Torsion CHAPTER 7.1 INTRODUCTION

Transcription

1 CHAPTE 7 Torsion 7. INTODUCTION The orsion of circular shafs has been discussed in elemenar srengh of maerials. There, we were able o obain a soluion o his problem under he assumpion ha he cross-secions of he bar under orsion remain plane and roae wihou an disorion during wis. To observe his, consider he shee shown in Fig. 7.(a), subjec o shear sress. The shee deforms hrough an angle g, as shown in Fig. 7.(b). D B D lg D B B D l g C A C A A C (a) (b) (c) Fig. 7. Deformaion of a hin shee under shear sress and he resuling ube If he deformed shee is now folded o form a ube, he sides AB and CD can be joined wihou an disconinui and his joined face will assume he form of a fla heli, as shown in Fig. 7.(c). If g is he shear srain, hen from Hooke s law τ g = (7.) G where G is he shear modulus. Owing o his srain, poin D moves o D [Fig. 7.(b)], such ha DD = lg. When he shee is folded ino a ube, he op face BD in Fig. 7.(c), roaes wih respec o he boom face hrough an angle q* = l γ (7.) r

2 where r is he radius of he ube. Subsiuing for g from Eq. (7.) q* = τ l G r θ * τ or = l Gr Also, he momen abou he cenre of he ube is T = r pr or T = i.e. π τ τ I = r r 3 r p T I p Torsion 3 (7.3) τ = (7.4) r where l p is he second polar momen of area. Equaions (7.3) and (7.4), herefore, give T τ Gθ* = = (7.5) Ip r l he familiar equaions from elemenar srengh of maerials. Now one can sack a series of ubes, one inside he oher and for each ube, Eq. (7.5) would be valid. These sacked ubes can form he secion of a solid (or a hollow) shaf if he op Gθ* face of each ube has he same roaion Gq *, i.e. if is he same for each l ube. Therefore, he raio τ r is he same for each ube, or in oher words, varies linearl wih r. Furher, if T is he orque on he firs ube wih polar momen of ineria I p, T he orque on he second ube wih polar momen of ineria I p, ec., hen T T Tn T + T Tn = = T... = = = I I I I + I I I p p pn p p pn p where T is he oal orque on he solid (or hollow) shaf and I p is is polar momen of ineria. From he above analsis we observe ha for circular shafs, he cross-secions remain plane before and afer, and here is no disorion of he secion. Bu, for a non-circular secion, his will no longer be valid. In he case of circular shafs, he shear sresses are perpendicular o a radial line and var linearl wih he radius. We can see ha boh hese canno be valid for a non-circular shaf. For, if he shear sress were alwas perpendicular o he radius OB [Fig. 7.(a)], i would have a componen perpendicular o he boundar. This is obviousl inadmissible since he surface of he shaf is unloaded and a shear sress canno cross an unloaded boundar. Hence, a he boundar, he shear sress mus be angenial o he boundar. Furher, b he same argumen, he shear sress a he corner of a recangular secion mus be zero, since he shear sresses on boh he verical faces are zero, i.e. boh boundaries are unloaded boundaries [Fig. 7.(b)]. In order o solve he orsion problem in general, we shall adop S. Venan s semi-inverse mehod. According o his mehod, displacemens u, u and u z are

3 3 Advanced Mechanics of Solids B O = 0 O B 0 = (c) = 0 Fig. 7. (b) (a) (a) Figure o show ha shear sress mus be angenial o boundar; (b) shear sress a he corner of a recangular secion being zero as shown in (c). assumed. The srains are hen deermined from srain-displacemen relaions [Eqs (.8) and (.9)]. Using Hooke s law, he sresses are hen deermined. Appling he equaions of equilibrium and he appropriae boundar condiions, we r o idenif he problem for which he assumed displacemens and he associaed sresses are soluions. 7. TOSION OF GENEAL PISMATIC BAS SOLID SECTIONS We shall now consider he orsion of prismaic bars of an cross-secion wised b couples a he ends. I is assumed here ha he shaf does no conain an holes parallel o he ais. In Sec. 7., mulipl-conneced secions will be discussed. On he basis of he soluion of circular shafs, we assume ha he crosssecions roae abou an ais; he wis per uni lengh being q. A secion a disance z from he fied end will, herefore, roae hrough q z. A poin P(, ) in his secion will undergo a displacemen rq z, as shown in Fig The componens of his displacemen are u = -rqz sin b u = rqz cos b O Fig. 7.3 qz P z T z P b (a) u rqz (b) P(, ) r qz b u (c) Prismaic bar under orsion and geomer of deformaion

4 Torsion 33 From Fig. 7.3(c ) sin b = r and cos b = r In addiion o hese and displacemens, he poin P ma undergo a displacemen u z in z direcion. This is called warping; we assume ha he z displacemen is a funcion of onl (, ) and is independen of z. This means ha warping is he same for all normal cross-secions. Subsiuing for sin b and cos b, S. Venan s displacemen componens are u = qz (7.6) u = qz u z = q(, ) (7.7) (, ) is called he warping funcion. From hese displacemen componens, we can calculae he associaed srain componens. We have, from Eqs (.8) and (.9), From Eqs (7.6) and (7.7) u e = u u, e =, e zz = z z u u u g = uz u uz +, γ z = +, γz = + z z e = e = e zz = g = 0 ψ g z = q + (7.8) ψ g z = q From Hooke s law we have s = ν E + E ε ( ) ( ) + ν ν + ν s = ν E + E ε ( ) ( ) + ν ν + ν s z = ν E + E ε ( ) ( ) zz + ν ν + ν = Gg, z = Gg z, z = Gg z where D= e + e + e zz Subsiuing Eq. (7.8) in he above se s = s = s z = = 0 z = Gq ψ + (7.9)

5 34 Advanced Mechanics of Solids ψ z = Gq The above sress componens are he ones corresponding o he assumed displacemen componens. These sress componens should saisf he equaions of equilibrium given b Eq. (.65), i.e. σ τ τ z + + = 0 z τ σ τz + + = 0 (7.0) z τ τ z z σ z + + = 0 z Subsiuing he sress componens, he firs wo equaions are saisfied idenicall. From he hird equaion, we obain i.e. ψ ψ Gθ + 0 = ψ ψ + = ψ = 0 (7.) Hence, he warping funcion is harmonic (i.e. i saisfies he Laplace equaion) everwhere in region [Fig. 7.3(b)]. Now le us consider he boundar condiions. If F, F and F z are he componens of he sress on a plane wih ouward normal n (n, n, n z ) a a poin on he surface [Fig. 7.4(a)], hen from Eq. (.9) n s + n + n z z = F n + n s + n z z = F (7.) n z + n z + n z s z = F z s Fig. 7.4 Ds n(n, n, o) (b) (a) Cross-secion of he bar and he boundar condiions D * * D Ds n

6 Torsion 35 In his case, here are no forces acing on he boundar and he normal n o he surface is perpendicular o he z-ais, i.e. n z 0. Using he sress componens from Eq. (7.9), we find ha he firs wo equaions in he boundar condiions are idenicall saisfied. The hird equaion ields ψ ψ Gθ n Gθ n = From Fig. 7.4(b) Subsiuing n = cos (n, ) = d ds, n ψ d ψ d 0 ds + = ds = cos (n, ) = d ds (7.3) (7.4) Therefore, each problem of orsion is reduced o he problem of finding a funcion which is harmonic, i.e. saisfies Eq. (7.) in region, and saisfies Eq. (7.4) on boundar s. Ne, on he wo end faces, he sresses as given b Eq. (7.9) mus be equivalen o he applied orque. In addiion, he resulan forces in and direcions should vanish. The resulan force in direcion is z d d = Gq ψ d d (7.5) The righ-hand side inegrand can be wrien b adding as ψ ψ ψ ψ = + + since = 0, according o Eq. (7.). Furher, + + = ψ ψ ψ ψ Hence, Eq. (7.5) becomes τ z ψ + + d d G ψ ψ = θ d d + + Using Gauss heorem, he above surface inegral can be convered ino a line inegral. Thus, ψ ψ τ z d d = Gθ Ú n + + n ds S = Gq ψ d ψ d Ú + + ds S ds ds = 0

7 36 Advanced Mechanics of Solids according o he boundar condiion Eq. (7.4). Similarl, we can show ha z d d = 0 Now coming o he momen, referring o Fig. 7.4(a) and Eq. (7.9) T = ( z z ) d d = G ψ ψ θ + + d d Wriing J for he inegral J = ψ ψ + + d d (7.6) we have T = GJq (7.7) The above equaion shows ha he orque T is proporional o he angle of wis per uni lengh wih a proporionali consan GJ, which is usuall called he orsional rigidi of he shaf. For a circular cross-secion, he quani J reduces o he familiar polar momen of ineria. For non-circular shafs, he produc GJ is reained as he orsional rigidi. 7.3 ALTENATIVE APPOACH An alernaive approach proposed b Prandl leads o a simpler boundar condiion as compared o Eq. (7.4). In his mehod, he principal unknowns are he sress componens raher han he displacemen componens as in he previous approach. Based on he resul of he orsion of he circular-shaf, le he nonvanishing sress componens be z and z. The remaining sress componens s, s, s z and are assumed o be zero. In order o saisf he equaions of equilibrium we should have τ τ z z τ τ = 0, z z = 0, + = 0 (7.8) z z If i is assumed ha in he case of pure orsion, he sresses are he same in ever normal cross-secion, i.e. independen of z, hen he firs wo condiions above are auomaicall saisfied. In order o saisf he hird condiion, we assume a funcion f (, ) called he sress funcion, such ha φ φ z =, τ z = (7.9) Wih his sress funcion (called Prandl s orsion sress funcion), he hird condiion is also saisfied. The assumed sress componens, if he are o be proper elasici soluions, have o saisf he compaibili condiions. We can subsiue hese direcl ino he sress equaions of compaibili. Alernaivel, we can deermine he srains corresponding o he assumed sresses and hen appl he srain compaibili condiions given b Eq. (.56). The srain componens from Hooke's law are e = 0, e = 0, e zz = 0 (7.0)

8 Torsion 37 Subsiuing from Eq. (7.9) g = 0, g z = G z, g z = G z g z = φ G, and g z = φ G From Eq. (.56), he non-vanishing srain compaibili condiions are (observe ha f is independen of z) γ z γ z + = 0 γ z γ z = 0 i.e. φ φ φ φ + = 0; + = 0 Hence, φ φ + = f = a consan F (7.) The sress funcion, herefore, should saisf Poisson's equaion. The consan F is e unknown. Ne, we consider he boundar condiions [Eq. (7.)]. The firs wo of hese are idenicall saisfied. The hird equaion gives φ φ n n = 0 Subsiuing for n and n from Eq. (7.3) φ d φ + d = 0 ds ds dφ i.e. = 0 (7.) ds Therefore, f is consan around he boundar. Since he sress componens depend onl on he differenials of f, for a simpl conneced region, no loss of generali is involved in assuming f = 0 on s (7.3) For a muli-conneced region (i.e. a shaf having holes), cerain addiional condiions of compaibili are imposed. This will be discussed in Sec On he wo end faces, he resulans in and direcions should vanish, and he momen abou O should be equal o he applied orque T. The resulan in direcion is z d d = φ d d

9 38 Advanced Mechanics of Solids = d = 0 φ d since f is consan around he boundar. Similarl, he resulan in direcion also vanishes. egarding he momen, from Fig. 7.4(a) T = ( z z ) d d = φ φ + d d = φ d d φ d d Inegraing b pars and observing ha f = 0 of he boundar, we find ha each inegral gives f d d Thus T = f d d (7.4) Hence, we observe ha half he orque is due o z and he oher half o z. Thus, all differenial equaions and boundar condiions are saisfied if he sress funcion f obes Eqs (7.), (7.3) and (7.4). Bu here remains an indeerminae consan in Eq. (7.). To deermine his, we observe from Eq. (7.9) φ φ + τ τ z z = = G γ z γ z = G u u uz uz + z + z = G u u z = G ( w z z ) where w z is he roaion of he elemen a (, ) abou he z-ais [Eq. (.5), Sec..8]. ( / z) (w z ) is he roaion per uni lengh. In his chaper, we have ermed i as wis per uni lengh and denoed i b q. Hence, φ φ + = f = Gq (7.5)

10 According o Eq. (7.9), z = φ φ, z = Torsion 39 Tha is, he shear acing in he direcion is equal o he slope of he sress funcion f (, ) in he direcion. The shear sress acing in he direcion is equal o he negaive of he slope of he sress funcion in he direcion. This condiion ma be generalised o deermine he shear sress in an direcion, as follows. Consider a line of consan f in he cross-secion of he bar. Le s be he conour line of f = consan [Fig. 7.5(a)] along his conour conour line f = cons. s n (n, ) D * Ds D Dn * s z n z (b) (c) (a) Fig. 7.5 Cross-secion of he bar and conour lines of f i.e. φ d d ds dφ = 0 ds (7.6a) φ d + = 0 d ds (7.6b) or d d z + τ ds z = 0 ds (7.6c) From Fig. 7.5(b) d ds = cos (n, ) = d dn and d d = cos (n, ) = ds dn where n is he ouward drawn normal. Therefore, Eq. (7.6c) becomes z cos (n, ) + z cos (n, ) = 0 (7.7a) From Fig. 7.5(c), he epression on he lef-hand side is equal o zn, he componen of resulan shear in he direcion n. Hence, zn = 0 (7.7b) This means ha he resulan shear a an poin is along he conour line of f = consan a ha poin. These conour lines are called lines of shearing sress. The resulan shearing sress is herefore zs = z sin (n, ) z sin (n, )

11 40 Advanced Mechanics of Solids = z cos (n, ) z cos (n, ) = d d z τ z dn dn φ d = d φ dn dn (7.8) or zs = φ n Thus, he magniude of he shearing sress a a poin is given b he magniude of he slope of f (, ) measured normal o he angen line, i.e. normal o he conour line a he concerned poin. The above poins are ver imporan in he analsis of a orsion problem b membrane analog, discussed in Sec TOSION OF CICULA AND ELLIPTICAL BAS (i) The simples soluion o he Laplace equaion (Eq. 7.) is = consan = c (7.9) Wih = c, he boundar condiion given b Eq. (7.4) becomes or i.e. d d = 0 ds ds d ds + = 0 + = consan where (, ) are he coordinaes of an poin on he boundar. Hence, he boundar is a circle. From Eq. (7.7), u z = qc. From Eq. (7.6) J = ( + ) d d = I p he polar momen of ineria for he secion. Hence, from Eq. (7.7) T = GI p q or q = T GI Therefore, u z = qc = p Tc GI p which is a consan. Since he fied end has zero u z a leas a one poin, u z is zero a ever cross-secion (oher han rigid bod displacemen). Thus, he crosssecion does no warp. The shear sresses are given b Eq. (7.9) as T z = Gq = I T z = Gq = I p p

12 Torsion 4 Therefore, he direcion of he resulan shear is such ha, from Fig. 7.6 τ an a = τ z z = Gθ = Gθ a r a Fig. 7.6 Torsion of a circular bar Hence, he resulan shear is perpendicular o he radius. Furher = τ z τ z + = T ( + ) I p Tr or = Ip where r is he radial disance of he poin (, ). Thus, all he resuls of he elemenar analsis are jusified. (ii) The ne case in he order of simplici is o assume ha = A (7.30) where A is a consan. This also saisfies he Laplace equaion. The boundar condiion, Eq. (7.4) gives, or i.e. or (A ) d ds (A ) d ds (A + ) d ds which on inegraion, ields (A + ) d ds = 0 (A + ) d ds = 0 (A ) d ds = 0 d ds (A + ) (A ) = 0 ( + A) ( A) = consan (7.3) This is of he form a + = b

13 4 Advanced Mechanics of Solids These wo are idenical if a b = A + A b a or A = b + a Therefore, he funcion b a = b + a represens he warping funcion for an ellipic clinder wih semi-aes a and b under orsion. The value of J, as given in Eq. (7.6), is J = ( + + A A ) d d = (A + ) d d + ( A) d d = (A + ) I + ( A) I Subsiuing I = 3 π ab 4 J = Hence, from Eq. (7.7) and I = 3 3 π ab a + b 3 π ab, one ges 4 T = GJq = Gq 3 3 π ab a + b or q = T a + b G 3 3 π ab The shearing sresses are given b Eq. (7.9) as (7.3) z = Gq ψ + d and similarl, a + b b a = T π ab b + a = T 3 π ab T z = 3 π ab (7.33a) (7.33b)

14 The resulan shearing sress a an poin (, ) is Torsion 43 / 4 4 / = τ z τ T z b a + = π ab (7.33c) To deermine where he maimum shear sress occurs, we subsiue for from a + =, or = a b b T giving = [a b 4 + a (a b ) ] / 3 3 π ab Since all erms under he radical (power /) are posiive, he maimum shear sress occurs when is maimum, i.e. when = b. Thus, ma occurs a he ends of he minor ais and is value is T 4 / ma = ( a b ) = T (7.34) 3 3 πab πab Wih he warping funcion known, he displacemen u z can easil be deermined. We have from Eq. (7.7) b Fig. 7.7 a depressed u z negaive elevaed u z posiive Cross-secion of an ellipical bar and conour lines of u z Tb ( a) u z = q = 3 3 π abg The conour lines giving u z = consan are he hperbolas shown in Fig For a orque T as shown, he conve porions of he crosssecion, i.e. where u z is posiive, are indicaed b solid lines, and he concave porions or where he surface is depressed, are shown b doed lines. If he ends are free, here are no normal sresses. However, if one end is buil-in, he warping is prevened a ha end and consequenl, normal sresses are induced which are posiive in one quadran and negaive in anoher. These are similar o bending sresses and are, herefore, called he bending sresses induced because of orsion. 7.5 TOSION OF EQUILATEAL TIANGULA BA Consider he warping funcion = A( 3 3 ) (7.35) This saisfies he Laplace equaion, which can easil be verified. The boundar condiion given b Eq. (7.4) ields ( 6A ) d ds (3A 3A + ) d ds = 0

15 44 Advanced Mechanics of Solids or i.e. Therefore, (6A + ) d ds + (3A 3A + ) d ds = 0 d 3 ds ( 3A A ) + + = 0 A(3 3 ) + + = b (7.36) where b is a consan. If we pu A = 6a and b = + a, 3 Eq. (7.36) becomes 6a (3 3 ) + ( + ) 3 a = 0 or ( 3 + a) ( a) ( a) = 0 (7.37) Equaion (7.37) is he produc of he hree equaions of he sides of he riangle shown in Fig The equaions of he boundar lines are C 3 a B o Fig. 7.8 D a a Cross-secion of a riangular bar and plo of z along -ais a = 0 on CD 3 + a = 0 on BC a = 0 on BD From Eq. (7.6) J = + + ( 3 3 ) ( 6 ) A A d d = 3 a a + + ( 3 3 ) ( 6 ) d A A d 0 3 a a a ( 3 3 ) ( 6 ) d A A d 3a 3 a = a = I 5 5 p (7.38)

16 Therefore, Torsion 45 T 5 q = = T (7.39) GJ 3 GIP I p is he polar momen of ineria abou 0. The sress componens are ψ z = Gθ + = Gθ ( 3A 3A + ) = G θ ( a ) a ψ and z = Gθ + (7.40) Gθ = ( a ) (7.4) a The larges shear sress occurs a he middle of he sides of he riangle, wih a value ma = 3 Gθ a (7.4) A he corners of he riangle, he shear sresses are zero. Along he -ais, z = 0 and he variaion of z is shown in Fig z is also zero a he origin TOSION OF ECTANGULA BAS The orsion problem of recangular bars is a bi more involved compared o hose of ellipical and riangular bars. We shall indicae onl he mehod of approach wihou going ino he deails. Le he sides of he recangular cross-secion be a and b wih he origin a he cenre, as shown in Fig. 7.9(a). T C B conve b o a concave D a A a (a) T (b) Fig. 7.9 (a) Cross-secion of a recangular bar (b) Warping of a square secion

17 46 Advanced Mechanics of Solids Our equaions are, as before, ψ ψ + = 0 over he whole region of he recangle, and ψ ψ n n = on he boundar. Now on he boundar lines = ±a or AB and CD, we have n = ± and n = 0. On he boundar lines BC and AD, we have n = 0 and n = ±. Hence, he boundar condiions become ψ = on = ±a ψ = on = ±b These boundar condiions can be ransformed ino more convenien forms if we inroduce a new funcion, such ha ψ = ψ In erms of, he governing equaion is ψ ψ + = 0 over region, and he boundar condiions become ψ = 0 on = ±a ψ = on = ±b I is assumed ha he soluion is epressed in he form of infinie series = X ( ) Y ( ) n= 0 n n where X n and Y n are respecivel funcions of alone and alone. Subsiuion ino he Laplace equaion for ields wo linear ordinar differenial equaions wih consan coefficiens. Furher deails of he soluion can be obained b referring o books on heor of elasici. The final resuls which are imporan are as follows: The funcion J is given b J = Ka 3 b For various b/a raios, he corresponding values of K are given in Table 7.. Assuming ha b > a, i is shown in he deailed analsis ha he maimum

18 Torsion 47 Table 7. b/a K K K shearing sress is a he mid-poins of he long sides = ±a of he recangle. On hese sides z = 0 and ma = K Ta J The values of K for various values of b/a are given in Table 7.. Subsiuing for J, he above epression can be wrien as ma = K Ta ab where K is anoher numerical facor, as given in Table 7.. For a square secion, i.e. b/a =, he warping is as shown in Fig. 7.9 (b). The zones where u z is posiive are shown b solid lines and he zones where u z is negaive are shown b doed lines. Empirical Formula for Squa Secions Equaion (7.3), which is applicable o an ellipical secion, can be wrien as T = π a b G = GA θ a + b 4π Ip ( a + b ) where A = pab is he area of he ellipse, and I p = A is he polar momen 4 of ineria. This formula is applicable o a large number of squa secions wih an error no eceeding 0%. If 4p is replaced b 40, he mean error becomes less han 8% for man secions. Hence, T GA θ = 40I p is an approimae formula ha can be applied o man secions oher han elongaed or narrow secions (see Secs 7.0 and 7.). 4

19 48 Advanced Mechanics of Solids 7.7 MEMBANE ANALOGY From he eamples worked ou in he previous secions, i becomes eviden ha for bars wih more complicaed cross-secional shapes, analical soluions end o become more involved and difficul. In such siuaions, i is desirable o resor o oher echniques eperimenal or oherwise. The membrane analog inroduced b Prandl has proved ver valuable in his regard. Le a hin homogeneous membrane like a hin rubber shee be sreched wih uniform ension and fied a is edge, which is a given curve (he cross-secion of he shaf) in he -plane (Fig. 7.0). z F D C F D F A B D F F b F A z B b + Db z Z + (a) (b) Fig. 7.0 Sreching of a membrane When he membrane is subjeced o a uniform laeral pressure p, i undergoes a small displacemen z where z is a funcion of and. Consider he equilibrium of an infiniesimal elemen ABCD of he membrane afer deformaion. Le F be he uniform ension per uni lengh of he membrane. The value of he iniial ension F is large enough o ignore is change when he membrane is blown up b he small pressure p. On face AD, he force acing is FD. This is inclined a an angle b o he -ais. an b is he slope of he face AB and is equal o z/. Hence, he componen of FD in z direcion is z ( ) F since sin b ª an ª b for small values of b. The force on face BC is also F D bu is inclined a an angle (b + Db) o he -ais. Is slope is herefore z ( ) z + and he componen of he force in z direcion is FD z z ( ) + Similarl, he componens of he forces FD acing on faces AB and CD are FD z z z and F +

20 Torsion 49 Therefore, he resulan force in z direcion due o ension F is F z F z z + F z + + z z F + z z = F + The force p acing upward on he membrane elemen ABCD is p D D, assuming ha he membrane deflecion is small. For equilibrium, herefore or F z z + p = z p + z = (7.43) F Now, if we adjus he membrane ension F or he air pressure p such ha p/f becomes numericall equal o Gq, hen Eq. (7.43) of he membrane becomes idenical o Eq. (7.5) of he orsion sress funcion f. Furher, if he membrane heigh z remains zero a he boundar conour of he secion, hen he heigh z of he membrane becomes numericall equal o he orsion sress funcion [Eq. (7.3)]. The slopes of he membrane are hen equal o he shear sresses and hese are in a direcion perpendicular o ha of he slope. The wising momen is numericall equivalen o wice he volume under he membrane [Eq. (7.4)]. 7.8 TOSION OF THIN-WALLED TUBES Consider a hin-walled ube subjeced o orsion. The hickness of he ube need no be uniform (Fig. 7.). Since he hickness is small and he boundaries are free, he shear sresses will be esseniall parallel o he boundar. Le be he magniude of he shear sress and he hickness. Consider he equilibrium of an elemen of lengh Dl, as shown. The areas of cu faces AB and CD are respecivel Dl and Dl. The shear sresses (complemenar shears) are and. For equilibrium in z direcion we should have τ l + τ l = 0 A C D B Z C A Dl D B Fig. 7. Torsion of a hin-walled ube

21 50 Advanced Mechanics of Solids or τ= τ = q, a consan (7.44) Hence, he quani is a consan. This is called he shear flow q, since he equaion is similar o he flow of an incompressible liquid in a ube of varing area. For coninui, we should have V A = V A, where A is he area and V he corresponding veloci of he fluid here. Consider ne he orque of he shear abou poin O [Fig. 7.(a)]. DF = q Ds Ds Dl d h O (b) Fig. 7. (a) Cross-secion of a hin-walled ube and orque due o shear The force acing on an elemenar lengh Ds of he ube is DF = τ s= q s The momen arm abou O is h and hence, he orque is DT = q sh= q A where DA is he area of he riangle enclosed a O b he base s. Hence, he oal orque is T = Σq A= qa (7.45) Where A is he area enclosed b he cenre line of he ube. Equaion (7.45) is generall known as he Bred Baho formula. To deermine he wis of he ube, we make use of Casigliano s heorem. eferring o Fig. 7.(b), he shear force on he elemen is Ds = q Ds. Because of shear srain g, he force does work equal o DU = ( τ s) δ = ( τ s) γ l = τ ( τ s) l G = = q l s G T l s 8A G (7.46) (7.47)

22 Torsion 5 using Eq. (7.45). The oal elasic srain energ is herefore T l ds U = (7.48) 8AG Hence, he wis or he roaion per uni lengh (Dl = ) is U q = = T ds T (7.49) 4A G Using once again Eq. (7.45) q q = ds AG (7.50) 7.9 TOSION OF THIN-WALLED MULTIPLE-CELL CLOSED SECTIONS We can eend he analsis of he previous secion o orsion of muliple-cell secions. Consider he wo-cell secion shown in Fig Dl q = Dl q = A A q 3 3 Fig. 7.3 (a) (b) Torsion of a hin-walled muliple cell closed secion Consider he equilibrium of an elemen a he juncion, as shown in Fig. 7.3(b). In he direcion of he ais of he ube τ l + τ l + τ33 l = 0 or = τ + τ33 q + q (7.5) q Fig. 7.4 Cell Cell A A q q O q Secion of a hin-walled muliple cell beam and momen ais i.e., q = 3 This is again equivalen o a fluid flow dividing iself ino wo sreams. Choose an momen ais, such as poin O (Fig. 7.4). The shear flow in he web can be considered o be made up of q and q, since q 3 = q q. The momen abou O due o q flowing in cell (wih web included) is [Eq. (7.45)] T = q A 3

23 5 Advanced Mechanics of Solids where A is he area of cell. Similarl, he momen abou O due o q flowing in cell (wih web included), wih A * as he area enclosed a O ouside cell, is T = * * q ( A + A ) q A The second erm wih he negaive sign on he righ-hand side is he momen due o he shear flow q in he middle web. Hence, he oal orque is T = T + T = q A + q A (7.5) A and A are he areas of cells and respecivel. Ne, we shall consider he wis. For coninui, he wis of each cell should be he same. According o Eq. (7.50), he wis of each cell is given b Gq = A Ú qds Le a = ds Ú for cell including he web Then, for cell a = Ú a = Ú ds ds for cell including he web for he web For cell Gq = ( aq a q ) (7.53) A Gq = ( a q a q ) (7.54) A Equaions (7.5) (7.54) are sufficien o solve for q, q and q. Eample 7. Figure 7.5 shows a wo-cell ubular secion whose wall hicknesses are as shown. If he member is subjeced o a orque T, deermine he shear flows and he angle of wis of he member per uni lengh. / a / a a Fig. 7.5 Eample 7.

24 Soluion For cell, ds a a a a 7a Ú = = = a For cell, ds a a a a 5a Ú = = = a For web, ds a Ú = = a From Eq. (7.53) Torsion 53 For cell, Gq = ( aq a q ) A = 7a ( q a q 7 ) = ( q q) a a For cell, Gq = ( a q a q ) A 5 q q = (5 q q ) a a Equaing, 7 q q = 5q 3 q or q = q 4 From Eq. (7.5) T = qa+ q A = a 3 7 ( q+ q 4 ) = aq a a = ( ) T \ q = and q 3 = T 7a 4a Gq = (5 q ) q a 5 q = q a 4 4a = ( ) T G 4a 7a T 8 ag or q = ( ) = 3 Eample 7. Figure 7.6 shows a wo-cell ubular secion as formed b a convenional airfoil shape, and having one inerior web. An eernal orque of 0000 Nm (0040 kgf cm) is acing in a clockwise direcion. Deermine he inernal shear flow disribuion. The cell areas are as follows: A = 680 cm A = 000 cm The peripheral lenghs are indicaed in Fig. 7.6.

25 54 Advanced Mechanics of Solids 0.09 cm 0.06 cm q 0.09 cm Cell s = 67 cm s = 33 cm s = 63 cm Cell s = 67 cm s = 48 cm q 0.09 cm Soluion For cell, a = For cell, a = 0.08 cm Fig. 7.6 Eample 7. Le us calculae he line inegrals Ú ds/ = = For web, a = 0.09 = 366 From Eqs (7.53) and (7.54) For cell, Gq = ( aq a q ) A = (483 q 366 ) q 680 =.89q 0.54q For cell, Gq = ( aq a q ) A = (409 q 366 q) 000 =.0 q 0.8 q Hence, equaing he above wo values.9 q 0.54 q =.0 q 0.8 q or,.37 q.74 q = 0 i.e. q =.36 q The orque due o shear flows should be equal o he applied orque. Hence, from Eq. (7.5) T = q A + q A or = q q 000 = 360 q q Subsiuing for q 0 6 = 360q q = 6800q

26 Torsion 55 q = N kgf N kgf q cm cm = cm cm 7.0 TOSION OF BAS WITH THIN ECTANGULA SECTIONS Figure 7.7 shows he secion of a recangular bar subjeced o a orque T. Le he hickness be small compared o he widh b. The secion consiss of onl one boundar and he value of he sress funcion f around A B his boundar is consan. Le f = 0. From Eq. (7.5) D b C φ φ + = Gq Eceping a he ends AD Fig. 7.7 Torsion of a hin recangular bar and BC, he sress funcion is fairl uniform and is independen of. Hence, we can ake f (, ) = f (). Therefore, he above equaion becomes φ = Gq Inegraing f = Gθ + a + a Since f = 0 around he boundar, one has f = 0 a = ± /. Subsiuing hese a = 0, G θ a = and From Eq. (7.9) f = Gq 4 z = φ = 0 4 (7.55) φ and z = = Gq (7.56a) These shears are shown in Fig Obviousl, he above equaions are no valid near he ends. The maimum shearing sresses are a he surfaces = ± /, and ( z ) ma = ± Gq (7.56b) From Eq. (7.4), T = φ d d b/ / = Gθ d d b/ / 4

27 56 Advanced Mechanics of Solids or 3 T = b Gθ 3 (7.57) The resuls are 3T q =, τ z = T, ( τ ) T 3 z ma =± G b b b (7.58) 7. TOSION OF OLLED SECTIONS The argumen leading o he approimaions given b Eqs (7.55) and (7.56) can be applied o an narrow cross-secion which has a relaivel small curvaure, as shown in Figs 7.8(a) (d). To see his, we imagine a 90 bend in he middle of he recangle shown in Fig. 7.7, so ha he secion becomes an angle. This secion has onl one boundar wih f = consan = 0. Eceping for he local effecs near he corner, he shape across he hickness will be similar o ha shown in Fig. 7.7, for he hin recangular secion. Hence, Eqs (7.55) and (7.57) can be applied, provided b is aken as he oal lengh of boh legs of he angle concerned and is he recangular coordinae in he direcion of he local hickness. b b b b (a) (b) b (c) b 3 b 3 Fig. 7.8 (d) Torsion of rolled secions In he case of a T-secion shown in Fig. 7.8(b), he lengh b = b + b if he hickness is uniform. If he hickness changes, as shown in Fig. 7.7(d), Eqs (7.55) and (7.57) become f = Gq 4 i ( i =,or 3)

28 Torsion and T = ( 3 3) 3 Gθ b + b + b (7.59) This is obained b adding he effec of each recangular piece. Eample 7.3 Analze he orsion of a closed ubular secion and he orsion of a ube of he same radius and hickness bu wih a longiudinal sli, as shown in Figs 7.9(a) and (b). T n r (a) (b) Fig. 7.9 (c) Eample 7.3 Torsion of a closed ubular secion and a sli ubular secion Soluion For he closed ube, if is he shear sress, we have from elemenar analsis T = ( πτ r). r= πrτ and θ= τ Gr 3 Therefore, T = πrgθ For he sli ube, here is onl one boundar and on his f = 0. According o Eq. (7.57) 3 3 T = b Gθ = πr Gθ 3 3 Furher, following he same analsis as for a hin recangular secion ma = Gθ The shear sress direcions in he sli ube are shown in Fig. 7.9(b). The raio of he orsional rigidiies is T ( π θ) ( πrgθ) T = 3 3 rg 3 = ( ) 3 r For a hin ube wih r/ = 0, ube (a) is 300 imes as siff as ube (b). If he sli ube is riveed along he lengh o form a closed ube of lengh l, as shown in Fig. 7.9(c), he force on he rives will be F = τl = Tl π r

29 58 Advanced Mechanics of Solids where for we have pu he value T = π r as for a non-sli ube. If here are n rives in a lengh l, hen he average force on each rive is F/n. Eample 7.4 (i) A 30- cm I beam (Fig. 7.0), wih flanges and wih a web.5 cm hick, is subjeced o a orque T = kgfcm (4900 Nm). Find he maimum shear sress and he angle of wis per 30 cm.5 cm uni lengh. (ii) In order o reduce he sress and he angle of wis,.5 cm hick fla 30 cm plaes are welded ono he sides of he secion, as shown b doed lines. Find he maimum shear sress and he angle of wis. Fig. 7.0 Eample 7.4 Soluion (i) Using Eq. (7.58) ma = 3T ( b ii) = 30 (5/4) + 30 (5/4) + ( 30.5 ) (5/4) = 097 kgf/cm (075 kpa) q = = 3T G 3 ( b ii) ( ) (5/4) + 30 (5/4) (5/4) G = 878/G radians per cm lengh (ii) When he wo side plaes are welded, he secion becomes a wo-cell srucure for which we can appl Eqs (7.5) (7.54). For each cell a = a = ds Ú = ( ).5 = T = qa + qa = 4qA 8.75 = 4τ or T = 3.5

30 Torsion 59 Therefore, = 50000/3.5 = 37.8 kgf /cm (3705 kpa) Gq = ( aq a q ) A Therefore q ( a a ) = A q = ( ) G = 0.06/G radians per cm lengh 7. MULTIPLY CONNECTED SECTIONS In Sec. 7. and 7.3, we considered he orsion of shafs wih secions which do no have holes. I is eas o eend he same analsis for he soluion of shafs, he cross-secions of which conain one or more holes. Figure 7. shows he secion of a shaf subjeced o a orque T. The holes have boundaries C and C. n A C C 0 T n C Fig. 7. Once again, as in Sec. 7.3, we assume ha z and z are he onl non-vanishing sress componens. The equaions of equilibrium ield τ z τ z τ = 0, = 0, z z Le f (, ) be a sress funcion, such ha φ z =, z = φ The non-vanishing srain componens are g z = φ τ z = G G and g z = φ τ z = G G Torsion of mulipl-conneced secions z τ z + = 0

31 60 Advanced Mechanics of Solids The compaibili condiions given b Eq. (.56) ield φ φ + = f = a consan F (7.60) So far, he analsis is idenical o ha given in Sec Considering he boundar condiions, we observe ha here are several boundaries and on each boundar he condiions given b Eq. (7.), Sec. 7. should be saisfied. Since each boundar is a free boundar, we should have φ φ n n = 0 Subsiuing for n and n from Eq. (7.3) φ d φ + d = 0 ds ds or dφ = 0 ds i.e. f for C i = K i (7.6) i.e. on each boundar f is a consan. Unlike he case where he secion did no conain holes, we canno assume ha f = 0 on each boundar. We can assume ha f = 0 on one boundar, sa on C 0, and hen deermine he corresponding values of K i on each of he remaining boundaries C i. To do his, we observe ha he displacemen of he secion in z direcion, i.e. u z = q (, ), from Eq. (7.7), mus be single valued. Consequenl, he value of d inegraed around an closed conour C i should be equal o zero, i.e. C dψ = ψ ψ Ú C d d 0 i + = Ú i From Eq. (7.9), and using he sress funcion and ψ ψ = φ τ z + = + Gθ Gθ (7.6) = φ τ z = (7.63) Gθ Gθ Hence, for he single valuedness of u z Gθ Ú φ C d φ d i + Ú C ( d d) i = 0 (7.64) The second inegral on he lef-hand side is equal o wice he area enclosed b he conour C i. This can be seen from Fig. 7.(a). C d = d+ d GKH HLG = area G GKHH area H HLGG Ú i = area enclosed b C i = A i

32 Torsion 6 K K C i G n H L L o G H (a) D Ds Dn D (b) Fig. 7. Evaluaion of he inegral around conour C i D Similarl, Ú d = LGK d+ Therefore, Ú ( ) C i KHL d = area L LGKK area K KHLL = A i d d = A i (7.65) The firs inegral in Eq. (7.64) can be wrien as φ φ C d d = φ d φ d Ú C ds i ds ds Ú i and from Fig. 7.(b) = Ú Ci φ d φ d + ds dn dn (7.66a) = φ Ú C ds (7.66b) i n where n is he ouward drawn normal o he boundar C i. Therefore, Eq. (7.64) becomes φ Ú C ds = GqA i (7.67) i n on each boundar C i. A i is he area enclosed b C i. The remaining equaions of Sec. 7.3 remain unalered, i.e. Eqs (7.4) and (7.5) are Torque T = φ d d φ φ and + = φ = Gθ The value of J defined in Eq. (7.6) can be obained for a mulipl-conneced bod in erms of he sress funcion f, as follows. Using Eq. (7.63) J = = ψ ψ + + d d φ φ + d d Gθ Gθ

33 6 Advanced Mechanics of Solids = φ φ d d Gθ + d d Gθ = φ ( φ) ( φ) Gθ Gθ C = φ d d + φ( d d) where we have made use of Gauss heorem and he subscrip C on he line inegral means ha he inegraion is o be performed in appropriae direcions over all he conours C i (i = 0,,,...,) shown in Fig. 7.. Since we have chosen f o be zero over he boundar C 0 J = φ d d ( d d) K ( d d) K... Gθ + Gθ + + C Gθ C where K, K,..., are he values of f on C, C..., And from Eq. (7.9) Gθ Gθ J = φ d d + ( KA + KA +...) where A i is he area enclosed b curve C i. Hence, J = φ d d KiAi Gθ + (7.68) Equaion (7.7), herefore, assumes he form T = GJθ = ( φ d d + KiAi) (7.69) For a solid shaf wih no holes, he above equaion reduces o Eq. (7.4). Eample 7.5 Analse he orsion problem of a hin-walled, muliple-cell closed secion, using equaions (7.8), (7.56) and (7.69). Assume uniform hickness. Soluion Consider he wo-celled secion shown in Fig According o Eq. (7.6), he sress funcion f is consan around each boundar. Pu n conour S C n conour S n C C 0 cell cell K φ = n f = 0 f = K f = K Fig. 7.3 Eample 7.5 f = 0

34 Torsion 63 f = 0 on boundar C 0, f = K on C and f = K on C. From Eq. (7.8), he φ resulan shear sress is given b zs = where n is he normal o he conour n of f, i.e. he line of shear sress. Since he hickness is small, he lines of shear sress follow he conours of he cells. Furher, since f = 0 on C 0 and f = K on C, we have for conour S of cell, For he web conour S φ 0 K = K = = τ n (7.70) φ K K = = τ (7.7) n and for conour S of cell φ 0 K = K = = τ (7.7) n From Eq. (7.67), for cell (S S ) + S = GqA where S is he peripheral lengh of cell including he web, and S he lengh of he web. Subsiuing for and or Similarl, for cell or ( ) K K K ( S S) S = GqA K S S = GqA (7.73) K (S S ) S = GqA K K K S S + S = GqA S S i.e. K K = GqA (7.74) From Eq. (7.69) T = ( d d ikiai) φ + Compared o A i, he area of he solid par of he ube secion is ver small and hence, he inegral on he righ-hand side can be omied. Wih his T = ( K A K A ) + (7.75) Equaions (7.73) (7.75) will enable us o solve for K, K and q. Eample 7.6 Using equaions (7.8) and (7.6), prove ha he shear flow is consan for a hin-walled ube (shown in Fig. 7.4) subjeced o orsion. Soluion Le S be he conour of he cenre line and s he hickness a an secion. According o Eq. (7.6), he sress funcion f is consan around

35 64 Advanced Mechanics of Solids s each boundar. Le f = 0 on C 0 and f = K on C. Then, from Eq. (7.8), a an secion C C 0 n Therefore, φ 0 K K = = = τ s n s s = K a consan s s Fig 7.4 Eample 7.6 i.e. q is consan. 7.3 CENTE OF TWIST AND FLEXUAL CENTE We have assumed in all he previous analses in his chaper ha when a wising momen or a orque is applied o he end of a shaf, he secion as a whole will roae and onl one poin will remain a res. This poin is ermed he cenre of wis. Similarl, i was saed in Sec. 6.5 ha here eiss a poin in he crosssecion, such ha when a ransverse force is applied passing hrough his poin, he beam bends wihou he secion roaing. This poin is called shear cenre or fleural cenre. Consider a clindrical rod wih one end firml fied so ha no deformaion occurs a he buil-in secion (Fig. 7.5). P = F Fig. 7.5 Cenre of wis and fleural cenre P = T For such a buil-in clinder, i can be shown ha he cenre of wis and he fleural cenre coincide. To see his, le he wising couple be T = P and he bending force be F = P. I is assumed ha P is applied a poin, which is he cenre of wis and P, hrough poin, he fleural cenre. Le d be he roaion caused a poin due o force P (= F) and le d be he deflecion (i.e. displacemen) of poin due o force P (= T). Bu d, he roaion, is zero since he force P is acing hrough he fleural cenre. Tha is, a = 0. Consequenl, from he reciprocal heorem, a = 0. Bu a is he deflecion (i.e. displacemen) of he fleural cenre due o orque. Since his is equal o zero, and since during wising, he onl poin which does no undergo roaion, i.e. deflecion, is he cenre of wis, he fleural cenre and he cenre of wis coincide. I is imporan o noe ha for his analsis o be valid i is necessar for he end o be firml buil-in.