curvature of central line = bending moment :- flexural rigidity,
and the special conditions at the supported end; there is no alteration
of this statement on account of the shears. As regards the special
condition at an end which is _encastree_, or built in, Saint-Venant
proposed to assume that the central tangent plane of the cross-section
at the end is vertical; with this assumption the tangent to the central
line at the end is inclined downwards and makes an angle s0 with the
horizontal (see fig. 18); it is, however, improbable that this condition
is exactly realized in practice. In the application of the theory to the
experimental determination of Young's modulus, the small angle which the
central-line at the support makes with the horizontal is an unknown
quantity, to be eliminated by observation of the deflection at two or
more points.
54. We may suppose the displacement in a bent beam to be produced by the
following operations: (1) the central-line is deflected into its curved
form, (2) the cross-sections are rotated about axes through their
centroids at right angles to the plane of flexure so as to make angles
equal to 1/2[pi] + s0 with the central-line, (3) each cross-section is
distorted in its own plane in such a way that the appropriate variable
anticlastic curvature is produced, (4) the cross-sections are further
distorted into curved surfaces. The contour lines of fig. 17 show the
disturbance from the central tangent plane, not from the original
vertical plane.
[Illustration: FIG. 17.]
55. _Practical Application of Saint-Venant's Theory._--The theory above
described is exact provided the forces applied to the loaded end, which
have W for resultant, are distributed over the terminal section in a
particular way, not likely to be realized in practice; and the
application to practical problems depends on a principle due to
Saint-Venant, to the effect that, except for comparatively small
portions of the beam near to the loaded and fixed ends, the resultant
only is effective, and its mode of distribution does not seriously
affect the internal strain and stress. In fact, the actual stress is
that due to forces with the required resultant distributed in the manner
contemplated in the theory, superposed upon that due to a certain
distribution of forces on each terminal section which, if applied to a
rigid body, would keep it in equilibrium; according to Saint-Venant's
principle, the stresses and strains due to such distributions of force
are unimportant except near the ends. For this principle to be exactly
applicable it is necessary that the length of the beam should be very
great compared with any linear dimension of its cross-section; for the
practical application it is sufficient that the length should be about
ten times the greatest diameter.
56. In recent years the problem of the bending of a beam by loads
distributed along its length has been much advanced. It is now
practically solved for the case of a load distributed uniformly, or
according to any rational algebraic law, and it is also solved for the
case where the thickness is small compared with the length and depth, as
in a plate girder, and the load is distributed in any way. These
solutions are rather complicated and difficult to interpret. The case
which has been worked out most fully is that of a transverse load
distributed uniformly along the length of the beam. In this case two
noteworthy results have been obtained. The first of these is that the
central-line in general suffers extension. This result had been found
experimentally many years before. In the case of the plate girder loaded
uniformly along the top, this extension is just half as great as the
extension of the central-line of the same girder when free at the ends,
supported along the base, and carrying the same load along the top. The
second noteworthy result is that the curvature of the strained
central-line is not proportional to the bending moment. Over and above
the curvature which would be found from the ordinary relation--
curvature of central-line = bending moment :- flexural rigidity,
there is an additional curvature which is the same at all the
cross-sections. In ordinary cases, provided the length is large compared
with any linear dimension of the cross-section, this additional
curvature is small compared with that calculated from the ordinary
formula, but it may become important in cases like that of suspension
bridges, where a load carried along the middle of the roadway is
supported by tensions in rods attached at the sides.
[Illustration: FIG. 18.]
57. When the ordinary relation between the curvature and the bending
moment is applied to the calculation of the deflection of _continuous
beams_ it must not be forgotten that a correction of the kind just
mentioned may possibly be requisite. In the usual method of treating the
problem such corrections are not considered, and the ordinary relation
is made the basis of the theory. In order to apply this relation to the
calculation of the deflection, it is necessary to know the bending
moment at every point; and, since the pressures of the supports are not
among the data of the problem, we require a method of determining the
bending moments at the supports either by calculation or in some other
way. The calculation of the bending moment can be replaced by a method
of graphical construction, due to Mohr, and depending on the two
following theorems:--
(i.) The curve of the central-line of each span of a beam, when the
bending moment M is given,[1] is identical with the catenary or
funicular curve passing through the ends of the span under a
(fictitious) load per unit length of the span equal to M/EI, the
horizontal tension in the funicular being unity.
(ii.) The directions of the tangents to this funicular curve at the ends
of the span are the same for all statically equivalent systems of
(fictitious) load.
When M is known, the magnitude of the resultant shearing stress at any
section is dM/dx, where x is measured along the beam.
[Illustration: FIG. 19.]
[Illustration: FIG. 20.]
58. Let l be the length of a span of a loaded beam (fig. 19), M1 and
M2 the bending moments at the ends, M the bending moment at a section
distant x from the end (M1), M' the bending moment at the same section
when the same span with the same load is simply supported; then M is
given by the formula
l - x x
M = M' + M1 ----- + M2 --,
l l
and thus a fictitious load statically equivalent to M/EI can be easily
found when M' has been found. If we draw a curve (fig. 20) to pass
through the ends of the span, so that its ordinate represents the
value of M'/EI, the corresponding fictitious loads are statically
equivalent to a single load, of amount represented by the area of the
curve, placed at the point of the span vertically above the centre of
gravity of this area. If PN is the ordinate of this curve, and if at
the ends of the span we erect ordinates in the proper sense to
represent M1/EI and M2/EI, the bending moment at any point is
represented by the length PQ.[2] For a uniformly distributed load the
curve of M' is a parabola M' = 1/2wx(l - x), where w is the load per
unit of length; and the statically equivalent fictitious load is
(1/12)wl^3/EI placed at the middle point G of the span; also the loads
statically equivalent to the fictitious loads M1(l - x)/lEI and
M2x/lEI are 1/2M1l/EI and 1/2M2l/EI placed at the points g, g' of
trisection of the span. The funicular polygon for the fictitious loads
can thus be drawn, and the direction of the central-line at the
supports is determined when the bending moments at the supports are
known.
[Illustration: FIG. 21.]
59. When there is more than one span the funiculars in question may be
drawn for each of the spans, and, if the bending moments at the ends
of the extreme spans are known, the intermediate ones can be
determined. This determination depends on two considerations: (1) the
fictitious loads corresponding to the bending moment at any support
are proportional to the lengths of the spans which abut on that
support; (2) the sides of two funiculars that end at any support
coincide in direction. Fig. 21 illustrates the method for the case of
a uniform beam on three supports A, B, C, the ends A and C being
freely supported. There will be an unknown bending moment M0 at B, and
the system[3] of fictitious loads is (1/12)wAB^3/EI at G the middle
point of AB, (1/12)wBC^3/EI at G' the middle point of BC, -1/2M0AB/EI
at g and -1/2M0BC/EI at g', where g and g' are the points of
trisection nearer to B of the spans AB, BC. The centre of gravity of
the two latter is a fixed point independent of M0, and the line VK of
the figure is the vertical through this point. We draw AD and CE to
represent the loads at G and G' in magnitude; then D and E are fixed
points. We construct any triangle UVW whose sides UV, UW pass through
D, B, and whose vertices lie on the verticals gU, VK, g'W; the point F
where VW meets DB is a fixed point, and the lines EF, DK are the two
sides (2, 4) of the required funiculars which do not pass through A, B
or C. The remaining sides (1, 3, 5) can then be drawn, and the side 3
necessarily passes through B; for the triangle UVW and the triangle
whose sides are 2, 3, 4 are in perspective.
[Illustration: FIG. 22.]
The bending moment M0 is represented in the figure by the vertical
line BH where H is on the continuation of the side 4, the scale being
given by
BH 1/2M0BC
-- = ----------- ;
CE (1/12)wBC^3
this appears from the diagrams of forces, fig. 22, in which the
oblique lines are marked to correspond to the sides of the funiculars
to which they are parallel.
In the application of the method to more complicated cases there are
two systems of fixed points corresponding to F, by means of which the
sides of the funiculars are drawn.