Civil Engineering Reference
In-Depth Information
since the end A is restrained against rotation. In a similar manner the rotation at B is
given by
M
B
M
θ
C
+
W
M
δ
2
+
M
A
M
θ
B
=
0
(v)
Solving Eqs (iv) and (v) for
M
A
gives
W
δ
2
θ
B
− δ
1
θ
C
θ
A
θ
C
−
M
A
=
θ
B
The fact that the arbitrary moment,
M
, does not appear in the expression for the
restraining moment at A (similarly it does not appear in
M
B
) produced by the load
W
indicates an extremely useful application of the reciprocal theorem, namely the
model analysis of statically indeterminate structures. For example, the fixed beam of
Fig. 15.25(c) could possibly be a full-scale bridge girder. It is then only necessary to
construct amodel, say, of perspex, having the same flexural rigidity,
EI
, as the full-scale
beam and measure rotations and displacements produced by an arbitrary moment,
M
,
to obtain the fixed-end moments in the full-scale beam supporting a full-scale load.
THEOREM OF RECIPROCAL WORK
Let us suppose that a linearly elastic body is to be subjected to two systems of
loads,
P
1
,
P
2
,
...
,
P
k
,
...
,
P
r
, and,
Q
1
,
Q
2
,
...
,
Q
i
,
...
,
Q
m
, which may be applied simul-
taneously or separately. Let us also suppose that corresponding displacements are
P
,1
,
P
,2
,
...
,
P
,
k
,
...
,
P
,
r
due to the loading system,
P
, and
Q
,1
,
Q
,2
,
...
,
Q
,
i
,
...
,
Q
,
m
due to the loading system,
Q
. Finally, let us suppose that the loads,
P
, produce displacements
Q
,1
,
Q
,2
,
...
,
Q
,
i
,
...
,
Q
,
m
at the points of application
and in the direction of the loads,
Q
, while the loads,
Q
, produce displacements
P
,1
,
P
,2
,
...
,
P
,
k
,
...
,
P
,
r
at the points of application and in the directions of the
loads,
P
.
Now suppose that the loads
P
and
Q
are applied to the elastic body gradually and
simultaneously. The total work done, and hence the strain energy stored, is then
given by
1
P
,1
)
1
P
,2
)
1
P
,
k
)
U
1
=
2
P
1
(
P
,1
+
+
2
P
2
(
P
,2
+
+···+
2
P
k
(
P
,
k
+
1
P
,
r
)
1
Q
,1
)
1
Q
,2
)
+···+
2
P
r
(
P
,
r
+
+
2
Q
1
(
Q
,1
+
+
2
Q
2
(
Q
,2
+
1
Q
,
i
)
1
Q
,
m
)
+···+
2
Q
i
(
Q
,
i
+
+···+
2
Q
m
(
Q
,
m
+
(15.54)
If now we apply the
P
-loading system followed by the
Q
-loading system, the total strain
energy stored is
1
1
1
1
1
1
U
2
=
2
P
1
P
,1
+
2
P
2
P
,2
+···+
2
P
k
P
,
k
+···+
2
P
r
P
,
r
+
2
Q
1
Q
,1
+
2
Q
2
Q
,2
1
1
P
1
P
,1
+
P
2
P
,2
+
P
k
P
,
k
+···+
P
r
P
,
r
(15.55)
+···+
2
Q
i
Q
,
i
+···+
2
Q
m
Q
,
m
+
