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84
are bending stresses when deformations are relatively small. As such, this method
is best suited to the design of redundant framed mild steel structures, continuous
or restrained beams and girders which carry load by virtue of flexural resistance.
2. Plastic design for steel structures is neither suggested for statically determinate
beams and girders nor for structures with pin-connected members.
3. The load factors given in IS: 800–1984 are used in the load combinations.
The advantage of the plastic analysis and design method lies in the fact that it
provides a simple analysis approach for redundant structures and results in smaller
size sections than those designed by WSD method. Therefore, the approach makes
use of the capacity of the steel (reserve strength) which is left unutilised in the
WSD method. Moreover, the failure mode can be visualised.
4/3!
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The idealised elasto-plastic stress–strain curve for an annealed mild steel specimen
in tension and the effect of residual stresses has been discussed in Section 1.7. The
stress-strain curve of Fig. 1.10 relates to the properties of mild steel. The flexural
behaviour of a beam can be expressed by plotting the moment–curvature (M–f) or
moment–rotation (M–q) curve. Basically, both represent the same shape for given
member and any of these may be used to express flexural response. However, the
response of beams in flexure is described better by the moment–curvature relation
and is shown in Fig. 3.1. It may be noted that the moment–rotation plot will also
be similar in shape, but with different slopes than that of moment–curvature plot.
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The M–f curve is the most appropriate curve for assessing the flexural response
of the beam as the slope of the curve gives directly the flexural rigidity.
Consider a rectangular beam of width b and depth d (Fig. 3.2). From the flexure
formula;
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or
or
M
E
=
I
R
1
M
=f=
R
EI
(1)
(EI = flexural rigidity)
Slope of M – f curve is EI.
where R = radius of curvature of the centroidal axis
1
f=
= curvature.
R
Refer Fig. 3.3.
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Relation between q and R can be expressed as
Rq = mp = nq = ef = initial length = constant (k)
Hence,
q=
1
k
= kf
= k¥
R
R
(2)
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q
k
where q = angle of rotation of the section.
From Eqs. (2) and (3),
or
f =
86
(3)
q
M
=
k
EI
M
EI
=
k
q
(4)
Slope of M–q curve is EI/k.
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For an introduction to the behaviour of a beam in flexure, a center loaded simple
I-shape beam under gradually increasing moment is shown in Fig. 3.4. Variations
of bending stress diagrams corresponding to the marked points on the idealised
stress–strain curve for mild steel of Fig. 3.5 are also shown in Fig. 3.4. The
maximum moment, outer fibre stress, and curvature occur at mid-span. In the
elastic range OC of the curve (Fig. 3.5), the stress is proportional to strain and
the response is linearly elastic. With the assumption that a plane section before
bending remains plane and normal to mid-plane after bending, the strain along
the depth of the section can be obtained and since stress is proportional to strain,
the stress can be determined from the curve. Since strains are proportional to the
distances from the neutral axis, the distribution of strain and stress over the depth
of the section are as shown in Fig. 3.4 (a). Under these elastic conditions the
moment of resistance for any point, say 1, on this part of the curve is given by
M1 = f1Ze
(5)
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This is known as flexure formula. Here, f1 is the extreme fiber stress and Ze is
the elastic section modulus (the first moment of area).
As the load on the beam is gradually increased, the outer fibre strain at the
place of maximum moment eventually will reach the yield strain value ey and
the stress f 2 becomes equal to f y at point 2 marked on the curve (Fig. 3.5). The
moment corresponding to this point is known as first yield moment and is given
by
M2 = My = f 2Ze = f y Z e
(6)