JOURNAL OF THEORETICAL AND APPLIED MECHANICS 42, 2, pp. 349-356, Warsaw 2004 ENERGETIC METHOD OF SOLVING THE STABILITY PROBLEM OF A SEMI-SPHERICAL SHELL LOADED WITH TORQUE1 Stefan Joniak Institute of Applied Mechanics, Poznań University of Technology e-mail: Stefan.Joniak@put.poznan.pl A thin-walled spherical shell is pivoted at both ends. The upper edge of the shell, loaded with a torque, may rotate around the shell axis. The problem of the loss of stability of the shell is solved with an energetic method. The change in the total energy of the shell while losing stabi- lity is determined. This requires the forms of the deflection and force functions to be assumed, according to actual boundary conditions. Co- efficients of the force function are determined from the solution to the inseparability equationwith theBubnov-Galerkinmethod. The stability equation of the shell is formulated as a result of application of the Ritz method to the total energy variation. It is an algebraic equation serving for determination of the critical load. It is equal to theminimal value of the load. The work ends with a numerical example. Key words: shells, non-linear stability 1. Introduction A thin-walled spherical shell being a subject of the analysis is shown in Fig.1. Its bottom edge is fixed and pivoted. The upper edge is also pivoted butmay rotate around the vertical axis of the shell. The upper edge is loaded with a torque. A non-linear problem of the loss of stability is considered. The problem is solved with an energetic method. For the assumed modes of the deflection and force function the Bubnov-Galerkin method is used in order to solve the inseparability equationand, afterwards, the total energy is calculated. 1 A part of this contribution was presented on the Xth Symposium ”Stability of Struc- tures” in Zakopane, September 8-12, 2003. 350 S.Joniak In order to formulate the stability equation the Ritz method is applied. The final goal of the work consists in the determination of the critical load. This, however, is possible only in the case of a numerical example, since the problem is of high complexity. Fig. 1. Scheme of the shell 2. Mathematical description of the problem 2.1. Total energy of the system The total energy of the systemwhile losing the stability amounts to V =U1+U2−L (2.1) where U1 is the energy of the membrane state, U2 – energy of the bending state, L – work of external forces and U1 = 1 Eh ∫ A [T21 +T 2 2 −2νT1T2+2(1+)S] dA U2 = Eh3 24(1−ν2) ∫ A [κ211+κ 2 22+2νκ11κ22+2(1−µ)κ 2 12] dA L= ∫∫ S ∂w ∂θ ∂w ∂ϕ dθdϕ Energetic method of solving the stability problem... 351 A is the middle surface of the shell, w – deflection function upon the loss stability, κ11, κ22, κ12 are variations in the main curvatures of the spherical shell (Mushtari andGalimov, 1957), T1, T2, S are forces of the bending state, Ψ denotes the force function and κ11 =− 1 R2 ∂2w ∂θ2 κ22 =− 1 R2 ( 1 sin2θ ∂2w ∂ϕ2 +cotθ ∂w ∂θ ) κ12 = 1 R2 sinθ ( cotθ ∂w ∂ϕ − ∂2w ∂ϕ∂θ ) T1 = 1 R2 ( 1 sin2θ ∂2Ψ ∂ϕ2 +cotθ ∂Ψ ∂θ ) T2 = 1 R2 ∂2Ψ ∂θ2 S= 1 R2 sinθ ( cotθ ∂Ψ ∂ϕ − ∂2Ψ ∂ϕ∂θ ) The force of the membrane state in (2.1) has the form S= M0 2πR2 sin2θ (2.2) 2.2. Inseparability equation In order to determine the forces of the bending state, the force functions must be known. This requires solving the equation of the inseparability of displacements. The inseparability equation is of the form (Mushtari andGali- mov, 1957) ∇ 2 ∇ 2Ψ−Eh(κ212−κ11κ22−κ11k22−κ22k11)= 0 (2.3) where ∇ 2 = 1 R2 ( cotθ ∂ ∂θ + ∂2 ∂θ2 + 1 sin2θ ∂2 ∂ϕ2 ) and kii are main curvatures. 3. Boundary conditions The following conditions should be met at the shell boundaries θ= θ1 w=0 Mθ =0 S= M0 2πR2 sin2θ1 T1 =0 (3.1) 352 S.Joniak and θ= π 2 w=0 Mθ =0 S = M0 2πR2 T1 =0 (3.2) 4. Deflection and force functions The following forms of the force and deflection functions were assumed Ψ = [bϕ+ csin(mϕ)]sin2θ (4.1) w= asin 2π(θ−θ1) π−2θ1 sin [2π(θ−θ1) π−2θ1 +mϕ ] sin2θ where a, b, c are constants, m – an integer number defining the mode of the loss of stability. The deflection function explicitly satisfies the first two conditions ((3.1) and (3.2)), while the third one is satisfied in the integral sense. On the other hand, the force function meets the third condition to a constant, without satisfying the condition required for the normal force T1. 5. Solution to the equation of inseparability Equation of inseparability (2.3) is solved with the Bubnov-Galerkin me- thod. The ortogonalization conditions are of the following form π 2 ∫ θ1 2π ∫ 0 F(θ,ϕ)ϕsin3θ dθdϕ=0 (5.1) π 2 ∫ θ1 2π ∫ 0 F(θ,ϕ)sin(mϕ)sin3θ dθdϕ=0 where F(θ,ϕ) denotes the left-hand side of equation (2.3), with force function components (4.1)1 applied as ortogonalization factors. Energetic method of solving the stability problem... 353 Once equations (5.1) and (5.2) are solved, the following expressions for constants b and c are obtained b=Eh(H1a 2+G1aR) c=Eh(H2a 2+G2aR) (5.2) where Hi,Gi are constants. The final form of the force function is as follows Ψ =Eha2 [( H1+G1 R a ) ϕ+ ( H2+G2 R a ) sin(mϕ) ] sin2θ (5.3) 6. Solution to the problem of the loss of stability The deflection and force functions should be substituted into the equation of total energy variation in order to calculate the variation itself. The total energy is a function of a. While losing the stability, the variation of the total energy takes the minimal value. According to the Ritz method, the condition for the minimum of the total energy variation has the form ∂V ∂a =0 (6.1) The implementation of the above expression leads to the formulation of an equation of the dimensionless torque M = M0 Eh3 =C1 (a h )2 +C2 R h a h +C3 (R h )2 +C4 (6.2) where Ci are constants dependingon θ1, the number m andPoisson’s ratio ν. Equation (6.2) is an answer to the problem of the loss of stability. It sho- uld serve for calculating the critical load. The critical load corresponds to the minimal value of the dimensionless load parameter M. It is determined for fixed dimensions of the shell, defined by the number m. For solving ortogo- nalization conditions (5.1) and equation (6.1) the Derive software was used. It includes procedures enabling transformations of algebraic expressions, dif- ferentiating, and integrating, thus enabling all the operations required by the above equations. 354 S.Joniak 7. Numerical example The critical load can be found only with a numerical method. This is a consequence of the complex structure of coefficients in equation (6.2). First of all, the dimensions of the shell should be assumed in such a way so that to allow for calculating the constants Ci corresponding to different values of the number m. This was carried out with the Derive program. The critical load was sought bymeans of a graphicalmethod. The plots should be drawn in the M−a/h coordinates for a series of the numbers m. The plots enabled finding theminimal value of the load M, i.e. Mcr. The value m corresponding to the minimal level of the load M is considered as the critical one mcr. The plots were drawnwith the help of theDerive forWindows software. Figure 2 shows plots of the dimensionless torque M as a function of a/h for a shell with the dimensions θ1 = π/6, R/h = 150, ν = 0.3 and different values of m. The minimal value of the load M =Mcr is found for m=mcr = 8. It should be noticed that for θ1 = π/10, and θ1 = 3π/14 the problem remains unsolved. This is certainly a result of the assumed forms of the deflection and force functions. Fig. 2. Diagrams m-a/h The critical loads of the shells of different dimensions are presented in Table 1.Apart fromthe Mcr values, thenumbers mcr areprovided inbrackets. Energetic method of solving the stability problem... 355 The comparison of the critical moments, contained in Table 1, allows one to come to an obvious conclusion that the growing ratio R/h decreases the resistance of the shell to the loss of stability. Table 1.Values of the critical moments R/h Mcr θ1 =π/12 θ1 =π/6 100 115.6 (8) 104.4 (8) 150 98.4 (10) 98.1 (10) 200 20.0 (11) 47.8 (11) The same problem was solved in the paper by Joniak (2003a) with the Bubnov-Galerkin method. The solution, however, was unsatisfactory, being valid only for delimited range of shell dimensions. On the other hand, the solution shown in the present paper is free of this fault. References 1. Avdonin A.S., 1969,Prikladnye metody razchoyota obolochek i tonkostennykh konstrukcii, Izd. ”Mashynostroenie”,Moskva 2. Joniak S., 2003a, Nieliniowe zagadnienie stateczności powłoki półkulistej ob- ciążonej momentem obrotowym,Mat. X Sympozjum Stateczności Konstrukcji, Zakopane, 193-198 3. Joniak S., 2003b, Nonlinear stability problem of spherical shell loaded with torque, Journal of Theoretical and Applied Mechanics, 41, 3, 537-544 4. Łukasiewicz S., 1976, Obciążenia skupione w płytach, tarczach i powłokach, PWNWarszawa 5. MushtariG.M.,GalimovK.Z., 1957,Nelineinaya teoria uprugikh obolochek, Tatkniizdat, Kazań 6. Nowacki W., 1979,Dźwigary powierzchniowe, PWN,Warszawa 7. Volmir A.S., 1967, Ustoychivost deformiruemykh system, Izd. ”Nauka”, Moskva 356 S.Joniak Metoda energetyczna rozwiązania problemu stateczności powłoki półkulistej obciążonej momentem obrotowym Streszczenie Cienkościenna powłoka półkulista jest podparta przegubowo na obu brzegach. Górny brzeg powłoki ma możliwość obrotu wokół osi powłoki; do tego brzegu przy- łożony jestmoment obrotowy.Rozpatrywany jest problemutraty stateczności tej po- włokimetodą energetyczną.Wyznacza się zmianę energii całkowitej powłoki podczas utraty stateczności. Wymaga to przyjęcia postaci funkcji ugięcia po utracie statecz- ności i funkcji sił odpowiednich do warunków brzegowych.Współczynniki funkcji sił wyznacza się z rozwiązania równania nierozdzielności metodą Bubnowa-Galerkina. Równanie utraty stateczności powstaje po zastosowaniu do całkowitej zmiany ener- gii metody Ritza. Jest to równanie algebraiczne, z którego wyznacza się obciążenie krytyczne; odpowiada onaminimalnej wartości parametru obciążenia. Manuscript received December 11, 2003; accepted for print February 24, 2004