Previous class:
One dimensional elements (formulations and general procedure);
Review of CEE360/CEE 470 (Structural Analysis)
This class: One Dimensional Elements (cont’d)
I-Clickers
Bar and beam elements of arbitrary orientation
Assembly of elements
Review of CEE360/CEE 470 (Structural Analysis)
Remarks on modeling
Textbook sections
Chapter 2 (2.1 to 2.7, 2.9, 2.12)
CEE570 / CSE 551 Class #4
1
2
Bar / beam element
2
2v
)( 2u
1
1v
)( 1u
Direct approach to assemble stiffness
matrix: give unit d.o.f, calculate
corresponding loads at each node
2
2
1
1
2
2
1
1
44434241
34333231
24232221
14131211
M
F
M
F
v
v
kkkk
kkkk
kkkk
kkkk
Review of previous class
3
Review of Beam Theories
Beam theory planar cross-section assumption
____________ beam – neglect interlayer shear
deformation
n
n
_____________beam – consider interlayer shear
t
Bernoulli-Euler
Timoshenko
4
1D element of arbitrary orientation
Describes element behavior
eeefuK
Defined under base vector coordinate
F1
'
2u
'
1u
F2
x2v
Fy2
2u
1v
1u
Fx1
Fx1
Fy1
Local coord. Global coord.
fdK fdK
xx
5
1D element of arbitrary orientation
F1
'
2u
'
1u
F2
x
Local coord.
2v
F2y
2u
1v
1u
F2x
F1x
F1y
Global coord.
fdK fdK x x
'
2
'
1
u
ud
2
1
F
Ff
2
2
1
1
v
u
v
u
d
2
2
1
1
y
x
y
x
F
F
F
F
f
22K 44K
6
1D element of arbitrary orientation
2
2
1
1
'
2
'
1
00
00
v
u
v
u
sc
sc
u
udTd
Find relationship ,
then compute K }{}{ ff }{}{ dd
cosc
sins
Transformation matrix, T
fTffTf T
1
2
3
42222444
TKTKT
fdK fdK
7
1D element of arbitrary orientation
Find relationship ,
then compute K }{}{ ff }{}{ dd
Direct approach: obtain the same result
22
22
22
22
44
scsscs
csccsc
scsscs
csccsc
kK
2v
F2y
2u
1v
1u
F2x
F1x
F1y
Global coord.
fdK x
Bar element in 3D space:
111
111
000
000
nml
nmlT
k = EA/L
8
1D element of arbitrary orientation
Bar element in 3D space:
111
111
000
000
nml
nmlT
Beam element in 3D space:
Direction cosines of local x’ w.r.t. global x,y,z
A0000A0000A0000A
T1212
333
222
111
nml
nml
nml
A
9
1D element of arbitrary orientation
Coordinate transformation from local coordinate (x’,y’,z’) to
global coordinate (x,y,z)
Rotational transformation does not alter intrinsic element properties
(just change of base)
See Prof. Paulino’s CEE471 (former cee361) web page:
http://cee.uiuc.edu/paulino/cee361
A vector V can be expressed in terms of components uvw in global system xyz
or in terms of components u’v’w’ in local system x’y’z’
10
• Element-level force-displacement relations
eeefuK
Structure nodal equilibrium & displacement
compatibility (assembly and BC)
sssfuK
• Solve system of linear equations
• Compute gradients (stress, strain)
sssfKu \
u~,
Recall: General Computational Procedure
11
Assembly of elements (general concept)
Discretization Assembly
eeefuK sss
fuK
Each structural node must be in equilibrium;
Assembly of elements and loads produces a
set of equations stating “equilibrium” (F=0)
Structural loads come from:
• Element deformation
• Initial stress
• External loads (distributed and concentrated)
12
Assembly of elements
Structural loads applied on nodes include:
• __________________
• ________________ (d.o.f=0)
• __________________
node element
Load applied by element to
structure nodes
dK rdK
Load applied by element to
maintain node equilibrium
dKr
er
P
Structural nodes in equilibrium:
0Prr
elsels N
iie
N
i
i
11
Element deformation
Structural support
Direct external loads
13
Assembly of elements
Structural nodes in equilibrium:
0Prr
elsels N
iie
N
i
i
11
dKr
Internal
force
Reaction
force External
force
RDK Global equilibrium:
elsN
i
i
1
kK
elsN
iie
1
rPR
- {Internal force} =
{reaction force} +
{external force}
14
Applying boundary condition and solving
4
6
fx1=H1
fy1=V1
0
fy2=10
0
0
0
0
0
0
0
fy6=V6
k11
k21
k31
…
…
…
…
…
…
…
…
kn1
1
2
3
5
P=10
V6
H1
V1
unknown
kn1
kn2
kn3
…
…
…
…
…
…
…
…
knn
u1=0
v1=0
u2
v2
u3
v3
u4
v4
u5
v5
u6
v6=0
…
…
…
…
…
…
…
…
…
…
…
…
k13
k23
k33
…
…
…
…
…
…
…
…
kn3
known
unknown
known
k12
k22
k32
…
…
…
…
…
…
…
…
kn2
• Without applying BCs, the stiffness matrix is
________.
• At the dofs where displacements are known,
reactions are not known and vice-versa.
• The system can be reduced by removing the
highlited rows and columns (see below)
singular
15
Interpretation
r
f
r
f
rrrf
frff
f
f
u
u
KK
KK
rfrffff uKfuK
rrrfrfr uKuKf
• Partitioning the system
corresponding to free dofs
and restrained (prescribed)
dofs (denoted by f and r
subscripts)
• Solve for unknown (free)
displacements by these
equations:
(all right-hand side terms
are known)
• Calculate unknown
reactions by these
equations:
(all right-hand side terms
are known)
16
Assembly and Structural Node Numbers
Structural nodes in equilibrium
1 1
2 2
1 2
3
1
2 3
1 4
3 2
Local numbering Structural numbering
3
2
1
987
654
321
11
d
d
d
aaa
aaa
aaa
dk
3
2
1
987
654
321
22
d
d
d
bbb
bbb
bbb
dk
17
Assembly and Structural Node Numbers
1 1
2 2
1 2
3
1
2 3
1 4
3 2
Local numbering Structural numbering
3928173
3625142
3322111
dadadar
dadadar
dadadar
+ =
Contribution to nodal force in
local numbering by element1:
294817
264514
234211
2
4
1
DaDaDar
DaDaDar
DaDaDar
s
s
s
Contribution to nodal force in
structural numbering by element1:
18
Assembly and Structural Node Numbers 3
1
2
1 4
3 2
(Element [K])
1
2
1 2
3
1
2
Local numbering
+ =
294817
264514
234211
2
4
1
DaDaDar
DaDaDar
DaDaDar
s
s
s
Contribution to nodal force in
structural numbering by element 1:
4
3
2
1
564
897
231
0
0000
0
0
4
3
2
1
D
D
D
D
aaa
aaa
aaa
r
r
r
r
s
s
s
s
987
654
321
1
aaa
aaa
aaa
k
s1k (structural) HW (optional): Repeat the procedure for element #2
19
Assembly and Structural Node Numbers 3
1
2
1 4
3 2
(Element [K])
1
2
1 2
3
1
2
Local numbering
+ =
987
654
321
1
aaa
aaa
aaa
k
123
456
789
564
897
231
21
0
0
0
0000
0
0000
0
0
bbb
bbb
bbb
aaa
aaa
aaa
sskkK
987
654
321
2
bbb
bbb
bbb
k
20
Remarks on Modeling
Choice of appropriate elements
Truss use bar element; not beam element
Inappropriate refinement only adds trouble
Limit of “exact solution” of beam element
Explore sections 2.12, 2.13
A B C D E F G H
Solution of FEM Linear Systems
Relationship between structural node numbering
and available solution schemes
Direct versus Iterative Solvers
Next class
21