Shells with membrane behavior

Transcription

1 Chaper 3 Shells wih membrane behavior In he presen Chaper he sress saic response of membrane shells will be addressed. In Secion 3.1 an inroducory example emphasizing he difference beween bending and membrane sresses for he simplified geomery of a spherical shell undergone o an inernal pressure load is presened. Nex, in Secion 3.2 he general heory of he membrane shells having an axiallysymmeric geomery ogeher wih an axially symmeric load is presened also including several specific geomery cases. Finally, in Secion 3.3 he case of a generic non-symmeric load is carried ou also considering some special applicaion cases. 3.1 General issues on membrane v.s. bending behavior of shell srucures In he following shell srucures will be assumed having consan hickness and small compared o he oher geomeric dimensions, represened by local radius of curvaure of he average surface. In order o describe he geomery of a shell srucure, he geomery of he average surface and hickness are o be described. Generally, a shell is considered o be a hin shell when he raio beween is hickness, indicaed wih, and he smaller radius of curvaure, indicaed wih r, is lower han 1/20, alhough in many examples of ineres in he aeronauical field his raio can be also of order of In order o sudy he behavior of a shell srucure, wo differen approach can be considered. The easies ype considers as dominan dominan behavior he membrane behavior and his can be assumed when here are no load or srucural disconinuiies on he shell srucure. The mos general shell behavior include also he bending behavior and his allows also 41

2 42 o rea he disconinuiy in he sress field caused by he effec of load or srucural disconinuiy as well. I is imporan o noe ha he mos general way does no have he purpose of improving he soluion offered by he membrane approach, bu i is able o deal wih differen problems and, more precisely, he effecs of disconinuiy in he load or in he srucure, ha he membrane heory can no describe. I is also observed ha he equilibrium condiions in he membrane model are sufficien o describe and close he sress problem so avoiding he inroducion of furher relaions like consiuive or kinemaic relaionships. The fundamenal hypoheses a he basis of he sudy of a hin-walled shell in he case of small deflecions are (in analogy wih he plae models): i) he hickness is small compared o he smaller radius of curvaure r << 1 ii) he deflecion is small compared o he hickness of he shell, w << 1 ( r < 1 ) 20 iii) he normal o he average surface plane secions remain plane and hose are sill normal o he deformed average surface, so his hypohesis requires ha he shear sress ε xz and ε yz (z is he direcion normal o he shell surface) are negligible, as well as sher he deformaion ε zz iv) normal sress σ zz is negligible compared o he membrane sresses. Figure 3.1: Par of a spherical shell subjeced o a pressure

3 43 In order o discuss he differen role of membrane in-plane sress σ m wih respec o he bending in-plane sress σ f, le us consider as an example he case of a porion of a spherical shell of hickness and radius r subjeced o a uniform pressure of inensiy p, as indicaed in Fig.3.1. The equilibrium condiion requires ha he sum of verical forces is zero and allows o derive he force o he membrane per uni of lengh, N φ, by he relaion: or 2πr 0 N φ sinφ φ 0 p cos φ 2πr sin φ r d φ = 0 N φ = pr 0 2 sin φ = sin φ pr 2 sin φ = pr 2 For reasons of symmery, which concern he srucure (a spherical shell) and he load (pressure), one has a unique value for he sress given by: σ θ = σ φ = σ m = pr 2 Nex, one can evaluae he deformaion wih he use of consiuive laws, which in case of isoropic maerial provide he consiuive relaionship: ε m = 1 E (σ m νσ m ) = (1 ν) pr 2E This deformaion value corresponds o a variaion of he radius of he sphere given by relaionship: r = r (1 + ε m ) and his variaion of he radius leads o a variaion of he curvaure given considering a Taylor series expansion for ε m, by: χ = 1 r 1 r = 1 ( ) 1 r ε m Then, by using he equaion for ε m, one has: χ = (1 ν)p 2E ε m r In he case of spherical shell under consideraion one can derive he local bending momen, in analogy wih he case of he plae, wih he consiuive relaion: wih he bending siffness given by: M = D(χ + νχ) = D(1 + ν)χ = D(1 ν 2 ) p 2E D = E 3 12(1 ν 2 )

4 44 and, herefore, he bending momen is given by: M = p2 24 Thus, one can calculae he bending sress from: σ f = M I z max = 12M 3 2 = 12p = p 4 in which I = 3 12 and y = 2. The relaionship beween he membrane sress and he bending sress is σ m σ f = pr 2 4 p = 2r I is hus eviden ha he membrane sress is much more imporan because, in he basic assumpions of he heory considered, he raio beween radius of curvaure and hickness of he shell is very large and consequenly also he relaionship beween he membrane sress and he bending sress is large as well. The case considered here is a special load case for a special geomery. However,he obained resul and commen are very general. I is observed ha he hin shells are also subjeced o problems of criical buckling load when subjeced o compressive loads. In order o esimae he exen of he criical sress, one can use a relaion of ype: σ cr = k E r where E indicaes he modulus of elasiciy of he shell maerial and k indicaes a consan whose value depends on several facors, bu a value ha may be indicaive can be k = I is o be noed ha, due o he ofen very small value of he raio beween he hickness of he shell and radius of curvaure, he value of he criical buckling sress indicaed by he previous relaion is very low compared o he ensile srengh of he maerial. If one considers he case of a spherical shell made by aluminum alloy (2024 or 7075 ype) and wih he raio r = 10 3 one has: σ cr = = 17.5MP a which is in fac a very low value compared o he ensile srengh of he lighly alloy. 3.2 Axial symmeric shells undergone o axial symmeric loads Le us consider a very special case of shells from he geomerical poin of view, which are described by an average surface obained by he roaion of a curve, denoed as meridian curve,

5 45 around an axis which lies in he plane o which he curve belongs. In general, he maerial lines ha describe he surface of he undeformed shell are no Caresian. I will show in he following he wo naural coordinaes associaed o hese lines wih φ (meridian coordinaed in he revoluion shells) and ϑ (longiude coordinae). As indicaed in Fig.3.2,a poin on Figure 3.2: Geomery of he revoluion shell he surface of he shell is idenified by coordinaes (ϑ, φ) and an infiniesimal surface elemen, indicaed wih ABCD, is idenified by meridian and parallel incremen dφ and dϑ. The curvaure radius bending radii shown in Figure wih r 0 and r 1 are relaed o he parallel (ϑ direcion) and meridian (φ direcion) curves respecively, whereas r 2 represens he disance beween he generic surface maerial poin and he inersecion beween he curvaure radius

6 46 and he roaion axis. The radii r 1 and r 2 are relaed by he relaionships: wih r 0 = r 2 sin φ l AC = r 0 dϑ l CD = r 1 dφ ds = r 0 r 1 dϑdφ = r 1 r 2 sinφdϑdφ Noe ha he radii of curvaure of he shell r 0, r 1, and r 2 which describe he geomery, are known for specific geomery as funcion of he maerial co-ordinaes (φ, ϑ). This ype of revoluion shell, is considered he basis of he following consideraions on he shells as hey can keep a simple geomeric descripion while referring o srucures of ineres pracical in he field of aeronauics. In he case of a shell of revoluion o which is applied a load in axial symmery, axial symmeric load, you have as unknowns only wo membrane forces, per uni of lengh, N ϑ, N φ, while here are no shear sress. In order o deermine he wo unknowns membrane forces, one can consider wo equaions of equilibrium in he normal direcion, z, and in he direcion φ. In he Fig. 3.3, is considered in deail he shell elemen ABCD already indicaed in Fig Due o he axial symmery, he forces o he membrane and he load does no have variaions along he direcion ϑ. The exernal forces applied, which are forces per uni area, are indicaed wih p φ and p z = p r (p z is in he normal ouward direcion). In order o evaluae he force balance in he normal direcion, i is necessary o deermine he componens he forces applied in his direcion and one has he following equaion (see Fig. 3.3): which becomes: p z r 0 r 1 dϑdφ + N φ r 0 dϑdφ + N ϑ r 1 sin φdϑdφ = 0 N φ r 0 + N ϑ r 1 sin φ + p z r 0 r 1 = 0 If one divides by r 0 r 1 and considers ha r 0 = r 2 sin φ, one obains: N φ r 1 + N ϑ r 2 = p z (3.1) By imposing he equilibrium in he direcion φ of he applied forces, one has φ (N φr 0 ) dφdϑ N 0 r 1 dφdϑcosφ + p φ r 1 dφr o dϑ = 0

7 47 Figure 3.3: Elemen of revoluion shell: forces balance or, considering loads axialsymmery and hus including he ordinary derivaive wih respec o

8 48 φ, 1 d dφ (N φr 0 ) N ϑ r 1 cosφ + p φ r 1 r o = 0 (3.2) Now, considering N ϑ as given by Eq. 3.1, subsiuing his o Eq. 3.2 muliplied by sin φ, one Figure 3.4: Revoluion shell under he angle φ obains: d dφ (N φr 0 ) sin φ + N φ r 0 cos φ = p φ r 1 r 2 sin φ sin φ p z r 1 r 2 sin φ cos φ The lef-hand erm of his equaion can be wrien as: and hen is obained 1 N φ (φ) = r 0 (φ) sin φ d dφ (N φr 0 sin φ) = d ( Nφ r 2 sin 2 φ ) dφ φ φ 1 r 1 ( φ)r 2 ( φ) (p z cos φ + p φ sin φ) sin φd φ F 2πr 0 (φ) sin φ (3.3) 1 Noe ha, insead of his second equaion of equilibrium, one could use he one obained by considering a porion of he shell seen under an angle φ as indicaed in Fig.3.4. If one indicaes wih F he verical resulan of all he exernal loads ha are applied o he shell and i is considered ha for reasons of symmery he forces N φare consan on he edge, one has he balance equaion: and herefore: 2πr 0N φ sin φ + F = 0 F N φ = 2πr 0 sin φ This equilibrium equaion is only a differen global way of expressing he equilibrium condiion locally given by Eq. 3.2.

9 49 where φ 1 is he exreme value of he coordinae φ in correspondence of which he exernal load boundary is applied and so giving a posiive verical force downwards and equal o F. Indeed, in his case wih p z = p φ = 0, one has F N φ = 2πr 0 sin φ The equilibrium equaions in he direcion of z and φ allow o obain he unknowns N ϑ and N φ and from hese relaed sresses. The membrane sresses, in he case of a shell of revoluion wih axisymmeric load, are derived from he equaions of equilibrium according z and seconds φ. Cases of special shell geomeries are now considered in he following Spherical pressured shell In his case r 1 = r 2 = r and he equilibrium in normal direcion z becomes (p φ = 0): N φ + N ϑ = p z r Consider an inernal pressure of inensiy p such ha p z = p. Moreover, he balance equaion in φ becomes: Therefore N φ = The membrane sress is hen: p r sin 2 φ N φ = p r 2 φ 0 r 2 cos φ sin φd φ = pr 2 N ϑ = p z r N φ = pr p r 2 = pr 2 σ φ = σ ϑ = N = pr 2 From he consiuive equaions in he case of isoropic maerial, he srain is obained by: ε φ = ε ϑ = σ ϑ E (1 ν) = p r (1 ν) E 2 Noe ha we can also calculae he variaion of he radius of he spherical shell wih: [ r = r (1 + ε ϑ ) = r 1 + p r ] (1 ν) E 2

10 50 Figure 3.5: Conical shell Conical shell In his case indicaed in Fig.3.5, he angle φ is consan and he radius r 1 goes o infinie. From he condiion of equilibrium in z direcion: one has: N φ r 1 + N ϑ r 2 = p z N ϑ = r 2 p z = p z r 0 sin φ while he condiion of equilibrium occurs only a he end of he conribuion due o possible boundary erms: F N φ = 2πr o sin φ Cylindrical shell of circular secion This case can be raced back o he previous conical shell if you pu he angle φ = π 2. From he equaion equilibrium in he direcion z one has (r 1 e r 2 = r 0 = r ): N ϑ = p z r 0 = p z r and by he condiion of equilibrium in he direcion φ one has (noe ha no inegral conribuions of Eq. 3.3 are presen for his geomery): N φ = F = F 2πr 0 2πr

11 51 In he case of cylindrical shell, closed a he ends, and pressurized wih an inernal pressure p as in he case of a fuselage one has: p = p z F = πr 2 0p Thus, he axial sress in he circumferenial direcion is given by σ ϑ = N ϑ = pr 0 = pr whereas in he axial direcion σ xx = σ φ = N φ = πr2 0 p 2πr 0 = pr 0 2 = pr 2 The circumferenial axial srain may be obained: ε ϑϑ = ε ϑ = 1 E (σ ϑ νσ φ ) = 1 E ( pr0 ν pr ) 0 = pr ( 0 1 ν ) 2 E 2 and he axial srain: ε xx = ε φ = 1 E (σ φ νσ ϑ ) = ( ) 1 2 ν pr0 E In he case of a cylindrical fuselage wih a radius r 0 =4m, wih a hickness = m and wih a pressure p=10 5 Pa, he axial sress componens are σ ϑ = pr 0 = 200MP a σ φ = σ ϑ 2 = 100MP a wih corresponding srains ε ϑ = pr ( 0 1 ν ) = E 2 ε φ = Spherical dome loaded by is own weigh Consider a homogeneous semi-spherical dome of radius r and wih ai hickness, simply suppored o he boundary and subjeced o is own weigh, as shown in Fig.3.6. M is he oal mass of he dome, S is oal area and g he acceleraion of graviy. Thus, he weigh force per uni area in he graviaional field direcion has an inensiy q = Mg. Thus, he componens of S he applied load are:

12 52 Figure 3.6: Geomery of a spherical dome p ϑ = 0 p φ = q sin φ p z = q cos φ applying he Eq. 3.3, one has: N φ (φ) = 1 r sin 2 φ φ 0 r 2 q(cos 2 φ + sin 2 φ) sin φd φ = r q(1 cos φ) sin 2 φ From he condiion of equilibrium according wih z: = r q 1 + cos φ one obains: N ϑ r N ϑ = p z r N φ = q cos φr + + N φ r = p z By he membrane sress componens N ϑ, N φ are obained: σ φ = N φ σ ϑ = N ϑ r ( ) q 1 + cos φ = 1 r q 1 + cos φ cos φ r q = (1 + cos φ) ( ) = r q 1 (1 + cos φ) cos φ In he case of he hemispherical dome one has for σ φ a behavior as a funcion of he angle φ ha sars from he value for φ = 0: σ φ = r q 2

13 53 and reaches φ = 90 o σ φ = r q Regarding σ ϑ i has a course always in funcion of he angle φ ha sars from: and arrives a: σ ϑ (0 o ) = r q 2 σ ϑ (90 o ) = r q as indicaed in Fig.3.7, σ ϑ changes sign and hen vanishes for a value of he angle φ obained Figure 3.7: Performance of sress σ ϑ e σ φ by: cos φ = cos φ φ = 51 o 50 Since he sress componens have been deermined, he deformaions can be obained by: ε ϑϑ = 1 E (σ ϑϑ νσ φφ ) ε φφ = 1 E (σ φφ νσ ϑϑ )

14 54 Consider as an example he case of a hemispherical cemen dome wih he following daa: q = 1.725KN/m 2 r = 35m = m E = 20GP a σ ϑcomp = 21MP a The σ φ is always compression, while he σ ϑ may eiher be compression, when φ < 51 o 50 or racion, when φ > 51 o 50. The maximum value of compression is given by: σ φcomp = qr = = 0.805MP a As can be seen he maximum value of compression is much smaller han ha shown for he maerial. However, as regards he ensile behavior, his is deermined from σ ϑ ha has he same value of he compressive sress. The maximum ensile srengh is known by he specific characerisics of he maerial and, in he case of cemen, i can be very limied. Regarding he compression, he criical load could be esimaed as limi case by he formula: σ cr = KE r where he coefficien K can be pu K = 0.25 in order o obain for his example: σ cr = = 10.71MP a also in his case is σ φcomp σ cr Open dome a he op Consider again a dome wih a radius r and hickness, bu open a he op for a corner φ 0 wih a reinforcemen ring and a load per uni lengh, P (ie from Eq. 3.3, F = 2πr P sin φ 0 ) which acs on he ring as shown in Fig.3.8. While sill accoun ha, as in he previous case, a weigh force per uni of spherical area acs equal o q = Mg, he componens of he weigh are: S p ϑ = 0 p φ = q sin φ p z = q cos φ

15 55 Figure 3.8: Open dome a he op One has, by applying l Eq. 3.3: N φ (φ) = 1 φ r sin 2 r 2 q ( cos 2 φ + sin 2 φ ) F sin φd φ φ φ 0 2πr sin 2 φ = r q sin 2 φ (cos φ 0 cos φ) P sin φ 0 sin 2 φ Then one ge: From: N φ = σ φ = qr ( cos φ0 cos φ sin 2 φ ) P sin φ 0 sin 2 φ N φ + N ϑ = p z r one has (con p z = q cos φ): N ϑ = σ ϑ = pr ( cos φ0 cos φ sin 2 φ One can hen calculae he deformaion ε ϑ from: ) cos φ + P sin φ 0 sin 2 φ ε ϑ = 1 E (σ ϑ νσ φ ) and you can evaluae he deformaion of he radius r = r (1 + ε ϑ ).

16 Pressure Tanks Many pressure vessels are consruced wih a cylindrical body and end caps ha are shells of revoluion (can be considered as a special case of semi-spheres or ellipsoids). In all cases hese end caps mus wihsand an inernal pressure p is consan and perpendicular o he wall. One has p ϑ = 0 p φ = 0 p z = p I is ineresing o poin ou ha, in all hese cases, one can obain N φ by Eq. 3.3 as which becomes: N φ (φ) = N φ (φ) = 1 [ φ ] r 2 sin 2 r 1 ( φ)r 2 ( φ)(p z cos φ + p φ sin φ) sin φd φ φ 0 1 r 2 sin 2 φ φ 0 r 1 r 2 p cos φ sin φd φ = p r0 r 2 sin 2 r 0 d r 0 φ 0 In fac: r 2 sin φ = r 0 dr 0 = ds cos φ ; ds = r 1 dφ ; dr 0 = r 1 cos φdφ Then: N φ = While from he equilibrium equaion: p r0 2 r 2 sin 2 φ 2 = p r2 2 sin2 φ 2 sin 2 φ = p 2 r 2 N φ r 1 + N ϑ r 2 = p one has: N ϑ = pr 2 N φ r 2 r 1 herefore: N ϑ = pr 2 p r 2 ( 2 = pr 2 1 r ) 2 (2r 1 r 2 ) = pr 2 2 r 1 2r 1 2r 1

17 57 As a special case for he hemispherical shell, one has r 1 = r 2 = r and hen: N φ = p 2 r σ φ = p r 2 N ϑ = pr (2r r ) 2r = p 2 r σ ϑ = p 2 In he case of an ellipsoidal cap, Fig.3.9, one has: r Figure 3.9: Ellipsoidal cap r 1 = r 2 = a 2 b 2 (a 2 sin 2 φ + b 2 cos 2 φ) 3 2 a 2 (a 2 sin 2 φ + b 2 cos 2 φ) 1 2 (corre- There are wo ypical siuaions for φ = 0 (corresponding o he poin 0) and φ = π 2 sponding o he poin A,A ): Per φ = 0 r 1 = a2 b ; r 2 = a2 b

18 58 hen: N ϑ = p a2 b N φ = p 2 r 2 = p a 2 2 b a 2 b 2 a2 b = p a 2 2 b σ ϑ = σ φ = p 2 a 2 b 1 Per φ = π 2 r 1 = b2 a ; r 2 = a hen: N φ = p 2 a σ φ = p a 2 ( N ϑ = p a 1 a a ) 2b 2 = p a ) ) (1 a2 2b 2 σ ϑ = p a (1 a2 1 2b 2 Noe ha N φ is always posiive, while N ϑ can be negaive if goes back o a = b = r. In he case in which is a = 2b = r one has: Perφ = 0 N φ = p 4b 2 2 b N ϑ = p 2 = 2pb = pr a 2 b = pr a 2 > 1. The case of he sphere 2b2 Per φ = π 2 N φ = p 2 a = p 2 r ( ) N ϑ = pr 1 4b2 2b 2 = pr In he case of a very flaened cap, for example, a = 10b = r one has: For φ = 0 N φ = p 100b 2 = 50pb = 5pr 2 b N ϑ = p 100b 2 = 5pr 2 b

19 59 For φ = π 2 N φ = p 2 r ( ) N ϑ = pr 1 100b2 2b 2 = 49pr Toroidal ank wih circular secion wih inernal pressure Consider now a oroidal ank of circular cross secion, radius of he orus b, subjeced o an inernal pressure p. If reference is made o he par of he shell defined by he angle φ as Figure 3.10: Toroidal ank wih circular secion indicaed in Fig.3.10, for he balance of verical forces one has: whence Bu one has he posiions: and hen i has: From: 2πr 0 N φ sin φ = πp(r 2 0 b 2 ) N φ = p(r 0 + b)(r 0 b) 2r 0 sin φ r 0 b = r sin φ r 0 + b = 2r 0 r sin φ N φ = pr (r 0 + b) 2r 0 N φ r 1 + N ϑ r 2 = p z

20 60 puing p z = p r 1 = r r 2 sin φ = r 0 is obained N ϑ = r 2 p r 2 r N φ = r 2 p r 2p(r 0 + b) 2r 0 namely: Finally, he sresses are: N ϑ = r 2 p r 2p(2r 0 r sin φ) r 0 = r 2 sin φpr 2r 0 = pr 2 N ϑ N φ = σ ϑ = pr 2 = σ φ = pr 2 r 0 + b r 0 = pr 2 ( r ) sin φ + 2b r sin φ + b Consider as an example a oroidal ank of circular cross secion buil in ligh alloy wih he following daa: r = 200mm ; b = 1.5m ; p = P a = m One has σ ϑ = = 50MP a ( σ φ = pr r ) sin φ + 2b 2 r sin φ + b by: φ = 90 o φ = 0 o σ φ = pr 2 σ φ = pr = 100MP a ( r ) + 2b r = 94MP a + b

21 61 If r b i can be wrien: σ ϑ = pr 2 σ φ = pr = 2σ ϑ ε ϑ = 1 E (σ ϑ νσ φ ) = 1 E (σ ϑ 2νσ ϑ ) = σ ϑ (1 2ν) E ε φ = 1 E (σ φ νσ ϑ ) = 1 E (σ ϑ(2 ν) Therefore he oroidal ank pressurized behaves in a similar manner o cylindrical ank wih a radius equal o r. 3.3 Axial-symmeric shells undergone o no axial-symmeric load In he more general case he shell of revoluion is subjeced o a load condiion which does no have axial symmery; in his case, he unknowns are sill N ϑ and N φ and more cuing forces per uni lengh N φϑ and N ϑφ in he case of hin shell are equal and herefore have hree unknowns which can be deermined by imposing he hree equilibrium equaions according o he axes z, ϑ, φ. The load per uni area has he hree componens p z, p ϑ, p φ. Fig.3.11 indicaes he elemen of he shell ABCD wih he forces. If you consider he balance of forces in he direcion ϑ i has ha he conribuion of exernal forces is given by he componen p ϑ for he area elemen r 0 dϑr 1 dφ; i has hen he conribuion due o variaion according o ϑ di N ϑ ha resuls N ϑ ϑ r 1dϑdφ; i hen has he componen due o shear forces acing on he sides AB and CD of he elemen of he shell which form an angle dϑ and hen have he resuling indicaed ha he direcion angen o a parallel; finally, one mus consider he conribuion of he cuing forces acing on he sides AC and BD which is given by: N ϑφ dr 0 dφ dφdϑ + N ϑφ φ r 0dφdϑ = φ (r 0N ϑφ ) dϑdφ he balance equaion in ϑ direcion resuls: which finally becomes: N ϑφ r 1 cos φdφdϑ + N ϑ ϑ r 1dφdϑ + φ (r 0N ϑφ ) dϑdφ + p ϑ r 0 r 1 dϑdφ φ (r 0N ϑφ ) + N ϑ ϑ r 1 + N ϑφ r 1 cos φ + p ϑ r 0 r 1 = 0 (3.4)

22 62 Figure 3.11: Shell of revoluion wih no axisymmeric load You mus proceed in a similar manner wih regard o he balance equaion according wih φ in which compared o he case of axial symmery mus consider he erm N ϑφ ϑ r 1dϑdφ which derives from he difference of he cuing forces acing on he sides AB and CD of he elemen of he shell and is obained: φ (r 0N φ ) + N ϑφ ϑ r 1 N ϑ r 1 cos φ + p φ r 0 r 1 = 0 (3.5) Finally, as regards he hird equaion of equilibrium in he z-direcion is observed ha he conribuion of shear forces is zero and he equaion is hen seen previously in he axisymmeric case: N φ r 1 + N ϑ r 2 = p z (3.6) We hus have he hree equilibrium equaions ha allow he calculaion of forces in membrane N ϑ, N φ, N φϑ in he case of a shell of revoluion wih a no axisymmeri load.

23 Cylindrical shell of circular secion I is now considered a paricular case from geomerical poin of view: he cylinder of circular secion. An elemen of his ype of shell is shown in Fig.3.12 and is consiued by wo generarices and wo planes normal o he cylinder axis away dx; considering a load p x,p ϑ, p z you have as unknowns he forces per lengh uni N x, N ϑ, N xϑ and heir incremens. Figure 3.12: Cylindrical shell of circular secion Considering he force balance in he x direcion, one has p x dxrdϑ + N x x dxrdϑ + N ϑx ϑ dxdϑ = 0 For he balance according o ϑ one obains: p ϑ dxrdϑ + N ϑ ϑ dxdϑ + N xϑ x dxrdϑ = 0 And finally, according o he balance along z one has: p z rdxdϑ + N ϑ dxdϑ = 0

24 64 This yields he sysem of equaions: N ϑ = p z r N xϑ x + 1 N ϑ r ϑ = p ϑ N x x + 1 N xϑ r ϑ = p x We noe ha he equaions of his sysem can be solved in sequence saring from he firs; in fac, once assigned o he exernal load, one can derive immediaely in finie erms he firs unknown N ϑ and hen he oher wo unknowns for inegraion by he relaions: x ( N xϑ = p ϑ + 1 ) N ϑ r d x + f 1 (ϑ) ϑ N x = x 0 x x 0 ( p x + 1 r N xϑ ϑ ) d x + f 2 (ϑ) where he funcions f 1 (ϑ) e f 2 (ϑ) should be evaluaed depending on he boundary condiions of he problem Cylindrical ank Consider a cylindrical ank wih a consan secion of radius r suppored a he ends, bu free o move in he x direcion, wih a lengh l as indicaed in Fig The ank is compleely filled wih a liquid of specific weigh γ = ρg (ρ is he mass densiy and g he graviy acceleraion) he pressure of which depends on he angle of posiion indicaed wih ϑ. Figure 3.13: cylindrical duc The pressure in he ank is deermined by he law which considers he hydrosaic pressure equal o he weigh of he liquid column a he poin considered; in he case of he ank, he heigh of

25 65 he column varies from zero, corresponding o ϑ = 0, up o 2r, corresponding o ϑ = π. The exernal forces per uni area, acing on he ank are hen: p z = γr (1 cosϑ) p ϑ = 0 (3.7) p x = 0 dove γ := ϱg.from he firs of 3.7 one has: By using he expression of N ϑ one has: and he second of he equaions 3.7 becomes: N xϑ = N ϑ = p z r = γr 2 (1 cos ϑ) 1 r N ϑ ϑ = 1 r γr 2 sin ϑ = γr sin ϑ x By using he expression of N xϑ one has: 0 γr sin ϑd x + f 1 (ϑ) = γr x sin ϑ + f 1 (ϑ) (3.8) 1 N xϑ r ϑ Then he hird equaion 3.7 becomes: = x cos ϑ γr r + 1 df 1 r dϑ N x (x, ϑ) = γx2 2 cos ϑ x r df 1 dϑ + f 2(ϑ) The condiion of he ank boundary ha leaves freedom of movemen in he x direcion is ranslaed ino: x = ± l 2 N x = 0 γ l2 8 cos ϑ l df 1 2r dϑ + f 2(ϑ) = 0 γ l2 8 cos ϑ + l df 1 2r dϑ + f 2(ϑ) = 0 Summing and subracing hese wo equaions are obained f 2 (ϑ) = γ l2 8 cos ϑ df 1 (ϑ) dϑ = 0

26 66 From he second of hese equaions is derived: f 1 (ϑ) = cos = c I is observed from he equaion 3.8 ha he funcion f 1 (ϑ),and hen he consan c, represens he value of N xϑ in he cenral poin of he ank, bu since i is no presen on a load of orsion, his shear force mus be zero and herefore he consan c mus be zero. The soluion of he problem is hen given by he equaions: N ϑ = γr 2 (1 cos ϑ) (3.9) N xϑ = γr x sin ϑ N x = γ cos ϑ ( ) x 2 l2 2 4 By consiuive equaions in he case of an isoropic maerial, he deformaion is obained: ε xx = 1 E (σ xx νσ ϑϑ ) which can be wrien in erms of forces per uni lengh as: ε xx = 1 E (N x νn ϑ ) From he inegraion of he kinemaic equaion is hen obained: u(x) = 1 E x 0 (N x νn ϑ ) d x = 1 E x while sresses are o be deermined by he equaions 3.9: 0 [ γ cos ϑ ( ) ] x 2 l2 νγr 2 (1 cos ϑ) d x 2 4 from which: σ ϑϑ = γr 2 (1 cos ϑ) σ xx = γ cosϑ ( ) x 2 l2 2 4 σ xϑ = γ r x sin ϑ σ ϑϑmax = 2γr 2 (ϑ = π) σ xxmax = γ l2 (x = 0, ϑ = π) 8 ( σ xϑmax = γ r l x = l 2 2, ϑ = π ) 2

27 Wind load on a spherical shell of revoluion In order o consider he load condiions, he pressure on one side of he shell and of depression on he opposie side, only he normal componen p z is aken ino accoun alhough, in realiy, here could be also fricion conribuion. Assuming ha he wind acing in he direcion of he meridian plane ϑ = 0 he componens of he pressure due o he wind are: p ϑ = 0 ; p φ = 0 p z = p sin φ cos ϑ where p is indicaed by he saic pressure of he wind. In he case of a spherical shell under he acion of he wind, one has: r 1 = r 2 = r r 0 = r sin φ The equaions 3.4, 3.5 e 3.6 become in he lodged loading condiions: φ (r 0N ϑφ ) + N ϑ ϑ r 1 + N ϑφ r 1 cos φ = 0 φ (r 0N φ ) + N ϑφ ϑ r 1 N ϑ r 1 cos φ = 0 N ϑ + N ϑ = p sin φ cos ϑ r 1 r 2 Wih he use of he hird of hese equaions ha can be wrien as: N φ r 0 + N ϑ r 1 sin φ = pr 0 r 1 sinφ cos ϑ locaed in he firs wo equaions, we can eliminae N ϑ from hem and he obained sysem is: ( ) N φ 1 dr 0 + φ r 0 dφ + co φ N φ + r 1 N ϑφ r 0 ϑ = pr 1 cos ϑ cos φ ( N ϑφ 1 dr 0 + φ r 0 dφ + r ) 1 co φ N ϑφ 1 N φ r 2 sin φ ϑ = pr sin ϑ N ϑ = pr 0 cos ϑ N φr 0 r 1 sin φ These equaions allow o derive he membrane forces in a shell of revoluion ha is subjec o wind load.

28 68