unknown stresses are not more than three in number, and the three conditions of equilibrium, page 17, enable us to calculate the stress on each; or using the bending moment and shearing force diagrams, we can readily apply the conditions given on page 18 to as many sections as necessary, and so design the whole beam. The sections chosen for this purpose should lie just to one side of the joints.

For any given system of loading, a beam is said to be of uniform strength when the cross section of each piece is such, that the ratio of the ultimate or proof resistance of the material and the tension or compression which it has to bear is constant; an allowance being made, if necessary, for pieces which also act as small beams.

The term Flanged Girder is employed to denote all girders consisting of a web and one flange, or of two flanges connected together either by a continuous web or by open lattice work; in bridges one at least of the flanges is usually straight, and also horizontal.

In figs. 8 and 9, page 16, the stress at A is horizontal and is denoted by Pa; if the flange is thin, as is often the case in iron bridges, this stress is sensibly constant; and since the intensity of the stress to which a piece is exposed should not exceed the strength of its material, we have

Pa ≤f....

..(1.) where f is the working or proof stress as may be desired; and if t amount of stress on the horizontal flange, and S the cross sectional area of that flange, then

=

When there is only one set of triangles, as is shown in the lowest of the three systems of bracing, fig. 87, and also in the two upper systems if we neglect the counter-bracing, the amount of stress on the straight boom may be found thus ;Take a cross section at a point, say F, just on either side of a joint in the boom opposite the straight boom (in fig. 87, both booms are straight); take moments round the point F, and since we neglect the counter-brace, the only member not passing through the point F is the upper boom; the product t.h gives the moment where h is the depth of the beam at

the point; this is equal to the bending moment, and we have by substituting the value of t given in equation 2,

t.h = S.f.h

= M.........

·(3.)

The quantity M is different at different points of the beam, and ƒ is a constant quantity; if the above equation is to be fulfilled for every point, we make S.h vary as M; in practice one of these two factors is generally kept constant, and the other is made to vary.

If we make h constant, then both booms are horizontal, and S varies as the bending moment; hence the bending moment diagram gives, upon a suitable scale, the area of the boom at each point. The vertical component of the stress on any diagonal is the amount of the shearing force at that end of the diagonal where the shearing force is greatest; hence if be the stress on a diagonal, F the shearing force at the end of the diagonal where it is greatest and as given by the shearing force diagram, and the angle made by the diagonal with the vertical, then T = F sec 0........

.(4.)

The depth h is chosen from th to th of the span to ensure stiffness; in fig. 87, h is taken at 3 feet, that is th of span, and that figure shows how the stresses on the booms and on the diagonals are found. Between the points K and L, counterbracing is required; this is accomplished in the two upper girders by introducing pairs of diagonals between these points, both being ties or both struts as the case may be; the one diagonal resists the positive, the other resists the negative shearing force. In the Warren girder, the third in the figure, one diagonal resists both the positive and negative shearing force, the one being applied suddenly after the other; each diagonal should be designed to bear the stress due to the sum of these stresses. The lowest girder shown in the figure is one with thin flanges and a thin continuous web; part of the bending moment is resisted by the web and part of the shearing is resisted by the flanges, these parts however are small; practically the flanges are considered, as in the case of open beams, to resist the whole of the bending moment, and the web is considered to resist the whole of the shearing force; an approximate result

is obtained if we consider that the shearing force is uniformly distributed over the cross sectional area of the web.

If we make S constant, then h varies as M; and, fig. 88,

the elevation of the beam will correspond with the bending moment diagram; ho the depth of the beam at the centre is, as in the previous case, to be taken sufficient to ensure stiffness. The curved boom will bear a share of the shearing force; this compounded with the stress on the horizontal boom at the same section will give the resultant stress on the curved boom. It is usual in practice to make the area of the curved and of the straight booms uniform, and to make the diagonals sufficiently strong to resist the whole of the shearing force as in the previous case.

Where the curved member slopes considerably, as in the Bowstring Girder, it is made sufficiently strong to bear the whole shearing force, the diagonals being intended for another purpose, viz. to distribute partial loads in a sensibly uniform manner.

The theoretical elevations reduce to a height zero at the ends, and so give no material to resist the shearing force at the point where it is greatest; sufficient material is generally allowed at the ends either by making the span of the girder exceed the clear span, or by departing from the theoretical form along a tangent near the end. Further, whatever the curves may be, and, as we have seen, they are generally parabolas, they are usually replaced by circles which nearly coincide therewith; when the figure passes from one curve to another, the passage is made along a tangent as will be seen on some of the figures. Approximate forms consisting entirely of straight lines enveloping the bending moment diagram, are sometimes adopted.

MOMENT OF RESISTANCE TO BENDING OF RECTANGULAR AND TRIANGULAR CROSS SECTIONS.

The Moment of Resistance to Bending we have defined as the moment of the total stress upon the cross section about any point in it; and this we have shown, figs. 7, 8, 9, to be equal to the couple which is the moment of the normal stress on the cross section.

The stress, fig. 8, might be artificially produced by building on the portion O'A, columns of a material tending to gravi

tate towards the left; and on the portion O'B, columns gravitating towards the right. These columns standing on very small bases, being of uniform density, and of the proper height to produce the intensity at each point, will, if we suppose them to become one solid, form a wedge with a stepped or notched sloping surface; the more slender the columns are, the more accurately do they give the stress at each point, and the smaller are the notches on the wedge; hence two right wedges exactly represent the normal stress on the cross section. Such a stress is called an uniformly varying stress. Taking the density of the wedges as unity, the height of one will be expressed by p., and the other by P.; the volumes of the wedges will give the two normal forces, fig. 10, the resultants of the thrusts and tensions respectively; these forces are equal, the volumes of the two wedges must therefore be equal, and the position of the neutral axis of the cross section is thus determined. Further, each wedge, instead of distributing its weight over its base, may be supposed to stand on the point below its centre of gravity; this enables us to find the positions of the normal forces, fig. 10, and gives us the arm of the couple; if we multiply the volume of either wedge by this arm, we have M the moment of resistance to bending. For a

Rectangular cross section, the neutral axis is at the centre since the wedges are equal, and p. equals p.; and if the common volumes of the right wedges be represented by V,

then

V = 1 Pa × bh = 1 p.bh ;

where b is the breadth and h is the depth of the beam, as shown in fig. 1. Each wedge stands on a rectangular base, so that the point on the cross section below its centre of gravity is distant from O' by two-thirds of O'A, or } h; hence the arm of the couple is 3h, and

By increasing the bending moment we can increase M till Pa becomes equal to f, the resistance of the material to direct tension or thrust, but no further; because if p. becomes