fem prob and answers
TRANSCRIPT
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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 ( )=
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G.P.Nikishkov. Introduction to the Finite Element Method 2
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
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G.P.Nikishkov. Introduction to the Finite Element Method 3
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
=
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G.P.Nikishkov. Introduction to the Finite Element Method 4
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
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G.P.Nikishkov. Introduction to the Finite Element Method 5
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
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G.P.Nikishkov. Introduction to the Finite Element Method 6
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
=
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G.P.Nikishkov. Introduction to the Finite Element Method 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
=