Civil Engineering Reference
In-Depth Information
In this case
θ
A
= δ =
0 so that, from Eq. (16.42)
1
2
M
BA
Therefore one-half of the applied moment,
M
BA
, is carried over to A so that the
COF
M
AB
=
=
1/2. Now from Eq. (16.43) we have
2
M
BA
−
L
6
EI
M
BA
2
θ
B
=−
so that
4
EI
L
θ
B
M
BA
=−
from which (see Eq. (16.32))
4
EI
L
K
BA
=
(
=
K
AB
)
Case 2: A simply supported, B simply supported, moment
M
BA
applied at B
This situation arises when we release the beam at an internal support (B) and the
adjacent support (A) is an outside support which is pinned and therefore free to
rotate. In this case the moment,
M
BA
, does not affect the moment at A, which is
always zero; there is, therefore, no carry over from B to A.
From Eq. (16.43)
L
6
EI
2
M
BA
(
M
AB
=
θ
B
=−
0)
which gives
3
EI
L
θ
B
M
BA
=−
so that
3
EI
L
K
BA
=
(
=
K
AB
)
Case 3: A and B simply supported, equal moments
M
BA
and
−
M
AB
applied at B and A
This case is of use in a symmetrical beam that is symmetrically loaded and would apply
to the central span. Thus identical operations will be carried out at each end of the
central span so that there will be no carry over of moment from B to A or A to B. Also
θ
B
=−
θ
A
so that from Eq. (16.41)
2
EI
L
θ
B
M
BA
=−
and
2
EI
L
K
BA
=
(
=
K
AB
)
