Theory of Plates and Shells(Second Edition)(2-2) .pdf
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1、THEORY OF PLATES AND SHELLS S. TIMOSHENKO Professor Emeritus of Engineering Mechanics Stanford University S. WOINOWSKY-KRIEGER Professor of Engineering Mechanics Laval University SECOND EDITION MCGRAW-HILL BOOK COMPANY Auckland Bogota Guatemala Hamburg Lisbon London Madrid Mexico New Delhi Panama Pa
2、ris San Juan SSo Paulo Singapore Sydney Tokyo CHAPTER 9 PLATES OF VARIOUS SHAPES 62. Equations of Bending of Plates in Polar Coordinates. In the discussion of symmetrical bending of circular plates polar coordinates were used (Chap. 3). The same coordinates can also be used to advan- tage in the gen
3、eral case of bending of circular plates. If the r and 0 coordinates are taken, as shown in Fig. 136a, the relation between the polar and cartesian coordinates is given by the equations from which it follows that Using these expressions, we obtain the slope of the deflection surface of a plate in the
4、 x direction as A similar expression can be written for the slope in the y direction. To obtain the expression for curva- ture in polar coordinates the second derivatives are required. Repeating twice the operation indicated in expression (c), we find FIG. 136 In a similar manner we obtain With this
5、 transformation of coordinates we obtain Repeating this operation twice, the differential equation (103) for the deflection surface of a laterally loaded plate transforms in polar coordi- nates to the following form: (191) When the load is symmetrically distributed with respect to the center of the
6、plate, the deflection w is independent of 0, and Eq. (191) coincides with Eq. (58) (see page 54), which was obtained in the case of sym- metrically loaded circular plates. Let us consider an element cut out of the plate by two adjacent axial planes forming an angle dd and by two cylindrical surfaces
7、 of radii r and r + dr, respectively (Fig. 1366). We denote the bending and twisting moments acting on the element per unit length by Mr, Mty and Mrt and take their positive directions as shown in the figure. To express these moments by the deflection w of the plate we assume that the x axis coin- c
8、ides with the radius r. The moments Mn Mt, and Mri then have the same values as the moments MZJ My, and Mxy at the same point, and by substituting 6 = 0 in expressions (d), (e), and (/), we obtain (192) In a similar manner, from formulas (108), we obtain the expressions for the shearing forces1 (193
9、) where Aw is given by expression (g). In the case of a clamped edge the boundary conditions of a circular plate of radius a are In the case of a simply supported edge In the case of a free edge (see page 87) The general solution of Eq. (191) can be taken, as before, in the form of a sum W = WQ + Wi
10、 (k) in which W0 is a particular solution of Eq. (191) and Wx is the solution of the homogeneous equation (194) This latter solution we take in the form of the following series:2 (195) in which RQ, Ri, . . . , R, Ri1 . . . are functions of the radial distance r only. Substituting this series in Eq.
11、(194), we obtain for each of these functions an ordinary differential equation of the following kind: The general solution of this equation for m 1 is Rm = Amrm + Bmrm + Cmrm+2 + Dmr+2 (I) 1 The direction of Q r in Fig. 1366 is opposite to that used in Fig. 28. This explains the minus sign in Eq. (1
12、93). 2 This solution was given by A. Clebsch in his “Theorie der Elasticitat fester Korper,“ 1862. For m 0 and m = 1 the solutions are Ro = A0 + BQr2 + C0 log r + DQr2 log r and #1 = A1T + BiH + C1/-1 + JV log r W Similar expressions can be written for the functions Rm. Substituting these expression
13、s for the functions R7n and Rm in series (195), we obtain the general solution of Eq. (194). The constants Am, Bm, . . . , Dm in each particular case must be determined so as to satisfy the boundary- conditions. The solution Ro, which is independent of the angle 0, repre- sents symmetrical bending o
14、f circular plates. Several particular cases of this kind have already been discussed in Chap. 3. 63. Circular Plates under a Linearly Varying Load. If a circular plate is acted upon by a load distributed as shown in Fig. 137, this load can always be divided into two parts: (1) a uniformly distribute
15、d load of intensity i(p2 + Pi) and (2) a linearly varying load having zero intensity along the diameter CD of the plate and the intensities p and +P at the ends A and B of the diameter AB. The case of uniform load has already been discussed in Chap. 3. We have to consider here only the nonuniform lo
16、ad represented in the figure by the two shaded triangles.1 The intensity of the load q at any point with coordinates r and 0 is FIG. 137 The particular solution of Eq. (191) can thus be taken in the following form: This, after substitution in Eq. (191), gives Hence As the solution of the homogeneous
17、 equation (194) we take only the term of series (195) that contains the function Ri and assume W1 = (Air + Bxrz + CiT-1 + D1T log r) cos B (c) 1 This problem has been discussed by W. Fliigge, Bauingenieur, vol. 10, p. 221, 1929. Since it is advantageous to work with dimensionless quantities, we intr
18、o- duce, in place of r, the ratio “ 5 With this new notation the deflection of the plate becomes w = W0 +W1 = j5 (p5 + Ap + Bp* + Cp-1 + Dp log P) cos B (d) where p varies from zero to unity. The constants A9 B, . . . in this expression must now be determined from the boundary conditions. Let us beg
19、in with the case of a simply supported plate (Fig. 137). In this case the deflection w and the bending moment Mr at the boundary vanish, and we obtain (w),=i = 0 (Mr)P=i = 0 (e) At the center of the plate (p = 0) the deflection w and the moment Mr must be finite. From this it follows at once that th
20、e constants C and D in expression (d) are equal to zero. The remaining two constants A and B will now be found from Eqs. (e), which give Since these equations must be fulfilled for any value of 0, the factors before cos B must vanish. This gives and we obtain Substituting these values in expression
21、(d), we obtain the deflection w of the plate in the following form: For calculating the bending moments and the shearing forces we substi- tute expression (/) in Eqs. (192) and (193), from which It is seen that (ikfr)max occurs at p = 1/V3 and is equal to The maximum value of Mt occurs at and is equ
22、al to The value of the intensity of the vertical reaction at the boundary is1 The moment of this reaction with respect to the diameter CD of the plate (Fig. 137) is This moment balances the moment of the load distributed over the plate with respect to the same diameter. As a second example, let us c
23、onsider the case of a circular plate with a free boundary. Such a condition is encountered in the case of a circular foundation slab supporting a chimney. As the result of wind pressure, a moment M will be transmitted to the slab (Fig. 138). Assuming that the reac- tions corresponding to this moment
24、 are distributed following a linear law, as shown in the figure, we obtain the same kind of loading as in the previous case; and the general solution can be taken in the same form (d) as before. The boundary conditions at the outer boundary of the plate, which is free from forces, are FIG. 138 The i
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