Civil Engineering Reference
In-Depth Information
In Eq. (iii) the displacement of the centroids of the triangles in panels A, B and C is
1/3 while the displacement of the centroids of the rectangular portions of panels B and
C is 1/2. Eq. (iii) simplifies to
VW
(
w
)
=
96
−
8
x
(iv)
Equating Eqs (ii) and (iv)
4
m
1
1
x
+
=
96
−
8
x
from which
2
12
x
x
2
−
m
=
(v)
1
+
x
For a maximum, (d
m
/d
x
)
=
0, i.e.
x
2
)
(1
+
x
)(12
−
2
x
)
−
(12
x
−
0
=
(1
+
x
)
2
which reduces to
x
2
+
2
x
−
12
=
0
from which
x
=
2
.
6m (the negative root is ignored)
Then, from Eq. (v)
m
=
13
.
6 kNm
/
m
In some cases the relationship between the ultimate moment
m
and the dimension
x
is complex so that the determination of the maximum value of
m
by differentiation is
tedious. A simpler approach would be to adopt a trial and error method in which a
series of values of
x
are chosen and then
m
plotted against
x
.
In the above we have calculated the internal virtual work produced by an ultimate
moment of resistance which acts along a yield line (Fig. 19.4). This situation would
occur if the direction of the reinforcement was perpendicular to the direction of the
yield line or if the reinforcement was isotropic (see Eq. (19.3)). A more complicated
case arises when a band of reinforcement is inclined at an angle to a yield line and the
slab is not isotropic.
Consider the part of a slab shown in Fig. 19.6 in which the yield line AB is of length
L
and is inclined at an angle
α
to the axis of rotation. Suppose also that the direction of
the reinforcement
m
is at an angle
β
to the normal to the yield line.
