CHAPTER 7

CONTINUOUS RECTANGULAR PLATES

52. Simply Supported Continuous Plates. Floor slabs used in build-ings, besides being supported by exterior walls, often have intermediatesupports in the form of beams and partitions or in the form of columns.In the first case we have to deal with proper continuous plates; in thecase of columns without intermediate beams we have to deal with flatslabs. The floor slab is usually subdivided by its supports into several

FIG. 110

panels. Only continuous plates with panels of rectangular shape will beconsidered in this chapter.

We begin with a case allowing a rigorous solution by methods alreadyused in the foregoing chapter. A rectangular plate of width b and lengtha\ + a% + a3, supported along the edges and also along the intermediatelines ss and U9 as shown in Fig. 110, forms a simply supported continuousplate over three spans. We suppose that the intermediate supportsneither yield to the pressure in the transverse direction nor offer anyresistance to the rotation of the plate with respect to the axes ss and U.With these assumptions, the bending of each span of the plate can bereadily investigated by combining the known solutions for laterallyloaded, simply supported rectangular plates with those for rectangularplates bent by moments distributed along the edges.

Let us begin with the symmetrical case in which

Q>\ = a2 = &3 = a

and the middle span is uniformly loaded while the side spans are withoutload (Fig. 1106). Considering the middle span as a simply supportedrectangular plate and using expression (b) of Art. 44 (see page 198), weconclude that the slope of the deflection surface along the edge X2 = a/2 is

where ft* = mwa/2b. Owing to the continuity of the plate, bendingmoments Mx are distributed along the edges X2 = + a/2. From sym-metry it is seen that these moments can be represented by the followingseries:

The deflections W\ produced by these moments can be obtained fromEq. (173), and the corresponding slope along the edge X2 = a/2 [seeEq. (e), page 198] is

From the condition of continuity we conclude that the sum of expres-sions (a) and (c) representing the slope of the plate along the line X2 = a/2must be equal to the slope along the same line of the deflection surfaceof the plate in the adjacent span. Considering this latter span as asimply supported rectangular plate bent by the moments (6) distributedalong the edge xz = —a/2, we find the corresponding deflection W2 ofthe plate by using Eq. (176) (see page 185), from which follows

The corresponding slope along the edge Xz = — a/2 is

The equation for calculating the coefficients E7n is

Since this equation holds for any value of y, we obtain for each value ofm the following equation:

from which

It is seen that Em decreases rapidly as m increases and approaches thevalue —2qb2/Tr3m3. Having the coefficients E7n calculated from (g), weobtain the values of the bending moments Mx along the line U fromexpression (b). The value of this moment at y = 0, that is, at themiddle of the width of the plate, is

Taking, as an example, b — a, we have $m = mir/2, and the formula (g)gives

The bending moments at the center of the middle span can be readilyobtained by combining bending moments of a simply supported plate,bent by uniform load, with moments corresponding to the deflections W\.Taking, for example, a — b and v = 0.2, which is a convenient value for

concrete, we get for the first of these moments the values of

(see Table 8, page 120) and for the second moments the values

Therefore

If a side span is uniformly loaded, as shown in Fig. 110c, the deflectionsurface is no longer symmetrical with respect to the vertical axis of sym-metry of the plate, and the bending moment distributions along the linesss and tt are not identical. Let

To calculate the coefficients Em and Fm we derive two systems of equa-tions from the conditions of continuity of the deflection surface of theplate along the lines ss and tt, Considering the loaded span and usingexpressions (a) and (e), we find that the slope of the deflection surfaceat the points of the support ss, for a\ = a* = a3 = a, is

Considering now the middle span as a rectangular plate bent by themoments Mx distributed along the lines ss and tt and given by the series(h), we find, by using Eq. (175) (see page 184),

From expressions (i) and (j) we obtain the following system of equationsfor calculating coefficients Em and Fm:

where the following notations are used:

The slope of the deflection surface of the middle span at the supportingline U1 by using expression (J)1 is

This slope must be equal to the slope in the adjacent unloaded spanwhich is obtained from expression (c) by substituting Fm for Em. In thisway we find the second system of equations which, using notations (I)1

can be written in the following form:

From this equation we obtain

Substituting in Eq. (Ic)1 we find

Substituting in each particular case for Ami Bmi and Cm their numericalvalues, obtained from Eqs. (I)1 we find the coefficients Em and Fm; andthen, from expressions (Zi), we obtain the bending moments along thelines ss and tt. Take, as an example, b = a. Then fim = ra7r/2, and wefind from Eqs. (Z)

A1 = -0.6677 B1 = -1.1667 Ci = -0.7936A3 = -0.9983 Bz = -1.0013 C3 = -0.9987

For m larger than 3 we can take with sufficient accuracy

Substituting these values in Eq. (o), we obtain

The moment at the middle of the support ss is

For the middle of the support tt we obtain

Having the bending moments along the lines of support, the deflectionsof the plate in each span can readily be obtained by superposing onthe deflections produced by the lateral load the deflections due to themoments at the supports.

The bending moments in the panels of the continuous plate can beobtained in a similar manner. Calculating, for example, the momentsat the center of the middle span and taking v = 0.2, we arrive at the

values

The equations obtained for threespans can readily be generalized andexpanded for the case of any number ofspans. In this way an equation similarto the three-moment equation of con-tinuous beams will be obtained.1 Let

p us consider two adjacent spans i andi + 1 of the length a» and a,-+i, respec-

tively (Fig. 111). The corresponding values of the functions (Z) aredenoted by Ain, B)n, Cin and AjJ-1, UjJ"1, CjJ"1. The bending moments alongthe three consecutive lines of support can be represented by the series

1 This problem in a somewhat different way was discussed by B. G. Galerkin; see his"Collected Papers," vol. 2, p. 410, Moscow, 1953.

Considering the span i + 1 and using expressions (a) and (j), we find

In the same manner, considering the span i, we obtain

From the condition of continuity we conclude that

Substituting expressions (p) and (q) in this equation and observing thatit must be satisfied for any value of y, we obtain the following equationfor calculating E*-1, E]n, and E*+1:

(177)

Equations (/c) and (m), which we obtained previously, are particular casesof this equation. We can write as many Eqs. (177) as there are inter-mediate supports, and there is no difficulty in calculating the momentsat the' ntermediate supports if the ends of the plate are simply supported.The left-hand side of Eq. (177) holds not only for uniform load but alsofor any type of loading that is symmetrical in each span with respect tothe x and y axes. The right-hand side of Eq. (177), however, has adifferent value for each type of loading, as in the three-moment equa-tion for beams.

The problem of continuous plates carrying single loads can be treatedin a similar manner. In the particular case of an infinite number ofequal spans with a single load applied at any point of only one span, thedeflection of the plate may be obtained by resolving an equation with

finite differences for the unknown coefficient Ein as functions of theindex i.1

If the intermediate supports are elastic, the magnitude of the coeffi-cients E]n is governed by the five-term equations, similar to the five-moment equations of the theory of continuous beams.2 The torsionalrigidity of supporting beams, tending to reduce the rotations of the platealong the support, can also be taken into account in considering thebending of continuous plates.3

As the simplest example of a continuous plate carrying a concentrated load, let usconsider an infinitely long plate simply supported along the sides x = 0, x = a, con-tinuous over the support y — 0, and submitted to a concentrated load P at some point% = £> V = V (Fig. 112a). The load and boundary conditions under considerationcan be readily satisfied by superposition of cases shown in Fig. 1126 and c. In the

case of Fig. 1126 each panel of the plate issimply supported along the line y = Q, andthe elastic surface is given by the expression±Wi/2, in which the sign must be chosenaccording to whether y is greater or lessthan zero, W\ denotes the deflections (a) ofArt. 51, and \y\ < \q\. In the case shown inFig. 112c, each panel is clamped along theedge y = 0, and the corresponding deflec-tions are w/2, w being given by expression(h) in Art. 51. We have therefore

FIG. 112

and the moments along the edge y = 0 become equal to one-half of the clampingmoments of a semi-infinite plate with one edge built in, these latter moments beinggiven by expression (j) of Art. 51.

53. Approximate Design of Continuous Plates with Equal Spans.4

The layout of a floor slab usually involves continuity not only in onedirection, as assumed in Art. 52, but rather in two perpendicular direc-tions. A continuous slab of this kind is shown in Fig. 113. The spansand the thickness of the plate are equal for all rectangular panels. Each

1 See S. Woinowsky-Krieger, Ingr.-Arch., vol. 9, p. 396, 1938.2 Continuous plates on elastic beams were considered by V. P. Jensen, Univ. Illinois

Bull. 81, 1938, and by N. M. Newmark, Univ. Illinois Bull. 84, 1938.3 See K. Girkmann, "Flachentragwerke," 4th ed., p. 274, Vienna, 1956.4 The method given below is substantially due to H. Marcus; see his book "Die

vereinfachte Berechnung biegsamer Platten," Berlin, 1929. The coefficients of Tables51 to 56 are, however, based on solutions considered in Chap. 6 and on the value ofPoisson's ratio v = 0.2, whereas Marcus uses for the same purpose a simplified theoryof rectangular plates and assumes ^ = O .

panel may carry a dead load go and,possibly, a live load p, both distri-buted uniformly over the area ofthe panel, the largest intensity ofthe load being q = g0 + p.

Let us begin with the computationof bending moments at the inter-mediate supports of the floor plate.Calculations show that these mo-ments depend principally on theloading of the two adjacent panels,and the effect of loading panelsfarther on is negligible. I t is justi-fiable, therefore, to calculate themoments on supports by assumingthe load q uniformly distributed FlG- 1 1 3

over the entire floor slab (Fig. 114a). Neglecting, at first, the rotationsof the plate along the intermediate supports, each panel in Fig. 114a willhave the same conditions as a rectangular plate clamped along the inter-mediate supports and simply supported at the external boundary of thefloor slab.

FIG. 114

The maximum bending moments for plates with such boundary con-ditions have been tabulated (see Tables 51 to 56). Six possible combi-nations of simply supported and built-in edges of a rectangular plate areshown at the head of these tables. The direction of the x and y axes ineach panel of the slab (Fig. 113) must be chosen in accordance with Figs.116 to 121; span a must be measured in the direction of the x axis andspan b in the direction of the y axis of the respective panel. The sixcases shown in Figs. 116 to 121 may be numbered 1 to 6, and the corre-sponding indices are attached to the coefficients of Tables 51 to 56.

To illustrate the application of the tables, let us calculate the bendingmoment at the middle of the support tw (Fig. 113). We calculate forthis purpose the clamping moment of both panels adjacent to the sup-port. For panel 2 we have to use the formula

M2y = 82qP (a)

and Table 52, I being the smaller of spans a and b of the panel. In asimilar manner we obtain the clamping moment of panel 6 from theexpression

M6x = y6ql* (b)

by making use of Table 56. The moment in question now is given withsufficient accuracy by

Mtw = i(Miv + M6x) (c)

and the moments on other intermediate supports are obtainable in a sim-ilar manner.

It should be noted that Eq. (c) expresses nothing else than a moment-distribution procedure in its simplest form, i.e., a procedure in which the"carried-over" moments from other supports, as well as any differencein the stiffness values of both adjacent panels, are neglected. Such asimplified procedure is far more justified in the case of a continuous platethan in the case of a continuous beam.

Next, let us consider the bending moments at the center of panel 6(Fig. 113) as an example. The load distribution most unfavorable forthese moments can be obtained by superposition of loads shown in Fig.1146 and c.

The contribution of the uniformly distributed load g0 + p/2 to thevalues of the moments is obtained by use of Table 56, which gives

I denoting the smaller of both spans of panel 6.Let us consider now the effect of the checkerboard loading as shown in

Fig. 114c. The boundary conditions of each panel here are the same as

those of a simply supported plate, and the moments at the center arereadily computed by means of Table 51 for case 1. The load +p/2acting in panel 6 yields

and the largest moments at the center of panel 6 are

In order to calculate the largest nega-tive moments at the same point wehave only to alter the sign of the loadin Fig. 114c. Still using results (d) and(e)j we then have

M6x = M6', - Mi', . ,

Mey = m y - MZ W

As a second example of the application ofthe approximate method, let us computethe bending moments of the continuousplate shown in Fig. 115, which was treatedrigorously in Art. 52.

First we choose the direction of the x andy axes in accordance with Figs. 117 and 118.Assuming next a load q = qo + p uniformlydistributed over the entire surface of theplate (Fig. 1156) and using the coefficientsgiven in Tables 52 and 53 for cases 2 and 3,with b/a = 1, we obtain at the center of thesupport ss the moment

the procedure being the same as in the foregoing example [Eq. (c)]. Using the rigoroussolution, the numerically largest moment at ss is produced by the load distributionshown in Fig. 115c. Superposing the bending moment obtained on page 231 uponthose calculated on page 234, the exact minimum value of the moment M,, provesto be

Mat = -[0.0381 (g0 + v) + 0.0424(g0 + p) - 0.0042g0]o2

or M88 = -(0.0805g0 + 0.0763p)a2 (i)

Putting, for instance, q0 = q/3, p = 2q/3, the result (i) yields -0.0777ga2as comparedwith the value — 0.0769ga2 obtained by the approximate method.

Finally, let us calculate the largest bending moment at the center of the middlepanel, the most unfavorable load distribution being such as shown in Fig. 115d.

FIG. 115

FIG. 116

TABLE 51. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 1

v = 0.2, I = the smaller of spans a and b

Center of plate

b /a FactorMx = aiql2 My = /3i^2

ai 0i

O 0.0250* 0.12500.5 0.0367 0.09990.6 0.0406 0.08680.7 0.0436 0.0742 ?62

0.8 0.0446 0.0627

0.9 0.0449 0.0526

I

1.0 0.0442 0.04421.1 0.0517 0.04491.2 0.0592 0.04491.3 0.0660 0.04441.4 0.0723 0.04391.5 0.0784 0.0426

qa2

1.6 0.0836 0.04141.7 0.0885 0.04021.8 0.0927 0.03911.9 0.0966 0.03782.0 0.0999 0.0367oo 0.1250 0.025Of

* Mm&x = 0.0364gb2 at 0.486 from the short edge.t Mmai = 0.0364ga2 at 0.48a from the short edge.

Combining the load in accordance with Fig. 115e and / and using the coefficients aand p of Tables 53 and 51, we arrive at the following expressions for these moments:

FIG. 117

TABLE 52. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 2v — 0.2; I = the smaller of spans a and b

„ ^ - w Middle ofCenter of plate ^ , ,fixed edgeb/a Factor

Mx = a2ql2 My = p2ql2 Mv = M 2

«2 $2 ^2

0 0.0125 0.0625 -0.12500.5 0.0177 0.0595 -0.12100.6 0.0214 0.0562 -0.11560.7 0.0249 0.0514 -0.1086 qb*0.8 0.0272 0.0465 -0.10090.9 0.0294 0.0415 -0.0922

1.0 0.0307 0.0367 -0.08401.1 0.0378 0.0391 -0.09161.2 0.0451 0.0404 -0.09831.3 0.0525 0.0415 -0.10401.4 0.0594 0.0418 -0.10841.5 0.0661 0.0418 -0.1121

qa*1.6 0.0722 0.0414 -0.11481.7 0.0780 0.0408 -0.11721.8 0.0831 0.0399 -0.11891.9 0.0879 0.0390 -0.12042.0 0.0921 0.0382 -0.1216

oo 0.1250 0.0250* -0.1250

* Mmax = 0.0387ga2 at 0.80a from the built-in edge.

It is of interest to verify the foregoing approximate values by use of the resultsobtained on pages 232 and 234. Distributing the load again as shown in Fig. 115dand interchanging the indices x and y in the results mentioned above, we have

Mx = O.O317(go + p)a2 - (0.0051 -f 0.0051)q0a2

= (O.O215go + 0.0317p)a2

My - O.O375(go + p)a2 - (0.0039 + 0.0039)g0o2 ( }

= (0.0297g0 + 0.0375p)a2

Setting again g0 = 9/3 and p — 2q/3, we obtain for the moments the exact valuesof 0.0283ga2 and 0.0349qa2, respectively. Eqs. (j) yield for the same moments theapproximate values of 0.029lqa* and 0.0358ga2

FIG. 118

TABLE 53. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 3

v = 0.2, I = the smaller of spans a and b

^ , j. i . Middle ofCenter of plate - , ,F fixed edge

b/a Factor

Mx = a3ql* My = №* Mv = S3^2

«3 03 ^3

0 0.0083* 0.0417 -0.08330.5 0.0100 0.0418 -0.08420.6 0.0121 0.0410 -0.08340.7 0.0152 0.0393 -0.0814 qb*0.8 0.0173 0.0371 -0.07830.9 0.0196 0.0344 -0.0743

1.0 0.0216 0.0316 -0.0697L I ' 0.0276 0.0349 -0.07871.2 0.0344 0.0372 -0.08681.3 0.0414 0.0391 -0.09381.4 0.0482 0.0405 -0.09981.5 0.0554 0.0411 -0.1049

qa*1.6 0.0620 0.0413 -0.10901.7 0.0683 0.0412 -0.11221.8 0.0741 0.0408 -0.11521.9 0.0795 0.0401 -0.11742.0 0.0846 0.0394 -0.1191

oo 0.1250 0.025Of -0.1250

* Mm a x = 0.0174qb2 at 0.306 from the supported edge,t Mm a x = 0.0387ga2 at 0.80a from the built-in edge.

The largest error of the approximate method ensues from the fact that the largestpositive moments do not always occur at the center of the panel. This is especiallyfar from being true in the case of distinctly oblong rectangular panels. If 6, forexample, is much larger than a, the largest moment Mv occurs near the short side ofthe rectangular plate. Some values of these largest moments are given in footnotes tothe tables, and they should be considered as the least possible values of the corre-sponding columns, regardless of the actual ratio b/a.

It should be noted, finally, that in the unsymmetrical case 4 neither Mx nor My

FIG. 119

TABLE 54. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 4*

v = 0.2, I = the smaller of spans a and b

Center of plate Middle of fixed edge __~ '

b/a Factor

Mx = a*ql2 My = /34(^2 Mx = y*ql2 Mv = 6*ql2 Mm>x = e4ql*

<X4 /S4 74 54 €4

0.5 0.0191 0.0574 -0.0787 -0.1180 0.06620.6 0.0228 0.0522 -0.0781 -0.1093 0.05700.7 0.0257 0.0460 -0.0767 -0.0991 0.0501 qb*0.8 0.0275 0.0396 -0.0746 -0.0882 0.0430

0.9 0.0282 0.0336 -0.0715 -0.0775 0.03631.0 0.0281 0.0281 -0.0678 -0.0678 0.03051.1 0.0330 0.0283 -0.0766 -0.0709 0.03581.2 0.0376 0.0279 -0.0845 -0.0736 0.0407

1.3 0.0416 0.0270 -0.0915 -0.0754 0.04521.4 0.0451 0.0260 -0.0975 -0.0765 0.04911.5 0.0481 0.0248 -0.1028 -0.0772 0.0524 ga2

1.6 0.0507 0.0236 -0.1068 -0.0778 0.0553

1.7 0.0529 0.0224 -0.1104 -0.0782 0.05861.8 0.0546 0.0213 -0.1134 -0.0785 0.06081.9 0.0561 0.0202 -0.1159 -0.0786 0.06362.0 0.0574 0.0191 -0.1180 -0.0787 0.0662

* The authors are indebted to the National Research Council of Canada for a grantwhich greatly facilitated the computation of the table.

is the largest bending moment at the center of the plate. Table 54 shows, however,that th\3 difference between Mmax and the largest of the values of Mx and Mv does notexceed 10 per cent of the latter values and that the general procedure described onpage 238 is justified in case 4 as well.

For the purpose of the design of isolated panels without continuity (Fig. 119),Table 54 contains the values of the largest moments Mm&x acting at x = 0.1a, y = 0.16;for rectangular plates the direction of <rmax is practically that of the shorter span andfor square plates that of the diagonal x = — y. For the sake of a greater securitythose values of Mn^x may also be used in calculating continuous panels of oblong shape.

FIG. 120

TABLE 55. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 5*v = 0.2, I = smaller of spans a and b

Center of plate Middle of fixed edge

b/a FactorMx = abql* My - /3&9/2 Mxybql* Mv = S5^2

«6 06 TS ^5

0.5 0.0206 0.0554 -0.0783 -0.1140.6 0.0245 0.0481 -0.0773 -0.1020.7 0.0268 0.0409 -0.0749 -0.09070.8 0.0277 0.0335 -0.0708 -0.0778 q

0.9 0.0274 0.0271 -0.0657 -0.06581.0 0.0261 0.0213 -0.0600 -0.05471.1 0.0294 0.0204 -0.0659 -0.05661.2 0.0323 0.0192 -0.0705 -0.0573

1.3 0.0346 0.0179 -0.0743 -0.05741.4 0.0364 0.0166 -0.0770 -0.05761.5 0.0378 0.0154 -0.0788 -0.05691.6 0.0390 0.0143 -0.0803 -0.0568 t

qa*1.7 0.0398 0.0133 -0.0815 -0.05671.8 0.0405 0.0125 -0.0825 -0.05671.9 0.0410 0.0118 -0.0831 -0.05662.0 0.0414 0.0110 -0.0833 -0.0566

x 0.0417 0.0083 -0.0833 -0.0566

* The data of this table are due substantially to F. Czerny, Bautech.-Arch., vol. 11,p. 33, W. Ernst & Sohn, Berlin, 1955.

The method given in this article is still applicable if the spans, the flexural rigidities,or the intensity of the load differs only slightly from panel to panel of the continuousplate. Otherwise more exact methods should be used.

It should be noted, however, that the application of the rigorous methods to thedesign of continuous floor slabs often leads to cumbersome calculations and that theaccuracy thus obtained is illusory on account of many more or less indeterminablefactors affecting the magnitude of the moments of the plate. Such factors are, forexample, the flexibility and the torsional rigidity of the supporting beams, the restrain-

FIG. 121

TABLE 56. BENDING MOMENTS FOR UNIFORMLY LOADED PLATES IN CASE 6v — 0.2, I = the smaller of spans a and b

Center of plate Middle of fixed edge

b/a I Factor

Mx = a6ql2 Mu - /S6^' Mx = 76gZ2 My = 8«ql2

«6 06 7« 5e

0 0.0083 0.0417 -0.0571 -0.08330.5 0.0118 0.0408 -0.0571 -0.08290.6 0.0150 0.0381 -0 .0571 -0.07930.7 0.0178 0.0344 -0.0569 -0 .0736 qb*0.8 0.0198 0.0299 -0.0559 -0 .06640.9 0.0209 0.0252 -0 .0540 -0 .05881.0 0.0213 0.0213 -0 .0513 -0 .05131.1 0.0248 0.0210 -0 .0581 -0 .05381.2 0.0284 0.0203 -0 .0639 -0.05541.3 0.0313 0.0193 -0.0687 -0 .05631.4 0.0337 0.0181 -0 .0726 -0 .05681.5 0.0358 0.0169 -0 .0757 -0.0570

qa*1.6 0.0372 0.0157 -0.0780 -0.05711.7 0.0385 0.0146 -0.0799 -0 .05711.8 0.0395 0.0136 -0 .0812 -0.05711.9 0.0402 0.0126 -0 .0822 -0.05712.0 0.0408 0.0118 -0.0829 -0.0571=o 0.0417 0.0083 -0 .0833 -0.0571

ing effect of the surrounding walls, the anisotropy of the plate itself, and the inaccuracyin estimating the value of such constants as the Poisson ratio v.

However, we can simplify the procedure of calculation by restricting the Fourierseries, representing a bending moment in the plate, to its initial term or by replacingthe actual values of moments or slopes along some support of the plate by theiraverage values or, finally, by use of a moment distribution procedure.1

64. Bending of Plates Supported by Rows of Equidistant Columns—(Flat Slabs). If the dimensions of the plate are large in comparison with

1 For such methods see C. P. Siess and N. M. Newmark, Univ. Illinois Bull. 43,1950, where a further bibliography on the subject is given. See also the paper ofH. M. Westergaard, Proc. Am. Concrete Inst., vol. 22, 1926, which contains valuableconclusions regarding the design of continuous floor slabs.

the distances a and b between the columns (Fig. 122) and the lateral loadis uniformly distributed, it can be concluded that the bending in allpanels, which are not close to the boundary of the plate, may be assumedto be identical, so that we can limit the problem to the bending of onepanel only. Taking the coordinate axes parallel to the rows of columnsand the origin at the center of a panel, we may consider this panel as auniformly loaded rectangular plate with sides a and 6. From symmetrywe conclude that the deflection surface of the plate is as shown by thedashed lines in Fig. 1226. The maximum deflection is at the center ofthe plate, and the deflection at the corners is zero. To simplify theproblem we assume that the cross-sectional dimensions of the columnsare small and can be neglected in so far as deflection and moments at

FIG. 122

the center of the plate are concerned.1 We then have a uniformly loadedrectangular plate supported at the corners, and we conclude from sym-metry that the slope of the deflection surface in the direction of thenormal to the boundary and the shearing force are zero at all pointsalong the edges of the plate except at the corners.2

Proceeding as in the case of a simply supported plate (Art. 30), wetake the total deflection w in the form

where

1 In this simplified form the problem was discussed by several authors; see, forexample, A. Nadai, tlber die Biegung durchlaufender Platten, Z. angew. Math.Mech., vol. 2, p. 1, 1922, and B. G. Galerkin, "Collected Papers," vol. 2, p. 29, Mos-cow, 1953.

2 The equating to zero of the twisting moment Mxy along the boundary follows fromthe fact that the slope in the direction of the normal to the boundary is zero.

represents the deflection of a uniformly loaded strip clamped at the endsy = ±6/2 and satisfies the differential equation (103) of the plate as wellas the boundary conditions

The deflection W2 is taken in the form of the series

each term of which satisfies the conditions (c). The functions Ym mustbe chosen so as to satisfy the homogeneous equation

AAw2 = 0 (e)

and so as to make w satisfy the boundary conditions at the edgesy = ±6/2. Equation (e) and the conditions of symmetry are satisfiedby taking series (d) in the form

where the constants A0, Am, and Bm are to be determined from theboundary conditions along the edge y = 6/2. From the condition con-cerning the slope, viz., that

we readily find that

in which, as before,

Considering now the boundary condition concerning the shearing force,we see that on a normal section nn (Fig. 1226) of the plate infinitelyclose to the boundary y = 6/2, the shearing force Qy is equal to zero atall points except those which are close to the column, and at these pointsQy must be infinitely large in order to transmit the finite load %qab to thecolumn (Fig. 122c) along an infinitely small distance between x = a/2 — cand x = a/2 + c. Representing Qv by a trigonometric series which, from

symmetry, has the form

and observing that

we find, by applying the usual method of calculation, that

where P = qab is the total load on one panel of the plate. Substitutingthese values of the coefficients Co and Cm in series (i), the required bound-ary condition takes the following form:

Substituting expression (a) for w and observing that the second term inparentheses vanishes, on account of the boundary condition dw/dy = 0,we obtain

from which, by using expression (/), we find that

Solving Eqs. (g) and (j) for the constants Am and B7n, we obtain

The deflection of the plate takes the form

The constant A0 can now be determined from the condition that thedeflection vanishes at the corners of the plate. Hence

The deflection at any point of the plate can be calculated by using expres-sions (I) and (ra). The maximum deflection is evidently at the center ofthe plate, at which point we have

Values of this deflection calculated for several values of the ratio b/a aregiven in Table 57. Values of the bending moments (Mx)̂ o1W=O and(My)x=sQ,yz=Q calculated by using formulas (101) and expression (I) fordeflection are also given. It is seen that for b > a the maximum bend-

T A B L E 57. D E F L E C T I O N S AND M O M E N T S AT T H E C E N T E R OF A P A N E L (Fig. 122)

v = 0.2

qb4

b/a W = a D M* = ^ 2 Mv = Piqb'Z

a fJ ffi

1 0.00581 0.0331 0.03311.1 0.00487 0.0261 0.03521.2 0.00428 0.0210 0.03631.3 0.00387 0.0175 0.03751.4 0.00358 0.0149 0.03841.5 0.00337 0.0131 0.03872.0 0.00292 0.0092 0.0411» 0.00260 0.0083 0.0417

ing moment at the center of the plate does not differ much from themoment at the middle of a uniformly loaded strip of length b clampedat the ends.

Concentrated reactions are acting at the points of support of the plate, and themoments calculated from expression (Z) become infinitely large. We can, however,assume the reactive forces to be distributed uniformly over the area of a circle repre-senting the cross section of the column. The bending moments arising at the centerof the supporting area remain finite in such a case and can be calculated by a pro-cedure similar to that used in the case of rectangular plates and described on page 147.With reference to Fig. 122, the result can be expressed by the formulas1

In these expressions

g «_ e-rbia} anci c denotes the radius of the circle, supposed to be small comparedwith spans a and b of the panel. Carrying out the required calculations, we canreduce Eqs. (o) to the form

in which a and /3 are coefficients given for several values of the ratio b/a in Table 58.

TABLE 58. VALUES OF COEFFICIENTS a AND 0 IN EQS. (p) FOR MOMENTS

ON SUPPORT

b/a 1 1.1 1.2 1.3 1.4 1.5 2.0

a 0.811 0.822 0.829 0.833 0.835 0.836 0.838

0 0.811 0.698 0.588 0.481 0.374 0.268 -0.256

The bending moments corresponding to the centers of columns of rec-tangular cross section also can be calculated by assuming that the reac-tions are uniformly distributed over the rectangles, shown shaded in Fig.

1 Given by A. Nadai in his book "Elastische Platten," p. 154, Berlin, 1925.

TABLE 59. BENDING MOMENTS AND LARGEST SHEAR FORCE OP A SQUAREPANEL OF A UNIFORMLY LOADED PLATE (Fig. 123)

v - 0.2

u/a -m k W)*-9—lt •» faa* (3/)x_v-o — 0iqa* (Mx)x-a/t,v-o — P*qa* (My)x-.an,v-o - Ptqa* Qmax - yqa

0 fii fh fit y

0 - « 0.0331 -0.0185 0.0512 «0.1 -0.196 0.0329 -0.0182 0.0508 2.730.2 -0.131 0.0321 -0.0178 0.04890.3 -0.0933 0.0308 -0.0170 0.0458 0.8420.4 -0.0678 0.0289 -0.0158 0.04150.5 -0.0487 0.0265 -0.0140 0.0361 0.419

123, that represent the cross sections of the columns.1 In the case ofsquare panels and square columns we have u/a = v/b = k, and themoments at the centers of the columns and at the centers of the panelsare given by the following formulas:

The values of these moments, to-gether with values of moments athalf a distance between columns, ob-tained from the same solution andcalculated for various values of k andfor v = 0.2, are given in Table 59.

It is seen that the moments at thecolumns are much larger than themoments at the panel center and thattheir magnitude depends very muchon the cross-sectional dimensions ofthe columns. The moments at thepanel center remain practically con-stant for ratios up to k = 0.2. Hence the previous solution, obtained on

1 This case was investigated by S. Woinowsky-Krieger; see Z. angew. Math. Mech.,vol. 14, p. 13, 1934. See also the papers by V. Lewe, Bauingenieur, vol. 1, p. 631,1920, and by K. Frey, Bauingenieur, vol. 7, p. 21, 1926.

FIG. 123

the assumption that the reactions are concentrated at the panel corners,is sufficiently accurate for the central portion of the panel.

An approximate calculation of moments given by Eq. (q) in the form of a series canalso be made by means of expressions (p). Using for this purpose Eq. (c), Art. 37,we substitute

i.e., the radius of a circle equivalent to the given square area u by u, in Eqs. (p). Inthe particular case of square panels numerical results obtained in this manner are butslightly different from those given in the second column of Table 59.

The shearing forces have their maximum value at the middle of thesides of the columns, at points m in Fig. 123. This value, for the case ofsquare panels, depends on the value of the ratio k and can be representedby the formula Q = yqa2. Several numerical values of the factor y aregiven in Table 59. It is interesting to note that there is a difference ofonly about'10 per cent between these values and the average valuesobtained by dividing the total column load qa2(l — /c2) by the perimeter

4/ca of the cross section of the column.Uniform loading of the entire plate

gives the most unfavorable conditionat the columns. To get the maxi-mum bending moment at the centerof a panel, the load must be distri-buted as shown by the shaded areasin Fig. 124a. The solution for thiscase is readily obtained by combiningthe uniform load distribution of in-tensity q/2 shown in Fig. 1246 withthe load q/2 alternating in sign inconsecutive spans shown in Fig. 124c.The deflection surface for the lattercase is evidently the same as thatfor a uniformly loaded strip of lengtha simply supported at the ends.

Taking, as an example, the case of square panels and using the values inTable 57, we find for the center of a panel (Fig. 124a):

FIG. 124

From Table 59 we conclude, furthermore, that

(M,)_o.y^/a = iq ' 0.0512a2 + T\qa2 = 0.0881?a2

The foregoing results are obtained in assuming that the plate is free torotate at the points of support. Usually the columns are in rigid con-nection with the plate, and, in the case of the load distribution shown inFig. 124, they produce not only vertical reactions but also couples with arestraining effect of those couples on the bending of the panels. A frameanalysis extended on the flat slab and the columns as a joint structuretherefore becomes necessary in order to obtainmore accurate values of bending moments underalternate load.1

The case in which one panel is uniformlyloaded while the four adjacent panels are notloaded is obtained by superposing on a uniformload q/2 the load q/2, the sign of which alter-nates as shown in Fig. 125. In this latter caseeach panel is in the same condition as a simplysupported plate, and all necessary informationregarding bending can be taken from Table 8. Taking the case of asquare panel, we find for the center of a panel that

O)x=y=o = \q • 0.00581 ~ + ^q • 0.00406 ^ = 0.00494 ^

(Mx)w«o = (My)x=v=o = ^q • 0.0331a2 + ± g • 0.0479 ^ a 2 = 0.0387?a2

The case of an infinitely large slab subjected to equal concentratedloads centrally applied in all panels can be handled substantially in thesame manner as in the preceding case, i.e., by using the double periodicityin the deflections of the plate.2

The problem of bending of a uniformly loaded flat slab with skewpanels has also been discussed.3

55. Flat Slab Having Nine Panels and Slab with Two Edges Free.So far, an infinite extension of the slab has always been assumed. Nowlet us consider a plate simply supported by exterior walls, forming thesquare boundary of the plate, together with four intermediate columns(Fig. 126). From symmetry we conclude that a uniformly distributed

1 The procedure to be used is discussed in several publications; see, for instance,H. Marcus, "Die Theorie elastischer Gewebe," p. 310, Berlin, 1932.

2 This problem was discussed by V. Lewe in his book "Pilzdecken und anderetragerlose Eisenbetonplatten," Berlin, 1926, and also by P. Pozzati, Riv. math. Univ.Parma, vol. 2, p. 123, 1951.

3 See V. I. Blokh, Doklady Akad. Nauk S.S.S.R., n. s., vol. 73, p. 45, 1950.

FIG. 125

load of intensity q produces equalcolumn reactions R, which we mayconsider as redundant in the givenstatically indeterminate structure.Removing all columns, we obtain asimply supported square plate carry-ing merely the given load q. Thedeflections WQ produced by this loadat the center of the columns caneasily be calculated by means of thetheory given in Chap. 5. Next, re-moving the load q and distributing aload R = 1 (acting downward) uni-formly over each area u by u, weobtain some new deflections Wi at thesame points x = ±a/2, y — ±a/2as before. From the condition that inthe actual case these points do not de-flect, we conclude that Wo — Rw\ = 0,which yields R = Wo/wi. Now itremains only to combine the effectof the uniform load q with the effectof four known reactions on the bend-ing moments of the square plate ofthe size 3a by 3a.

In the case of a partial loading,such as shown in Fig. 1266 and c, wehave to superpose one-half of themoments previously obtained on themoments of a simply supported platewith the area a by 3a, carrying auniformly distributed load ±<//2.Calculations of this kind carried outby Marcus1 led to the values of bend-ing moments given in Table 60. Thereaction of a column is R — 1.196ga2

in this case. The bending of an in-finite plate which is supported notonly along both its parallel sides

1 "Die Theorie elastischer Gewebe"; seealso Lewe, op. cit. The case of a square plate with one intermediate support wasdiscussed by N. J. Nielsen, "Bestemmelse af Spaendinger I Plader," p. 217, Copen-hagen, 1920.

FIG. 126

TABLE 60. COEFFICIENTS /3 FOR CALCULATION OF BENDING MOMENTS M = Pqa2

OF A SIMPLY SUPPORTED SQUARE PLATE WITH FOUR INTERMEDIATE

COLUMNS (Fig. 126)

u/a = 0.25, v = 0.2

Load a Load 6 Load cPoint -

a a Mx My Mx My Mx My

1 0 0 0.021 0.021 -0 .048 -0.004 0.069 0.0252 0.5 0 -0.040 0.038 -0.020 0.019 -0.020 0.0193 1.0 0 0.069 0.025 0.093 0.027 -0.024 -0.0024 0 0.5 0.038 -0.040 -0.036 -0.036 0.074 -0.0045 0.5 0.5 -0.140 -0.140 -0.070 -0.070 -0.070 -0.070

6 1.0 0.5 0.074 -0.004 0.092 0.014 -0 .018 -0.0187 0 1.0 0.025 0.069 -0.028 0.017 0.052 0.0528 0.5 1.0 -0.004 0.074 -0.002 0.037 -0.002 0.0379 1.0 1.0 0.053 0.053 0.066 0.044 -0 .013 0.009

but also by one or several rows of equidistant columns1 can be discussedin a similar manner.

The case of bending of a long rectangular plate supported only by thetwo parallel rows of equidistant columns (Fig. 127) can also be solvedwithout any difficulty for several types of loading. We begin with the casein which the plate is bent by the moments My represented by the series

Since there is no lateral load, the deflection surface of the plate can betaken in the form of the series

the coefficients of which are to be determined from the following boundaryconditions:

1TWs problem has been considered by K. Grein, "Pilzdecken," Berlin, 1948.

and from the condition that the deflection vanishes at the columns.Substituting series (b) in Eas. Cc). we find that

Combining this solution with solution (I), Art. 54, we can investigate thebending of the plate shown in Fig. 127a under the action of a uniformly

FIG. 127

distributed load. For this purpose we calculate the bending momentsMy from expression (Z) by using formula (101) and obtain

Equating this moment to the moment (a) taken with the negative sign,we obtain the values of M0 and Em which are to be substituted in Eqs.(d) for the constants Ai, Am, and Bm in expression (b). Adding expres-sion (b) with these values of the constants to expression (J), Art. 54, weobtain the desired solution for the uniformly loaded plate shown in Fig.127a.

Combining this solution with that for a uniformly loaded and simplysupported strip of length b which is given by the equation

we obtain the solution for the case in which the plate is bent by the loaduniformly distributed along the edges of the plate as shown in Fig. 1276.

56. Effect of a Rigid Connection with Column on Moments of the Flat Slab. Indiscussing the bending of a flat slab it has always been assumed that the columnreactions are concentrated at some points or distributed uniformly over some areascorresponding to the cross section of the columns or their capitals. As a rule, however,concrete slabs are rigidly connected with the columns, as shown in Fig. 128.

In discussing moments at such rigid joints, let us begin with the case of a circularcolumn and let c be the radius of its cross section. The calculation of bending

moments using expression (Z) in Art. 54 shows1 that, in the case of a square panel(o = b) and small values of c/a, the bending moments in the radial direction practicallyvanish along a circle of radius e = 0.22a (Fig. 122a). Thus the portion of the platearound the column and inside such a circle is in the state of an annular plate simplysupported along the circle r = 0.22a and clamped along the circle r = c, with atransverse displacement of one circle with respect to the other. Hence the maximum

Middle line of panel Middle line of panel

Mid

dle

line

of p

anel

Mid

dle

lin

e o

f p

anel

FIG. 129 FIG. 130

bending stress around the column can be obtained by using formulas (75), previouslyderived for circular plates (see page 61), and combining cases 3 and 8 in Fig. 36.

A more elaborate discussion of the same problem is due to F. Tolke.2 Numericalresults obtained by F. Tolke for a square panel and c/a = 0.1 (Fig. 129) are given inTable 61, together with values of bending moments calculated for the same case on the

1 Such calculations were made by A. Nddai; see his book "Elastische Platten," p.156, Berlin, 1925.

2 F. Tolke, Ingr.-Arch., vol. 5, p. 187, 1934.

basis of the customary theory. It is seen that a rigid connection between slab andcolumn tends to increase numerically the moments on support and to reduce the posi-tive moments of the slab.

TABLE 61. COEFFICIENTS £ FOR CALCULATION OF BENDING MOMENTS M — Pqa2

OF A UNIFORMLY LOADED SQUARE PANEL OF A FLAT SLAB

v = 0.2

Bendingmoment

Mx = MyMx

MyMx = My

Mx

MrMr

Location

x = a/2, y = a/2x = a/2, y - 0x = a/2, y = 0x = 0, y = 0

x = w/2, 2/ - 0s = u/2, y = M/2

r = c

Circular column(Fig. 129)

Rigidconnection

withcolumn

0.02920.0399

-0 .0161

-0 .1682

Customarytheory

0.03230.0494

-0 .0179- 0 . 1 4 3

-0 .0629

Square column(Fig. 130)

Rigidconnection

withcolumn

0.02640.0348

-0 .0146

-0 .0626OO

Customarytheory

0.03210.0487

-0 .0178-0 .131-0 .0803-0 .0480

The same table also gives moments for a flat slab rigidly connected with a columnof a square cross section1 (Fig. 130). The infinitely large stresses occurring at thecorners of columns in this case are of a highly localized character. Practically, theyare limited by a cracking of concrete in tension and a local yielding of the steelreinforcement.

From this discussion we may conclude that (1) the actual values of bending momentsof a flat slab at the columns generally lie between the values given in Table 61 for therigid connection and those given by the usual theory, and (2) circular columns securea more uniform distribution of clamping moments than columns with a square-shapedsupporting area.2

1 See S. Woinowsky-Krieger, / . Appl. Mechanics, vol. 21, p. 263, 1954.1 See T. Haas, "Conception et calcul des planchets a dalles champignon," Paris,

1950. The distribution of stresses in a flat slab has been investigated experimentallyby M. Ro§ and A. Eichinger, Proc. Congr. Concrete and Reinforced Concrete, Lie"ge,1930; by R. Caminade and R. L'Hermite, Ann. inst. tech. bdtiment et trav. publ.,February, 1936; and more recently by J. G. Hageman, Ingenieur, vol. 65, June, 1953.