Remembering that in this case the centre bending moment ∑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

1 . . . H = wL²/8y,

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

H = wx²/2y,

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.  Chain Loaded uniformly along a Horizontal Line. 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 = wx²/2y : 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

R² = H²+V² = w²x4/4y²+w²x²,

or,

2 . . . R = wx√(1+x²/4y²).

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+2y²/3x.

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) {y²+2yδy+(∆y)²} = x+2y²/3x+4yδy/3x+2∆y²/3x,

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.  Beam bent by external loads. 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

d²y
line
dx²
=M
line
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
line
dx
=1
line
EI
∫Mdx
y =1
line
EI
∫∫Mdx².

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(a²-x²), where a is the half span. Then the deflection at the centre is the value of y for x = a, and is

δ =5
line
24
wa4
line
EI
.

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

R = EI/M.

Fig. 72.  Graphic Method of finding Deflection 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 ½l 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 = l²/10,000h,

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.)

[1] 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.