10. Plates of various shapes
CHAPTER PLATES OF VARIOUS SHAPES 62 Equations of Bending of Plates in Polar Coordinates In the discussion of symmetrical bending of circular plates polar coordinates were used (Chap 3) The same coordinates can also be used to advan- tage in the general case of bending of circular plates If the r and coordinates are taken, as shown in Fig 136a, the relation between the polar and cartesian coordinates is given by the equations = arctan xÙ re = z? +? (a) from which it follows that or _ Ox = r _ ogg or Y Oy 06 = ly _ _ sn6 Táw a r r _ gine (b) 06 2x _ cosé ÔU r? fr Using these expressions, we obtain the slope of the deflection surface of a plate in the z direction as ow du dx or oroax , dw dé | 06 Ox ow Ow = 3, cos — an Sim (c) A similar expression can be written Fia for the 136 slope in the y direction To obtain the expression for curva- ture in polar coordinates the second derivatives are required twice the operation indicated in expression (c), we find đạt Ox? = (jr o0s or — „in _ ows, đạp (FF 39/\ð ở? — gpn 0050 — cos os — Repeating 152 sin 0) 06 sin @cos@ , dwsin? r Tạp dw sin @ cos v6 —y 0°w sin? Flag lt oe 282 OO) _ n a similar manner we obtain d2w sin 6cos , dw cos? r Tp T ngập Đạp = Ow Cw ore IN iy? cos 26 6ö? 0’w > ốp SnỚc0s0 T238 xôu | d*w cos* _ dw sin @ cos | 0?w 283 SHAPES VARIOUS Ox PLATES 06 r? dwcos 26 Ôu sin cos0 r Or 062 r? d%wsin cos r? — g0? (e) (F) With this transformation of coordinates we obtain _Oa?80 Aw aw dw , lodw , dw (9) r or cư 00t ror 2y Tôn + Repeating this operation twice, the differential equation (103) for the leflection surface of a laterally loaded plate transforms in polar coordijates to the following form: saw = (= 4,18 4! a(S tage ts = ot) ar?! por | 72002) \ ar? | rér ` r2 06? (191) When the load is symmetrically distributed with respect to the center of the plate, the deflection w is independent of 6, and Eq (191) coincides with Eq (58) (see page 54), which was obtained in the case of sym- metrically loaded circular plates Let us consider an element cut out of the plate by two adjacent axial planes forming an angle dé and by two cylindrical surfaces of radii r and r + dr, respectively (Fig 136b) We denote and twisting the bending moments acting on the element per unit length by M,, M,, and M,, and To express these take their positive directions as shown in the figure moments by the deflection w of the plate we assume that the z axis coinsides with the radius r The moments M,, M,, and M,, then have the same values as the moments M,, M,, and M,, at the same point, and by substituting @ = in expressions (d), (e), and (f), we obtain -_ 0? 0?w 0? —P lõm +? ray),, + Âm Mẹ = =Đ | M,= ween —D(Ñ + ay? | ” ax? M„=q=?)Ð ö?u Ox See ied "a = — ee low , l 6° oe9 Se) ác + ae) | ror ` r?00? ô%p Dd € droog + dw or? (199) r? 5) In a similar manner, from formulas (108), we obtain the expressions for 284 THEORY OF PLATES AND SHELLS the shearing forces! Q; = —D z= (Aw) and 0, = —D d(Aw) 30 (193) where Aw is given by expression (g) In the case of a clamped edge the boundary conditions of a circular plate of radius a are aw (32) ()„—u, = O = (h) CM.);—a = (2) In the case of a simply supported edge (W)r—a = In the case of a free edge (see page 87) _ (M,) roa _ — V — | _ (2 oM r 0g rt _ ) _~ , (7) The general solution of Eq (191) can be taken, as before, in the form of a sum W= Wo + Wi (k) in which wp is a particular solution of Eq (191) and w;, is the solution of the homogeneous equation | ở? 1a @? 0*w, (5+ zxz + nan) (+ Ow, án Tạ 0?) ¬ ốp) (194) This latter solution we take in the form of the following series :? w, = Rot > m=1 Ri, cos mé + > Ri, sin mé (195) m=] in which Ro, Ri, , Ri, Ry, are functions of the radial distance ronly Substituting this series in Eq (194), we obtain for each of these functions an ordinary differential equation of the following kind: dt ,1d mm (d2R„ , 1dR„ m?R„\ _ (+ sấy — m) (ấn Tự HT et) =o The general solution of this equation for m > is Rm = Anr™ + Burm + Car™t? + Dar-™t* (1) The direction of Q, in Fig 1360 is opposite to that used in Fig 28 This explains the minus sign in Eq (193) | ? This solutlon was given by A Clebsch in his ““Theorie der Elasticität fester Körper,'” 1862 | PLATES For m = and m = OF VARIOUS 285 SHAPES the solutions are (m) Ryo = Ao t+ Bor? + Co log r + Dor? log r m Ry = Aư + Byr? + Cyr + Dyr log r and Substituting ‘Similar expressions can be written for the functions Ri these expressions for the functions R,, and R,, in series (195), we obtain the general solution of Eq (194) The constants Am, Bn, , Dm in Chap | each particular case must be determined so as to satisfy the boundary The solution fo, which is independent of the angle 6, repreconditions Several particular cases sents symmetrical bending of circular plates of this kind have already been discussed in | Plates under a Linearly If a circular plate is acted 63 Circular Varying Load C upon by a load distributed as shown in Fig 137, this load can always be divided into ¢wo parts: (1) a uniformly distributed load of intensity 4(p2 + pi) and (2) a linearly varying load having zero intensity along the diameter CD of the plate and the intensities —py and +p at the ends A and B of the diameter AB has already been have to consider load represented The intensity = Pi The case of uniform load 137 Fic m™ discussed in Chap We IG, here only the nonuniform in the figure by the two shaded triangles.! of the load q at any point with coordinates r and @ 1s r cos | g=T— The particular Po (a) solution of Eq (191) can thus be taken in the following vi form: w= r®> cos A pro’ ae This, after substitution in Eq (191), gives : _ Hence | A= 192D _ pr® cos Wo = Toad (b) As the solution of the homogeneous equation (194) we take only the term of series (195) that contains the function A, and assume wy, = (Agr + Bir? + Cir! + Dir log r) cos (c) This problem has been discussed by W Fligge, Bauingenieur, vol 10, p 221, 1929 286 THEORY OF PLATES AND SHELLS p — Sis Since it is advantageous to work with dimensionless quantities, we introduce, in place of r, the ratio With this new notation the deflection of the plate becomes wW = Wo + Wi = Foor (96 + Ap + Bo + Co! + Dp log p) cos (A) where p varies from zero to unity The constants A, B, in this expression must now be determined from the boundary conditions Let us begin with the case of a simply supported plate (Fig 137) In this case the deflection w and the bending moment M, at the boundary vanish, and we obtain (w)p1 =O (M,),-1 = — (@ At the center of the plate (p = 0) the deflection w and the moment M, must be finite From this it follows at once that the constants C and D in expression (d) are equal to zero The remaining two constants A and B will now be found from Eas (e), which give (W)p-a = qoap (1 + + B) cos = CM,),—ì = pa’ [4(5 + — T09 v) + 2(3 + v)B] cos = Since these equations must be fulfilled for any value of 6, the factors before cos must vanish This gives 4(5 + and we obtain B= 1+A+B=0 ») + 2(5 + v)B =0 _ 2(5 + ») s+ yp Substituting these values in expression of the plate in the following form: y = pee A=/T?+ 3+ pv (d), we obtain the deflection w th [7 + » — (3 + »)p?] eos (f) For calculating the bending moments and the shearing forces we substitute expression (f) in Eqs (192) and (193), from which M, = Be (5 + v)p(1 — p®) cos M, = Sư p[( + z)(1 + 3») — (1 + 5r)(3 + »)p%] cos (9) PLATES OF VARIOUS SHAPES 287 » [2(5 + ») — 9( + »)ø!] cos 0, =53+ Q,= — WS LD p[2(5 -} ») — 3( + v)p?] sin It is seen that (Af,)max occurs at p = 1/+/3 and is equal to 2(5 + v) (MF) The maximum value o= M,) = pa°(5 ms 72 4⁄3 + of J/, occurs at V(b + v1 + 3r)/V380 + 5r)(3 + ) and is equal to CU _ pa? (5 + »)(1 + 3») ve — 7) mm + y The value of the intensity of the vertical reaction at the boundary is! —V The moment — —Q, of this reaction plate (Fig 137) is a /2 PO + with OMr, r 06 pa respect cos to the nos a2 cos 6d0 diameter CD of the = This moment balanees the moment of the load distributed over the plate with respect to the same diameter As a second example, let us consider the case of a circular plate with a free boundary Such a condition is encountered in the case of a circular foundation slab supporting a chimney a moment M will be transmitted to the slab (Tig 1388) Assuming that the reac- tions corresponding to this moment are distributed following a linear law, as shown in the figure, we obtain the same : As the result of wind pressure, > b C¬ Q -7 - " | | | OO -p +— Fig 1388 —+ p kind of loading as in the previous case; and the general solution can be taken in the same form (d) as before The boundary conditions free from forces, are Q22 =0 at the outer boundary of the plate, which (V)nr= (0-352) =0 1s (0 The inner portion of the plate of radius b is considered absolutely rigid It is also assumed that the edge of the plate is clamped along the circle The reaction in the upward direction is taken as positive THEORY 288 AND PLATES SHELLS for p = b/a = B the following boundary Hence of radius b OF conditio must be satisfied: € Op )| p=B ( p ) /p=A Substituting expression (d) in Eqs (¢) and (7), we obtain equations for the determination of the constants: 4(5 + v) + 2( + »)B + 2(1 — v)C + (+ y)D 4(17 + ») + 2(3 + z)B + 2(1 — v)C — (3 — v)D 464 + 282B — 28-?Œ + D From these equations pa C= -2 tt vy) + (3 +») 42 + »)8* - (3 + (3+ ») + (1 J the followin; =0 = =0 (1 — ?)8°*(3 + 8*) + (1 — ?)8' p_~ ta y)8*3 +) — ?n)Ø! Substituting these values in expression (d) and using Eqs (192) and (193), The we can obtain the values of the moments and of the shearing forces The corresponding term constant A does not appear in these equations in expression (d) represents the rotation of the plate as a rigid body with respect to the — Z VM oo Kì ù (a) of the given moment M and the reactions of M ý ‡—à | (P > = gop | Be , _ &P E‹ = g.p|ứ R= ' ` pc — Pb Hy = — ‘sep | r @ + b?)(a? — r?) > b , (a? + r?)(a? — 6?) +B) log gt 18mD|r Pbù3 Da? + b?) log at 11 ¡_ 2° — a?b? ap Rn —= Smữn Pb— 1)xÐ | a me + |ứn (2a? — b?)r? — _ by [2(a? — b?)r À 2a? a‘? (a2 — b?)?r3 — 4r 1)b ma? _— Ro = nm Sm(m _ 1)rD gầm a + (m —— l)r — m(m — 1) br? mm +1+- hog a, p5 '9E p albA — b? Ì » | + _ m—l,, ae meso) | im — 20?— mat + 2, b3m~® ms m —_- |} - tae b2 wild) a 2m | This series is analogous to the series that was used in the case of continuous plates (see p 248) ... derivative and for the PLATES integral of series OF VARIOUS (k) by the use of which SHAPES the moments 301 and the deflections of a plate | ean be calculated The deflection of the plate at the center... Berlin, 1929 PLATES OF VARIOUS SHAPES 293 annular plates For circular plates having no hole and carrying but one eccentric load, simpler solutions can be obtained by the method of complex variables,!... M, = Bqa? 1M = B.qa? Œ) PLATES OF VARIOUS SHAPES 297 Several values of these factors for points in which a, 8, and @; are numerical factors taken on the axis of symmetry of a sector are given in