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The normal slope
w
,a is assumed to vary linearly
w
= (
1
ξ)w
+ ξw
(13.153)
,a
,ai
,a j
Use the relationships of Eqs. (13.128), (13.129), (13.152), and (13.153) to find
w
w ,a
w ,s
=
L ij v ij
(13.154)
where
yj ] T
=
w
θ
θ
w
θ
θ
v ij
[
i
xi
yi
j
xj
and
N B 1
sN B 3
cN B 3
N B 2
sN B 4
cN B 4
0
c
(
1
ξ)
s
(
1
ξ)
0
c
ξ
s
ξ
L ij =
(
N B 1 , ξ )/ ij
(
sN B 3 , ξ )/ ij
(
cN B 3 , ξ )/ ij
(
N B 2 , ξ )/ ij
(
sN B 4 , ξ )/ ij
(
cN B 4 , ξ )/ ij
Substitute Eqs. (13.151) and (13.154) into Eq. (13.150) to obtain
= σ T C ij v ij
U Si j
(13.155)
where
ij
ij 1
0
R ij L ij dS
R ij (ξ)
C ij
=
=
L ij
(ξ)
d
ξ
0
The matrix C ij is of order of 9
6 and the integration can be performed in closed form.
Summation of the integrals U Si j for the three sides completes the integral of the second
term on the right hand side of Eq. (13.140)
×
σ T Cv
U 2
=
S (w
q a
w
,a m a
w
,s m t
)
dS
=
U S 12
+
U S 23
+
U S 31
=
(13.156)
with Cv
=
C 12 v 12 +
C 23 v 23 +
C 31 v 31 .
Note that C is a 9
×
9 matrix and v , which contains
all of the nodal variables, is a 9
1 vector. Substitution of Eqs. (13.145) and (13.156) into Eq.
(13.140) (without the last term) leads to
×
1
2 σ T B
H =−
+ σ T Cv
σ
(13.157)
Since, ultimately, the displacements are the desired variables, the expression of the stiffness
equation must be in terms of v , i.e.,
σ should be expressed in terms of v . Note that the stress
σ are independent of v . In order for Π H
parameters
to be stationary, set the first variation
to be equal to zero [Chapter 2, Eq. (2.82)]
δ Π H = Π H
σ δ σ + Π H
0or Π H
Π H
δ
v
=
σ =
0
and
=
0
v
v
Then
Π H
σ =−
B
σ
+
Cv
=
0
This leads to
B 1 Cv
σ
=
(13.158)
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