Remembering that in this case the centre bending moment ∑wl will be equal to wL&SUP2;/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

1 . . . H = wL&SUP2;/8y,

or, calling x the distance from the vertex to the point of support,

H = wx&SUP2;/2y,

The value of H is equal to the maximum tension on the bottom flange, or compression on the top flange, of a girder of equal span, equally and similarly loaded, and having a depth equal to the dip of the suspension bridge. Fig. 70.

Consider any other point F of the curve, fig. 70, at a distance x from the vertex, the horizontal component of the resultant (tangent to the curve) will be unaltered; the vertical component V will be simply the sum of the loads between O and F, or wx. In the triangle FDC, let FD be tangent to the curve, FC vertical, and DC horizontal; these three sides will necessarily be proportional respectively to the resultant tension along the chain at F, the vertical force V passing through the point D, and the horizontal tension at O; hence

H : V = DC : FC = wx&SUP2;/2y : wx = x/2 : y,

hence DC is the half of OC, proving the curve to be a parabola.

The value of R, the tension at any point at a distance x from the vertex, is obtained from the equation

R&SUP2; = H&SUP2;+V&SUP2; = w&SUP2;x4/4y&SUP2;+w&SUP2;x&SUP2;,

or,

2 . . . R = wx√(1+x&SUP2;/4y&SUP2;).

Let i be the angle between the tangent at any point having the co-ordinates x and y measured from the vertex, then

3 . . . tan i = 2y/x.

Let the length of half the parabolic chain be called s, then

4 . . . s = x+2y&SUP2;/3x.

The following is the approximate expression for the relation between a change ∆s in the length of the half chain and the corresponding change ∆y in the dip: -

s+∆s = x+(2/3x) {y&SUP2;+2yδy+(∆y)&SUP2;} = x+2y&SUP2;/3x+4yδy/3x+2∆y&SUP2;/3x,

or, neglecting the last term,

5 . . . ∆s = 4y∆y/3x,

and

6 . . . ∆y = 3x∆s/4y.

From these equations the deflection produced by any given stress on the chains or by a change of temperature can be calculated. Fig. 71.

36. Deflection of Girders. - Let fig. 71 represent a beam bent by external loads. Let the origin O be taken at the lowest point of the bent beam. Then the deviation y = DE of the neutral axis of the bent beam at any point D from the axis OX is given by the relation

 d&SUP2;y dx&SUP2; = M EI

where M is the bending moment and I the amount of inertia of the beam at D, and E is the coefficient of elasticity. It is usually accurate enough in deflection calculations to take for I the moment of inertia at the centre of the beam and to consider it constant for the length of the beam. Then

 dy dx = 1 EI ∫Mdx
 y = 1 EI ∫∫Mdx&SUP2;.

The integration can be performed when M is expressed in terms of x. Thus for a beam supported at the ends and loaded with w per inch length M = w(a&SUP2;-x&SUP2;), where a is the half span. Then the deflection at the centre is the value of y for x = a, and is

 δ = 5 24 wa4 EI .

The radius of curvature of the beam at D is given by the relation

R = EI/M. Fig. 72.

37. Graphic Method of finding Deflection. - Divide the span L into any convenient number n of equal parts of length l, so that nl = L; compute the radii of curvature R, R, R for the several sections. Let measurements along the beam be represented according to any convenient scale, so that calling L and l the lengths to be drawn on paper, we have L = aL; now let r, r, r be a series of radii such that r = R/ab, r = R/ab, etc., where b is any convenient constant chosen of such magnitude as will allow arcs with the radii, r, r, etc., to be drawn with the means at the draughtsman's disposal. Draw a curve as shown in fig. 72 with arcs of the length l, l, l, etc., and with the radii r, r, etc. (note, for a length &FRAC12;l at each end the radius will be infinite, and the curve must end with a straight line tangent to the last arc), then let v be the measured deflection of this curve from the straight line, and V the actual deflection of the bridge; we have V = av/b, approximately. This method distorts the curve, so that vertical ordinates of the curve are drawn to a scale b times greater than that of the horizontal ordinates.

Thus if the horizontal scale be one-tenth of an inch to the foot, a = 120, and a beam 100 ft. in length would be drawn equal to 10 in.; then if the true radius at the centre were 10,000 ft., this radius, if the curve were undistorted, would be on paper 1000 in., but making b = 50 we can draw the curve with a radius of 20 in. The vertical distortion of the curve must not be so great that there is a very sensible difference between the length of the arc and its chord. This can be regulated by altering the value of b. In fig. 72 distortion is carried too far; this figure is merely used as an illustration.

38. Camber. - In order that a girder may become straight under its working load it should be constructed with a camber or upward convexity equal to the calculated deflection. Owing to the yielding of joints when a beam is first loaded a smaller modulus of elasticity should be taken than for a solid bar. For riveted girders E is about 17,500,000 lb per sq. in. for first loading. W.J.M. Rankine gives the approximate rule

Working deflection = δ = l&SUP2;/10,000h,

where l is the span and h the depth of the beam, the stresses being those usual in bridgework, due to the total dead and live load.

(W. C. U.)

 For the ancient bridges in Rome see further Rome: Archaeology, and such works as R. Lanciani, Ruins and Excavations of Ancient Rome (Eng. trans., 1897), pp. 16 foll.