13. Bending of plates under the combined action of lateral loads and forces in the middel plane of the plate

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BENDING OF PLATES UNDER THE COMBINED ACTION OF LATERAL LOADS AND FORCES IN THE MIDDLE

PLANE OF THE PLATE

90 Differential Equation of the Deflection Surface In our previous discussion it has always been assumed that the plate is bent by lateral loads only If in addition to lateral loads there are forces acting in the middle plane of the plate, these latter forces may have a considerable effect on the bending of the plate and must be considered in deriving the corresponding differential equation of the deflection surface Proceed- ing as in the case of lateral loading (see Art 21, page 79), we consider the equilibrium of a small element cut from the plate by two pairs of planes parallel to the zz and yz coordinate planes (Fig 191) In addition to the forces discussed in Art 21 we now have forces acting in the middle plane of the plate We denote the | magnitude of these forces per unit length by mS N., N,, and Nz, = Ny,z, aS shown in the figure Projecting these forces on the x and y axes and assuming that there are no body forces or

| | Net lì on tangential forces acting in those directions at | I ° , the faces of the plate, we obtain the following | Ny equations of equilibrium: = pene ở ONz , 9Ne _ g Nx "SN + ôNxy dx OX Oy (216) —+>> y ò ỳ 1` 0 \ ON zy + ONy =0 NOT Nyx + a dy Ox oy

` Ny + Fay These equations are entirely independent of (b) the three equations of equilibrium considered Fic 191 in Art 21 and can be treated separately, as

will be shown in Art 92

In considering the projection of the forces shown in Fig 191 on the z axis, we must take into account the bending of the plate and the resulting small angles between the forces Nz and N, that act on the opposite sides of the element As a result of this bending the projection

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of the normal forces N, on the z axis gives

N dz) & + —, dw , ð? 5x5 is) dy

After simplification, if the small quantities of higher than the second order are neglected, this projection becomes —N: dụ 2+ ( n, 2 ® ae dy oN oe (a) In the same way the projection of the normal forces N, on the z axis gives nN, 2 ae oat dy + ae ae dy (b)

Regarding the projection of the shearing forces N,, on the z axis, we observe that the slope of the deflection surface in the y direction on the two opposite sides of the element is dw/dy and dw/dy + (d?w/dx dy) dz Hence the projection of the shearing forces on the z axis is equal to

Ow, 4, ON xy OW

Nev ae ay YF ox ay OY

An analogous expression can be obtained for the projection of the shearing forces N, = Nz, on the z axis The final expression for the projection of all the shearing forces on the z axis then can be written as

oN Ow ¡ vđy + ane ow

2N dx dy + oy 72 ox dx dy (c)

“3s iy

Adding expressions (a), (b), and (c) to the load g dx dy acting on the element and using Eqs (216), we obtain, instead of Eq (100) (page 81), the following equation of equilibrium: OM, (07M, , OA, - 0?w dx? mm ^ ôU? ~ - (4 + NSS +N Ge + 2N san ) Substituting expressions (101) and (102) for A/,, AL,, and Äƒz„, we obtain aw dtw dw oat T ” Ga? ay? * ays | q+ N.S WIN Woy, Jw (217) mm tăm '” Ox Oy

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If there are body forces! acting in the middle plane of the plate or tangential forces distributed over the surfaces of the plate, the differential equations of equilibrium of the element shown in Fig 191 become ON ONey + —— Xx = Ox oY + 0 (218) ONey ON, =0 Ox oy —

Here X and Y denote the two components of the body forces or of the tangential forces per unit area of the middle plane of the plate

Using Eqs (218), instead of Eqs (216), we obtain the following differential equation? for the deflection surface:

“— + N,— +2N.,—— —-X— - + vaya T ” ax day ox ay

(219)

04w 0° 04w 1 ð? g?tu ö? ow Ow

Equation (217) or Eq (219) together with the conditions at the boundary (see Art 22, page 83) defines the deflection of a plate loaded

° *T—_x laterally and submitted to the action of forces in the f— ime 3 middle plane of the plate

va -* P 91 Rectangular Plate with Simply Supported „ _Ÿ Edges under the Combined Action of Uniform f€ ~-~~@ -==== >| Lateral Load and Uniform Tension Assume y that the plate is under uniform tension in the tra, 192 z direction, as shown in Fig 192 The uniform lateral load g can be represented by the trigonometric series (see page 109) Mar nny oy > 9m —— sin b (a) Equation (217) thus becomes O4w 04w 04w N, 0?w ant | 7 3gt ay? * ay! D oz? = a » » — sin — sin =e (b)

This equation and the boundary conditions at the simply supported edges

1 An example of a body force acting in the middle plane of the plate is the gravity force in the case of a vertical position of a plate

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will be satisfied if we take the deflection w in the form of the series

Mrz ney

w= Omn Sin —— sin |= (c)

Substituting this series in Eq (b), we find the following values for the coefficients Amn:

Onn = 16g - (d)

6n -

Damn a? 6? a’ Da?

in which m and n are odd numbers 1, 3, 5, , and dm, = O1f m or n or both are even numbers Hence the deflection surface of the plate is

1 _ mrn nry

5 3 >] Sin —— sin —- (e)

m n 2 Nm | a b

“ (2 + is) T a’ Da?

Comparing this result with solution (131) (page 110), we conclude from the presence of the term N,m?/x?Da? in the brackets of the denominator that the deflection of the plate is somewhat diminished by the action of the tensile forces N, This is as would be expected

By using M Lévy’s method (see Art 30) a solution in simple series may be obtained which is equivalent to expression (e) but more con- venient for numerical calculation The maximum values of deflection and bending moments obtained in this way! for v = 0.3 can be represented in the form qb" a Eh3 (TM) max — 8q? (M,) max —= 8g)? (f) Wmax — The constants a, 8, and 8: depend upon the ratio a/b and a parameter _ N,b? Y ™ 42D

and are plotted in Figs 193, 194, and 195

If, instead of tension, we have compression, the force N, becomes

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negative, and the deflections (e) become larger than those of the plate bent by lateral load only It may be seen also in this case that at cer- tain values of the compressive force Nz the denominator of one of the terms in series (e) may vanish This indicates that at such values of N, the plate may buckle laterally without any lateral loading

92 Application of the Energy Method The energy method, which was previously used in discussing bending of plates by lateral loading (see Art 80, page 342), can be applied also to the cases in which the 0.14 _——— _ O12 0.10 / 1/1 „ Z 25 — \AA AL ™ ` 0.04 Wok 0.02 ` - me W ® g — b Fia 193

lateral load is combined with forces acting in the middle plane of the plate To establish the expression for the strain energy corresponding to the latter forces let us assume that these forces are applied first to the unbent plate In this way we obtain a two-dimensional problem which can be treated by the methods of the theory of elasticity.1 Assuming that this problem is solved and that the forces N., N,, and N., are known at each point of the plate, the components of strain of the middle plane of the plate are obtained from the known formulas representing Hooke’s

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law, 012., 1 1 hE (N, — vN,) Cụ — hi (Ny — „N,) Ne, (a) Yzu — hG es = The strain energy, due to stretching of the middle plane of the plate, 1s then | V¥i= AS J (Nez + Nyey + N xyz) du dy | = | [N2 + N2 — 2vN,N, + 2(1 + r)N?, | dà dụ (220) where the integration is extended over the entire plate

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0.02 we TT | —- 0.06 0.05 YY Ree 0.04 ——— _ F—— _ † mm B; 0.03 yal —————— YO _— ` _—— —— „z3 L7 —~ TT” Nh WwW + 0 b tg 195

however, we have to take it into consideration, since this small strain in combination with the finite forces Nz, N,, Nz, may add to the expression for strain energy some terms of the same order as the strain energy of bending The x, y, and z components of the small displacement that a point in the middle plane of the plate experiences during bending will be ak 4x15 0 A x | Y —9; = 9W dx K Ox U Fh~ \ z ut gu dx Fig 196 , and w, respectively Considering a linear element AB of that plane in the z direction, it may be seen from Fig 196 that the elongation of the element

due to the displacement u is equal to

(du/dx) dx The elongation of the same element due to the displacement w is 4(dw/dx)? dx, as may be seen from the com- denoted by u, 2,

parison of the length of the element A,B, in Fig 196 with the length of its projection on the z axis

tion of an element taken in the middle plane of the plate is

€

Thus the total unit elongation in the z direc-

Ou 1 fow\?

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Similarly the strain in the y direction is

,_ v , 1 (aw)?

| fy = » + 5 (0y) (222)

Considering now the shearing strain in the middle plane due to bending, we conclude as before (see Fig 23) that the shearing strain due to the displacements u and v is du/dy + dv/dx To determine the shearing strain due to the displacement w we take two infinitely small linear elements OA and OB in the x and y directions, asshown in Fig 197 Because of displacements in the z direction these elements come to the positions O1A; and O,B; The difference between the angle 7/2 and the angle A,0,B, is the shearing strain corresponding to the displacement w To determine this difference we con-

sider the right angle B;O+4;, in which 0 dx

B.O, is parallel to BO Rotating the 3 | plane B.O,A, about the axis 0,A, by 0, the angle dw/dy, we bring the plane y

B,0,Ai into coincidence with the 2, A Ow gs

plane B,O,A;* and the point B: to posi- % ox

tionC Thedisplacement B.C'is equal cá + ow to (dw/dy) dy and is inclined to the ver- i wy Ol

tical B.B, by the angle dw/dz Hence Ox be ; BiC is equal to (dw/dx)(dw/dy) dy, Fic 197 and the angle CO,B,, which repre-

sents the shearing strain corresponding to the displacement w, is (dw/dx)(dw/dy) Adding this shearing strain to the strain produced by the displacements wu and v, we obtain >< > = { ds \ ——>ke-~—=— a D> \ a, >< ——> a = Ow Ow Vey int hp + dx by (223)

Formulas (221), (222), and (223) represent the components of the additional strain in the middle plane of the plate due to small deflections Considering them as very small in comparison with the components «:, €,, and yz, used in the derivation of expression (220), we can assume that the forces N., Ny, Nz, remain unchanged during bending With this assumption the additional strain energy of the plate, due to the strain produced in the middle plane by bending, is

Vo = Sf(Nae, + Nye, + Nevyi,) dx dy

Substituting expressions (221), (222), and (223) for «¿, «,, and y;,, we

* The angles dw/dy and dw/dzx correspond to small deflections of the plate and are

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finally obtain

ve= ff] N + vy 28 + Na (% + án) | đe dy Oy Ox

-LIxŒ}+w(3) ceEB]zx m

It can be shown, by integration by parts, that the first integral on the right-hand side of expression (224) is equal to the work done during bend- - ing by the forces acting in the middle plane of the plate Taking, for example, a rectangular plate with the coordinate axes directed, as shown in Fig 192, we obtain for the first term of the integral

b a Ou b a b a ON

EM: = | du — [fous dxd

Proceeding in the same manner with the other terms of the first integral in expression (224), we finally find

Ov Ou Ov

Jo Lr [NS + esp tM (Gy + az) Jee a b

=Í (val + [Nay )+ (|! + Nay ) as

0

b fa „+ ÔN - aN 0N,

“LÍ, u (SE + vt) av ay — [ I 0 ( su) ae dy The first integral on the right-hand side of this expression is evidently equal to the work done during bending by the forces applied at the edges = Qand z = a of the plate Similarly, the second integral is equal to the work done by the forces applied at the edges y = Oandy = b The last two integrals, by virtue of Eqs (218), are equal to the work done during bending by the body forces acting in the middle plane These integrals each vanish in the absence of such corresponding forces

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uced by the lateral forces is

Ow Ow Ow OW

=zJJ|x v-(2) + w, (3) + OMe ox By sy | dụ

g?› ? 0?w 07w 6? \/

+ JJ Z + am) —?0 95 aự — (say) |p mem (225)

.pplying the principle of virtual displacement, we now give a variation 6w o the deflection w and obtain, from Eq (225),

Ow Ow aw aw

r= 58 ff [ve GB) +055) + 20 Se ay | eae 0*w 0N, tw atu aw \’

+72 || (Ss +=) “ag =» [Be ay? - (25,) |} egy (226)

‘he left-hand side in this equation represents the work done during the irtual displacement by the lateral load, and the right-hand side is the orresponding change in the strain energy of the plate The application f this equation will be illustrated by several examples in the next article 93 Simply Supported Rectangular Plates under the Combined Action f Lateral Loads and of Forces in the Middle Plane of the Plate Let us egin with the case of a rectangular plate uniformly stretched in the direction (ig 192) and carrying a concentrated load P at a point with oordinates and 7 The general expression for the deflection that satis- es the boundary conditions is co ©Ò TW] sin sỶ b (a) _ Marx w= Amn Sin m=1,2,3, n=1,2,3,

‘o obtain the coefficients ann in this series we use the general equation 226) Since N, = N., = 0 in our case, the first integral on the right- and side of Eq (225), after substitution of series (a) for w, is

©© 0

"mm m=l1n=1

‘he strain energy of bending representing the second integral in Kq 225) is [see Eq (d), page 343]

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To obtain a virtual deflection dw we give to a coefficient @m,n, 2 increase 5Qm,n, The corresponding deflection of the plate is

MyWr sin Nyw.H y b | The work done during this virtual displacement by the lateral load P is ồu = ồđ„,„, SIn (5.83 ‹ố — Tù17TTỊ 7 Sn (d) The corresponding change in the strain energy consists of the two terms which are sã I i ) dx dy 2-2 ~ i (3 Na.” cok oman, Ti Na "an bdmin, (Ê) aV Í 2\2 and 5V = OV sa OOm,n, = 4 ab Da» (Ga +3) Ôn, P6Qm,n, SIN Substituting expressions (d) and (e) in Eq (226), we obtain mE Minn QÙ mr? n Tp =7 Ms tmm Cạn Oman 2\2 + © Dames (7 +1) 5Qm,n, ~ My P6dm,n, S10 from which Mart nin 4P sin ame sin Oem, = —— — Ø abz*D l + m) + mir ?a?Ù

Substituting these values of the coefficients an,n, in expression (a), we find the deflection of the plate to be 2 00 mr & NTN _ 4P sin a sin _b_ TH _ _ TLTỤ = Tp me na\? mE N sin — — sin [= (9) mal n=l (= +4) T 722D

If, instead of the tensile forees W;, there are compressive forces of the same magnitude, the deflection of the plate is obtained by substituting

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The smallest value of Nat which the denominator of one of the terms in expression (h) becomes equal to zero is the critical value of the compressive force N, It is evident that this critical value is obtained by taking n = 1 Hence

2772 2 1 2 2D 2

m? b? a

where m must be chosen so as to make expression (227) a minimum Plotting the factor

mb a \?

k= (= + =)

against the ratio a/b, for various integral values of m, we obtain a system of curves shown in Fig 198 The portions of the curves that must be 10 3 m=l 8 7 kề 5 4 3 2 1 0 O 1 4⁄2 2 q 3 4 5 b Fic 198

used in determining k are indicated by heavy lines It is seen that the factor k is equal to 4 for a square plate as well as for any plate that can be subdivided into an integral number of squares with the side 6 It can also be seen that for long plates k remains practically constant at a value of 4.* Since the value of m in Eq (227) may be other than 1 for oblong plates, such plates, being submitted to a lateral load combined with compression, do not generally deflect! in the form of a half wave in the direction of the longer side of the plate If, for instance, a/b = 2, 4, the respective elastic surface becomes markedly unsymmetrical with respect to the middle line x = a/2 (Fig 192), especially so for values of N, close to the critical value (Nz) cr

By using the deflection (g) produced by one concentrated load, the

* A more detailed discussion of this problem is given in S Timoshenko, ‘‘Theory of Elastic Stability,’’ p 327, 1936

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deflection produced by any lateral load can be obtained by superposition Assuming, for example, that the plate is uniformly loaded by a load of intensity g, we substitute q dé dn for P in expression (g) and integrate the expression over the entire area of the plate In this way we obtain the same expression for the deflection of the plate under uniform load as has already been derived in another manner (see page 381)

If the plate laterally loaded by the force P is compressed in the middle plane by uniformly distributed forces N, and N,, proceeding as before we obtain — = sin mene i sin nen 4P a b abx4D m2 n?\° m?N, n2N, m=1 n=] az + có — mr nay w= sin — sin > (2) b2 ra2D x?b?D The critical value of the forces Nz and WN, is obtained from the condition! m?(Na)er , 22(Ny)cr — (m2, n\? aD + wD ~\art 0)

where m and n are chosen so as to make N, and N, a minimum for any given value of the ratio N,/N, In the case of a square plate submitted to the action of a uniform pressure p in the middle plane we have a = b

and N, = N, = p Equation (7) then gives 2D | Der = “ (m2 + 17) min | (È) The critical value of p is obtained by taking m = n = 1, which gives 2m?D cr — a? (228)

In the case of a plate in the form of an isosceles right triangle with simply supported edges (lig 161) the deflection surface of the buckled plate which satisfies all the “nes conditions is?

WX Lí

w= alsin — sin 2 ong +- sin 272 ~ sin Tử

a a a

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94 Circular Plates under Combined Action of Lateral Load and Tension or Com-

pression Consider a circular plate (Fig 199) submitted to the simultaneous action

of a symmetrical lateral load and a uniform compression N, = N; = N in the middle

plane of the plate Owing to the slope ¢ of the deformed plate (Fig 27) the radial compression N gives a transverse component N dy/dr which we have to add to the

shearing force Q (Fig 28) due to the lateral load

Hence the differential equation (54) becomes ở? ld k? ] ry i%4(2-4)6- -£ @ dr? r dr a? r? D in Which _ Na? k2 D 6)

In the case of a circular plate without a hole! the

solution of Eq (a) is of the form k: | p= CW; (*) + Yo (c) Fic 199

a

where J; is the Bessel function of the order one, go a particular solution of Eq (a) depending on Q, and C; a constant defined by the boundary conditions of the plate Let us take as an example a rigidly clamped? plate carrying a uniform load of

intensity g Then, as a particular solution, we use _ qra? _ qr f6 92D 9N and therefore dw kr qr = — = CJ¡| — Ì —_—— d | ° dự mỹ Œ) 2N “ It follows, by integration, that Cia kr qr? = — J,{ — }] —- — C „ww (=) AN TC (e)

where Jo 1s the Bessel function of the order zero and C2 a second constant Having calculated Ci from the condition ¿ = 0 onr = a, and C2 from the condition w = 0 on r = a, we obtain the final solution?

kr

(5) 20 | gear =

nS 2:57 1(k) D —— 48D — Ớ)

The deflections (ƒ) beeome infinite for J¡(k) = 0 Denoting the zeros of the func-

tion J; in order of their magnitude by ji, j2, we see that the condition k = j; 1 In the case of a concentric hole a term proportional to a Bessel function of second

kind must be added to expression (c) The inner boundary must be submitted then

to the same compression JN, or else the problem becomes more complex because of the inconstancy of stresses N, and N:

? The case of an elastic restraint without transverse load has been discussed by H Reismann, J Appl AMfechanics, vol 19, p 167, 1952

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defines the lowest critical value

No = | (9)

of the compressive stress N Now, for the function Ji(k) we have the expression

100 =8(1—#) (1-8) 0

in which 7; = 3.83171, j2 = 7.01559, As k <j7¡ we can negleet the terms k2/j? beginning with the second parentheses Observing, furthermore, that k? _N peo Ne by virtue of Eqs (b) and (g) we have, approximately, k Ji(k) = 2 (1 — @) (7) N where a= N (9) Making use of the expression (i), it can be shown that, approximately,’ We a

where w, is the deflection due to the load q alone Cases with other boundary conditions and other laws of distribution of the lateral load may be handled in like manner In the general case of a symmetrical lateral load combined with compression we can put, approximately, for the center of the plate (r = 0) (wq)o l1 —œ (k) (dw) _ (aw) _ 1 bee (1 dus 7 dr dr? 0 l —œ r dr 0 and on the boundary (r = a) Ldw\ _1+ee (1 dw, r dr a l1 —œ r dr a (1) dw _ l+c”a dw, dr? 1—ea dr? a

where w, relates to a plate carrying the given lateral load alone and @ = N/Ner has the following meaning: Wo = Na? For a simply supported plate: a = 190D " (m) F or a clamped plate: ] d plate: © = 468D = Na

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TABift 82 VALUES OF CONSTANTS IN APPROXIMATE Expressions (k) AND (1) y = 0.3 ee Boundary

Case Load distribution conditions ^ Constants 1 | Uniform edge couples Simply supported | co = 0.305 c’ = —0.270 c” = —1.219 2 | Uniform load Simply supported | co = 0.0480 c =c” = —0.0327 3 Clamped Co = 0.308 c” = —0.473 4 | Central uniform load over area} Simply supported _ 2.153 of radius ea c= —It+ 1 — 1.3Ìn< c =c” = 0.205 5 Clamped 1.308 Co = — — Ìn e c’’ = 0.0539 the former value being valid for vy = 0.3 The values of the constants co, c’, and c’’ are given in Table 81 |

If the circular plate is subjected to a lateral load combined with a uniform tension N, instead of compression, then we have, approximately, (w) _ (We) ro r=0 1 + œ where a is the absolute value of )/W; As for the curvatures, a factor 1 l1 +(1+c)œ

instead of the factor (1 + ca)/(1 — a) must be used in expressions (k) and (1), the

constant c having the meaning of co, c’, and c’’, respectively

95 Bending of Plates with a Small Initial Curvature.! Assume that a plate has some initial warp of the middle surface so that at any point there is an initial deflection wo which is small in comparison with the thickness of the plate If such a plate is submitted to the action of transverse loading, additional deflection w, will be produced, and the total deflection at any point of the middle surface of the plate will be Wo + Ww; In calculating the deflection w,; we use Eq (103) derived for flat plates This procedure is justifiable if the initial deflection wo is

1 See S Timoshenko’s paper in Mem Inst Ways Commun., vol 89, St Petersburg,

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small, since we may consider the initial deflection as produced by a fictitious load and apply the principle of superposition.! If in addition to lateral loads there are forces acting in the middle plane of the plate, the effect of these forces on bending depends not only on w, but also on wo To take this into account, in applying Eq (217) we use the total deflection WwW = Wo + 0u On the right-hand side of the equation It will be remem- bered that the left-hand side of the same equation was obtained from expressions for the bending moments in the plate Since these moments depend not on the total curvature but only on the change in curvature of the plate, the deflection w, should be used instead of w in applying that side of the equation to this problem Hence, for the case of an initially curved plate, Eq (217) becomes 01% 01 d4w, I 0?2(wo + Wi) ô+4 Tổ dx? dy? T dyt D la + Ne dx? 0?(wo + 401) dy? - 8?(wo + un) + 2N,, Het | (230) + Ny

It is seen that the effect of an initial curvature on the deflection is equivalent to the effect of a fictitious lateral load of an intensity

g7 Wo ở? Wo 07Wo

“9x2 ¥ “Oy? *’ Ox Oy

Thus a plate will experience bending under the action of forces in the xy plane alone provided there is an initial curvature

Take as an example the case of a rectangular plate (Fig 192), and assume that the initial deflection of the plate is defined by the equation

_ TỦ Vì TY

t0 = đi Sin ~~ sin = (a)

If uniformly distributed compressive forces Nz are acting on the edges of this plate, Eq (230) becomes

311 ‘ d2 11 ¬ 1 œ1? WH WY 07w,

0+2 +2 dx? dy? + éyt D (W g? sin a sin b Ne aye 0) Let us take the solution of this equation in the form + N + 2N 72 Ty w, = A sin a sin — b (c) Substituting this value of w, into Eq (b), we obtain auiN; A= r?D a2\? PT

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With this value of A expression (c) gives the deflection of the plate pro- Juced by the compressive forces Nz Adding this deflection to the initial Jeflection (a), we obtain for the total deflection of the plate the following 2xpression: " — Q11 TU wy w= Wo + Wi Toe SP | sn FS (đ) in which a r?D We a? (e) a (: + ‘) The maximum deflection will be at the center and will be Winx = TH l—ea (f)

This formula is analogous to that used for a bar with initial curvature.’ In a more general case we can take the initial deflection surface of the rectangular plate in the form of the following series:

0 ce

mnrx ny

Wo = Amn SI —— sin = b (g)

m=1 n=1

Substituting this series in Iq (230), we find that the additional deflection at any point of the plate is co co - mar TTÌ WwW, = » » Omn SIN — Sin — (h) m=1 n=1 in which Oman — 2 D = 2 (2) 7 (m +") _N, a? mm b?

It is seen that all the cocfficients ba, increase with an increase of N; Thus when N, approaches the critical value, the term in series (h) that corresponds to the laterally buckled shape of the plate [see Eq (227)] becomes the predominating one We have here a complete analogy with the case of bending of initially curved bars under compression

The problem can be handled in the same manner if, instead of compression, we have tension in the middle plane of the plate In such a cuse it is necessary only to change the sign of N, in the previous equations Without any difficulty we can also obtain the deflection in the case when there are not only forces N, but also forces N, and Nz, uniformly distributed along the edges of the plate

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