Civil Engineering Reference
In-Depth Information
Yield line
a
cos
a
m
1
m
1
a
a
m
2
m
a
m
2
a
sin
a
m
t
a
F
IGURE
19.3
Determination of the
ultimate moment
along a yield line
Slab
Reinforcement
(a)
(b)
ULTIMATE MOMENT ALONG A YIELD LINE
Figure 19.3(a) shows a portion of a slab reinforced in two directions at right angles;
the ultimate moments of resistance of the reinforcement are
m
1
per unit width of slab
and
m
2
per unit width of slab. Let us suppose that a yield line occurs at an angle
α
to the reinforcement
m
2
. Now consider a triangular element formed by a length
a
of
the yield line and the reinforcement as shown in Fig. 19.3(b). Then, from the moment
equilibrium of the element in the direction of
m
α
, we have
m
α
a
=
m
1
a
cos
α
(cos
α
)
+
m
2
a
sin
α
(sin
α
)
i.e.
m
1
cos
2
α
m
2
sin
2
α
m
α
=
+
(19.1)
Now, from the moment equilibrium of the element in the direction of
m
t
m
t
a
=
m
1
a
cos
α
(sin
α
)
−
m
2
a
sin
α
(cos
α
)
so that
(
m
1
−
m
2
)
m
t
=
sin 2
α
(19.2)
2
Note that for an isotropic slab, which is one equally reinforced in two perpendicular
directions,
m
1
=
m
2
=
m
, say, so that
m
α
=
m
t
=
0
(19.3)
INTERNAL VIRTUAL WORK DUE TO AN ULTIMATE MOMENT
Figure 19.4 shows part of a slab and its axis of rotation. Let us suppose that at some
point in the slab there is a known yield line inclined at an angle
α
to the axis of rotation;
