07 CST Elements

SME 3033 FINITE ELEMENT METHOD

Two-Dimensional Problems Using CST Elements (Initial (Initial notes notes are designed designed by Dr. Nazri Kamsah) Kamsah)


8-1 Introduction A thin plate of thickness thickness t , with a hole in the middle, is subjected to a uniform traction load, T as shown. shown. This 3-D plate can be analyzed as a two-dimensional problem.

y T x

2-D problems generally fall into two stress and plane strain. categories: plane stress stress problem A plane stress

a) Plane Stress

stress problem, where the normal and The thin plate can be analyzed as a s a plane stress shear stresses perpendicular to the x-y plane are assumed to be zero, i.e.  z

 0;  xz  0;  yz  0

The nonzero stress components are  x

 0;  y  0;  xy  0


8-1 Introduction A thin plate of thickness thickness t , with a hole in the middle, is subjected to a uniform traction load, T as shown. shown. This 3-D plate can be analyzed as a two-dimensional problem.

y T x

2-D problems generally fall into two stress and plane strain. categories: plane stress stress problem A plane stress

a) Plane Stress

stress problem, where the normal and The thin plate can be analyzed as a s a plane stress shear stresses perpendicular to the x-y plane are assumed to be zero, i.e.  z

 0;  xz  0;  yz  0

The nonzero stress components are  x

 0;  y  0;  xy  0


y

b) Plane Strain A dam subjected subjected to uniform pressure and a pipe under a uniform internal pressure can be analyzed in twodimension as plain strain problems. The strain components perpendicular to the x-y plane are assumed to be zero, i.e.  z  0;  xz  0;  yz  0

x A dam subjected to a uniform un iform pressure

z

y

Thus, the nonzero strain components are  x ,  y , and  xy .  x

x

 0;  y  0;  xy  0 Pipe under a uniform internal pressure

z


8-2 General Loading Condition A two-dimensional body can be subjected subjected to three types of forces: a) Concent Concentrat rated ed forces forces,, P x & P y at a point, i; b) Body forces, es, f b,x & f b,y acting at its centroid ;

c) Tract ractio ionn forc force, e, T (i.e. force per unit length), acting along a perimeter .


The 2-dimensional body experiences a deformation due to the applied loads.

At any point in the body, there are two components of displacement, i.e. u = displacement in x-direction; v = displacement in y-direction.


Stress-Strain Relation Recall, at any point in the body, there are three components of strains, i.e.  u 

  x     x      v       y         y   xy   u v       y x 

The corresponding stress components at that point are

 x        y     xy 


The stresses and strains are related through,

    D  where [D] is called the material matrix, given by

1 v E  D   v 1   2  1 v  0 0

0



0 1v 2

 

for plane stress problems and v 1  v E  D      v 1 v 1  v  1  2v   0  0

for plane strain problems.

0 



0  1  v 2 


8-3 Finite Element Modeling The two-dimensional body is transformed into finite element model by subdividing it using triangular elements. Note: 1. Unfilled region exists for curved boundaries, affecting accuracy of the solution. The accuracy can be improved by using smaller elements. 2. There are two displacement components at a node. Thus, at a node j, the displacements are: Q2 j 1 in x-direction Q2 j in y -direction


Finite element model of a bracket.


8-4 Constant-Strain Triangle (CST) Consider a single triangular element as shown. The local node numbers are assigned in the counterclockwise order.

The local nodal displacement vector for a single element is given by,

q   q1 ,

q2 , ...,

T

q6 

Within the element, displacement at any point of coordinate ( x, y), is represented by two components, i.e. u in the xdirection and v in the y-direction. Note: We need to express u and v in terms of the nodal displacement components, i.e. q1 , q2 , …, q6 .


8-5 Displacement Functions Displacement components u and v at any point ( x, y) within the element are related to the nodal displacement components through

u  N 1q1  N 2 q3  N 3 q5

(i)

v  N 1q2  N 2 q4  N 3 q6 where N 1, N 2 and N 3 are the linear shape functions, given by N1   ;

N2   ;

N 3  1   

(ii)

in which  and  are the natural coordinates for the triangular element. Substituting Eq.(ii) into Eq.(i) and simplifying, we obtain alternative expressions for the displacement functions, i.e.

u  q1  q5   q3  q5   q5 v  q2  q6   q4  q6   q6

(iii)


Eq.(i) can be written in a matrix form as,

u   N q where

 N1 0  N     0 N1

q   q1 ,

N2

0

N 3

0

0

N2

0

N 3 

q2 , ...,



T

q6 

For the triangular element, the coordinates ( x, y) of any point within the element can be expressed in terms of the nodal coordinates, using the same shape functions N 1 , N 2 and N 3. We have, x  N1 x1  N2 x2  N3 x3 y  N1 y1  N2 y2  N3 y3

This is called an isoparametric representation.


Substituting for N i using eq. (ii), we get

x   x1  x3     x2  x3   x3 y   y1  y3     y2  y3   y3 Using the notation, xij = xi – x j and yij = yi – y j, the above equations can then be written as

x  x13  x23  x3 y  y13  y23  y3 Note: The above equations relate the x- and y-coordinates to the  - and  coordinates (the natural coordinates). We observe that,

x13  x1  x3 y23  y2  y3


8-6 The Shape Functions

The shape functions for the triangular element are illustrated in the figures. Recall, we have N1   ; N 2   ; N 3  1   

Also,

N 1 + N 2 + N 3 = 1


Area Coordinate Representation The shape functions can be physically represented by area coordinates,

N 1  N 2  N 3 

where A is the area of the triangular element, i.e. A = A1 + A2 + A3

A1 A A2 A A3 A

; ;


Exercise 8-1 Consider a triangular element shown below. Evaluate the shape functions N 1 , N 2, and N 3 at an interior point P .

y x

The triangular element for solution.


Solution x  N1 x1  N 2 x2  N 3 x3  1.5N1  7N 2  4N 3  3.85 y  N1 y1  N 2 y2  N 3 y3  2N1  3.5N 2  7N 3  4.8

Using the notation, xij = xi – x j and yij = yi – y j, the above become x   x1  x3     x2  x3   x3  2.5  3  4  3.85 y   y1  y3     y2  y3   y3  5  3.5  7  4.8

Simplifying the equations yields, 2.5  3  0.15 5  3.5  2.2

Solving the equations simultaneously, we obtain  = 0.3 and h = 0.2. Thus, the shape functions for the triangular element are, N 1  0.3

N 2  0.2

N 3  0.5


8-7 Area of the Triangular Element The area, A of any arbitrarily oriented straight-sided triangular elements can be determined using a formula A 

1 2

det  J 

where [J ] is a square matrix called the Jacobian, given by

 x13  J     x23

y13 



y23 

The determinant of the Jacobian [ J ] is det  J   x13 y23  x23 y13

Note: “l l” represents the “magnitude of”. Most computer software use counterclockwise order of local node numbering, and use det [ J ] for computing the area of the triangular element.


8-8 Strain-Displacement Matrix The strains within the triangular element are related to the components of the nodal displacement by a relation

    Bq where [B] is a (3 x 6) rectangular matrix called the strain-displacement matrix, given by

 y23 1  0  B   det  J   x32

0

y31

0

y12

x32

0

x13

0

y23

x13

y31

x21

0 

  y12 

x21

Note: For the given magnitude of { q}, the strains within the element depend only on [B] matrix, which in turns depends on the nodal coordinates, which are constant. Hence the strains are the same everywhere within the element, thus the name constant-strain triangle (CST).


Exercise 8-2 Consider a triangular element in Exercise 8-1. a) Write the Jacobian matrix; b) Find the determinant of the Jacobian matrix; c ) Compute the area of the triangular element; d) Establish the strain-displacement matrix for the element.

y x

The triangular element for solution.


7-9 Potential Energy Approach The total potential energy of a 2-D body, discretized using triangular elements, is given by

 e

1

   D  tdA  2 T

e

T

   u  f tdA   u T tdL e L T

e

 ui P i T

i

The first term represents the sum of internal strain energy of all elements, U e. For a single element, the internal strain energy is Ue 

1

   D  t  dA  2 T

e


8-10 Element Stiffness Matrix We will derive the stiffness matrix of a triangular element using the potential energy approach. Recall, the internal strain energy of an element, U e is given by Ue 

1

    D t  dA T

2

(i)

e

The strains { } are related to nodal displacements { q} by,

    Bq

(ii)

Substituting Eq.(ii) into Eq.(i), we get Ue 

1 2

 q  B   D  B qt  dA T

T

(iii)

e

Taking all constants in Eq.(iii) out of the integral we obtain, Ue 

1 2

q  B  D  B  t  e dA q T

T

(iv)


Note that,

 dA  A , i.e. the area of the triangular element. e

e

Substituting this into eq.(iv) and further simplifying, we get, U e 

1 2

qT t e Ae  BT  D B q

(v)

The internal strain energy of the element can now be written as Ue 

1 2

T

e

q  k  q

(vi)

From eq.(vi) we identify the stiffness matrix [ k ]e of the triangular (CST) element as, e

T

 k   te Ae  B

 D   B

Note: Since there are 6 DOFs for a given element, [ k ]e will be a (6 x 6) rectangular symmetric matrix.


Exercise 8-3 Determine the stiffness matrix for the straight-sided triangular element of thickness t = 1 mm, as shown. Use E = 70 GPa, n = 0.3 and assume a plane stress condition. Solution Element stiffness matrix is given by e T  k   te Ae  B  D   B

(i)

where, t e  1 mm

Ae 



1 2 1 2

det  J  

1 2

23.75

2 Ae  11.875 mm

x13 y23  x23 y13 (Dimension is in mm)


The strain-displacement matrix, [B] is given by

 y23 1 0  B  det  J    x32

0

y31

0

y12

x32

0

x13

0

y23

x13

y31

x21

0 

  y12 

x21

0 7 2 0 2  3.5 0  3.5  7 1   0 47 0 1.5  4 0 7 1.5   23.75   4  7 3.5  7 1.5  4 7  2 7 1.5 2  3.5  0 5 0 0  1.5  3.5 1   B 0 3 0 2.5 0 5.5       23.75   3 3.5 2.5 5 5.5 1.5 


The transpose of [ B] matrix is, 0 3  3.5  0  3 3.5    0 2.5 1  5 T  B    5  2.5 23.75  0 1.5 0 5.5    5.5 1.5  0

For a plane stress condition, the material’s matrix [D] is given by

 1 n E  n 1  D  2  1 n  0 0

 3  70 10 0  2 1 0.3  1 1 n   2 0

 1 0.3  0.3 1  0 0

  0  1 1  0.3  2 0


Substituting all the terms into eq.(i) we have, 0 3   3.5  0  3  3.5     1 0.3 0  3 0  2.5 70  10  1  5 e  k   1 11.875  0 . 3 1 0     5  1  0.32    2.5 23.75  0  0 0 0.35   1.5  0 5.5   5.5  1.5   0

 1.5 0 5 0 0   3.5 1      0 3 0 2.5 0 5.5  23.75    3  3.5  2.5  5.5  1.5 5


Multiplying and simplifying, we obtain q1

q2

q3

q4

q5

q6

2.494 1.105 2.409 0.425 0.085 0.68    2.152 0.233 0.223 0.873 2.374    4.403 1.316 1.994 1.549  e 4  k   10    2.429 1.741 2.652     2.079 0.868   5.026   symmetry

Note: Connectivity with the local DOFs is shown.


8-11 Element Force Vector We will derive the force vector for a single element, which is contributed by a) body force, f and b) traction force, T . We need to convert both f and T into the equivalent nodal forces. Note: The concentrated forces can be included directly into the global load vector, appropriate DOF direction.

a) Body Force Suppose body force components, f x and f y, act at the centroid of a triangular element. The potential energy due to these forces is given by,



e

T

u

 f t  dA  te e uf x  vf y  dA

(i)


u  N 1q1  N 2 q3  N 3 q5

Recall,

v  N 1q2  N 2 q4  N 3 q6 1 Ni dA  Ae e 3



Also,

Substituting the above into eq.(i), we get

 u

T

e

T

e

 f  t  dA  q  f 

where { f }e is the element body force vector, given by e

 f 

te Ae

 f x ,  3

fy,

fx ,

fy,

fx ,

f y 

Note: Physical representation of force vector { f }e is shown.

T


b) Traction Force Suppose a linearly varying traction components act along edge 1-2 of a triangular element.

The potential energy due to the traction force is,



e

T

u

(ii)

T tdL  l  uT x  vTy  tdL 12

Using the relations, u  N1q1  N 2q3 v  N1q2  N 2q4 T x  N1Tx1  N 2Tx 2

with,

T y  N1Ty1  N 2Ty 2 1 Also, l N dl  l12 , 12 3



2 1

l12 

1 N dl  l12 , l12 3



2 2

2

 x2  x1    y2  y1  1 N1N 2dl  l 12 l 12 6



2


Substitution into eq.(ii) yields,

 u

T

e

T tdL   q1,

q2 ,

T

e

q3 , q4  T 

where {T }e is the equivalent nodal force vector due to traction force, given by

T e 

t e l 12 6

2T

x1  T x 2 

2T

 y1  T y 2

Note:

The physical representation of the nodal force vector {T }e is shown.

T x1  2T x2  T y1  2T y 2 T

(iii)


Special Case: If the traction forces are uniform, then

T x1  Tx 2  Tx ;

Ty1  Ty 2  Ty

Thus, the nodal force vector in eq.(iii) becomes e

T  

tel 12

T

T x , Ty , Tx , Ty  2

(iv)


8-12 Concentrated Force The concentrated force term can be easily considered by having a node at the point of application of the force. If concentrated load components P x and P y are applied at a point i, then T

ui Pi  Q2i 1Px  Q2i Py Thus, P x and P y, i.e. the x and y components of { P }i get added to the ( 2i - 1)th component and (2i)th components of the global force vector, { F }.

Note: The contribution of the body, traction and concentrated forces to the global force vector, { F } is represented by,

 F  

 e

e

e



 f   T   P 


Exercise 8-4 Consider a portion of finite element model of a plate as shown. A uniform traction force of 2 kN/m2 acts along the edges 7-8 and 8-9 of the model. Determine the equivalent nodal forces at nodes 7, 8, and 9.


Solution We will consider the two edges, 7-8 and 8-9 separately, and then merge the final results. cos  sin  

20 15 15 25

 

4

T x  T cos  2 

5 3

T y  T sin   2 

5

4

5 3 5

 1.6 kN/m  1.6 N/mm

 1.2 kN/m  1.2 N/mm

For edge 7-8 (edge 1-2 local) l 12 

100  852  20  402  25 mm

t e  10 mm

Equivalent nodal forces due to uniform traction force T = 2 kN/m2 is,

T   1

1025 2

1.6

1.2 1.6 1.2  200 150 T

200 150 

T

N


For edge 8-9 (edge 1-2 local) l 12 

85  702  40  602  25 mm

t e  10 mm

Equivalent nodal forces due to uniform traction force T = 2 kN/m2 is, 2

T   2

10  25  2

T

1.6 1.2 1.6 1.2 

T    200 150

T

200 150  N

These loads add to global forces F 13, F 14 , …,F 18 as shown.


8-13 Strains and Stress Calculations a) Strains The strains in a triangular element are,

    x   du dx  e    dv       y    dy     du dv   xy   dy  dx     e

    Bq Note: We observed that { }e depends on the [B] matrix, which in turn depends only on nodal coordinates ( x ,i yi), which are constant. Therefore, for a given nodal displacements {q}, the strains {  }e within the element are constant.

Hence the triangular element is called a constant-strain triangle.


b) Stresses The stresses in a triangular element can be determined using the stress-strain relation,

 x  e e e       y    D    D Bq    xy  Note: 1. Since the strains { }e are constant within the element, the stresses are also the same at any point in the element.

2. Stresses for plane stress problem differ from those for plane strain problem by the material’s matrix [D]. 3. For interpolation purposes, the calculated stresses may be used as the values at the centroid of the element.

4. Principal stresses and their directions are calculated using the Mohr circle.


Example 8-1 Consider a thin plate having thickness t = 0.5 in. being modeled using two CST elements, as shown. Assuming plane stress condition, (a) determine the displacements of nodes 1 and 2, and (b) estimate the stresses in both elements.


Solution Element connectivity Local Nodes Element No

1

2

3

1

1

2

4

2

3

4

2

For plane stress problem, the materials matrix is given by

 1 n  0 E   n 1 0  D  2  1 n   0 0 12 1 n   0.25 0   1  0  D  32 106 0.25 1   0 0 0.375


Element 1 A1 

Area of element,

1 2

det  J  

1 2

 6   3 in 2

The strain-displacement matrix,

 y23 1  0  B   det  J   x32

0

y31

0

y12

x32

0

x13

0

y23

x13

y31

x21

 2 0 3 0 2 0  1   x21  0 3 0 3 0 0  6   3 2 3 0 0 2 y12  0

Multiplying matrices [ D][ B] we get,

(1)

 D  B

0  1.067 0.4 0 0.4 1.067  0  107  0.267 1.6 0 1.6 0.267   0.6 0.4 0.6 0 0.4  0


The stiffness matrix is given by, (1)

T

 k   t1 A1  B1  D B1 Substitute all parameters and multiplying the matrices, yields Q1

(1)

 k 

Q2

Q3

Q4

Q7

Q8

0.5 0.45 0.2 0.533 0.3  0.983     1.4 0.3 1.2 0.2 0.2    0.3 0.45 0 0 7  10    1.2 0.2 0    symmetric 0.533 0    0.2  

Note: Connectivity with global DOFs are shown.


Element 2 A2 

Area of element,

1 2

det  J  

1 2

 6   3 in 2

The strain-displacement matrix is

 y23 1  0 B     det  J   x32

0

y31

0

y12

x32

0

x13

0

y23

x13

y31

x21

 2 0 0 0 2 0  1  0 3 0 3 0 0 x21   6   3 2 3 0 0 2 y12  0

Multiplying matrices [ D][ B] we get,

(2 )

 D  B

0 0.4 1.067 0  1.067 0.4  107  0.267 1.6 1.6 0.267 0  0  0.6 0 0.4  0.4 0.6 0


The stiffness matrix is given by, (2 ) T  k   t2 A2  B2  D B2

Substituting all parameters and multiplying the matrices yield Q5

(2)

 k 

Q6

Q7

Q8

Q3

Q4

0.5 0.45 0.2 0.533 0.3  0.983   1.4 0.3 1.2 0.2 0.2      0.3  0.45 0 0 7  10    1.2 0.2 0    symmetric 0.533 0    0.2  

Note: Connectivity with global DOFs are shown.


Write the global system of linear equations, [ K ]{Q} = { F }, and then apply the boundary conditions: Q2 , Q5 , Q6 , Q7 , and Q8 = 0.

The reduced system of linear equations are,

 0.983 0.45 0.2 Q1   0      7  10  0.45 0.983 0 Q3    0        0.2 0 1.4   Q4  1000 Solving the reduced SLEs simultaneously yields,

 Q1   1.913      5 Q 0.875 in.   3   10 Q  7.436  4  

07 CST Elements

Recommend Documents