09. Plates on elastic foundation

09. Plates on elastic foundation

ON PLATES CHAPTER ELASTIC FOUNDATION A laterally 57 Bending Symmetrical with Respect to a Center loaded plate may rest on an elastic foundation, as in the case of a conWe begin the discussion of such crete road, an airport runway, ora mat problems with the simplest assumption that the intensity of the reaction This of the subgrade is proportional to the deflections w of the plate The constant k, expressed intensity is then given by the expression kw in pounds per square inch per inch of deflection, is called the modulus of The numerical value of the modulus depends largely the foundation on the properties of the subgrade; in the case of a pavement slab or a mat of greater extension this value may be estimated by means of the | diagram in Table 62 TABLE OF VALUES 62 THE MopuLUS SUBGRADE OF Modulus “k” in Ib/sq in./ in T T General T soil rating os subgrade, subbase or base Good | Best Excellen† : Poor Very poor subgrode | cupgrode | Feirto good subgrode | supgrode| P-Poorly groded ' G-Grovel S ~ Sand H- High M-Mo,very fine sand silt | , compressibili†y 0-0 less thon 0.1mm _ Organic W- Weill graded + P + | OH CH l Ỉ MH Ỉ OL | CL GP GF F - Fines, material ~ P000 §Ubb0§€ [ pose |bgse | L - Low to med compressibility C€C- Cloy («80 | I J Joo | so | | | Lebo ¬T ML SF SP GC cw + L c 9W SC m J Let us begin with the case of a circular plate in which the load is disIn using Eq (58), tributed symmetrically with respect to the center The table should not be regarded as a Based on Casagrande’s soil classification information see Trans Am Soc Civ further For tests bearing plate for substitute Geotechnique, vol 5, p 297, 1955 Terzaghi, K also See 1948 901, p 113, Engrs., vol (Harvard Soil Mechanics Series, no 51) 259 260 THEORY OF PLATES AND SHELLS we add the load —kw, due to the reaction of the subgrade, to the given lateral load g Thus we arrive at the following differential equation for the bent plate: d°® , 1d\(d , ldu\ _ q— ku E Ty i) & Te 7) “Dp (178) In the particular case of a plate loaded at the center with a load P,* q is equal to zero over the entire surface of the plate except at the center By introducing the notation D Eq (178) becomes d? (a) “le + ldđ\/d2 , 1du (sat oa) (Ga +23) +w=0 6) Since k is measured in pounds per cubic inch and D in pound-inches, the quantity / has the dimension of length To simplify our further discussion it is advantageous to introduce dimensionless quantities by using the following notations: + r Then Eq (6) becomes d? lở đ3z , dz (5+z2)(2+222)+77—0 Using the symbol A for d? @) ld +” we then write AAz+z =0 (2) This is a linear differential equation of the fourth order, the general solution of which can be represented in the following form: Z£ —= Ai) + A2X o(x) + A3X 3(x) + A4X 4(2) (f) where Ai, , A, are constants of integration and the functions Xi, , X4 are four independent solutions of Eq (e) We shall now try to find a solution of Eq (e) in the form of a power * This problem was discussed by H Hertz, Wiedemann’s Ann Phys u Chem., vol 22, p 449, 1884; see also his “‘Gesammelte Werke,”’ vol 1, p 288, 1895, and A Féppl, ‘‘Vorlesungen tiber technische Mechanik,” vol 5, p 103, 1922 It is worth noting that Hertz’s investigation deals with the problem of a floating plate rather than with that of a piate on an elastic foundation Thus, in this case the assumption regarding the constancy of & is fulfilled, k being the unit weight of the liquid PLATES series find ON ELASTIC FOUNDATION Let a,x” be a term of this series A(anz") and Then, by differentiation, we = n(n — 1)d„+"—~? + nd„+"~? AA(a„z") 261 = n(n — 2)?d„x"~° = n’anx"-? To satisfy Eq (e) it is necessary that each term a," in the series have a corresponding term a,—42"~* such that m„?(n — 2)?q„+"~* + Anat”* = (9) Following this condition, all terms cancel when the series is substituted in Eq (e); hence the series, if it is a convergent one, represents a particular From Eq (g) it follows that solution of the equation _ An—4 Observing also that AA(ao) = and AA(a2x?) = (2) we can conclude that there are two series satisfying Eq (e), v7z., Xi(z) +3 — ø5rA? = +Š 29:42:01.82 T 212 — and Xo(x) = 2? — +8 42-62 53747 EF BF 102 (9) | 210 * 12 42-62-82 _ 10? | 42 - 62 - 832- % 14 102 - 122 - 142? +} se It may be seen from the notations (c) that for small values of the distance r, that is, for points that are close to the point of application of the load P, the quantity x is small, series and (7) are rapidly convergent It may be seen also that the consecutive derivatives of series (7) remain finite at the point of application of the load (x = 0) This indicates that these series alone are not sufficient to represent the stress conditions at the point of application of the load where, as we know from previously discussed cases, the bending moments become infinitely large For this reason the particular solution X3 of Eq (e) will be taken in the following form: X3= in which F3(x) power series is a function X1 log x of x which By differentiation we find d?X, + F;(%) can again be represented (k) by a 262 THEORY OF PLATES AND SHELLS and substituting X; for z in Eq (e), we obtain 4X1 | dx Since X, satisfies Eq (e) and is represented by the first of the series ( j) we obtain the following equation for determining F(z): AAF;(z) + _ 40 2-3-4 6-7-8: 4! 1011-12-28 Ì01-27:07.81!— 22-47-07-82.107 1222 F3(2) = = dat = -4(- “92 42 42 | ) 0) Taking #';(+) 1n the form of the series F3(x) = bart + bez? + Dye? + and substituting this series in Eq bs, bie, that - | (m) (1), we determine the coefficients bu, SO that the resulting equation will be satisfied AA(b¿z1) Observing = 47+ 2?- bg we find, by equating to zero the sum of the terms that not contain z, that 4+ ° » 2° + bs = 2-3-4 4° aaa 28:4 = "94-44 128 Iquating to zero the sum of the terms containing x‘, we find 25 og = — 1,769,472 —¬ — In general, we find n?(n — 2)? - 82? Thus the third particular solution of Eq X = X1 log x + jog — (e) is 20 —_ ; log z + °s vs ứ¿ (e) is obtained in a similar F a(x)= XoXe!log x + 4.598 a6 14:56 - gee (4 fo „8 1769.4722 T The fourth particular integral X, of Eq manner by taking Xe=4 X21 - - + 10-9-8 2) 24 e - 10! (0) X PLATES ON ELASTIC FOUNDATION 263 By substituting the particular solutions (7), (nr), and (0) in expression (f) we obtain the general solution of Eq (e) in the following form: 2= A, (1 - 78 sp t+ orrgr gg I +10 "` + mn") +8 +A:|(1= gẼn đựng 88T CÔ 25 +° log + 535 +19 _“ằ-x: ma - 62 - 8?- 10? an )log + 5955 , 442308 + | (P) It remains now to determine in each particular case the constants of integration Ay, , Asso as to satisfy the boundary conditions Let us consider the case in which the edge of a circular plate of radius a is entirely free Making use of expression (52) for the radial moments and expression (55) for the radial shear force Q,, we write the boundary conditions as d?w C- d + {d’w , ldw + (Ge In addition to these two dw r ee) Ts r =) conditions = (q) =8 we have two more conditions that hold at the center of the plate; viz., the deflection at the center of the plate must be finite, and the sum of the shearing forces distributed over the lateral surface of an infinitesimal circular cylinder cut out of the plate at its center must balance the concentrated force P From the first of these two conditions it follows that the constant A; in the general solution (p) vanishes The second condition gives ([,° 2z Qrd0) +P=0 (r) or, by using notation (a), d {fd’w , ldw “” —Ƒ}H md“” |(Fe +I) 2re + P — =O | (s) where « is the radius of the infinitesimal cylinder Substituting lz for w in this equation and using for z expression (p), we find that for an infinitely small value of x equal to «/l the equation reduces to —k a 2mec+ P =0 264 THEORY OF PLATES from which | | AND SHELLS P Ae SRP Having the values of the constants 4; and 4, the remaining two constants A, and A» can be found from Eqs (g) For given dimensions of the plate and given moduli of the plate and of the foundation these equations furnish two linear equations in A, and As Let us take, as an example, a plate of radius a = in and of such rigidity that D PP = sin t= We apply at the center a load P such that = 102-1075 ~ Sxkiề Using this value of A, and substituting lz for w, we find, by using expression (p) and taking « = a/l = 1, that Eqs (q) give 0.500A; + 0.250A2 0.687A, — 8.483A2 = 4.062A, = 11.09A, 4.062 - 102 - 10-5 11.09 - 102 - 10-5 These equations give A, = 86-10-4 = Ay = — 64-1078 Substituting these values in expression (p) and retaining only the terms that contain # to a power not larger than the fourth, we obtain the following expression for the deflection: | w=lz=5 E - 10-4 (: — 92.42 ) — 64 - 1075? + 102 - 10-52? log >| The deflection at the center (x = 0) is then max and the deflection at the boundary = 43 - 107% in (z = 1) is Wmin = 39.1 - 107? in The difference of these deflections is comparatively small, and the pressure distribution over the foundation differs only slightly from a uniform distribution If we take the radius of the plate two times larger (a = 10 in.) and retain the previous values for the rigidities D and k, z becomes equal to at the boundary, and Eqs (g) reduce to These equations give 0.826.4 + 1.9804; 2.66541 — 5.745A, A, = 3.93A,4 = 400 - 1075 Az = 1.208A, 16.3/7A;a —1.08A, = —105- 1075 (w) PLATES ON ELASTIC FOUNDATION 265 The deflection is obtained from expression (p) as w == ly lz == ö 1400 - - 10 I0—5 x4 — 22 A2 +6 105 - 10 10—5 | + x2? — —— 576 — 102-107 10 + |] toe (« — + Zs) | * 3456" Ì The deflections at the center and at the boundary of the plate are, respectively, Wmax = 2.1072 in and Wmin = 0.88 - 107? in It is thus seen that, if the radius of the plate is twice as large as the quantity 1, the distribution of pressure over the foundation is already far from uniform The applica- tion of the strain energy method to the problem of bending of a plate on elastic subgrade will be shown in Art 80 58 Application to the Functions of Bessel of the Problem Circular Plate The general solution (f) of Eq (e) in the preceding article can also be represented in terms of Bessel functions (e) a new variable § = z a/ i ; To this end we introduce into Eq thus we arrive at the equation AAz—=z=0 (a) in which the symbol A’ stands for d? + ld dị? E dé Now Eq (a) is equivalent to equation A’(A’z +2) and also to Hence Eq A’(A’z (a) — — z) + (A’z +2) = (b) (A’z = (c) —z) is satisfied by the solutions of the Bessel differential equation đ?z A’ z+zZ dz = —— — qe’+ —tat? =0 (d)ở = (e) as well as by the solutions of the equation Alzg Z— — đ?z de = —— + dz — de —— — Z which is transformable into Eq (d) by substituting ¢ for & tion of Eqs (d) and Thus the combined solu- (e) can be written as = Bilo(x Vi) + Balo(zi Vi) + BsKo(a Vi) + BiKo(ai Vi) I, and Ky being Bessel functions of the first and second kind, respectively, imaginary argument, whereas B,, Bo, are arbitrary constants (f) and of ‘The argument x being real, all functions contained in Eq (f) appear in a complex form To single out the real part of the solution, it is convenient to introduce four other functions, first used by Lord Kelvin and defined by the relations? Ina V +1) = berx ttbeiz Ko(x V #+t‡) = ker z ‡ ¿zkelz (g) See, for Instance, G N Watson, ““Theory of Bessel Funections,”” p 81, Cambridge, 1948 266 THEORY OF PLATES AND SHELLS Setting, furthermore, B, + B; + B, B, = where the new constants Ci, C2, the deflections of the plate: Ct/l B, — B, = —€¿zt/Ì C,/l B; _— B, = —C32/l are real, we obtain the following expression for w= C, ber x -} C; bei z + C3 keixz + Ci, keraz All functions herein contained are tabulated functions, argument (h) real for real values of the For small values of the argument we have ber z = — z1/64 + - - bei z = z?/4 — z°/2,304 + ker z = — logx +log2—y+7277/16+ kei z = im which (i) — (2/4) log z — z/4 + (1 + log — +)z2/4 + y = 0.5772157 - - - is Euler’s constant and - log — + = 0.11598 - - - For large values of the argument the following asymptotic expressions hold: be TA e5 ~ rtu™~ TT — ~ e7 T / 2x ez a/ kei x ~— in which o = 2/ 3⁄2 The general solution (h) can be a circular plate, with or without constants C, corresponding in the be determined in each particular €OS§ | AJ Wnt bel ~ ke —— +/m sinfo ( COS — V⁄2z/z — — 5) + [| a (9) T _ sin({o m{ịø +—+~8 used for the analysis of any symmetrical bending of a hole, resting on an elastic foundation The four most general case to four boundary conditions, must case.? 1See ‘‘Tables of Bessel Functions Jo(z) and J;(z) for Complex Arguments,”’ Columbia University Press, New York, 1943, and ‘Tables of Bessel Functions Yo(z) and Y,(z) for Complex Arguments,’’ Columbia University Press, New York, 1950 We have ber « = Re [Jo(xe*™/*)] kerz - = bei x = — Re [Yo(xe®™/4)] — im L8 — Im [Jo(ze**/4)] [J o(xe*™!4)] T kei z = im [VY o(xe*™/4)] — Re [J o(ze**/4)] ? Many particular oook “Kreisplatten Z4(x) = tables of functions solutions of this problem auf elastischer Unterlage,” Zi(z) = ber z, Z2(z) = are given Berlin, — bei 2, by F Schleicher in his 1926, which also contains Z3(x) = —(2/xr) —(2/x) ker x as well as the first derivatives of those functions kei x, and An abbrevi- ated table of the functions Z and their first derivatives is given in Art 118, where they are denoted by the symbol y PLATES ON ELASTIC FOUNDATION 267 We shall confine ourselves to the case of an infinitely extended plate carrying & single load P at the point x = Now, from the four functions forming solution (h), the first two functions increase indefinitely with increasing argument in accordance with Eqs (j); and the function ker x becomes infinitely large at the origin, as we can — conclude from Ings (4) Accordingly, setting C; = C2 = C, = 0, solution (h) 1s reduced to w = C3 kei x (k) In order to determine the constant C3, we caleulate, by means of l2qs (2), the shearing force [see Eqs (193)] Dd (dw 1dờ\ 22 (2412) a= CD(1 xz (-š+ `) 13 - As x decreases, the value of Q, tends to C3;D/l’x = C3D/l’r distributing the load P uniformly over the circumference Q, = —P/2xr Equating On the other hand, upon with radius r, we have both expressions obtained for Q,, we have C3; = Pl? — ứ 2) 2rD Substitution of C; into Eq (k) yields, finally, the complete solution of Hertz’s problem in the form PP - 3p w= and the corresponding reaction (179) kei x of the subgrade ¬ is given by p = kw = wD a The variation of these quantities along a meridional section through the deflection surface of the plate is shown in Fig 131, together with similar curves based on a theory which will be discussed in Art 61 At the origin we have kei x = —7/4 and the deflection under the load becomes w max =~ Pl? 3D — (180) 180 For the reaction of the subgrade at the same point we obtain p max = P Sự 7, 181 (181) If we take an infinitely large plate with the conditions of rigidity and loading assumed on page 264, the deflection under the load becomes 10ax = —— = —— = lA, = (3.14)(5)(102 - 10-5) = 0.016 in as compared with the value of 0.02 in obtained for a finite circular plate with the | | radius a = 2) The distribution of the bending moments due to the concentrated load is shown in 268 THEORY O O0 OF lí | Ke 0.15 SHELLS VA | | Z 0.10 AND a | 0.05 PLATES ew D G6 > r/lo= Xo > r/L=x arp (Elst, solid) ° ~w 12 (Hertz]) L=1.241 lo 0.192 -~ ——0.125 (Hertz) (a) / | — 0.05} | ⁄/ : Le a ~~p sp O r/lo= > — => r/L=x Lễ Xo ‘Sp = (Elast solid) (Hertz) 0.10 L=0.8061lo 0.15 —-—04125 (Hertz) 0.192 (b) 0 M P 005H | { X=X ° “Mt - /=0.3 0.10Ƒ 0.15 a } | | (c) Fic 131 Fig 131c It is seen that the radial moments become negative at some distance from the load, their numerically largest value being about —0.02P The positive moments are infinitely large at the origin, but at a small! distance from the point of application of the load they can be easily calculated by taking the function kei z in the form (2) Upon applying formulas (52) and (53) to expression (179), we arrive at the results As compared with the characteristic length = a/ D/k M, Pf (oe a+» > — 4x P[ 2I rT -5a-% - 7) (1+ »)(log—r —-y]+50 - ») a An 269 FOUNDATION 21 M,=—| A comparison ELASTIC ON PLATES Eqs of the foregoing expressions with (90) and | | ¬ (182) | (91) shows that the stress condition in a plate in the vicinity of the load in Hertz’s case is identical with that of a simply supported circular plate with a radius a = Qle-¥ = 1.1231, except for a moment M = M ;=“— tr (1 — »), which is superimposed on the moments of the 7T cireular plate Let us consider now the case in which the load P is distributed over the area of a The bending moments at the center circle with a radius c, small in comparison with of a circular plate carrying such a load are M, = M P = — 4z (m) | + y) log = + la Cc By This results from Eq (83), if we neglect there the term c?/a? against unity obtain we v), — —P/8r(1 substituting a = 2le~7 into Eq (m) and adding the moment at the center of the loaded circle of the infinitely large plate the moments or Mex P = q + »)P Muax = = 47 ;) —+> + Cc (n) P (108 L + 0616) GQ tu 4x (183) C Stresses resulting from Eq (183) must be corrected by means of the thick-plate theory Such a corrected stress formula is given on in the case of a highly concentrated load page 275 In the case of a load uniformly distributed over the area of a small rectangle, we may The equivalent of a square area, in particular, is a proceed as described in Art 37 circle with the radius c = 0.57u, u being the length of the side of the square (see page Substituting this into Eq (183) we obtain 162) Muax The effect of any 59 Rectangular group = + + _P (1 of concentrated ỉ u — ¬+- (o) 1177) loads on the deflections of the infinitely large plate can be calculated by summing up the deflections produced by each load separately example of time along a beam of foundation the elastic of the tube and Continuous Plates on Elastic Foundation An a plate resting on elastic subgrade and supported at the same a rectangular boundary is shown in Fig 132, which represents a rectangular tubular cross section pressed into an elastic The bottom plate of the beam, loaded by by the loads P reactions of the foundation, is supported by the vertical sides and by the transverse diaphragms indicated in the figure by 270 THEORY OF PLATES AND SHELLS dashed lines It is assumed again that the intensity of the reaction p at any point of the bottom plate is proportional to the deflection w at that point, so that p = kw, k being the modulus of the foundation In accordance with this assumption, the differential equation for the deflection, written in rectangular coordinates, becomes ——— mem ì —Ÿ—x vị aw0+ jy dw, aw _ ạD _ EmD Tổ Ox? dy? T 0t —* aI where q, as before, is the intensity of the lateral load ble | = P Let us begin with the case shown in Fig 132 If wo denotes the deflection of the edges of the bottom plate, and w the deflection of this plate P with respect to the plane of its boundary, f the intensity of the reaction of the foundation at any point is k(wo — w), and Eq (a) becomes Wo i (a) AAw = (Wo — w) (b) + Taking the coordinate axes as shown in the 16 132 figure and assuming that the edges of the plate parallel to the y axis are simply supported and the other two edges are clamped, the boundary conditions are O*w (W)r-0,2-a = O (40)y_.s/ () = Ú _ ow OY | = =0 Jy—+b/2 (c) (d) The deflection w can be taken in the form of a series: eo ° _ 4k =v => MTL sm mit m=1,35, \Á gÁ a © i marx Ym Sinwy ot + + D (2) m=1,3,5, The first series on the right-hand side is a particular solution of Eq (b) representing the deflection of a simply supported strip resting on an elastic foundation The ous equation second series is the solution of the homogene- AAw + w =0 (f) Hence the functions Y,, have to satisfy the ordinary differentia] equation ve — 9.9 ye 44 k + (BE +5) va | (9) PLATES ON ELASTIC FOUNDATION 271 Using notations — 262 = Vu and taking = D [Lm —_ + ÁN! + nạ of Eq the solution following four roots: 8B + + = À " (h) 2y2 = Vu$ + M4 — Hệ (g) in the form —8 +? e’¥, we B—1Y | (2) for r the obtain —8 — The corresponding four independent particular solutions of Eq (g) are eBmy sin Ymy eFmY sin YmY eFm’ COS YmY e8mY COS YmY (3) which can be taken also in the following form: cosh Bn COS YmY cosh Bny SIN YmY sinh BnYy COS YmY sinh Bny SIN YmY (Ie) From symmetry it can be concluded that Y,, in our case is an even function of y Hence, by using integrals (k), we obtain Ym = Am cosh Buy COS Ymy¥ + Br sinh Bry SIN YnY and the deflection of the plate is eo w= marx | 4kg sin —— a Da › m=1,3,5, + mir4 k m(™ + 5) A,, cosh Bny COS Ymy + Bm sinh Bmy sin yny | () This expression satisfies the boundary conditions (c) To satisfy conditions (d) we must choose the constants A, and B,, so as to satisfy the : equations Ahwo _ Dr : I m wsg* + D)dD mix k + A,, cosh Bmb cos —— Ymb 98-2 + B,, sinh Bmb sin Ym mb = (m) VmD > (AmBm + B„y„) sinh ——7 — (AmYm — BmB8m) cosh Pn? sin Yn = Substituting these values of A, and B,, in expression (1), we obtain the required deflection of the plate The problem of the plate with all four edges simply supported can be solved by using Eq (a) Taking the coordinate axes as shown in Fig 59 272 THEORY OF PLATES AND SHELLS ‘page 105) and using the Navier solution, the deflection of the plate is ec eo | w= m=1 Jr similar manner Max Amn Sin —— n=] Từn sin —% a (n) b let the series oo oo q = Gmn Sin —— m=1 sin > (0) n=1 represent the distribution of the given load, and the series p= kw = D>, bmn sin gin MY (p) represent the reaction of the subgrade Substituting the series (n) in the left-hand side and the series (0) and (p) in the right-hand side of Kq (a), we obtain Amn = m2 na (q) AS an example, let us consider the bending of the plate by a force P concentrated at some point (£7) In such a case _4P Onn = > by Eq (b) on page 111 liq (n) we finally obtain mré my ©© = — ?t= rr= nin : «Sin = (r) By substitution of expressions (g) and (r) into 4P a | mrt sin | TT ry b iD &a? + m) +h — Mann sin —— 6° bề nry sin —— (s) ° Having the deflection of the plate produced by a concentrated force, the deflection produced by any kind of lateral loading is obtained by the method of superposition Take, as an example, the case of a uniformly distributed load of the intensity g Substituting q dé dy for P in expression (s) and integrating between the limits and a and between and b, we obtain | oo 00 _ Mnrx nary w= 16g SH Ta BR TC = m=1,3,5, n=1,3,5, mm ?+ b? | (t) PLATES ON ELASTIC FOUNDATION 273 When k is equal to zero, this deflection reduces to that given in Navier solution (131) for the deflection of a uniformly loaded plate.? Let us consider now the case represented in Fig 133 A large plate which rests on an elastic foundation is loaded at equidistant points along We shall take the coordinate axes as shown 1n the x axis by forces P.* PO” ` — (v) ) m= 2,4,6, In order to express the constants A‘, in terms of the magnitude of loads P, we consider the shearing force Q, acting along the normal section of the plate through the x axis From symmetry we conclude that this force vanishes at all points except the points of application of the loads P, at which points the shearing forces must give resultants equal to —P/2 It was shown in the discussion of a similar distribution of shearing forces in Art represented by the series 54 (see page 248) that the shear forces can be “ _f (—1)”/? cos — m = 2,4,6, The shearing force, as calculated from expression (v), is (0? , 9°? Gy = — Đạp (ấm + au? Joc : —-— =-¿ 2a / A, BmYm(B% » 2D MUTE , 2) cos + 72.) — m= 2,4,6, Comparing these two expressions for the shearing force, we find P(- Am ~ or, by using notations (2), Al 20 DBn¥m 1)™/2 (Bin + v2.) P(—1)"”? ™ — aDr Vrt + nt Substituting this in expression (v), we finally obtain w= Wo + —ak | (7 NI dế vàpe G08 TU a | Ym COS Ymy + Bn SIN Ymy) (w) The maximum deflection is evidently under the loads P and is obtained by substituting x = a/2, y = in expression (w), which gives Pr Wyryrax — 2/2 — P)? ak + ak ` > m= 2,4,6, Ym ——————E VM + yt 184) 184 PLATES ON ELASTIC FOUNDATION 275 ~The deflection in the particular case of one isolated load P acting on an infinitely large plate can also be obtained by setting a = © in formula Insucha case the first term in the formula vanishes, and by using (184) notations (2) we obtain = PA? oD A/D ok », 2x [VME m= 2,4,6, a ah = oh M+ tr, Pr VaR —9 sal, Using the substitution mm_ Mw we find Unax \* + pA =H a, | Qu Yu? +1 PA = ————— V4⁄9xÈ % du 4⁄21 + 10? = PA? OT), 8k (185) With this magnitude of the deflecin accordance with the result (180) tion, the maximum pressure on the elastic foundation is (D) max = | k“max —= PX? 8- P = Ie |k (186) The maximum tensile stress is at the bottom of the plate under the point The theory developed above gives an infinite of application of the load value for the bending moment at this point, and recourse should be had In the above-mentioned to the theory of thick plates (see Art 26) investigation by Westergaard the following formula for calculating maximum tensile stress at the bottom of the plate is established by using the thick-plate theory: a (ø.)„ = 0.275(1+ ») logio a (x) Here h denotes the thickness of the plate, and b =C \/1.6c2 + h? — 0.675h whence < 1.724h when c > 1.724h where c is the radius of the circula: area over which the load P is assumed to be uniformly distributed For c = the case of the concentrated | force is obtained In the case of a square loaded area u by u, we have to replace c by 57u (see page 162) ˆ The case of equidistant loads P applied along the edge of a semi-infinite The plate, as shown in Fig 134, ean also be treated in a similar way plate the of bottom the at stress final formula for the maximum tensile 276 THEORY OF PLATES AND SHELLS under the load when the distance a is large is (oz) max = 0.529(1 + 0.547) mã oe Ca — 0471 (0) where ö 1s calculated as In the previous case, and e is the radius of the semicircular area over which the load P is assumed to be uniformly distributed Formulas (x) and (y) have proved very useful in the design of concrete roads, in which case the circle of radius c represents the area ° ° of contact of the wheel tire with the road surface ø [7 = Ụ E2 ⁄ mm 60 Plate Carrying Rows of Equidistant Columns As a last example, let us consider an infinite plate or mat resting on w ° elastic‘ subgrade and carrying equidistant and equal loads P, each load being distrib- 75 Ao ⁄7 | OVAVA ụ In ụ uted uniformly over the area u by v of a rectangle, as shown in Fig 135 The | >In 2⁄4 V7 q- h | `2 pf Ae +Ạ ị ! X a Ị Ị y “mm | J ⁄ Lis ⁄ “Oo "tp ~d ——-~ “|p d~-z⁄4Z mn(Dybn + F) amx COS cos B PnY Ớ) The bending moments of the plate are now obtained by the usual differentiation, and n the pressure between the plate and the subgrade is found by multithe distributioof (f) by the modulus k expression of plication = corresponds toa uniformly distributed reaction of the k case _ The particular subgrade, 7.e., to the case of a ‘‘reversed flat slab” uniformly loaded with q = P/ab It is seen from Eq (f) that the introduction of the modulus tends to reduce the deflections and also the bending moments of the plate The case of a rectangular plate of finite dimensions resting on an elastic foundation and submitted to the action of a concentrated load has been discussed by H Happel.? The Ritz method (see page 344) has been used to determine the deflections of this plate, and it was shown in the particular example of a centrally loaded square plate that the series representing the deflection converges rapidly and that the deflection can be calculated with sufficient accuracy by taking only the first few terms of the series.® Due to V Lewe, Math Z., vol 6, -1949 The problem of experimentally; see p 1, 1935; and that Bauingenieur, vol 3, p 453, 1923 See also F Halbritter, Bautechnik, vol 26, p 181, p 203, 1920 | a square plate on an elastic foundation has also been investigated the paper by J Vint and W N Elgood, Phil Mag., ser 7, vol 19, by G Murphy, Jowa Siaie Coll Eng Expt Sta Bull 135, 1937 278 THEORY Of PLATES AND SHELLS 61 Bending of Plates Resting on a Semi-infinite Elastic Solid So far, the settling of the subgrade at some point of its surface has been assumed as proportional to the pressure between the plate and the subgrade at the same point, and consequently as independent of the pressure elsewhere This is correct in the case of a floating plate, considered by Hertz (see page 260), but in the case of a coherent subgrade such a hypothesis approximates but crudely the actual behavior of the subgrade; a better approximation can sometimes be obtained on the basis of the following assumptions: The foundation has the properties of a semi-infinite elastic body The plate rests on the subgrade without friction A perfect contact between the plate and foundation also exists in the case of a negative mutual pressure This last supposition appears arbitrary; however, a negative pressure between plate and subgrade actually is compensated, more or less, by the weight of the plate The elastic properties of the elastic foundation may be characterized, if isotropy is assumed, by a Young modulus E, and a Poisson ratio vy The approximate numerical values! of these constants, depending on the nature of the subgrade and based on results of dynamical tests, are given in Table 63, together with the value of the constant ko Eo = 2(1 —— »?) v2) | (a) used in the following TABLE 63 VALUES oF ELAsTIc NATURE Subgrade CoNSTANTS Medium sand Sand and gravel Liassic plastic clay (air-slaked) Sandstone DEPENDING ON FOUNDATION Eo, psi Clay Loess and clay Lime OF Vo 11,000 13,000 0.17 0.42 ko, psi 5,700 7,900 14,000-18, 500 40,000 33-0.23 0.31 7, 900-9 , 800 22,000 165, 000-190, 000 0.32-0.38 92,000-110,000 38 , 000 1,600,000 0.44 0.26 23 , 500 860, 000 We restrict the further consideration to the case of an infinitely large plate in a state of axial symmetry Using polar coordinates r, 6, we can write the plate equation as DAAw(r) = g(r) — p(r) (0) where q(r) denotes the given surface loading and p(r) the reaction of the subgrade Let Ko(r,p,¢) be the deflection at the point (7,0) of the subgrade surface due to a normal unit load applied on this surface (p,¢) The form of the ‘influence function” K, depends merely upon the nature of the foundation Making use of some properties of the Bessel functions, it can be shown? that Eq (b) is satisfied by the expression _ mr) = | Due Berlin, 2'The Intern ” O(a) K(a)J o(ar)a da + DatK(a) (c) to E Schultze and H Muhs, ‘‘ Bodenuntersuchungen fir Ingenieurbauten,”’ 1950 See also Veréffentl Degebo, Heft 4, p 37, 1936 solution of the problem in this general form is due to D L Holl, Proc Fifth Congr Appl Mech., Cambridge, Mass., 1938  ... load separately example of time along a beam of foundation the elastic of the tube and Continuous Plates on Elastic Foundation An a plate resting on elastic subgrade and supported at the same a... —-y]+50 - ») a An 269 FOUNDATION 21 M,=—| A comparison ELASTIC ON PLATES Eqs of the foregoing expressions with (90) and | | ¬ (182) | (91) shows that the stress condition in a plate in the vicinity... first series on the right-hand side is a particular solution of Eq (b) representing the deflection of a simply supported strip resting on an elastic foundation The ous equation second series is