Civil Engineering Reference
In-Depth Information
the value of
S
K
with the unit load immediately to the left of K. Thus, with the load
between A and K the
S
K
influence line is the line a
2
g in Fig. 20.1(d) so that, when the
unit load is at C, the value of
S
K
is equal to the ordinate c
2
f.
With the unit load between K and B the shear force at K is given by
S
K
=−
R
A
(or
S
K
=
R
B
−
1)
Substituting for
R
A
from Eq. (20.1) we have
L
−
x
S
K
=−
(
a
≤
x
≤
L
)
(20.4)
L
Again
S
K
is a linear function of load position. Therefore when
x
=
L
,
S
K
=
0 and when
x
=
a
, i.e. the unit load is immediately to the right of K,
S
K
=−
(
L
−
a
)/L which is the
ordinate kh in Fig. 20.1(d).
FromFig. 20.1(d) we see that the gradient of the line a
2
g is equal to [(
a
/
L
)
−
0]/
a
=
1/
L
1/
L
. Thus
the gradient of the
S
K
influence line is the same on both sides of K. Furthermore,
gh
+
(
L
−
a
)/
L
]/(
L
−
a
)
=
and that the gradient of the line hb
2
is equal to [0
=
kh
+
kg or gh
=
(
L
−
a
)/
L
+
a
/
L
=
1.
M
K
influence line
The value of the bending moment at K also depends upon whether the unit load is to
the left or right of K. With the unit load at C
M
K
=
R
B
(
L
−
a
)
(or
M
K
=
R
A
a
−
1(
a
−
x
))
which, when substituting for
R
B
from Eq. (20.2) becomes
(
L
−
a
)
x
M
K
=
(0
≤
x
≤
a
)
(20.5)
L
From Eq. (20.5) we see that
M
K
varies linearly with
x
. Therefore, when
x
=
0,
M
K
=
0
and when
x
=
a
,
M
K
=
(
L
−
a
)
a
/
L
, which is the ordinate k
1
j in Fig. 20.1(e).
Now with the unit load between K and B
M
K
=
R
A
a
which becomes, from Eq. (20.1)
L
a
−
x
M
K
=
(
a
≤
x
≤
L
)
(20.6)
L
Again
M
K
is a linear function of
x
so that when
x
=
a
,
M
K
=
(
L
−
a
)
a
/
L
, the ordinate k
1
j
in Fig. 20.1(e), and when
x
0. The complete influence line for the bending
moment at K is then the line a
3
jb
3
as shown in Fig. 20.1(e). Hence the bendingmoment
at K with the unit load at C is the ordinate c
3
i in Fig. 20.1(e).
=
L
,
M
K
=
