Civil Engineering Reference
In-Depth Information
x
1
(point 1). We then calculate the displacement,
a
22
, at B due to a vertically downward
unit load at B. The total displacement at B due to the unit load at
x
1
and the reaction
R
B
is then
a
21
−
a
22
R
B
=
0
(i)
since the support at B is not displaced. In Eq. (i) the term
a
22
R
B
is negative since
R
B
is in the opposite direction to the applied unit load at B.
Both the flexibility coefficients in Eq. (i) may be obtained from a single unit load
application since, from the reciprocal theorem (Section 15.4), the displacement at B
due to a unit load at
x
1
is equal to the displacement at
x
1
due to a unit load at B.
Therefore we apply a vertically downward unit load at B.
The equation for the displaced shape of the beam is that for a simply supported beam
carrying a central concentrated load. Therefore, from Eq. (iv) of Ex. 13.5
1
48
EI
(4
x
3
3
L
2
x
)
=
−
v
(ii)
or, for the beam of Fig. 20.21(a)
x
12
EI
(
x
2
v
=
−
48)
(iii)
At B, when
x
=
4m
32
3
EI
v
B
=−
=
a
22
(iv)
Furthermore, the displacement at B due to the unit load at
x
1
(
=
displacement at
x
1
due to a unit load at B) is from Eq. (iii)
x
1
12
EI
(
x
1
−
v
x
1
=
=
a
21
48)
(v)
Substituting for
a
22
and
a
21
in Eq. (i) we have
x
1
32
3
EI
R
B
=
12
EI
(
x
1
−
48)
+
0
from which
x
1
128
(
x
1
−
R
B
=−
48)
(0
≤
x
1
≤
4
.
0m)
(vi)
Equation (vi) gives the influence line for
R
B
with the unit load between A and B; the
remainder of the influence line follows from symmetry. Eq. (vi) may be checked since
we know the value of
R
B
with the unit load at A and B. Thus fromEq. (vi), when
x
1
=
0,
R
B
=
0 and when
x
1
=
4
.
0m,
R
B
=
1 as expected.
If the support at B were not symmetrically positioned, the above procedure would be
repeated for the unit load on the span BC. In this case the equations for the deflected
shape of AB and BC would be Eqs (xiv) and (xv) in Ex. 13.6.
