Torsional rigidity of circular cylinder. Prism of any shape con simple twist, amount of the tangential force in either set of planes is a pe unit of area, if n be the rigidity of the substance. Hence ther is no force between parts of the substance lying on the two ses of any element of any circular cylinder coaxal with the bound: cylinder or cylinders; and consequently no force is required the cylindrical boundary to maintain the supposed state of stric And the mutual action between the parts of the substance on the two sides of any normal plane section consists of force in this plane, directed perpendicular to the radius through each pois and amounting to nτr per unit of area. The moment of this de tribution of force round the axis of the cylinder is (if do denote an element of the area) nrffdor, or the product of nr into the moment of inertia of the area round the perpendicular to its plan through its centre, which is therefore equal to the moment of the couple applied at either end. 702. Similarly, we see that if a cylinder or prism of ary strained to a shape be compelled to take exactly the state of strain above specified (§ 701) with the line through the centres of inertia of the normal sections, taken instead of the axis of the cylinder, the mutual action between the parts of it on the two sides of any normal section will be a couple of which the moment will be expressed by the same formula, that is, the product of the rigidity, into the rate of twist, into the moment of inertia of the section round its centre of inertia. requires tractions The only additional remark required to prove this is, that if the forces in the normal section be resolved in any two rectangular directions, OX, OY, the sums of the components, being respectively nsfxdo and nτffydo, each vanish by the property (§ 230) of the centre of inertia. 703. But for any other shape of prism than a solid or on its sides. symmetrical hollow circular cylinder, the supposed state of strain will require, besides the terminal opposed couples, force parallel to the length of the prism, distributed over the prismatic boundary, in proportion to the distance along the tangent, from each point of the surface, to the point in which this line is cut by a perpendicular to it from the centre of inertia of the normal section. To prove this let a normal section of the prism be represented in the annexed diagram. Let PK, representing the shear at any point, P, close to the prismatic boundary, be resolved into PN and PT respectively along the Traction on normal and tangent. The whole shear, PK, being equal to T, its component, PN, is equal to Tr sine or T.PE. The corresponding component of the required stress is nτ.PE, and involves (§ 661) equal forces in the plane of the diagram, and in the T N plane through TP perpendicular to it, each amounting to nr.PE per unit of area. sides of prism constrained to a simple twist. correction strain pro mere twist applied to An application of force equal and opposite to the distribution thus found over the prismatic boundary, would of course alone produce in the prism, otherwise free, a state of strain which, compounded with that supposed above, would give the state of strain actually produced by the sole application of balancing couples to the two ends. The result, it is easily St. Venant's seen (and it will be proved below), consists of an increased to give the twist, together with a warping of naturally plane normal duced by sections, by infinitesimal displacements perpendicular to them- ing couples selves, into certain surfaces of anticlastic curvature, with equal the ends. opposite curvatures in the principal sections (§ 130) through every point. This theory is due to St. Venant, who not only pointed out the falsity of the supposition admitted by several previous writers, that Coulomb's law holds for other forms of prism than the solid or hollow circular cylinder, but discovered fully the nature of the requisite correction, reduced the determination of it to a problem of pure mathematics, worked out the solution for a great variety of important and curious cases, compared the results with observation in a manner satisfactory and interesting to the naturalist, and gave conclusions of great value to the practical engineer. 704. We take advantage of the identity of mathematical conditions in St. Venant's torsion problem, and a hydrokinetic problem first solved a few years earlier by Stokes,' to give 1 "On some cases of Fluid Motion."-Camb. Phil. Trans. 1843. Hydrokinetic analogue to torsion problem. Solution of torsion problem. the following statement, which will be found very useful a estimating deficiencies in torsional rigidity below the amo calculated from the fallacious extension of Coulomb's law: 705. Conceive a liquid of density n completely filling & closed infinitely light prismatic box of the same shape with as the given elastic prism and of length unity, and let a coupl be applied to the box in a plane perpendicular to its leng The effective moment of inertia of the liquid1 will be equal to the correction by which the torsional rigidity of the elasti prism calculated by the false extension of Coulomb's law, mt be diminished to give the true torsional rigidity. Further, the actual shear of the solid, in any infinitely the plate of it between two normal sections, will at each point be, when reckoned as a differential sliding (§ 172) parallel to thei planes, equal to and in the same direction as the velocity of the liquid relatively to the containing box. 706. To prove these propositions and investigate the mathematical equations of the problem, we first show that the conditions of the case (§ 699) are verified by a state of strain compounded of (1.) a simple twist round the line through the centres of inertia, and (2.) a distorting of each normal section by infinitesimal displacements perpendicular to its plane: ther find the interior and surface equations to determine this warping and lastly, calculate the actual moment of the couple to which the mutual action between the matter on the two sides of any normal section is equivalent. Taking OX, OY in any normal section through O any convenient point (not necessarily its centre of inertia), and OZ perpendicular to them, let x+a, y+ß, z+y be the co-ordinates of the position to which a point (x, y, z) of the unstrained solid is displaced, in virtue of the compound strain just described. Thus y will be a function of x and y, without z; and, if the twist (1.) be denoted by 7 according to the simple twist reckoning of § 120, we shall have x+a=xcos (TZ)-ysin (TZ), y+B=xsin (7)+ycos(T2) (7) Hence, for infinitely small values of z, 1 That is, the moment of inertia of a rigid solid which, as will be proved in Vol. II, may be fixed within the box, if the liquid be removed, to make its motions the same as they are with the liquid in it. Adhering to the notation of §§ 670, 693, only changing to Saxon Equations dy of strain, stress, and internal (9), equilibrium. P=0, Q=0, R=0, S=n(rx+dy), T=n(−ty+dy), U=0 (10). And with the notation of § 698, (8) and (9), dx (11). (12), the equations of internal equilibrium [§ 698 (6)] are all satisfied. have, by § 662 (1), F=0, G=0, H=Tsino+Scos dy dp or eliminating T and S by (10), and introducing to denote made zero. the rate of variation of y in the direction perpendicular to the dy H = n{(cos + dx dy sin 4)-7(y sin p—x cosp)} To find the mutual action between the matter on the two sides Couple reof a normal section, we first remark that, inasmuch as each of the traction in two parts of the compound strain considered (the twist and the warping) separately fulfils the conditions of $ 700, we must have ffTdxdy=fx Hds, and ff Sdxdy=fy Hds (15). Hence when the prescribed surface condition H=0 is fulfilled, we (16), and there remains only a couple N=ff (Sx—Ty)dxdy=n+ff(x2+y3)dxdy—nff(y- -x )dxdy (17), dy dy dx dy in the plane of the normal section. That condition, by (14), gives dy cos+dy sin +=(y sin 4—xcos 4) normal section. We shall see in Vol. 11. that (12) and (18) are differential equa- Hydro tions which determine a function, y, of x, y, kinetic ap dy dy plication of such that and dx dy are the components of the velocity of a perfect liquid initially at torsional equations Hydrokinetic application of torsional equations. St. Venant's invention of solvable cases. rest in a prismatic box as described in § 705, and set in motion: dy nτff(x2+y3)dxdy over N. -y dx Also, a and b in (9) are the com ponents, parallel to OX and OY, of the velocity of the l relatively to the box, since -ry and Tx are the components: the velocity of a point (x, y) rotating in the positive direct a round OZ with the angular velocity T. Hence the propost (§ 705) to be proved. 707. M. de St. Venant finds solutions of these equations two ways:-(A.) Taking any solution whatever of (12), he fi a series of curves for each of which (18) is satisfied, and a one of which, therefore, may be taken as the boundary of a prism to which that solution shall be applicable: and (B) E the purely analytical method of Fourier, he solves (12), subjec to the surface equation (18), for the particular case of a rect angular prism. (A.) For this M. de St. Venant finds a general integral of the boundary condition, viewed as a differential equation in terms of the two variables x, y, thus:-Multiplying (18) by ds, and replacing sin øds and cos øds by their values dy and — dr, we hav In this the first two terms constitute a complete differential of 1 function of x and y, independent variables; because y satisfes (12). Thus, denoting this function by u, we have for every point in the boundary. It is to be remarked that, because we have, from (20), d dy d dy -= dy dx dx dy d'u, d'u or u also, as 7, fulfills the equation V3u = 0. A function. algebraically homogeneous as to x, y, which satisfies this equation is [Appendix B. (a)] a spherical harmonic independent of |