az% v +K) P = P an (X + Pv2w + PZ = The internal motions specified by these equa tions are vibratory since they arise from small elastic displacements within the medium. The fact that they can be verified experimentally furnishes complete evidence of the correctness of Hooke's law, of our assumption that the displacements are small, and of the validity of the analysis.
Applications. We shall now solve a few typical problems in order to show the use of the foregoing equations. Most problems un fortunately give rise to extremely difficult partial differential equations; the general methods of integration are fully discussed in the works by Love and Riemann-Weber cited at the end of this article.
(1) A cylinder of density p and length 1 is suspended from one end and hangs verti cally.
Take the Y-axis vertical with the origin 1 below the upper end. Since Yy, the tensile stress at any point, is due to the weight of the material below that point, Yy = gpy; the five remaining stresses are zero. There are no surface forces except at the upper end where the entire weight of the cylinder is uniformly distributed over the supporting surface; there the internal and external stresses balance. Hence.
= gPY, esx =au yaws, Now .- ay E (a) ' 2E where v. is a function of a and s because the derivative is partial. Since there is no shear, v must satisfy as au atv av x y - F =0, au aft aW as whence . (b) ay ax ' 733 as ' By differentiation, a1/48's. aiw ais au Ow 'WY But ax as so that alve app =-_ ax' as' E (c) The value ve + +ax+cs+k . . . . (d) will satisfy equation (c) ; substitution in (a) gives v = -I- ore -1- ax -1- a k 2E At the upper end v=0 and k = when x0 0 , y=1 , z=0 . 2E The solution must be correct when the rod is rigid; in this case E= 00 and v=0 so that a =0,.c=0.
gn v= 2F_ ( ± ex' + est) . . .. (e) The formula obtained in books on the strength of materials It% gP which is therefore correct only along the axis; it is, however, approximately true at any point when the cylinder is very long compared with the radius.
Integrating (b), substituting (d), and sup posing as above that E= co, we get t'VeY W E if the upper end is free to contract.
(2) A straight uniform rod is twisted by couples applied at the ends.
Saint-Venant was the first to solve the gen eral problem in his great memoir on torsion, 1855, although Coulomb had previously, 1784, succeeded in finding the twisting moment offered by a circular cylinder. The following is a brief sketch of Saint-Venant's method.
Let the cylinder, of any cross-section, have its axis along Z. Since there is no shear on the mantle, X1 =0 and Ov au = . . . (a) . .
Ox ay As there are no external normal forces the normal strains vanish and Ouav atv = _ _ _0 . . . (b) ax ay az Equations (5) with become aXs . . . (c) az Oz Ou , Ow where + . (d) ax ay az By equation (b) tv does not contains and by (c) and (d) u and v are linear functions of z. All the above equations will be satisfied only by , , . (e) where 0 is a function of x, y. The values of u and v show that since 0 (e AO and y/x, the displacement in the plane of the cross-section is (1) normal to the radius vector; (2) proportional to the radius vector; (3) proportional to the distance of the cross section from the origin. Therefore radial
straight lines remain straight and of constant length and the boundary of any section is not distorted in transverse planes. If 0 is not zero these lines will be warped in the direction of the cylinder axis; 0 is therefore the warp ing function. r is the angle of twist.
Equations (e) substituted in (d) give _ + . . (I) As there is no shear on the mantle the resultant of Xsand Ys must be normal to the boundary of ay any section, i.e. Ox , . from (f) + x) d x d (g) Y g This is the differential equation of the boundary curve of any section.
The twisting moment equals the sum of the moments of the shear forces on any section, i.e., =_-/irf (x2 ± + ax dy . . (h) The angle of twist, found by differentiating (f) alf and eliminating is axd r =__ it la _ax.\ (i)k ar The differential equation of a circle is xdx= ydy; equation (g) reduces to this when 0 is constant. As there is no lengthening of the cylinder there is no translition of a cross-sec tion and = 0. Now if 0 =ax-1- by (g) is the equation of a circle, but as the centre of the circle is at the axis of the cylinder it will be found that a and b must vanish. Hence for a circular section ’ = 0 and cross-sections remain plane. In this case equation (h) gives the well-known engineers' formula.
Equation (g) will represent the ellipse e 0 al provided 0 + a* xY Since w =24, the contour lines found by giv ing 0 a series of constant values are equilateral hyperbolas; in the first and third quadrants of the ellipse the displacements will be negative and in the other quadrants positive. Equation (h) gives m=_-/trir -- (3) Vibrations in an infinite elastic medium.
If there are no external forces, X=Y=Z=0. Let all quantities in the XY plane be constant so that the same state exists throughout that plane; then the x and y derivatives in equa tions (90 are zero, whence Ow OA aiw Os ' Os Likewise alie aft, aitv , == The equations of motion thus reduce to p _ at. ah, all p a P (X + 214) ahv. . . . (b) atl Equations (a) are satisfied, as substitution will verify, by any function of ± t. This function must be periodic for otherwise the displacement would become infinite in course of time or would remain as a permanent set. Experiment contradicts both of these supposi tions, hence the motion is vibratory. Equation (b) is satisfied by a similar function. If at a point s ds at a time t dt the displace ment is in the same phase and of the same magnitude as it was at Is at a time s + ds + c (t + dt) s + a where for (a) and cl X -I- 21' for (b); dz c ' which is the velocity of propagation of the dis turbance: not the velocity of a material point but of a state of motion. Equation (b) defines the longitudinal, and (a) the transverse waves in an infinite elastic medium.
Bibliography.Love,