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From Eq. (4.94), by finding the derivatives
d
2
dx
2
and
d
3
dx
3
w/
w/
from
d
w/
dx
=−
θ
+
V
/
k
s
GA,
it is possible to form
w
(
x
)
=
Rz
(
x
)
,
which relates the deflection
w(
x
)
and its
derivatives to the state variables
z
(
x
)
, where
10
0
0
0
−
1
1
/
k
s
GA
0
R
=
(4.115)
η
0
0
−
1
/
EI
0
ζ
−
η
−
1
/
EI
+
η/
k
s
GA
0
An expression for the transfer matrix is obtained by substituting Eq. (4.114) into
z
(
x
)
=
R
−
1
w
(
x
).
Thus,
R
−
1
Q
R
−
1
Q
z
(
x
)
=
(
x
)
w
(
0
)
=
(
x
)
Rz
(
0
)
or
R
−
1
Q
U
i
z
a
z
b
=
()
Rz
a
=
(4.116)
This relationship, i.e.,
U
i
=
R
−
1
Q
()
R
,
forms the basis for a very useful method for
finding the transfer matrix.
EXAMPLE 4.7 Transfer Matrix for a Beam Element with Axial Force
For a beam element with no foundations and no shear deformation effects,
λ
=
0 (1)
To develop
U
i
using Eq. (4.110), it is necessary to set up
R
and
R
−
1
,
and to calculate the
elements of
Q
. From Eq. (4.115),
0
,
η
=
0
,
ζ
=
N
/
EI,
1
/
k
s
GA
=
10 0
0
1
0
0
0
0
−
1
0
0
0
−
1
0
0
R
−
1
=
=
R
(2)
00 0
−
1
/
EI
0
−
ζ
EI
0
−
EI
0
ζ
−
1
/
EI
0
0
0
−
EI
0
From Eqs. (4.112) and (4.113), with
a
2
=
ζ
,
L
−
1
s
2
L
−
1
s
3
s
e
1
=
=
=
cos
a
s
4
+
ζ
s
2
+
ζ
a
2
cos
a
a
3
sin
a
e
0
=−
a
sin
a
e
=−
e
=
−
1
−
2
L
−
1
1
1
a
e
2
=
=
sin
a
s
2
+
ζ
L
−
1
1
s
=
1
1
a
1
a
2
(
e
3
=
sin
ax dx
=
1
−
cos
a
)
s
2
+
ζ
0
1
a
2
(
=
a
2
−
1
a
3
1
a
3
(
e
4
=
1
−
cos
ax
)
dx
sin
a
=
a
−
sin
a
)
(3)
0
For the case of a beam element with no axial force, i.e.,
ζ
=
0
,
the functions
e
i
,i
=−
2
,
−
1
,
0
,
1
,
...
,
4
,
can be obtained by going to the limit in (3) as
a
→
0
.
These
e
i
relations are substituted
in Eq. (4.114) to form
Q
().
To simplify
Q
, note that
e
1
+
ζ
e
3
=
1
,e
2
+
ζ
e
4
=
,e
0
+
ζ
e
2
=
0
,e
−
1
.
Now we can proceed to find the transfer matrix. First establish
R
−
1
Q
+
ζ
e
1
=
0
,
and,
e
−
2
+
ζ
e
0
=
0
()
as
1
e
3
e
4
0
−
1
−
e
2
−
e
3
R
−
1
Q
()
=
(4)
0
−
ζ
EI
0
−
EI
0
0
−
EIe
1
−
EIe
2
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