fem prob and answers

7/29/2019 FEM Prob and Answers

1/7

G.P.Nikishkov. Introduction to the Finite Element Method 1

Problems with solutions

Problem 1

Obtain shape functions for the one-dimensional quadratic element:

Solution

With shape functions, any field inside element is presented as:

.

At nodes the approximated function should be equal to its nodal value:

.

Since the element has three nodes the shape functions can be quadratic polynomials (with

three coefficients). The shape functionN1 can be written as:

.

Unknown coefficients i are defined from the following system of equations:

The solution is: . Thus the shape functionN1 is equal to:

.

The equation system for the determination of coefficients for the shape function

-1 0 1

1 2 3

u ( ) Niui i, 13= =

u 1( ) u1 u 0( ) u2 u 1( ) u3=,=,=

N1 1 2 32

+ +=

N1 1( ) 1 2 3+ 1= =

N1 0( ) 1 0= =

N1 1( ) 1 2 3+ + 0= =

1 0 2 1 2 3 1 2=,=,=

N11

2--- 1 ( )=


2/7


can be written as:

The solution gives and the expression forN2 is: .

It is easy to obtain that the shape functionN3 is equal to: .

Problem 2

Prove that should be preserved at any point inside element.

Solution

The finite element should represent exactly the constant field u = c:

.

Problem 3

Write down a matrix [A] which relates local (element) and global (domain) node enumeration

for the following finite element mesh:

N2 1 2 32

+ +=

N2 1( ) 1 2 3+ 0= =N2 0( ) 1 1= =

N2 1( ) 1 2 3+ + 0= =

1 1 2 0 3 1=,=,= N2 1 2

=

N31

2--- 1 +( )=

Ni 1=

c N1c N2c + + Nic= =

Node orderfor an element

1 2

34

1 2 3

4 5 6

7 8

e1 e2

e3


3/7


Solution

The matrix [A] relates element and global nodal values in the following way:

,

where {Q} is a global vector of nodal values and {Qd} is vector containing element vectors.

The explicit rewriting of the above relation looks as follows:

Problem 4

Evaluate the Jacobian matrix [J] for the four-node element (square with sides equal to 1

rotated by 45 degrees):

Solution

The Jacobian matrix includes the following entries:

.

The components of the Jacobian matrix are calculated using coordinates of element nodes:

Qd{ } A[ ] Q{ }=

Q1

Q2

Q5

Q4

Q2

Q3

Q6

Q5

Q5

Q6

Q8

Q7

1 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 1 0 0 0 0

0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0

Q1

Q2

Q3

Q4

Q5

Q6

Q7

Q8

=

x

y

45

J[ ] x y x y

=


4/7


The shape functions of the linear element are:

,

where , are local coordinates and i, i are their values at nodes. Global coordinates ofnodes are: .

Derivatives of shape functions in respect to the local coordinate are:

.

The first entry of the Jacobian matrix can be calculated as:

.

Other entries can be calculated in the similar manner. The Jacobian matrix for the square lin-

ear element rotated by 45 degrees is equal to:

.

For elements of a simple form (parallel sides) derivatives of global coordinates in respect to

local coordinates can be evaluated as finite differences. For the considered element the esti-

mation is:

.

Problem 5

Calculate nodal equivalents of a distributed load with constant intensity applied to the side of

a two-dimensional quadratic element:

x

Ni

xi y

Ni

yi =,=

Ni1

4--- 1 i+( ) 1 i+( )=

x1 0 y1, 0 x2; 2 2 y2, 2 2 x3 0 y3,=; 2 x4; 2 2 y4, 2 2= = = = = = =

Ni 1

4

---i 1 i+( )=

x

Ni

xi 14--- 1 ( ) 0 1 ( )2

2------- 1 +( ) 0 1 +( ) 2

2-------

+ + 2

4-------= = =

J[ ] 2 4 2 4

2 4

2 4

=

x x

------

2

2------- 2 2

4-------= = =

-1 0 1

1 2 3

l = 1

p = 1

x


5/7


Solution

The nodal equivalent of the distributed load is calculated as:

or

.

The shape functions for the one-dimensional quadratic elements are:

.

The value of nodal forces at nodes 1, 2 and 3 are defined by integration:

Problem 6

Show the fill of the global stiffness matrix for the finite element mesh shown below. Each

node has one degree of freedom.

p{ } N[ ]Tpd

dx d1

1

=

p{ }

p1

p2

p3 N1

N2

N3

pd

dx

ddx,

1

1

x------1

2---= = = =

N112--- 1 ( ) N2 1

2 N3

12--- 1 +( )=,=,=

p11

2--- 1 ( )1

2--- d

1

1

16---= =

p2 1 2

( )12--- d

1

1

23---= =

p3 12--- 1 +( )12--- d1

1

16---= =

Node orderfor an element

1 2

34

1 2 3

4 5 6

7 8

e1 e2

e3


6/7


Solution

The global stiffness matrix has a size of 8 by 8. Location of nonzero values in the global stiff-

ness matrix is defined by pairs of global node numbers for elements. The algorithm of assem-

bly of the element stiffness matrix k into the global stiffness matrix K is given by the

following pseudo-code:

do i=1,n

do j=1,n

K[C[i],C[j]] = K[C[i],C[j]] + k[i,j]

end do

end do

Here C[i] is ith global node number of the element. For example, the element e1 gives con-

tributions to rows 1, 2, 5 and 4. Each of these four rows contains entries from element e1 at

columns 1, 2, 5 and 4. The total fill of the global stiffness matrix is shown below where non-

zero values are denoted by 1:

Problem 7

Describe the structure of the global stiffness matrix from the previous problem using sparse

row-wise format.

Solution

The matrix structure in the sparse row-wise format is represented by the two arrays:

col[pcol[N+1]] = column numbers for nonzero entries of the matrix;

pcol[N+1] = pointers to the beginning of each row.

These arrays are:

col[] = 1,2,4,5, 1,2,3,4,5,6, 2,3,5,6,

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

pcol[] = 0, 4, 10, 14, 18, 26, 32, 36, 40

K[ ]

1 1 0 1 1 0 0 0

1 1 1 1 1 1 0 0

0 1 1 0 1 1 0 0

1 1 0 1 1 0 0 0

1 1 1 1 1 1 1 1

0 1 1 0 1 1 1 1

0 0 0 0 1 1 1 1

0 0 0 0 1 1 1 1

=


7/7


Problem 8

Apply boundary conditionx2 = 0.5 to the following equation system:

Use both the explicit method and the method of large number.

Solution

According to the explicit method, the column 2 of the matrix is multiplied by x2 = 0.5 and is

moved to the right-hand side. Then the equation 2 is replaced by the x2 = 0. The system after

the application of the boundary condition looks like:

Using large number we need to change just the equation with the prescribed value of the

unknown:

1 1 01 2 1

0 1 1

x1

x2

x3 0

0

1

=

1 0 0

0 1 0

0 0 1

x1

x2

x3 0.5

0.5

1.5

=

1 1 0

1 2 1020 1

0 1 1

x1

x2

x3 0

1 1020

1

=

fem prob and answers

Documents