that of M. de St. Venant to the torsion of prisms 1.' To one end of a long straight prismatic rod, wire, or solid or hollow cylinder of any form, a given couple is applied in a plane perpendicular to the length, while the other end is held fast: it is required to find the degree of twist produced, and the distribution of strain and stress throughout the prism. The conditions to be satisfied here are that the resultant action between the substance on the two sides of any normal section is a couple in the normal plane, equal to the given couple. Our work for solving the problem will be much simplified by first establishing the following preliminary propositions : 670. Let a solid (whether aeolotropic or isotropic) be so acted on by force applied from without to its boundary, that throughout its interior there is no normal traction on any plane parallel or perpendicular to a given plane, XoY, which implies, of course, that there is no distorting stress with axes in or parallel to this plane, and that the whole stress at any point of the solid is a simple distorting stress of tangential forces in some direction in the plane parallel to XOY, and in the plane perpendicular to this direction. Then (1) The interior distorting stress must be equal, and similarly directed, in all parts of the solid lying in any line perpendicular to the plane XOY. (2) It being premised that the traction at every point of any surface perpendicular to the plane XOY is, by hypothesis, a distribution of force in lines perpendicular to this plane; the integral amount of it on any closed prismatic or cylindrical surface perpendicular to Xoy, and bounded by planes parallel to it, is zero. (3) The matter within the prismatic surface and terminal planes of (2) being supposed for a moment ($ 584) to be rigid, the distribution Y of tractions referred to in (2) constitutes a couple whose moment, divided by P the distance between those terminal planes, is equal to Q P the resultant force of the s tractions area of either, and whose plane is T parallel to the lines of these resultant forces. In other words, the moment of the 0 X distribution of forces over the prismatic surface referred to in (2) round any line (OY or OX) in the plane XOY, is equal to the sum of the components (T or S), perpendicular to the same line, of the traction in either of the terminal planes multiplied by the distance between these planes. 1 Mémoires des Savants Etrangers, 1855. “De la Torsion des Prismes, avec des Considérations sur leur Flexion,' etc. A To prove (1) consider for a moment as rigid ($ 584) an infinitesimal prism, AB (of sectional area w), perpendicular to XOY, and having plane ends, ►TW A, B, parallel to it. There being no forces on its sides (or cylindrical boundary) perpendicular to its length, its equilibrium so far as motion in the direction of any line (OX), perpendicular to its length, requires that the components of the tractions on its ends be equal and in opposite directions. Hence, in the notation § 633, the distorting - stress components, T, must be equal at A and B; Twe and so must the stress components S, for the B same reason. To prove (2) and (3) we have only to remark that they are required for the equilibrium of the rigid prism referred to in (3). 671. For a solid or hollow circular cylinder, the solution of § 669 (given first, we believe, by Coulomb) obviously is that each circular normal section remains unchanged in its own dimensions, figure; and internal arrangement (so that every straight line of its particles remains a straight line of unchanged length), but is turned round the axis of the cylinder through such an angle as to give a uniform rate of twist equal to the applied couple divided by the product of the moment of inertia of the circular area (whether annular or complete to the centre) into the rigidity of the substance. 672. Similarly, we see that if a cylinder or prism of any shape be compelled to take exactly the state of strain above specified ($ 671) 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. 673. But for any other shape of prism than a solid or 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 (page 248). Let PK represent the shear at any point, P, close to the prismatic boundary, be resolved into PN and PT respectively along the normal and tangent. The whole shear, PK, being equal to tr, its component, PN, is equal to trsin w or 7.PE. The corresponding component of the required stress is nt.PE, and involves ($ 632) equal forces in the plane of the diagram, and in the plane through TP perpendicular to it, each amounting to nt.PE per unit of area. An application of force equal and opposite to the distribution thus found over the prismatic bounK dary, would of course alone T produce in the prism, other wise 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 seen (and it will be proved below), consists of an increased twist, together with a warping of naturally plane normal sections, by infinitesimal displacements perpendicular to themselves, into certain surfaces of anticlastic curvature, with equal opposite curvatures in the principal sections ($ 122) 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. 674. 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 the following statement, which will be found very useful in estimating deficiencies in torsional rigidity below the amount calculated from the fallacious extension of Coulomb's law :: 675. Conceive a liquid of density n completely filling a closed infinitely light prismatic box of the same shape within as the given elastic prism and of length unity, and let a couple be applied to the box in a plane perpendicular to its length. The effective moment of inertia of the liquid 2 will be equal to the correction by which the torsional rigidity of the elastic prism calculated by the false extension of Coulomb's law, must be diminished to give the true torsional rigidity. Further, the actual shear of the solid, in any infinitely thin plate of 1. On some cases of Fluid Motion,' Cambridge Philosophical Transactions, 1843. 2 That is the moment of inertia of a rigid solid which, as will be proved in Vol. 11., 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. it between two normal sections, will at each point be, when reckoned as a differential sliding (§ 151) parallel to their planes, equal to and in the same direction as the velocity of the liquid relatively to the containing box. 676. St. Venant's treatise abounds in beautiful and instructive graphical illustrations of his results, from which we select the following: (1) Elliptic cylinder. The plain and dotted curvilineal arcs are ' contour lines' (coupes topographiques) of the section as warped by torsion; that is to say, lines in which it is cut by a series of parallel planes, each perpendicular to the axis. These lines are equilateral hyperbolas in this case. The arrows indicate the direction of rotation in the part of the prism above the plane of the diagram. (2) Equilateral triangular prism. — The contour lines are shown as in case (1); the dotted curves being those where the warped section falls below the plane of the diagram, the direction of rotation of the part of the prism above the plane being indicated by the bent arrow. (3) This diagram shows a series of lines given by St. Venant, and more or less resembling squares. Their common equation containing only one constant a. It is remarkable that the values a=0.5 and a=-{(12-1) give similar but not equal curvilineal squares (hollow sides and acute angles), one of them turned through half a right angle relatively to the other. Everything in the diagram outside the larger of these squares is to be cut away as irrelevant to the physical problem; the series of closed curves remaining exhibits figures of prisms, for any one of which the torsion problem is solved algebraically. These figures vary continuously from a circle, inwards to one of the acute-angled squares, and outwards to the other : each, except these extremes, being a continuous closed curve with no angles. The curves for a=0•4 and a=-0.2 approach remarkably near to the rectilineal squares, partially indicated in the diagram by dotted lines. (4) This diagram shows the contour lines, in all respects as in the cases (1) and (2) for the case of a prism having for section the figure indicated. The portions of curve outside the continuous closed curve are merely indications of mathematical extensions irrelevant to the physical problem. |