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with
0
−
1
1
/
k
s
GA
0
0
0
−
0
0
0
1
/
EI
A
=
P
=
p
z
0
k
0
0
0
0
k
∗
−
N
1
0
As can be observed in Eq. (4.90b), the loading elements
z
i
can be computed if the
transfer matrix
U
i
is available. Since, for a complete solution, it is necessary only to find
U
i
,
we can set
p
z
equal to zero in Eqs. (4.94) and (4.95).
Thus, we seek to solve the homogeneous differential equations
d
z
dx
=
Az
(4.96)
The solution can be in the form [Eq. (4.90c)]
U
i
z
0
=
e
A
z
0
U
i
z
a
z
=
or
z
b
=
(4.97)
A function of a square matrix
A
of order
n
is equal to a polynomial in
A
of order
n
−
1
.
Hence, for the 4
×
4 matrix
A
for a beam,
e
A
=
2
3
c
0
I
+
c
1
A
+
c
2
(
A
)
+
c
3
(
A
)
(4.98)
The Cayley-Hamilton Theorem (“a matrix satisfies its own characteristic equation”) per-
mits
A
of Eq. (4.98) to be replaced by its characteristic values
λ
i
.
Thus,
e
λ
i
=
2
3
,
c
0
+
c
1
λ
i
+
c
2
(λ
i
)
+
c
3
(λ
i
)
i
=
1
,
2
,
3
,
4
(4.99)
represents four equations that can be solved for the functions
c
0
,
c
1
,
c
2
, and
c
3
.
Substitution
of these values into Eq. (4.98) leads to the desired transfer matrix.
To find the characteristic values
λ
i
,
substitute
e
λ
i
into the homogeneous first order
governing equations, yielding the characteristic equation
|
I
λ
i
−
A
|=
0
(4.100)
With
A
given by Eq. (4.95)
λ
i
1
−
(
1
/
k
s
GA
)
0
=
0
λ
i
0
−
(
1
/
EI
)
|
I
λ
i
−
A
|=
0
(4.101)
−
k
0
λ
i
0
−
(
k
∗
−
)
−
λ
i
0
N
1
The roots
λ
1
,
λ
2
,
λ
3
,
λ
4
=
(
±
n
1
,
±
in
2
)
,
of this determinant are readily found to be
n
1
,
2
=
(((ζ
+
η)
2
/
−
λ)
1
/
2
±
(ζ
−
η)/
)
1
/
2
4
2
(4.102)
where
k
∗
)/
λ
=
k
/
EI,
η
=
k
/(
k
s
GA
)
,
ζ
=
(
N
−
EI
Substitute these characteristic values into Eq. (4.99), giving
e
n
1
=
2
3
c
0
+
c
1
n
1
+
c
2
(
n
1
)
+
c
3
(
n
1
)
e
−
n
1
=
2
3
c
0
−
c
1
n
1
+
c
2
(
n
1
)
−
c
3
(
n
1
)
(4.103)
e
in
2
=
c
0
+
ic
1
n
2
−
c
2
(
n
2
)
2
−
ic
3
(
n
2
)
3
e
−
in
2
=
2
3
c
0
−
ic
1
n
2
−
c
2
(
n
2
)
+
ic
3
(
n
2
)
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