Transcript Finite Element Method

The Finite Element Method
A Practical Course
CHAPTER 4
FEM FOR FRAMES
CONTENTS





INTRODUCTION
FEM EQUATIONS FOR PLANAR FRAMES
– Equations in local coordinate system
– Equations in global coordinate system
FEM EQUATIONS FOR SPATIAL FRAMES
– Equations in local coordinate system
– Equations in global coordinate system
CASE STUDY
REMARKS
INTRODUCTION

Frame members are loaded axially and
transversely.

It is capable of carrying, axial, transverse
forces, as well as moments.

Frame elements are applicable for the analysis
of skeletal type systems of both planar frames
(2D frames) and space frames (3D frames).

Known generally as the beam element or
general beam element in most commercial
software.
FEM EQUATIONS FOR PLANAR
FRAMES

Consider a planar frame element
 d1   u1 
d   v 
 2  1 
 d   
d e   3    z1 
 d 4   u2 
 d5   v2 
   
d 6   z 2 


 diplacement components at node 1




 diplacement components at node 2

Y, V

y, v
x, , u
node 2
(u2, v2, z2)
z
node 1
(u1, v1, z1)
?
l=2a
X, U
z
Equations in local coordinate system
y, v
 Truss
x, , u
+ beam
node 2
(u2, v2, z2)
z
node 1
(u1, v1, z1)
 d1   u1 
d   v 
 2  1 
 d   
d e   3    z1 
 d 4   u2 
 d5   v2 
   
d 6   z 2 
Truss
l=2a
From the truss element,
d1  u1
d 4  u2

Beam
k truss
e
(Expand to 6x6)
 AE
2a









0
0  AE
2a
0
0
0
0
0
AE
2a
sy.
0 0   d1  u1
0 0 
0 0

0 0   d 4  u2
0 0

0 
Equations in local coordinate system
y, v
x, , u
node 2
(u2, v2, z2)
From the beam element
z
(Expand to 6x6)
node 1
(u1, v1, z1)
l=2a
d 2 (v1 ) d3 ( z1 ) d5 (v2 ) d 6 ( z 2 )
k beam
e
0











0
0
0
3 EI z
3 EI z
2 a3
2 a2
2 EI z
0  32EIa3z
a

0  32EIa2z
0
sy.
0
0
3 EI z
2 a3

0 
3 EI z 
2 a 2   d 2  v1
EI z 
 d 3   z1
a

0 

 32EIa2z   d5  v2
2 EI z 
  d 6   z 2
a
Equations in local coordinate system
 AE
2a



truss
ke  





0  AE
2a
0
0
0
0
0
0
AE
2a
sy.
 AE
2a



ke  




0
0
0

0
0

0
0
0
0
0
0
+
0
0
3 EI z
3 EI z
3
2
2a
2a
2 EI z
a
k beam
e

AE
2a
0
0
AE
2a
sy.
0









0
0
3 EI z
3 EI z
3
2
2a
a
 32EIa 2z
0
2a3
0  32EIa2z
0
3 EI z
sy.
 32EIa 3z
0
0  32EIa3z
0
0
3 EI z
2a
2 EI z
0
2 a3
0 
3 EI z 
2a2 
EI z

a

0 
 32EIa 2z 

2 EI z

a
0 
3 EI z 
2 a2 
EI z 
a

0 

 32EIa2z 
2 EI z 

a
Equations in local coordinate system

Similarly so for the mass matrix
0 35 0
0 
70 0


78
22
a
0
27

13
a


2
2
8a
0 13a  6a 
Aa 
me 


70 0
0 
105 

sy.
78  22a 

2 
8
a



And for the force vector,
 f x a  f sx1 
 f a f 
sy1 
 y
 f y a2  m 

s1 
fe   3

f
a

f
sx 2 
 x
 f y a  f sy 2 
 f a2

y
 3  ms 2 
Equations in global coordinate system

Coordinate transformation
Similar to trusses
Y
x
d e  TD e
0
0
0
0
0
0
0
1
0
0
0
0 lx
mx
0
0
0 ly
0 0
my
0
0
0
0

0
0

1
u1
D2j
z1
D3i 0
mx
my
D3j
X
D3i-1
global node i
local node 1
l x
l
y
0
T

,

0
0

 0
z2
x
u2
D3j -2
global node j
local node 2
v1
where
 D3i  2 
D 
 3i 1 
 D3i 
De  

D
3
j

2


 D3 j 1 


 D3 j 

o
D3j - 1
v2
y
D3i -2
2a
Equations in global coordinate system
Y
y
Direction cosines in T:
lx  cos( x, X )  cos  
mx  cos( x, Y )  sin  

z2
D3j
X
le
Y j  Yi
v1
D3i-1
global node i
local node 1
z1
D3i 0
le
u1
D2j

D3i -2
Y j  Yi
le
X j  Xi
le  ( X j  X i )2  (Y j  Yi )2
x
u2
D3j -2
global node j
local node 2
X j  Xi
l y  cos( y, X )  cos(90   )   sin   
my  cos( y, Y )  cos  
x
o
D3j - 1
v2
le
(Length of element)
2a
Equations in global coordinate system
Finally, we have
K e  T k eT
T
Me  T meT
T
Fe  T f e
T
l x
l
y
0
T
0
0

 0
mx
my
0
0
0
0
0
0
0
1
0
0
0
0 lx
mx
0
0
0 ly
0 0
my
0
0
0
0

0
0

1
FEM EQUATIONS FOR
SPATIAL FRAMES

Consider a spatial frame element
v2
 d1   u1 
d   v 
 2  1 
 d3   w1 
   
 d 4   x1 
 d5   y1 
   
 d 6   z1 
de      
 d 7   u2 
 d8   v2 
   
 d9   w2 
d   
 10   x 2 
 d11   y 2 
d   
 12   z 2 



Displacement

 components at
 node 1






 Displacement
 components at
 node 2



u2
y2
v1
2
x2
w2
y1
z2
1
u1
x1
w1
z1
y
x
?
z
Equations in local coordinate system











ke  










Truss + beam
u1
v1
w1  x1
 y1  z 1




AE
2a


u2
v2
w2
 x2
 y2






0
0
0
0
0
0
0
0 
3 EI z 
2 a2 

0 
0 

0 

EI z

a

0 
3 EI z 
2 a2

0 

0 

0 
2 EI z 

a
0
0
0
0
0
 AE
2a
3 EI z
0
0
0
3 EI z
0
2 a3
3 EI y
2 a3
0
GJ
2a
3 EI y
2 a2
0
2 EI y
a
2 a2
0
0
3 EI z
2 a3
0
0
0
0
0
0
0
2 EI z
a
0
AE
2a
3 EI z
2 a2
0
3 EI z
2 a3
sy.
3 EI y
2 a3
0
3 EI y
2a
2
0
 GJ
2a
0
3 EI y
2 a2
0
EI y
a
0
0
0
0
0
0
0
0
0
3 EI y
2a
3
0
GJ
2a
3 EI y
2 a2
0
2 EI y
a
 z2
v2
u2
y2
v1
2
x2
w2
y1
z2
1
u1
x1
w1
z1
y
x
z
Equations in local coordinate system
0
0
0 35 0
0
0
0
0 
70 0 0

78 0
0
0
22a 0 27
0
0
0
 13a 

78 0  22a 0
0 0
27
0
13a
0 


2
2
70rx
0
0
0 0
0
 35rx
0
0 


8a 2
0
0 0  13a
0
 6a 2
0 
Aa 
8a 2 0 13a
0
0
0
 6a 2 


me 
70 0
0
0
0
0 
105 
78
0
0
0
 22a 


78
0
22a
0 


2
sy
.
70
r
0
0
x


2
8a
0 


8a 2 
Ix
where r 
A
2
x
Equations in global coordinate system
D6j-4
D6j-1
d8
d11 d7
y
d10
2
3
d12
x
D6i-4
D6j-5
d9 D6j-2
D6j
D6i-1
d2
d5 d1
d4
1
d6
z
D6i-5
D6j-3
Y
y
d3 D6i-2
x
D6i
X
D6i-3
z
Z
Equations in global coordinate system

Coordinate transformation
d e  TD e
where
 D6 i  5 
D 
 6i  4 
 D6 i  3 


D
 6i  2 
 D6i 1 


D
 6i 
De  

D
6
j

5


 D6 j  4 


D
 6 j 3 
 D6 j  2 


D
 6 j 1 
D 
 6j 
T3
0
, T   0

0
0
T3
0
0
0
T3
0
0
0
0 
0

T3 
l x
T3  l y
l z
mx
my
mz
nx 
n y 
n z 
Equations in global coordinate system
Direction cosines in T3
l x

T3  l y
l z
mx
my
mz
nx 

ny 
n z 
lx  cos( x, X ), mx  cos( x, Y ), nx  cos( x, Z )
l y  cos( y, X ), my  cos( y, Y ), n y  cos( y, Z )
lz  cos( z , X ), mz  cos( z , Y ), nz  cos( z , Z )
Equations in global coordinate system

Vectors for defining location and orientation of
2
frame element in space





V1  X 1 X  Y1Y  Z1Z




V2  X 2 X  Y2Y  Z 2 Z




V3  X 3 X  Y3Y  Z 3 Z
 
V3  V1
 



V2  V1  X 21 X  Y21Y  Z 21Z
 



V3  V1  X 31 X  Y31Y  Z 31Z
 
2
2
l  2a  V2  V1  X 21
 Y212  Z 21
X kl  X k  X l 

Ykl  Yk  Yl 
Z kl  Z k  Z l 
y

y
3
 
V2  V1

x

z
V3

V2
Y
1
V1
 
 
(V2  V1 )  (V3  V1 )
X
k, l = 1, 2, 3
Z
Equations in global coordinate system

Vectors for defining location and orientation of
frame element in space (cont’d)

 
(
V
V ) X  Y  Z 

x  2 1  21 X  21 Y  21 Z
2a
2a
2a
V2  V1
  X 21
l x  cos( x, X )  x  X 
2a
  Y
mx  cos( x, Y )  x  Y  21
2a
  Z 21
nx  cos( x, Z )  x  Z 
2a
 
 
 (V  V )  (V3  V1 )
z  2 1
 
(V2  V1 )  (V3  V1 )

z
 
V3  V1
y
2
3

x
 
V2  V1

z
V3

V2
Y
1
V1
X
 
 
(V2  V1 )  (V3  V1 )



1
{(Y21Z 31  Y31Z 21 ) X  ( Z 21 X 31  Z 31 X 21 )Y  ( X 21Y31  X 31Y21 ) Z }
2 A123
A123  (Y21Z 31  Y31Z 21 ) 2  (Z 21 X 31  Z 31 X 21 ) 2  ( X 21Y31  X 31Y21 ) 2
Z
Equations in global coordinate system

Vectors for defining location and orientation of
frame element in space (cont’d)
1
lz  z  X 
(Y21Z 31  Y31Z 21 )
2 A123
mz  z  Y 
 
V3  V1
1
( Z 21 X 31  Z 31 X 21 )
2 A123
1
nz  z  Z 
( X 21Y31  X 31Y21 )
2 A123
  
y  zx
l y  mz nx  nz mx
m y  nz l x  l z nx

y
2
3

x
 
V2  V1

z
V3

V2
Y
1
V1
X
 
 
(V2  V1 )  (V3  V1 )
n y  l z mx  mz l x
Z
Equations in global coordinate system
Finally, we have
K e  T k eT
T
Me  T meT
T3
0
T
0

0
0
0
T3
0
0
T3
0
0
0
0 
0

T3 
T
Fe  T f e
T
l x
T3  l y
l z
mx
my
mz
nx 
n y 
n z 
CASE STUDY

Finite element analysis of bicycle frame
CASE STUDY
Young’s modulus,
E GPa
Poisson’s ratio,
69.0
0.33

74 elements (71 nodes)
Ensure connectivity
CASE STUDY
Horizontal load
Constraints in all directions
CASE STUDY
M = 20X
CASE STUDY
Axial stress
-9.68 x 105 Pa
-6.264 x 105 Pa
-6.34 x 105 Pa
9.354 x 105 Pa
-6.657 x 105 Pa
-1.214 x 106 Pa
-5.665 x 105 Pa