Thin plates with large deflection - Buckling instability and response

Transcription

1 Chpter 2 Thin pltes with lrge deflection - Buckling instility nd response In the present chpter some issue on the theory of thin pltes with lrge deflections will e ddressed. This llows to study oth structurl stility nd response prolems for this clss of structure. 2.1 Generl equtions of isotropic plte with lrge deflection In this section to generl model for the coupled in-pln/out-of-plne prolem for flt plte will e ddressed Kinemtic reltionships From the hypothesis ε zx = ε zy = ε zz = 0 one hs: thus: u(x, y, z = u 0 (x, y z w v(x, y, z = v 0 (x, y z w w(x, y, z = w 0 (x, y ε xx = u 0 z 2 w 2 ε yy = v 0 z 2 w 2 γ xy = u 0 + v 0 2z 2 w From now on w(x, y, z = w 0 (x, y the zero suscript will not e used ny more. (2.1 (2.2 25

2 Constitutive lw for linerly isotropic mterils σ xx = σ yy = E 1 ν 2 (ε xx + νε yy = E ( u0 1 ν 2 ( v0 σ xy = Gγ xy = E 1 ν 2 (ε yy + νε xx = E E 2(1 + ν 1 ν 2 ( u0 + v 0 The moments per unit length re defined s follows: M y := M x := M xy := zσ xx dz = zσ yy dz = zτ xy dz = + ν v ( o E 2 w z 1 ν ν 2 w 2 + ν u ( o E 2 w z 1 ν ν 2 w 2 z E 2 w 1 + ν [ ( z 2 E 2 w dz 1 ν 2 ( E z 2 dz 1 ν 2 ( E z 2 dz 1 + ν The forces per unit lenght re defined s follows: N x = N y = N xy = σ xx dz = σ yy dz = τ xy dz = 2 + ν 2 w 2 (2.3 ] Et 3 = 12(1 ν 2 (w xx + νw yy (w yy + νw xx = D(w yy + νw xx w xy = D(1 νw xy Et ( u0 1 ν 2 + ν v 0 Et ( v0 1 ν 2 + ν u 0 Et 1 + ν ( v0 + u 0 In ddition there re integrl definitions without relted constitutive lws ut expressed y the locl equilirium rund x nd y xis (see next Eqs. 2.7 nd 2.8 given y: Q x = Q y = σ xz = M y σ yz = M x + M xy + M xy = D(w xx + νw yy x D(1 νw xyy = D 2 w = D 2 w (2.4 (2.5 ( Equilirium Rottions round the x nd y xes x xis: M x + M xy = Q y (2.7

3 27 y xis: M y + M xy = Q x (2.8 NB: the effects of the position of the deformed configurtion re higher order effects for the ove equilirium equtions. Figure 2.1: Equilirium to rottions Trnsltions long x nd y directions By hypothesis memrne distriuted lods re not considered. It is used the following nottion for descriing the locl plte deformtion (see Fig. 2.4: β = w β = w + 2 w 2 dx β = w + 2 w dy furthermore cosβ 1, cosβ 1, cosβ 1 (see Fig. 2.4 so the equilirium long the x xis cn e written in the form: (2.9 N x dy + (N x + N x dxdy N xydx + (N xy + N xy dydx = 0 (2.10 thus is otined: N x + N xy = 0 (2.11

4 28 long the y xis for the equilirium is defined similrly: γ = w γ = w + 2 w 2 dy γ = w + 2 w dx (2.12 with cosγ 1, cosγ 1, cosγ 1. One cn write: N y dx + (N y + N y dydx N xydy + (N xy + N xy dy (2.13 nd N y + N xy = 0 (2.14 Figure 2.2: Equilirium long the (x,y trnsltions Trnsltions long z direction The lnce is composed of prt due to sher forces per unit length Q nd prt relting to the forces per unit length N. The representtions of the stte of stress Q from the deformed

5 29 configurtion to e projected long the z direction must e multiplied y the cosines of the ngles. For smll ngles these cosines re pproximted to one then the equilirium is nlogous to the undeformed configurtion: Q x dy + (Q x + Q x dxdy Q ydx + (Q y + Q y dydx + p(x, ydxdy +... = 0 (2.15 where p(x, y is the distriuted lod directed long z. Figure 2.3: Equilirium to trnsltion long the z xis. Q components. The components of N to e projected long the direction of z must e multiplied y the sines of the ngles, the ngles themselves re considered in sustitution of the sines due to their smll vlues. It is considered the rottion of n infinitesiml prt of the plte round the xis y of the quntities β, β, β : N x dyw x + (N x + N x dx(w x + w xx dxdy N xy w x dx + (2.16 (N xy + N xy dydx(w x + w xy dy = (2.17 = N x w xx dxdy + N x w xdxdy + N xy w xy dxdy + N xy w xdxdy (2.18 Considering the similr contriution otined y the rottion out the x xis nd summing ech other these contriutions one otins: Q x dy + (Q x + Qx dxdy Q ydx + (Q y + Qy dydx + p(x, ydxdy (2.19 ( +N x w xx dxdy + N xy w xy dxdy + w Nx x + Nxy dxdy + (2.20 N y w yy dxdy + N xy w xy dxdy + w y ( Ny Due to the equtions 2.11 nd 2.14 it cn e simplified to: Q x + Q y + N 2 w x 2 + 2N xy w + N y + Nxy dxdy = 0 ( w = p(x, y (2.22 2

6 30 Figure 2.4: Equilirium to trnsltion long the z xis. N components Finl Equtions Tking the derivtive with respect to x nd y of Eqs. 2.7 nd 2.8 nd sustituting into the Eq. 2.22, even fter using the definitions of moments per unit length 2.4, one hs for the homogenous plte D 4 2 w w N x 2 2N 2 w xy N 2 w y + ϱt ẅ = p(x, y ( where the inertil force hs een lso included in the lnce in z direction. Finlly, the prolem is closed y using the Eqs nd 2.14 together with the constitutive Eqs. 2.5 which define the memrne prolem for the unknowns u nd v. 2.2 Anlysis of two-edge uckled rectngulr plte In the present section the stility nd response nlysis of two-edge uckled rectngulr plte is presented Buckling stility nlysis For the in-plne memrne prolem, the oundry conditions re the sme s in the Figure 2.5.

7 31 Figure 2.5: Constrints nd lods pplied to the plte. Considering the Eqs nd 2.14 one cn otin: N x (x, y = N N y (x, y = 0 N xy (xy = 0 These lso stisfy the oundry conditions. In ddition, the equtions 2.5 ecomes: N = Et 1 ν 2 (u 0x + νv 0y 0 = Et 1 ν 2 (v 0y + νu 0x 0 = E 1 + ν (u 0y + v 0x (2.24 (2.25 Figure 2.6 Multipling the first y ν nd sutrcting it to the second: thus: νn = Et 1 ν 2 ( u 0xν ν 2 v 0y + v 0y + νu 0x (2.26 v 0 = N Et ν (2.27

8 32 From the first minus the second y ν simplifying: Thus: N = Et 1 ν 2 [u 0x + νv 0y νv 0y ν 2 u 0x ] (2.28 u o = N Et (2.29 v 0 (x, y = νn Et y + f(x u 0 (x, y = N (2.30 Et x + g(y From the third hving to e df dx = dg dy we hve tht this mkes sense only if f(x = c 1 nd g(y = c 2. c 1 is determined y imposing tht for symmetry resons v(x, /2 = 0 nd the sme for c 2 eing u(0, y = 0, nd therefore: v 0 (x, y = νn ( y Et 2 u 0 (x, y = N (2.31 Et x Next, the following out-of-plne prolem hs to e solved: D 4 w + N 2 w 2 + w ρh 2 t 2 = 0 (2.32 B.C. of simply supported plte I.C. ddressing s generic prolem of dynmic response The prolem will e solved using the eigenfunctions method. Indeed, the structurl opertor is now defined s: with w = 0 t the edges nd: L := D 4 +N 2 2 ( w 2 = 0 2 w 2 = 0 for x = 0, for y = 0, The eigenfunctions Φ mn (x, y nd the eigenvlues λ mn such tht: (2.34 LΦ mn = λ mn Φ mn (2.35 re still orthogonl nd the λ mn re rel, ut the opertor is not necessry positive or, with positive eigenvlues: in fct, in this cse one hs the following functions stisfying the given oudnry conditions: ( mπx Φ mn (x, y = C sin sin ( nπy (2.36

9 33 nd thus, in order to e the eigenfunctions of the given prolem they hve to stispy lso Eq. 2.35, or [ (mπ 2 ( nπ ] 2 2 λ mn = D + N ( mπ 2 m, n = 1, 2, 3... (2.37 with N modifying the positiveness of the λ mn. It is worth to pointing out tht the eigenvlues λ mn depend on the prmeter N ut the eigenfunctions Φ mn (x, y. Now, the generic dynmic free-response prolem defined y Eq cn e solved with the eigenfunctions method so writing: Φ mn (λ mn w mn + ρhẅ mn = 0 λ mn w mn + ρhẅ mn = 0 (2.38 m,n where w(x, y = Φ mn (x, yw mn (t (2.39 m,n The following infinite free virtion prolems hve to e solved ẅ mn + ωmnw 2 mn = 0 w mn (0 = < w(x, y, 0, Φ mn(x, y > < Φ mn (x, y, Φ mn (x, y > ẇ mn (0 = < w(x, y, 0, Φ mn (x, y > < Φ mn (x, y, Φ mn (x, y > m, n = 1, 2,... (2.40 with ω 2 mn := λmn ρh. The poles s mn of second order differentil equtions re given y s 2 mn + ω 2 mn = 0 s mn = ±jω mn (2.41 nd give oscilltory solution, when perturtion is ssigned y mens of the initil conditions, only if λ mn > 0 once the prmeter N might e incresed. However, with the increse of N y 2.37 ll λ mn decrese, so the system poles tend to move on the imginry xis of the complex plne towrds the origin. For ll m, n N cr such tht ωmn 2 = 0, or [ (mπ 2 ( nπ λ mn = D + ] 2 2 ( mπ 2 N cr = 0 (2.42 the lowest of these, for ll m, n, is the criticl (zero pole. Supposing to hve found this, then this is the condition of doule chrcteristic exponent (pole in the origin of the complex plne of the poles. For λ cr rs = 0 from the 2.40 one gets: ẅ cr rs = 0 = w cr rs(t = A rs t + B rs (2.43

10 34 Thus, re-uilding the complete solution y the eigenfunctions one hs: w(x, y, t = w mn (tφ mn (x, y (2.44 m,n = (A rs t + B rs Φ cr rs(x, y + m,n m,n r,s [A mn cos(ω mn t + B mn sin(ω mn t] Φ rs (x, y It is noted tht, in order to determine N cr, inertil forces re not essentil ecuse it is necessry to consider the condition for hving zero-eigenvlue λ mn for the structurl opertor, or, zeropoles in the chrcteristic eqution mn ( s 2 + ω 2 mn = 0. In other words, N is such tht to put equl to zero the known term of the criticl-elementry chrcteristic eqution s 2 + λ mn = 0 nd in this procedure, inertil prmeters ρt re clerly not influent. Now, the engineering prolem is to find the minimum vlue for N with respect to m, n (y incresing N from zero, which verifies zero eigenvlue condition. Note tht this prolem depends on the plte geometry, nmely, y its spect rtio /. Thus, for given geometry,, one should find for ll m, n choices, which (m, n-pir gives the lowest criticl vlue for N. Indeed, setting the Eq to zero, one otins: [ ( D mπ 2 ( + nπ ] 2 2 [ ( m 2 ( N cr (m, n = ( mπ 2 = Dπ 2 + n ] 2 2 ( m 2 (2.45 It is pprent tht ny optiml pir hs n = 1 (n = 0 is not cceptle. For m dependency one hs: [ N cr (m = Dπ 2 m + ] [ m 2 = Dπ2 m 2 ρ + ρ ] 2 = Dπ2 m 2 k m (2.46 [ m with ρ := / nd k(ρ, m := ρ + ρ ] 2. In order to find the optiml m which minimize the [ m m positive quntity ρ + ρ ] once ρ is ssigned, two inequlities must e solved: m m ρ + ρ m < m ρ ρ m + 1 m ρ + ρ m < m 1 + ρ (2.47 ρ m 1 Once solved, one gets: { ρ 2 < m(m + 1 ρ 2 > m(m 1 Therefore, if the spect rtio ρ is ssigned, m vlue hs to stisfy: (2.48

11 35 Figure 2.7: Generl ehviour of k(m Figure 2.8: Plots of functions k(ρ, m. m(m 1 < ρ 2 < m(m + 1 (2.49 Moreover, once m is ssigned, the minimum with respect to ρ of the function k m (ρ is given y: dk m dρ = 0 = d dρ All the otined results re presented in Figs. 2.7 nd 2.8. [ m ρ + ρ ] 2 ( m = 2 m ρ + ρ [ m m ρ ] = ρ = m (2.50 m

12 36 Note lso tht if the limit cse ρ << 1 is present, for m=1 one cn ssume: k = ( 1 2 ρ + ρ 1 ρ 2 (2.51 or, 1 k ρ2, which implies tht N cr Dπ2 2. This mens tht in this limit cse the pltes prcticlly ehves like ucked em. Figure 2.9: Two cses of pltes with respectively ρ = m = 2 nd ρ = m = Response of uckled plte to sttic pressure lod If the plte is structurlly stle, n effective response prolem cn e crried out. The lod is in this cse comined lod with pressure field p(x, y = p 0 nd n in-plne uckling lod in the x direction given y N x := N. The finl eqution for the sttic prolem is 4 w = 1 D ( p 0 N 2 w 2 For the ssumed geometry nd oundry conditions the eigenfunction pproch cn e used expnding the pressure lod which uses only the odd indexes in this cse (see Section 1.4.2: p(x, y = m=1 n=1 p mn sin mπx nπy sin = 16p 0 π 2 m=1 n=1 1 mπx sin mn sin nπy m = n = 1, 3, 5, nd therefore for the solution w(x, y one hs: w(x, y = m=1 n=1 w mn sin mπx nπy sin Thus m=1 n=1 { [ (m w mn π 4 2 D n2 2 N ] } ( m 2 p mn sin mπx D π sin nπy = 0

13 37 now, s lwys, it is necessry to set to zero the term inside the rces to otin the reltions etween the unknown coefficients w mn nd those lredy known p mn : w mn = 16p [ 0 ( ] π 6 Dmn m 2 2 ( + n2 2 N m 2 2 D π Thus the deflection w(x, y is: 16p 0 w(x, y = [ ( π 6 Dmn m 2 ( + n 2 ( N m ] 2 2 sin mπx D π sin nπy It is pprent tht the presence of t memrne lod N, rings to n decrese of the deflection w(x, y in the cse the lod is trction lod nd to n increse of the deflection (nmely, decrese of the plte stiffness in the cse of compression lod, s it my look intuitive Response of uckled plte hving n initil deflection Let us consider now the cse of plte, in generl lod cse conditions in the plin nd norml, in which there is n initil known deflection w 0 (x, y too. 1 s sum of the initil one nd tht one due to the loding cse: w(x, y = w 0 (x, y + w 1 (x, y The totl deflection cn e written where w 1 (x, y is the elstic plte displcement with respect to the initil deflection w 0 (x, y. The plte eqution ecomes: D 4 2 w w 1 = p + N x 2 + 2N 2 w xy + N 2 w y 2 it is pointed out tht the initil deflection w 0 (x, y does not occur in the left hnd side term of the eqution ecuse it derives from the vrition of the flexurl moments. From the plte eqution one cn see how the presence of the initil deflection w 0 (x, y cuses n effect similr to tht of side lod p given y the: p = N x 2 w N 2 w 0 xy + N y 2 w 0 2 For exmple let us consider rectngulr plte supported on its order in the cse the initil deflection cn e expressed y: w o (x, y = m=1 n=1 mn sin mπx nπy sin ( Oviously the initil deflection w 0(x, y must e smller with respect to the plte thickness s stted y min hypothesis.

14 38 where the mn coefficients cn e evluted là Fourier once w 0 (x, y is given. So they re known once the initil deflection is given. If uniform uckling lod on the edges x = is pplied, then one hs for the xcomponent of the in-plne stress N x (x, y = N. So, if the pressure lod p(x, y is not considered one hs: D 4 w 1 = N 2 w 1 2 N 2 w 0 2 If lso the unknown w 1 (x, y is represented s n expnsion of the prolem eigenfunction one hs w 1 (x, y = m n w mn sin mπx nπy sin Thus, pplying the procedure of the eigenfunction method one hs for the components w mn : w mn = D 2 π 2 m 2 mn N [ ( m 2 ( + n ] 2 2 N Note tht solution is otined if the ucklung condition for the prmeter N = N cr, see Section 2.2.1, N CR = D2 π 2 m 2 [ (m 2 + ( n is voided. Indeed, s the component w mn ecomes lrger with the increse of the compression 2 ] 2 lod N, they my rech n infinite vlue when the criticl uckling for the plte is set. 2.3 Some generl issues on uckling nlysis for spcediscretized structurl models Whenever n nlytic solution is not ville, spce-discretized pproch hs to e tken into ccount. Specificlly, the linerized uckling prolem cn e generlly descried in PDE form of type L (1 u N L (2 u + ϱü = f + B.C. + I.C. (2.53 where L (1 represents the structurl opertor ssocited to the locl elstic properties of the continuum wheres L (2 represents the linerized contriution on the deformed configurtion of the stress component loclly orthogonl to the displcement ut t the undeformed configurtion due to the compressive lod N. In the flt plte cse (see Section these 3D opertors re sclr nd given y L (1 = D 4 L (2 = 2 (2.54 2

15 39 whenever the uckling lod is in x direction. Next, ssuming generl spce-discretiztion, for instnce, tht given y the finite element pproch (so introducing the ht spce functions one ssumes u(x, t N = ˆψn (xˆq n (t (2.55 n 1 Thus, sustituting to Eq one hs N (L (1 ˆψn NL (2 ˆψn ˆq n + n=1 N ϱ ˆψ n ˆqn = f (2.56 n=1 Then, projecting on ˆψ m one hs N L (1 ˆψn NL (2 ˆψn, ˆψ m ˆq n + n N n ϱ ˆψ n, ˆψ m ˆqn = f, ˆψ m (2.57 which reduces to (ˆK N ˆKd ˆq + ˆM ˆq = ê (2.58 with: ˆK nm := L (1 ˆψn, ˆψ m = ˆK dnm := L (2 ˆψn, ˆψ m = V V L (1 ˆψn ˆψ m dv L (2 ˆψn ˆψ m dv (2.59 So the limit uckling lod for N, nmely, N cr, is tht which give zero poles for the dynmic system given y Eq Therefore, it cn e evluted directly solving the following eigenprolem nd considering the lowest rel eigenvlue N cr nd the ssocited eigenvector w cr ] [0 2 ˆM + (ˆK Ncr ˆKd w cr = 0 (ˆK Ncr ˆKd w cr = 0 (2.60 Then, the criticl uckling mode shpe ϕ cr (x cn e lso reuilt in the physicl spce using Eq nd the eigenvector w cr components so otining ϕ cr (x = ˆψn (xw ncr (2.61 As finl comment for the numericl procedure emploied for the evlution of the differentil mtrix ˆK d, it is worth to pointing out tht this evlution is performed in the neighourhood of sttic (typiclly memrne like, for exmple, given y the lod vector p. Thus, ˆKd is evluted on the se of the reference sttic prolem ˆKq s = p giving the reference solution q s. This informtion llows to evlute the differntil mtrix ˆK d to perform the uckling stility nlysis.

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