**288**

**BRIDGES**

to the point where the resultant meets the plane of the cross section. The following considerations allow the maximum intensity of stress to be approximately calculated in most of the cases which are practically met with. 1. Let the cross section (fig. C) have an axis of sym metry XX 1} and let YYj be an axis pass ing through the centre of gravity of the sur face at right angles to XXj ; the axis YYj will be called the neutral axis. 2. Let the resultant stress pass through some point A in the axis XX 1} at a distance x from the axis YY V 3. Let the material . ,-,_-,, be such that its modulus of elasticity is constant under all the intensities of stress which result from the given total stress ; then calling the whole force P, the area of the cross section S, the mean intensity of stress p , the maximum intensity of stress p v the maximum distance of any point of the cross section from the axis x v and representing by I the moment of inertia (vide 9) of the surface of the cross section round the axis

YY,, we observe—first, thatP A-g-i and secondly, that the non-axial force may be assumed to produce a uniformly varying stress, the maximum intensity of which will occur at the distance x t from the neutral axis on the same side as x is taken. This uniformly varying stress is equivalent to a uniformly distributed stress of the intensity p and a couple of the moment al, where a is the constant rate of increase of the stress. We also know by the principles of mechanics that the force P applied at a distance X Q from the centre might be replaced by an equal force P at the centre and a couple of the moment ~Px Q . Hence we have—

P.r 1 P# = al, or a p . But the maximum intensity of stress due to the couple will Pa; x be xfl, i.e., -^M , and the maximum stress p l maybe con sidered as consisting of two parts, first, the mean intensity of stress p , and secondly, the maximum stress due to the couple. Hence we have—

2. . I A

This equation shows that the maximum intensity of stress increases very rapidly as x increases, and it must be borne in mind that the ultimate strength of any member of a structure is determined, not by the mean stress, but by the maximum stress on any part, for when the part most strained yields the structure is weakened thereby, and this failure must continue to extend until the whole yields.

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Fig. 6.

§9. *Note on the Value of* I, *the Moment of Inertia of the Cross* *Section*.—Suppose the cross section divided into a very
large number of rectangles, so small that the distance x of
the centre of each from the neutral axis may be taken as
sensibly expressing the distance of every part of the
rectangle ; then I is the sum of the products of the areas
of all the elementary rectangles each multiplied into the
square of its distance from the neutral axis ; or calling the
area of the elementary rectangle [ *missing text* ]—

1 I = "

The subjoined table gives the values of I for some simple geometrical forms ; the axis in all cases passes through the centre of gravity of the surface.

*Moment of Inertia*? I.

Surface. Neutral Axis. I. Circle, radius r Square, side d Square, side d liectangle with sides d and b Triangle, base b, height d Diameter Parallel to one side Diagonal Parallel to b Parallel to b ITTT* Jod* J*d* JsbiP TfW

Whenever a cross section can be conceived as obtained by the addition or subtraction of one surface to or from another, both surfaces having a common neutral axis, the value of I for that cross section is got by adding or subtracting the value of I for one surface to or from that for the other ; thus for a ring surface, with external and internal radii r and r v the value of I is—

- (**-!*).

The value Ij of the moment of inertia of any plane surface S round an axis in its own plane parallel to the neutral axis, and at a distance x from that axis, is given by the equation—

9 T _T_L.C 2 " J-j -L -T- Oc , where I is the moment of inertia of the surface round a neutral axis, i.e., round a parallel axis passing through the centre of gravity of the section. We can thus obtain the value of the moment of inertia I of complicated cross sections whenever we can divide these into rectangles, circles, or triangles ; for then, calling s 1? s. 2 , s. A . &c., the sur faces of each elementary part, x v x 2 , x z the distances of the centre of each part from the neutral axis of the whole cross section, and I,, !, I //y the moments of inertia of each element calculated round its own neutral axis, we have, the moment of inertia of the whole round its neutral axis—

3 T = 21 + 2* ^ 2 .

§10. *Specific Gravity of* *Materials*.—In order to calculate the load due to the superstructure of a bridge, and the
stability due to the weight of the abutments and anchorages
of arches and suspension bridges, it is necessary to know the
specific gravity of the materials employed specific gravity
being for the purposes of the present article defined as the
weight of the material in Ibs. per cubic foot. The follow
ing are the most useful numbers:—

*Weight per Cubic Foot of Different Materials*.

Name of Material. Weight of cubic foot in Ibs. Water (pure) at 39 "4. Falir 62 425 Basalt 187-3 Brick 100 to 135 Brickwork (ordinary) 112 Chalk 117 to 174 Clay 120 Granite 164 to 172 Limestone and Marble 169 to 175 Masonry 116 to 144 Mortar 86 to 119 Mud 102 Sandstone 130 to 157 Sand (damp) 118 Sand (dry) 88 6 Asphalt 156 Concrete (ordinary) 119 Concrete in cement 137 Earth 77 to 125 Slate 175 to 181 Trap 170 Cast-iron (average) 444 "Wrought Iron (average) 480 Eed Pine 30 to 44 Larch 31 to 35

O.a-t (European) 43 to 62