# Page:EB1911 - Volume 17.djvu/989

970
[STATICS
MECHANICS

with corresponding bars parallel, but without complete geometric similarity. In the case of fig. 52 it may be shown that an equivalent condition is that the six points A, B, C, D, E, F should lie on a conic (M. W. Crofton). This is fulfilled when the opposite sides of the hexagon are parallel, and (as a still more special case) when the hexagon is regular. ' .

When a frame has a critical form it may be in a state of stress independently of the action of extraneous forces; moreover, the stresses due to extraneous forces are indeterminate, and may be infinite.

For suppose as before t at one of the bars is removed. If there are no extraneous forces the equation of virtual work

reduces to S . 5s=o, where S is the stress in the removed bar, and Bs is the change in the distance between the joints which it connected. In a critical form we

have 6s=o, and the equation is satisfied FIG 53 by an arbitrary value of S; a consistent system of stresses in the remaining bars can then be found by preceding rules. Again, when extraneous forces P act on the joints, the equation is >:(P.ap)+s.as=o,

where bp is the displacement of any joint in the direction of the corresponding force P. If E(P. ép) =o, the stresses are merely indeterminate as before; but if E(P . 6p) does not vanish, the equation cannot be satisfied by any finite value of S, since és =o. This means that, if the material of the frame were absolutely unyielding, no finite stresses in the bars would enable it to withstand the extraneous forces. With actual materials, the frame would yield elastically, until its configuration is no longer “ critical.” The stresses in the bars would then be comparatively very great, although finite. The use of frames which approximate to a critical form is of course to be avoided in practice.

A brief reference must suffice to the theory of three dimensional frames. This is important from a technical point of view, since all structures are practically three-dimensional. We may note that a frame of n joints which is just rigid must have 3n-6 bars; and that the stresses produced in such a frame by a given system of extraneous forces in equilibrium are statically determinate, subject to the exception of “ critical forms." § ro. Statics of Inexlensible Chains.-The theory of bodies or structures which are deformable in their smallest parts belongs properly to elasticity (q.'v.). The case of inextensible strings or chains is, however, so simple that it is generally included in expositions of pure Statics. It is assumed that the form can be sufficiently represented by a plane curve, that the stress (tension) at any point P of the curve, between the two portions which meet there, is in the direction of the tangent at P, and that the forces on any linear element 53 must satisfy the conditions of equilibrium laid down in § I. It follows that the forces on any finite portion will satisfy the conditions of equilibrium which apply to the case of a rigid body (§ 4).

We will suppose in the first instance that the curve is plane. It is often convenient to resolve the forces on an element PQ (=5s) in the directions of the

tangent and normal respectively.

W' T, ” If T, T + ar be the tensions at P, Q, and 51,0 be the angle between

the directions of the curve at

these points, the components

Q' of the tensions along the tangent at P give (T+5T) cos uP-T,

or 6T, ultimately; whilst for the

component along the normal at

P we have (T-l-5T) sin' 5¢, or

T5¢, or T55/p, where p is the radius of curvature. Suppose, for example, that we have a light string stretched over a smooth curve; and let R5s denote the normal pressure (outwards from the centre of curvature) on cis. The two resolutions give 5T = o, T61// = Rés, or

FIG. 54.

l'=const., R='l'/p. (I)

The tension is constant, and the pressure per unit length varies as the curvature.

Next suppose that the curve is “rough ”; and let F5s be the tangential force of friction on 55. We have 5T == F55 = o, TW = Rés, where the upper or lower sign is to be taken according to the sense in which F acts. We assume that in limiting equilibrium we have F =/JR, everywhere, where y. is the coefficient of friction. If the string be on the point of slipping in the direction in which 1// increases, the lower sign is to be taken; hence 5T = F5s =/.¢T6//, whence

• V T=T0erL~l/, (2)

if To be the tension corresponding to 1L=o. This illustrates the resistance to dragging of a rope coiled round a post; e.g. if we put lI.='3, 1]/=21r, we find for the change of tension in one turn T/T0=6~ 5. In two turns this ratio is squared, and so on.

Again, take the case of a string under gravity, in contact with a smooth curve in a vertical plane. Let uP denote the inclination to the horizontal, and 'w5s the weight of an element 6s. (The tangential and normal components of w5s are -206; sin (L and -~w6s cos // Hence 6'I' =w5s sin 1/1, TW =w6s cos gl/-|-Rés. (3) If we take rectangular axes Ox, Oy, of which Oy is drawn vertically upwards, we have 5y=sin xl/ 5s, whence 6T='w6y. If the string be uniform, w is constant, and T =wy -l~ Const- = 'wfy-yo), (4)

say; hence the tension varies as the height above some fixed level (yo). The pressure is then given by the formula R=T%-5-w cos il/. (5)

In the case of a chain hanging freely under gravity it is usually convenient to formulate the conditions of equilibrium of a Bnite portion PQ. The forces on this reduce to three, viz. the weight of PQ and the tensions at P, Q. Hence these three forces will be concurrent, and their ratios will be given by a triangle of forces. In particular, if we consider a length AP beginning at the lowest point A, then resolving horizontally and vertically we have

T cos 1//=T0, T sin ¢=W, (6)

where Tn is the tension at A, and W is the weight of PA. The former equation expresses that the horizontal tension is constant.

If the chain be uniform we have W=ws, where s is the arc AP: hence ws=T0 tan x// If we write T0=wa, so that a is T

P

To ~ A 5 ' N o

W

Fic.. 55.

the length of a portion of the chain whose weight would equal the horizontal tension, this becomes s=a tan il/ (7)

This is the “ intrinsic ” equation of the curve. If the axes of x and y be taken horizontal and vertical (upwards), we derive x =a log (sec ¢»-l-tan 1//), y'= a sec 1//. '(8) Eliminating I# we obtain the Cartesian equation y =a cosh 5 (9)

of the common catenary, as it is called (fig. 56). The omission of the additive arbitrary constants of integration in (8) is equivalent to a special choice of the origin O of co-ordinates; viz. O is at a distance a vertically below the lowest point (// = o) of the curve. The horizontal line through O is called the directrix. The relations

r ==asinh§ , y'==02+Si» T=ToS¢C 1/'=wy| (10)