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on the boundary of the cross-
section will be obtained first. Constant elements and linear elements will be used in the
computation. For the constant element, the warping function over the whole element is
assumed to be constant and is equal to the value of the warping function at the node which
is located at the center of the element. Thus the warping function is approximated by
To find the torsional constant, the warping function
ω
ω
=
ω
i
(1)
on each element
i,
where
ω
i
are nodal values of the warping function of the
i
th node
(element). Substitute (1) into Eq. (14) of Example 9.2 to obtain
M
M
1
H
ij
1
G
ij
q
i
c
ω
+
ω
=
j
=
1
,
2
,
...
M
(2)
j
i
=
=
i
i
where
r
k
a
k
r
2
H
ij
G
ij
=−
=−
dS,
ln
rdS
S
i
S
i
c
is given in Eq. (14) of Example 9.2,
r
is the distance from node
j
to node
i, r
k
is the
component of
r
in the
x
k
,k
2
,
3
,
direction,
a
k
is the direction cosine of t
he
outer normal
of the element with respect to the
x
k
axis,
M
is the number of elements, and
q
i
=
=
−
x
3
a
2
(Eq. (3) of Example 9.2) at the nodal points. Since the shape function for the constant
elements is unity, the matrix
H
ij
in Eq. (9.34) becomes a single number
H
ij
in (2). Add
c
to
H
ii
so that
H
ij
becomes
H
ij
x
2
a
3
Also change the notation
G
ij
to
G
ij
. After every
.
ω
i
is spanned
and the element matrices assembled, (2) becomes
HV
=
GP
(3)
where
N
]
T
=
ω
1
ω
···
ω
V
[
2
q
N
]
T
P
=
[
q
1
q
2
···
H
11
H
12
···
H
1
N
.
.
.
H
=
and
···
H
N
1
H
N
2
···
H
NN
G
11
G
12
···
G
1
N
.
.
.
G
=
···
G
N
1
G
N
2
···
G
NN
where
N
is the number of nodes. Note that for the torsion problem, the quantity
q,
the
derivative of the warping function with respect to the outer normal, on the boundary is
known, and
B
0
.
To avoid evaluating the diagonal elements of
H
, the technique of Eq. (9.39) can be used,
i.e.,
=
M
H
ii
=−
H
ij
(4)
j
=
1
j
=
i
ω
The solution of (3) results in the warping function
on the boundary.
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