Civil Engineering Reference
In-Depth Information
Substituting in Eq. (iii) for
M
and
∂
M
/
∂
R
B
we have
L
R
B
(
L
x
)
3
d
x
w
2
(
L
x
)
2
−
−
−
=
0
(iv)
0
from which
3
wL
8
R
B
=
which is the result obtained in Ex. 13.19.
The algebra in the above solutionwould have been slightly simplified if we had assumed
an origin for
x
at the end B of the beam. Equation (iv) would then become
L
R
B
x
2
2
x
3
d
x
w
−
=
0
0
which again gives
3
wL
8
Having obtained
R
B
, the remaining support reactions follow from statics.
R
B
=
An alternative approach is to release the structure so that it becomes statically
determinate by removing the support at B (by inspection the degree of statical inde-
terminacy is 1 so that one release only is required in this case) and then to calculate
the vertical displacement at B due to the applied load using, say, the unit load method
which, as we have seen, is based on the principle of virtual work or, alternatively, com-
plementary energy. We then calculate the vertical displacement at B produced by
R
B
acting alone, again, say, by the unit load method. The sum of the two displacements
must be zero since the beam at B is supported, so that we obtain an equation in which
R
B
is the unknown.
It is not essential to select the support reaction at B as the release. We could, in
fact, choose the fixing moment at A in which case the beam would become a simply
supported beam which, of course, is statically determinate. We would then determine
the moment at A required to restore the slope of the beam at A to zero.
In the above, the released structure is frequently termed the
primary structure
.
Suppose that the vertical displacement at the free end of the released cantilever due
to the uniformly distributed load
v
B,0
. Then, from Eq. (iii) of Ex. 15.9 (noting that
M
A
in that equation has been replaced by
M
a
here to avoid confusion with the bending
moment at A)
L
M
a
M
1
EI
v
B,0
=
d
x
(v)
0
in which
w
2
(
L
x
)
2
M
a
=−
−
M
1
=−
1(
L
−
x
)
