Advanced Finite Element Methods for Engineers

Exercise 10

Prof. Dr.-Ing. Mikhail Itskov

Department of Continuum Mechanics, RWTH Aachen University, Germany

WiSe 2018/19
Overview

Repetition
Quadratic shape functions
Task 1

2 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Quadratic shape functions

For piecewise defined quadratic shape functions e

holds in element e i j k
û = aiNie + aj Nje + ak Nke in Ωe 1
Nie Nke
where ai, aj and ak are the values which corre- 0
sponds to the three to e corresponding nodes. Nje

3 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Quadratic shape functions

For piecewise defined quadratic shape functions e

holds in element e i j k
û = aiNie + aj Nje + ak Nke in Ωe 1
Nie Nke
where ai, aj and ak are the values which corre- 0
sponds to the three to e corresponding nodes. Nje

The parts of the shape functions in element e calculate as:

(xj − x) (xk − x)
Nie(x) =
(xj − xi) (xk − xi)
(xi − x) (xk − x)
Nje(x) =
(xi − xj ) (xk − xj )
e (xi − x) (xj − x)
Nk (x) =
(xi − xk ) (xj − xk )

3 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Overview

Repetition
Quadratic shape functions
Task 1

4 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Given is a linear-elastic bar. Calculate the stiffness matrix.

l
Divide the bar into three elements of length 3 and use piecewise quadratic shape
functions and the Galerkin method.

5 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Given is a linear-elastic bar. Calculate the stiffness matrix.

l
Divide the bar into three elements of length 3 and use piecewise quadratic shape
functions and the Galerkin method.
∂ 2u
Help: EA 2 = −q(x)
∂x

5 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1

x
l/3 2l/3 l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1

N1
x
l/3 2l/3 l

9 18
N1 = 1 − x + 2 x2 0≤x≤ l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2

N1
x
l/3 2l/3 l

9 18
N1 = 1 − x + 2 x2 0≤x≤ l
3
l l
12 36
N2 = x − 2 x2 0≤x≤ l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2

N1 N3 N3
x
l/3 2l/3 l

9 18
N1 = 1 − x + 2 x2 0≤x≤ l
3
l l
12 36
N2 = x − 2 x2 0≤x≤ l
3
l l
(
18 2 l
l2
x − 3l x 0≤x≤ 3
N3 = 21 18 2 l 2l
6− l
x + l2
x 3
≤x≤ 3

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2 N4

N1 N3 N3
x
l/3 2l/3 l

9 18
N1 = 1 − x + 2 x2 0≤x≤ l
3
l l
12 36
N2 = x − 2 x2 0≤x≤ l
3
l l
(
18 2 l
l2
x − 3l x 0≤x≤ 3
N3 = 21 18 2 l 2l
6− l
x + l2
x 3
≤x≤ 3

36 36
N4 = −8 + x − 2 x2 l
3
≤x≤ 2l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2 N4

N1 N3 N3 N5 N5
x
l/3 2l/3 l
(
15 18 2 l 2l
9 18 3− l x + l2 x 3 ≤ x≤ 3
N1 = 1 − x + 2 x2 0≤x≤ l N5 =
l l 3
15 − 33l x + 18
l2 x
2 2l
3 ≤x ≤l
12 36
N2 = x − 2 x2 0≤x≤ l
3
l l
(
18 2 l
l2
x − 3l x 0≤x≤ 3
N3 = 21 18 2 l 2l
6− l
x + l2
x 3
≤x≤ 3

36 36
N4 = −8 + x − 2 x2 l
3
≤x≤ 2l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2 N4 N6

N1 N3 N3 N5 N5
x
l/3 2l/3 l
(
15 18 2 l 2l
9 18 3− l x + l2 x 3 ≤ x≤ 3
N1 = 1 − x + 2 x2 0≤x≤ l N5 =
l l 3
15 − 33l x + 18
l2 x
2 2l
3 ≤x ≤l
12 36
N2 = x − 2 x2 0≤x≤ l
3 60 36
l l
N6 = −24 + x − 2 x2 2
3
l ≤x≤l
(
18 2 l
l l
l2
x − 3l x 0≤x≤ 3
N3 = 21 18 2 l 2l
6− l
x + l2
x 3
≤x≤ 3

36 36
N4 = −8 + x − 2 x2 l
3
≤x≤ 2l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2 N4 N6

N1 N3 N3 N5 N5 N7
x
l/3 2l/3 l
(
15 18 2 l 2l
9 18 3− l x + l2 x 3 ≤ x≤ 3
N1 = 1 − x + 2 x2 0≤x≤ l N5 =
l l 3
15 − 33l x + 18
l2 x
2 2l
3 ≤x ≤l
12 36
N2 = x − 2 x2 0≤x≤ l
3 60 36
l l
N6 = −24 + x − 2 x2 2
3
l ≤x≤l
(
18 2 l
l l
l2
x − 3l x 0≤x≤ 3
N3 =
6− 21
x + 18 2
x l
≤x≤ 2l
27 18
l l2 3 3
N7 = 10 − x + 2 x2 2
3l ≤x≤l
l l
36 36
N4 = −8 + x − 2 x2 l
3
≤x≤ 2l
3
l l

6 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
The equation system Ka = r for the weighted residual approach is known. The
weights for the Galerkin method are Wi = Ni:
Z n
! Z
X
Wi L aj Nj dΩ = − Wi [L (u0) + p] dΩ i = 1, 2, ..., n
Ω j=1 Ω
Z n
! Z
X
Ni L aj Nj dΩ = − Ni [L (u0) + p] dΩ i = 1, 2, ..., n
Ω j=1 Ω

7 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
The equation system Ka = r for the weighted residual approach is known. The
weights for the Galerkin method are Wi = Ni:
Z n
! Z
X
Wi L aj Nj dΩ = − Wi [L (u0) + p] dΩ i = 1, 2, ..., n
Ω j=1 Ω
Z n
! Z
X
Ni L aj Nj dΩ = − Ni [L (u0) + p] dΩ i = 1, 2, ..., n
Ω j=1 Ω

q d2
Rearranging and insertion of the problem p = EA , L= dx2 , u0 = 0 leads to
Zl n
! Zl
X q
Ni aj Nj′′ dx = − Ni dx i = 1, 2, ..., n
j=1
EA
0 0
 l 
n Z Zl
 NiNj′′ dx aj = − Ni q dx
X
i = 1, 2, ..., n
j=1
EA
0 0

7 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Rb ′
Rb
With integration by parts AB dx = AB|ba − A′B dx follows
a a
 l 
n Z Zl
X q
 Ni Nj′′ dx aj =− Ni dx i = 1, 2, ..., n
j=1
|{z} |{z} EA
0 A B′ 0

8 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Rb ′
Rb
With integration by parts AB dx = AB|ba − A′B dx follows
a a
 l 
n Z Zl
X
′′ q
 N i N j dx  aj = − N i dx i = 1, 2, ..., n
j=1
|{z} |{z} EA
0 A B′ 0
 l 
n Zl Zl
X q
Ni Nj′  − Ni′ Nj′ dx aj = − Ni
 
 |{z} dx i = 1, 2, ..., n
j=1
|{z} |{z} |{z} EA
A B 0 A′ B 0
0

8 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Rb ′
Rb
With integration by parts AB dx = AB|ba − A′B dx follows
a a
 l 
n Z Zl
X
′′ q
 N i N j dx  aj = − N i dx i = 1, 2, ..., n
j=1
|{z} |{z} EA
0 A B′ 0
 l 
n Zl Zl
X q
Ni Nj′  − Ni′ Nj′ dx aj = − Ni
 
 |{z} dx i = 1, 2, ..., n
j=1
|{z} |{z} |{z} EA
A B 0 A′ B 0
0
 l 
n   n Z Zl
 Ni′ Nj′ dx aj  = − Ni q dx
X   l
X
NiNj′ 0 aj − i = 1, 2, ..., n
j=1 j=1
EA
| {z } 0 0
l
[u′ Ni ]0

8 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Rb ′
Rb
With integration by parts AB dx = AB|ba − A′B dx follows
a a
 l 
n Z Zl
X
′′ q
 N i N j dx  aj = − N i dx i = 1, 2, ..., n
j=1
|{z} |{z} EA
0 A B′ 0
 l 
n Zl Zl
X q
Ni Nj′  − Ni′ Nj′ dx aj = − Ni
 
 |{z} dx i = 1, 2, ..., n
j=1
|{z} |{z} |{z} EA
A B 0 A′ B 0
0
 l 
n   n Z Zl
 Ni′ Nj′ dx aj  = − Ni q dx
X   l
X
NiNj′ 0 aj − i = 1, 2, ..., n
j=1 j=1
EA
| {z } 0 0
l
[u′ Ni ]0
 l 
n Z Zl
X
 Ni′ Nj′ dx aj  q l
− =− Ni dx − [u′ Ni]0 i = 1, 2, ..., n
j=1
EA
0 0

8 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
 l 
n Z Zl
 Ni′Nj′ dx aj  = − Ni q dx − [u′Ni]l
X
− 0
j=1
EA
0 0
leads in matrix notation to
Zl Zl
q l
Kji = − Ni′Nj′ dx aj , rj = − Ni dx − [u′Ni]0
EA
0 0

9 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution

Ka = r
with
 
ZlN1′ N1′ · · · N7′ N1′
K = −  ... ... ..  dx
.
′ ′ ′ ′
0 N1 N7 · · · N7 N7
 
a1
a =  ... 
a7
   ′ 
Zl N1 u (l)N1(l) − u′ (0)N1(0)
q  ..  ..
r =− . dx −  . 
EA
0 N7 u′ (l)N7(l) − u′ (0)N7(0)

10 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution

Ka = r
with
 
ZlN1′ N1′ · · · N7′ N1′
K = −  ... ... ..  dx
.
′ ′ ′ ′
0 N1 N7 · · · N7 N7
 
a1
a =  ... 
a7
   ′ 
Zl N1 u (l)N1(l) − u′ (0)N1(0)
q  ..  ..
r =− . dx −  . 
EA
0 N7 u′ (l)N7(l) − u′ (0)N7(0)

E, A and q can be dependent of x and can in general not be taken out of the integral. In our case only q has this
dependency.

10 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Due to the boundary condition u(0) = 0 follows directly a1 = 0. The system can
be reduced by eliminating the first row and column.

11 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Due to the boundary condition u(0) = 0 follows directly a1 = 0. The system can
be reduced by eliminating the first row and column.
 
Z l N2′ N2′ · · · N7′ N2′
K = −  .. ... ..  dx
0 N2′ N7′ · · · N7′ N7′
 
a2
a =  .. 
a7
   ′ ′

Zl
q N. 2 u (l)N2(l) − u (0)N2(0)
..
r =− . dx −  
EA N u′(l)N7(l) − u′(0)N7(0)
0 7

11 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
y
1
N2 N4 N6

N1 N3 N3 N5 N5 N7
x
l/3 2l/3 l
(
15 18 2 l 2l
9 18 3− l x + l2 x 3 ≤ x≤ 3
N1 = 1 − x + 2 x2 0≤x≤ l N5 =
l l 3
15 − 33l x + 18
l2 x
2 2l
3 ≤x ≤l
12 36
N2 = x − 2 x2 0≤x≤ l
3 60 36
l l
N6 = −24 + x − 2 x2 2
3
l ≤x≤l
(
18 2 l
l l
l2
x − 3l x 0≤x≤ 3
N3 =
6− 21
x + 18 2
x l
≤x≤ 2l
27 18
l l2 3 3
N7 = 10 − x + 2 x2 2
3l ≤x≤l
l l
36 36
N4 = −8 + x − 2 x2 l
3
≤x≤ 2l
3
l l

12 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Compon. a b Value Compon. a b Value
K22 K44
K23, K32 K45, K54
K24, K42 K46, K64
K25, K52 K47, K74
K26, K62 K55
K27, K72 K56, K65
K33 K57, K75
K34, K43 K66
K35, K53 K67, K76
K36, K63 K77
K37, K63

13 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Compon. a b Value Compon. a b Value
1 1
K22 0 l K44
3 3l l
1 1 2
K23, K32 0 3l K45, K54 l l
3 3
K24, K42 - - K46, K64 - -
K25, K52 - - K47, K74 - -
K26, K62 - - 1
K55 3l l
K27, K72 - - 2
K56, K65 3l l
2 2
K33 0 3l K57, K75 3l l
1 2 2
K34, K43 3l 3l K66 3l l
1 2 2
K35, K53 3l 3l K67, K76 3l l
K36, K63 - - 2
K77 3l l
K37, K63 - -

13 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Compon. a b Value Compon. a b Value
1
K22 0 l − 16l K44 1
− 16l
3 3l l
1 8 1 2 8
K23, K32 0 3l l K45, K54 l l
3 3 l
K24, K42 - - 0 K46, K64 - - 0
K25, K52 - - 0 K47, K74 - - 0
1
K26, K62 - - 0 K55 3l l − 14l
K27, K72 - - 0 2 8
K56, K65 3l l l
2
K33 0 3l − 14l K57, K75 2
3l l − 1l
1 2 8 2
K34, K43 3l 3l l K66 3l l − 16l
1 2
K35, K53 3l 3l − 1l K67, K76 2
3l l 8
l
2
K36, K63 - - 0 K77 3l l − 7l
K37, K63 - - 0

13 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Task 1

Solution
Finally follows for the stiffness matrix
 
−16 8 0 0 0 0
 8 −14 8 −1 0 0
 
 0 8 −16 8 0 0 1
K=  
 0 −1 8 −14 8 −1 l

 0 0 0 8 −16 8 
0 0 0 −1 8 −7

14 of 15 Advanced Finite Element Methods for Engineers | Prof. Dr.-Ing. Mikhail Itskov |
Department of Continuum Mechanics, RWTH Aachen University, Germany | WiSe
2018/19
Thank you for your attention!

Any questions?