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and eventually to
L
0
δw
L
0
δw
i
v
dx
i
v
EI
EI
w
=−
[
δw
V
+
δθ
M
]
S
u
+
S
p
+
[
δ
V
w
+
δ
M
θ
]
S
u
+
S
p
+
w
dx
(9.2)
Substitution of Eq. (9.2) into Eq. (9.1) gives
L
0
δw
L
0
δw
i
v
EI
w
dx
−
p
z
dx
−
[
δw
V
+
δθ
M
]
S
u
+
[
δ
V
w
+
δ
M
θ
]
S
p
−
[
δw
V
+
δθ
M
]
S
p
+
[
δ
V
w
+
δ
M
θ
]
S
u
=
0
(9.3)
Combine the prescribed and reaction forces on
S
p
and
S
u
into a single bracket. Treat the
displacements similarly. Then, Eq. (9.3) can be written as
L
0
δw
L
0
δw
i
v
EI
w
dx
−
p
z
dx
−
[
δw
V
+
δθ
M
]
S
u
+
S
p
+
[
δ
V
w
+
δ
M
θ
]
S
u
+
S
p
=
0
or
L
0
δw
L
0
δw
i
v
EI
w
dx
=
p
z
dx
+
δw
L
V
L
+
δθ
L
M
L
−
δw
0
V
0
−
δθ
0
M
0
−
δ
V
L
w
L
−
δ
M
L
θ
L
+
δ
V
0
w
0
+
δ
M
0
θ
0
(9.4)
where
δw
=
δw
|
δ
=
δ
|
L
,
V
0
V
0
,
etc.
L
x
=
x
=
δw
Choose
such that the differential equation
i
v
EI
δw
=
δ(ξ
,x
)
(9.5)
is satisfied. Here,
is the Dirac delta function, which is useful in the representation of
concentrated loads, with the fundamental property
+∞
−∞
δ(ξ
δ(ξ
,x
)
)
=
,x
dx
1
Also, the sifting property of the delta function for a finite domain would be
L
f
(
x
)δ(ξ
,x
)
dx
=
f
(ξ )
(9.6)
0
on the beam and
another is a point
x
which corresponds to the integration variable
x
in Eq. (9.4). Substitution
of Eq. (9.5) into Eq. (9.4) will result in an expression for the displacement at point
Note that
δw
is a function involving two points, one is a source point
ξ
ξ
,
so that
the left-hand side of Eq. (9.4) is a function of
,
and so is the right-hand side. Let the solution
to Eq. (9.5), i.e., the
kern function
(the
fundamental solution
or
Green's function
), be represented
as [Stern, 1989]
ξ
1
12
EI
[sgn
δw
=
w
∗
(ξ
3
=
w
∗
,x
)
=
(
x
−
ξ)
]
(
x
−
ξ)
d
dx
w
∗
(ξ
1
4
EI
[sgn
2
=
θ
∗
δθ
=−
,x
)
=−
(
x
−
ξ)
]
(
x
−
ξ)
(9.7)
d
2
dx
2
w
∗
(ξ
1
2
[sgn
M
∗
δ
M
=−
EI
,x
)
=−
(
x
−
ξ)
]
(
x
−
ξ)
=
d
3
dx
3
w
∗
(ξ
1
2
[sgn
V
∗
δ
V
=−
EI
,x
)
=−
(
x
−
ξ)
]
=
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