Appendix A Plane Strain Theory for an Orthotropic Cylinder with Pure

Lekhnitskii (1963) derives a general theory for the state of in an elastic body bounded by a cylindrical surface and possessing cylindrical 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 (A.1) or r oe r

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 (A.2) [ Be l= a12 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 (A.5) D +1Xz=a33 (Ar sin(}+ Brcos (}+C) , 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). Since we are restricting attention to cos nO expansions, we may drop the sin 0 term on the right-hand side of (A.13). The presence of the cos 0 term requires that the case n = 1 be handled separately. Case n = 1. Since for n = 1, C = 0 and therefore q2 = 0, we have f1 =A1rP' +A2 rP 2 +A3 r+~rlogr+ P1r +A1r3 , (A.26) where - tXrl Pl=- . /322 (tXl + tX2) B(a13 -a23) (A.27) 210 Plain Strain Theory for an Orthotropic Cylinder with Pure Bending

The~ term can be shown (Timoshenko and Goodier 1970) to lead to a term in the expressions for displacements of the type ecos e which corresponds to multi valued displacements for closed cylinders. Thus we take A4 = 0. The stresses becomes

(Jr = [(P1 -1)A1rP' - 2 + (P2 -1)A2rP 2 - 2 + 2Q1r + Pd cos()

(Jo = [P1 (P1 -1)A1rP' - 2 + P2(P 2 -1)A2rP 2 - 2 + 6Q1r + 2Pd cos() (A.28) r.6 = [(P1 -1)A1rP' - 2 + (P2 -1)A2rP 2 - 2 + 2Q1r+ Pd sin()

Cases n;;;;2

We have

4 fn = L Akrpk + Onr2 , (A.29) k=1 where

Q = ar,n (A.30) " P22 [(n2 -2)0(2 +Q(d and for the stresses

Ak(Pk-n2)rPk-2+(2-n2)Qn] cosn8 (J·=Lt1

2 (J6 = [ ktl AkPk(Pk -1)rPk- + 2Q0 J cos n() (A.31)

2 r.6 =[ I Akn(Pk-1)rPk- +nQ"J sinn8. k=1 Appendix B Computer Program for Computation of Residual Growth Stresses in Radially Inhomogeneous Cylinders

PROGRAM GSTRESS(INPUT,OUTPUT) C*************************** C COMPUTES RESIDUAL STRESSES IN A RADIALLY INHOMO C OTHOTROPIC CYLINDER DUE TO A PRESCRIBED SET OF C (STO AND SZO) GROWTH STRESSES INDUCED IN A NEW LAYER C OF GROWTH(N.LT.40) (JUNE 13,1984 VERSION) C USES PLANE STRAIN THEORY TO SET UP STIFFNESS C CONNECTING RADIAL STRESSES TO RADIAL DISPLACEMENT UNKNOWNS. C USES AXIAL EQUILIBRIUM TO FIND EZ. C CAN BE USED WITH 'BORING' PROGRAM (JUNE 1984) TO FIND RESIDUAL C STRESSES DUE TO MEASURED SURFACE STRAINS(ETM,EZM) AT OUTER C RADIUS. c COMMON/RINGS/ R(40),DR COMMON/ECONST/ TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/COEFF/C1E(40),C2E(40),C1S(40),C2S(40) COMMON/RSTR/ST(3,40,40),SZ(3,40,40) COMMON /TRIDIAG/ A(40),B(40),C(40),D(40) COMMON/DISPL/U(40) c C INITIALIZE VARIABLES c NPTS=5 JL=2 JH=JL+NPTS RING=10. DR=.1 N=JH N1=JL R(1)=0.00001 R(JL)=RING DO 4 K=1 ,NPTS KK=JL+K 4 R(KK)=R(KK-1 )+DR PRINT 123,(K,R(K),K=1, 12) 123 FORMAT(1X,I5,F6.2) JHP1=JH+1 c C GET ELASTIC CONST FOR INHOMO MATERIAL c CALL INHOMO(JHP1) c C DERIVE MATRICES FOR RESIDUAL STRESSES c DO 7 J=JL ,JH CALL GLAYER ( J) 7 CONTINUE 212 Computer Program for Computation of Residual Growth Stresses c C RESPONSE TO GROWTH STRESSES IN JTH RING c 17 CONTINUE c C TEST PROGRAM FOR TI MATERIAL c CALL TEST(JH,JL) STOP END SUBROUTINE INHOMO(N) C********************************* C PRODUCES EA(6,40),EC(6,40) ELASTIC CONSTS IN C IN EACH RING(K=1 ,N) c COMMON/RINGS/R(40),DR COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) DIMENSION E(6) DO 17 K=1,N c C ORTH CONST BEECH C ER,ET,EZ,-PRT,-PZR,-PZT c TC(1)=2.24 TC( 2 )= 1. 1 4 TC(3)=13.7 TC(4)=0.75 TC(5)=0.45 TC(6)=0.45 E(1 )=1./TC( 1) E(4)=1./TC(2) E(6)=1./TC(3) E(2)=-TC(4)/TC(1) E(3)=-TC(5)/TC(3) E(5)=-TC(6)/TC(3) DA=E(1)*E(4)*E(6)+2.*E(2)*E(3)*E(5) DA=DA-E(3)**2*E(4)-E(2)**2*E(6)-E(5)**2*E(1) EC(1,K)=(E(4)*E(6)-E(5)**2)/DA G(1,1,K)=EC(1,K) EC(2,K)=(E(3)*E(5)-E(2)*E(6))/DA G(1,2,K)=EC(2,K) G(2,1,K)=EC(2,K) EC(3,K)=(E(2)*E(5)-E(3)*E(4))/DA G(1,3,K)=EC(3,K) G(3,1,K)=EC(3,K) EC(4,K)=(E(1 )*E(6)-E(3)*E(3))/DA G(2,2,K)=EC(4,K) EC(5 ,K)=(E(3 )*E(2)-E(1 )*E(5 ))IDA G(2,3,K)=EC(5,K) G(3,2,K)=EC(5,K) EC(6,K)=(E(1)*E(4)-E(2)**2)/DA G(3,3,K)=EC(6,K) DO 19 L=1,6 19 EA(L,K)=E(L) Computer Program for Computation of Residual Growth Stresses 213

17 CONTINUE J=1 PRINT 111,J,(EA(K,J),K=1,6) PRINT 111 , J , ( EC ( K, J ) , K= 1 , 6 ) 111 FORMAT(1X,I5,6F10.4) 27 CONTINUE RETURN END SUBROUTINE SOLTRI(NLOW,NHIGH) C******************************* C SOLVES FOR DISP U(K) c COMMON/DISPL/U(40) COMMON/TRIDIAG/A(40),B(40),C(40),D(40) DIMENSION RS(40),S{40) N1=NLOW N2=NHIGH U(1)=0. NP=N1+1 NQ=N2-1 W=B(N1) RS(N1 )=C(N1 )/W S(N1)=D(N1)/W DO 10 K=NP,NQ W=B(K)-RS(K-1 )*A(K) RS(K)=C(K)/W S(K)=(D(K)-S(K-1 )*A(K))/W 10 CONTINUE W=B(N2)-A(N2)*RS(NQ) S(N2)=(D(N2)-S(NQ)*A(N2))/W c C UPPER TRI FORM COMPLETED C NOW BACK SUBST c U(N2)=S(N2) DO 50 J=2 ,N2 L=N2-J+1 50 U(L)=S(L)-RS(L)*U(L+1) RETURN END SUBROUTINE ELEM(L,E,F) C***************************** C 2X2 STIFFNESS MATRIX E,F:S=Q*U+F*EZ C FOR LTH ELEMENT c COMMON/RINGS/R(40),DR COMMON/ECONST/ TC(6),EA(6,40),EC(6,40),G{3,3,40) DIMENSION E(2,2),F(2) DO 7 I=1,2 F(I)=O. DO 7 J=1 ,2 7 E(I,J)=O. R1=R(L) R2=R(L+1) 214 Computer Program for Computation of Residual Growth Stresses

CALL PARAM(L,XL,ZETA) IF(L.EQ.1) THEN E(2,2)=(XL*EC(1, 1 )+EC(2, 1 ))/R2 F(2)=EC(3, 1)+ZETA*EC(1, 1 )*(1.-XL) RETURN ELSE DK=R1**XL*R2**(-XL)-R2**XL*R1**(-XL) PHK=(R1**XL*R2**(-XL)+R2**XL*R1**(-XL) + )IDK AA=XL*(PHK-2.*R2/(R1*DK)) BB=XL*(PHK-2.*R1/(R2*DK)) E ( 1 , 1 )=-(EC( 1 ,L )*XL*PHK+EC( 2 ,L)) /R 1 E(1,2)=2.*EC(1,L)*XL/(R1*DK) E(2,1)=2.*EC(1,L)*XL/(R2*DK) E(2,2)=(EC(2,L)-EC(1 ,L)*XL*PHK)/R2 F(1 )=-EC(3,L)+EC( 1 ,L)*ZETA*(-1 ,+AA) F(2)=EC(3,L)+EC(1 ,L)*ZETA*(1,+BB) CPRINT 333,((E(I,J),J=1,2),I=1,2) 333 FORMAT(1X,2F12.4) RETURN END IF END SUBROUTINE SETABC(NLOW,NHIGH) C**************************** C SETS UP TRIDIAG SYSTEM c COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/TRIDIAG/A(40),B(40),C(40),D(40) DIMENSION E(2,2),F(2) NM1=NHIGH-1 K=NLOW CALL ELEM(K,E,F) B( 1) =E ( 1 , 1 ) C(1)=E(1,2) A(2)=E(2, 1) B(2)=E(2,2) DO 17 K=2 ,NM1 CALL ELEM(K,E,F) B(K)=B(K)+E(1,1) C(K)=E(1 ,2) A(K+1)=E(2,1) B(K+1)=E(2,2) 17 CONTINUE RETURN END c SUBROUTINE GSTRBC(J,UNST,UNEZ) C************************** C PUTS IN BOUND COND FOR GROWTH STRESS PROB COMMON/RINGS/R(40),DR COMMON/TRIDIAG/A(40),B(40),C(40),D(40) COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/DISPL/U(40) DIMENSION E(2,2),F(2) Computer Program for Computation of Residual Growth Stresses 215

JM1 =J-1 U(1 )=0. K=1 CALL ELEM(K,E,F) D(1 )=-F( 1 )*UNEZ D(2)=-F(2)*UNEZ DO 17 K=2 , JM1 CALL ELEM(K,E,F) D(K)=D(K)-F(1 )*UNEZ 17 D(K+1)=-F(2)*UNEZ IF(UNST.GT.UNEZ) D(J)=-UNST*DR/R(J) RETURN END SUBROUTINE GLAYER(J) C*************************** C FOR EACH LAYER J FORMS RESID. STRESS MATRICES C (K=1 ,J) DUE TO STO AND SZO c COMMON/RINGS/R(40),DR COMMON/COEFF/C1E(40),C2E(40),C1S(40),C2S(40) COMMON/DISPL/U(40) COMMON/TRIDIAG/A(40),B(40),C(40),D(40) COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/RSTR/ST(3,40,40),SZ(3,40,40) DIMENSION USAV(2,40),FE(40),FS(40),STE(3),SZE(3) CALL SETABC(1,J) JM1=J-1 UNEZ=1. UNST=O DO 27 IU=1 ,2 CALL GSTRBC(J,UNST,UNEZ) CPR INT 123, (A( K) ,B( K), C( K), D( K), K=1 ,J) 123 FORMAT(1X,4E12.4) CALL SOLTRI(2,J) DO 23 K=1 ,J 23 USAV(IU,K)=U(K) UNEZ=O. 27 UNST=1. DO 4320 K=2 ,J CALL EXACT(K,J,US,UE,KZ,FEK,FSK) C PRINT 345,K,R(K),UE,USAV(1,K),US,USAV(2,K) 4320 CONTINUE 345 FORMAT(1X,I5,F5.1,4F12.8) CALL ZFORCES(1 ,JM1,USAV,FE,FS) DO 4406 KZ=1 ,JM1 CALL EXACT(K,J,US,UE,KZ,FEK,FSK) C PRINT 789,KZ,R(KZ),FEK,FE(KZ),FSK,FS(KZ) 4406 CONTINUE 789 FORMAT(1X,I5,F5.1,4F16.10) CALL EZ(FE,FS,EZT,EZZ,JM1) C PRINT 333,EZT,EZZ 333 FORMAT(1X,2F10.6) CALL DELSTR(1,JM1 ,EZT,EZZ) DO 41 4 I=2 ,J 216 Computer Program for Computation of Residual Growth Stresses

CALL EXSTR(I,I,J,STE,SZE) PRINT 147,I,R(I),(ST(L,J,I),STE(L),L=1,3) 41 4 CONTINUE 147 FORMAT(1X,I5,F5.1,6F8.3) DO 415 I=2,J CALL EXSTR(I,I,J,STE,SZE) PRINT 147,I,R(I),(SZ(L,J,I),SZE(L),L=1,3) 415 CONTINUE RETURN END SUBROUTINE PARAM(K,XL,ZETA) COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) XL=SQRT(EC( 4 ,K) /EC( 1, K)) IF(EC(1,K).EQ.EC(4,K)) THEN ZETA=O. ELSE ZETA=(EC(5,K)-EC(3,K))/(EC(1,K)-EC(4,K)) END IF RETURN END SUBROUTINE EZ(FE,FS,EZT,EZZ,NS) COMMON/RINGS/R(40),DR DIMENSION FE(40),FS(40) NP=NS+1 SUME=O. SUMS=O. D017K=1,NS SUMS=SUMS+FS( K) 17 SUME=SUME+FE(K) EZT =-SUMS/ SUME EZZ=-2.*R(NP)*DR/SUME RETURN END SUBROUTINE ZFORCES(NLOW,NHIGH,USAV,FE,FS) C************************** C COMPUTE EZ AND INCR IN STRESSES DUE TO JTH LAYER c COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/RINGS/R(40),DR COMMON/COEFF/C1E(40),C2E(40),C1S(40),C2S(40) DIMENSION USAV(2,40),FE(40),FS(40) N2=NHIGH Nl=NLOW DO 100 K=N1 ,N2 R1=R(K) R2=R(K+1) U1E=USAV(1,K) U2E= USA V ( 1 , K + 1 ) U1 S=USAV(2, K) U2S=USAV(2,K+1) CALL PARAM(K,XL,ZETA) IF(K.EQ.N1) THEN C1E(K)=U2E*R2**(-XL)-ZETA*R2**(1.-XL) C2E(K)=O. Computer Program for Computation of Residual Growth Stresses 217

C1S(K)=U2S*R2**(-XL) C2S(K)=O. C PRINT 226,K,C1E(K),C2E(K),C1S(K),C2S(K) XX=(XL*EC(3,K)+EC(5,K))/(XL+1.) YY=EC(6,K)+ZETA*(EC(3,K)+EC(5,K)) FE(K)=XX*R2*(U2E-ZETA*R2)+YY*R2**2/2. FS(K)=XX*R2*U2S ELSE DE=R1**XL*R2**(-XL)-R2**XL*R1**(-XL) XX=R2**(-XL)*(U1E-ZETA*R1) YY=R1**(-XL)*(U2E-ZETA*R2) C1E(K)=(XX-YY)/DE XX=R1**XL*(U2E-ZETA*R2) YY=R2**XL*(U1E-ZETA*R1) C2E(K)=(XX-YY)/DE SE=C1E(K)*(EC(3,K)*XL+EC(5,K)) TE=C2E(K)*(EC(5,K)-XL*EC(3,K)) WE=EC(6,K)+ZETA*(EC(3,K)+EC(5,K)) E1=SE* (R2** ( 1. +XL )-R 1** ( 1. +XL) )I( XL+1 . ) E2=TE*(R2**(1.-XL)-R1**(1.-XL))/(1.-XL) E3=WE*(R2**2-R1**2)/2. FE(K)=E1 +E2+E3 XX=R2**(-XL)*U1S-R1**(-XL)*U2S C1S(K)=XX/DE XX=R 1**XL*U2S-R 2**XL*U 1S C2S(K)=XX/DE SS=C1S(K)*(EC(3,K)*XL+EC(5,K)) TS=C2S(K)*(EC(5,K)-XL*EC(3,K)) S1=(R2**(1.+XL)-R1**(1.+XL))*SS S2=(R2**(1.-XL)-R1**(1.-XL))*TS FS(K)=S1/(1.+XL)+S2/(1.-XL) END IF 100 CONTINUE C PRINT 226,(K,C1E(K),C2E(K),C1S(K),C2S(K),K=1,NHIGH) 226 FORMAT(1X,I5,4F8.4) RETURN END SUBROUTINE TEST (N ,N 1) C**~********************** C TEST ON TI CYLINDER WITH CONSTANT G.S. c COMMON/RINGS/R(40),DR COMMON/RSTR/ST(3,40,40),SZ(3,40,40) DIMENSION SRA(40),STA(40),SZA(40),STG(40),SZG(40) ST0=-0. 7 SZ0=7. DO 6 K=2,N STG(K)=STO 6 SZG(K)=SZO DO 10 J=N1,N SUMR=O. SUMT=O. SUMZ=O. DO 8 I=N1 ,N 218 Computer Program for Computation of Residual Growth Stresses

SUMR=SUMR+ST ( 1,I ,J )*STG( I )+SZ( 1 ,I ,J )*SZG(I) SUMT=SUMT+ST(2,I,J)*STG(I)+SZ(2,I,J)*SZG(I) 8 SUMZ=SUMZ+ST(3,I,J)*STG(I)+SZ(3,I,J)*SZG(I) SRA ( J )=SUMR STA(J)=SUMT 1 0 SZA (J) =SUMZ ROUT=R (N) DO 20 K=N1 ,N RM=R (K) AA=ALOG(RMIROUT) SRZ=STO*(AA) SRT=STO*AA STT=STO*(AA) SZT=SZ0*(2.*AA) 20 PRINT 111,K,SRT,SRA(K),STT,STA(K),SZT,SZA(K) 111 FORMAT(1X,I5,6F10.2) RETURN END SUBROUTINE EXACT(I,N,US,UE,KZ,FEK,FSK) c ***************************************************************** COMMON/RINGS/R(40),DR COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) CALL PARAM(I,XL,ZETA) X1=EC(1 ,I)*XL+EC(2,I) X2=ZETA* ( EC ( 1 ,I) +EC( 2, I)) +EC( 3, I) C1E=-R(N)**(1.-XL)*X2/X1 C1S=-DR*R(N)**(-XL)/X1 US=C1S*R(I)**XL UE=C1E*R(I)**XL+ZETA*R(I) KP=KZ+ 1 P1=(EC(3,KZ)*XL+EC(5,KZ))*C1E P2=ZETA*(EC(3,KZ)+EC(5,KZ))+EC(6,KZ) Q1=C1S*(EC(3,KZ)*XL+EC(5,KZ)) ZZ=(R(KP)**(XL+1.)-R(KZ)**(XL+1.))/(XL+1.) YY=P2*(R(KP)**2-R(KZ)**2)/2. FEK=P1*ZZ+YY FSK=Q1*ZZ RETURN END SUBROUTINE EXSTR(IP,IE,N,STE,SZE) COMMON/RINGS/R(40),DR COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) DIMENSION STE(3),SZE(3) CALL PARAM(IE,XL,ZETA) AA=1.+XL A 1=G( 1 , 1 ,IE) *XL+G( 1 ,2, IE) A2=G(1 ,3,IE)+(G(1, 1 ,IE)+G(1 ,2,IE) )*ZETA A3=G(1 ,3,IE)*XL+G(2,3,IE) A4=G( 3, 3,IE)+ ZETA* ( G( 1 , 3, IE) +G( 2, 3, IE)) A5=A4*0.5-A2*A3/(AA*A1) A6=A3/(A1*AA) EZT=A6*DR/(A5*R(N)) C1=-(DR+R(N)*A2*EZT)/(A1*R(N)**XL) C2=0. CALL STRESS(IP,IE,C1 ,C2,EZT,STE) 111 FORMAT(1X,2I5,4F12.6) Computer Program for Computation of Residual Growth Stresses 219

EZZ=-2.*DR/(A5*R(N)) C1=-A2*EZZ*R(N)/(A1*R(N)**XL) C2=0. CALL STRESS(IP,IE,C1 ,C2,EZZ,SZE) RETURN END SUBROUTINE STRESS(KP,KE,C1 ,C2,EZ,SS) COMMON/RINGS/R(40),DR COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) DIMENSION SS(3),F(3) CALL PARAM(KE,XL,ZETA) RP=R(KP)**(XL-1.) RM=R(KP)**(-XL-1.) F(1 )=XL*(C1*RP-C2*RM)+ZETA*EZ F(2)=C1*RP+C2*RM+ZETA*EZ F(3)=EZ DO 17 I =1, 3 SS(I)=O. DO 17 K=1, 3 17 SS(I)=SS(I)+G(I,K,KE)*F(K) RETURN END SUBROUTINE DELSTR(NPL,NPH,EZT,EZZ) COMMON/RSTR/ST(3,40,40),SZ(3,40,40) COMMON/ECONST/TC(6),EA(6,40),EC(6,40),G(3,3,40) COMMON/RINGS/R(40),DR COMMON/COEFF/C1E(40),C2E(40),C1S(40),C2S(40) DIMENSION STD(3),SZD(3) N1=NPL N2=NPH N2M1 =N2-1 N2P1 =N2+1 DO 200 K=N1 ,N2 C1=C1S(K)+C1E(K)*EZT C2=C2S(K)+C2E(K)*EZT CALL STRESS(K,K,C1,C2,EZT,STD) 444 FORMAT(1X,2I5,6F10.4) C1=EZZ*C1E(K) C2=EZZ*C2E(K) CALL STRESS(K,K,C1 ,C2,EZZ,SZD) DO 71 I=1 ,3 ST(I,N2P1,K)=STD(I) 71 SZ(I,N2P1,K)=SZD(I) 200 CONTINUE K=N2P1 C1=C1S(N2)+C1E(N2)*EZT C2=C2S(N2)+C2E(N2)*EZT CALL STRESS(K,N2,C1 ,C2,EZT,STD) C1=C1E(N2)*EZZ C2=C2E(N2)*EZZ CALL STRESS(K,N2,C1 ,C2,EZZ,SZD) DO 73 I=1 ,3 ST(I,N2P1,K)=STD(I) 73 SZ(I,N2P1,K)=SZD(I) RETURN END Appendix C Eigenfunctions for a Finite Strip

Consider the semi-infinite region -1 sxs 1 -t/rosys oo if solutions of Eq. (6.55) are sought in the form c/J=X(x)e-'-Y , (C.l) it follows that d4 X d 2X dx4 +2A2C1 dx2 +A.4C2X=O (C.2) where C1 = (2 fJ12 + /366)/2 fJ22 (C.3) C2=fJu/fJ22 · (C.4) The solution of Eq. (C.2) can be written as X= (AcosA.b1x + B cosA.b2x + C sin A.b1x + D sin A.b2x)e-'-Y , (C.5) where bi=C1 +(Ci-C2)112 (C.6) b~=C1-(Ci-C2) 112 (C.7) It may be noted that for isotropic materials b1 and b2 approach 1 and for this case of repeated roots the solution (C.5) takes a modified form (Timoshenko and Goodier 1970). In the present case the symmetry of the stress states with respect to the center results in dropping the terms inC and Din Eq. (C.5). The boundary conditions at the edges x = ± 1 are O"x=!xy=O · (C.8) These homogeneous equations yield a spectrum of eigenvalues A.i which correspond to nontrivial solutions of the equations.

1 [ cos b A.i cos b2A.i J [Aj] [OJ (C.9) b1 sin b1A.i b2 sin b2A.i Bj = 0 The jth eigenfunction is given by

Fi (A.i, x) =cos b1A.ix + Bi cos b2A.ix , (C.10) where Bi = -cos b1A.i/cos b2A.i . (C.11) It is easily verified that if A.i is a root, so is - A.i as well as ± A.j, the complex conjugates of ± A.i. Eigenfunctions for a Finite Strip 221

The general solution can be written as

00 4J+ = L aiFi(A.i,x)e-;.JY , (C.12) j=l where the summation is over all the eigenvalues in the right half of the complex plane. It is assumed that the eigenvalue A.i =0 is excluded from (C.12), and thus 4J+ represents the solution due to self equilibrated stresses on the finite end of the strip. By an analogous procedure, it follows that the decaying solutions in the region -1::;;x::;;1, -oo

00 4J-= L eiFi(A.i,x)e;.JY. (C.13) j=l

Thus, for the finite strip of present interest, i.e., -1::;;; x::;;; 1, -t/r0 ::;;; y ::;;;tjr0 , which represents the intersections of the two semi-infinite regions considered above, the solution can be written as

00 4J= L Fi(A.i,x)(A1ie;.JY+A2ie-;.JY), (C.14) j=l where A1i and A2i are constants to be determined from the boundary conditions at y= ±t/r0 • For convenient reference expressions for the stresses and displacements are given: ux = L A.J(cos b1A.ix + Bi cos b2A.ix)(A1ie;.JY +A2ie-;.JY) (C.15)

(C.16) txy = L A.Jeb1 sin b1A.ix + Bib2 sin b2A.ix)(A1ie;.JY -A2ie-;.JY) (C.17) j

=" 1 . [p (sin b1A.ix) B· sin b2A.ix) u L.... II.J 11 b + J b j 1 2

-P12 (b1 sin b1A.ix + Bi sin b2A.ix) J(A 1ie;.JY + A2ie-;.JY) (C.18) v = L A.i [p12 (cos b1A.ix + Bi cos b2A.ix) j - P22 (br cos b1 A.ix +Bib~ cos b2A.ix)](A1ie;.JY-A2ie-;.JY) (C.19)

It may be shown that a uniform stress, u~, satisfies Eq. (6.55) and may be added to the above set of stresses.

For a particular set of prescribed boundary stresses along y= ±t/r0 , Eqs. (C.16) and (C.17) may be used to satisfy the boundary stresses by a least-squares technique. The minimization of the boundary residuals yields a linear system [C]{A} = {P} , (C.20) where [C) is a symmetric positive , {A} a vector of coefficients consisting of A1i and A2i and {P} a vector calculated from the prescribed boundary conditions. Appendix D End Problem of an Anisotropic Strip

A semi-analytic finite element method for solving plane strain and stress problems in rectangular strips is presented here. A more complete version of the method is given in Bandyopadhyay (1978).

D.l Transformation of the Matrix

The elastic constants in the cartesian (x, y, z) are found from the elastic constants for the cylindrically orthotropic case using (Lekhnitskii 1963)

[C] = [B]T [Cc) [B) , (D.1) where [C] is the elasticity matrix in the Cartesian system, i.e.,

{a}= [C]{e} , (D.2) and [Cc] is the elasticity matrix in the cylindrical system

(D.3) where

{ a}T = { 11xxl1yyl1zz 'Lxy Lyz 0 zx}

{ e} T = { Gxx GyyGzz Yxy Yyz Yzx} (D.4) { ac}T = { a,,a66Uzz 'Lrz L6z Lr6}

{ec}T = {e,,e66BzzYrzY6zYr6} •

The transformation matrix [B) takes the form

sin2 rjJ cos2 rjJ 0 sinrjJcos rjJ 0 0 cos2 rjJ sin2 rjJ 0 -sin rjJcos rjJ 0 0 0 0 1 0 0 0 (D.5) 0 0 0 0 cos rjJ sinrjJ 0 0 0 0 -sinrjJ cosrjJ sin2rjJ -sin 2rjJ 0 cos2rjJ 0 0 where rjJ is the angle between the radial direction and the positive y direction. Eigenvalue Problem 223

D.2 Plane Strain Problem

The rectangular strip is defined in Fig. 6.36. Normalization with respect to the stem radius follows by taking x=x/r0 , y=y/ro , z=rzo/ro (D.6) t=t/r0 , e=e/r0 , b=6/r0 , where the hatted variables are dimensional and the unhatted are non-dimensional. A semi-infinite plane strain problem is to be solved for the region Rboo defined by

-b~x~b

O~y~oo . Two sets of eigenfunctions are superimposed to solve the problem in the finite region Rbt defined by

-b~x~b

-t~y~t The only nonzero strain-displacement relations for the plane strain case are au Bxx=ox ' au ov (D.7) Yxr=oy +ox ' where u and v are displacement components in the x and y direction respectively. The stress-strain relations are: {u}=[C']{e} , (D.8) where

{u}T = { O'xxO'yy 'Lxy}

{eY = { BxxByy Yxy}

[C'] =[~:: ~:: ~::] ' c13 C23 c33 with the Cij defined by (D.1).

D.3 Eigenvalue Problem

The basic set of functions (eigenfunctions) used to solve the end problem are derived in terms of the displacements. Approximate solutions are found by the minimum potential energy principle. For the region Rboo normalized displacements are 224 End Problem of an Anisotropic Strip written u(x, y)= U(x)exp( -A.y) (D.9) v(x,y)=V(x)exp(-A.y), where U(x) and V(x) are unknown eigenfunctions and the A.'s are determined by requiring that the homogeneous boundary conditions x = ± b

O"xx·u=O (D.10) 'rxy·v=O be satisfied. The total potential energy II can be written (Timoshenko and Goodier 1970)

00 b II= 1/2 f f (uxxllxx +uyyllyy +rxyYxy)dxdy-f (r~y · U +u~yv)dx , (D.11) 0 -b L where r~y and u~y are end values and L represents both ends. The equilibrium condition is given by (Timoshenko and Goodier 1970) bii=O . (D.12)

Using Eqs. (D. 7) and (D.8), the potential energy can be expressed as a functional of the displacements. The calculation of bii is similar to that used in Bandyopadhyay (1978) (Chap. 2.3). After eliminating the derivatives with respect to y using integration-by-parts, one obtains

b +A. f {Cl2 (U'bV-VbU')+Cl3 (U'bU-UbU') -b

+Ch(V'bV -VbV')+C33 (V'bU -UbV')}dx (D.13)

+A-2 1{ -C2 2 VbV-C23 (UbV+VbU)-ChUbU}dx]

+[[ {(rxy-r~y)bU +(uyy-u~y)bV}dx ]=0 , where primes denote differentiation with respect to x.

D.4 Finite Element Discretization

Following a procedure developed in Bandyopadhyay (1978) (Chap. 2), the region is discretized in the x direction by using thin elements, and low-order polynomials are assumed to give good approximations to the displacements. The elastic properties are assumed to be constant over each element. Plane Stress Problem 225

The displacements and their x-derivatives are taken as the nodal variables. Thus eight "degrees of freedom" are obtained for each element. Inside each element the displacements are taken as

U(x)] rU(x1) U'(x1) U(x2) U'(x2)] [tP1(x)] r = tP2(x) V(x) V(x1) V'(x1) V(x2) V'(x2) tP3 (x) , (D.14) tfJ4(x) where the c/J's are hermite cubic interpolation polynomials (Becker et al. 1981). These polynomials are defined by

1 tPtfJ2(x)](x) [10 [ (D.15) I[J3 (~) 0 tfJ4(x) 0 where

17=(x-x1)/P (D.16)

P=x2 -xl . The displacements and their x-derivatives are continuous at the interelement boundary. The element "stiffness" matrices are computed by numerical integration from Eq. (D.13) using the limits (x1, x2) rather then (-b, b). Assembling the element stiffness matrices and imposing the displacement boundary conditions, if any, leads to a quadratic matrix eigenvalue problem of the form

(D.17)

The eigenvalues A.K which correspond to nontrivial solutions of (D.17) are found by a search technique similar to the one described in Chapter 6.3.1.5. For a specific end problem 't"xy and/or u and Uyy and/or v are prescribed along the edges y = ± t. These boundary conditions can be satisfied in a least-squares sense by taking linear combinations of the eigenfunctions.

D.5 Plane Stress Problem

If the thickness 2 t of the strip is assumed to be small enough so that the zero normal and shear stresses on the y-faces remain negligible over the entire thickness, then we have a plane stress problem (Timoshenko and Goodier 1970). The nondimensional coordinates (x, z) are used. Thus a strip of normalized length 21 has the stresses applied at the ends z = ± 1. 226 End Problem of an Anisotropic Strip

For this case au ow Gxx = OX ' Gzz = OZ (D.18)

(D.19) and (D.20) where {aY ={O'xxO'zz!xz} {t:}T ={ExxGzzYxz} ' with ell 2 ~ [C ]~ [ ~13 (D.21) ell= ell+ ~1 c12 + ~3c14 e13 = c13 + ~2c12 + ~4c14 e33 = c33 + ~2c23 + ~4c34 (D.22) e66=(CssC66 -C~6)/Css

~1 =(C14C24 -C12C44)jD

~2 = (C24C34 -C44C23)/D (D.23) ~3 = (C12C24 -C22C14)/D

~4 = ( C23 C24 - C22 c34)/D D=C22C44 -C~4 and the Cii are defined by Eq. (D.1). The homogeneous side boundary conditions at the ends x = ± b are: O'xx·u=O (D.24) rxz. v=O . The solution procedure for the plane stress problem is almost exactly the same as described above for the plane strain problem. Of course y is replaced by z, v by w, t by 1 and [C 1] by [C2 ] to complete the full correspondence. In order to gain some idea as to the accuracy of these finite element solutions, the eigenvalues for an orthotropic strip are compared with very accurate solutions for both the plane strain and plane stress cases. By using the special case of a Plane Stress Problem 227

Table 0.1. Comparison of eigenvalues for an orthotropic strip

FEM with 42 elements Exact•

Plane strain A1 2.8128 +i 0.69416 2.8128 +i 0.69417 As 11.0063 11.0062 A10 21.1914 21.2798 A2o 45.2827 +i 0.69399 45.2821 +i 0.69423 A3o 67.1190 67.1077

Plane stress A1 1.3715 1.3717 As 4.6340+i 0.17386 4.6338+i 0.17354 A10 10.3087 10.3091 ~0 19.8107 +i 0.27754 19.8009+i 0.27866 A3o 30.1683 30.1428

• Vendhan and Archer 1977 diametrical strip, a comparison can be made between the exact eigenvalues obtained by Vendhan and Archer (1977) (Table D.1). The excellent agreement suggests that the FEM used with a sufficient number of elements should give results for the off• diametral problems. Appendix E Analysis of Surface Strain Changes Caused by Removal of Inside Layers of an Orthotropic Cylinder Containing Residual Stresses

Starting with the differential equation (5.61)

rcj/'+

and the boundary conditions -(JoA 1 L1(Jr(x)= Jc- 1 (X""- -1) (E.3) L1(Jr(1) = 0 , (E.4) it follows that

L1(Jr(1) =c1 +c2 = 0 . (E.6) Thus L1(Jr(X) = c1 (xA- 1 -X-A- 1) = -o=r(X)

c1 = -o=r(x)/(xA- 1 _X_A_ 1) , (E.7) so that

(E.10) and using Eq. (E.9)

- U.(x) [ A -1 ( 1 1) -A -1 ( 1 1)] A X" 1 -x A 1 (! +A +Q A- +a33Ll(Jz · (E.11) Analysis of Surface Strain Changes Caused by Removal of Inside Layers 229

Overall longitudinal equilibrium requires that Fz+AFz=O , (E.12) where

(E.13)

1 AFz= J Auz(e)ede (E.14)

Using the last equation in (5.59) and (E.13) yields upon integration

a32 [X2 -1 1-XH1 Fz= -uzx2lnx+- ue ---+ A. -flnxJ . (E.15) a33 1 -x 1 - From (E.11) it is possible to solve for Auz to obtain

Au = Aaz + a32 o'r(x) [ ;.-1(A.+ 1)+ -;.-1(A.- 1)] z 2-1 -2-1 e e , (E.16) a33 a33 X -x where as with earlier applications of plane strain theories Auz is a constant. Using (E.16), the integration in (E.14) reduces to AF - 1 - x2 Aaz a32 - ( ) 2 Ll z------O"r X X ' (E.17) 2 a33 a33 Now substituting for Fz and AFz from (E.15) and (E.17) into (E.12) and solving for Aaz gives (since Aaz is constant)

Aaz(1)=F1(x)(a32u8 +a33 uz), (E.18) where F 1 (x) is defined by Eq. ( 5 .49), which completes the derivation of the second of Eq. (5.66). An expression for Aa8 (1) can be found by using the stress-strain relation (4.1) to obtain

Aa8 (1) = a21 Aur(1) + a22 Au8 (1) + a23 Auz(1) , (E.19) but according to Eq. (5.62b), the first term drops out. From Eq. (5.65) Aue(1)= -Mr(x)2/(x"-1-x--<-1) , (E.20) and from Eq. (4.1) Auz(1) = [Aaz -a32Aue(1)]/a33 , (E.21) and substitution of (E.20) and (E.21) into (E.19) gives

Aa8 (1)=(a22 -a23a32ja33)Aue(1)+a23Aaz/a33 . (E.22) Using Eq. (5.59) to find o'r(x) and inserting into (E.20) gives

ue (X2 - 1 -1) Aue(1)= -2A. A.-1 (x" 1_X ;. 1) ' (E.23) 230 Analysis of Surface Strain Changes Caused by Removal of Inside Layers or

Mo(1)= F2 (x)8o , (E.24) where

2), xv. -xHl Fz (x) = 1 -2 xu -1 (E.25) Combining (E.18) and (E.24) yields

(E.26) which completes the derivation of the first equation in (5.66). References

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Alnus, spiral grain 94 normal wood 32 Amelanchier, spiral grain 93 swelling 50 Anisotropy tension wood 33, 40 cell wall 50 types 54, 55 cylindrical 73, 95, 205 wall 44, 50-59 shrinkage 50 youngwood 3 Annual rings Cellulose active shortening 2 framework 50 microfibril angle variation 54-56 tension hypothesis 65 Center of gravity Banding, reduction of end-splitting crown 61 168, 169 stem 30 Bark Checks (see also Splits) pressure 51, 67-69, 97 avoidance by cutting procedure 204 removal 18 boards 49 strain 59 cell wall 44 stress 59 control by grooves 168 Beams standing trees and felled logs 3, 6 cantilever 59 Compatibility deflections 34 condition 11, 189 Bending equation 68, 98, 115, 206 action of gravitational loads 19 geometric 7 deformations 152 Compression moment 19,20,29, 30,61,198,201 failures 89 off-diametral plank 198 gradient in studs 121 response to strip removal 153 moduli of elasticity 19 strains 15, 152, 153 quartersawn lumber 122 stresses 20, 61, 62, 122, 151 recovery movements 2 Board, quartersawn 120-125 stress change and microfibril angle 56 Bowing 120 surface stress measurements 56 Butt swell 59 tangential stress 6 zones 125 Cambial Compression wood 3, 27, 35, 43, 50-56, 61, activity 59 62, 143 area 27 Contraction layers 54 cell 50 production 56 diametral planks 194 zone 16 gelatinous fibers 56 Carya ovata, surface stresses 45 plank strips 9, 13-15 Cell reference circle on stem surface 27 allignment 151 reference lengths on prismatic elements anatomy 28 24 compression wood 3, 52 Conversion contraction 50 geometric distortions and thinning development 67 practice 48 differentiation 29, 66 log-to-plank strain relief 186 distortion 51 residual stresses in lumber 74 lignification 51 Core samples low density 30 diameter changes 27 238 Subject Index

Core samples experiments 126, 127 residual stress effects 49 theory for thin case 128 Cracks (see also Checks and Splits) thick pieces 128 drying 66 wedges removed 130 frost 151 Drying stresses 11, 120 Crook 120, 121 Cross-cuts Eccentric growth 65 critical distance from cut 17 Eigenvalue method disks 126, 130 gage length changes 4 asymmetric stem stress due to crosscut 181-183 strain reduction in strips 151 end stresses near crosscut 160-164 Crown plank stresses areas 49 diametral 189-192 bending moment 19 changing size 59 off-diametral 198-203 Elasticity, general linear theory 68 gravitationalload 19 moment action 54 Elastic moduli fibril 50 new growth 30 position 16, 29 inelastic model 89 unbalanced development 54 longitudinal, tangential, radial 42, 50, Cryptomeria japonica 129 internal stresses by thin-layer-removal rectangular, in strip 199 relation to cell type 54 method 135 strips 3, 10 microfibril angle variation 56 studs 120 peripheral strain variation 61 surface variation, leaning tree 61 seasonal strain changes 56 Electrical strain gages 20, 21, 40 variability of surface stresses 54 End splitting reduction 167 Curvature Environmental influences, residual stress 48 change in stem 83, 84 Eperua, stresses and anatomical features plank 2, 122, 200-202 prismatic piece 16 25,26 Equalizing times, residual stress reduction stem 1, 2 120 strip 8, 14, 93 Eucalyptus stud 121 camaldulenis, strain relief, diametral tree form 29 planks 93 cypellocarpa, surface stresses 121 Decay in response to cutting delegatensis, silvicultural practice and asymmetric stem stresses 181 residual strains 49 characteristic length 155-158 gigantea distance from end face 158 plank stripping 9, 14 end stresses 160, 164, 196 radial strain gradient 3, 6 off-diametral plank 201 nitens (see delegatensis) response parameter 160, 183-186 regnans Density (see Tree stand density) regrowth trees, stresses 48 basic 16 strain changes due to crosscut 4 cell wall 30 strain variation with height 45 changes with stem height 45 surface strains, standing tree 18 Diametral plank tree form 28 expansion upon removal from stem 6 Extensometer residual stress pattern 91-93 mechanical 16 Dimensional changes two-way sensor 28 cells inclined to stem axis 96 Exterior-layer removal 147 cell wall 50 lumber cut from stems 121 Direction of lean, relation to peripheral Fagus sylvatica surface strains 34 radial checks 3 Disks, crosscut from logs stress gradients 3, 6 residual stresses variation of elastic modulus 88 Subject Index 239

Failure Magnolia obovata, tensile stress 65 criterion 168 Marker points 8, 22, 147 ratio 173, 174 Microfibril angle Finite elements surface strain 39, 53-56 general 164,172,204 transition angle 52, 56 semi -analytic 186, 197, 222 Microfibrillar structure 55 three-dimensional 186 Modulus of elasticity 5, 120 Fourier Movements, stems and branches 2 components 77, 136, 184 expansion 77, 83, 178 Normal wood 3, 30, 31, 44, 53, 62, 63 series 76, 110-112, 135, 151, 171, 178 Off-diametral plank 197-199 Fraxinus Opposite wood 30, 60, 62, 63 americana, stress variation with height 45 Origin, residual stresses 3, 7, 32, 50, 65 excelsior, plank stresses 192 Orthotropy cylindrical 66, 84, 94, 154, 189, 197, 222 Gage (see also Anisotropy, cylindrical) length 4, 16, 24 elastic model 66 points 16, 17 surface strain 20, 40, 142 Pinus Gelatinous caribaea, strain relief 22, 26 fibers 32,40,44, 50, 56,64 densiflora, internal stress distribution 143 layer 64, 65 radiata, surface strains 40 Geometric taeda, stud warping 120 compatibility (see compatibility) thunbergii, surface strain 45 distortions upon conversion 48 Pith 3-6, 9, 11, 12, 52, 54, 73, 76, 88-91, Grain 103,104,115,116,120,122, 124,125,130, direction, surface stress 151 135,142,143,161,166-168,173, 188,193, inclined 93, 97, 103 200,204 Gravitational Plank loads 19, 61 cutting 116 moment 19, 61 extension, after cutting 116 strains 60, 61 -stripping 1, 3, 16, 52, 93 stresses 20, 61-63 -to--strip 116 Populus Hardwood studs 119 angulata, microfibril angles 55 Heartwood deltoides, residual stresses 48 brittle 89 euramericana, residual stresses 48 checks 166,168,203,204 maximuwicziix p. beolinensis, surface Heat treatments, stress changes 6 strains 61 Helical fissures 52, 56 nigra, residual stresses 48 Hooke's law trichocarpa, spiral grain 94 simple 5, 9, 10 Pyrus, spiral grain 94 generalized 66, 108, 205 Quercus gilva, surface stresses 64 Increment core myrsinaefolia, surface stresses 31 diameter changes 27 rubra, internal residual stresses 155, 168, samples 49 204 Inelastic behavior 88 serrata, surface stresses 31 In-tree Radial longitudinal stress 115, 195 equilibrium 68, 84, 98 128 transverse strains inhomogeniety 84 Rate of thinning, residual strains 49 Kerf, longitudinal 121 Reaction wood (see Compression wood; Tension wood) Leaning stems 2, 29, 30, 34, 44, 49, 59 Regulation, longitudinal stress generation Least squares 14, 23, 24 30 Lignification 2, 50-52, 59, 65 Reorienting stem 39, 55 240 Subject Index

Self-equilibrating stress states 6, 7, 64, 71, in-tree 10, 130, 199 115, 160, 162,181,188,190, 195, 196, 198 pre-stressing 66 Shakes (see also Checks and Splits) profiles 88, 178 heart 154, 174, 184 shear ring 184 crosscut 161, 162 Shattering, diametral planks 187, 196 end face 155 Shiia sieboldi, microfibril angles 55 growth increment 77 Silvicu1tural practice, surface strains 49 interior stem 79 Spiral grain 93, 94, 97, 103 longitudinal variation 113 Splits (see also Checks) off-diametral plank 198 board 44 spiral grain 104 end 45 singular 81 log 3, 49 yield 91 reduction by steaming 4 young conifer stems 52 Stem curvature changes Tectona grandis, single hole method 26 movement 29 Tension wood 3, 33 reorienting 35 Thin layer removal method 114, 125, 130, Steppedcuts 169,170 137, 141, 143, 144, 159 Strain Three-dimensional accumulation 6 finite element analysis 186 changes oblique cuts 173 diametral plank 8 stress analysis 171, 172, 186 released element 147 Tilia, spiral grain 93 generation Transformation, growth strain 96 seasonal changes 56 Transition zone, reaction wood 31, 35 surface 50 Transverse stress relief, planks gradient 3, 11, 14 diametral 114, 188, 196 log-to-plank 116 off-diametral 197-199 log-to-strip 116 Tree form 1, 16, 30 measurement devices 114 Tree stand density 49 plane 11,23, 66, 68, 72, 76, 84,94,95, Tsuga canadensis, stress generation 51 115, 154, 187-193, 197-203 Twisting action profile 35, 82 angle of twist 96 relief moment 99, 100, 151 diametral planks 186 spiral grain 94 end face 147 Two-dimensional strain exterior machining 147 field 59 near single hole 22 sensor 28 stimu1us 61 Stress Upward curvature, tree stem 2 accumulation 74, 102, 112 clamping action 167 Vertical orientation 29 concentration 2, 10, 154, 155 Viscoelastic effects 89 constitutive relations 162, 189 differentiation of reaction wood 30 Warping, lumber 120 function 69, 70, 206 Water potential effects 45 gelatinous fibers 32 Wedges, crosscut disk 126 gradients 3, 6 Weight effects 2, 59 hypothesis 30 Wind 48,59