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Example
Wall
1k 2k
3k11,uP 22 ,uP
Obtain expressions for the displacements (u1, u2) of the rigid blocks, in terms of spring stiffnesses k1, k2, k3 and prescribed static loads P1, P2.
Compatibility:
Equilibrium: 121 sFPP =+ 322 ss FFP +=
No rotation of blocks
Constitutive: 111 ukFs = ( )1222 uukFs −= ( )1233 uukFs −=
Example
Wall
1k 2k
3k11,uP 22 ,uP
Combining, we have:
1
211 k
PPu +=
+
++=321
21
12
11kkk
PkPu
Example
Wall
1k 2k
3k11,uP 22 ,uP
Now consider each spring as a 2-noded finite element, as follows:
On each spring, we have:
010 ,uF
1k
111,uF 121,uF
2k
222 ,uF 131,uF
3k
232 ,uF
Geometry
Example
121,uF
2k
222 ,uF
131,uF
3k
232 ,uF
010 ,uF
1k
111,uF
( )10110 uukF −=
( )01111 uukF −=
=
−
−
11
10
1
0
11
11
FF
uu
kkkk
=
−
−
22
21
2
1
22
22
FF
uu
kkkk
=
−
−
32
31
2
1
33
33
FF
uu
kkkk
stiffness matrix displacement
vector
loading/force vector
Example
121,uF
2k
222 ,uF
131,uF3k
232 ,uF
010 ,uF
1k
111,uF
1312111 PFFF =++
23222 PFF =+ 11,uP 22 ,uP010 ,uF
=
+−−−−++−
−
2
1
10
2
1
0
3232
323211
11
0
0
PPF
uuu
kkkkkkkkkk
kk
FKu =
Example
121,uF
2k
222 ,uF
131,uF3k
232 ,uF
010 ,uF
1k
111,uF
2
1
0
3232
323211
11
0
0
uuu
kkkkkkkkkk
kk
→→→
+−−−−++−
−=K
1
0
11
11
uu
kkkk
→→
−
−
2
1
22
22
uu
kkkk
→→
−
−
2
1
33
33
uu
kkkk
→→
−
−
Direct Stiffness Method
Example 11,uP
22 ,uP010 ,uF
=
+−−−−++−
−
2
1
10
2
1
0
3232
323211
11
0
0
PPF
uuu
kkkkkkkkkk
kk
FKu =
FKu 1−=K-1 does not exist as det(K)=0
1k
2k
3k1 2 3
1 2 3 1
2
3
Example
11,uP 22 ,uP010 ,uF
=
+−−−−++−
−
2
1
10
2
1
0
3232
323211
11
0
0
PPF
uuu
kkkkkkkkkk
kk
=
+−−−−++
2
1
2
1
3232
32321
PP
uu
kkkkkkkkk
K-1 exists: FKu 1−=
Example Wall
1k 2k
3k( ) 111 ,, uutP ( ) 222 ,, uutP
Now consider blocks subject to acceleration ü1, ü2, and acceleration ü0 at the wall. Mass of blocks are m1 and m2.
121,uF
2k
222 ,uF 131,uF
3k
232 ,uF010 ,uF
1k
111,uF
( ) 001110 umuukF e =−+
0u
( ) 1110111 umuukF =−+
0ume 11um 11um 22um 11um 22um
1m 2m
( ) 1112221 umuukF =−+
( ) 2221222 umuukF =−+
( ) 1112331 umuukF =−+
( ) 2221332 umuukF =−+
Example
1121 ,, uuF
2k
2222 ,, uuF
1131 ,, uuF 3k
2232 ,, uuF
0010 ,, uuF
1k
1111 ,, uuF
Assembling:
( )( )( )
=
+−−−−++−
−+
tPtPtF
uuu
kkkkkkkkkk
kk
uuu
mm
m we
2
1
2
1
0
3232
323211
11
2
1
0
2
1
0
0
000000
em
1m 2m
FKuuM =+
111 ,, uuP 222 ,, uuP 00 ,, uuFw
em 1m 2m
Example 111 ,, uuP 222 ,, uuP 00 ,, uuFw
( )( )( )
=
+−−−−++−
−+
tPtPtF
uuu
kkkkkkkkkk
kk
uuu
mm
m we
2
1
2
1
0
3232
323211
11
2
1
0
2
1
0
0
000000
em 1m 2m
lumped mass matrix
111 ,, uuP 222 ,, uuP
1m 2m
StationaryWall
What if P1 and P2 are zero, while ü1 and ü2 are non-zero?
prescribed quantities known/measured
quantity
Example
111 ,, uuP 222 ,, uuP
1m 2m
StationaryWall
=
+−−−−++
+
00
00
2
1
3232
32321
2
1
2
1
uu
kkkkkkkkk
uu
mm
0KuuM =+ Free vibration equation
( ) 0uIKM 1 =−− 2ω
If u is periodic, equation may be re-configured as:
Eigenvalue problem
Modal frequency Mode shape
Degrees of Freedom at a Node
y
z x
The freedom of movement of a nodal point, if that point is unconstrained
For a typical structural node in a 3D space, subject to mechanical forces only, We may identify six degrees of freedom (x, y, z, θx, θy, θz)
Each nodal point may have nf degrees of freedom
F(t)
x
y
z
Summary Any structure may be broken up into nodes and elements
A system of n nodes with nf degrees of freedom on each node:
generates an eigenvalue problem with a maximum of nnf modal frequencies and nnf mode shapes
Problem is solved at the nodes and approximated across elements
Essential (displacement) boundary conditions must always be satisfied
generates an nnf x 1 displacement vector generates an nnf x 1 loading/force vector
generates an nnf x nnf stiffness matrix generates an nnf x nnf mass matrix