370 SHELLS 个五 Figure 8.8:The loads and the membrane forces on a cylinder. about the x-axis yield (Fig.8.8) N2nR=N 2Ny P:2R (8.10) (Nzy R)2nR=T. where R is the radius of the wall's reference surface.From these equations,the membrane forces are N Nx=2R Ny P:R Nsy =2R (8.11) The strains corresponding to these membrane forces are calculated by Eq.(8.4). The axial u°，radial w°，and circumferential v°displacements are °=ed=xe+ugw°=Regv°= yo,dx xy+vo. (8.12) where ue and ve represent rigid-body motion. 8.2.2 Built-In Ends As we noted previously,near boundary supports the membrane theory is inaccu- rate,and the forces,moments,and displacements of the shell must be calculated by other means.In the following,we consider thin-walled circular cylinders built-in at each end.The cylinder is subjected to pressure pa,axial load N.and torque Figure 8.9:Cylinder built-in at both ends subjected to pressure p:,axial load N,and torque T. k 2R
370 SHELLS Nxy Nx N pz Ny x x T Figure 8.8: The loads and the membrane forces on a cylinder. about the x-axis yield (Fig. 8.8) Nx2π R = N 2Ny = pz2R (8.10) (NxyR) 2π R = T , where R is the radius of the wall’s reference surface. From these equations, the membrane forces are Nx = N 2π R Ny = pzR Nxy = T 2π R2 . (8.11) The strains corresponding to these membrane forces are calculated by Eq. (8.4). The axial uo, radial wo, and circumferential vo displacements are uo = ) o xdx = xo x + uo o wo = Ro y vo = ) γ o xydx = xγ o xy + vo o, (8.12) where uo o and vo o represent rigid-body motion. 8.2.2 Built-In Ends As we noted previously, near boundary supports the membrane theory is inaccurate, and the forces, moments, and displacements of the shell must be calculated by other means. In the following, we consider thin-walled circular cylinders built-in at each end. The cylinder is subjected to pressure pz, axial load N, and torque T pz L N N T T 2R x Figure 8.9: Cylinder built-in at both ends subjected to pressure pz, axial load N, and torque T
8.2 CYLINDRICAL SHELLS 371 Figure 8.10:Forces and moments inside the wall of a thin cylinder. (Fig.8.9).The pressure may vary linearly along the cylinder's axis, Pz=pao+xpa, (8.13) where Pao and Pa are specified constants.(For the pressure distribution shown in Fig.8.9,pa is negative.)For the applied loads(Fig.8.9),neither the stresses nor the strains vary circumferentially.Accordingly,the equilibrium equations are4 dN dx =0 (8.14) d (RNy+My)=0 (8.15) dx Ny dM =Pa (8.16) R dx2 v.=dM (8.17) dx dMy Vy=dx (8.18) where the forces Nr,Ny,Ny,the moments Mr,Mry,and the transverse shear forces V and V are illustrated in Figure 8.10. The strains and the curvatures of the reference surface are5 du e- wo du° (8.19) dx e= R y89= dx dPw° wo Kx= Ky= 2dv° Kxy=一 (8.20) dx2 R dx where°，v°，andw°are the axial,.circumferential,.and radial displacements of the reference surface. The force-strain relationships are identical to those of laminated plates and are given by Eq.(8.3). The equilibrium equations(Egs.8.14-8.18),the strain-displacement relation- ships (Egs.8.19-8.20),and the force-strain relationships (Eq.8.3)provide the forces,moments,and displacements of the wall.In the following we reduce these equations to readily usable forms. 4Ibid,Pp.205-206. 5 Ibid.,p.211
8.2 CYLINDRICAL SHELLS 371 Ny Nxy Nx Mxy Mxy My Mx Vy Vx Figure 8.10: Forces and moments inside the wall of a thin cylinder. (Fig. 8.9). The pressure may vary linearly along the cylinder’s axis, pz = pz0 + xpz1, (8.13) where pz0 and pz1 are specified constants. (For the pressure distribution shown in Fig. 8.9, pz1 is negative.) For the applied loads (Fig. 8.9), neither the stresses nor the strains vary circumferentially. Accordingly, the equilibrium equations are4 dNx dx = 0 (8.14) d dx (RNxy + Mxy) = 0 (8.15) Ny R − d2Mx dx2 = pz (8.16) Vx = dMx dx (8.17) Vy = dMxy dx , (8.18) where the forces Nx, Ny, Nxy, the moments Mx, Mxy, and the transverse shear forces Vx and Vy are illustrated in Figure 8.10. The strains and the curvatures of the reference surface are5 o x = duo dx o y = wo R γ o xy = dvo dx (8.19) κx = −d2wo dx2 κy = −wo R2 κxy = − 2 R dvo dx , (8.20) where uo, vo, and wo are the axial, circumferential, and radial displacements of the reference surface. The force–strain relationships are identical to those of laminated plates and are given by Eq. (8.3). The equilibrium equations (Eqs. 8.14–8.18), the strain–displacement relationships (Eqs. 8.19–8.20), and the force–strain relationships (Eq. 8.3) provide the forces, moments, and displacements of the wall. In the following we reduce these equations to readily usable forms. 4 Ibid., pp. 205–206. 5 Ibid., p. 211
372 SHELLS Table 8.2.The parameters required in the equations 2- [a]= [as]= A6- B12- R A6+ A-管 A2- B11 6- [a2l= 6- B6+ [a= B16- 2D16 [A= 「H H21 H22 [a]-[aallas]-'[az] g- [aallas]- i= 2五=-(h1+H)5=0f无=P0-装后=p1 The starting point of the analysis is the integration of the first two equilibrium equations (Eqs.8.14 and 8.15).These integrations yield D=Nx ,M」 D.=Nxy+R (8.21) where D,D2 are as yet unknown constants.By substituting the force-strain rela- tionships (Eq.8.3)into Eq.(8.21)and by introducing Eqs.(8.19)and (8.20)into the resulting equations,we obtain a=}+} (8.22) The matrices [az]and [a3]are given in Table 8.2. From Eq.(8.22)the derivatives of u°andu°are (8.23) The internal forces Ny and Mr may be expressed as (Eq.8.3) 公 A26 B12 B22 B61 Bu B12 B16 (8.24) Di D12 Kx
372 SHELLS Table 8.2. The parameters required in the equations [a1] = / A22 − B22 R B12 B12 − D12 R D110 [a3] = / A11 A16 − 2B16 R A16 + B16 R A66 − B66 R − 2D66 R2 0 [a2]= / A12 − B12 R B11 A26 − D26 R2 B16 + D16 R 0 [a4] = / A12 A26 − 2B26 R B11 B16 − 2D16 R 0 [H] = H11 H12 H21 H22! = [a1] − [a4][a3] −1[a2] g = g1 g2 = [a4][a3] −1 D1 D2 f1 = H22 f2 = − 1 R (H21 + H12) f3 = H11 R2 f4 = pz0 − g1 R f5 = pz1 The starting point of the analysis is the integration of the first two equilibrium equations (Eqs. 8.14 and 8.15). These integrations yield D1 = Nx D2 = Nxy + Mxy R , (8.21) where D1, D2 are as yet unknown constants. By substituting the force–strain relationships (Eq. 8.3) into Eq. (8.21) and by introducing Eqs. (8.19) and (8.20) into the resulting equations, we obtain D1 D2 = [a2] 1 wo R −d2wo dx2 6 + [a3] 1 duo dx dvo dx 6 . (8.22) The matrices [a2] and [a3] are given in Table 8.2. From Eq. (8.22) the derivatives of uo and vo are 1duo dx dvo dx 6 = −[a3] −1 [a2] 1 wo R −d2wo dx2 6 + [a3] −1 D1 D2 . (8.23) The internal forces Ny and Mx may be expressed as (Eq. 8.3) Ny Mx = A12 A22 A26 B12 B22 B26 B11 B12 B16 D11 D12 D16! o x o y γ o xy κx κy κxy . (8.24)
8.2 CYLINDRICAL SHELLS 373 By substituting Eqs.(8.19)and(8.20)into Eq.(8.24),we obtain }-}+ (8.25) The matrices [a]and [a4]are given in Table 8.2.Substitution of Eq.(8.23)into Eq.(8.25)results in =}+ (8.26) where [H]and g are given in Table 8.2.This equation can be written as P2w =H1R-H2+81 (8.27) w dPw° M,=1R-d+82, (8.28) where H,H2,H1,H2.g1,g2are the elements of the matrix [and the vectorg. By introducing Eqs.(8.27)and (8.28)into the third equilibrium equation(Eq.8.16) we obtain d4w°.edPw° h4+hd+后w°=店+x5, (8.29) where fi.....,fs are given in Table 8.2.We note that f contains the two as yet unknown constants D and D2. Solution of this fourth-order differential equation yields the radial displace- ment of the reference surface5 wo=(e-ix[Ci cos(Bx)+C2 sin(Bx)]+e-(L-[C3 cos(B(L-x)) +cm(L-加+[房6+ (8.30) where L is the length of the cylinder and A and B are the real and imaginary parts of the roots of the characteristic polynomial, λ=Re(y) -五+√经-4iB where (8.31) B Im(y) 2f Equation (8.30)is the solution of interest.This equation contains six unknown constants D,D2,C1-C4.The constant Di is given by Eqs.(8.21)and(8.11) N D =Nx 27R (8.32) The second equality in this equation is written by virtue of the fact that M is per unit length,N is the total force,and 2m R is the circumference;D2 is given by 6 E.Kreyszig.Advanced Engineering Mathematics.7th edition.John Wiley Sons,New York,1993, Pp.136-144
8.2 CYLINDRICAL SHELLS 373 By substituting Eqs. (8.19) and (8.20) into Eq. (8.24), we obtain Ny Mx = [a1] 1 wo R −d2wo dx2 6 + [a4] 1duo dx dvo dx 6 . (8.25) The matrices [a1] and [a4] are given in Table 8.2. Substitution of Eq. (8.23) into Eq. (8.25) results in Ny Mx = [H] 1 wo R −d2wo dx2 6 + g1 g2 , (8.26) where [H] and g are given in Table 8.2. This equation can be written as Ny = H11 wo R − H12 d2wo dx2 + g1 (8.27) Mx = H21 wo R − H22 d2wo dx2 + g2, (8.28) where H11, H12, H21, H22, g1, g2 are the elements of the matrix [H] and the vector g. By introducing Eqs. (8.27) and (8.28) into the third equilibrium equation (Eq. 8.16) we obtain f1 d4wo dx4 + f2 d2wo dx2 + f3wo = f4 + x f5, (8.29) where f1,...., f5 are given in Table 8.2. We note that f4 contains the two as yet unknown constants D1 and D2. Solution of this fourth-order differential equation yields the radial displacement of the reference surface6 wo = {e−λx[C1 cos (βx) + C2 sin (βx)] + e−λ(L−x) [C3 cos (β(L− x)) + C4 sin (β(L− x))]} + 1 f3 ( f4 + x f5) ! , (8.30) where Lis the length of the cylinder and λ and β are the real and imaginary parts of the roots of the characteristic polynomial, λ = Re(γ ) β = Im(γ ) where γ = 7889− f2 + $ f 2 2 − 4 f1 f3 2 f1 . (8.31) Equation (8.30) is the solution of interest. This equation contains six unknown constants D1, D2, C1–C4. The constant D1 is given by Eqs. (8.21) and (8.11) D1 = Nx = N 2π R . (8.32) The second equality in this equation is written by virtue of the fact that Nx is per unit length, Nis the total force, and 2π R is the circumference; D2 is given by 6 E. Kreyszig, Advanced Engineering Mathematics. 7th edition. John Wiley & Sons, New York, 1993, pp. 136–144
374 SHELLS Figure 8.11:The shear force and the twist moment. Eq.(8.21).The total torque acting at the edge of the cylinder is(Fig.8.11) =(Nsy R)27R+Mxy2n R. (8.33) Thus,from Eqs.(8.21)and (8.33)D2 is D2=2R (8.34) The constants C-Ca are obtained from the boundary conditions,which state that at a built-in end the radial displacement and its slope are zero as follows: dw w°=0 =0 atx=0 dx dwo (8.35) w°=0 =0 atx=L. dx The derivative of wo is given in Table 8.4.With the displacement given in Eq.(8.30),the boundary conditions above give 4 1 0 Y13 -入 B Y3 4 f+Lfs (8.36) 31 Y 0 C3 Y41 Yi where Yii are listed in Table 8.3.Equations(8.36)provide C1-C4.The displacement w is calculated with the constants D,D,C-C thus determined. Table 8.3.The parameters in Eq.(8.36) Yis =e-AL cos BL Yia =e-AL sin BL Y =e-L cos BL+Bsin BL) Ya=e-L (-B cos BL+sin BL) Ys1=e-AL cos BL Ys2=e-AL sin BL Ya =-e-L cos BL+B sin BL) Y=eL (B cos BL-sin BL)
374 SHELLS Nxy Mxy Figure 8.11: The shear force and the twist moment. Eq. (8.21). The total torque acting at the edge of the cylinder is (Fig. 8.11) T = (NxyR) 2π R + Mxy2π R. (8.33) Thus, from Eqs. (8.21) and (8.33) D2 is D2 = T 2π R2 . (8.34) The constants C1–C4 are obtained from the boundary conditions, which state that at a built-in end the radial displacement and its slope are zero as follows: wo = 0 dwo dx = 0 at x = 0 wo = 0 dwo dx = 0 at x = L. (8.35) The derivative of wo is given in Table 8.4. With the displacement given in Eq. (8.30), the boundary conditions above give 1 0 Y13 Y14 −λ β Y23 Y24 Y31 Y32 1 0 Y41 Y42 λ β C1 C2 C3 C4 = − f4 f3 f5 f3 f4+Lf5 f3 f5 f3 , (8.36) where Yi j are listed in Table 8.3. Equations (8.36) provide C1–C4. The displacement wo is calculated with the constants D1, D2, C1–C4 thus determined. Table 8.3. The parameters in Eq. (8.36) Y13 = e−λL cos βL Y14 = e−λL sin βL Y23 = e−λL (λ cos βL+ β sin βL) Y24 = e−λL (−β cos βL+ λ sin βL) Y31 = e−λL cos βL Y32 = e−λL sin βL Y41 = −e−λL (λ cos βL+ β sin βL) Y42 = e−λL (β cos βL− λ sin βL)