1d elements - indian institute of technology...
TRANSCRIPT
1D ELEMENTS
• Simplest type of FE problems• All object are 1D• All forces are 1D• All stresses / strains are 1D
BASICS
• u=u(x) : Deformations• ε=ε(x) : Strain• σ=σ(x) : Stress• f=f(x) : Body forces• T=T(x) : Tractive forces• P=P(x) : Point loads• σ=Eε, ε=du/dx
INDIVIDUAL ELEMENTS
x1 x2 :Coordinatesξ=-1 ξ=1 : Local coordinatesq1 q2 : Deformations
(At the nodes)
q2:Deformation
q1 within the element
Linear Interpolation N1, N2: Shape functions
node1 node2
LINEAR SHAPE FUNCTIONS
N1 N2
u = N1 q1 + N2 q2 = (1 - ξ) / 2 * q1+ (1 + ξ) / 2 * q2
Relationship between local and global coordinates-
ξ = -1+ 2*(x-x1)/(x2-x1)
N1 =(1 - ξ) / 2 N2 =(1 + ξ) / 2
STRAIN:
ε = du/dx
= du/dξ * dξ/dx
u=N1q1+N2q2
dξ/dx = 2/(x2-x1)
= 2/Le
du/dξ = (-q1+q2)/2
therefore-
ε = 1/le [-1 1][q1 q2]T
= Bq
N1 =(1 - ξ) / 2 N2 =(1 + ξ) / 2
Where B is the element strain matrix and
B== 1/le [-1 1]
As B is constant, this element is CONSTANT STRAIN ELEMENT
That means strain inside the element does not vary.
STRESS:
σ = EBq
P.E. approach
↓↓↓↓
−−−∈
−−−∈=Π
∑∑ ∑ ∫∫∫∑
∑∫∫∫
iie e L
T
L
T
L
T
e
iiL
T
L
T
L
T
PQTdxuAdxFuAdx
PuTdxuAdxFuAdx
σ
σ
21
21
STRAIN ENERGY FORCED TRACTIVE POINT FIELD FORCE FORCE LOADS
Total energy of the body is the sum of P.E, of all Elements.
ELEMENT STRAIN ENERGY
qEA qle
T2
1
1111
⎥⎦
⎤⎢⎣
⎡−
−=
∫ ∈=L
Te AdxU σ2
1
∫=L
TT EBqAdxBq21
qBdxBEAqL
TT⎥⎦⎤
⎢⎣⎡= ∫2
1
[ ] [ ]
∫
∫
⎥⎦
⎤⎢⎣
⎡−
−=
⎥⎦⎤
⎢⎣⎡ −−=
qdxl
EAq
qdxEAq
e
T
L
Tl
Te
11111
1111.
221
12
12
CONTRIBUTION OF THE FORCES TO PE
BODY FORCES
dxNfAqfAdxNq TTTT∫ ∫=
⎥⎦
⎤⎢⎣
⎡=
∫∫
dxNdxN
fAq T
2
1
node1 node2
1N1
2/2/)(.
12
1
eLxxtriangletheunderareadxN
=−==∫
x1x2
2tf 2tf
⎥⎦
⎤⎢⎣
⎡=
11
2eT lfAq
vectorforcebodyelementfq eT ==
⎥⎦
⎤⎢⎣
⎡=11
2eT fAlq
Thus, body force on element gets split equally attwo nodes.
TRACTIVE FORCES
2tT 2tT
∫ ∫= TdxNqTdxu TTT
⎥⎦
⎤⎢⎣
⎡=
∫∫
dxNdxN
TqT
2
1
eT
eT
Tq
Tlq
=
⎥⎦
⎤⎢⎣
⎡=
11
2[TOTAL TRACTIVE FORCE ON THE ELEMENT CAN BE ASSUMED TO BE SPLIT EQUALLY AT THE NODES]
∑−−−=Π PqTqfqqkq Te
Te
Te
Te 2
1
∑ ∑ ∑−=Π=Π eT
eT
e Fqqkq21
⎥⎦⎤
⎢⎣⎡ +⋅+=
2222eeee
eTlfAlTlfAlF
POTENTIAL ENERGY
Potential energy for an element is-
Total Potential energy-
CONSIDER A SET OF ELEMENTS
Element No. Node 1 Node 2
1 1 2
2 2 3
3 3 4
4 4 5
5 5 6
1 e1 2 e2 3 e3 4 e4 5 e5 6
Q1 Q2 Q3 Q4 Q5 Q6
ELEMENT STIFFNESS MATRIX
[ ] [ ] [ ] [ ]TTTT qqqqqqqq 54433221
[ ]TnqqqqQ −−−= 321
L⎥⎦
⎤⎢⎣
⎡−
−⎥⎦
⎤⎢⎣
⎡−
−⎥⎦
⎤⎢⎣
⎡−
−1111
1111
1111
3
33
2
22
1
11
lAE
lAE
lAE
GLOBAL DISPLACEMENT MATRIX
ELEMENT DISPLACEMENT MATRIX
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−+−
−+−
−−−−+−
−−−
=
n
nn
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
nlAE
lAE
K
.
.
.
00
0
0
00
4
44
3
33
3
33
3
33
3
33
2
22
2
22
2
22
2
22
1
11
1
11
1
11
1
11
2 L2 3 i Li I+11 L1
GLOBAL STIFFNESS MATRIX1 2 3 4 5 6 .......... n
K = 1⎡k111 k1
12 0 0 0 0 ..... 0⎤2⎜k1
21 k122+k2
11 k212 0 0 0 ..... 0 ⎜
3⎜0 k221 k2
22+k311 k3
12 0 0 ..... 0 ⎜4⎜0 0 k3
21 k322+k4
11 k412 0 ..... 0 ⎜
5..
n ⎣0 0 ....... km21 km
22⎦where
ki11, ki
12, ki21, ki
22: stiffness matrix elements of element number i.
PROPERTIES OF K
• Symmetric• Banded• Can be sparse (if numbering is not
appropriate)• Is N X N (where N is the number of nodes
in a 1 D problem)
THE RESULTING K MATRIX: SPARSE AND NON-BANDED
1 2 3 4 5 6
K = 1⎡k111 0 0 0 0 k1
12 ⎤2⎜0 k2
22+k311 k3
12 0 0 k212 ⎜
3⎜0 k321 k4
11+k322 k4
12 0 0 ⎜4⎜0 0 k4
21 k422+k5
11 k512 0 ⎜
5⎜0 0 0 k521 k5
22 0 ⎜6⎣k1
21 k212 0 0 0 k1
22 + k211⎦
ASSEMBLING GLOBAL FORCE MATRICES FROM ELEMENT MATRICES
∑∑ +=++= PFP)T(f ie
ie
ie
Where : force on node 1 of element i: force on node 2 of element I: point load on node number i
T441
e32
e
331e
22e
212e
12e
111e
)...]PF(F
)PF)(FPF)(FP[(FF
++
+++++=
i1eF
i2eFiP
[ ][ ]
MATRIXFORCEGLOBALfffffffF
fifiOR
MATRIXFORCEELEMENT
TllfATllfA
T
iiiiii
...
2222
41323122211211
21
+++=
⎥⎦⎤
⎢⎣⎡ ++
TOTAL PE USING GLOBAL MATRICES
∑ ∑ ∑ ∑−−−=∏ iieTeT
eT uPTqfqqkq ******2/1
iieT
eT uPFqqkq∑ ∑ ∑−−= *****2/1
FQQKQ TT ****2/1 −=
where
Q = [q1 q2 q3 q4 .... qn]T
K : Global stiffness matrix
F : Global Force matrix
( ) ∫ ∑∫∫ =−−−∈s
T
v
T
v
T PTdSfdVdV
APPROACHSGALERKIN
0
'
φφσφσ
( )
( )
( )[ ]T
BN
dxd
x
21 φφ
φφ
φφ
φφ
=Ψ
Ψ=⇒∈Ψ=
=⇒∈
=
DONEWORKINTERNAL
( )
[ ]
Ψ=
Ψ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
−=
Ψ=
Ψ=
Ψ=∈
∫
∫∫
eT
e
eT
Tee
T
Te
T
eTTT
kq
lEAq
BBlEAq
dxBBEAq
dxAEBBqdV
1111
φσν
eT
eeT
eeT
TeeT
Te
T
eTTT
F
lAf
dN
dNlAf
dNlAf
dxNfA
dxfANfdV
Ψ=
⎥⎦
⎤⎢⎣
⎡Ψ=
⎥⎥⎦
⎤
⎢⎢⎣
⎡Ψ=
Ψ=
Ψ=
Ψ=
∫∫
∫
∫
∫ ∫
11
2
2
2
2
1
ξ
ξ
ξ
φν
Virtual work dueTo body forces-
[ ]Teee
TT TlTwhereTTdx
DONEWORKTRACTIVESIMILARLY
112
=Ψ=∫ν
φ
Thus Galerkin's Equation Becomes
( )
∑ ∑∑∑ ∑
∫ ∫ ∫
=Ψ−Ψ⇒
=Ψ−Ψ−Ψ−Ψ
=Σ−−−∈
e ee
Te
T
Te
Te
Te
T
T
S S
TTT
FkqPTfkq
Ptdsfdvdv
00
0φφφφσν
[ ]nii ΨΨΨΨ + LLLL 11
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎦
⎤⎢⎣
⎡ Ψ+
Ψ−Ψ−Ψ
+
++
.
.
.
1
1
11
i
ii
iii
i
iii
i
iii
i
iii
Q
lAE
lAE
lAE
lAE
LL
Q
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
lAE
i
ii
i
ii
i
ii
i
ii
i
ii
i
ii
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
+−
−+
−−
+−
−
+
++
−
−−
1
11
1
11
2
22
1
11
1
11
1
11
1
11
TO SOLVE THE PROBLEM• Find Deformations Qi’s
1. Define Boundary conditions2. Apply minimization of PE
• Find strains ε = B * qFind stresses σ = E ε = E B q
SOLVING FOR Q
[ ][ ]
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
=
=
=
nnnn
n
n
Tn
Tn
KKK
KKKKKK
K
FFFF
QQQQ
21
22221
11211
.....21
.....21
.
.....
FQKQQ
FQFQFQFQ
QKQQKQQKQQKQ
QKQQKQQKQQKQQKQQKQQKQQKQ
TT
nn
nnnnnnnnnnn
nn
nn
−=
+++−
⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜
⎝
⎛
+++
++++++++
=Π
21
).....(
.......
..........
21
332211
32211
22323222221212
11313121211111
BOUNDARY CONDITIONS
CONSTRAINTPOINTSINGLE−
...],0,0..[
....
52
2,1 21
==
==
QQge
aQaQ pp
LET US CONSIDER AS AN EXAMPLE
Q1 = a1
(
)...()........
.
.
...
...21
332211
2211
23231313
22323222221212
11313121211111
nn
nnnnnnnn
nn
nn
FQFQFQFaQKQaKQaKQ
QKQaKaQKQQKQQKQaKQ
QKaQKaQKaaKa
BECOMESA
++++−+++
+++++++++
+++++=Π
:.. PRINICIPLEEPMINIMUM
.0).....
.....(21
0
0).....
.....(21
0
0
33333232113
3333322131
3
22332222112
2323222121
2
⎪⎩
⎪⎨⎧
=−+++
++++=
∂Π∂
⎪⎩
⎪⎨⎧
=−+++
++++=
∂Π∂
=∂
Π∂
FQKQKKQaK
QKQKKQaKQ
FQKQKKQaK
QKQKKQaKQ
Q
nn
nn
nn
nn
i
⎪⎩
⎪⎨⎧
=−+++
+++++=
∂Π∂
0).....
.....(21
0
.
332211
332211
inniiii
niniii
i FQKQKKQaK
QKQKKQaKQ
111
11
1313
1212
3
2
432
3343332
2242322
113322
13133333322
12122233222
.
...
.....
...
............
...................
FQK
aKF
aKFaKF
Q
KKKK
KKKKKKKK
aKFKQKQKQ
aKFKQKQKQaKFKQKQKQ
nnnnnnnn
n
n
iiinnii
nn
nn
=⇒
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−=+++
−=+++−=+++
K’= OBTAINED FROM K BY DELETING 1ST ROW AND COLUMN
Q’= OBTAINED FROM Q BY DELETING Q1
F’ = OBTAINED FROM F BY DELETING F1 AND SUBTRACTING Ki1a1FROM FI
IF INSTEAD OF Q1 = a1, WE HAD Qi=ai THE SAME STEPS SHALL BE CARRIED PUT BY DOING THESE OPERATIONS ON ith ROW AND COLUMN.
REACTION AT THE SUPPORT (NODE 1 IF Q1 = a1)
113132121111
111313212111
........
FQKQKQKQKRRFQKQKQKQK
nn
nn
−++++=⇒+=++++
ELIMINATION METHOD
ENALTY APPROACH
•
ATTACH A SPRING OF STIFFNESS C
DEFLECT THE FIXED END BY a1
DEFLECTION OF NODE 1-Q1
e1 e5e2 e3 e4wallC
q1
a1
11, aQC →∞→
11 aQ −=δ
( ) [ ]SPRINGofEPaQCU s ..21 2
11 −=
.
....
...(21
)(21
21)(2
1
22323222221212
11313121211111
211
211
+++++
+++++
+−=
−+−=Π
nn
nn
TT
QKQQKQQKQQKQ
QKQQKQQKQQKQ
aQC
FQKQQaQC
-DEFLECTION OF SPRING
AS
)...()...
332211
22323222221212
nn
nn
FQFQFQFQQKQQKQQKQQKQ
++++−++++
0.....0
0.....0
0.....0
0
333333221313
223232221212
11313221111111
=−+++→=∂
Π∂
=−+++→=∂
Π∂
=−++++−→=∂Π∂
=∂Π∂
FQKQKKQQKQ
FQKQKKQQKQ
FQKQKQKQKCaCQQ
Q
nn
nn
nn
i
0.....0 332211 =−+++→=∂
Π∂nnnnnnn
n
FQKQKKQQKQ
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡ +
=
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡ +
nnnnnnn
n
n
F
FF
CaF
Q
QQQ
KKKK
KKKKKKKKKKKCK
.
...
....
..
..
..
3
2
11
3
2
1
321
3333231
2232221
13131211
Multi point constraints-
FQQQCKQQ
approachpenaultybyitsolvecanweaaexamplefor
Tpp
Tm
pp
−−++=Π
=−
=+
)(21
21
2
02211
21
02211
βββ
βββ
Since C has very large value, P.E. takes minimum valueWhen (β1Qp1+ β1Qp1-β0) is minimum.
⎥⎥⎦
⎤
⎢⎢⎣
⎡
++
++→
⎥⎥⎦
⎤
⎢⎢⎣
⎡22222112
21212
111
2212
2111
βββ
βββ
CkCk
CkCk
kk
kk
pppp
pppp
pppp
pppp
The modified stiffness and force matrices are-
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
+
+→
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
202
101
2
1
ββ
ββ
CF
CF
F
F
p
p
p
p
Reactions at support are given by-
)(
)(
0221122
0221111
ββββ
ββββ
−+−=
−+−=
ppp
ppp
QQCR
QQCR
STEPS INVOLVED IN SOLVING A 1D FE PROBLEM
• Make the Geometric Model
• Make an FE Mesh
• Define the Loading
• Develop Element and Global Matrices
• Define Boundary Conditions
• Develop Modified Global Matrices
• Solving Using Numerical Techniques
A SIMPLE PROBLEM
6” 5.25”l1=12”
12” Elem 1:1-2
P P=100lbE=3x107 psi l2=12”ρ=0.2836lb/in3
12” t=1” Elem_2:2-3
3” 3.75”
k1 = AE /L ⎡1 -1 ⎤ = 15.75 * 107 / 12 ⎡1 -1 ⎤⎣-1 1 ⎦ ⎣-1 1 ⎦
k2 = AE /L ⎡1 -1 ⎤ = 11.25 * 107 / 12 ⎡1 -1 ⎤⎣-1 1 ⎦ ⎣-1 1 ⎦
f1 = ρ A L / 2 [1 1]T = 5.25 * 12 *0.2836 / 2 [1 1]T
f2 = ρ A L / 2 [1 1]T = 3.75 * 12 *0.2836 / 2 [1 1]T
GLOBAL MATRICES
K = 3x107 / 12 ⎡ 5.25 -5.25 0 ⎤⎜-5.25 9.00 -3.75 ⎜⎣0 -3.75 3.75⎦
F = [8.93 15.31+100 6.38]T
BOUNDARY CONDITIONS
Q1 = 0
MODIFIED MATRICESK’= 3x107 / 12 ⎡ 5.25 -5.25 0 ⎤
⎜-5.25 9.00 -3.75 ⎜⎣0 -3.75 3.75⎦
= 3x107/12⎡9 -3.75 ⎤⎣-3.75 3.75⎦
F’= [8.93 (15.31+100-K21*a1) (6.38-K31*a1)]T
= [115.31 6.38]T
Q’ = [Q2 Q3]T