readily find (and the result of course is verified also by diffe- Flat circu 100 + Σ 1 é =o ((i + 2)2 − ¿a -1 2 (A cos ið +3, sin ið) e ̃(i−2)3) – † (¤ ̧ cos 0 + B ̧ sin 0) de3 + v′ (13), v' being any solution of (11), which may be conveniently taken lar ring the only case hitherto solved. lar plate, loaded by pairs of 656. Although the problem of fulfilling arbitrary boundary Rectangu conditions has not yet been solved for rectangular plates, there held and is one remarkable case of it which deserves particular notice; diagonal not only as interesting in itself, and important in practical corners. application, but as curiously illustrating one of the most difficult points corners, becomes strained to a condition of uniform anticlastic curvature throughout, with the sections of no-flexure parallel to its sides, and therefore with sections of equal opposite maximum curvature in the normal planes inclined to the sides at 45°. This follows immediately from § 648, if we suppose the corners rounded off ever so little, and the forces diffused over them. Rectangu lar plate, held and loaded by diagonal pairs of corners. Transition to finite dicated. Or, in each of an infinite number of normal lines in the edge AB, let a pair of opposite forces each equal to P be applied; which cannot disturb the plate. These, with halves of the single forces P in the dissimilar directions at the corners A and B, constitute a diffused couple over the whole edge AB, amounting in moment per unit of length to P, round axes perpendicular to the plane of the edge. Similarly, the other halves of the forces P at the corners A, B, with halves of those at C and D and introduced balancing forces, constitute diffused couples over the edges CA and DB; and the remaining halves of the corner forces at C and D, with introduced balancing forces, constitute a diffused couple over CD; each having P for the amount of moment per unit length of the edge over which it is diffused. Their directions are mutually related in the manner specified in § 638 (2), and thus taken all together, they constitute an anticlastic stress of value = P. Hence (§ 642) the result is uniform anticlastic strain amounting to P/k, and having its axes inclined at 45° to the edges; that is to say (§ 639), a flexure with maximum curvatures on the two sides of the tangent plane each equal to P/k, and in normal sections in the positions stated. 657. Few problems of physical mathematics are more flexures in curious than that presented by the transition from this solution, founded on the supposition that the greatest deflection is but a small fraction of the thickness of the plate, to the solution for larger flexures, in which corner portions will bend approximately as developable surfaces (cylindrical, in fact), and a central quadrilateral part will remain infinitely nearly plane; and thence to the extreme case of an infinitely thin perfectly flexible rectangle of inextensible fabric. This extreme case may be easily observed and experimented on by taking a carefully cut rectangle of paper (§ 145), supporting it by fine threads attached to two opposite corners, and kept parallel, while two equal weights are hung by threads from the other corners. Transmission of force elastic solid. 658. The definitions and investigations regarding strain of through an SS 154-190 constitute a kinematical introduction to the theory of elastic solids. We must now, in commencing the elementary dynamics of the subject, consider the forces called into play sion of force through the interior of a solid when brought into a condition of Transmis strain. We adopt, from Rankine*, the term stress to designate through an such forces, as distinguished from strain defined (§ 154) to express the merely geometrical idea of a change of volume or figure. elastic solid. ous stress. 659. When through any space in a body under the action Homogeneof force, the mutual force between the portions of matter on the two sides of any plane area is equal and parallel to the mutual force across any equal, similar, and parallel plane area, the stress is said to be homogeneous through that space. In other words, the stress experienced by the matter is homogeneous through any space if all equal similar and similarly turned portions of matter within this space are similarly and equally influenced by force. mitted surface in 660. To be able to find the distribution of force over the Force transsurface of any portion of matter homogeneously stressed, we across any must know the direction, and the amount per unit area, of the elastic solid. force across a plane area cutting through it in any direction. Now if we know this for any three planes, in three different directions, we can find it for a plane in any direction, as we see in a moment by considering what is necessary for the equilibrium of a tetrahedron of the substance. The resultant force on one of its faces must be equal and opposite to the resultant of the forces on the three others, which is known if these faces are parallel to the three planes for each of which the force is given. tion of a 661. Hence the stress, in a body homogeneously stressed, is specificacompletely specified when the direction, and the amount per unit stress; area, of the force on each of three distinct planes is given. It is, in the analytical treatment of the subject, generally convenient to take these planes of reference at right angles to one another. But we should immediately fall into error did we not remark that the specification here indicated consists not of nine but in by six independent reality only of six independent elements. For if the equili- elements. brating forces on the six faces of a cube be each resolved into three components parallel to its three edges OX, OY, OZ, we have in all 18 forces; of which each pair acting perpendicularly * Cambridge and Dublin Mathematical Journal, 1850. Relations between pairs of tangential tractions necessary for equili brium. Specification of a indepen dent elements: longitudinal and three simple shearing on a pair of opposite faces, being equal and directly opposed, balance one another. The twelve tangential components that remain constitute three pairs of couples having their axes in the T Y direction of the three edges, each of which must separately be in equilibrium. The diagram shows T the pair of equilibrating couples having OY for axis; from the consideration of which we infer that the forces on the faces (zy), parallel to OZ, are equal to the forces on the faces (yx), parallel to OX. Similarly, we see that the forces on the faces (yx), paral X lel to OY, are equal to those of the faces (xz), parallel to OZ; and that the forces on (cz), parallel to OX, are equal to those on (zy), parallel to OY. 662. Thus, any three rectangular planes of reference being stress; by six chosen, we may take six elements thus, to specify a stress: P, Q, R the normal components of the forces on these planes; and S, three simple T, U the tangential components, respectively perpendicular to stresses, OX, of the forces on the two planes meeting in OX, perpendicular to OY, of the forces on the planes meeting in OY, and perpendicular to OY, of the forces on the planes meeting in OY; each of the six forces being reckoned per unit of area. A normal component will be reckoned as positive when it is a traction tending Simple lon- to separate the portions of matter on the two sides of its plane. and shear P, Q, R are sometimes called longitudinal stresses, sometimes simple normal tractions, and S, T, U shearing stresses. stresses. gitudinal, ing, stresses. Force across any surface in terms of rectangular specification of stress. From these data, to find in the manner explained in § 660, the force on any plane, specified by l, m, n, the direction-cosines of its normal; let such a plane cut OX, OY, OZ in the three points X, Y, Z. Then, if the area XYZ be denoted for a moment by A, the areas YOZ, ZOX, XOY, being its projections on the three rectangular planes, will be respectively equal to Al, Am, An. Hence, for the equilibrium of the tetrahedron of matter bounded by those four triangles, we have, if F, G, H denote the com ponents of the force experienced by the first of them, XYZ, per Force across any surface in terms of rectangular specification of and the two symmetrical equations for the components parallel to stress. OY and OZ. Hence, dividing by A, we conclude These expressions stand in the well-known relation to the ellipsoid Px2 + Qy2+ Rz2+2(Syz + Tzx + Uxy) = 1.........(2), according to which, if we take x= lr, y=mr, z = nr, and if λ, μ, v denote the direction-cosines and p the length of the quadric. 663. For any fully specified state of stress in a solid, a stressquadric surface may always be determined, which shall represent the stress graphically in the following manner:— To find the direction, and the amount per unit area, of the force acting across any plane in the solid, draw a radius perpendicular to this plane from the centre of the quadric to its surface. The required force will be equal to the reciprocal of the product of the length of this radius into the perpendicular from the centre to the tangent plane at the extremity of the radius, and will be perpendicular to this tangent plane. planes and stress. 664. From this it follows that for any stress whatever there Principal are three determinate planes at right angles to one another such axes of a that the force acting in the solid across each of them is precisely perpendicular to it. These planes are called the principal or normal planes of the stress; the forces upon them, per unit area, -its principal or normal tractions; and the lines perpendicular |