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
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THEORY OF PLATES AND SHELLS International Edition 1959 Exclusive rights by McGraw-Hill Book Co.— Singapore for manufacture and export. This book cannot be re-exported from the country to which it is consigned by McGraw-Hill. Copyright ©
1959 by McGraw-Hill, Inc.
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PREFACE
Since the publication of the first edition of this book, the application of the theory of plates and shells in practice has widened considerably, and some new methods have been introduced into the theory. To take these facts into consideration, we have had to make many changes and additions. The principal additions are (1) an article on deflection of plates due to transverse shear, (2) an article on stress concentrations around a circular hole in a bent plate, (3) a chapter on bending of plates resting on an elastic foundation, (4) a chapter on bending of anisotropic plates, and (5) a chapter reviewing certain special and approximate methods used in plate analysis. We have also expanded the chapter on large deflections of plates, adding several new cases of plates of variable thickness and some numerical tables facilitating plate analysis. In the part of the book dealing with the theory of shells, we limited ourselves to the addition of the stress-function method in the membrane theory of shells and some minor additions in the flexural theory of shells. The theory of shells has been developing rapidly in recent years, and several new books have appeared in this field. Since it was not feasible for us to discuss these new developments in detail, we have merely referred to the new bibliography, in which persons specially interested in this field will find the necessary information. S. Timoshenko S. Woinowsky-Krieger
NOTATION x, y, z Rectangular coordinates r, 0 Polar coordinates rx, ry Radii of curvature of the middle surface of a plate in xz and yz planes, respectively h Thickness of a plate or a shell q Intensity of a continuously distributed load p Pressure P Single load 7 Weight per unit volume (Tx, Radial stress in polar coordinates at, (re Tangential stress in polar coordinates r Shearing stress Txy, Txz, Tyz Shearing stress components in rectangular coordinates u, v, w Components of displacements e Unit elongation «*, «•/, fz Unit elongations in x, y, and z directions er Radial unit elongation in polar coordinates et, eo Tangential unit elongation in polar coordinates ttp, eo Unit elongations of a shell in meridional direction and in the direction of parallel circle, respectively yxy, Vxz, jyz Shearing strain components in rectangular coordinates 7r0 Shearing strain in polar coordinates E Modulus of elasticity in tension and compression G Modulus of elasticity in shear v Poisson's ratio V Strain energy D Flexural rigidity of a plate or shell Mx, My Bending moments per unit length of sections of a plate perpendicular to x and y axes, respectively Mxy Twisting moment per unit length of section of a plate perpendicular to x axis Mn, Mnt Bending and twisting moments per unit length of a section of a plate perpendicular to n direction Qx, Qy Shearing forces parallel to z axis per unit length of sections of a plate perpendicular to x and y axes, respectively Qn Shearing force parallel to z axis per unit length of section of a plate perpendicular to n direction JVx, Ny Normal forces per Unit length of sections of a plate perpendicular to x and y directions, respectively
Nxu
Shearing force in direction of y axis per unit length of section of a plate perpendicular to x axis Mr, Mt, MH Radial, tangential, and twisting moments when using polar coordinates Qr, Qt Radial and tangential shearing forces Nn Nt Normal forces per unit length in radial and tangential directions ri, r2 Radii of curvature of a shell in the form of a surface of revolution in meridional plane and in the normal plane perpendicular to meridian, respectively X 3 the calculations for a plate can be replaced by those for a strip without substantial error. T A B L E 8. NUMERICAL FACTORS a, /3, 7, 5, n FOR UNIFORMLY LOADED AND SIMPLY SUPPORTED RECTANGULAR
PLATES
v = 0.3
6/a
WUUi qa* ( M »>»« = a— = /8ga2
CAOmax = /ffiga2
«2*)ma* = 75a
(Qy)m« = 7iga
(F*) m a x = 5ga
(FJ^x = 5iga
/? — nqa2
a
j3
/Si
7
71
5
5i
n
1.0 1.1 1.2 1.3 1.4
0.00406 0.00485 0.00564 0.00638 0.00705
0.0479 0.0554 0.0627 0.0694 0.0755
0.0479 0.0493 0.0501 0.0503 0.0502
0.338 0.360 0.380 0.397 0.411
0.338 0.347 0.353 0.357 0.361
0.420 0.440 0.455 0.468 0.478
0.420 0.440 0.453 0.464 0.471
0.065 0.070 0.074 0.079 0.083
1.5 1.6 1.7 1.8 1.9
0.00772 0.00830 0.00883 0.00931 0.00974
0.0812 0.0862 0.0908 0.0948 0.0985
0.0498 0.0492 0.0486 0.0479 0.0471
0.424 0.435 0.444 0.452 0.459
0.363 0.365 0.367 0.368 0.369
0.486 0.491 0.496 0.499 0.502
0.480 0.485 0.488 0.491 0.494
0.085 0.086 0.088 0.090 0.091
2.0 3.0 4.0 5.0 00
0.01013 0.01223 0.01282 0.01297 0.01302
0.1017 0.1189 0.1235 0.1246 0.1250
0.0464 0.0406 0.0384 0.0375 0.0375
0.465 0.493 0.498 0.500 0.500
0.370 0.372 0.372 0.372 0.372
0.503 0.505 0.502 0.501 0.500
0.496 0.498 0.500 0.500 0.500
0.092 0.093 0.094 0.095 0.095
Expression (e) can be used also for calculating shearing forces and reactions at the boundary. Forming the second derivatives of this expression, we find
Substituting this in Eqs. (106) and (107), we obtain
For the sides x = 0 and y = — b/2 we find
These shearing forces have their numerical maximum value at the middle of the sides, where
The numerical factors y and 71 are also given in Table 8. The reactive forces along the side x = 0 are given by the expression
The maximum numerical value of this pressure is at the middle of the side (y = 0), at which point we find
where 5 is a numerical factor depending on v and on the ratio 6/a, which can readily be obtained by summing up the rapidly converging series that occur in expression (q). Numerical values of 8 and of Si, which corresponds to the middle of the sides parallel to the x axis, are given in Table 8. The distribution of the pressures (q) along the sides of a square plate is shown in Fig. 63. The portion of the pressures produced by the
Ratio f 64
FIG.
twisting moments Mxy is also shown. These latter pressures are balanced by reactive forces concentrated at the corners of the plate. The magnitude of these forces is given by the expression
The forces are directed downward and prevent the corners of a plate from rising up during bending. The values of the coefficient n are given in the last column of Table 8.
The values of the factors a, 0, 0i, 8 as functions of the ratio b/a are represented by the curves in Fig. 64. In the presence of the forces R, which act downward and are by no means small, anchorage must be provided at the corners of the plate if the plate is not solidly joined with the supporting beams. In order to determine the moments arising at the corner let us consider the equilibrium of the element abc of the plate next to its corner (Fig. 65) and let us introduce, for the same purpose, new coordinates 1, 2 at an angle of 45° to the coordinates x, y in Fig. 59. We can then readily verify that the bending moments acting at the sides ab and cb of the element are Mx = —R/2 and M2 = -f-#/2, respectively, and that the corresponding twisting moments are zero. In fact, using Eq. (39), we obtain for the side ac, that is, for the element of the edge, given by a = — 45°, the bending moment Mn — Mi cos2 a + M2 sin2 a = 0 in accordance with the boundary conditions of a simply supported plate. The magnitude of the twisting moment applied at the same edge element is obtained in like manner by means of Eq. (40). Putting ot = —45° we have Mnt = ^ sin 2*(MX - M2) - ^
FIG.
65
according to Eq. (r). Thus, the portion of the plate in the vicinity of the corner is bent to an anticlastic surface, the moments ±R/2 at the corner itself being of the same order of magnitude as the bending moments at the middle of the plate (see Table 8). The clamping effect of the corners of a simply supported plate is plainly illustrated by the distribution of the bending moments M1 and M2 of a square plate (Fig. 63). If the corners of the rectangular plate are not properly secured against lifting, the clamping becomes ineffective and the bending moments in the center portion of the plate increase accordingly. The values of (Mx) max and (M y ) max given in Table 8 must then be multiplied by some factor k > 1. The approximate expression1
may be used for that purpose. It should be noted that in the case of a polygonal plate with simply supported edges no single reactive forces arise at a corner point provided the angle between both adjacent sides of the plate is other than a right angle.2 Even in rectangular plates, however, no corner reactions are obtained if the transverse shear deformation is taken into account. In view of the strongly concentrated 1 Recommended by the German Code for Reinforced Concrete (1943) and basod on a simplified theory of thin plates due to H. Marcus; see his book " Die vereinfachte Berechnung biegsamer Platten," 2d ed., Berlin, 1925. 2 For a simple proof see, for example, H. Marcus, "Die Theorie e\astischer Gewebe," 2d ed., p. 46, Berlin, 1932.
reactive forces this shear deformation obviously is no longer negligible, and the customary thin-plate theory disregarding it completely must be replaced by a more exact theory. The latter, which will be discussed in Art. 39, actually leads to a distribution of reactive pressures which include no forces concentrated at the corners of the plate (see Fig. 81).
31. Simply Supported Rectangular Plates under Hydrostatic Pressure. Assume that a simply supported rectangular plate is loaded as shown in Fig. 66. Proceeding as in the case of a uniformly distributed load, we take the deflection of the plate in the form1
represents the deflection of a strip under the triangular load. This expression satisfies the differential equation
and the boundary conditions
The part Wi is taken in the form of a series
FIG.
66
where the functions Ym have the same form as in the preceding article, and m = 1, 2, 3, . . . . Substituting expressions (6) and (d) into Eq. (a), we obtain
where the constants Am and Bm are to be determined from the conditions
1
This problem was discussed by E. Estanave, op. cit. The numerical tables of deflections and moments were calculated by B. G. Galerkin, Bull. Polytech. InSt1 St. Petersburg, vols. 26 and 27, 1918.
From these conditions we find
In these equations we use, as before, the notation
Solving them, we find
The deflection of the plate along the x axis is
For a square plate a —
The deflection at the center of the plate is (uO—/i.iM> = 0.00203 2*£
(h)
which is one-half the deflection of a uniformly loaded plate (see page 116) as it should be. By equating the derivative of expression (g) to zero, we find that the maximum deflection is at the point x = 0.557a. This maximum deflection, which is 0.00206 q0a4/D, differs only very little from the deflection at the middle as given by formula (h). The point of maximum deflection approaches the center of the plate as the ratio b/a increases. For b/a = oo, as for a strip [see expression (b)], the maximum deflection is at the point x = 0.5193a. When b/a < 1, the point of maximum deflection moves away from the center of the plate as the ratio b/a decreases. The deflections at several points along the x axis (Fig. 66) are given in Table 9. It is seen that, as the ratio b/a increases, the deflections approach the values calculated for a strip. For b/a = 4 the differences in these values are about I^ per cent. We can always calculate the deflection of a plate for which b/a > 4 with satisfactory accuracy by using formula (6) for the deflection of a strip under triangular load. The bending moments Mx and My are found by substituting
TABLE 9. NUMERICAL FACTOR a FOR DEFLECTIONS OF A SIMPLY SUPPORTED RECTANGULAR PLATE UNDER HYDROSTATIC PRESSURE q = qox/a
b>a w = <xqoaA/D, y = 0 b/a
x = 0.25a
x = 0.50a
x = 0.60a
x = 0.75a
1 1.1 1.2 1.3 1.4
0.00131 0.00158 0.00186 0.00212 0.00235
0.00203 0.00243 0.00282 0.00319 0.00353
0.00201 0.00242 0.00279 0.00315 0.00348
0.00162 0.00192 0.00221 0.00248 0.00273
1.5 1.6 1.7 1.8 1.9
0.00257 0.00277 0.00296 0.00313 0.00328
0.00386 0.00415 0.00441 0.00465 0.00487
0.00379 0.00407 0.00432 0.00455 0.00475
0.00296 0.00317 0.00335 0.00353 0.00368
2.0 3.0 4.0 5.0 oo
0.00342 0.00416 0.00437 0.00441 0.00443
0.00506 0.00612 0.00641 0.00648 0.00651
0.00494 0.00592 0.00622 0.00629 0.00632
0.00382 0.00456 0.00477 0.00483 0.00484
expression (e) for deflections in Eqs. (101). the expression for Mx becomes
Along the x axis (y = 0)
The first sum on the right-hand side of this expression represents the bending moment for a strip under the action of a triangular load and is equal to (qQ/6)(ax — x3/a). Using expressions (/) for the constants Am and Bm in the second sum, we obtain
The series thus obtained converges rapidly, and a sufficiently accurate value of Mx can be realized by taking only the first few terms. In this
TABLE 10. NUMERICAL FACTORS 0 AND /3I FOR BENDING MOMENTS OF SIMPLY SUPPORTED RECTANGULAR PLATES UNDER HYDROSTATIC PRESSURE q — qox/a
v = 0.3, b > a Mx = /Sa2^0,
My = Pia2qQ,
y = 0
y = 0
b/a X =
X =
X =
X =
X =
X =
X =
X =
0.25a
0.50a
0.60a
0.75a
0.25a
0.50a
0.60a
0.75a
1.0 1.1 1.2 1.3 1.4
0.0132 0.0156 0.0179 0.0200 0.0221
0.0239 0.0276 0.0313 0.0346 0.0376
0.0264 0.0302 0.0338 0.0371 0.0402
0.0259 0.0289 0.0318 0.0344 0.0367
0.0149 0.0155 0.0158 0.0160 0.0160
0.0239 0.0247 0.0250 0.0252 0.0253
0.0245 0.0251 0.0254 0.0255 0.0254
0.0207 0.0211 0.0213 0.0213 0.0212
1.5 1.6 1.7 1.8 1.9
0.0239 0.0256 0.0272 0.0286 0.0298
0.0406 0.0431 0.0454 0.0474 0.0492
0.0429 0.0454 0.0476 0.0496 0.0513
0.0388 0.0407 0.0424 0.0439 0.0452
0.0159 0.0158 0.0155 0.0153 0.0150
0.0249 0.0246 0.0243 0.0239 0.0235
0.0252 0.0249 0.0246 0.0242 0.0238
0.0210 0.0207 0.0205 0.0202 0.0199
2.0 3.0 4.0 5.0 oo
0.0309 0.0369 0.0385 0.0389 0.0391
0.0508 0.0594 0.0617 0.0623 0.0625
0.0529 0.0611 0.0632 0.0638 0.0640
0.0463 0.0525 0.0541 0.0546 0.0547
0.0148 0.0128 0.0120 0.0118 0.0117
0.0232 0.0202 0.0192 0.0187 0.0187
0.0234 0.0207 0.0196 0.0193 0.0192
0.0197 0.0176 0.0168 0.0166 0 0165
_J
way the bending moment at any point of the x axis can be represented by the equation ( M , ) H = Pqoa2 (k) where /3 is a numerical factor depending on the abscissa x of the point. In a similar manner we get (My)y=0
= P1Q0O,2
(I)
The numerical values of the factors /3 and 0i in formulas (Zc) and (J) are given in Table 10. It is seen that for b g: 4a the moments are very close to the values of the moments in a strip under a triangular load. Equations (106) and (107) are used to calculate shearing forces. From the first of these equations, by using expression (j), we obtain for points on the x axis
The general expressions for shearing forces Qx and Qy are
(m)
(n)
The magnitude of the vertical reactions Vx and Vv along the boundary is obtained by combining the shearing forces with the derivatives of the twisting moments. Along the sides x = 0 and x — a these reactions can be represented in the form (o)
TABLE 11. NUMERICAL FACTORS 5 AND 5I FOR REACTIONS OF SIMPLY SUPPORTED RECTANGULAR PLATES UNDER HYDROSTATIC PRESSURE q — qox/a
v = 0.3, b > a Reactions 8qoa b
,
x =0
y
_ ~ "U
y~ 0.256
Reactions 5igo6
x =a
y
_n "U
y= 0.256
y = ±6/2
x= 0.25a
x= 0.50a
x= 0.60a
X = 0.75a
1.0 1.1 1.2 1.3 1.4
0.126 0.136 0.144 0.150 0.155
0.098 0.107 0.114 0.121 0.126
0.294 0.304 0.312 0.318 0.323
0.256 0.267 0.276 0.284 0.292
0.115 0.110 0.105 0.100 0.095
0.210 0.199 0.189 0.178 0.169
0.234 0.221 0.208 0.196 0.185
0.239 0.224 0.209 0.196 0.184
1.5 1.6 1.7 1.8 1.9
0.159 0.162 0.164 0.166 0.167
0.132 0.136 0.140 0.143 0.146
0.327 0.330 0.332 0.333 0.334
0.297 0.302 0.306 0.310 0.313
0.090 0.086 0.082 0.078 0.074
0.160 0.151 0.144 0.136 0.130
0.175 0.166 0.157 0.149 0.142
0.174 0.164 0.155 0.147 0.140
2.0 3.0 4.0 5.0 oo
0.168 0.169 0.168 0.167 0.167
0.149 0.163 0.167 0.167 0.167
0.335 0.336 0.334 0.334 0.333
0.316 0.331 0.334 0.335 0.333
0.071 0.048 0.036 0.029
0.124 0.083 0.063 0.050
0.135 0.091 0.068 0.055
0.134 0.089 0.067 0.054
and al^ng the sides y = ±6/2 in the form
in which 5 and 5i are numerical factors depending on the ratio b/a and on the coordinates of the points taken on the boundary. Several values of these factors are given in Table 11. The magnitude of concentrated forces that must be applied to prevent the corners of the plate rising up during bending can be found from the values of the twisting moments Mxy at the corners. Since the load is not symmetrical, the reactions Ri at x = 0 and y = ±6/2 are different from the reactions R2 at x = a and y = ±6/2. These reactions can be represented in the following form: Ri = niqoab
R2 = n2qoab
(q)
The values of the numerical factors n\ and n2 are given in Table 12. TABLE 12. NUMERICAL FACTORS nx AND n2 IN EQS. (ry) FOR REACTIVE FORCES Rl AND R2 AT THE CORNERS OF SlMPLY SUPPORTED RECTANGULAR PLATES UNDER HYDROSTATIC PRESSURE q = qox/a
v = 0.3, b > a b/a
m nj
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
3.0
4.0
5.0
0.026 0.026 0.026 0.026 0.025 0.024 0.023 0.022 0.021 0.021 0.020 0.014 0.010 0.008 0.039 0.038 0.037 0.036 0.035 0.033 0.032 0.030 0.029 0.028 0.026 0.018 0.014 0.011
Since a uniform load ^o is obtained by superposing the two triangular loads q = qox/a and qo(a — x)/a, it can be concluded that for corresponding values of b/a the sum n\ + n2 of the factors given in Table 12 multiplied by b/a must equal the corresponding value of n, the last column in Table 8. If the relative dimensions of the plate are such that a in Fig. 66 is greater than 6, then more rapidly converging series will be obtained by representing Wi and W2 by the following expressions: (r) (s) The first of these expressions is the deflection of a narrow strip parallel to the y axis, supported at y = ± 6/2 and carrying a uniformly distributed
TABLE 13. NUMERICAL FACTORS a FOR DEFLECTIONS OF SIMPLY SUPPORTED RECTANGULAR PLATES UNDER HYDROSTATIC PRESSURE q = gox/a
b O1 that is, below the x axis in Fig. 71. Putting, in particular, y = 0 we obtain the deflection of the plate along the x axis in the form
This series converges rapidly, and the first few terms give the deflections with sufficient accuracy. In the case of a load P applied at the center of the plate, the maximum deflection, which is at the center, is obtained by substituting x = £ = a/2 in expression (i). In this way we arrive at the result (147) Values of the numerical factor a for various values of the ratio b/a are given in Table 23. TABLE 23. FACTOR a FOR DEFLECTION (147) OF A CENTRALLY LOADED RECTANGULAR PLATE
b/a = a =
1.0
1.1
1.2
1.4
1.6
1.8
2.0
3.0
oo
0.01160 0.01265 0.01353 0.01484 0.01570 0.01620 0.016510.01690 0.01695
It is seen that the maximum deflection rapidly approaches that of an infinitely long plate1 as the length of the plate increases. The comparison of the maximum deflection of a square plate with that of a centrally loaded circular plate inscribed in the square (see page 68) indicates that the deflection of the circular plate is larger than that of the corresponding square plate. This result may be attributed to the action of the reactive forces concentrated at the corners of the square plate which have the tendency to produce deflection of the plate convex upward. The calculation of bending moments is discussed in Arts. 35 and 37. 35. Bending Moments in a Simply Supported Rectangular Plate with a Concentrated Load. To determine the bending moments along the central axis y = 0 of the plate loaded according to Fig. 71 we calculate the second derivatives of expression (146), which become
1 The deflection of plates by a concentrated load was investigated experimentally by M. Bergstrasser; see Forschungsarb., vol. 302, Berlin, 1928; see also the report of N. M. Newmark and H. A. Lepper, Univ. Illinois Bull, vol. 36, no. 84, 1939.
Substituting these derivatives into expressions (101) for the bending moments, we obtain
When b is very large in comparison with a, we can put
This series does not converge rapidly enough for a satisfactory calculation of the moments in the vicinity of the point of application of the load P, so it is necessary to derive another form of representation of the moments near that point. From the discussion of bending of a circular plate by a force applied at the center (see Art. 19) we know that the shearing forces and bending moments become infinitely large at the point of application of the load. We have similar conditions also in the case of a rectangular plate. The stress distribution within a circle of small radius with its center at the point of application of the load is substantially the same as that near the center of a centrally loaded circular plate. The bending stress at a point within this circle may be considered as consisting of two parts: one is the same as that in the case of a centrally loaded circular plate of radius a, and the other represents the difference between the stresses in a circular and those in a rectangular plate. As the distance r between FIG. 72 the point of application of the load and the point under consideration becomes smaller and smaller, the first part of the stresses varies as log (a/r) and becomes infinite at the center, whereas the second part, representing the effect of the difference in the boundary conditions of the two plates, remains continuous. To obtain the expressions for bending moments in the vicinity of the point of application of the load we begin with the simpler case of an infinitely long plate (Fig. 72). The deflection of such a plate can readily
be derived from expression (146) by increasing the length of the side 6, and consequently the quantity am = 7rnrb/2a, indefinitely, i.e., by putting
Substituting this into Eq. (146) the required deflection of the simply supported strip carrying a concentrated load P at x = £, y — 0 becomes1 (148) which holds for y > 0, that is, below the x axis (Fig. 72). The corresponding expressions for the bending moments and the twisting moment are readily obtained by means of Eqs. (101) and (102). We have
(149)
Once again using the quantity M = (Mx -f My)/(1 + v) introduced on page 92, we have (150) The moments (149) can be expressed now in terms of the function M in the following simple manner:
(151)
1 This important case of bending of a plate has been discussed in detail by A. Nadai; see his book "Elastische Platten," pp. 78-109, Berlin, 1925.
Summing up the series (150), we obtain the expression1
(152) and, using Eqs. (151), we are able now to represent the moments of the infinitely long plate in a closed form. Observing, furthermore, that AAw = 0 everywhere, except at the point (x = £, y = 0) of the application of the load, we conclude that the function M = —D Aw satisfies (except at the above-mentioned point) the equation AM — 0. By virtue of the second of the equations (111) the boundary condition M = O along the edges x = 0 and x = a is also satisfied by the function M. For the points along the x axis Eqs. (151) yield Mx = My and therefore
Using Eqs. (c) and Eq. (152) in the particular case of a load applied at the center axis of the strip, { = a/2, we obtain
a result which also can be obtained by summation of the series (b). Now let us return to the calculation of bending moments for points which are close to the point of application of the load but not necessarily on the x axis. In this case the quantities (x — J) and y are small and, using expression (152), we can put
Thus we arrive at the result
(153) 1 See, for instance, W. Magnus and F. Oberhettinger, "Formeln und Satze fur die speziellen Funktionen der mathematischen Physik," 2d ed., p. 214, Berlin, 1948.
in which represents the of application tution in Eqs. in the vicinity
distance of the point under consideration from the point of the load P. Now, using expression (153) for substi(151) we obtain the following expressions, valid for points of the concentrated load:
(154)
It is interesting to compare this result with that for a centrally loaded, simply supported circular plate (see Art. 19). Taking a radius r under an angle a to the x axis, we find, from Eqs. (90) and (91), for a circular plate
The first terms of expressions (154) and (e) will coincide if we take the outer radius of the circular plate equal to
Under this condition the moments Mx are the same for both cases. The moment My for the long rectangular plate is obtained from that of the circular plate by subtraction of the constant quantity 1 (1 — v)P/Anr. From this it can be concluded that in a long rectangular plate the stress distribution around the point of application of the load is obtained by superposing on the stresses of a centrally loaded circular plate with radius (2a/?r) sin (?r£/a) a simple bending produced by the moments My=
-(I
-
*>)P/4TT.
It may be assumed that the same relation between the moments of circular and long rectangular plates also holds in the case of a load P uniformly distributed over a circular area of small radius c. In such a case, for the center of a circular plate we obtain from Eq. (83), by neglecting the term containing c2,
1
We observe that x2 = r2 - y*.
Hence at the center of the loaded circular area of a long rectangular plate we obtain from Eqs. (154)
(155)
From this comparison of a long rectangular plate with a circular plate it may be concluded that all information regarding the local stresses at the point of application of the load P, derived for a circular plate byusing the thick-plate theory (see Art. 19), can also be applied in the case of a long rectangular plate. When the plate is not very long, Eqs. (a) should be used instead of Eq. (6) in the calculation of the moments Mx and My along the x axis. Since tanh am approaches unity rapidly and cosh am becomes a large number when m increases, the differences between the sums of series (a) and the sum of series (b) can easily be calculated, and the moments Mx and My along the x axis and close to the point of application of the load can be represented in the following form:
(156)
in which 71 and 72 are numerical factors the magnitudes of which depend on the ratio b/a and the position of the load on the x axis. Several values of these factors for the case of central application of the load are given in Table 24. Again the stress distribution near the point of application of the load is substantially the same as for a centrally loaded circular plate of radius (2a/ir) sin (ir£/a). To get the bending moments Mx and My near the load we have only to superpose on the moments of the
T A B L E 24. FACTORS 71 AND 72 I N E Q S . (156)
b/a
1.0
1.2
1.4
1.6
1.8
2.0
71 72
-0.565 +0.135
-0.350 +0.115
-0.211 +0.085
-0.125 +0.057
-0.073 +0.037
-0.042 +0.023
rj is obtained by substituting q di\ for qo and FIG. 75 y — K] for y in expression (i). The deflection produced by the entire load, at points for which y ^ v/2, is now obtained by integration as follows:
By a proper change of the limits of integration the deflection at points with y < v/2 can also be obtained. Let us consider the deflection along the x axis (Fig. 75). The deflection produced by the upper half of the load is obtained from expression (j) by substituting the quantity v/4 for y and for v/2. By doubling the result obtained in this way we also take into account the action of the lower half of the load and finally obtain
When v = oo 7 the load, indicated in Fig. 75, is expanded along the entire length of the plate, and the deflection surface is cylindrical. The corresponding deflection, from expression (k), is
Making { = u/2 = a/2 in this expression, we obtain
which represents the deflection curve of a uniformly loaded strip. The following expressions for bending moments produced by the load uniformly distributed along a portion u of the x axis are readily obtained from expression (i) for deflection w:
These moments have their maximum values on the x axis, where
In the particular case when £ = u/2 = a/2, that is, when the load is distributed along the entire width of the plate,
The maximum moment is at the center of the plate where
When u is very small, i.e., in the case of a concentrated load, we put
Then, from expression (n), we obtain
which coincides with expression (b) of the preceding article and can be expressed also in a closed form (see page 146). In the case of a load q uniformly distributed over the area of a rectangle (Fig. 75), the bending moments for the portion of the plate for which y ^ v/2 are obtained by integration of expressions (m) as follows:
(159)
The moments for the portion of the plate for which y < v/2 can be calculated in a similar manner. To obtain the moments along the x axis, we have only to substitute v/2 for v and v/4 for y in formulas (159) and
double the results thus obtained.
Hence
(160)
If values of the moments at the center of the loaded rectangular area are required, the calculation may also be carried out by means of expressions (167), which will be given in Art. 37. When v is very small, Eqs. (160) coincide with Eq. (n) if we observe that qv must be replaced in such a case by go. When v is very large, we have the deflection of the plate to a cylindrical surface, and Eqs. (160) become
The expressions for the deflections and bending moments in a plate of finite length can be obtained from the corresponding quantities in an infinitely long plate by using the method of images.1 Let us begin with the case of a concentrated force P applied on the axis of symmetry x of the rectangular plate with sides a and b in Fig. 76a. If we now imagine the plate prolonged in both the positive and the negative y directions and loaded with a series of forces P applied along the line mn at a disFIG. 76 tance b from one another and in alternate directions, as shown in Fig. 766, the deflections of such an infinitely 1 This method was used by A. Nadai (see Z. angew. Math. Mech., vol. 2, p. 1, 1922) and by M. T. Huber (see Z. angew. Math, Mech., vol. 6 ; p. 228, 1926).
long plate are evidently equal to zero along the lines AiBi, AB1 CD1 CiDi1 . . . . The bending moments along the same lines are also zero, and we may consider the given plate ABCD as a portion of the infinitely long plate loaded as shown in Fig. 766. Hence the deflection and the stresses produced in the given plate at the point of application 0 of the concentrated force can be calculated by using formulas derived for infinitely long plates. From Eq. (158) we find that the deflection produced at the x axis of the infinitely long plate by the load P applied at the point O is
The two adjacent forces P applied at the distances b from the point O (Fig. 766) produce at the x axis the deflection
in which, as before,
The forces P at the distance 2b from the point O produce at the x axis the deflection
and so on. The total deflection at the x axis will be given by the summation w = W1 + W2 + Wz + - - • (p) Observing that
we can bring expression (p) into coincidence with expression (146) of Art. 34. Let us apply the method of images to the calculation of the reactive force
acting at the corner D of the rectangular plate ABCD (Fig. 76) and produced by a load P at the center of this plate. Using Eqs. (151) and (152), we find that the general expression for the twisting moment of an infinitely long plate in the case of a single load becomes
Hence a load P concentrated at x = £ = a/2, y = 0 produces at x — 0 the twisting moment
Now, putting y = 6/2, 36/2, 56/2, . . . consecutively, we obtain the twisting moments produced by the loads ±P acting above the line DC. Taking the sum of these moments we obtain
To take into account the loads acting below the line DC we have to double the effect (s) of loads acting above the line DC in order to obtain the effect of all given loads. Thus we arrive at the final result
As for the reactive force acting downward at the point D, and consequently at the other corners of the plate, it is equal to R = — 2MXV, Mxy being given by Eq. (t). The method of images can be used also when the point of application of P is not on the axis of symmetry (Fig. 77a). The deflections and moments can be calculated by introducing a system of auxiliary forces as shown in thefigureand using the formulas derived for an infinitely long plate. If the load is distributed over a rectangle, formulas (167), which will be given in Art. 37, can be used for calculating the bending moments produced by actual and auxiliary loads. 37. Bending Moments in Simply Supported Rectangular Plates under a Load Uniformly Distributed over the Area of a Rectangle. Let us consider once more the practically important case of the loading represented in Fig. 78. If we proceed as described in Art. 33, wefindthat for small values of u/a and v/b the series representing the bending moments at the center of the loaded area converge slowly and become unsuitable for numerical computation. In order to derive more convenient formulas1 in this case let us introduce, in extension of Eq. (119), the following notation: 1 See S. Woinowsky-Krieger, Ingr.-Arch., vol. 21, p. 331, 1953.
(161) Hence
(162)
At first let us consider a damped circular plate of a radius a0 with a central load, distributed as shown in Fig. 78. The bending moments at the center of such a plate can be obtained by use of the Michell solution, for an eccentric single load. If u and v are small in
FIG.
77
FIG.
78
comparison with ao, the result, evaluated by due integration of expression (197) (p. 293), can be put in the form
in which
(163)
For a simply supported circular plate with the same radius a0 as before, we have to add a term P/4TT to Mx and Mv (see p. 68), i.e., a term P/%c(\ -f v) to M and nothing to N1 so that these latter quantities become
Finally, to obtain the corresponding expressions for an infinite strip (Fig. 75), we must assume a0 = 2a/ir sin Or£/a) and introduce an additional moment Mv = — (1 — J/)P/4TT (see p. 147). This latter operation changes the quantity M by - ( I — V)P/4TT(1 + v) and the quantity N by +P/4TT. Introducing this in Eqs. (6) we arrive at the result
(164)
The values of the factors
0.0964Mo&2/Z) 0.0620Af0fc2/D 0.0368Af 0a2/Z) 0.0280 Af0a2/£> 0.0174Af0a2/D
0.300M 0 0.387M 0 0.424M 0 0.394M 0 0.264Af0 0.153Af0
1.000M0 0.770M 0 0.476M 0 0.256M 0 0.046Af0 -0.010Af 0
Let us consider now the antisymmetrical case in which
In this case the deflection surface is an odd function of y, and we must put Bm = Cm = 0 in expression (/). Hence,
From the boundary conditions (c) it follows that
whence and
The constants An, are obtained from conditions (d), from which it follows that
Hence and
(174) We can obtain the deflection surface for the general case represented by the boundary conditions (d) from solutions (173) and (174) for the symmetrical and the antisymmetrical cases. For this purpose we split the given moment distributions into a symmetrical moment distribution Mfy and an antisymmetrical distribution M'yr, as follows:
( M ; W = {M'y)y=-m = Uf1(X) +J2(X)) « ) , = 6 / 2 = -(M'v')y=-b/2
=
Uf1(X)-J2(X)]
These moments can be represented, as before, by the trigonometric series
and the total deflection is obtained by using expressions (173) and (174) and superposing the deflections produced by each of the two foregoing moment distributions (Z). Hence
(175)
If the bending moments Mv = ^ E1n sin (mwx/l) are distributed only TO=I
along the edge y = 6/2, we have fo(x) = 0, E'm = K ' = $#»; and the deflection in this case becomes
(176) Solutions (173) to (176) of this article will be applied in the investigation of plates with various edge conditions. Moments M0 distributed along only one edge, say y = b/2, would produce, at the center of the plate, one-half the deflections and bending moments given in Table 28. In case of a simultaneous action of couples along the entire boundary of the plate, the deflections and moments can be obtained by suitable superposition of the results obtained above for a partial loading.1 42. Rectangular Plates with Two Opposite Edges Simply Supported and the Other Two Edges Clamped. Assume that the edges x = 0 and x = a of the rectangular plate, shown in Fig. 86, are simply supported and that the other two edges are clamped. The deflection of the plate under any lateral load can be obtained by first solving the problem on the assumption that all edges are simply supported and then applying bending moments along the edges y = ±b/2 of such a magnitude as to eliminate the rotations produced along these edges by the action of the lateral load. In this manner many problems can be solved by combining the solutions given in Chap. 5 with the FIG. 86 solution of the preceding article. Uniformly Loaded Plates.2 Assuming that the edges of the plate are simply supported, the deflection is [see Eq. (139), page 116] 1 Bending by edge couples was also discussed by H. Bay, Ingr.-Arch., vol. 8, p. 4, 1937, and by U. Wegner, Z. angew. Math. Mech., vol. 36, p. 340, 1956. 2 Extensive numerical data regarding rectangular plates with uniform load and sides simply supported or clamped in any combination may be found in a paper by F. Czerny; see Bautech.-Arch.t vol. 11, p. 33, Berlin, 1955.
and the slope of the deflection surface along the edge y — 6/2 is
To eliminate this slope and thus to.satisfy the actual boundary conditions we distribute along the edges y = ± 6/2 the bending moments My given by the series
and we determine the coefficients Em so as to make the slope produced by these moments equal and opposite to that given by expression (6). Using expression (173)l for the deflection produced by the moments, we find that the corresponding slppe along the edge y = 6/2 is
Equating the negative of this quantity to expression (6), we find that
Hence the bending moments along the built-in edges are
The maximum numerical value of this moment occurs at the middle of the sides, where x = a/2. Series (/) converges rapidly, and the maximum moment can be readily calculated in each particular case. For 1
From the symmetry of the deflection surface produced by the uniform load it can be concluded that only odd numbers 1, 3, 5, . . . must be taken for m in expression (173).
jcample, the first three terms of series (/) give — 0.070