1-D Finite Elements – Virtual Work Approach
U1
U2
U3
U4
P
1
2
3
q = force/length =constant
Um
l
mth element
Um+1
Um
Um+1
ξ
lm
1. Assume some variation of displacements in the element and write in
terms of the nodal displacements
Simplest case: linear variation
u x( m ) = U m +
(U m+1 − U m ) ξ
lm
⎛ ξ ⎞
⎛ξ ⎞
= U m ⎜ 1 − ⎟ + U m +1 ⎜ ⎟
⎝ lm ⎠
⎝ lm ⎠
H m(
m)
(ξ )
H m( +1) (ξ )
m
shape functions for displacement
If we let
H (j
m)
=0
j ≠ m, m + 1
then we can write the displacement
in each element in terms of all the nodal variables
M +1
u x = ∑ H (j m )U j
( m)
( m = 1,..., M )
j =1
( if we have M elements
we have M+1 nodes)
2. Compute the strains for each element in terms of the nodal
displacements
(m)
exx
∂u x( ) U m +1 − U m
=
=
∂x
lm
m
⎛ 1⎞
⎛1⎞
= U m ⎜ − ⎟ + U m +1 ⎜ ⎟
⎝ lm ⎠
⎝ lm ⎠
J m(
m)
(ξ )
J m( +1) (ξ )
m
shape functions for strain (constants in this case)
J (j
again, if we let
m)
=0
j ≠ m, m + 1
then we can write the strain
in each element in terms of all the nodal variables
M +1
exx = ∑ J (j m )U j
( m)
( m = 1,..., M )
j =1
3. Obtain virtual changes of the displacement and strain
M +1
δ u x = ∑ H (j m )δ U j
( m)
( m = 1,..., M )
j =1
M +1
δ exx = ∑ J (j m )δ U j
( m)
j =1
( m = 1,..., M )
4. Apply Principle of Virtual Work
If
∫ f δ u dV + ∑ P δ U
x
x
k
V
work done by: distributed
body force
k
= ∫ σ xxδ exx dV
V
is satisfied for all possible δU's
then equilibrium will be
(approximately) satisfied
concentrated
loads
⎧1 if load at node
For concentrated loads, let d j = ⎨
⎩0 otherwise
Then
∑ P δU
k
M +1
k
= ∑ Pj d jδ U j
j =1
and the principle gives
⎤
⎡ M +1 ( m )
⎤
⎡ M +1 ( m ) ⎤ ⎡ M +1 ( m )
⎪⎧
⎪⎫ M ⎪⎧
⎪⎫ M +1
⎨ ∫ E ⎢ ∑ J k U k ⎥ ⎢ ∑ J j δ U j ⎥ dVm ⎬ = ∑ ⎨ ∫ f x ⎢ ∑ H j δ U j ⎥ dVm ⎬ + ∑ Pj d jδ U j
∑
m =1 ⎪
⎦ ⎣ j =1
⎦
⎦
⎪⎭ m =1 ⎪⎩Vm ⎣ j =1
⎪⎭ j =1
⎩Vm ⎣ k =1
M
σ xx( m ) = Eexx( m )
δ exx( m )
δ u x( m)
⎫⎪ M ⎧⎪
⎫⎪ M +1
⎤
⎡ M +1 ( m )
⎤
⎡ M +1 ( m ) ⎤ ⎡ M +1 ( m )
⎪⎧
⎨ ∫ E ⎢ ∑ J k U k ⎥ ⎢ ∑ J j δ U j ⎥ dVm ⎬ = ∑ ⎨ ∫ f x ⎢ ∑ H j δ U j ⎥ dVm ⎬ + ∑ Pj d jδ U j
∑
m =1 ⎩
⎦ ⎣ j =1
⎦
⎦
⎪Vm ⎣ k =1
⎭⎪ m =1 ⎩⎪Vm ⎣ j =1
⎭⎪ j =1
M
Rewrite as:
M ⎡
⎧⎪ M +1 ⎡ M
⎫⎪
⎤
⎤
( m) ( m)
( m)
δ U j ⎨ ∑ ⎢ ∑ ∫ EJ j J k dVm ⎥ U k − ∑ ⎢ ∫ f x H j dVm ⎥ − Pj d j ⎬ = 0
∑
j =1
m =1 ⎢
⎥⎦
⎥⎦
⎪⎩ k =1 ⎢⎣ m =1 Vm
⎪⎭
⎣Vm
M +1
For this to be true for all variations of δUj
M ⎡
⎡M
⎤
⎤
( m) ( m)
( m)
⎢ ∑ ∫ EJ j J k dVm ⎥ U k = ∑ ⎢ ∫ f x H j dVm ⎥ + Pj d j
∑
k =1 ⎢
m =1 ⎢
⎥⎦
⎥⎦
⎣ m =1 Vm
⎣Vm
M +1
K (jk )
m
stiffness matrix
for each element
M
K jk = ∑ K (jk )
m =1
m
Rj
nodal forces produced
by distributed loads
total stiffness matrix
( j=1, 2, …,M+1)
M ⎡
⎡M
⎤
⎤
( m) ( m)
( m)
⎢ ∑ ∫ EJ j J k dVm ⎥ U k = ∑ ⎢ ∫ f x H j dVm ⎥ + Pj d j
∑
k =1 ⎢
m =1 ⎢
⎥⎦
⎥⎦
⎣ m =1 Vm
⎣Vm
M +1
M +1
∑K
k =1
jk
U k = R j + Pj d j
⎡
⎤
( m)
R j = ∑ ⎢ ∫ f x H j dVm ⎥
m =1 ⎢
⎥⎦
⎣Vm
M
M
K jk = ∑ K (jk )
m
m =1
M
K jk = ∑ ∫ EJ (j ) J k( ) dVm
( m)
m =1 Vm
m
( j=1, 2, …,M+1)
m
Now, consider our particular problem
U1
U2
U3
U4
q
P
F1
reaction at wall
1
⎡ K11
⎢K
⎢ 21
⎢ K 31
⎢
⎢⎣ K 41
2
K12
K13
K 22
K 23
K 32
K 42
K 33
K 43
3
K14 ⎤ ⎧U1 ⎫ ⎧ R1 ⎫ ⎧ F1 ⎫
K 24 ⎥⎥ ⎪⎪U 2 ⎪⎪ ⎪⎪ R2 ⎪⎪ ⎪⎪ 0 ⎪⎪
⎨ ⎬ = ⎨ ⎬+⎨ ⎬
K 34 ⎥ ⎪U 3 ⎪ ⎪ R3 ⎪ ⎪ 0 ⎪
⎥
K 44 ⎥⎦ ⎩⎪U 4 ⎭⎪ ⎩⎪ R4 ⎭⎪ ⎪⎩ P ⎪⎭
6. Obtain stiffness matrix for each element:
M
J m( m ) (ξ ) = −
K jk = ∑ ∫ EJ j J k dVm
( m)
( m) ( m)
m =1 Vm
3
= ∑ EAl J (j ) J k(
m
J m( m+1) (ξ ) =
m)
(1)
K jk :
− EA
⎤
0 0⎥
l
⎥
EA
0 0⎥
⎥
l
⎥
0
0 0⎥
0
0 0 ⎥⎦
7. Assemble total stiffness
matrix:
3
K jk = ∑ K (jk ) :
m =1
m
1
l
all other J (j m ) (ξ ) = 0
m =1
⎡ EA
⎢ l
⎢
⎢ − EA
⎢ l
⎢
⎢ 0
⎢⎣ 0
1
l
K (jk2) :
⎡0
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢⎣0
⎡ EA
⎢ l
⎢
⎢ − EA
⎢ l
⎢
⎢ 0
⎢
⎢
⎢ 0
⎢⎣
0
EA
l
− EA
l
0
− EA
l
2 EA
l
− EA
l
0
0⎤
⎥
− EA
0⎥
l
⎥
⎥
EA
0⎥
l
⎥
0
0 ⎥⎦
0
0
− EA
l
2 EA
l
− EA
l
⎤
0 ⎥
⎥
0 ⎥
⎥
⎥
− EA ⎥
l ⎥
EA ⎥
⎥
l ⎥⎦
K (jk3) :
⎡0
⎢0
⎢
⎢
⎢0
⎢
⎢0
⎢⎣
0
0
0
0
EA
l
− EA
l
0
0
0 ⎤
0 ⎥⎥
− EA ⎥
l ⎥⎥
EA ⎥
l ⎥⎦
U1
U2
U3
U4
q
P
F1
1
2
3
7. Obtain all equivalent nodal forces from distributed loads:
⎡
⎤
( m)
R j = ∑ ⎢ ∫ f x H j dVm ⎥
m =1 ⎢
⎥⎦
⎣Vm
M
=
∫fH
x
( 2)
j
Ad ξ
V2
l
= q ∫ H (j
0
2)
(ξ )dξ
H 2( 2) (ξ ) = 1 −
H 3( 2) (ξ ) =
ξ
l
ξ
l
all other H (j 2) (ξ ) = 0
R1 = 0
⎛ ξ
R2 = q ∫ ⎜1 −
l
0⎝
l
⎛ξ
R3 = q ∫ ⎜
l
0⎝
R4 = 0
l
ql
⎞
d
=
ξ
⎟
2
⎠
ql
⎞
d
=
ξ
⎟
2
⎠
Resulting system cannot be solved directly because we have not
eliminated rigid body motions
⎡ EA
⎢ l
⎢
⎢ − EA
⎢ l
⎢
⎢ 0
⎢
⎢
⎢ 0
⎢⎣
− EA
l
2 EA
l
− EA
l
0
0
− EA
l
2 EA
l
− EA
l
⎤ ⎧U1 ⎫
0 ⎥⎪ ⎪
⎥⎪ ⎪
⎪ ⎪
0 ⎥ ⎪U2 ⎪
⎥⎪ ⎪
⎥⎨ ⎬ =
− EA ⎥ ⎪ ⎪
U
l ⎥⎪ 3 ⎪
EA ⎥ ⎪ ⎪
⎥ ⎪U ⎪
l ⎥⎦ ⎪⎩ 4 ⎪⎭
⎧ F1 ⎫
⎧ 0 ⎫
⎪ ⎪
⎪
⎪
⎪ ⎪
⎪
⎪
⎪0⎪
⎪ql / 2 ⎪
⎪ ⎪
⎪
⎪
⎨
⎬ + ⎨ ⎬
⎪ql / 2 ⎪
⎪0⎪
⎪
⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎩ 0 ⎭
⎩P⎭
8. Eliminate all possible rigid body motions and solve the system of
equations
For this problem, if U j = 0 set
K jj = 1, K jk = 0
so that the jth equation gives U j = 0
stiffness matrix, also set
⎡1
⎢
⎢0
⎢
⎢
⎢0
⎢
⎢
⎢0
⎣
0
0
2 EA
l
− EA
l
− EA
l
2 EA
l
− EA
l
0
(k ≠ j)
and R j = Pj d j = 0
To preserve symmetry of the
K kj = 0
⎧U1 ⎫
0 ⎤⎪ ⎪
⎥⎪ ⎪
0 ⎥ ⎪U ⎪
⎥ ⎪⎪ 2 ⎪⎪
− EA ⎥ ⎨ ⎬ =
⎥⎪ ⎪
l ⎥ U3
⎪ ⎪
EA ⎥ ⎪ ⎪
⎥
l ⎦ ⎪U ⎪
⎪⎩ 4 ⎪⎭
⎧ 0 ⎫
⎧0⎫
⎪
⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎪ql / 2 ⎪
⎪0⎪
⎪
⎪
⎪ ⎪
⎨
⎬ + ⎨ ⎬
⎪ql / 2 ⎪
⎪0⎪
⎪
⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎪
⎪
⎪ ⎪
⎩ 0 ⎭
⎩P⎭