Civil Engineering Reference
In-Depth Information
In this example we require the
S
D
influence line so that we shall, in fact, need to
consider the value of
R
B
with the unit load on the span BC. Therefore from Eq. (xv)
in Ex. 13.6
1
12
EI
(
x
1
−
24
x
1
+
v
x
1
=−
144
x
1
−
128)
(4
.
0m
≤
x
1
≤
8
.
0m)
(vii)
Hence from Eq. (i)
1
128
(
x
1
−
24
x
1
+
R
B
=
144
x
1
−
128)
(4
.
0m
≤
x
1
≤
8
.
0m)
(viii)
A check on Eq. (viii) shows that when
x
1
=
4
.
0m,
R
B
=
1 and when
x
1
=
8
.
0m,
R
B
=
0
.
S
D
influence line
With the unit load to the left of D, the shear force,
S
D
, at D is most simply given by
S
D
=−
R
A
+
1
(ix)
where, by taking moments about C, we have
R
A
×
−
−
x
1
)
+
R
B
×
=
8
1(8
4
0
(x)
Substituting in Eq. (x) for
R
B
from Eq. (vi) and rearranging gives
1
256
(
x
1
−
R
A
=
80
x
1
+
256)
(xi)
whence, from Eq. (ix)
1
256
(
x
1
−
S
D
=−
80
x
1
)
(0
≤
x
1
≤
2
.
0m)
(xii)
Therefore, when
x
1
=
0,
S
D
=
0 and when
x
1
=
2
.
0m,
S
D
=
0
.
59, the ordinate d
1
gin
the
S
D
influence line in Fig. 20.21(c).
With the unit load between D and B
S
D
=−
R
A
so that, substituting for
R
A
from Eq. (xi)
1
256
(
x
1
−
S
D
=−
80
x
1
+
256)
(2
.
0m
≤
x
1
≤
4
.
0m)
(xiii)
Thus, when
x
1
=
2
.
0m,
S
D
=−
0
.
41, the ordinate d
1
f in Fig. 20.21(c) and when
x
1
=
4
.
0m,
S
D
=
0.
Now consider the unit load between B and C. Again
S
D
=−
R
A
