09. Plates on elastic foundation

09. Plates on elastic foundation

PLATES ON ELASTIC FOUNDATION

57 Bending Symmetrical with Respect to a Center A laterally loaded plate may rest on an elastic foundation, as in the case of a con- crete road, an airport runway, ora mat We begin the discussion of such problems with the simplest assumption that the intensity of the reaction

of the subgrade is proportional to the deflections w of the plate This intensity is then given by the expression kw The constant k, expressed

in pounds per square inch per inch of deflection, is called the modulus of the foundation The numerical value of the modulus depends largely

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

TABLE 62 VALUES OF THE MopuLUS OF SUBGRADE

do General soil rating os subgrade, subbase or base 0 | 1 Lebo | so | | («80 | I J Joo

Very poor subgrode | cupgrode | Feir to good subgrode | supgrode| P000 §Ubb0§€ [ pose |bgse

259

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

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

(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 dis- cussion it is advantageous to introduce dimensionless quantities by using the following notations:

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

series Let a,x” be a term of this series Then, by differentiation, we find

Eq (e); hence the series, if it is a convergent one, represents a particular solution of the equation From Eq (g) it follows that

_ An—4

Observing also that

we can conclude that there are two series satisfying Eq (e), v7z.,

+3 +Š Xi(z) = 1 — ø5rA? T 29:42:01.82

212

— 53747 EF BF 102 12

and (9)

+8 210 | Xo(x) = 2? — 42-62 * 42-62-82 10?

| 14 _ % +} se

42 - 62 - 832- 102 - 122 - 142?

It may be seen from the notations (c) that for small values of the dis- tance r, that is, for points that are close to the point of application of the load P, the quantity x is small, and series (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:

in which F3(x) is a function of x which can again be represented by a power series By differentiation we find

4 d?X,

and substituting X; for z in Eq (e), we obtain

4X1 |

2 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):

6-7-8: 4! 1011-12-28 Ì01-27:07.81!— 22-47-07-82.107 122 2 ) 0)

Taking #';(+) 1n the form of the series

F3(x) = bart + bez? + Dye? + - | (m)

and substituting this series in Eq (1), we determine the coefficients bu,

bs, bie, SO that the resulting equation will be satisfied Observing that

By substituting the particular solutions (7), (nr), and (0) in expression (f) we obtain the general solution of Eq (e) in the following form:

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 1 dw C- + r ee) = 0

d {d’w , ldw + (Ge Ts r =) =8

In addition to these two conditions 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 solu- tion (p) vanishes The second condition gives

—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

a

Let us take, as an example, a plate of radius a = 5 in and of such rigidity that

4 D t= PP = sin

We apply at the center a load P such that

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: |

4

92.42 w=lz=5 E - 10-4 (: — ) — 64 - 1075? + 102 - 10-52? log >|

The deflection at the center (x = 0) is then

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 2 at the boundary, and Eqs (g) reduce to

0.826.4 + 1.9804;

2.66541 — 5.745A, 1.208A, 16.3/7A;a

These equations give

A, = 3.93A,4 = 400 - 1075 Az = —1.08A, = —105- 1075 (w)

The deflection is obtained from expression (p) as

= ly = - I0—5 — — 10—5 | x2? — ——

w = lz = ö 1400 - 10 1 22 A2 105 - 10 + 576

+ 5 | 102-107 |] 2 — 6

+ 10 0 toe 2 (« 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 sub- grade will be shown in Art 80

58 Application of Bessel Functions to the Problem of the Circular Plate The

general solution (f) of Eq (e) in the preceding article can also be represented in terms

of Bessel functions To this end we introduce into Eq (e) a new variable § = z a/ i ;

thus we arrive at the equation

đ?z 1 dz

A’ z+zZ = —— + — — qe’ tat? =0 (d) ở

as well as by the solutions of the equation

đ?z 1 dz

Alzg Z— — 2 = —— de + de — —— — Z = (e) which is transformable into Eq (d) by substituting ¢ for & Thus the combined solu-

tion of Eqs (d) and (e) can be written as

I, and Ky being Bessel functions of the first and second kind, respectively, and of

imaginary argument, whereas B,, Bo, are arbitrary constants ‘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) = ber x ttbeiz

Ko(x V #+t‡) = ker z ‡ ¿z kel z

1 See, for Instance, G N Watson, ““Theory of Bessel Funections,”” p 81, Cambridge,

1948

(g)

w= C, ber x -} C; bei z + C3 keixz + Ci, keraz (h)

All functions herein contained are tabulated functions, real for real values of the

argument

For small values of the argument we have

ber z = 1 — z1/64 + - - -

ker z = — logx +log2—y+7277/16+ -

kei z = — (2/4) log z — z/4 + (1 + log 2 — +)z2/4 + - - -

im which y = 0.5772157 - - - is Euler’s constant and log 2 — + = 0.11598 - - -

For large values of the argument the following asymptotic expressions hold:

The general solution (h) can be used for the analysis of any symmetrical bending of

a circular plate, with or without a hole, resting on an elastic foundation The four constants C, corresponding in the most general case to four boundary conditions, must

be determined in each particular case.?

(9)

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

kerz = — 5 Re [Yo(xe®™/4)] — 5 im [J o(xe*™!4)]

- L8 T

kei z = 5 im [VY o(xe*™/4)] — 5 Re [J o(ze**/4)]

? Many particular solutions of this problem are given by F Schleicher in his

oook “Kreisplatten auf elastischer Unterlage,” Berlin, 1926, which also contains

tables of functions Zi(z) = ber z, Z2(z) = — bei 2, Z3(x) = —(2/xr) kei x, and

Z4(x) = —(2/x) ker x as well as the first derivatives of those functions 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

We shall confine ourselves to the case of an infinitely extended plate carrying & single load P at the point x = 0 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\ CD(1 xz a= 22 (2412) - 13 (-š+ `)

As x decreases, the value of Q, tends to C3;D/l’x = C3D/l’r On the other hand, upon distributing the load P uniformly over the circumference with radius r, we have

Q, = —P/2xr Equating both expressions obtained for Q,, we have

At the origin we have kei x = —7/4 and the deflection under the load becomes

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

(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

1 As compared with the characteristic length 1 = a/ D/k

A comparison of the foregoing expressions with Eqs (90) and (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

Stresses resulting from Eq (183) must be corrected by means of the thick-plate theory

in the case of a highly concentrated load Such a corrected stress formula is given on

The effect of any group of concentrated loads on the deflections of the infinitely

large plate can be calculated by summing up the deflections produced by each load separately

59 Rectangular and Continuous Plates on Elastic Foundation An example of a plate resting on elastic subgrade and supported at the same time along a rectangular boundary is shown in Fig 132, which represents

a beam of a rectangular tubular cross section pressed into an elastic foundation by the loads P The bottom plate of the beam, loaded by the elastic reactions of the foundation, is supported by the vertical sides

of the tube and by the transverse diaphragms indicated in the figure by

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

| = Let us begin with the case shown in Fig 132

vị If wo denotes the deflection of the edges of the

bottom plate, and w the deflection of this plate

P P with respect to the plane of its boundary, the

i f + 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

Hence the functions Y,, have to satisfy the ordinary differentia] equation