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Plane Strain Theory for an Orthotropic Cylinder with Pure Bending

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Plane Strain Theory for an Orthotropic Cylinder with Pure Bending Appendix A Plane Strain Theory for an Orthotropic Cylinder with Pure Bending Lekhnitskii (1963) derives a general theory for the state of stress in an elastic body bounded by a cylindrical surface and possessing cylindrical anisotropy where the stresses do not vary along the generator. In what follows, the analysis given in Archer (1976) for the transversely isotropic material model is extended to the orthotropic case (Archer 1985). The equations of equilibrium in cylindrical coordinates after dropping dependence on the axial coordinate, and the "torsion" equilibrium equation, and body force terms become ou, +! O't"re + u, -O"e =O or r oe r (A.1) O't"re +! OO"e + 2 't"re =O or r o(} r and the generalized Hooke's law for the orthotropic case e, [all [ Be l= a12 (A.2) Sz a13 where IX., IX9 , and IXz are prescribed strains as measured by strain relief experiments (Nicholson 1971, 1973a, Sasaki et al. 1978) and taken as independent ofz; u., ue, O"z, -r,e are stress components; e., Be, Bz, y9, are the strain components; r, (}, z the cylindrical coordinates; and all ... %6 the elastic constants. The shear stresses 't"ez and 't"rz have been uncoupled from the system (A.1-A.2) and are applicable to torsion problems which do not concern us here. The relevant strain-displacement relations are ou 1 ov u e,=or ' ee=~ oe+r (A.3) ow 1 ou ov v Bz= OZ ' l're=~ o(} +or-~ A modification of the Lekhnitskii (1963) derivation which includes the growth strains follows if we set (A.4) then since Sz and IXz are independent of z, we may write D +1Xz=a33 (Ar sin(}+ Brcos (}+C) , (A.5) 206 Plain Strain Theory for an Orthotropic Cylinder with Pure Bending where A and B correspond to "pure bending" stress terms and C to a "uniform axial" stress term. Thus the axial stress becomes (A.6) and u= U -z2/2(A sin 0+ B cos O)a33 +u0 cos 0+v0 sin() (A.7) v= V -z2/2(Acos () -B sin O)a33 -Uo sin 0+v0 cos 0+w3r , where (A.8) .. 1 ou ov v Yre=r oo+ar----r=P66"Cr8 . U and V depend only on rand() and the "rigid body" terms u0 , v0 , w3 have been separated out; and Pu =au -ai3/a33 , fJ12 = a12 -a13a23/a33 (A.9) fJ22 = a22 -a~3ja33 , /366 = a66 and &r = CXr -a13CXz/a33 (A.10) &e = CXe -a23cxz/a33 Now the pair of equilibrium equations (A.1) are satisfied if we write Fr Fee (F) Ur=-r-+7 ' Ue= Frr ' Tre=- r r() ' (A.11) where F is a "stress function" and ( )r = :r ( ), etc. A single equation for F follows if we eliminate U and V from (A.8) to obtain the compatibility equation 02 Be 1 02Br 2 OBe 1 OSr 1 02Yr9 1 OYrB (A.12) or+ r o0 2 +r ar--r ar-=r oro()+ r2 o() ' and then use (A.8) and (A.11) to obtain LF=2(a13 -a23) (A sinO+ B &8 r r coso)-(~+~~)orror (A.13) Plain Strain Theory for an Orthotropic Cylinder with Pure Bending 207 where (A.14) Restricting attention only to the cosine terms in the Fourier expansion of the growth strains gives 00 1Xr = I 1Xr,n cos ne n=O 00 IXe = I IXe,n cos ne (A.15) n=O 00 IXz = I IXz,n cos ne ' n=O where 1Xr,n' IXe,n, and IXz,n can be calculated from the measured or assumed distributions 1Xr(8), tX8 (8), and 1Xz{8). Thus &r =I &r,n cos ne (A.16) &o =I &e,n cos ne , where &r,n = 1Xr,n -a131Xz,n/a33 (A.17) &e,n =IXe,n -az31Xz,n/a33 If we seek solutions of (A.13) in the form 00 F= I f0 (r)cosn8 (A.18) n=O then each f" is obtained from the corresponding ordinary differential equation found by equating the coefficient of cos ne in (A.13) to zero. The case of n=O (Axisymmetric) is given in Chap. 4.1. For the n1h case the solution of the homogeneous equation (A.13) satisfies fn) n4fn n2fn) r,.v. +2 --nf~' 2 IX 1 (f~---+- f~ +1Xz (f~---+---2 f~ -- =0 , (A.19) r r2r3r4 r3rz ~ r4 where IX1 = (2 fJ12 + {366)/ f3zz (A.20) IXz =f311/f3zz · Solutions of (A.19) take the form 4 f0 = I Akrpk , (A.21) k=l 208 Plain Strain Theory for an Orthotropic Cylinder with Pure Bending TableA.l. Powers in stress function solution [Eq. (A21)] for various species• (green condition) Species n pl p2 p3 p4 Ash 0 2 0 1.675 0.325 2.614 -0.614 2 3.297 -1.297 1.882 0.118 3 3.877 -1.877 2.877 -0.877 Beech 0 2 0 1.664 0.336 1 2.501 -0.501 1 1 2 2.891 -0.891 2.054 -0.054 Birch 0 2 0 1.691 0.309 2.641 -0.641 1 2 3.358 -1.358 1.879 0.121 3 4.002 -2.002 2.842 -0.842 4 4.432 -2.432 4.020 -2.020 Maple 0 2 0 1.704 0.296 2.570 -0.570 1 2 3.083 -1.083 2.014 -0.014 Oak 0 2 0 1.635 0.365 1 2.564 -0.564 2 3.197 -1.197 1.867 0.133 3 3.688 -1.688 2.889 -0.889 Walnut 0 2 0 1.678 0.322 1 2.566 -0.566 1 2 3.124 -1.124 1.958 0.042 Yell ow poplar 0 2 0 1.628 0.372 2.781 -0.781 2 3.841 -1.841 1.663 0.337 3 4.977 -2.977 2.263 -0.263 4 6.150 -4.150 2.829 -0.829 Douglas fir 0 2 0 1.795 0.205 3.638 -1.638 1 2 5.764 -3.764 1.501 0.499 3 7.982 -5.982 1.911 0.089 4 10.231 -8.231 2.292 -0.292 Fir 0 2 0 1.665 0.335 2.747 -0.747 1 1 2 3.709 -1.709 1.737 0.263 3 4.728 -2.728 2.428 -0.428 4 5.780 -3.780 3.088 -1.088 Scots pine 0 2 0 1.667 0.333 1 3.543 -1.543 1 1 2 5.620 -3.620 1.433 0.567 3 7.784 -5.784 1.786 0.214 4 9.975 -7.975 2.114 -0.114 Sitka spruce 0 2 0 1.691 0.309 1 4.094 -2.094 1 1 2 6.809 -4.809 1.357 0.643 3 9.599 -7.599 1.642 0.358 4 12.411 -10.411 1.908 0.092 • For the first six species in the table, the roots for n values greater than those given in the table are complex. For these cases the solutions (A.21) take the form r•• cos bk In r and r•• sin bk In r. (Hildebrand 1976). Plain Strain Theory for an Orthotropic Cylinder with Pure Bending 209 Table A.2. Elastic constants (Hearrnon 1948, Table 2) modified to represent green condition [Hearmon 1948, Part 1(5)]. The Young's moduli and the rigidity moduli are given in MPa x 103 . Species Density EL ER ET VRT VLR VLT GLT GLR GTR Ash 0.67 12.9 0.71 0.32 0.92 0.26 0.77 0.60 0.95 0.16 Beech 0.75 11.2 1.05 0.46 0.98 0.26 0.77 0.71 1.14 0.28 Birch 0.62 13.3 0.52 0.25 1.01 0.28 0.64 0.61 0.84 0.11 Maple 0.59 8.15 0.71 0.35 1.07 0.26 0.75 0.74 0.87 0.17 Oak 0.66 4.32 1.01 0.39 0.83 0.19 0.75 0.51 0.92 0.23 Walnut 0.59 9.13 0.56 0.25 0.94 0.28 0.94 0.47 0.68 0.14 Yell ow poplar 0.38 7.91 0.42 0.16 0.91 0.18 0.58 0.45 0.51 0.066 Douglas fir 0.45-0.51 12.8 0.50 0.31 0.51 0.16 0.68 0.59 0.62 0.053 Fir 10.4 0.44 0.19 0.78 0.26 0.75 0.50 0.66 0.084 Scots pine 0.55 13.3 0.52 0.23 0.88 0.24 0.76 0.46 0.82 0.040 Sitka spruce 0.39 9.45 0.42 0.20 0.56 0.21 0.70 0.48 0.53 0.023 where P1 =1+qt, (A.22) P3=1 +cu , and q1 =B+B2-C (A.23) q2=B-B2-C , (A.24) with B=(1 +n2tX 1 +tX2)/2 (A.25) C =tX2 [1 +n2(n2 -2)] In Table A.1 the powers Pk are computed for a representative set of species. In each case the Young's moduli, Poisson's ratios, and rigidity moduli are taken from Hearmon (1948) (Table 2) and modified to reflect moisture conditions in the green condition (Table A.2). The general solution of(A.19) for the case of transversely isotropic materials as given in Archer (1976) can be recovered by setting tX1 = 2 and tX2 = 1 to obtain the four roots 1 ±(1 ±n).
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