The Torsion of Thin, Open Sections

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1 EM 424: Torsion of hin secions 26 The Torsion of Thin, Open Secions The resuls we obained for he orsion of a hin recangle can also be used be used, wih some qualificaions, for oher hin open secions such as shown in Fig. 1 Fig.1 For example, he effecive area momens for he cross secions shown can be calculaed as () a J eff = 1 3 b3 () b J eff = 1 3 b b () c J eff = 1 3 b b b Also, he maximum shear sress formula can sill be applied as τ max = T max J eff (1) where max is he larges hickness of he cross secion. However, his maximum shear sress occurs on he ouer edges of he hickes secion and does no accoun for he sress concenraions ha occur a re-enran corners such as hose marked wih a C in Fig. 1. A

2 EM 424: Torsion of hin secions 27 such locaions, he sresses depend on he local radius of curvaure of he corner and may be considerably larger han he value prediced from Eq. (1). Such sress concenraions can be aken ino accoun by finding eiher numerically or experimenally a sress concenraion facor, K, for each re-enran corner and hen examining all high sress poins and choosing he one wih he highes sress, i.e. τ max = K T J eff max (2) Anoher difference beween he behavior of general hin open secion and he hin recangular secion we considered previously is ha a general open secion can experience a significan warping displacemen, u x, along he cenerline of he cross secion, whereas he warping along he cenerline of he hin recangle is zero. You can experience his endency of hin open secions o warp by simply wising a hin rolled up piece of paper and see wha happens. To obain an expression for his cenerline warping, consider he displacemen along his line in he cross secion, u s, as shown in Fig. 2 s α z u s, σ xs e n r r e z O e y y Fig.2 Breaking his displacemen ino is componens along he y- and z-axes and using he Sain Venan assumpions for he y- and z- displacemens, we have

3 EM 424: Torsion of hin secions 28 u s = u y sinα + u z cosα = zφ()sinα x + yφ()cosα x = r φ() x (3) where we have used he fac ha he perpendicular disance from he cener of wis, O, o he line of acion of u s is given by r = r e n = ( ye y + ze z ) cosα e y + sinα e z = y cosα + zsinα ( ) (4) The srain, γ xs, a he cenerline hen is given by γ xs = u s x + u x s = r φ + u x s (5) and he corresponding shear sress, σ xs, is σ xs = Gγ xs = Gr φ + G u x s (6) However, jus as in he recangular secion we expec his cenerline sress and srain o be zero, so ha we can inegrae Eq. (6) under hese condiions o obain he warping displacemen, u x, along he cenerline as u x = φ s r +ω 0 0 = φ ω() s (7) where ω() s is called he secorial area and ω 0 is a consan of inegraion which jus produces a consan displacemen (no warping) of he cross secion. This secorial area is a funcion of he geomery of he cross secion and also depen on he choice of he saring poin for he inegraion (s = 0) as well as he locaion of he origin O. Noe ha a change of he saring poin merely changes ω( s) by a consan amoun. Alhough ω( s) also depen on he locaion of poin O, we are no free o choose his poin arbirarily. This is because he disance r used in defining he secorial area in Eq. (7) is measured

4 EM 424: Torsion of hin secions 29 from he cener of wis, which is a poin wih zero y- and z-displacemens abou which he cross secion appears o roae according o he Sain Venan assumpions on he deformaion. We have no ye deermined he locaion of his cener of wis. In general, he cener of wis is locaed a a specific posiion in he y-z plane called he shear cener for he cross secion, where he shear cener is defined o be he poin where a shear force, when applied o he cross secion, will produce bending only (no wising). To prove ha he cener of wis and he shear cener coincide, consider he cross secion shown below : (u ) z T d y S T S (θ) V z (a) V z (b) S S T (u y) V (θ) y T d z V y (c) (d)

5 EM 424: Torsion of hin secions 30 Consider firs cases (a) and (b). From he reciprocal heorem, he work done by he orque T acing hrough he roaion () θ Vz caused by he force V z is equal o he work done by V z acing hrough he verical displacemen, ( u z ) T, due o he orque T, i.e. V z ( u z ) T = T( θ) Vz If V z acs hrough he shear cener S d y = 0 above equaion shows ha u z reciprociy ( ) hen by definiion θ ( ) Vz = 0, so ha he ( ) T = 0. Similarly for cases (c) and (d) we have, from V y ( u y ) T = T( θ) Vy so ha if also d z = 0 hen () u y T = 0. Bu if boh y- and z- displacemens due o he orque T are zero a S, hen his is he cener of wis for ha orque, i.e he shear cener and cener of wis mus coincide. Laer, when we discuss he bending of unsymmerical cross secions by shear forces, we will show how o find he shear cener for a hin cross secion explicily. However, if boh he y- and z-axes are axes of symmery for he cross secion, he shear cener can be obained direcly since i coincides in ha case wih he cenroid of he cross secion. When he secorial area is compued using poin O a he shear cener and he locaion of he saring poin on he cross secion (s = 0) is also chosen such ha S ω = ω da = 0 (8) A hen he secorial area ha is calculaed is called he principal secorial area, ω p. Wih his choice he consan ω 0 value is fixed and he u x displacemen hen can be uniquely defined by Eq. (7). Summary

6 EM 424: Torsion of hin secions 31 For a general hin open cross secion, we have T = G φ J eff 1 3 J eff = b i 3 i i τ max = K T J eff max () u x ()= s φ ω p s Torsion of Thin, Closed Cross-Secions τ a a a b b K 1 K 2 c τ b τ c c Fig.6 The Prandl sress funcion approach can also be used direcly o obain he shear sresses in a hin closed secion. Consider, for example, he wo cell cross secion shown in Fig. 6. Since he walls are hin and we know he Prandl sress funcion is zero on he ouer boundary and consans on he inner boundaries, across secions such a-a, b-b, and c-c i is reasonable o assume ha he sress funcion simply varies linearly as shown in Fig. 7.

7 EM 424: Torsion of hin secions 32 Φ = K 1 Φ = K 1 Φ = K 2 Φ = K 2 a Φ = 0 a b b c c a b c Fig.7 Thus, we find he sresses a hese locaions are jus given by τ a = K 1 a, τ b = K 1 K 2 b, τ c = K 2 c (9) Since K 1, K 2 are consans, Eq. (9) shows ha he quaniy q = τ is jus a consan q 1 = K 1 q 1 q 2 q 2 = K 2 Fig. 8 in each wall. Furher, from Fig. (9) we see ha a a juncion of walls his q quaniy is conserved in he sense ha he sum of he q s flowing ino or ou of he juncion mus add up o zero (Fig. 10).

8 EM 424: Torsion of hin secions 33 q 1 q - q 2 1 -q 2 Fig. 9 For his reason, q is called he shear flow in he cross secion. I is he force/uni lengh acing in he cross secion. Alhough he shear sress may vary along he cell walls if he hickness varies, he shear flow is always a consan so ha i is a convenien quaniy o use o describe he orsion of hin closed secions. Given he shear flow, we can always find he corresponding shear sress a a given locaion by simply dividing he shear flow by he hickness a ha locaion. If we have a cross secion consising of muliple closed cells, hen he shear flow in each cell conribues o he oal orque. For example, consider he ih cell shown in Fig. 10.

9 EM 424: Torsion of hin secions 34 Fig.10 The conribuion o he oal orque T abou some poin P from he consan shear flow q i in his cell is T i where T i = q i r = q i r + r ABC CDA = q i ( ω ABC +ω CDA ) (10) ( ) = q i 2Ω ABC 2Ω CDA = 2q i Ω i where he radius r is he perpendicular disance from P o he cenerline of he hin cross secion. In Eq. (10), ω ABC and ω CDA are he secorial areas obained by inegraing around he curved segmen and sraigh segmen of he cell, respecively, However, he secorial area is defined as posiive when he area is swep ou in a counerclockwise direcion and negaive when he area is swep ou in a clockwise direcion, so ha ω ABC is posiive while ω CDA is negaive. The magniudes of hese secorial areas are wice he acual areas

10 EM 424: Torsion of hin secions 35 shown in Fig. 11 so ha heir difference is wice he oal area conained wihin he cenerline of he ih cell, i.e. (11) ω ABC + ω CDA = 2Ω ABC 2Ω CDA = 2Ω i which is he resul ha was used o obain he final form in Eq. (10). B Ω ABC C A C D A Ω CDA Ω = Ω Ω i - Ω i ABC CDA Fig.11 Equaion (10) is valid for each cell, so ha if we have N cells, he oal orque, T, is given by N T = T i = 2q i Ω i (12) i =1 N i =1 For a single cell, if he orque T is given, hen we can solve for he shear flow, q, in his cell in erms of he orque T and he area conained wihin he cenerline of he cell, Ω, as q = T 2Ω (13) bu for muliple cells, we need o obain addiional relaions o solve for he shear flows. These relaions are provided by he condiion ha he displacemen, u x, mus be single-

11 EM 424: Torsion of hin secions 36 valued. From Eq. (6), recall we found he sress on he cenerline of a secion could be expressed as Bu for he ih closed cell, σ xs = Gγ xs = Gr φ + G u x s σ xs = q (14) where q is he shear flow around he cell. Noe ha q here is he shear flow ha exiss in each cell, which may in some walls be a combinaion of several of he separae q i values defined earlier. Thus, we have Inegraing Eq.(15) around each cell, we find u x s ih cell u x s = q G φ r (15) = 0 = 1 G q φ r (16) ih cell Bu φ is a consan and he secorial area inegral jus gives wice he area conained in he ih cell, as before, so we find, finally ih cell 1 2GΩ i q = φ ih cell ( i = 1,..., N) (17) If he orque, T, is known, hen Eqs.(12) and (17) are N+1 equaions for he N+1 unknowns φ, q i ( i = 1,..., N) or, if φ is known, hese same equaions can be used o solve for he N+1 unknowns T,q i ( i = 1,..., N). For a single cell, Eq. (17) simply gives φ = T 4GΩ 2 (18) which shows ha we can again wrie he relaionship beween he orque and wis as T = GJ eff φ where he effecive polar area momen is J eff = 4Ω2 (19)

12 EM 424: Torsion of hin secions 37 For a general closed secion conaining muliple cells, we can also obain he warping displacemen along he cenerline of each cell by reurning o Eq. (15) and inegraing i o obain u x ()= s 1 s q G φ ω() s (20) 0 Bu since φ is he same for each cell, placing Eq.(17) ino Eq. (20) gives for each cell u x ()= s 1 s q ω( s) G 0 2GΩ i q (21) ih cell For a single cell, q is a consan ha is given in erms of he orque, T, by Eq. (13) so in ha case, Eq. (21) reduces simply o u x ()= s T s 2GΩ 0 ω( s) 2Ω (22) The maximum shearing sress for he orsion a single of a single or muliple cell based on he shear flow in he cross secion is, of course simply given by τ max = q max (23) bu, as in he case of he open secion, his does no accoun for he sress concenraions ha can exis around re-enran corners. If such sress concenraions are presen and characerized by heir sress concenraion (K) facors, hen Eq. (23) mus be replaced by τ max = K q max (24) Summary For he orsion of a general muli-cell hin closed cross secion The shear flows and eiher T or φ can be found by solving simulaneously he N+1 equaions N T = 2q i Ω i i =1

13 EM 424: Torsion of hin secions GΩ i q = φ ih cell ( i = 1,..., N) The warping of each cell is hen given by u x ()= s 1 s q ω( s) G 0 2GΩ i ih cell q For he orsion of a single hin closed cell q = T 2Ω where T = GJ eff φ J eff = 4Ω2 and he warping displacemen is given by u x ()= s T s 2GΩ 0 ω( s) 2Ω For eiher single or muli-cell closed secions, he maximum shear sress in he cross secion is given by τ max = K q max Torsion wih Resrain of Warping (Open Secions) The Sain Venan heory of orsion ha we have used up o now assumes ha he cross secion is free o warp (displace) in he x-direcion, and ha his warping displacemen is independen of he x-coordinae. This can be seen direcly if we recall ha for an open secion we had from Eq. (7) u x = φ ω (25)

14 EM 424: Torsion of hin secions 39 where φ was a consan and he secorial area funcion is independen of x. However, if a hin open secion is aached o a suppor ha prevens warping a ha suppor (such as requiring u x = 0 a a buil in suppor see Fig.), i is obvious ha he Sain Venan assumpions are incorrec. In his case we expec he angle of wis per uni lengh will vary wih disance x φ = φ () x ( ) and hence here will be a normal srain, e xx = u x / x z y x developed in he bar, This means ha a corresponding normal sress σ xx () x also will be developed. Because σ xx is he only significan normal sress, we have from Hooke s law and Eq. (25) σ xx = Ee xx = E u x x = Eω d 2 φ dx 2 (26) Since we are considering he problem of he pure orsion of a bar, if such normal sress is developed in he cross secion, i mus no produce any axial load or bending momens, so we mus have σ xx da = E φ ω da = 0 A A y A σ xx da = E φ yω da = 0 zσ xx da = E φ zω da = 0 A A A (27)

15 EM 424: Torsion of hin secions 40 where y and z are measured from he cenroid of he cross secion. We will see laer when we alk abou he bending of unsymmerical secions ha Eq. (27) will be saisfied idenically if he secorial area funcion used here is he principal secorial area funcion ω p defined earlier so ha σ xx = Eω p d 2 φ dx 2 (28) When a varying normal sress exiss in he hin cross secion, he changes of his normal sress will also induce a shear sress in he cross secion ha is uniform across he hickness of he cross secion, i.e a corresponding shear flow will be developed in he cross secion. This shear flow can be relaed direcly o he normal sress by considering he equilibrium of a small elemen of he open secion To give σ xx + σ xx x dx σ xx + qs ()+ q s dx qs ()dx = 0 (29) which, using Eq. (28), reduces o q s = σ xx x = E d 3 φ dx 3 ω p = E d 2 β dx 2 ω p (30) where β φ is he gwis per uni lengh. Inegraing his resul from a beginning of he open secion where q = 0, we find

16 EM 424: Torsion of hin secions 41 qs ()= E d 2 s β ω dx 2 p 0 = E d 2 β ω dx 2 p da (31) This shear flow arises from he effecs of an end consrain on he warping and generaes a oque conribuion, T q, which can be obained in he same form as he shear flow in a closed secion, namely T q = s =s f r q s =0 s f = qdω p 0 (32) where s = 0 and s = s f are he wo free en of he open secion. There is also a orque, T SV = GJ eff β which arises from he shear sresses in he open secion ha vary linearly across he hickness according o he Sain Venan heory, so ha he oal orque, T, is jus he sum of hese conribuions given by T = T SV + T q = GJ eff β + qdω p s f 0 (33) Inegraing he inegral in Eq. (33) by pars and using Eq. (30), we find s s= s f f q T = GJ eff β + qω p s= 0 ω p 0 s = GJ eff β E d 2 β ω 2 dx 2 p da A where he end conribuions vanish since he shear flow mus be zero here, and we have used he fac ha he area elemen of he cross secion can be wrien as da =. Leing J ω = ω 2 p da (34) A

17 EM 424: Torsion of hin secions 42 hen Eq. (34) can be wrien as a second order differenial equaion of he form d 2 β dx 2 k 2 β = k 2 T GJ eff (35) where k 2 = GJ eff EJ ω (36) When T is a consan he general soluion o Eq. (35) is β = C sinh( kx)+ Dcosh( kx)+ T (37) GJ eff The consans C and D mus be found for a paricular problem from he boundary condiions. For example: Fixed end: β = 0 (u x = 0) Unconsrained end: dβ dx = 0 (σ xx = 0) (38) Thus, if he cross secion is fixed a,say, x = 0 and unconsrained a x = L where a orque T is applied, we would have which gives β x = 0 = 0 D = T GJ eff dβ = 0 C = T anh( kl) dx x= L GJ eff (39) β = T GJ eff and, since φ = 0 a x = 0, L φ()= L β dx 0 [ anh( kl)sinh( kx) cosh( kx)+1] (40) = TL GJ eff ( ) anh kl 1 kl (41)

18 EM 424: Torsion of hin secions 43 The firs erm in Eq. (41) is he oal wis if warping was unresrained, so we see ha he effec of he end consrain is o reduce his wis and make he bar appear siffer. We can also solve for he normal sress in he bar since dβ σ xx = Eω p dx = Eω p Tk GJ eff which has is larges value a x = 0 given by σ xx x= 0 = Eω p [ anh( kl)cosh( kx) sinh( kx) ] (42) Tk anh( kl) (43) GJ eff

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