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w ,
θ ,M , and V are the responses of an infinite beam at point x due to a unit force
where
at point
ξ
and
1
when x
sgn
(
x
ξ) =
1
when x
w has been omitted.
An arbitrary integration constant in the expression for
can be interchanged, as Maxwell's reciprocal theorem [Chapter 3,
Eq. (3.38)] asserts that for linear problems the responses at point x due to a unit force at
point
Note that x and
ξ
ξ
are equal to the responses at point
ξ
due to a unit force at x . It should be noted that
when the positions of x and
ξ
are interchanged, x becomes the source point and
ξ
becomes
θ ,M , and V should
the integration variable. Then, the derivatives in the expressions for
be taken with respect to
= δw | ξ = L , etc.
Equation (9.4) will then become the expression for the displacement at point x .
With Eq. (9.5), the integral on the left-hand side of Eq. (9.4) becomes
L
0 δw
ξ
and dx of Eq. (9.4) should be changed to d
ξ
,
δw
L
L
0 δ(ξ
i
v
EI
w
dx
=
,x
)w(
x
)
dx
= w(ξ)
so that Eq. (9.4), with the help of Eq. (9.7), can be transformed to
L
0 w
V 0 w
M 0 θ
V L w
M L θ
w(ξ) =
,x
)
p z (
x
)
dx
+
+
0
0
L
L
w 0 V 0
θ 0 M 0
+ w L V L
+ θ L M L
(9.8)
with
V
V 0 ,
0
M
V L =
V
δ
V 0 = δ
V
| x = 0 =
, 0
) =
=
, 0
)
,
,L
)
,
etc.
The starred terms in Eq. (9.8) can be written as, for example,
1
2 [sgn
1
2
V 0 =
V
, 0
) =−
(
0
ξ)
]
=
1
2 [sgn
ξ) =− 2
M 0 =
M
, 0
) =−
(
0
ξ)
]
(
0
1
12 EI [sgn
1
12 EI (
w L = w
3
3
,L
) =
(
L
ξ)
]
(
L
ξ)
=
L
ξ)
Substitute these relationships into Eq. (9.8) to obtain
L
0 w
1
2 w 0 2 θ 0 +
1
2 w L + (
L
ξ)
2
w(ξ) =
,x
)
p z (
x
)
dx
+
θ L
3
12 EI V 0
2
4 EI M 0
3
2
ξ
ξ
+ (
L
ξ)
(
L
ξ)
V L
M L
(9.9)
12 EI
4 EI
Equation (9.9) is the expression for the deflection at point
ξ
along the beam axis. It can be
seen that the deflection is a function of
ξ
only and x is not involved. The slope should be
expressed as
θ(ξ) =−
dw
/
d
ξ.
The derivative of
w
with respect to
ξ
is given by
dw
d
w (ξ ) =
ξ =− θ(ξ)
2
4 EI V 0
2
1
2 θ
1
2 θ
ξ
ξ
2 EI M 0
(
L
ξ)
+ (
L
ξ)
2 EI
=−
V L
M L
0
L
4 EI
L
0 w
+
,x
)
p z (
x
)
dx
(9.10)
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