03. Pure bending ò plates.PDF

03. Pure bending ò plates.PDF

CHAPTER PURE BENDING OF PLATES Slope and Curvature of Slightly Bent Plates In discussing small deflections of a plate we take the middle plane of the plate, before bending occurs, as the xy plane During bending, the particles that were in the zy plane undergo small displacements w perpendicular to the zy plane and form the middle surface of the middle surface are called deflections Taking a normal section of the plate to the’xz plane (Fig 16a), we find plate These displacements of the of a plate in our further discussion parallel that the slope of the middle surface in the x direction ist; = 0w/dx Inthesame manner the slope in the y direction is 1, = dw/dy Taking now any direction an in the zy plane (Fig 16b) making an angle a with the z axis, we find that the difference in the deflections of the two adjacent points a and a, in the an direction is m dw = SY de + dy dw Owdx , dwdy _ dw dt dn ARE n | 4£ y and that the corresponding slope is on k2» dy dn de CSXT ow (b) ; -P aN Fic 16 |, SD a (a) To find the direction a; for which the slope is a maximum we equate to zero the derivative with respect to a of expression obtain tan a; = ow ay (a) /ow or In this way we (b) Substituting the corresponding values of sin a; and cos a; in (a), we obtain for the maximum slope the expression Ow _ J aw\’ dw (3) = (2) + G7) \’ ) By setting expression (a) equal to zero we obtain the direction for which 33 SHELLS AND PLATES OF THEORY 34 The corresponding angle a2 is deter| the slope of the surface is zero mined from the equation tan ag = — = (d) From Eqs (b) and (d) we conclude that tan a; tan ao = — Ì ‘which shows that the directions of zero slope and of maximum slope are perpendicular to each other In determining the curvature of the middle surface of the plate we In such a case observe that the deflections of the plate are very small the slope of the surface in any direction can be taken equal to the angle that the tangent to the surface in that direction makes with the zy plane, The and the square of the slope may be neglected compared to unity curvature of the surface in a plane parallel to the xz plane (Fig 16) is then numerically equal to 1_ _— (aw) _ _ ate Tx dx\dx] - © Ox? The minus We consider a curvature positive if it is convex downward as downward, convex sign is taken in Eq (e), since for the deflection shown in the figure, the second derivative 0°w/dx? is negative In the same manner we obtain for the curvature in a plane parallel to the yz plane | o= fdw 02w - (Se) — ) These expressions are similar to those used in discussing the curvature of a bent beam In considering the curvature of the middle surface in any direction an (lig 16) we obtain (ow Tn Substituting expression on \ On (a) for dw/dn and observing that — Ở ogg a t © sin On Ox ° Oy ° we find = — (2 cos | P — & Ì sọ? = Tz 0? a + Oo cos? a + œ — | zy Ow F sina) (5 gin 0? aay 2a + cos ow a + SY sin sin @ cos a + a + Ty sin? œ a) sin? «) (9) PURE BENDING OF PLATES 35 It is seen that the curvature in any direction at a point of the middle surface can be calculated if we know at that point the curvatures dẨddadaầ ?z and the quantity Ox? ] — ry, O° w = xu Or (h) Ov OY which is called the twist of the surface with respect to the x and y axes If instead of the direction an (ig 16b) we take the direction at perpendicular to an, the curvature in this new direction will be obtained from expression Thus we obtain (g) by substituting 7/2 + a@ fora — = Tt 1, — gin’ a 1, + — Tx sin 2a ] + — Try cos’? œ (2) Ty Adding expressions (g) and (z), we find ~4++=i441 a a Ty (34) rn which shows that at any point of the middle surface the sum of the curvatures in two perpendicular directions such as n and t is independent of the angle a ‘This sum is usually called the average curvature of the surface at a point | The twist of the surface at a with respect to the an and at directions is LL a (au T nt at an In calculating the derivative with respect to ¢, we observe that the direction at is perpendicular to an Thus we obtain the required derivative by substituting 7/2 + a for ain Hq (a) In this manner we find tL To cọs \ôc | 4+ = sin 2a(— gin — oy” ax So at) Ox? | ay? i sin 2a ( — *) Tx y in ˆ + dy vos O- + £08 2a Ox ~OY + cos 2a + Tx (7) In our further discussion we shall be interested in finding in terms of a the directions in which the curvature of the surface is a maximum or a minimum and in finding the corresponding values of the curvature We obtain the necessary equation for determining @ by equating the derivative of expression (g) with respect to a to zero, which gives © sin 2a + x Tx y cos 2a — sin 2a = Ty (k) 36 THEORY OF PLATES AND whence SHELLS — tan 2a = Vz — TT Tr (35) Ty SubstitutFrom this equation we find two values of a, differing by 7/2 ing these in Eq (g) we find two values of 1/r,, one representing the maximum and the other the minimum curvature at a point a of the surface These two curvatures are called the principal curvatures of the surface; and the corresponding planes naz and taz, the principal planes of curvature Observing that the left-hand side of Eq (k) is equal to the doubled value of expression (7), we conclude that, if the directions an and at (Fig 16) are in the principal planes, the corresponding twist 1/r,; zero We can use a circle, similar to Mohr’s circle representing stresses, to show how the curvature and the twist of a surface the angle a.* To simplify the discussion we assume that the planes zz and yz are taken parallel to the principal planes of at the point a Then Tot lay and we obtain from Eas (g) and (7) for any angle a ~ Fic = + cos? a + = sin? a "5 17 combined vary with coordinate curvature _9 ' T 1s equal to n Tìt x r y Tx Tụ e Taking the curvatures as abscissas and the twists as ordinates and constructing a circle on the diameter 1/rz — 1/r,, as shown in Fig 17, we see that the point A defined by the angle 2a has the abscissa OB = 00 + 0B = 5(2 + tY + (2 = and the ordinate — Tz (7z l., cos? a + —sin*a Ty # 2) 008 20 y Tụ — AB l(1 1\ = AG — =) sin 2a r7; Ty Comparing these results with formulas (36), we conclude that the coordi* See S Timoshenko, ‘‘Strength of Materials,” part I, 3d ed., p 40, 1955 PURE BENDING OF PLATES 37 nates of the point A define the curvature and the twist of the surface for any value of the angle a It is seen that the maximum twist, represented by the radius of the circle, takes place when a = 7/4, i.e., when we take two perpendicular directions bisecting the angles between the principal planes In our example the curvature in any direction is positive; hence the surface is bent convex downward If the curvatures 1/r, and 1/r, are both negative, the curvature in any direction is also negative, and we have a bending of the plate convex upward Surfaces in which the curvatures in all planes have like signs are called synclastic Sometimes we shall deal with surfaces in which the two principal curvatures have opposite signs A saddle is a good example Such surfaces are called anticlastic The circle in Fig 18 represents a particular case of such surfaces when Ý Lf Ý Ý ££ $ Ý My Lf $ $ fff M X ⁄ M, t =f F y "nt X Fie 18 Tf ị fff M f ¢ 2f.TF ¢ ¢ Y Zz Fig 19 1/r, = —1/rz It is seen that in this case the curvature becomes zero for a = 7/4 and for a = 37/4, and the twist becomes equal to +1/rz 10 Relations between Bending Moments and Curvature in Pure Bending of Plates In the case of pure bending of prismatic bars a rigorous solution for stress distribution is obtained by assuming that cross sections of the bar remain plane during bending and rotate only with respect to their neutral axes so as to be always normal to the deflec- tion curve Combination of such bending in two perpendicular directions brings us to pure bending of plates Let us begin with pure bending of a rectangular plate by moments that are uniformly distributed along the edges of the plate, as shown in Fig 19 We take the zy plane to coincide with the middle plane of the plate before deflection and the x and y axes along the edges of the plate as shown The z axis, which is then perpendicular to the middle plane, is taken positive downward We denote by M, the bending moment per unit length acting on the edges parallel to the y axis and by M, the moment per unit length acting on the edges parallel to the z axis These moments we consider positive when they are directed as shown in the figure, 7.e., when they produce compression SHELLS AND PLATES OF THEORY 38 The thickness in the upper surface of the plate and tension in the lower of the plate we denote, as before, by h and consider it small in comparison with other dimensions Let us consider an element cut out of the plate by two pairs of planes parallel to the xz and yz planes, as shown in Fig 20 Since the case shown — in Fig 19 represents the combination of two uniform bendings, the stress conditions are identical in all elements, as shown in Fig 20, and we have Assuming a uniform bending of the plate that during bending of the plate the lateral a Zao sides of the element remain plane and rotate I ù { | Lt +-7"-4 ñLZ jf |xý x z + v b7 eee} about the neutral axes nn so as to remain nor- h £ plate, it can be concluded that the middle plane of the plate does not undergo any extension during this bending, and the middle mal to the deflected middle surface of the surface is therefore the neutral surface.1 1/r, and Let 1/r, denote, as before, the curva- tures of this neutral surface in sections parallel to the xz and yz planes, Then the unit elongations in the x and y directions of an respectively elemental lamina abcd (Fig 20), at a distance z from the neutral surface, are found, as in the case of a beam, and are equal to €x Using now Hooke’s the lamina abed are law [Eq Øy = —= z Tz Cụ ee T— Z Ty ( —— 5], the corresponding (1), page iz = a ) stresses In (0) a(=+>2) These stresses are proportional to the distance z of the lamina abcd from the neutral surface and depend on the magnitude of the curvatures of the bent plate The normal stresses distributed over the lateral sides of the element in Fig 20 can be reduced to couples, the magnitudes of which per unit length evidently must be equal to the external moments M, and M, In this way we obtain the equations J —h/2 + dụ dz = I, dụ J h/2 h/2 (c} oz dx dz = M, dz It will be shown in Art 13 that this conclusion is accurate enough if the deflections of the plate are small in comparison with the thickness h PURE BENDING OF PLATES 39 Substituting expressions (b) for oz and o,, we obtain M, = D(2 1 d*w +»2) ö?) - (5 ay? oe 07w 2) (38) where D is the flexural rigidity of the plate defined by Eq denotes small deflections of the plate in the z direction (3), and w Let us now consider the stresses acting on a section of the lamina abcd parallel to the z axis and inclined to the z and y axes If acd (Fig 21) represents a portion of the lamina cut by such a section, the stress acting on the side ac can be found by means of the equations of statics Resolving this stress into a normal component ¢, and a shearing component Tn, X the magnitudes of these eomponents are obtained by projecting the forces acting on the element acd on the n and ¿ directions respectively, gives the known equations On = 06, C0S’a Trt = 4(o, — Ox) +, sin sin’? a which (d) 2a in which @ is the angle between the normal n and the x axis or between the direction ¢ and the y axis (Fig 21a) if measured in a clockwise direction Considering all laminas, such as acd in the plate, the normal stresses cn give the section ac of the plate, the magnitude of M, = Ƒ h/2 h/2 The angle is considered positive | Vig 21b, over the thickness of bending moment acting on the which per unit length along ac œ + Mu sin? œ ơ„z dz = M,„ cos? (39) The shearing stresses 7,, give the twisting moment acting on the section OF THEORY 40 SHELLS AND PLATES ac of the plate, the magnitude of which per unit length of ac is Mu _f£h/2 = tz dz —h/2 = Asin 2a(M, — M,) (40) The signs of M, and M,, are chosen in such a manner that the positive values of these moments are represented by vectors in the positive directions of n and ¢ (Fig 21a) if the rule of the right-hand screw is used Fora = 7/2 or 34/2, we When a is zero or 7, Eq (39) gives M, = M obtain M, = M, The moments M,, become zero for these values of a Thus we obtain the conditions shown in Fig 19 Equations (39) and (40) are similar to Eqs Mn (36), and by using them the bending and twisting moments for any value can be readily calculated of a We can also use the graphical method for the same purpose and find the values of M, and M,, from Mohr’s circle, which can be constructed as shown in the previous article by takThe diameter of the circle will ing M, as abscissa and M nt as ordinate Then the coordinates OB and be equal to M, — M,, as shown in Fig 22 AB of a point A, defined by the angle 2a, give the moments M, and Mu Fig 22 respectively Let us now represent M, and M, as functions of the curvatures and Substituting in Eq (39) for twist of the middle surface of the plate M, and M, their expressions (37) and (38), we find M, = D (= cos? a + + sin? «) # U Using the first of the equations + vD ( zx sin? a + + cos? «) y (36) of the previous article, we conclude that the expressions in parentheses represent the curvatures of the middle surface in the n and ¢ directions respectively M, = D(+ 1 +92) = Hence 0*w —D(Fa+ 07w * an) (41) To obtain the corresponding expression for the twisting moment Mn, let us consider the distortion of a thin lamina abcd with the sides ab and ad parallel to the n and ¢ directions and at a distance z from the middle During bending of the plate the points a, b, c, and d plane (Fig 23) The components of the displacement of undergo small displacements the point a in the n and ¢ directions we denote by u and v respectively Then the displacement of the adjacent point d in the n direction is u + (du/dt) dt, and the displacement of the point b in the ¢ direction is Owing to these displacements, we obtain for the shear0 + (dv/dn) dn PURE BENDING ing strain Yn OF ou = PLATES 4] Ov + a7 The corresponding shearing stress is (e) | Ou , Ov mu = (H+ 2) (f) From Fig 236, representing the section of the middle surface made by the normal plane through the n axis, it may be seen that the angle of rotation in the counterclockwise direction of an element pq, which initially was perpendicular to the zy plane, about an axis perpendicular to the nz plane is equal to —dw/dn Owing to this rotation a point of the d Z ` + y / ⁄ > DỊ ` by * OL (b) I> ur dt (a) “1 Fig 23 element at a distance z from the neutral surface has a displacement in the n direction equal to ya on Considering the normal section through the ¢ axis, it can be shown that the same point has a displacement in the ¢ direction equal to Ụ = —Z ow ot Substituting these values of the displacements % and v in expression (f), | we find tet = —2Gz 0? On ol (42) and expression (40) for the twisting moment becomes M ne an b/2 lu, Taiz dz = Gh? aw Ondt - 2q — aw "an ot — (43) AQ THEORY OF FLATES AND SHELLS It is seen that the twisting moment for the given perpendicular directions n and tis proportional to the twist of the middle surface corresponding to When the n and ¢ directions coincide with the z and those directions y axes, there are only bending moments M, and M, acting on the sections perpendicular to those axes (Fig 19) Hence the corresponding twist 1s zero, and the curvatures 1/r, and 1/r, are the principal curvatures of the They can middle surface of the plate moments bending the if (38) and Eqs (37) curvature in any other direction, defined calculated by using the first of the equations Fig 17 readily be calculated from The M, and M, are given by an angle a, can then be (36), or it can be taken from Regarding the stresses in a plate undergoing pure bending, it can be concluded from the first of the equations (d) that the maximum normal stress acts on those sections parallel to the xz or yz planes The magni- tudes of these stresses are obtained from Eas (6) by substituting z2 = h/2 and by using Eqs (37) and (38) In this way we find (oz) max — 6M, h? (ơy) max — 6M h? : (44) If these stresses are of opposite sign, the maximum shearing stress acts in the plane bisecting the angle between the xz and yz planes and is equal to T max (oz _ — 3(M„—h2 M,) Cy) (45) If the stresses (44) are of the same sign, the maximum shear acts in the plane bisecting the angle between the xy and xz planes or in that bisecting the angle between the xy and yz planes and is equal to 4(cy) max OF $(oz) max; depending greater on which of the two principal stresses (oy) max OF (Gz)max 18 11 Particular Cases of Pure Bending In the discussion of the previous article we started with the case of a rectangular plate with uniformly distributed bending moments acting along the edges To obtain a general case of pure bending of a plate, let us imagine that a portion of any shape is cut out from the plate considered above (Fig 19) by a cylindrical or prismatic surface perpendicular to the plate ing of this portion will remain unchanged The conditions of bend- provided that bending and twisting moments that satisfy Eqs (39) and (40) are distributed along the boundary of the isolated portion of the plate Thus we arrive at the case of pure bending of a plate of any shape, and we conclude that pure bending is always produced if along the edges of the plate bending moments M, and twisting moments M,, are distributed in the manner given by Eqs (39) and (40) Let us take, as a first example, the particular case in which M,=M,=M PURE BENDING OF PLATES 43 Jt can be concluded, from Eqs (39) and (40), that in this case, for a plate of any shape, the bending moments are uniformly distributed along the entire boundary and the twisting moments vanish Irom Eggs (387) and (38) we conclude that mm DỤ +7) đề z.e., the plate in thĩs case is bent to a spherical surface the curvature ot which is given by Eq (46) In the general case, when M,= Then, from lỏqs Af, is different from AZ,, we put M, (37) and and 11, = 1: (38), we find Ô”t0 _ My — pl, đt ST DỤ — y9 ady? - — ÂF,D(1— —vÀU,v?) and in addition * 0°w Ou dy - 0) Integrating these equations, we find w= where define Ci, the vMs y _ M= 2D(1 — p})'` p2 — Mỹ: 2D(@ — 1À1T y> + Cy + Coy + C's — „3 (c) Co, and C3; are constants of integration These constants plane from which the deflections w are measured If this plane is taken tangent to the middle surface of the plate at the origin, the constants of integration must be equal to zero, and the deflection surface is given by the equation MEN yg ~Me, ~ 2p — 3)" 2p0 = »3” gy In the particular case where 1/; = Aly = AZ, we get from Eq — s đfIệy! + `) 2D(1 + n) | (d) (6) t.e., a paraboloid of revolution instead of the spherical surface given by Inq (46) The inconsistency of these results arises merely from the use of the approximate expressions 0?w/d.2? and 0?w/dy? for the curvatures 1/r, and 1/r, in deriving Eq (e) These second derivatives of the deflections, rather than the exact expressions for the curvatures, will be used also in all further considerations, in accordance with the assump- tions made in Art This procedure greatly simplifies the fundamental equations of the theory of plates 44 THEORY OF PLATES AND SHELLS Returning now to Eq (d), let us put M, principal curvatures, from Eqs (a), are tT Ta = —M, In this case the | _ — M, dx? D(1 — v) Ở) and we obtain an anticlastic surface the equation of which is yn = My | Straight lines parallel to the xz axis become, after bending, parabolic curves convex downward (Fig 24), whereas straight lines in the y direcAlong the lines bisecting the tion become parabolas convex upward angles between the x and y axes we have x = y, or x = —y; thus deflections along these lines, as seen from Eq (g), are zero All lines parallel to these bisecting lines before bending remain straight during bending, A rectangle abcd bounded by such lines rotating only by some angle will be twisted as shown in Fig 24 Imagine normal sections of the plate x along lines ab, bc, cd, and ad From Eqs (39) and (40) we conclude that - bending moments along these sections are zero and that twisting moments along sections ad and be are equal to Fig 24 M, and along sections ab and cd are equal to —M; Thus the portion abcd of the plate is in the condition of a plate undergoing pure bending produced by twisting moments uniThese twisting moments formly distributed along the edges (Fig 25a) are formed by the horizontal shearing stresses continuously distributed This horizontal stress distribution can be over the edge [Eq (40)] replaced by vertical shearing forces which produce the same effect as To show this, let the edge ab be — the actual distribution of stresses If divided into infinitely narrow rectangles, such as mnpq in Fig 25b A is the small width of the rectangle, the corresponding twisting couple is M.A and can be formed by two vertical forces equal to M, acting along This replacement of the distributed the vertical sides of the rectangle horizontal forces by a statically equivalent system of two vertical forces cannot cause any sensible disturbance in the plate, except within a distance comparable with Proceeding in the forces M, acting another and only the thickness of the plate,! which is assumed small same manner with all the rectangles, we find that all along the vertical sides of the rectangles balance one two forces M, at the corners a and are left Making This follows from Saint Venant’s principle; see S Timoshenko ‘Theory of Elasticity,” 2d ed., p 33, 1951 and J N Goodier, PURE BENDING OF PLATES 45 the same transformation along the other edges of the plate, we conclude that bending of the plate to the anticlastic surface shown in lig 25a can Such an be produced by forces concentrated at the corners! (Jig 25c) experiment is comparatively simple to perform, and was used for the experimental verification of the theory of bending of plates discussed In these experiments the deflections of the plate along the line bod above.2 (Fig 24) were measured and were found to be in very satisfactory agreement with the theoretical results obtained from Eq (g) Some discrepancies were found only near the edges, and they were more pro- OM, Fig 25 nounced in the case of comparatively thick plates, as would be expected from the foregoing discussion of the transformation of twisting couples along the edges As a last example let us consider the bending of a plate (Fig 19) toa cylindrical surface having its generating line parallel to the y axis such a case 02w/dy? = 0, and we find, from Eqs (37) and (38), in ư*› Ơ”+ It is seen that to produce bending of the plate to a cylindrical surface we must apply not only the moments JZ, but also the moments ly Without these latter moments the plate will be bent to an anticlastic The first of equations (h) has already been used in Chap in surface discussing the bending of long rectangular plates to a cylindrical surface Although in that discussion we had a bending of plates by lateral loads and there were not only bending stresses but also vertical shearing stresses This transformation of the force system acting along the edges was first suggested by Lord Kelvin and P G Tait; see ‘‘Treatise on Natural Philosophy,”’ vol 1, part 2, p 203, 1883 Such experiments were made by A N&dai, Forschungsarb., vols 170, 171, Berlin, 1915; see also his book ‘‘Elastische Platten,” p 42, Berlin, 1925 We always assume very small deflections or else bending to a developable surface The case of bending to a nondevelopable surface when the deflections are not smali will be discussed later; see p 47 46 THEORY OF PLATES AND SHELLS acting on sections perpendicular to the z axis, it can be concluded from a comparison with the usual beam theory that the effect of the shearing forces is negligible in the case of thin plates, and the equations developed for the case of pure bending can be used with sufficient accuracy for lateral loading 12 Strain Energy in Pure Bending of Plates If a plate is bent by uniformly distributed bending moments M, and M, (Fig 19) so that the vz and yz planes are the principal planes of the deflection surface of the plate, the strain energy stored in an element, such as shown in Fig 20, is obtained by calculating the work done by the moments M, dy and M, dx on the element during bending of the plate Since the sides of the element remain plane, the work done by the moments M, dy is obtained by taking half the product of the moment and the angle between the corresponding sides of the element after bending Since —0?w/dz* represents the curvature of the plate in the xz plane, the angle corresponding to the moments M, dy is — (02w/dx?) dx, and the work done by these moments is — M, = dx dy An analogous expression is also obtained for the work produced by the moments M,dz Then the total work, equal to the strain energy of the element, is | qv = — (Me + My Sa) de dy Substituting for the moments their expressions (37) and (38), the strain energy of the elements is represented in the following form: 3”*”\7 ”u`\? „0 >w an Since in the case of pure bending the curvature is constant over the entire surface of the plate, the total strain energy of the plate will be obtained if we substitute the area A of the plate for the elementary area dx dy in expression o Then 8” 0?w\? 0? V=3DA (=oa) + (=) + oe Ô” (47) If the directions x and y not coincide with the principal planes of curvature, there will act on the sides of the element (Fig 20) not only the bending moments M, dy and M, dz but also the twisting moments The strain energy due to bending moments is repreM,, dy and M,,dx sented by expression (a) In deriving the expression for the strain energy due to twisting moments M,, dy we observe that the corresponding angle of twist is equal to the rate of change of the slope dw/dy, as x varies, PURE BENDING OF PLATES Al multipled with dx; hence the strain energy due to M,, dy is 0° which, applying Eq (483), becomes ] Đ(1 — ) (3 *w 5y) \° dx dy The same amount of energy will also be produced by the couples M,, dz, so that the strain energy due to both twisting couples is DA — v) am) dx dy —(®) Since the twist does not affect the work produced by the bending moments, the total strain energy of an element of the plate is obtained by adding together the energy of bending (a) and the energy of twist (6) Thus we obtain al — 02w \? O2w —— — 5>| (3) + ($5) — \? + )?w4 oe2® | a dy + D(1 — r) (=F oy dx dy or aw , dw\? = = —+—) al D (5 + sat) | -—20 (1 - 0*w 07w —=—s ¿ E dy? *w (2 > | \? lÌ k dv dy (48) The strain energy of the entire plate is now obtained by substituting the area A of the plate for the elemental area dz dy Expression (48) will be used later in more complicated cases of bending of plates | 13 Limitations on the Application of the Derived Formulas In discussing stress distribution in the case of pure bending (Art 10) it was assumed that the middle surface is the neutral surface of the plate This condition can be rigorously satisfied only if the middle surface of the bent plate is a developable surface Considering, for instance, pure bending of a plate to a cylindrical surface, the only limitation on the application of the theory will be the requirement that the thickness of the plate be small in comparison with the radius of curvature In the problems of bending of plates to a cylindrical surface by lateral loading, discussed in the previous chapter, it is required that deflections be small in comparison with the width of the plate, since only under this condition will the approximate expression used for the curvature be accurate enough If a plate is bent to a nondevelopable surface, the middle surface undergoes some stretching during bending, and the theory of pure bend- 48 THEORY OF PLATES AND SHELLS ing developed previously will be accurate enough only if the str corresponding to this stretching of the middle surface are small in parison with the maximum bending stresses given by Eqs (44) or, is equivalent, if the strain in the middle surface is small in compa with the maximum bending strain n/2r,;, This requirement put additional limitation on deflections of a plate, viz., that the deflectic of the plate must be small in comparison with its thickness h To show this, let us consider the bending of a circular plate by | ing couples M uniformly distributed along the edge The deflection face, for small deflections, is spherical with radius r as defined by Eq Let AOB (Fig 26) represent a diametral section of the bent circular Ị a its outer radius before bending, and the deflection at the middle assume at first that there is no stretching of the middle surface o plate in the radial direction In such a case the arc OB must be equ the initial outer radius a of the plate The angle ¢ and the radius the plate after bending are then given by the following equations: => a a b=rsin ¢ It is seen that the assumed bending of the plate implies a compre strain of the middle surface in the circumferential direction “The m: tude of this strain at the edge of the plate Ee = a—-b a re—rsing rey For small deflections we can take ) sng =o — & which, substituted in Eq (a), gives Yo represent this strain as a function of the maximum observe that § = r(1 Hence — cos g) ~ yeyp? a—28 Substituting in Eq (b), we obtain € — oO Or deflection ¿ PURE BENDING OF PLATES s 49 This represents an upper limit for the circumferential strain at the edge of the plate It was obtained by assuming that the radial strain is zero Under actual conditions there is some radial strain, and the circumferential compression is somewhat smaller! than that given by Eq (49) From this discussion it follows that the equations obtained in Art 10, on the assumption that the middle surface of the bent plate is its neutral surface, are accurate provided the strain given by expression (49) is small in comparison with the maximum bending strain h/2r, or, what is equivalent, if the deflection is small in comparison with the thickness h of the plate A similar conclusion can also be obtained in the more general case of pure bending of a plate when the two principal curvatures are not equal.2 Generalizing these conclusions we can state that the equations of Art 10 can always be applied with sufficient accuracy if the deflections of a plate from its initial plane or from a true developable surface are small in comparison with the thickness of the plate 14 Thermal Stresses in Plates with Clamped Edges Equation (46) for the bending of a plate to a spherical surface can be used in calculating thermal stresses in a plate for certain cases of nonuniform heating Assume that the variation of the temperature through the thickness of the plate follows a linear law and that the temperature does not vary in planes parallel to the surfaces of the plate In such a case, by measuring the temperature with respect to that of the middle surface, it can be concluded that temperature expansions and contractions are proportional to the distance from the middle surface Thus we have exactly the same condition as in the pure bending of a plate to a spherical surface If the edges of the nonuniformly heated plate are entirely free, the plate will bend to a spherical surface Let a be the coefficient of linear expansion of the material of the plate, and let t denote the difference in temperature of the upper and lower faces of the plate The difference between the maximum thermal expansion and the expansion at the middle surface is at/2, and the curvature resulting from the nonuniform heating can be found from the equation t from which h “5 at rT Oh | 44) (50) This bending of the plate does not produce any stresses, provided the 1'This question is discussed later; see Art 96 2See Kelvin and Tait, op czt., vol 1, part 2, p 172 $It is assumed that deflections are small in comparison with the thickness of the rlate 50 THEORY OF PLATES AND SHELLS edges are free and deflections are small in comparison with the thickness of the plate _ Assume now that the middle plane of the plate is free to expand but, that the edges are clamped so that they cannot rotate In such a case the nonuniform heating will produce bending moments uniformly distributed along the edges of the plate The magnitude of these moments is such as to eliminate the curvature produced by the nonuniform heating [Eq (50)], since only in this way can the condition at the clamped edge be satisfied Using Eq (46) for the curvature produced by the bending moments, we find for determining the magnitude M of the moment per unit length of the boundary the equation! M at Di+yv) ih : M = ae v) from which (b) The corresponding maximum stress can be found from Eqs (44) and is equal to | 6M _ batD(1 + ») Omax — b2 Substituting for D its expression — (3), we finally obtain ath Cmax = FT y) It is seen that the stress is proportional to the coefficient of expansion a, to the temperature difference ¢ between the two the plate, and to the modulus of elasticity # The thickness plate does not enter into formula (51); but since the difference peratures usually increases in proportion to the thickness of the can be concluded that greater thermal stresses are to be expected plates than in thin ones (51) thermal faces of h of the ¢ of template, it in thick The effect of pure bending upon the curvature of the entire plate thus is equivalent but opposite in sign to the effect of the temperature gradient Now, if the plate remains, in the end, perfectly plane, the conditions of a built-in edge are evidently satisfied along any given boundary Also, since in our case the bending moments are equal everywhere and in any direction, the clamping moments along that given boundary are always expressed by the same Eq (b)  ... and the twist becomes equal to +1/rz 10 Relations between Bending Moments and Curvature in Pure Bending of Plates In the case of pure bending of prismatic bars a rigorous solution for stress distribution... developed for the case of pure bending can be used with sufficient accuracy for lateral loading 12 Strain Energy in Pure Bending of Plates If a plate is bent by uniformly distributed bending moments M,... during bending and rotate only with respect to their neutral axes so as to be always normal to the deflec- tion curve Combination of such bending in two perpendicular directions brings us to pure bending