6.1 Plate Theory - Auckland
Section 6.16.1 Plate Theory6.1.1PlatesA plate is a flat structural element for which the thickness is small compared with thesurface dimensions. The thickness is usually constant but may be variable and ismeasured normal to the middle surface of the plate, Fig. 6.1.1lateral loadmiddle surface of plateFig. 6.1.1: A plate6.1.2Plate TheoryPlates subjected only to in-plane loading can be solved using two-dimensional planestress theory1 (see Book I, §3.5). On the other hand, plate theory is concerned mainlywith lateral loading.One of the differences between plane stress and plate theory is that in the plate theory thestress components are allowed to vary through the thickness of the plate, so that there canbe bending moments, Fig. 6.1.2.MFig. 6.1.2: Stress distribution through the thickness of a plate and resultant bendingmomentPlate Theory and Beam TheoryPlate theory is an approximate theory; assumptions are made and the general threedimensional equations of elasticity are reduced. It is very like the beam theory (see Book1although if the in-plane loads are compressive and sufficiently large, they can buckle (see §6.7)Solid Mechanics Part II120Kelly
Section 6.1I, §7.4) – only with an extra dimension. It turns out to be an accurate theory provided theplate is relatively thin (as in the beam theory) but also that the deflections are smallrelative to the thickness. This last point will be discussed further in §6.10.Things are more complicated for plates than for the beams. For one, the plate not onlybends, but torsion may occur (it can twist), as shown in Fig. 6.1.3Fig. 6.1.3: torsion of a plateAssumptions of Plate TheoryLet the plate mid-surface lie in the x y plane and the z – axis be along the thicknessdirection, forming a right handed set, Fig. 6.1.4.zyxFig. 6.1.4: Cartesian axesThe stress components acting on a typical element of the plate are shown in Fig. 6.1.5.z zx zz yz yy xxy xyxFig. 6.1.5: stresses acting on a material elementSolid Mechanics Part II121Kelly
Section 6.1The following assumptions are made:(i) The mid-plane is a “neutral plane”The middle plane of the plate remains free of in-plane stress/strain. Bending of the platewill cause material above and below this mid-plane to deform in-plane. The mid-planeplays the same role in plate theory as the neutral axis does in the beam theory.(ii) Line elements remain normal to the mid-planeLine elements lying perpendicular to the middle surface of the plate remain perpendicularto the middle surface during deformation, Fig. 6.1.6; this is similar the “plane sectionsremain plane” assumption of the beam theory.undeformedline element remainsperpendicular to mid-surfaceFig. 6.1.6: deformed line elements remain perpendicular to the mid-plane(iii) Vertical strain is ignoredLine elements lying perpendicular to the mid-surface do not change length duringdeformation, so that zz 0 throughout the plate. Again, this is similar to an assumptionof the beam theory.These three assumptions are the basis of the Classical Plate Theory or the KirchhoffPlate Theory. The second assumption can be relaxed to develop a more exact theory (see§6.10).6.1.3Notation and Stress ResultantsThe stress resultants are obtained by integrating the stresses through the thickness of theplate. In general there will bemoments M:out-of-plane forces V:in-plane forces N:Solid Mechanics Part II2 bending moments and 1 twisting moment2 shearing forces2 normal forces and 1 shear force122Kelly
Section 6.1They are defined as follows:In-plane normal forces and bending moments, Fig. 6.1.7: h / 2Nx h / 2 xx dz,Ny h / 2 h / 2Mx z xx yy h / 2 h / 2dz , M y h / 2dz z (6.1.1)yydz h / 2y yyNyMy xxxNxMxFig. 6.1.7: in-plane normal forces and bending momentsIn-plane shear force and twisting moment, Fig. 6.1.8: h / 2N xy h / 2xyM xy dz , h / 2 z xy(6.1.2)dz h / 2M xyM xyy xy xyN xyN xyxFig. 6.1.8: in-plane shear force and twisting momentOut-of-plane shearing forces, Fig. 6.1.9: h / 2V x zx dz , h / 2Solid Mechanics Part II h / 2V y yz dz(6.1.3) h / 2123Kelly
Section 6.1 yz zxy yz zxVyxVxFig. 6.1.9: out of plane shearing forcesNote that the above “forces” and “moments” are actually forces and moments per unitlength. This allows one to have moments varying across any section – unlike in the beamtheory, where the moments are for the complete beam cross-section. If one considers anelement with dimensions x and y , the actual moments acting on the element areM x y, M y x, M xy x, M xy y(6.1.4)and the forces acting on the element areV x y, V y x, N x y, N y x, N xy x, N xy y(6.1.5)The in-plane forces, which are analogous to the axial forces of the beam theory, do notplay a role in most of what follows. They are useful in the analysis of buckling of platesand it is necessary to consider them in more exact theories of plate bending (see later).Solid Mechanics Part II124Kelly
Section 6.26.2 The Moment-Curvature Equations6.2.1 From Beam Theory to Plate TheoryIn the beam theory, based on the assumptions of plane sections remaining plane and thatone can neglect the transverse strain, the strain varies linearly through the thickness. Inthe notation of the beam, with y positive up, xx y / R , where R is the radius ofcurvature, R positive when the beam bends “up” (see Book I, Eqn. 7.4.16). In terms ofthe curvature 2 v / x 2 1 / R , where v is the deflection (see Book I, Eqn. 7.4.36), onehas xx y 2v x 2(6.2.1)The beam theory assumptions are essentially the same for the plate, leading to strainswhich are proportional to distance from the neutral (mid-plane) surface, z, and expressionssimilar to 6.2.1. This leads again to linearly varying stresses xx and yy ( zz is alsotaken to be zero, as in the beam theory).6.2.2 Curvature and TwistThe plate is initially undeformed and flat with the mid-surface lying in the x y plane.When deformed, the mid-surface occupies the surface w w( x, y ) and w is the elevationabove the x y plane, Fig. 6.2.1. w xinitialpositionyFig. 6.2.1: Deformed PlateThe slopes of the plate along the x and y directions are w / x and w / y .Solid Mechanics Part II125Kelly
Section 6.2CurvatureRecall from Book I, §7.4.11, that the curvature in the x direction, x , is the rate of changeof the slope angle with respect to arc length s, Fig. 6.2.2, x d / ds . One finds that x 2 w / x 2 1 w / x 2 (6.2.2)3/2Also, the radius of curvature R x , Fig. 6.2.2, is the reciprocal of the curvature, R x 1 / x .wsRx xFig. 6.2.2: Angle and arc-length used in the definition of curvatureAs with the beam, when the slope is small, one can take tan w / x andd / ds / x and Eqn. 6.2.2 reduces to (and similarly for the curvature in the ydirection) x 2w1 2 ,R x x y 2w1 2R y y(6.2.3)This important assumption of small slope, w / x, w / y 1 , means that the theory tobe developed will be valid when the deflections are small compared to the overalldimensions of the plate.The curvatures 6.2.3 can be interpreted as in Fig. 6.2.3, as the unit increase in slope alongthe x and y directions.Solid Mechanics Part II126Kelly
Section 6.2wACDy x yxw w 2 w x x x 2CBw w x w yAA w 2 w y y y 2BxyFigure 6.2.3: Physical meaning of the curvaturesTwistNot only does a plate curve up or down, it can also twist (see Fig. 6.1.3). For example,shown in Fig. 6.2.4 is a plate undergoing a pure twisting (constant applied twistingmoments and no bending moments).z M xyy M xyxFigure 6.2.4: A twisting plateIf one takes a row of line elements lying in the y direction, emanating from the x axis, thefurther one moves along the x axis, the more they twist, Fig. 6.2.4. Some of these lineelements are shown in Fig. 6.2.5 (bottom right), as veiwed looking down the x axistowards the origin (elements along the y axis are shown bottom left). If a line element atposition x has slope w / y , the slope at x x is w / y x ( w / y ) / x . Thismotivates the definition of the twist, defined analogously to the curvature, and denoted by1 / Txy ; it is a measure of the “twistiness” of the plate:1 2w Txy x ySolid Mechanics Part II127(6.2.4)Kelly
Section 6.2wABCyDxDCx 2 y x x ywBA x 2 x y x ywCADB yyFigure 6.2.5: Physical meaning of the twistThe signs of the moments, radii of curvature and curvatures are illustrated in Fig. 6.2.6.Note that the deflection w may or may not be of the same sign as the curvature. Notealso that when M x 0, 2 / x 2 0 , when M y 0, 2 / y 2 0 .Rx 0Mx 0zx 2 / x 2 0 2 / x 2 0Mx 0zxRx 0Figure 6.2.6: sign convention for curvatures and momentsOn the other hand, for the twist, with the sign convention being used, whenM xy 0, 2 / x y 0 , as depicted in Fig. 6.2.4.Principal CurvaturesConsider the two Cartesian coordinate systems shown in Fig. 6.2.7, the second ( t n )obtained from the first ( x y ) by a positive rotation . The partial derivatives arising inSolid Mechanics Part II128Kelly
Section 6.2the curvature expressions can be expressed in terms of derivatives with respect to t and nas follows: with w w x, y , an increment in w is w w w x y x y(6.2.5)Also, referring to Fig. 6.2.7, with n 0 , x t cos , y t sin (6.2.6)Thus w w w cos sin t x y(6.2.7)Similarly, for an increment n , one finds that w w w sin cos n x y(6.2.8)Equations 6.2.7-8 can be inverted to get the inverse relations w w w cos sin x t n w w wsin cos n y t(6.2.9)ynt to y x xFigure 6.2.7: Two different Cartesian coordinate systemsThe relationship between second derivatives can be found in the same way. For example, 2w w w cos sin cos sin 2 n n x t t 222 w w w cos 2 2 sin 2 2 sin 2 t n t nSolid Mechanics Part II129(6.2.10)Kelly
Section 6.2In summary, one has 2w 2 2 sin 2 cos x 2 t 2 2w 2 2 cos 2 sin y 2 t 2 2 sin 2 n 2 2 sin 2 n 2 2 t n 2 t n(6.2.11) 2 2 2w 2 sin cos 2 2 cos 2 x y t n t nand the inverse relations 2w 2 2 2 22 cossinsin2 x y t 2 x 2 y 2 2w 2 2 2 22 sincossin2 x y n 2 x 2 y 2(6.2.12) 2 2 2 2w sin cos 2 2 cos 2 x y t n x yor11111 cos 2 sin 2 sin 2 RtRxRyTxy1111 sin 2 cos 2 sin 2 RnRxRyTxy(6.2.13) 111 1 sin cos cos 2 R TtnTxy y Rx These equations which transform between curvatures in different coordinate systems havethe same structure as the stress transformation equations (and the strain transformationequations), Book I, Eqns. 3.4.8. As with principal stresses/strains, there will be someangle for which the twist is zero; at this angle, one of the curvatures will be theminimum and one will be the maximum at that point in the plate. These are called theprincipal curvatures. Similarly, just as the sum of the normal stresses is an invariant(see Book I, Eqn. 3.5.1), the sum of the curvatures is an invariant2:1111 R x R y Rt Rn(6.2.14)If the principal curvatures are equal, the curvatures are the same at all angles, the twist isalways zero and so the plate deforms locally into the surface of a sphere.12these equations are valid for any continuous surface; Eqns. 6.2.12 are restricted to nearly-flat surfaces.this is known as Euler’s theorem for curvaturesSolid Mechanics Part II130Kelly
Section 6.26.2.3 Strains in a PlateThe strains arising in a plate are next examined. A comprehensive strain-state will be firstexamined and this will then be simpilfied down later to various approximate solutions.Consider a line element parallel to the x axis, of length x . Let the element displace asshown in Fig. 6.2.8. Whereas w was used in the previous section on curvatures to denotedisplacement of the mid-surface, here, for the moment, let w( x, y, z ) be the generalvertical displacement of any particle in the plate. Let u and v be the correspondingdisplacements in the x and y directions. Denote the original and deformed length of theelement by dS and ds respectively.The unit change in length of the element (that is, the exact normal strain) is, usingPythagoras’ theorem,p q pqds dS u v w xx 1 1dSpq x x x 222(6.2.15)q y w x xp ww( x, y ) x xw x, y mid-surface p v x, y vv( x, y ) x x u x, y q x v x xx uu( x, y ) x xFigure 6.2.8: deformation of a material fibre in the x directionIn the plate theory, it will be assumed that the displacement gradients are small: u, x u, y u, z v, x v, y v, z w, zof order ε 1 say, so that squares and products of these terms may be neglected.However, for the moment, the squares and products of the slopes will be retained, as theymay be significant, i.e. of the same order as the strains, under certain circumstances:22 w w , , x y Solid Mechanics Part II131 w w x yKelly
Section 6.2Eqn. 6.2.15 now reduces to2 xx 1 2 u w 1 x x (6.2.16)With 1 x 1 x / 2 for x 1 , one has (and similarly for the other normal strains) xx yy zz u 1 w x 2 x 2 v 1 w y 2 y w z2(6.2.17)Consider next the angle change for line elements initially lying parallel to the axes, Fig.6.2.9. Let be the angle r p q , so that / 2 is the change in the initial rightangle rpq . z w z z r v z zr u z zr r q z s p q q w x xq p x x u x x v x xFigure 6.2.9: the deformation of Fig. 6.2.8, showing shear strainsTaking the dot product of the of the vector elements p q and p r :Solid Mechanics Part II132Kelly
Section 6.2cos p q r r q q r r q q p r p q p r u u v v w w x z z x x z x z x z x z x z 222222 u v w w u v x 1 z 1 z z z x x x (6.2.18)Again, with the displacement gradients u / x , v / x , u / z , v / z , w / z of order2ε 1 (and the squares w / x at most of order ε 1 ), u v v w x z x z x z u w v v z x z x cos x z z x x z(6.2.19)For small , sin cos , so (and similarly for the other shear strains)1 u v w w xy 2 y x x y 1 u w 2 z x xz 1 v(6.2.20) w yz 2 z y The normal strains 6.2.17 and the shear strains 6.2.20 are non-linear. They are the startingpoint for the various different plate theories.Von Kármán StrainsIntroduce now the assumptions of the classical plate theory. The assumption that lineelements normal to the mid-plane remain inextensible implies that zz w 0 z(6.2.21)This implies that w w x, y so that all particles at a given x, y through the thickness ofthe plate experinece the same vertical displacement. The assumption that line elementsperpendicular to the mid-plane remain normal to the mid-plane after deformation thenimplies that xz yz 0 .The strains now readSolid Mechanics Part II133Kelly
Section 6.2 xx yy zz u 1 w x 2 x 2 v 1 w y 2 y 021 u v(6.2.22) w w xy 2 y x x y xz 0 yz 0These are known as the Von Kármán strains.Membrane Strains and Bending StrainsSince xz 0 and w w( x, y ) , one has from Eqn. 6.2.20b, w w u u 0 ( x, y ) u ( x, y , z ) z x z x(6.2.23)It can be seen that the function u 0 ( x, y ) is the displacement in the mid-plane. In terms ofthe mid-surface displacements u 0 , v0 , w0 , then,u u0 z w0 w, v v0 z 0 , w w0 x y(6.2.24)and the strains 6.2.22 may be expressed as2 xx u 0 1 w0 2 w0 z x 2 x x 22 2 w0 v0 1 w0 z y 2 y y 2 v 1 w w01 u 0 0 02 y x 2 x y yy xy(6.2.25) 2 w0 z x y The first terms are the usual small-strains, for the mid-surface. The second terms,involving squares of displacement gradients, are non-linear, and need to be consideredwhen the plate bending is fairly large (when the rotations are about 10 – 15 degrees).These first two terms together are called the membrane strains. The last terms,involving second derivatives, are the flexural (bending) strains. They involve thecurvatures.When the bending is not too large (when the rotations are below about 10 degrees), onehas (dropping the subscript “0” from w)Solid Mechanics Part II134Kelly
Section 6.2 u 0 2w z 2 x x v 2w 0 z 2 y y xx yy1 u(6.2.26) v 2w xy 0 0 z2 y x x ySome of these strains are illustrated in Figs. 6.2.10 and 6.2.11; the physical meaning of xx is shown in Fig. 6.2.10 and some terms from xy are shown in Fig. 6.2.11. w xu0 zp zpmid-surfacea u u0 z xx w x w 2 w x x x 2q q a w u0 2w z 2 x x x xw q0b w w x x bu0 x u0 x xFigure 6.2.10: deformation of material fibres in the x directionb a p v0 a, pzq c w 2w z x y x yv0 v0 z v0 x x w yb, qFigure 6.2.11: the deformation of 6.2.10 viewed “from above”; a , b are thedeformed positions of the mid-surface points a, bSolid Mechanics Part II135Kelly
Section 6.2Finally, when the mid-surface strains are neglected, according to the final assumption ofthe classical plate theory, one has xx z 2w, x 2 yy z 2w, y 2 xy z 2w x y(6.2.27)In summary, when the plate bends “up”, the curvature is positive, and points “above” themid-surface experience negative normal strains and points “below” experience positivenormal strains; there is zero shear strain. On the other hand, when the plate undergoes apositive pure twist, so the twisting moment is negative, points “above” the mid-surfaceexperience negative shear strain and points “below” experience positive shear strain; thereis zero normal strain. A pure shearing of the plate in the x y plane is illustrated in Fig.6.2.12.topmid-surfaceybottomxFigure 6.2.12: Shearing of the plate due to a positive twist (neative twisting moment)CompatibilityNote that the strain fields arising in the plate satisfy the 2D compatibility relation Eqn.1.3.1:2 2 xy 2 xx yy 2 x y y 2 x 2(6.2.28)This can be seen by substituting Eqn. 6.25 (or Eqns 6.26-27) into Eqn. 6.2.28.6.2.4 The Moment-Curvature equationsNow that the strains have been related to the curvatures, the moment-curvature relations,which play a central role in plate theory, can be derived.Solid Mechanics Part II136Kelly
Section 6.2Stresses and the Curvatures/Twist in a Linear Elastic PlateFrom Hooke’s law, taking zz 0 , xx 111 xx yy , yy yy xx , xy xyEEEEE(6.2.29)so, from 6.2.27, and solving 6.2.29a-b for the normal stresses, 2wz 2 x 2wE z1 2 y 2 xx yy xy E1 2 2w y 2 2w x 2 (6.2.30)E 2wz1 x yThe Moment-Curvature EquationsSubstituting Eqns. 6.2.30 into the definitions of the moments, Eqns. 6.1.1, 6.1.2, andintegrating, one has 2wM x D 2 x 2wM y D 2 yM xy D 1 2w y 2 2w x 2 (6.2.31) 2w x ywhereD Eh 312 1 2 (6.2.32)Equations 6.2.31 are the moment-curvature equations for a plate. The momentcurvature equations are analogous to the beam moment-deflection equation 2 v / x 2 M / EI . The factor D is called the plate stiffness or flexural rigidity andplays the same role in the plate theory as does the flexural rigidity term EI in the beamtheory.Stresses and MomentsFrom 6.30-6.31, the stresses and moments are related through xx Solid Mechanics Part IIM yzM xy zMxz,, yyxyh 3 / 12h 3 / 12h 3 / 12137(6.2.33)Kelly
Section 6.2Note the similarity of these relations to the beam formula My / I with I h 3 / 12times the width of the beam.6.2.5 Principal MomentsIt was seen how the curvatures in different directions are related, through Eqns. 6.2.11-12.It comes as no surprise, examining 6.2.31, that the moments are related in the same way.Consider a small differential element of a plate, Fig. 6.2.13a, subjected to stresses xx , yy , xy , and corresponding moments M x , M y , M xy given by 6.1.1-2. On anyperpendicular planes rotated from the orginal x y axes by an angle , one can find thenew stresses tt , nn , tn , Fig. 6.2.13b (see Fig. 6.2.5), through the stresstransformatrion equations (Book I, Eqns. 3.4.8). Then M t z tt dz cos 2 z xx dz sin 2 z yy dz sin 2 z xy dz cos 2 M x sin 2 M y sin 2 M xy (6.2.34)and similarly for the other moments, leading toM t cos 2 M x sin 2 M y sin 2 M xyM n sin 2 M x cos 2 M y sin 2 M xy(6.2.35)M tn cos sin M y M x cos 2 M xyAlso, there exist principal planes, upon which the shear stress is zero (right through thethickness). The moments acting on these planes, M 1 and M 2 , are called the principalmoments, and are the greatest and least bending moments which occur at the element.On these planes, the twisting moment is zero. xy nn xy tt tn xxFigure 6.2.13: Plate Element; (a) stresses acting on element, (b) rotated elementSolid Mechanics Part II138Kelly
Section 6.2Moments in Different Coordinate SystemsFrom the moment-curvature equations 6.2.31, { Problem 1} 2 2 Mt D 2 2 n t 2 2 M n D 2 2 t n 2 M tn D 1 t n(6.2.36)showing that the moment-curvature relations 6.2.31 hold in all Cartesian coordinatesystems.6.2.6 Problems1. Use the curvature transformation relations 6.2.11 and the moment transformationrelations 6.2.35 to derive the moment-curvature relations 6.2.36.Solid Mechanics Part II139Kelly
Section 6.36.3 Plates subjected to Pure Bending and Twisting6.3.1Pure Bending of an Elastic PlateConsider a plate subjected to bending moments M M 1 0 and M M 2 0 , with noxyother loading, as shown in Fig. 6.3.1.M1yM2M2M1xFigure 6.3.1: A plate under Pure BendingFrom equilibrium considerations, these moments act at all points within the plate – theyare constant throughout the plate. Thus, from the moment-curvature equations 6.2.31, onehas the set of coupled partial differential equationsM1 2w 2w 2 ν 2 ,D x yM 2 2w 2w 2w 2 ν 2 , 0 D x y y x(6.3.1)Solving for the derivatives, 2 w M 1 νM 2, x 2D(1 ν 2 ) 2 w M 2 νM 1, y 2D(1 ν 2 ) 2w 0 x y(6.3.2)Integrating the first two equations twice gives 1w 1 M 1 νM 2 21 M 2 νM 1 2x f1 ( y ) x f 2 ( y ), w y g1 ( x) y g 2 ( x)22 D(1 ν )2 D(1 ν 2 )(6.3.3)and integrating the third shows that two of these four unknown functions are constants: w G ( x), x1 w F ( y) y (6.3.4)f1 ( y ) A, g1 ( x) Bthis analysis is similar to that used to evaluate displacements in plane elastostatic problems, §1.2.4Solid Mechanics Part II140Kelly
Section 6.3Equating both expressions for w in 6.3.3 gives1 M 1 νM 2 21 M 2 νM 1 2x Ax g 2 ( x) y By f 2 ( y )22 D(1 ν )2 D(1 ν 2 )(6.3.5)For this to hold, both sides here must be a constant, C say. It follows thatw 1 M 1 νM 2 2 1 M 2 νM 1 2x y Ax By C2 D(1 ν 2 )2 D(1 ν 2 )(6.3.6)The three unknown constants represent an arbitrary rigid body motion. To obtain valuesfor these, one must fix three degrees of freedom in the plate. If one supposes that thedeflection w and slopes w / x, w / y are zero at the origin x y 0 (so the origin ofthe axes are at the plate-centre), then A B C 0 ; all deformation will be measuredrelative to this reference. It follows thatw M2[(M 1 / M 2 ) ν ]x 2 [1 ν (M 1 / M 2 )]y 222 D(1 ν )[](6.3.7)Once the deflection w is known, all other quantities in the plate can be evaluated – thestrain from 6.2.27, the stress from Hooke’s law or directly from 6.2.30, and moments andforces from 6.1.1-3.In the special case of equal bending moments, with M 1 M 2 M o say, one hasw Mox2 y22 D(1 ν )()(6.3.8)This is the equation of a sphere. In fact, from the relationship between the curvatures andthe radius of curvature R,Mo 2w 2w 2 2D(1 ν ) x y R D(1 ν ) constantMo(6.3.9)and so the mid-surface of the plate in this case deforms into the surface of a sphere withradius given by 6.3.9, as illustrated in Fig. 6.3.2.Figure 6.3.2: Deformed plate under Pure Bending with equal momentsSolid Mechanics Part II141Kelly
Section 6.3The character of the deformed plate is plotted in Fig. 6.3.3 for various ratios M 2 / M 1 (forν 0.3 ).M 2 / M 1 1.5M 2 / M1 3M 2 / M 1 1.5M 2 / M 1 3Figure 6.3.3: Bending of a PlateWhen the curvatures 2 w / x 2 and 2 w / y 2 are of the same sign 2, the deformation iscalled synclastic. When the curvatures are of opposite sign, as in the lower plots of Fig.6.3.3, the deformation is said to be anticlastic.Note that when there is only one moment, M y 0 say, there is still curvature in bothdirections. In this case, one can solve the moment-curvature equations to getMxMx 2w 2w 2w , ν,w x 2 ν y 2 )(22222 x y x2 (1 ν ) D(1 ν ) D(6.3.10)which is an anticlastic deformation.In order to get a pure cylindrical deformation, w f (x) say, one needs to apply momentsM x and M y νM x , in which case, from 6.3.6,2or principal curvatures in the case of a more complex general loadingSolid Mechanics Part II142Kelly
Section 6.3Mx 2x2Dw (6.3.11)The deformation for M 2 / M 1 3 in Fig. 6.3.3 is very close to cylindrical, since thereM x νM y for typical values of ν .6.3.2Pure Torsion of an Elastic PlateIn pure torsion, one has the twisting moment M M 0 with no other loading, Fig.xy6.3.4. From the moment-curvature equations,0 2w 2w 2w 2wM 2wν,0ν, D(1 ν ) x y x 2 y 2 y 2 x 2(6.3.12)so that 2w 0, x 2 2w 0, y 2 2wM x yD(1 ν )(6.3.13)MyMxFigure 6.3.4: Twisting of a PlateUsing the same arguments as before, integrating these equations leads tow MxyD(1 ν )(6.3.14)The middle surface is deformed as shown in Fig. 6.3.5, for a negative M xy . Note thatthere is no deflection along the lines x 0 or y 0 .The principal curvatures will occur at 45o to the axes (see Eqns. 6.2.13):Solid Mechanics Part II143Kelly
Section 6.3M1 ,R1D(1 ν )M1 R2D(1 ν )(6.3.15)yxR2Figure 6.3.5: Deformation for a (negative) twisting momentSolid Mechanics Part II144Kelly
Section 6.46.4 Equilibrium and Lateral LoadingIn this section, lateral loads are considered and these lead to shearing forces V x , V y , in theplate.6.4.1The Governing Differential Equation for Lateral LoadsIn general, a plate will at any location be subjected to a lateral pressure q , bendingmoments M x , M y , M xy and out-of-plane shear forces V x and V y ; q is the normal pressureon the upper surface of the plate:z h / 2 0, q( x, y ), z h / 2 zz ( x, y ) (6.4.1)These quantities are related to each other through force equilibrium.Force EquilibriumConsider a differential plate element with one corner at ( x, y ) (0,0) , Fig. 6.4.1,subjected to moments, pressure and shear force. Taking force equilibrium in the verticaldirection (neglecting a possible small variation in q, since this will only introduce higherorder terms): Fz V y V V y x V y y x V x y V x x x y q x y 0 x y (6.4.2)qM xy y y M xy x x xV x x x yM y y y V y y y M x x x Fig. 6.4.1: a plate element subjected to moments, pressure and shear forcesEqn. 6.4.2 gives the vertical equilibrium equation V x V y q x ySolid Mechanics Part II145(6.4.3)Kelly
Section 6.4This is analogous to the beam theory equation p dV / dx (see Book I, §7.4.3).Next, taking moments about the x axis: M M y M xy y x x y M x M M y M xyyyx xy y x V y y x y y / 2 V x y yV y x y y V y y V y y / 2 V x x x y y y / 2 q x y 0 x (6.4.4)Using 6.4.3, this reduces to (and similarly for moments about the x-axis), M x M xy x y M xy M yVy x yVx (6.4.5)These are analogous to the beam theory equation V dM / dx (see Book I, §7.4.3).Relations directly from the Equations of EquilibriumThe equilibrium relations 6.4.3, 6.4.5 can also be derived directly from the equations ofequilibrium, Eqns. 1.1.9, which encompass the force balances: xx yx zx 0 x z y xy yy zy 0 z y x xz yz zz 0 z y x(6.4.6)Taking the first of these (which ensures equilibrium of forces in the x direction),multiplying by z and integrating over the plate thickness, gives h / 2 h / 2 z h / 2 h / 2 yx zx xxdz zdz zdz 0 z y x h / 2 h / 2 h / 2 h / 2 h / 2 h / 2 z dzz dz z yx xxzx h / 2 zx dz 0 x h / 2 h / 2 y h / 2 (6.4.7)and, since the shear stress zx must be zero over the top and bottom surfaces, one hasEqn. 6.4.5a. Applying a similar procedure to the second equilibrium equation gives Eqn.Solid Mechanics Part II146Kelly
Section 6.46.4.5b. Finally, integrating directly the third equilibrium equation without multiplyingacross by z, one arrives at Eqn. 6.4.3.Now, eliminating the shear forces from 6.4.3, 6.4.5 leads to the differential equation 2 M xy 2 M y 2M x 2 q x y x 2 y 2(6.4.8)This equation is analogous to the equation 2 M / x 2 p in the beam theory. Finally,substituting in the moment-curvature equations 6.2.31 leads to1q 4w 4w 4w 2 4224D y x y x(6.4.9)This is sometimes called the equation of Sophie Germain after the French investigatorwho first obtained it in 18152. This partial differential equation is solved subject to theboundary conditions of the problem, i.e. the fixing conditions of the plate (see below).Again, when once an expression for ( x, y ) is obtained, the strains, stresses, forces andmoments follow.Note that the differential equation 6.4.8 with q 0 is trivially satisfied in the simple purebending and torsion problems considered earlier.Eqn. 6.4.9 can be succinctly expressed as 4w qD(6.4.10)where 2 is the Laplacian, or “del” operator: 2 2 2 x 2 y 2(6.4.11)Note that the Laplacian
Also, the radius of curvature Rx, Fig. 6.2.2, is the reciprocal of the curvature, Rx 1/ x. Fig. 6.2.2: Angle and arc-length used in the definition of curvature As with the beam, when the slope is small, one can take tan w/ x and d /ds / x and Eqn. 6.2.2 reduces to (and similarly for the curvature in the y direction) 2 2 2
CVR Odometer and Plate Type Updates - EVR Policies . Plate Types . Plate Type . Plate Types. There are select plates that can now be issued as a new plate through EVR. Just like it works for special plates on EVR, these plate types will go through the . Centralized Plate Distribution Process. You will select the plate type during a New Plate .
For plates - Thin & thick plates – Thin plate b smallest side Thick plate t 20 b t 20 b 2000 1 10 1 b t t rd_mech@yahoo.co.in Ramadas Chennamsetti 5 Small deflections – Thin plate theory – Kirchoff’s Classical Plate Theory (KCPT) Thick plate theory – Reissner – Mindlin Plate Theory (MPT) 5 w
PROPERTY PROFILE REPORT 1 Sample Road, Albany, Auckland, 0632 Prepared on 08 July 2020 PLESAMPLE. 1 Sample Road, Albany, Auckland, 0632 3 1 2 210m2 612m2 Property Details . Email Us: help@corelogic.co.nz 1 Sample Road, Albany, Auckland, 0632 CoreLogic Property Profile Report
bistro bambina 268 ponsonby rd ponsonby 1011 central auckland blue elephant thai restaurant 237 parnell road parnell 1052 central auckland . cibo 91 st georges bay road parnell 1052 central auckland . jervois rd wine bar & kitchen 170 jervois rd herne bay 1011 central auckland
Dec 11, 2014 · Mob: 021 969 4444 Email: George.Hawkins@aucklandcou ncil.govt.nz Ken Penney 146e Great South Road Manurewa, Auckland 2012 Mob: 021 287 2244 Email: Ken.Penney@aucklandcouncil.govt.nz Angela Cunningham –Marino C/- Auckland Council Private Bag 92300 Auckland 1142
proceedings Proceedings Integrating the OpenSky Network into GNSS-R Climate Monitoring Research † Mike Laverick1,* , Delwyn Moller 2, Christopher Ruf 3, Stephen Musko 3, Andrew O'Brien 4, Ryan Linnabary 4, Wayne Thomas 5, Chris Seal 1 and Yvette Wharton 1 1 Centre for eResearch, The University of Auckland, Auckland 1010, New Zealand; c.seal@auckland.ac.nz (C.S.); y.wharton@auckland.ac.nz .
THE CompLETE pARTS REFEREnCE 5 TABLE S-4 the LEGO plates (continued) LABEL NAME COLOR J plate 2x4 orange, black K plate 2x6 dark azure L plate 2x8 dark grey m plate 2x12 grey n plate 4x4 black o plate 6x10 black p plate 4x4 corner round black Q plate 4x4 round grey R wing 4x4 black S wing 3x8 right black T wing 3x8 left black h
2x7 Matrix Plate, Straight SP-2630 2x7 Matrix Plate, Curved SP-2631 4-Hole Straight Plate 43-1005 6-Hole Straight Plate with Gap SP-2703 7-Hole Straight Plate 43-1007 16-Hole Straight Plate 43-1016 16-Hole Straight Plate 42-1016. PLATES – SILVER. 6-Hole Double Y Plate, Small SP-2705. MANDIBLE
