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As indicated in Fig. 4.14, the distributed load is represented analytically as
p
0
L
p
z
(
x
)
=
(
L
−
x
)
(1)
Then, from Eq. (4.119),
L
0
b
w
=−
p
z
(
x
)
U
(
L
−
x
)
dx
w
V
0
L
dx
3
p
0
L
4
30
EI
p
0
L
−
(
L
−
x
)
=−
(
L
−
x
)
=
(2)
3!
EI
0
.
where
in Eq. (4.119) has been set equal to
L
4.5.3 Response of Simple Beams
The transfer matrix developed in Sections 4.5.1 and 4.5.2 can be used to find the displace-
ment and other state variables along uniform beams. For prescribed loading, Eq. (4.90a) can
be employed to provide these variables. The boundary conditions are utilized to evaluate
the state variables at the left end, i.e., to find
z
a
.
EXAMPLE 4.10 Beam with Linearly Varying Loading
Return to the beam of Fig. 4.14, which was treated in Example 4.5 using a stiffness matrix.
The boundary conditions for this fixed-hinged beam are
w
x
=
0
=
0
,
θ
x
=
0
=
0
,
w
x
=
L
=
0
,
M
x
=
L
=
0
(1)
z
i
to evaluate
z
a
. Let
x
a
correspond to the
location of the left end of the beam so that the elements of
z
a
are the state variables at the
origin, i.e.,
x
U
i
z
a
+
These conditions can be inserted in
z
b
=
L
so that the elements of
z
b
are the state
variables at the right end of the beam. Then, using the transfer matrix of Eq. (4.86),
=
x
a
=
0
,
of the beam. Also, set
x
b
=
L
3
6
EI
L
2
2
EI
p
0
L
4
w
=
0
1
−
L
−
−
w
=
0
/
30
EI
x
=
L
x
=
0
p
0
L
3
V
01
L
2
2
EI
L
EI
θ
=
0
−
/
8
EI
=
+
V
M
−
p
0
L
/
2
001
0
p
0
L
2
M
=
0
−
/
3
00
L
1
U
i
z
i
z
b
=
z
a
+
b
for this loading
is given by Eq. (2) of Example 4.9. The unknown initial parameters
V
x
=
0
,M
x
=
0
of the
above expression are evaluated by solving the equations
where the loading function column is calculated using Eq. (4.113). In fact,
w
w
x
=
L
=
0
,M
x
=
L
=
0
.
Thus, with
V
x
=
0
=
V
0
,M
x
=
0
=
M
0
,
V
0
L
3
M
0
L
2
p
0
L
4
w
x
=
L
=
0
=−
/
6
EI
−
/
2
EI
+
/
30
EI
(2)
p
0
L
2
M
x
=
L
=
0
=
V
0
L
+
M
0
−
/
3
Cramer's rule or another solution technique gives
p
0
L
2
=
(
/
)
=
(
−
/
)
V
0
2
5
p
0
L,
M
0
1
15
(3)
The initial shear force and bending moment in Sign Convention 1 have been utilized here.
As expected, the signs of
V
0
and
M
0
of (3) differ from those in Example 4.5, where Sign
Convention 2 applies.
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