De Pascalis
					, 
						Riccardo

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Università del Salento: ESE - Salento University Publishing

Chapter 6Euler buckling for compressiblecylindersOne of the first, and most important, problems to be tackled by the theoryof linear elasticity is that of the buckling of a column under an axial load. UsingBernoulli’s beam equations, Euler found the critical load of compressionNcr leadingto the buckling of a slender cylindrical column of radius B and length L. Asrecalled by Timoshenko and Gere [124], Euler looked at the case of an ideal columnwith (built in)-(free) end conditions. What is now called “Euler buckling”, or the“fundamental case of buckling of a prismatic bar” [124] corresponds to the case ofa bar with hinged-hinged end conditions. The corresponding critical load is givenbyNcrπ3B2=E4(BL)2, (6.1)where E is the Young modulus. The extension of this formula to the case ofa thick column is a non-trivial, even sophisticated, problem of non-linear three-dimensional elasticity. In general, progress can be only made by using reductive(rod, shells, etc.) theories. However, there is another choice of boundary condi-tions for which the criterion (6.1) is valid: namely, the case where both ends are“guided” or ”sliding” (the difference between the two cases lies in the shape of thedeflected bar, which is according to the half-period of a sine in one case and of acosine in the other case). In exact non-linear elasticity, there exists a remarkablethree-dimensional analytical solution to this problem (due to Wilkes [129]) whichdescribes a small-amplitude (incremental) deflection superimposed upon the largehomogeneous deformation of a cylinder compressed between two lubricated platens.In this case, the Euler formula can be extended to the case of a column with finitedimensions, for arbitrary constitutive law.The exact incremental solution allows for an explicit derivation of Euler’s for-mula at the onset of non-linearity, which combines third-order elastic constantswith a term in (B/L)4. Goriely et al. [47], showed that for an incompressiblecylinder,Ncπ3B2=E4(BL)2− π296(203E + 9A)(BL)4, (6.2)where A is Landau’s third-order elasticity constant. This formula clearly shows109110 Chapter 6. Euler buckling for compressible cylindersthat geometrical non-linearities (term in (B/L)4) are intrinsically coupled to phys-ical non-linearities (term in A) for this problem. (For the connection betweenEuler’s theory of buckling and incremental incompressible nonlinear elasticity, seethe early works of Wilkes [129], Biot [13], Fosdick and Shield [41], and the referencescollected in [47].)Now, in third-order incompressible elasticity, there are two elastic constants:the shear modulus µ (= E/3) and A (see (2.45)). In third-order compressible elas-ticity, there are five elastic constants: λ and µ, the (second-order) Lame´ constants(or equivalently, E and ν, Young’s modulus and Poisson’s ratio), and A, B, and C,the (third-order) Landau constants (see (2.43)). Euler’s formula at order (B/L)2,equation (6.1), involves only one elastic constant, E. It is thus natural to askwhether Poisson’s ratio, ν, plays a role in the non-linear correction to Euler for-mula of (B/L)4, the next-order term. The final answer is found in our recent work[26] as formula (6.32) below, which shows that the non-linear correction involvesall five elastic constants.6.1 Finite compression and bucklingIn this section, we recall the equations governing the homogeneous compressionof a cylinder in the theory of exact (finite) elasticity and we also use the form ofsome incremental solutions that is, of some small-amplitude static deformationswhich may be superimposed upon the large compression and which indicate theonset of instability for the cylinder.The mathematical method used to determine the solutions for incremental so-lution is another example for the semi-inverse method. In fact our ansatz for thecomponents of the mechanical displacement will be in looking for incremental staticsolutions that are periodic along the circumferential and axial directions, and haveyet unknown radial variations (see formula (6.10)).6.1.1 Large deformationWe take a cylinder made of a hyperelastic, compressible, isotropic solid withstrain energy function W = Wˆ (λ1, λ2, λ3) say, with radius B and length L in itsundeformed configuration. We denote by (R,Θ, Z) the coordinates of a particle inthe cylinder at rest, in a cylindrical system.Then we consider that the cylinder is subject to the following deformation,r = λ1R, θ = Θ, z = λ3Z, (6.3)where (r, θ, z) are the current coordinates of the particle initially at (R,Θ, Z), λ1 isthe radial stretch ratio and λ3 is the axial stretch ratio. Explicitly, λ1 = b/B andλ3 = l/L, where b and l are the radius and length of the deformed cylinder, respec-tively. The physical components of F , the corresponding deformation gradient,are:F = Diag (λ1, λ1, λ3) , (6.4)showing that the deformation is equi-biaxial and homogeneous (and thus, univer-sal).Chapter 6. Euler buckling for compressible cylinders 111The (constant) Cauchy stresses required to maintain the large homogeneouscompression are given by formula (1.42),Ti = J−1λiWi, i = 1, 2, 3 (no sum), (6.5)where Wi ≡ ∂W/∂λi. In our case, T1 = T2 because the deformation is equi-biaxial,and T1 = T2 = 0 because the outer face of the cylinder is free of traction. HenceT1 = T2 = λ−11 λ−13 W1 = 0, T3 = λ−21 W3. (6.6)Note that we may use the first equality to express one principal stretch in termsof the other (provided, of course, that inverses can be performed).6.1.2 Incremental equationsNow we recall the equations governing the equilibrium of incremental solutions,in the neighborhood of the finite compression. They read in general as (1.80)div ΣT = 0, (6.7)whereΣ is the incremental nominal stress tensor. It is related to u, the incrementalmechanical displacement, through the incremental constitutive law (1.81), that incomponent form can be written asΣji = A0jilkuk,l, (6.8)where the comma denotes partial differentiation with respect to the current coordi-nates and A0 is the fourth-order tensor of incremental elastic moduli. Its non-zerocomponents, in a coordinate system aligned with the principal axes of strain, aregiven by (1.84). Note that here, some of these components are not independentone from another because λ1 = λ2 and T1 = T2 = 0. In particular, we find thatA01212 = A02121 = A01221, A02323 = A01313 = A01331 = A02332,A02222 = A01111, A02233 = A01133, A03232 = A03131, (6.9)A01122 = A01111 − 2A01212.6.1.3 Incremental solutionsWe look for incremental static solutions that are periodic along the circum-ferential and axial directions, and have yet unknown radial variations. Thus ouransatz for the components of the mechanical displacement is the same as Wilkes’s[129]:ur = Ur(r) cosnθ cos kz,uθ = Uθ(r) sinnθ cos kz, (6.10)uz = Uz(r) cosnθ sin kz,where n = 0, 1, 2, . . . is the circumferential mode number ; k is the axial wavenum-ber ; the subscripts (r, θ, z) refer to components within the cylindrical coordinates(r, θ, z); and all upper-case functions are functions of r alone.112 Chapter 6. Euler buckling for compressible cylindersDorfmann and Haughton [31] show that the following displacements U (1), U (2),and U (3) are solutions to the incremental equations (6.7),U (1)(r), U (2)(r) =[I ′n(qkr),−nqkrIn(qkr),−(A01111q2 −A03131)q(A01313 +A01133) In(qkr)]T,(6.11)andU (3)(r) =[1rIn(q3kr),−q3knI ′n(q3kr), 0]T, (6.12)where q = q1, q2 in turn and In is the modified Bessel function of order n. Here q1,q2, and q3 are the square roots of the roots q21, q22 of the following quadratic in q2:A01313A01111q4 + [(A01133 +A01313)2 −A01313A03131−A03333A01111]q2 +A03333A03131 = 0, (6.13)and of the root of the following linear equation in q2A01212q2 −A03131 = 0, (6.14)respectively.From (6.8) we find that the incremental traction on planes normal to the axialdirection has components of the same form as that of the displacements, namelyΣrr = Srr(r) cosnθ cos kz,Σrθ = Srθ(r) sinnθ cos kz, (6.15)Σrz = Srz(r) cosnθ sin kz,say, where again the functions Srr, Srθ, Srz are functions of r alone. Then we findthat the traction solutions corresponding to the solutions (6.11) and (6.12) aregiven byrS(1)(r), rS(2)(r) =[2A01212I ′n(qkr)−(2A01212 n2qkr+ qkrA01111 − krA01133 (A01111q2 −A03131)q (A01313 +A01133))In(qkr),2nA01212(In(qkr)qkr− I ′n(qkr)),−krA01313(A01111q2 −A03131A01313 +A01133 + 1)I ′n(qkr)]T,(6.16)andrS(3)(r) =[2A01212(In(q3kr)r− q3kI ′n(q3kr)),A01212(2q3knI ′n(q3kr)−(2nr+q23k2rn)In(q3kr)),−kA01313In(q3kr)]T. (6.17)Chapter 6. Euler buckling for compressible cylinders 113The general solution to the incremental equations of equilibrium is thus of theformrS(r) =[rS(1)(r)∣∣∣rS(2)(r)∣∣∣rS(3)(r)] c, (6.18)where S ≡ [Srr, Srθ, Srz]T , and c is a constant three-component vector. Note thatwe use the quantity rS for the traction (instead of S), because it is the Hamiltonianconjugate to the displacement in cylindrical coordinates [117].Now when the cylinder is compressed (by platens say), its end faces should stayin full contact with the platens so that the first incremental boundary condition isuz = 0, on z = 0, l, (6.19)which leads to [31, 47]k = mπ/l, (6.20)for some integer m, the axial mode number. From (6.15), we now see that on thethrust faces, we haveΣrz = 0, on z = 0, l, (6.21)which means that the end faces of the column are in sliding contact with thethrusting platens. In other words, in the limit of a slender column, we recover theEuler strut with sliding-sliding, or guided-guided end conditions. In Figure 6.1,we show the first two axi-symmetric and two asymmetric modes of incrementalbuckling.The other boundary condition is that the cylindrical face is free of incrementaltraction: S(b) = 0. This gives∆ ≡ det[bS(1)(b)∣∣∣bS(2)(b)∣∣∣bS(3)(b)] = 0. (6.22)6.2 Euler buckling6.2.1 Asymptotic expansionsWe now specialize the analysis to the asymmetric buckling mode n = 1, m = 1,corresponding to the Euler buckling with guided-guided end conditions, in thelimit where the axial compressive stretch λ3 is close to 1 (the other modes arenot reached for slender enough cylinders). To this end, we only need to considerthe third-order elasticity expansion of the strain energy density (2.43). (Note thatthere are other, equivalent, expansions based on other invariants, such as the onesproposed by Murnaghan [86], Toupin and Bernstein [125], Bland [17], or Eringenand Suhubi [37], see Norris [91] for the connections.)To measure how close λ3 is to 1, we introduce ǫ, a small parameter proportionalto the slenderness of the deformed cylinder,ǫ = kb = πb/l. (6.23)114 Chapter 6. Euler buckling for compressible cylindersn=0m=1n=0m=2n=1m=1n=1m=2Figure 6.1: First two axi-symmetric and two asymmetric modes of buckling for acompressed strut with guided-guided end conditions: n is the circumferential modenumber and m the axial mode number. For slender enough cylinders, the n = 1,m = 1 mode is the first mode of buckling.Then we expand the radial stretch λ1 and the critical buckling stretch λ3 in termsof ǫ up to order M ,λ1 = λ1(ǫ) = 1 +M∑p=1αpǫp +O(ǫM+1), λ3 = λ3(ǫ) = 1 +M∑p=1βpǫp +O(ǫM+1),(6.24)say, where the α’s and β’s are to be determined shortly. Similarly, we expand ∆in powers of ǫ,∆ =Md∑p=1dpǫp +O(ǫMd+1),and solve each order dp = 0 for the coefficients αp and βp, making use of thecondition T1 = 0. We find that αp and βp vanish identically for all odd values ofp, and that λ1 and λ3, up to the fourth-order in ǫ, are given byλ1 = 1 + α2ǫ2 + α4ǫ4 +O(ǫ6), λ3 = 1 + β2ǫ2 + β4ǫ4 +O(ǫ6), (6.25)Chapter 6. Euler buckling for compressible cylinders 115with α2 and α4 given byα2 =ν4, (6.26)andα4 = −ν(1 + ν)32−(1 + ν)(1− 2ν)16E[ν2A+ (1− 2ν + 6ν2)B + (1− 2ν)2C]− νβ4, (6.27)whereinβ2 = −14, (6.28)andβ4 =29 + 39ν + 8ν296(1 + ν)+116E[(1− 2ν3)A+ 3(1− 2ν)(1 + 2ν2)B + (1− 2ν)3C] .Note that we switched from Lame´ constants to Poisson’s ratio and Young’s modulusfor these expressions, using the connections (1.53).6.2.2 Onset of nonlinear Euler bucklingThe analytical results presented above are formulated in terms of the currentgeometrical parameter ǫ, defined in (6.23). In order to relate these results to theclassical form of Euler buckling, we introduce the initial geometric slendernessB/L. Recalling that ǫ = πb/l, λ3 = l/L, and b = λ1B, we find thatǫλ3 = πλ1(B/L). (6.29)We expand ǫ in powers of B/L, and solve (6.29) to obtainǫ = π(B/L) + (α2 − β2)π3(B/L)3 +O((B/L)4)= π(B/L) + (1 + ν)(π3/4)(B/L)3 +O ((B/L)4) . (6.30)Second, we wish to relate the axial compression to the current axial load N .To do so, we integrate the axial stress over the faces of the cylinder,N = −2π∫ b0rT3dr = −πb2T3 = −πλ21B2T3, (6.31)because T3 is constant, given by (6.6)2.Finally, in order to write the nonlinear buckling formula, we expand λ1 andλ3 in (6.31), using (6.25), and then expand ǫ in powers of the slenderness (B/L),116 Chapter 6. Euler buckling for compressible cylindersusing (6.30). It gives the desired expression for the first non-linear correction toEuler formula,Ncπ3B2=E4(BL)2− π296δ NL(BL)4, (6.32)whereδ NL = 213 + 12ν − 2ν2(1 + ν)E+12[(1− 2ν3)A+ 3(1− 2ν)(1 + 2ν2)B + (1− 2ν)3C] . (6.33)We now check this equation against its incompressible counterpart (6.2). The-oretical considerations and experimental measurements [22, 29, 94, 131], show thatin the incompressible limit, E and A remain finite,ν → 1/2, (1− 2ν)B → −E/3, (1− 2ν)3C → 0. (6.34)It is then a simple exercise to verify that (6.32) is indeed consistent with (6.2) inthose limits.6.2.3 ExamplesTable 6.1: Lame´ constants and Landau third-order elastic moduli for five solids (109 N·m−2)material λ µ A B CPolystyrene 1.71 0.95 −10 −8.3 −10.6Steel Hecla 37 111 82.1 −358 −282 −177Aluminium 2S 57 27.6 −228 −197 −102Pyrex glass 13.5 27.5 420 −118 132SiO2 melted 15.9 31.3 −44 93 36To evaluate the importance of the non-linear correction, we computed the crit-ical axial stretch ratio of column buckling for two solids. In Table 6.1, we list thesecond- and third-order elastic constants of five compressible solids, as collectedby Porubov [101] (in the Table we converted the “Murnaghan constants” given byPorubov to Landau constants A,B, C). Figure 6.2 shows the variations of λ3 withthe squared slenderness (B/L)2, for pyrex and silica (two last lines of Table 1).NotesThe present analysis can be considered an extension of the article [47] fromincompressible solids to compressible solids. It provides an asymptotic formulafor the critical value of the load for the Euler buckling problem, with guided-guided (sliding-sliding) end conditions. This formula was checked both in theincompressible limit and on particular cases against the exact value of the bucklingChapter 6. Euler buckling for compressible cylinders 117 Euler2Exact Euler  4(B/L)2λ3(a)EulerEuler24Exact(B/L)2λ3(b)Figure 6.2: Comparison of the different Euler formulas obtained by expanding theexact solution to order 2 (classical Euler buckling formula, plot labeled “Euler2”)and to order 4 (plot labeled “Euler4”), for pyrex (a) and for silica (b).obtained from the exact solutions. Not surprisingly it reinforces the universal andgeneric nature of the Euler buckling formula, as the correction is small for mostsystems even when nonlinear elastic effects and nonlinear geometric effects aretaken into account. It would be of great interest to see if these effects could beobserved experimentally.We note that Goriely et al. [47] write the equations for the linearized problemof instability using the Stroh formalism [122]. They adapt the work of Shuvalov[117] on waves in anisotropic cylinders to develop a Stroh-like formulation of theproblem. From a computational perspective, the Stroh formalism is particularlywell suited and well behaved (Biryukov [14]; Fu [42]) and if numerical integrationwas required it would provide an ideal representation of the governing equation.Here we could have also presented the linearized problem of instability for thecompressible cylinder within the Stroh formalism. We have omitted this formalismbecause it was not essential for our discussion.

Euler buckling for compressible cylinders

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Euler buckling for compressible cylinders

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