(b) View parallel to EF (c) View parallel to AEFIGURE 9. Analysis of two-way slab by mechanism method.Calculation Procedure:- Draw line GH perpendicular to AE at E; express distances b
and c in terms of a
Consider a slab to be reinforced in orthogonal directions. If the reinforcement in one di-
rection is identical with that in the other direction, the slab is said to be isotropically rein-
forced; if the reinforcements differ, the slab is described as orthogonally anisotropic. In
the former case, the capacity of the slab is identical in all directions; in the latter case, the
capacity has a unique value in every direction. In this instance, assume that the slab size is
excessive with respect to balanced design, the result being that the failure of the slab will
be characterized by yielding of the steel.
In a steel beam, a plastic hinge forms at a section; in a slab, a plastic hinge is assumed
to form along a straight line, termed a yield line. It is plausible to assume that by virtue of
symmetry of loading and support conditions the slab in Fig. 9a will fail by the formation
of a central yield line EF and diagonal yield lines such as AE, the ultimate moment at
these lines being positive. The ultimate unit moment mu is the moment acting on a unit
length.
Although it is possible to derive equations that give the location of the yield lines, this
procedure is not feasible because the resulting equations would be unduly cumbersome.
The procedure followed in practice is to assign a group of values to the distance a and to
determine the corresponding values of mu. The true value of mu is the highest one ob-
tained. Either the static or mechanism method of analysis may be applied; the latter will
be applied here.
Expressing the distances b and c in terms of a gives tan a = 6/a = AE/b = cl(AE)\ b =
aAE/6; c = 6AEIa. - Find the rotation of the plastic hinges
Allow line EF to undergo a virtual displacement A after the collapse load is reached.
(a) Plan of slab