F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

Energy dissipation in a yield line 305
Therefore, total energy dissipation for yield lines AE, DE, BF, CF and EF
[
al b/2] (1 - 2a)/
= 4m b/2 + ar + Zm b/2
2m(1 + 2A. 2 a) ( h 1 /)
= A.a w ere~~.= b
The work equation is therefore
2m(1 :a 2A. 2 a) = q:l (3 _ Za)
q/ 2 (3 - 2a)a ( /)
m = 12 . 1 + 2A.2a where A. = b (8.4-3)
The minimum required value of m is the maximum as given by the above
work equation.
gives
dm _ 0
da -
a = ~ 2 ( + ~(1 + 3A.2) - 1) (8.4-4)
On substitution of this value of a into the work equation,
qlz (~(1 + 3A.2) - 1)2 A
m = 24 . A_4 ns.
Recognizing that a2 = (~(1 + 3A.2) - 1f/4A.\ the answer may also be
expressed as
q/2
m --a 2
- 6
where a has the value determined by eqn (8.4-4).
Comments
In this example, a full algebraic procedure was followed, in which the value
of a corresponding to dml da = 0 was determined and substituted back into
the work equation to obtain the value of m(max). In practice, m(max) is
frequently obtained directly from the work equation, i.e. eqn (8.4-3), by
calculating the values of m for various trial values of a. Usually, this
arithmetic method of trial and error yields a close approximation to the
correct answer in a comparatively short time; another advantage of the
arithmetic method over the algebraic method is that gross mistakes are
easily noticed.
Example 8.4-3
A rectangular slab is of length 10 m and width 5 m, and is isotropically
reinforced with top and bottom reinforcements such that the yield moment
of resistance ism per metre width of slab, both for positive and for negative