrix, u the displacement vector inside each element and d the elemental nodal displacement vector. Here we assume that u depends on the nodal values of u only, and v on nodal values of v only.From strain-displacement relation, the strain vector can be derived as:

ε=Du=( DN ) d Or ε=Bd

where B=( DN ) is the strain-displacement matrix and D is derivative matrix (see below).

Consider the strain energy stored in an element and take the 2D as an example:

U =12∫V

[ σT ε ]dV =12∫V

(σ xx ε xx+σ yy ε yy+σ xy ε xy )dV

¿12∫V

( Εε )T ε dV =12∫V

εT Εε dV=12 ∫V

(Βd )T Ε( Βd )dV

Since nodal displacement vector d is independent on the elemental coordinate dV.

U =dT ( 12∫V

ΒT ΕΒ dV )d=dT kd

From this, we obtain the general formula for the element stiffness matrix,

Remarks: Note that unlike 1-D cases, E here is a matrix that is given by the stress-strain relation

(generalised Hooke’s rule) The stiffness matrix k is symmetric since E is symmetric. Also note that for given the material property E, the behaviour of k depends on the B

matrix only, which in turn on the shape functions. Thus, the quality of finite elements in representing the behaviour of a structure is entirely determined by the choice of shape functions.

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Week 10 MECH3361

Constant Strain Triangle (CST or T3)This is the simplest 2-D element, which is also called linear triangular element.For this element, we have three nodes at the vertices of the triangle, which are numbered around the element in the counter-clockwise direction. Each node has “two degrees of freedom” (can move in both x and y directions). The displacements u and v are assumed to be linear functions within the element, that is,

u=u( x , y )=b1+b2 x+b3 yv=v ( x , y )=b4+b5 x+b6 y

where bi (i = 1, 2, ..., 6) are constants to be determined. From these, the strains are found to be,

ε xx=∂u∂ x

= ∂∂ x (b1+b2 x+b3 y )=b2

ε yy=∂ v∂ y

= ∂∂ y (b4+b5 x+b6 y )=b6

2 ε xy=( ∂u∂ y

+∂ v∂ x )= ∂

∂ x (b4+b5 x+b6 y )+ ∂∂ y (b1+b2 x+b3 y )=b3+b5

Obvious the strains are constant throughout the element. Thus, we have the name “constant strain triangle” (CST) element.Displacement functions should satisfy the following six equations,

u1=u( x=x1 , y= y1 )=b1+b2 x1+b3 y1

v1=v ( x=x1 , y= y1 )=b4+b5 x1+b6 y1u2=u( x=x2 , y= y2)=b1+b2 x2+b3 y2

v2=v ( x=x2 , y= y2 )=b4+b5 x2+b6 y2

u3=u( x=x3 , y= y3 )=b1+b2 x3+b3 y3

v3=v ( x=x3 , y= y3 )=b4+b5 x3+b6 y3Solving these six equations, we can find the coefficients b1, b2, ..., b6 in terms of nodal displacements and coordinates. Substituting these coefficients into displacement equations and rearranging the terms, we obtain:

{uv }=[N1 0 N2 0 N3 00 N1 0 N2 0 N3 ]{

u1

v1

u2

v2

u3

v3

} (linear distribution)

where the shape functions (note that they are linear functions of x and y) are

N1=12 A [( x2 y3−x3 y2)+( y2− y3) x+( x3−x2) y ]

N2=12 A [( x3 y1−x1 y3)+( y3− y1) x+( x1−x3 ) y ]

N3=12 A [( x1 y2−x2 y1)+( y1− y2) x+( x2−x1 ) y ]

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Week 10 MECH3361

where

A=12

det [1 x1 y1

1 x2 y2

1 x3 y3] is the area of the triangle

Using the strain-displacement relation, we have,

{εxxε yy

ε xy}=Bd= 1

2 A [ y23 0 y31 0 y12 00 x32 0 x13 0 x21

x32 y23 x13 y31 x21 y12]{

u1

v1

u2

v2

u3

v3

}where xij = xi - xj and yij = yi - yj (i, j = 1, 2, 3). Again, we see constant strains within the element (because variables x and y disappeared). From stress-strain relation (Hooke’s law), we see that stresses obtained in the CST element should also be constant.

Applying general formula for stiffness matrix, we obtain it for the CST element,

in which t is the thickness of the element. Notice that k for CST is a 6 by 6 symmetric matrix. The matrix multiplication can be carried out by a computer program.

Applications of the CST Element: Use in the areas where the strain gradient is small. Use in the mesh transition areas (fine mesh to coarse mesh) to accommodate different

mesh topologies. Avoid using CST in stress concentration or other crucial areas in the structure, such as

edges of holes and corners. Recommended for quick and preliminary FE analysis of 2-D problems.

Linear Strain Triangle (LST or T6)This element is also called quadratic triangular element.

There are six nodes on this element: three corner nodes and three mid-side nodes. Each node has two degrees of freedom (DOF) as before. The displacements (u, v) are assumed to be quadratic functions of (x, y),

u=b1+b2 x+b3 y+b4 x2+b5 xy+b6 y2

v=b7+b8 x+b9 y+b10 x2+b11 xy+b12 y2

where bi (i=1,2,…12) are the constants to be determined.

Thus the strains are found to be:

ε xx=∂u∂ x

= ∂∂ x (b1+b2 x+b3 y+b4 x2+b5 xy+b6 y2)=b2+2 b4 x+b5 y

ε yy=∂ v∂ y

= ∂∂ y (b7+b8 x+b9 y+b10 x2+b11 xy+b12 y2)=b9+b11 x+2 b12 y

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Week 10 MECH3361

2 ε xy=(∂u∂ y

+∂ v∂ x )=∂

∂ x (b1+b2 x+b3 y+b4 x2+b5 xy+b6 y2)+∂∂ y (b7+b8 x+b9 y+b10 x2+b11 xy+b12 y2 )

¿ (b3+b5)+ (b3+2 b10 ) x+(2 b6+2b11) yObviously all these strains all are in linear functions. Thus, we have the “linear strain triangle” (LST), which may provide better accuracy than the CST.

In a natural coordinate system, the six shape functions for the LST element can be defined:N1=ξ (2 ξ−1) , N2=η(2η−1 ) , N 3=(1−ξ−η) [2(1−ξ−η)−1 ]N4=4 ξη , N 5=4 η(1−ξ−η ) , N 6=4 (1−ξ−η )ξ

Displacement can be written as:u=∑

i=1

6

N i ui , v=∑i=1

6

N i v i (quadratic distribution)

which means that for given nodal displacements, one can know the displacement at any point within the element. In other words, we use shape functions and nodal displacement to express displacement field (function).

Remarks: Note that the element stiffness matrix is still given by k e=∫

VBT E B dV

, but BT EB is quadratic w.r.t. x and y. In general, the integral has to be computed numerically.

Bi-Linear Quadrilateral Element (Q4)

There are four nodes at the corners of the quadrilateral shape. In the natural coordinate system (, ), the four shape functions are given as,

N1=14(1−ξ )(1−η )

, N2=

14(1+ξ )(1−η)

,

N3=14(1+ξ )(1+η)

, N4=

14(1−ξ )(1+η )

Mapping from Cartesian coordinate to natural coordinate systems can be done as follows:

x=∑i=1

4

N i (ξ ,η )x i , y=∑i=1

4

N i( ξ , η) y i,

Displacement can be written as: u=∑

i=1

4

N i(ξ , η)ui , v=∑i=1

4

N i(ξ ,η )vi (bilinear distribution)

Strain: ε xx=

∂u∂ x

= ∂∂ x (∑i=1

6

N i( ξ , η )ui)ε yy=

∂ v∂ y

= ∂∂ y (∑i=1

6

N i(ξ ,η )v i)2 ε xy=( ∂u

∂ y+∂ v

∂ x )= ∂∂ x (∑i=1

6

N i(ξ , η)v i)+ ∂∂ y (∑i=1

6

N i(ξ ,η)ui)

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Week 10 MECH3361

We cannot directly obtain the results because u and v are defined as functions of natural coordinate and rather than as a function of x and y. We can however use chain rule:

{∂ u∂ξ∂ u∂ η

}=[∂ x∂ξ

∂ y∂ξ

∂ x∂η

∂ y∂η ]

⏟[ J ]

{∂u∂ x∂u∂ y

} or

{∂u∂ x∂u∂ y

}=[ ∂ x∂ ξ

∂ y∂ ξ

∂ x∂ η

∂ y∂ η ]

⏟[ J ]−1−1

{∂u∂ ξ∂u∂η

}=[ J11¿ J 12

¿

J 21¿ J 22

¿ ]{∂u∂ ξ∂u∂η

}where [J] is called Jacobian matrix.

Coefficients in [J] can be obtained as:∂ x∂ ξ

=∑(∂ N i

∂ ξx i) , ∂ y

∂ ξ=∑( ∂ N i

∂ ξy i)

Thus ε x=

∂u∂ x

=J 11¿ ∂ u

∂ ξ+J12

¿ ∂ u∂ η

where J11¿ , J12

¿

are the coefficients in the first row of the inverse Jacobian matrix [J]-1. ∂u∂ξ

=∑ (∂ N i

∂ ξui) , ∂ u

∂ η=∑ (∂ N i

∂ ηu i)

Thus we change the integration from Cartesian coordinate to natural coordinate as

k e=∫V

BT E B dV =∫−1

1

∫−1

1

BT EB t|J|dξdη

which needs to use numerical integration method.

According to Gaussian rule of integration, one can have:

I=∫−1

1

∫−1

1

φ ( ξ , η ) dξdη=∑i=1

n

∑j=1

m

W i W j φ (ξ i , η j )Gaussian integration points:

33

33

33

33

x

y

Gaussian points and weights:m,n , Wi

11 0.022 ±√3/3=±0 .5774 1

330

±√0 .6=±0 .77460.88880.5555

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Week 10 MECH3361

The stress, strain results at the Gaussian point should be most accurate in the entire element.

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Week 10 MECH3361

Quadratic Quadrilateral Element (Q8)

x

y

x

y

mapping

1 2

3

4

5

6

7

8

1 2

3

5

6

7

8

4

N1=−14

(1−ξ )(1−η )(1+ξ+η)

N2=−14

(1+ξ )(1−η)(1−ξ+η)

N3=−14

(1+ξ )(1+η)(1−ξ−η)

N4=−14

(1−ξ )(1+η)(1+ξ−η )

N5=12

(1−ξ2)(1−η)

N6=12

(1+ξ )(1−η2 )

N7=12

(1−ξ2)(1+η )

N8=12

(1−ξ )(1−η2)

Mapping from Cartesian coordinate to natural coordinate systems: x=∑

i=1

8

N i x i , y=∑i=1

8

N i y i,

The displacement field is given by: u=∑

i=1

8

N i u , v=∑i=1

8

N i vi (quadratic distributions)

The stiffness matrix:

k e=∫V

BT E B dV =∫−1

1

∫−1

1

BT EB t|J|dξdη

which needs to use numerical integration method as

I=∫−1

1

∫−1

1

φ ( ξ , η ) dξdη=∑i=1

n

∑j=1

m

W i W j φ (ξ i , η j )

Remarks: Q4 and T3 are usually used together in a mesh with linear elements. Q8 and T6 are usually applied in a mesh composed of quadratic elements. Quadratic elements are preferred for stress analysis, because of their high accuracy

and the flexibility in modeling complex geometry, such as curved boundaries.

Example 8.7 Sketch the displacement distribution of the following elements (CST, LST, Q4 and Q8) and explain the difference (final exam 2010, 2011, 2012):

Solution:CST element (T3): linear distribution of displacement and constant in the strain.

4-nodel element (Q4): bilinear distribution of displacements and bilateral linear in the strain

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Week 10 MECH3361

x

y

u

i

j

k

uj ui

uk x

y

i

j

k

vj

vi

vk

v

vl

l

3-node CST element 4-nodel bilinear element

LST element (6-node): quadratic distribution of displacement and linear distribution in the strain

8-node element: quadratic distribution of displacements and linear distribution in the strain

x

y

u

i

j

k

uj ui

uk x

y

i

j

k

vj

vi

vk

v

vl

l

Quadratic surfaceQuadratic surface

6-node LST element 8-nodel quadratic element

Example 8.8For the 8-node element in Example 8.7, if 2×2 Gaussian points are considered, go on to determine the displacements of all the Gaussian points if the nodal displacement vectors are given as below (exam 2010, 2012):

{uv }=[1 0 −1 −1 0 1 2 12 1 −1 −1 1 0 −1 −2 ]

Soln:The displacement functions as per shape functions Ni and nodal displacements ui and vi can be expressed as:

u=∑i=1

4

N i(ξ ,η)ui , v=∑i=1

4

N i(ξ ,η )v i

Shape functions for 8-node elements can be written in terms of natural coordinate:

N1 (ξ , η )=−14

(1−ξ )(1−η )(1+ξ+η)

N2 (ξ , η )=−14

(1+ξ )(1−η)(1−ξ+η)

N3 (ξ , η )=−14

(1+ξ )(1+η)(1−ξ−η)

N4 (ξ , η)=−14

(1−ξ )(1+η)(1+ξ−η )

N5 (ξ , η )=12

(1−ξ2 )(1−η)

N6 (ξ , η )=12

(1+ξ )(1−η2 )

N7 (ξ , η )=12

(1−ξ2)(1+η )

N8 (ξ , η )=12

(1−ξ )(1−η2)

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Week 10 MECH3361

Gaussian points can be defined in the natural coordinates as (refer to Table on Page 6)Gaussian point #1: =0.5774, =0.5774; Gaussian point #2:=-0.5774, =0.5774; Gaussian point #3: =-0.5774, =-0.5774; Gaussian point #4:=0.5774, =-0.5774;

Therefore for {u}={1 0 −1 −1 0 1 2 1 }T :

u(ξ ,η)=u(0 .5774 ,0 . 5774 )=∑i=1

8

N i ui= N1 u1+ N2u2+N3 u3+N 4u4+N5u5+N6u6+N7 u7+N 8u8

¿−0 .25 (1−ξ )(1−η)(1+ξ+η )u1−0 .25(1+ξ )(1−η )(1−ξ+η )u2−0.25(1+ξ )(1+η )(1−ξ−η)u3

−0 .25(1−ξ )(1+η)(1+ξ−η )u4+0 .5(1−ξ2 )(1−η)u5+0 .5(1+ξ )(1−η2 )u6+0. 5(1−ξ2)(1+η )u7

+0 . 5(1−ξ )(1−η2 )u8

¿−0 .25×(1−0 .5774 )(1−0 . 5774 )(1+0 .5774+0 .5774 )×1−0 .25×(1+0 .5774 )(1−0 . 5774 )(1−0 .5774+0 .5774 )×0−0 .25×(1+0 .5774 )(1+0 .5774 )(1−0 .5774−0 .5774 )×(−1)−0 .25×(1−0 . 5774)(1+0 . 5774 )(1+0 .5774−0 .5774 )×(−1)+0 . 5×(1−0 .57742)(1−0 . 5774)×(0 )+0 .5×(1+0 .5774 )(1−0 . 57742 )×(1)+0 . 5×(1−0 .57742)(1+0. 5774 )×(2)+0 .5×(1−0 .5774 )(1−0 .57742 )×(1 )∴ u(0 .5774 , 0 .5774 )=1 .6923

For {v }={2 1 −1 −1 1 0 −1 −2 }T :

v ( ξ , η)=v (0 . 5774 , 0 .5774 )=∑i=1

8

N i vi= N1v1+N 2 v2+N 3 v3+ N4 v4+N5 v5+N6 v6+N7 v7+ N8 v8

¿−0 .25×(1−0 .5774 )(1−0 . 5774 )(1+0 .5774+0 .5774 )×2−0 . 25×(1+0 .5774 )(1−0 . 5774 )(1−0 . 5774+0 .5774 )×1−0 . 25×(1+0 .5774 )(1+0 .5774 )(1−0 .5774−0 .5774 )×(−1)−0 . 25×(1−0 . 5774)(1+0 . 5774 )(1+0 .5774−0 .5774 )×(−1)+0 . 5×(1−0 .57742 )(1−0 . 5774)×(1 )+0. 5×(1+0 .5774 )(1−0 .57742)×(0 )+0 . 5×(1−0 .57742 )(1+0. 5774 )×(−1 )+0 .5×(1−0 .5774 )(1−0. 57742 )×(−2)v (0 . 5774 , 0 .5774 )=−0 .9553Similarly, displacements in other Gaussian points can be found using Matlab code as:xi=0.5774; eta=0.5774;u1=1; u2=0; u3=-1; u4=-1; u5=0; u6=1; u7=2; u8=1;v1=2; v2=1; v3=-1; v4=-1; v5=1; v6=0; v7=-1; v8=-2;N1=-(1-xi)*(1-eta)*(1+xi+eta)/4; N2=-(1+xi)*(1-eta)*(1-xi+eta)/4;N3=-(1+xi)*(1+eta)*(1-xi-eta)/4; N4=-(1-xi)*(1+eta)*(1+xi-eta)/4;N5=(1-xi*xi)*(1-eta)/2; N6=(1+xi)*(1-eta*eta)/2;N7=(1-xi*xi)*(1+eta)/2; N8=(1-xi)*(1-eta*eta)/2;u=N1*u1+N2*u2+N3*u3+N4*u4+N5*u5+N6*u6+N7*u7+N8*u8v=N1*v1+N2*v2+N3*v3+N4*v4+N5*v5+N6*v6+N7*v7+N8*v8Use the Matlab code, we can find the following results:

u(0 .5774 , 0 .5774 )=1.6923 , v (0.5774 , 0 . 5774 )=−0 . 9553u(−0 . 5774 , 0 .5774 )=1 . 6218 , v (−0 . 5774 , 0 .5774 )=−1.7956u(−0 .5774 , −0.5774 )=1 .3075 , v (−0 .5774 , −0 .5774 )=−0 .3778u(0 . 5774 , −0 .5774 )=1 .0445 , v (0. 5774 , −0 .5774 )=0 .1290

10