04. Symmetrjcal bending crcular plates.PDF

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CHAPTER 3

SYMMETRICAL BENDING OF CIRCULAR PLATES

15 Differential Equation for Symmetrical Bending of Laterally Loaded Circular Plates.! If the load acting on a circular plate is symmetrically distributed about the axis perpendicular to the plate through its center, the deflection surface to which the middle plane of the plate is bent will also be symmetrical In all points equally distant from the center of the plate the deflections will be the same, and it is sufficient to consider deflections in one diametral section through the axis of symmetry (Fig 27) Let us take the origin of coordinates

O at the center of the undeflected plate and B

denote by r the radial distances of points

in the middle plane of the plate and by w

their deflections in the downward direction

The maximum slope of the deflection sur-

face at any point A is then equal to —dw/dr,

and the curvature of the middle surface of - O "

the plate in the diametral section rz for a

where ¢ is the small angle between the normal to the deflection surface

at A and the axis of symmetry OB From symmetry we conclude that

1/r, 18 one of the principal curvatures of the deflection surface at A

The second principal curvature will be in the section through the normal

AB and perpendicular to the rz plane Observing that the normals, such

as AB, for all points of the middle surface with radial distance r form a conical surface with apex B, we conclude that the length AB is the radius

of the second principal curvature which we denote by 7; Then, from

1 The solution of these problems of bending of circular plates was given by Poisson;

see ‘‘Memoirs of the Academy,”’ vol 8, Paris, 1829

ol

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Having expressions (a) and (b) for the principal curvatures, we can obtain the corresponding values of the bending moments assuming that relations (37) and (38), derived for pure bending, also hold between these moments and the curvatures.! Using these relations, we obtain

Equations (52) and (53) contain only one variable, w or ¢, which can

be determined by considering the equilibrium of an element of the plate

such as element abcd in Fig 28 cut out from the plate by two cylindrical sections ab and cd and by two diametral sections ad and bc The couple acting

on the side cd of the element 1s

ab of the element Denoting by Q the shearing force per unit length of

(see p 69).

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SYMMETRICAL BENDING OF CIRCULAR PLATES 53

the cylindrical section of radius r, the total shearing force acting on the side cd of the element is Qz đ6, and the corresponding force on the side ab is

equilibrium of the element abcd:

of these equations is simplified if we observe that they can be put in the following forms:

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observing that

Q2rr = [ g2nr đr

ro lở rêu =+ "gr dr dr|rdr\’ dr)/| Bị"

Differentiating both sides of this equation with respect to r and dividing

where 1s a constant of integration to be found later from the conditions

at the center and at the edge of the plate Multiplying both sides of

Eq (c) by r, and making the second integration, we find

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SYMMETRICAL BENDING OF CIRCULAR PLATES _ 55

Circular Plate with Clamped Edges In this case the slope of the deflec-

tion surface in the radial direction must be zero for r = 0 and r = a

Hence, from Eq (59),

sion for the slope:

Having expression (61) for the slope, we obtain now the bending moments M, and M; by using expressions (52) and (53), from which

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Substituting 7 = a in these expressions, we find for the bending moments

at the boundary of the plate

From expressions (65) and (66) it is seen that the maximum stress is at

_ ga _

= Tepa av (7)

Adding this to the deflections (62) of the clamped plate, we find for the plate with a simply supported edge

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SYMMETRICAL BENDING OF CIRCULAR PLATES od

w = g(a* — r?) ũ + ve _ ) (67)

64D 1+ yp

Substituting r = 0 in this expression we obtain the deflection of the plate

(Or) max — (Ø¿)max — h- -_ 8h? (71)

To get the maximum stress at any distance r from the center we must add to the stress calculated for the plate with clamped edges the con-

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corresponding to the pure bending shown in Fig 300 The same stress 1s obtained also from Fig 29 by measuring the ordinates from the horizontal! axis through O; It may be seen that by clamping the edge a more favorable stress distribution in the plate is obtained

17 Circular Plate with a Circular Hole at the Center Let us begin

with a discussion of the bending of a plate by the moments M, and M;

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of the plate must be considered Assume, for example, that the plate in lig 31 is supported along the outer edge Then w = QO for r = a, and

we find, from (6),

Cia? a*(a®M, — 67M) _4 ~~ 21 + ») D(a? — 6?)

C3; =

In the particular case when M, = O we obtain

As a second example we consider the case of bending of a plate by shearing forces Qo uniformly distributed along the inner edge (Fig 32) The shearing force per unit length of a

where P = 2rbQ, denotes the total load _——_

applied to the inner boundary of the

plate Substituting this in Eq (57) and integrating, we obtain

and +0 = =p (Vor 7 og — -1) 4 C; log a + C3 (f)

The constants of integration will now be calculated from the boundary conditions Assuming that the plate is simply supported along the outer edge, we have

Tor the inner edge of the plate we have

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Substituting expressions (e) and (f) in Eqs (g) and (h), we find

we find the slope and the deflection at any point of the plate shown in Fig 32 For the slope at the inner edge, which will be needed in the further discussion, we obtain _

10 =

Meee Que does not affect the deflection of the plate

bk " Combining the loadings shown in Figs 31

ra and 32, we can obtain the solution for the

case of a plate built in along the inner edge M/ and uniformly loaded along the outer edge Fie 2a (Fig 33) Since the slope at the built-in

edge is zero in this case, using expressions (72) and (j), we obtain the following equation for determining the bending moment M;, at the built-in edge:

Da — \(@ — b) (ï+i+7z)* Sr xp |2le7 —

~T+z TT ø— ba '98 1a)

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SYMMETRICAL BENDING OF CIRCULAR PLATES a 61 from which

constants of integration are given by expressions (7)

By using the same method of superposition we can obtain also the

solution for the case shown in Tig 34, in which the plate is supported

along the outer edge and carries a uniformly distributed load In this

case we use the solution obtained in the previous article for the plate

without a hole at the center Considering the section of this plate cut by the cylindrical surface of radius b and perpendicular to the plate, we find that along this section there act a shearing force Q = mqb?/2rb = qb/2

and a bending moment of the inten-

Hence to obtain the stresses and de- Fig 34

flections for the case shown in Fig 34,

we have to superpose on the stresses and deflections obtained for the plate

without a hole the stresses and deflections produced by the bending moments and shearing forces shown in Fig 35 These latter quantities are obtained from expressions (72), (73), (e), and (f), with due attention

being given to the sign of the applied shears and moments

Several cases of practical importance

1 The calculations for cases 1 to 8 inclusive were made by A M Wahl and G Lobo, Trans ASME, vol 52, 1930 Further data concerning symmetrically loaded circular plates with and without a hole may be found in K Beyer, ‘‘Die Statik im Stahlbe- tonbau,”’ 2d ed., p 652, Berlin, 1948 |

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TABLE 3 COEFFICIENTS k AND k; IN Eas (75) AND (76) FOR THE TEN

Cases SHOWN IN Fic 36

0.519 0.491 0.0183 0.0313 0.0062 0.0064 0.414 0.0249 0.0242 0.0139

1.48 2.04 1.04 0.74 0.71

0.405 1.440 0.753 0.454 0.480

0.672 0.902 0.0938 0.1250 0.0329 0.0237 0.664 0.0877 0.0810 0.0575 0.703

1.880 1.205 0.673 0.657

0.734 1.220 0.295 0.291

|0.110

0.062 0.824 0.209 0.172 0.130

2.17 4.30 2.99 1.45 2.23

2.08

0.9355

1.514 1.021 0.710

0.724 1.300 0.448 0.417 0.179

0.092 0.830 0.293 0.217 0.162

2.34 5.10 3.69 1.59 2.80 1.13 2.19 1.745 1.305 0.730

0.704 1.310 0.564 0.492 0.234

0.114 0.813 0.350 0.238 0.175

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The maximum deflections in the same cases are given by formulas of the type

max = kia Eh? or max = ki: Eh? (76)

The coefficients k, are also given in Table 3

When the ratio a/b approaches unity, the values of the coefficients k and

k, in Eqs (75) and (76) can be obtained with sufficient accuracy by considering a radial strip as a beam with end

conditions and loading as in the actual plate 1 b> |

The effect of the moments M; on bending is f | 3 | 1

18 Circular Plate Concentrically Loaded Mm =m,

Dividing the plate into two parts as shown M | | h\

in Fig 37b and c, it may be seen that the : , (c)

inner portion of the plate is in the condition Fic 37

of pure bending produced by the uniformly

distributed moments M, and that the outer part is bent by the moments

M, and the shearing forces Q; Denoting by P the total load applied,

The inner portion of the plate is bent to a spherical surface, the curvature

of which is given by expression (46) Therefore the corresponding slope

at the boundary is

dw _ M 1Ö

(7) 7 D (1 + v) (c)

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Equating expressions (b) and (c), we obtain

1+ »)P log -

( (1 — v)P(a? — 6?)

(W) rap ~ gep | 6= b3) ( +54 Ty a? ) + 20 log 1 (e)

To find the deflections of the inner portion of the plate, we add to the deflection (e) the deflections due to pure bending of that portion of the plate In this manner we obtain

¬ ae 11—va?—b? 1 by

b b? — +? lu — Pa — py (+) log :

G | <- b ->| ) tained by superposing on the deflec-

outer edge of the plate (Fig 38) and

of such a, magnitude that the slope of the deflection surface at that edge

is equal to zero From expression (77) the slope at the edge of a simply supported plate is"

(3) _ — 4rmD1-+»w a (J)

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The slope produced by the moments Mz is

Deflections produced by this moment are

Adding these deflections to the deflections (77) and (78) we obtain for the

outer portion of a plate with a built-in edge

bounded by a circle of radius c (Fig 39) Expression (77) is used to

obtain the deflection at any point of the unloaded portion of the plate

(a>*r>c) The deflection produced by an elementary loading dis-

tributed over a ring surface of radius 6 and width db (see Fig 39) is obtained by substituting P = 2mbq db in that expression, where g 1s the intensity of the uniform load Integrating the expression thus obtained with respect to b, we obtain the deflection

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The maximum bending moment is at the center and is found by using

expression (d) Substituting 2rbq db for P in this expression and inte-

zrating, we find

_ c(l1—va?—b? 1+, Ò Mow = 4 |, ( 4 ¬5 5 log 2) bad

where, as before, P denotes the total load xec?q.*

Expression (81) is used to obtain the bending moments M, and M; at any point of the unloaded outer portion of the plate Substituting this expression in the general formulas (52) and (53), we find

_ (+ »)P a, (l—»v)Pc’?/l 1

M, = 4m log T + 16x (7 2) (84)

— '2n2

M.= £| (+ ») log 2 +1-—- | - Cope +a) (85)

* This expression applies only when c is at least several times the thickness h The case of a very small c is discussed in Art 19.

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The maximum values of these moments are obtained at the circle r = c, where

_(4+»)P, a, 0 — )P@ — )

_ ?P a _ — OU — ø)P(œ + c°) -

NM, = An lí + v) log ° + 1 / “ren 16xa2 (87)

The same method of ealeulating dofleetions and moments can be used also for any kind of symmetrical loading of a circular plate

The deflection at the center of the plate can easily be calculated also for any kind of unsymmetrical loading by using the following consideration Owing to the complete symmetry of the plate and of its boundary con-

ditions, the deflection produced at its center by an isolated load P depends only on the magnitude of the load and on its radial distance from the

center This deflection remains unchanged if the load P is moved to

another position provided the radial distance of the load from the center

remains the same ‘The deflection remains unchanged also if the load P

is replaced by several loads the sum of which is equal to P and the radial

distances of which are the same as that of the load P From this it

follows that in calculating the deflection of the plate at the center we can replace an isolated load P? by a load P uniformly distributed along a circle the radius of which is equal to the radial distance of the isolated load lor the load uniformly distributed along a circle of radius 6 the deflection

at the center of a plate supported at the edges is given by Eq (78) and is

P | 34+ y

(W)rmo = | Ty)

This formula gives the deflection at the center of the plate produced by

an isolated load P at a distance b from the center of the plate Having this formula the deflection at the center for any other kind of loading

can be obtained by using the method of superposition.! It should be noted that the deflections and stresses in a circular plate with or without

a hole can be efficiently reduced by reinforcing the plate with either con- centric? or radial ribs In the latter case, however, the stress distribution

is no longer symmetrical with respect to the center of the plate

19 Circular Plate Loaded at the Center The solution for a concen- trated load acting at the center of the plate can be obtained from the

(a2 — b?) — b? log A (i)

1'This method of calculating deflections at the center of the plate was indicated by Saint Venant in his translation of the ‘‘Théorie de ]’élasticité des corps solides,’’ by Clebsch, p 363, Paris, 1883 The result (7) can also be obtained by applying Max- well’s reciprocal theorem to the circular plate

This case is discussed by W A Nash, J Appl Mechanics, vol 15, p 25, 1948 See also C B Biezeno and R Grammel, ‘“‘Technische Dynamik,”’ 2d ed., vol 1, p 497, 1953.

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