5.1 Introduction
The form of the bending moment diagram of a continuous beam in which the tension is alternated between the top and bottom fibres between the regions of supports and spans, respectively (see Figure 3.14), suggests that such a beam behaves as consisting of inter- mediate simply supported beams resting on cantilevers, extending from the internal supports up to the points where the bending moments become null. The length of these cantilevers constitutes a percentage of the corresponding span, which, for internal spans having a constant depth, is of the order of 20%, depending on the type of loading, as well as the span ratios (Figure 5.1).
For beams of large spans, as are usually applied in bridges, it is statically purposeful to follow a parabolic change in depth in all spans, with the maximum depth at the supports and the minimum depth at the midspan sections. In this way there is an increase in the moments at the supports (see Section 3.2.10) and, according to the equilibrium requirement (suspension of the moment diagram for the simply supported beam between the values at the supports), a reduction in the midspan moments. Moreover, the cantilever lengths are increased compared with those of the corresponding beam with constant depth (see Figure 5.1). Thus, in the case of a constant depth, the bending moment diagram is independent of the moment of inertia of the beam, as explained in Section 3.2.5.
The aforementioned change in depth leads to a proportional distribution of self- weight that, for a support-to-midspan depth ratio of 3 (a common value for bridges with large spans), results in a significant increase (up to 35%) in the lengths of the considered ‘cantilevers’ of the internal spans. Therefore, the ‘simply supported beams’ are limited to 30% of the span lengths, with far smaller bending moments than those of the ‘cantilevers’ (see Figure 5.1). The fact that the thus resulting support moments do not differ substantially from those of the cantilevers extending up to the midspan
gave rise to the construction of concrete bridges according to the so-called balanced
cantilever method, namely the construction of progressively smaller prestressed cantilever segments from both sides of the two free-standing supports (piers) until the midspan, and thereafter the restoration of the beam continuity (see Section 5.5.2). This depth variation in the ‘cantilevers’ contributes favourably to the shear response, as explained in Section 4.4.
5.2 Steel beams
As, mainly for reasons of economy, steel beams are nowadays designed using the ‘plastic theory’ rather than the ‘elastic’ one, it is advisable to be familiar with this means of design, as applied to a statically indeterminate continuous beam, by considering first the behaviour of a fixed-ended beam and then that of a fixed simply supported beam. The bending (and shear) response at any span can be described using the basic charac- teristics for either beam type.
5.2.1 Fixed-ended beam
A fixed-ended beam under a uniform load q develops the bending moment diagram
shown in Figure 5.2. From the figure and the discussion above it is obvious that there
must be two cantilevers with length 0.21 L. The moment of the support q L2/12 is
L ~0.30 · L ~0.35 · L ~0.35 · L L ~0.60 · L ~0.20 · L ~0.20 · L
double the maximum moment of the midspan, which is of course the span moment of
the ‘fictitious’ simply supported beam of length 0.58 L. It is assumed that the cross-
section of the beam is symmetrical with respect to the horizontal axis of bending, and
accordingly it has the same bending resistance Mpl for moments causing tension
either at the top fibres or at the bottom ones.
For some loadq1the bending moment will first reach the plastic resistance value at the
supports. Thus q1 L2/12¼ Mpland q1¼ 12 Mpl/L2. For this load a plastic hinge will be
developed at both ends of the beam, as discussed in Section 4.1.2 (Figure 5.3). Obviously, the beam does not collapse under this load, and is still in a position to accept more loading. During the imposition of any additional load, the plastic hinges at the two ends rotate
under the constant moment Mpl, which can be considered as being externally applied.
q
0.21 · L 0.58 · L 0.21 · L
q · L2/12 q · L2/8
Figure 5.2 Load-bearing action of a fixed-ended beam
Virtual loading for the determination of rotation at the plastic hinge
q1 qu Mpl Mpl Mpl Mpl Mpl qu = 16 · Mpl/L2 qu · L2/8 1 1 1 1
The diagram of the bending moment corresponding to each loadq is obtained by hanging
the diagram for the simply supported beamq L2/8 from the constant points that corre-
spond to Mpl. Clearly the beam will collapse when the load q takes the value qu at
which a bending momentMplis developed at the midspan of the beam. In this case, an
additional plastic hinge will be created at the midspan (first-degree mechanism) and no further increase in load will be possible, as discussed in Section 4.1.2 for a simply supported beam. Thus it may be written thatqu L2/8 Mpl¼ Mpland, consequently,qu¼ 16 Mpl/
L2(see Figure 5.3).
The problem now of designing a beam under a certain service loadqseris thatMplshould
be defined such that the beam may sustain, or even better collapse at, a loadqu¼ qser
(see Section 4.6). According to the last equation the value sought forMplis given by
Mpl¼ ( qser) L2/16
It should be pointed out that designing a beam for the ‘collapse’ load qserby following
the elastic moment diagram means that a value of ( qser) L2/12 must be selected for
Mpl, which is of course higher than above value, and consequently represents an
uneconomical solution. Precisely for this reason it was pointed out in Section 4.6 that in the equation
S R (a)
the design value S does not necessarily correspond to the elastic solution for the load
qser. Thus, in order to design a fixed-ended beam with lengthL, so that the ultimate
bending resistance is exploited at the ends as well as at the midspan, one should consider the bending moment diagram for the simply supported beam that ensures balance with the applied load qserand then suitably shift its base until equal partsMplare cut at
both ends as well as the midspan (see Figure 5.3). Thus, it is required that ( qser) L2/8¼ Mplþ Mpl
This result coincides with the preceding one.
It is important though to point out that, according to the static theorem of plastic analysis (see Section 6.6.2), the satisfaction of the above inequality (a) at any point
on the beam ensures that the value qseris less than or equal to the ultimate load,
provided that the value S belongs to a bending moment diagram which is in
equilibrium with the external load qser. The confirmation that ( qser)< qu, can be
made when selecting both Mpl¼ ( qser) L2/16 and Mpl¼ ( qser) L2/12. It is, of
course, clear that while in the first case the ultimate loadqucoincides with qser, in
the second case, according to the above equation, it results in qu¼ 16 Mpl/L2¼ 1.33 ( qser)
It is purposeful at this point to find also the correspondingMplthat would result if the
design was made on the basis of the allowable stressfy/. As the section at the beam end
is the most unfavourable one, according to Section 4.1.2 My/ ¼ qser L2/12. As
Mpl¼ a My, it follows that the selected section on which the load qser causes the
Mpl¼ a ( qser) L2/12. This value of the moment resistance leads to a collapse load
equal to qu¼ 16 Mpl/L2¼ a 1.33 ( qser). For a¼ 1.15, which is appropriate for
I-shaped cross-sections, qu¼ 1.53 ( qser). This result shows that the safety factor
determined according to the method of permissible stresses is practically equal to 1.53 . Finally, another point of interest is the rotation developed in the plastic hinge at the instant of collapse. This rotation, which must be developed in order to realise the aforementioned moment diagram, is limited by the plastic deformability (ductility) of the material, which for steel (as experimentally shown) is of the order of 0.1 rad. This rotation can be determined directly from the existing moment diagram for the instant of collapse by applying the principle of virtual work.
A self-equilibrating system of unit moments, applied at the ends of the simply supported beam, is selected as the virtual load. According to the principle of virtual work, and taking into account the above (see Figure 5.3):
2 1 ’ ¼ ð MM EI ds¼ Mpl L 3 EI ¼ 2 a fy L 3 E h (see Section 4.1.2)
For the valuesa¼ 1.15 and fy¼ 4.2 105kN/m2, it results that’ ¼ 0.77 10—3 (L/h). It
is concluded that for practical values of the ratioL/h, i.e. values of 25—30, the resulting value of’ is feasible.
The resulting value of the angle’ for the elastic diagram of bending moments must be
equal to zero, as the diagram under consideration is supposed to satisfy, besides the equilibrium, the compatibility of the deformations. Such a requirement cannot, of course, be satisfied for the adopted design diagram, as elastic behaviour has been exceeded and plastic behaviour has already been established.
5.2.2 Fixed simply supported beam
A fixed simply supported beam under a uniform load q develops the bending moment
diagram shown in Figure 5.4. It can be seen that there is a cantilever part of length q q · L2/8 q · L2/8 max M 0.25 · L 0.75 · L 0.375 · L
0.25 L, with a fixed end moment equal to q L2/8. The maximum span moment corre-
sponding to the bending moment of a ‘fictitious’ simply supported beam of length 0.75 L
occurs at a distance 3 L/8 from the free end, and is equal to q L2/14.2.
The beam at first develops a plastic hinge at the fixed end under the loadq1, so that
Mpl¼ q1 L2/8, and it can be loaded further, up to the collapse under the total loadqu.
Under this load, and with Mpl applied externally at the henceforth ‘hinged’ end, the
beam develops its maximum span moment, which is also equal toMpl(Figure 5.5). It
is found that the loadquthat leads to collapse isqu¼ 12 Mpl/L2.
The same design procedure used for the fixed-ended beam can be equally applied to the fixed simply supported beam also. On the basis of the bending moment diagram of
the simply supported beam under an external load qser, a position of the rotating
baseline around the simply supported end is sought at which the cut maximum value of the span is equal to the one corresponding to the fixed end (see Figure 5.5). These
equal values are obviously the plastic moment Mpl, and according to the above this
results inMpl¼ ( qser) L2/12. q1 qu L Mpl Mpl Mpl M M 1 1
Virtual loading for the determination of rotation at the plastic hinge
qu = 12 · Mpl/L2
qu · L2/8
5 · L/12 5 · L/12
(γ · qser) · L2/8