dC LP
-- = - -- + La = 0,
dl l²
/*
P = al² = G;
that is, when the cost of one pier is equal to the cost erected of the main
girders of one span. Sir Guilford Molesworth puts this in a convenient but
less exact form. Let G be the cost of superstructure of a 100-ft. span
erected, and P the cost of one pier with its protection. Then the economic
span is l = 100[root]P/[root]G.
30. _Limiting Span._--If the weight of the main girders of a bridge, per
ft. run in tons, is--
w_3 = (w_1+w_2)lr/(K-lr)
according to a formula already given, then w_3 becomes infinite if k-lr =
0, or if
l = K/r,
[v.04 p.0554] where l is the span in feet and r is the ratio of span to
depth of girder at centre. Taking K for steel girders as 7200 to 9000,
Limiting Span in Ft.
r = 12 l = 600 to 750
= 10 = 720 to 900
= 8 = 900 to 1120
[Illustration: FIG. 58.]
[Illustration: FIG. 59.]
[Illustration: FIG. 60.]
In a three-span bridge continuous girders are lighter than discontinuous
ones by about 45% for the dead load and 15% for the live load, if no
allowance is made for ambiguity due to uncertainty as to the level of the
supports. The cantilever and suspended girder types are as economical and
free from uncertainty as to the stresses. In long-span bridges the
cantilever system permits erection by building out, which is economical and
sometimes necessary. It is, however, unstable unless rigidly fixed at the
piers. In the Forth bridge stability is obtained partly by the great excess
of dead over live load, partly by the great width of the river piers. The
majority of bridges not of great span have girders with parallel booms.
This involves the fewest difficulties of workmanship and perhaps permits
the closest approximation of actual to theoretical dimensions of the parts.
In spans over 200 ft. it is economical to have one horizontal boom and one
polygonal (approximately parabolic) boom. The hog-backed girder is a
compromise between the two types, avoiding some difficulties of
construction near the ends of the girder.
[Illustration: FIG. 61.]
[Illustration: FIG. 62.]
Most braced girders may be considered as built up of two simple forms of
truss, the king-post truss (fig. 61, a), or the queen-post truss (fig. 61,
b). These may be used in either the upright or the inverted position. A
_multiple truss_ consists of a number of simple trusses, e.g. Bollman
truss. Some timber bridges consist of queen-post trusses in the upright
position, as shown diagrammatically in fig. 62, where the circles indicate
points at which the flooring girders transmit load to the main girders.
_Compound_ trusses consist of simple trusses used as primary, secondary and
tertiary trusses, the secondary supported on the primary, and the tertiary
on the secondary. Thus, the Fink truss consists of king-post trusses; the
Pratt truss (fig. 63) and the Whipple truss (fig. 64) of queen-post trusses
alternately upright and inverted.
[Illustration: FIG. 63.]
[Illustration: FIG. 64.]
A combination bridge is built partly of timber, partly of steel, the
compression members being generally of timber and the tension members of
steel. On the Pacific coast, where excellent timber is obtainable and steel
works are distant, combination bridges are still largely used (Ottewell,
_Trans. Am. Soc. C.E._ xxvii. p. 467). The combination bridge at Roseburgh,
Oregon, is a cantilever bridge, The shore arms are 147 ft. span, the river
arms 105 ft., and the suspended girder 80 ft., the total distance between
anchor piers being 584 ft. The floor beams, floor and railing are of
timber. The compression members are of timber, except the struts and bottom
chord panels next the river piers, which are of steel. The tension members
are of iron and the pins of steel. The chord blocks and post shoes are of
cast-iron.
[Illustration: FIG. 65.]
33. _Graphic Method of finding the Stresses in Braced Structures._--Fig. 65
shows a common form of bridge truss known as a _Warren girder_, with lines
indicating external forces applied to the joints; half the load carried
between the two lower joints next the piers on either side is directly
carried by the abutments. The sum of the two upward vertical reactions must
clearly be equal to the sum of the loads. The lines in the diagram
represent the directions of a series of forces which must all be in
equilibrium; these lines may, for an object to be explained in the next
paragraph, be conveniently named by the letters in the spaces which they
separate instead of by the method usually employed in geometry. Thus we
shall call the first inclined line on the left hand the line AG, the line
representing the first force on the top left-hand joint AB, the first
horizontal member at the top left hand the line BH, &c; similarly each
point requires at least three letters to denote it; the top first left-hand
joint may be called ABHG, being the point where these four spaces meet. In
this method of lettering, every enclosed space must be designated by a
letter; all external forces must be represented by lines _outside_ the
frame, and each space between any two forces must receive a distinctive
letter; this method of lettering was first proposed by O. Henrici and R. H.
Bow (_Economics of Construction_), and is convenient in applying the theory
of reciprocal figures to the computation of stresses on frames.
34. _Reciprocal Figures._--J. Clerk Maxwell gave (_Phil. Mag. 1864_) the
following definition of reciprocal figures:--"Two plane figures are
reciprocal when they consist of an equal number of lines so that
corresponding lines in the two figures are parallel, and corresponding
lines which converge to a point in one figure form a closed polygon in the
other."
Let a frame (without redundant members), and the external forces which keep
it in equilibrium, be represented by a diagram constituting one of these
two plane figures, then the lines in the other plane figure or the
reciprocal will represent in direction and magnitude the forces between the
joints of the frame, and, consequently, the stress on each member, as will
now be explained.
Reciprocal figures are easily drawn by following definite rules, and afford
therefore a simple method of computing the stresses on members of a frame.
The external forces on a frame or bridge in equilibrium under those forces
may, by a well-known proposition in statics, be represented by a closed
polygon, each side of which is parallel to one force, and represents the
force in magnitude as well as in direction. The sides of the polygon may be
arranged in any order, provided care is taken so to draw them that in
passing round the polygon in one direction this direction may for each side
correspond to the direction of the force which it represents.
[Illustration: FIG. 66.]
This polygon of forces may, by a slight extension of the above definition,
be called the _reciprocal figure_ of the external forces, if the sides are
arranged in the same order as that of the joints on which they act, so that
if the joints and forces be numbered 1, 2, 3, 4, &c., passing round the
outside of the frame in one direction, and returning at last to joint 1,
then in the polygon the side representing the force 2 will be next the side
representing the force 1, and will be followed by the side representing the
force 3, and so forth. [v.04 p.0555] This polygon falls under the
definition of a reciprocal figure given by Clerk Maxwell, if we consider
the frame as a point in equilibrium under the external forces.
Fig. 66 shows a frame supported at the two end joints, and loaded at each
top joint. The loads and the supporting forces are indicated by arrows.
Fig. 67a shows the reciprocal figure or polygon for the external forces on
the assumption that the reactions are slightly inclined. The lines in fig.
67 a, lettered in the usual manner, correspond to the forces indicated by
arrows in fig. 66, and lettered according to Bow's method. When all the
forces are vertical, as will be the case in girders, the polygon of
external forces will be reduced to two straight lines, fig. 67 b,
superimposed and divided so that the length AX represents the load AX, the
length AB the load AB, the length YX the reaction YX, and so forth. The
line XZ consists of a series of lengths, as XA, AB ... DZ, representing the
loads taken in their order. In subsequent diagrams the two reaction lines
will, for the sake of clearness, be drawn as if slightly inclined to the
vertical.
[Illustration: FIG. 67.]
If there are no redundant members in the frame there will be only two
members abutting at the point of support, for these two members will be
sufficient to balance the reaction, whatever its direction may be; we can
therefore draw two triangles, each having as one side the reaction YX, and
having the two other sides parallel to these two members; each of these
triangles will represent a polygon of forces in equilibrium at the point of
support. Of these two triangles, shown in fig. 67 c, select that in which
the letters X and Y are so placed that (naming the apex of the triangle E)
the lines XE and YE are the lines parallel to the two members of the same
name in the frame (fig. 66). Then the triangle YXE is the reciprocal figure
of the three lines YX, XE, EY in the frame, and represents the three forces
in equilibrium at the point YXE of the frame. The direction of YX, being a
thrust upwards, shows the direction in which we must go round the triangle
YXE to find the direction of the two other forces; doing this we find that
the force XE must act down towards the point YXE, and the force EY away
from the same point. Putting arrows on the frame diagram to indicate the
direction of the forces, we see that the member EY must pull and therefore
act as a tie, and that the member XE must push and act as a strut. Passing
to the point XEFA we find two known forces, the load XA acting downwards,
and a push from the strut XE, which, being in compression, must push at
both ends, as indicated by the arrow, fig. 66. The directions and
magnitudes of these two forces are already drawn (fig. 67 a) in a fitting
position to represent part of the polygon of forces at XEFA; beginning with
the upward thrust EX, continuing down XA, and drawing AF parallel to AF in
the frame we complete the polygon by drawing EF parallel to EF in the
frame. The point F is determined by the intersection of the two lines, one
beginning at A, and the other at E. We then have the polygon of forces
EXAF, the reciprocal figure of the lines meeting at that point in the
frame, and representing the forces at the point EXAF; the direction of the
forces on EH and XA being known determines the direction of the forces due
to the elastic reaction of the members AF and EF, showing AF to push as a
strut, while EF is a tie. We have been guided in the selection of the
particular quadrilateral adopted by the rule of arranging the order of the
sides so that the same letters indicate corresponding sides in the diagram
of the frame and its reciprocal. Continuing the construction of the diagram
in the same way, we arrive at fig. 67 d as the complete reciprocal figure
of the frame and forces upon it, and we see that each line in the
reciprocal figure measures the stress on the corresponding member in the
frame, and that the polygon of forces acting at any point, as IJKY, in the
frame is represented by a polygon of the same name in the reciprocal
figure. The direction of the force in each member is easily ascertained by
proceeding in the manner above described. A single known force in a polygon
determines the direction of all the others, as these must all correspond
with arrows pointing the same way round the polygon. Let the arrows be
placed on the frame round each joint, and so as to indicate the direction
of each force on that joint; then when two arrows point to one another on
the same piece, that piece is a tie; when they point from one another the
piece is a strut. It is hardly necessary to say that the forces exerted by
the two ends of any one member must be equal and opposite. This method is
universally applicable where there are no redundant members. The reciprocal
figure for any loaded frame is a complete formula for the stress on every
member of a frame of that particular class with loads on given joints.
[Illustration: FIG. 68]
[Illustration: FIG. 69]
Consider a Warren girder (fig. 68), loaded at the top and bottom joints.
Fig. 69 b is the polygon of external forces, and 69 c is half the
reciprocal figure. The complete reciprocal figure is shown in fig. 69 a.
The method of sections already described is often more convenient than the
method of reciprocal figures, and the method of influence lines is also
often the readiest way of dealing with braced girders.
35. _Chain Loaded uniformly along a Horizontal Line._--If the lengths of
the links be assumed indefinitely short, the chain under given simple
distributions of load will take the form of comparatively simple
mathematical curves known as catenaries. The true catenary is that assumed
by a chain of uniform weight per unit of length, but the form generally
adopted for suspension bridges is that assumed by a chain under a weight
uniformly distributed relatively to a horizontal line. This curve is a
parabola.
Remembering that in this case the centre bending moment [Sigma]wl will be
equal to wL²/8, we see that the horizontal tension H at the vertex for a
span L (the points of support being at equal heights) is given by the
expression