Professor Terje Haukaas University of British Columbia, Vancouver terje.civil.ubc.ca. St. Venant Torsion

Transcription

1 S. Venan Torsion Torque in srucural members is carried by S. Venan orsion, and possible warping orsion. In S. Venan orsion he orque is carried by shear sresses; in warping orsion he orque is carried by axial sresses. This documen focuses solely on S. Venan orsion. In paricular, his documen emphasizes he use of a sress funcion o obain he soluion. From his perspecive, he naming of he heory afer dhémar Jean Claude Barré de Sain-Venan ( ) is perhaps somewha misleading. This is because S. Venan formulaed he problem in erms of an unknown funcion ha represens he axial displacemen a any poin of he cross-secion. I was Ludwig Prandl ( ) who reformulaed he problem in erms of an unknown sress funcion, which is he one addressed in his documen. Two primary objecives are idenified in his documen; he firs is o compue he cross-secional consan for S. Venan orsion; he oher is o compue sresses in he cross-secion for given orque. xisymmeric Cross-secion Consider firs he raher simple case of an axisymmeric cross-secion, e.g., a circular cross-secion. Equilibrium By considering an infiniesimally shor elemen wih lengh, subjeced o a disribued orque wih inensiy mx, one obains: m x = dt () Secion Inegraion Inegraion of shear sresses muliplied by heir disance o he cenre yiel: T = r o (2πr) τ r dr (2) where ri and ro are he inner and ouer radii of he cross-secion. r i Maerial Law Hooke s law for shear sresses and srains rea: τ = G γ where G is he shear modulus, G=E/(2(+n)). (3) Kinemaics The relaionship beween shear srain, g, and he roaion of he cross-secion, f, is obained by expressing he lengh of he line segmen idenified by an arrow in Figure in wo ways: Boh ()(g ) and (df)(r) expresses his lengh, which lea o he kinemaics equaion S. Venan Torsion Updaed February 4, 209 Page

2 ! r Figure : Kinemaics for an axisymmeric cross-secion subjeced o orsion. d! dφ = γ r (4) m x m x = dt m x = GJ d2 φ 2 T = GJ dφ ϕ T = r o r i T (2πr) τ r dr τ = G r dφ dφ = γ r τ τ = G γ γ Figure 2: Governing equaions for axisymmeric cross-secions. Differenial Equaion The differenial equaion is obained by combining all he previous equaions, which are summarized in Figure 2: where he following definiion has been made: J = r o r i m x = G J d2 φ 2 2πr 3 dr = π 2 r 4 4 ( o r i ) (5) (6) S. Venan Torsion Updaed February 4, 209 Page 2

3 J is he cross-secional consan for S. Venan orsion, and is someimes denoed Ip in oher lieraure. If he equilibrium equaions are omied hen he differenial equaion rea: T = GJ dφ General Soluion Inegraing he differenial equaion wice yiel he general soluion: (7) φ = m x 2 GJ x2 + C x + C 2 (8) Shear Sress To obain an expression for he shear sress in erms of he sress resulan, he maerial law and kinemaics equaions are firs combined: τ = G r dφ Then he following differenial equaion ha encompasses all of maerial law, kinemaics, and sress resulan is considered: T = GJ dφ Subsiuion of Eq. (0) ino Eq. (9) yiel he sough sress: (9) (0) τ = T J r () The expression shows ha he shear sress increases ouwar, proporional o he radius, i.e., he disance from he cenre o he considered fibre. rbirary Open Cross-secions General equaions for S. Venan orsion are esablished here, and hey are valid for boh massive and hin-walled cross-secions. Subsequen pages specializes he equaions o paricular cross-secion ypes. Equilibrium and Prandl s Sress Funcion The equilibrium beween exernally applied disribued orque, mx, and sress resulan, T, expressed in Eq. () remains valid in all S. Venan orsion. However, i is he inernal equilibrium of sresses a a poin ha requires special aenion. ccording o mechanics of soli, angular momenum equaions yield he equaliy of shear sress pairs. Conversely, equilibrium of linear momenum a any poin in he maerial impose he following condiion: S. Venan Torsion Updaed February 4, 209 Page 3

4 σ xx,x + τ yx,y + τ zx,z = 0 σ ij,i = 0 τ xy,x + σ yy,y + τ zy,z = 0 τ xz,x + τ yz,y + σ zz,z = 0 (2) where inernal body forces are negleced. Insead of working direcly wih hese sress componens, Prandl inroduced a new funcion ha plays a cenral role in he heory of orsion. The funcion is called Prandl s sress funcion, denoed j(y,z). By iself, he funcion has no physical meaning, bu is all-imporan feaure is ha sresses are derived from i. Specifically, Prandl s sress funcion is defined such ha τ xy = ϕ z = ϕ,z τ xz = ϕ y = ϕ,y I is observed ha he shear sress in one direcion is obained by differeniaing he sress funcion in he perpendicular direcion. I is also noed ha j varies only wih he crosssecion coordinaes y and z bu i does no vary wih x. Subsiuion of Eq. (3) ino Eq. (2) yiel: σ xx,x + τ yx,y + τ zx,z = 0 +ϕ,zy ϕ,yz = 0 τ xy,x + σ yy,y + τ zy,z = ϕ,zx = 0 τ xz,x + τ yz,y + σ zz,z = ϕ,zx = 0 because ) he sress funcion does no vary wih x; 2) all axial sresses are zero in S. Venan heory; and 3) he shear sress yz is zero because he shear srain gyz is zero when he cross-secion is assumed o reain is shape. s a resul, he use of Prandl s sress funcion as a measure of sress auomaically saisfies he equilibrium equaions. Boundary Condiions for he Sress Funcion The shear sress on he surface of a srucural member is obviously zero. This ranslaes ino a boundary condiion for he sress funcion. To formulae his boundary condiion mahemaically, le s be he coordinae ha follows he edge of he cross-secion, le r be he axis ha is perpendicular o s, and le a denoe he angle beween he y-axis and he edge of he cross-secion, as illusraed in Figure 3. (3) (4) z! xz s! xr! z cos(! ) = dy sin(! ) = dz! xy y! dz dy y S. Venan Torsion Updaed February 4, 209 Page 4

5 Figure 3: Shear sress perpendicular o he edge of he cross-secion (lef) and relaionship beween differenials (righ). Because here are no sresses on he free surface, i is required ha, on he free surface: τ xr = 0 By he definiion of he sress funcion, he shear sress in he r-direcion is obained by differeniaing he sress funcion in he s-direcion: τ xr = ϕ s Thus, he fac ha here are no sresses on he free surface ranslaes ino he following boundary condiion for he sress funcion: ϕ s = 0 noher way of deriving he same resul is o use decomposiion of he shear sresses xy and xz. Then, he condiion of zero shear sress perpendicular o he edge of he crosssecion rea: τ xr = τ xz cos(α ) τ xy sin(α ) = ϕ y y s + ϕ z z s = ϕ s = 0 In shor, he derivaive of he sress funcion mus be zero along he edge of he crosssecion, which implies ha he sress funcion mus be consan. For convenience and wihou loss of generaliy, his consan is se equal o zero. When observing he simpliciy of his boundary condiion i is noed ha S. Venan s formulaion of he heory, in erms of a funcion ha describes he axial displacemen in he cross-secion, inroduces a more complicaed equaion for his boundary condiion. Wih Prandl s sress funcion one simply nee o assure a consan value along he edge of he cross-secion. Secion Inegraion Inegraion of shear sresses muliplied by heir disance o he cenre yiel: T = ( τ xz y τ xy z)d When he sresses are formulaed in erms of Prandl s sress funcion hen i akes he form: T = ϕ y y + ϕ z z d This inegral is simplified furher by invoking inegraion by pars. The firs erm in he inegrand is parially inegraed in he y direcion and he second erm in z direcion: (5) (6) (7) (8) (9) (20) S. Venan Torsion Updaed February 4, 209 Page 5

6 T = ϕ y y d ϕ z z d = ϕ y dγ ϕ d ϕ z dγ ϕ d = ϕ y dγ ϕ z dγ + ϕ d + ϕ d 0 = 2 ϕ d 0 where he boundary inegrals vanish because he sress funcion is zero around he ouer edges of he cross-secion. Eq. (2) is a cornersone of he orsion heory ha is presened on he following pages, and i is imporan o undersand ha he facor 2 is generally valid. For cerain cross-secions, such as he open hin-walled cross-secions, equilibrium of he shear flow appears o give a orque ha is half he value of Eq. (2). To address his puzzle i is firs noed ha Eq. (2) consiss of wo equal conribuions from shear sress in wo perpendicular direcions. For cerain simplified sress funcions, such as he one used for hin-walled open cross-secions, he sress a he shor en is negleced, which causes he apparen anomaly. In acualiy here are shear sresses a hose locaions; in fac, hose conribuions double he orque because heir momen arm is large. In Chaper 0 of heir book on heory of elasiciy, Timoshenko and Goodier acknowledge ha he small shear sresses ha are someimes negleced can have an appreciable effec because heir momen arm is subsanial. They also sugges furher reading by menioning ha he quesion was cleared up by Lord Kelvin in Kelvin and Tai s Naural Philosophy, Vol. 2, page 267 (Timoshenko and Goodier 969). Maerial Law Hooke s law for a 3D maerial poin relaes axial sresses o axial srains, and shear sresses o shear srains. However, no every maerial law equaion is necessary for Sain Venan orsion. When considering kinemaics i will become apparen ha all axial srains are zero. In urn, all axial sresses are zero, and hence he maerial law equaions for axial srains/sresses are no needed. For shear srains/sresses, he maerial law rea: (2) τ xy = G γ xy τ xz = G γ xz (22) τ yz = G γ yz Kinemaics s a fundamenal kinemaics posulaion, i is assumed ha he cross-secion reains is shape during orsion: v = φ z w = φ y I is also assumed ha he axial displacemen in he cross-secion, u, only varies wih y and z. In oher wor, u is independen of x. The original formulaion by S. Venan goes (23) S. Venan Torsion Updaed February 4, 209 Page 6

7 o greaer lenghs o characerize he funcion u(y,z). However, his is circumvened in he heory formulaed by Prandl in erms of he sress funcion inroduced above. Wih hese assumpions, he kinemaics for general 3D problems, firs for axial srains, rea: ε x = du = 0 ε y = dv dy = 0 ε z = dw dz = 0 where he firs equaion equals zero because u is independen of x, he second and hird equaions equal zero because no deformaion of he shape of he cross-secion is allowed. For shear srains, he general kinemaics equaions are: γ xy = dv + du dy = dφ z + du dy γ xz = du dz + dw = du dz + dφ y γ yz = dw dy + dv dz = 0 where he las equaion is zero because he cross-secion does no change shape. Differenial Equaion Kinemaics and maerial law equaions combined wih Prandl s sress funcion in Eq. (3) yiel: ( ) ( ) ϕ,z = τ xy = G γ xy = G φ,x z + u,y ϕ,y = τ xz = G γ xz = G u,z + φ,x y These wo equaions can be combined ino one. By differeniaing he firs wih respec o z and he second wih respec o y he quaniy u,yz becomes a common quaniy ha faciliaes he merger, which yiel: ϕ,yy +ϕ,zz = 2 G φ,x This is he general differenial equaion for Sain Venan orsion of members wih arbirary cross-secions, when Prandl s sress funcion is employed. Pu anoher way, his is he differenial equaion ha governs he sress funcion. However, i is noed ha i does no include he secion inegraion equaion in Eq. (2), which is he oher key equaion in S. Venan orsion heory. General Expression for J The preceding derivaions yielded he differenial equaion in Eq. (27) and he secion inegraion equaion in Eq. (2). Boh are formulaed in erms of Prandl s sress funcion, j. Hence, he challenge in S. Venan orsion is no o deermine he roaion from Eq. (24) (25) (26) (27) S. Venan Torsion Updaed February 4, 209 Page 7

8 (27), bu o deermine j. Once he sress funcion is deermined, he sresses are obained from Eq. (3). The sress funcion also deermines he cross-secional consan for S. Venan orsion, J. The general expression for J as a funcion of j is esablished by combining Eq. (27) and Eq. (2) ino one equaion of he form of Eq. (7). Specifically, subsiuion of T from Eq. (2) ino he lef-hand side of Eq. (7) and subsiuion f,x from Eq. (27) ino he righ-hand side of Eq. (7) yiel: Solving for J yiel: 2 ϕ d = GJ ϕ,yy +ϕ,zz 2 G T J = 4 V 2 ϕ where V is he volume under he sress funcion, wrien as an inegral in Eq. (2), and Ñ 2 j is shorhand for j,yy+j,zz. Figure 4 summarizes he governing equaions for S. Venan orsion. m x φ,x φ (28) (29) m x = dt T T = GJ dφ ϕ,yy +ϕ,zz = 2 G φ,x γ xy = φ,x z + u,y γ xz = u,z +φ,x y T = 2 ϕ d ϕ τ xy = ϕ,z τ xz = ϕ,y τ xy τ xz τ xy = G γ xy τ xz = G γ xz γ xy γ xz Gives σ ij,i = 0 Figure 4: Governing equaions for arbirary open cross-secions. S. Venan Torsion Updaed February 4, 209 Page 8

9 lernaive Expression for J Laer in his documen, Eq. (29) is employed o deermine J for specific cross-secion ypes. For compleeness, an alernaive formulaion of Eq. (29) is presened by firs inegraing he differenial equaion in Eq. (27): where G is an arbirary closed pah in he cross-secion, and G is he area wihin ha pah. Inegraion by pars yiel Inroducion of sresses from Eq. (3) yiel Subsiuion back ino Eq. (30) yiel Employing his version of he differenial equaion o subsiue ino he righ-hand side of Eq. (7), while sill subsiuing he expression for T from Eq. (2) ino he lef-hand side of Eq. (7) yiel Solving for J yiel: ϕ,yy +ϕ,zz d = 2 G φ,x d Γ ϕ,yy +ϕ,zz d = ϕ,y dz ϕ,z dy Γ Γ Γ ( ) dz ( ϕ,y dz ϕ,z dy) = ( τ xz dz + τ xy dy) = τ xz + τ dy xy = Γ Γ Γ = 2 G φ ' Γ Γ 2 ϕ d = GJ 2 G Γ Γ Γ (30) (3) (32) (33) (34) J = 4 V Γ Γ (35) Thin-walled Open Cross-secions Consider a hin-walled cross-secion, such as a wide-flange beam and an L-shaped cross-secion. In he plane of he cross-secion he pars may be sraigh or no, bu everywhere he hickness is significanly smaller han he lengh of he par of he crosssecion. Furhermore, here are no cells, i.e., caviies, in he cross-secion. Finally, le he cross-secional coordinae r be defined o always be perpendicular o he longiudinal coordinae, s, of any cross-secion par. Then consider he sress funcion ϕ(r,s) = k 4 r 2 2 (36) S. Venan Torsion Updaed February 4, 209 Page 9

10 In wor, his is a quadraic pillow wih magniude k ha follows he longiudinal direcion of all cross-secion pars. Wherever a cross-secion par en here is a minor error; he sress funcion is no zero a ha edge. However, Eq. (36) remains appropriae as long as is small. The sress funcion in Eq. (36) is uilized o evaluae V and Ñ 2 j in order o evaluae Eq. (29): /2 b/2 V ϕ d = k 4 r 2 2 dr = 2 k b 3 /2 b/2 2 ϕ ϕ,ss +ϕ,rr = 8k 2 Shear Sress The sress funcion is ϕ(r,s) = k 4 r 2 2 The sresses are compued from he sress funcion: = ϕ,r = 8k r where k is deermined from he value of he orque, T: where he volume under he sress funcion is s a resul, he shear sress in he s-direcion is 2 τ xr = ϕ,s = 0 T = 2 ϕ d = 2 V V = 2 k b 3 J = 3 3 b (37) (38) (39) (40) (4) = 8 r 2 i k = where he subscrip i is inroduced o idenify he hickness, i, a he locaion where he sress is compued, while. b in he las parenhesis is summed over he enire crosssecion because i is par of he compuaion of he volume under he sress funcion. I is noed ha he shear sresses, and equivalenly he shear flow, circulae in he crosssecion. The flow is parallel o he longiudinal direcion of each cross-secion par, wih opposie direcion on each side of he mid-line. I is larges a he edges of he crosssecion. Composie Cross-secions Cross-secions wih several pars like he one addressed above are deal wih as follows. The oal orque, T, is carried by superposiion: 8 r 2 i 3 T 4 b (42) S. Venan Torsion Updaed February 4, 209 Page 0

11 T = T i where Ti is he orque in each par. The expression for he orque in each par of he crosssecion is subsiued from Eq. (7): T = GJ i dφ i However, for he cross-secion o reain is shape, all pars of he cross-secion mus roae by he same angle f, which yiel: T = GJ i dφ = dφ ( GJ i ) dφ GJ Hence, i is concluded ha he oal orsional siffness GJ is obained by summing conribuions GJi from all pars of he cross-secion. Furhermore, i is inferred ha he orque on each par of he cross-secion is relaive o he orsional siffness of ha par: (43) (44) (45) T i = GJ i dφ i = GJ i T ( GJ i ) (46) rbirary Cross-secions cross-secions have one or more cells and carry orque far beer han open crosssecions. If one draws a coninuous line around he circumference of he cross-secion, hen here will be more han one line: One will ouline he exernal circumference and he oher(s) will ouline he openings (cells). When posulaing a sress funcion under hese circumsances hen he required consan value around he edge may be differen around he differen edges. Hence, a sress funcion wih more han one unknown is needed. In urn, more equaions are needed o deermine he unknowns. Coninuiy equaions come o rescue. They express ha he ne axial displacemen around any cell mus be zero: du = 0 (47) du ε ε 2 Figure 5: Conribuions o oal shear srain in an infiniesimal elemen. d v S. Venan Torsion Updaed February 4, 209 Page

12 To idenify du, firs le s denoe an axis ha follows he closed in he counerclockwise direcion, and le v denoe he displacemen in he s-direcion. Similar o Eq. (25), and wih srain conribuions shown in Figure 5, kinemaics yield he following equaion ha conains du: Rearranging and subsiuing maerial law (xs=g. gxs) yiel s a resul he coninuiy requiremen in Eq. (47) rea Nex, he displacemen v in he s-direcion is expressed in erms of he cross-secion roaion, f, and he disance from he cenre of roaion o he angen line of he s-axis a any locaion along he closed : The coninuiy equaion now rea: γ xs = ε xs, + ε xs,2 = du + d v du = G d v G d v = 0 G dφ h = v = φ h Wih reference o Figure 6, he las inegral is expressed in erms of he area wihin he closed, denoed. Specifically, he produc h. is wice he area of he shaded region in Figure 6, herefore: G dφ h = 0 (48) (49) (50) (5) (52) h 2 (53) z s h 2!!h Figure 6: Evaluaion of he inegral of h.. y S. Venan Torsion Updaed February 4, 209 Page 2

13 s a resul, he coninuiy equaion rea: G dφ 2 = 0 (54) By invoking Eq. (7) o express df/ he final version of he coninuiy equaion is: = T J 2 (55) This equaion includes kinemaics and maerial law, bu no he secion inegraion equaion, and herefore serves a similar role o ha of he differenial equaion in Eq. (27). Thin-walled Cross-secions wih One Cell The assumed sress funcion has zero ampliude around he exernal edge, and ampliude K around he inernal edge. I has a quadraic variaion in beween, wih M denoing he heigh of he parabola above he 0-o-K ampliude. Leing he s-coordinae run along he mid-line, wih an r-coordinae running perpendicular, he sress funcion rea: ϕ(s,r) = K Two equaions are required o deermine K and M. To his end, wo coninuiy equaions are esablished. One of hese could be replaced by he differenial equaion expressed in Eq. (29), bu an imporan poin abou he magniude of M is beer made wih wo coninuiy equaions. The shear sress needed for hese equaions is: The coninuiy equaion in Eq. (55) expressed along he mid-line where r=0 is: The coninuiy equaion along he inner edge where r=/2 is: K Combining Eqs. (58) and (59) yiel: 2 + r + M = ϕ r = K 8Mr 2 Γ i K = T Γ m J 2 m 4M 2r = T Γ i J 2 i 2 (56) (57) (58) (59) S. Venan Torsion Updaed February 4, 209 Page 3

14 The line inegral of / is readily deermined in he fashion L/ + L2/ where L is he lengh of he cross-secion par wih hickness and so forh. The observaion is now made ha for hin-walled cross-secions he wo las fracions are boh approximaely equal o uniy, hence M 0. Consequenly, he sress funcion has only one unknown, K: which implies ha he sress in he s-direcion is K/, which in urn implies ha he shear flow around he cell is K. In conras wih he derivaions ha produced he general expression for J in Eq. (29), he coninuiy equaion in Eq. (58) akes he place of he differenial equaion in Eq. (27). The oher ingredien is he secion inegraion in Eq. (2), which for his sress funcion rea: Subsiuion ino he coninuiy equaion in Eq. (58) yiel wha is someimes called Bred s formula or Bred s second formula: Shear Sress Here he sress funcion is M = K 4 i m ϕ(s,r) = K To compue he shear sress for hese cross-secions i is necessary o deermine he value of he consan K in he sress funcion in erms of he applied orque, T. The secion inegraion equaion yiel: In urn, he shear sress in he s-direcion is obained, by he definiion of he sress funcion, by differeniaing he sress funcion in Eq. (6): Γ m Γ i 2 + r T = 2 ϕ d = 2 K m J = 4 2 m Γ m ϕ(s,r) = K 2 + r T = 2 ϕ d = 2 K m K = T 2 m (60) (6) (62) (63) (64) (65) = ϕ,r = K (66) S. Venan Torsion Updaed February 4, 209 Page 4

15 I is here noed ha he shear flow for a closed cross-secion is enirely differen from ha of an open one. In he closed cross-secion he shear sress is consan hrough he hickness and flows around he cell in one large loop. In fac, Eq. (66) reveals ha he shear flow is consan and equal o K around he cell: Composie Cross-secions To derive an expression for he orsional siffness GJ for composiion cross-secions, which was done earlier for open cross-secions, he coninuiy equaion in Eq. (52) is revisied. However, his ime G may vary and canno be pulled ou of he inegral. Hence, he coninuiy equaion is: q s = = K = dφ G h (67) (68) This ime, subsiuion of Eq. (7) yiel: = T GJ 2 G i i ( ) Subsiuion of he sress funcion from Eq. (6) yiel: K = T GJ ( 2 G i i ) (69) (70) gain combining he coninuiy equaion wih he sress resulan equaion in Eq. (62) yiel GJ = 2 m ( 2 i G i ) Γ m (7) Thin-walled Cross-secions wih Muliple Cells bove, he sress funcion in Eq. (56) was suggesed for a one-cell cross-secion. However, i was found ha for hin-walled cross-secions M is small. This resuled in he sress funcion in Eq. (6), which varies linearly from zero a he ouer edge o K a he inner edge. s a resul, K is he value of he shear flow around he cell and cross-secions wih more han one cell may have a differen shear flow around each cell. In oher wor, each cell is associaed wih he sress funcion in Eq. (6), bu each cell has a differen value of Ki, where i idenifies each cell by a number. The soluion approach is again o demand coninuiy around each cell, expressed earlier in Eq. (55): = T J 2 m (72) S. Venan Torsion Updaed February 4, 209 Page 5

16 where is he area wihin he closed ha is raced around he cell. Provided ha xs=j,r=ki/ i is sraighforward o combine Eqs. Eq. (6) and (72). However, cauion mus be exercised in in he compuaion of xs for he wall ha separaes he cells. There, he shear flow from boh cells conribues. Specifically, he shear flow is K around Cell, excep in he wall ha is adjacen o Cell 2, where he shear flow is K K2. Similarly, he shear flow around Cell 2 is K2, excep in he wall ha is adjacen o Cell, where he shear flow is K2 K. s a resul, coninuiy around Cell according o Eq. (72) requires: Γ m, K K 2 = T J 2 m, W,2 (73) where Gm,i is he line around he enire Cell i a mid-wall, W,2 is line along he wall ha separaes Cell and 2, and m,i is he area wihin Cell i, measured from he line a midwall. Coninuiy around Cell 2 requires: Γ m,2 K 2 K = T J 2 m,2 W,2 (74) The consans Ki can be pulled ou of he inegrals, hus he wo equaions (73) and (74) in he wo unknowns K and K2 can be wrien in marix form: Γ m, W,2 Inversion of he coefficien marix yiel he soluion: W,2 Γ m,2 K K 2 = T J 2 m, m,2 (75) K K 2 = 2 T J Γ m, W,2 s before, he coninuiy equaions are combined wih he secion inegraion equaion, which was provided in Eq. (62) for he sress funcion a hand: W,2 Γ m,2 m, m,2 (76) T = 2 ϕ(y,z)d = 2 K m, + 2 K 2 m,2 = 2 K K 2 T m, m,2 (77) Subsiuing Eq. (76) ino (77) and solving for J yiel S. Venan Torsion Updaed February 4, 209 Page 6

17 J = 4 Γ m,! W,2 W,2! Γ m,2 m, m,2 T m, m,2 (78) Shear Sress Here he coninuiy equaions, formulaed for each cell, are gahered in a linear sysem of equaions ha yield he consans, Ki, for he sress funcion in each cell: K K 2 = 2 T J Γ m, W,2 ccording o Eq. (67), Ki is he shear flow around each cell. Hence, once Ki are deermined from Eq. (76), he shear sress is compued by dividing Ki by he wall hickness. In walls beween cells, he shear flow conribuions from he adjacen cells are subraced o yield coninuous shear flow. W,2 Γ m,2 m, m,2 (79) References Timoshenko, S., and Goodier, J. N. (969). Theory of elasiciy. McGraw-Hill. S. Venan Torsion Updaed February 4, 209 Page 7