Aircraft Structures. CHAPTER 8. Thin-walled beams. Active Aeroelasticity and Rotorcraft Lab. Prof. SangJoon Shin

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1 Aircraft Structures HAPTER 8. Thin-walled beams Prof. SangJoon Shin Active Aeroelasticity and Rotorcraft Lab.

2 8. Basic equation for thin-walled beams Typical aeronautical structures Typical aeronautical structures --- Light-weight, thin walled, beam-like structure complex loading environment (combined axial, bending, shearing, torsional loads) losed or open sections, or a combination of both : profound implications for the structural response (shearing and torsion) Thin-walled beams : specific geometric nature of the beam will be exploited to simplify the problem s formulation and solution process 8. : closed section 8. : open section 8. : combination of both 8.4 : multi-cellular section

3 8. Basic equation for thin-walled beams 8.. The thin wall assumption : geometry of the section, along the mid-thickness of the wall s : length along the contour, orientation along t(s) : wall thickness The thin wall assumption --- wall thickness is assumed to be much smaller than the other representative dimensions. ts () b ts () h ts () b h,, (8.)

4 8. Basic equation for thin-walled beams 8.. The thin wall assumption The thin-walled beam must also be long to enable the beam theory to be a reasonable approximation b h L 4

5 8. Basic equation for thin-walled beams 8.. Stress flows The stress components acting in the plane of the cross-section are assumed to be negligible as compared to the others.,, Only non-vanishing components : axial stress transverse shear stress, It is preferable to use the stress components parallel and normal to.,, rather than artesian components. n s 5 dx dx n cos sin ds ds dx dx s sin cos ds ds dx cos ds dx,sin ds (8.a) (8.b) Sign convention for s

6 8. Basic equation for thin-walled beams 8.. Stress flows Principle of reciprocity of shear stress normal shear stress n must vanish at the two edges of the wall because the outer surfaces are stress free. No appreciable magnitude of this stress component can build up since the wall is very thin. n vanishes through the wall thickness. The only non-vanishing shear stress component :, tangential stress s Inverting Eq. (8.a), (8.b), and s dx ds, 0 n s dx ds (8.) 6

7 8. Basic equation for thin-walled beams 8.. Stress flows Thin-walled beams : It seems reasonable to assume that is uniformly distributed across the wall thickness since the wall is very thin. s oncept of stress flow nx (, s) ( x, sts ) ( ) f ( x, s) ( x, s) t( s) s (8.4a) (8.4b) n : axial stress flow, axial flow f : shearing stress flow, shear flow Only necessary to integrate a stress flow along, instead of over an area, to compute a force. 7

8 8. Basic equation for thin-walled beams 8.. Stress resultant Integration over the beam s cross-sectional area integration along curve Infinitesimal area of the cross-section da tds axial force N ( x ) da tds nds A Axial flow bending moments M ( x ) nx ds M ( x) nxds (8.5) (8.6) shear forces dx V( x) f ds ds dx V( x) f ds ds (8.7) 8

9 8. Basic equation for thin-walled beams 8.. Stress resultant Torque about origin O, M ( x ) r fds O rp xi xi ds dx i dx i P : position vector of point P : increment in curvilinear coord. ( ) ( ) O ( dx dx M x x dx x dx fi x ) x fi ds ds ds At point P o, dx ro x cos x sin x x ds dx ds (8.8) 9

10 8. Basic equation for thin-walled beams 8.. Stress resultant Magnitude of the torque M ( x ) fr ds O O, O P r r (8.9) --- torque = magnitude of the force X perpendicular distance from the point to the line of action of the force Torque about an arbitrary point K, of the cross-section Mk( x) fr kds (8.0) and, dx r ( x x )cos ( x x )sin r x x ds k k k O k k dx ds rk : perpendicular distance from K to the line of action of the shear flow (8.) 0

11 8. Basic equation for thin-walled beams 8..4 Sign conventions variable s, () x s s a l () s x s b l l a b,, The perpendicular distance from O, to the tangent curve, denote ro, becomes dx dx s b s a ab ro x x a b ds ds l l l l l variable s, (8.) s ' x ( s') a l, r o becomes, x () s b s ' l dx dx s ' b s' a ab r' O x x a b ds ' ds ' l l l l l (8.)

12 8. Basic equation for thin-walled beams 8..4 Sign conventions The sign convention for the torque is independent of the choice of the curvilinear variable, s s: counterclockwise, s : clockwise f '( s') f( s) r' ( s') r ( s) O O However, the resulting torque is unaffected by this choice. M ( x ) fr ds f ' r ' ds ' O O O

13 8. Basic equation for thin-walled beams 8..5 Local equilibrium equation A differential element of the thin-walled beam --- all the forces acting along axis i n f nds n dxds fdx f ds dx x s 0 After simplification, n x f s 0 (8.4) Any change in axial stress flow, n, along the beam axis must be equilibrated by a corresponding change in shear flow, f, along curve that defines the cross-section

14 8. Bending of thin-walled beams A thin-walled beam subjected to axial forces and bending moments --- Euler-Bernoulli assumptions are applicable for either open or closed cross-sections - Assuming a displacement field in the form of Eq. (6.) Stain field given by Eq. (6.a) (6.c) --- axial stress distribution, from Eq. (6.5) N xh xh xh xh S H H c c c c E M M (8.5) S H A EdA, Ex da c A H H H ( H ) c c c c, H Ex da, H Ex A xda c A - axial flow distribution using Eq. (8.4a) c c c c N( x) x() s H x() s H x() s H x() s H nx (, s) Ests ( ) ( ) M( x) M( x) S H H (8.6) 4

15 8. Shearing of thin-walled beams Bending moments in the thin-walled beams are accompanied by transverse shear force give rise to shear flow distribution - evaluated by introducing the axial flow, given by Eq. (8.6) into the local equilibrium eqn., Eq. (8.4) f dn xh xh dm xh xh dm Et s S dx H dx H dx c c c c (8.7) - sectional equilibrium eqns, Eq. (6.6), (6.8), (6.0) substituting into (8.7), and assuming that p, q, q 0 c c c c f xh xh xh xh Ests ()() V V s H H (8.8) - Integration -> shear flow distribution arising from V, V c c c c s xh xh xh xh f () s c Et V 0 V ds H H (8.9) c: integration constant corresponding to the value at s = 0 5 The procedure to determine this depends on whether cross-section is closed or open.

16 8. Shearing of thin-walled beams H c, V, V - Since are function of x alone Q () s H Q () s H Q () s H Q () s H f () s c V V H H c c c c (8.0) where stiffness static moment or stiffness first constant s Q () s Ex () s tds Q() s Ex 0 () s tds 0 s (8.) --- static moments for the portion of the cross-section from s = 0 to s 6

17 8. Shearing of thin-walled beams 8.. Shearing of open sections Principle of reciprocity of shear stress,,, shear flow vanishes at the end points of curve Shear flow must vanish at point A and D since edges AE and DF are stress free. If the origin of s is chosen to be located at such a stress free edge, the integration constant c in Q () s H Q () s H Q () s H Q () s H f () s c V V H H c c c c must vanish. 7

18 8. Shearing of thin-walled beams 8.. Shearing of open sections Procedure to determine the shear flow distribution over cross-section. ompute the location of the centroid of the cross-section, and select a set of centroid axes, i and i, and compute the sectional centroidal bending stiffness c c c c H, H and H. (principal centroidal axes H 0 ). Select suitable curvilinear coord. s to describe the geometry of cross-section.. Evaluate the st stiffness moments using s Q() s Ex 0 () s tds s Q () s Ex () s tds 0 (8.) 4. f(s) is determined by Q () s H Q () s H Q () s H Q () s H f () s c V V H H c c c c (8.0) 8

19 8. Shearing of thin-walled beams 8.. Evaluation of stiffness static moments homogeneous, thin-walled rectangular strip oriented at an angle s 0 s s 0 Q () s Ex tds E ( d ssin ) tds Est( d sin ) (8.) Young s modulus the area of strip coord. of the centroid of the local area similar result of the other stiffness static moment, s Q() s Est( d cos ) (8.) Since the strip is made of a homogeneous material, E factors Out of integral. Q() s E x 0 tds s Area static moment 9

20 8. Shearing of thin-walled beams 8.. Evaluation of stiffness static moments Thin-walled homogeneous circular arc of radius R ds Rd s d Q() s Ex 0 tds Et ( d 0 Rsin ) Rd EtR cos R Q s d R () EtR cos stiffness static moment = E centroid (8.4) area distance to the area Q s EAx Q () s EAx () Parallel axis theorem, but in this case, only the transport term remains since the static moment about the area centroid itself is zero, by definition. 0

21 8. Shearing of thin-walled beams 8.. Shear flow distributions in open sections Example 8. Shear flow distribution in a -channel - uniform thickness t, vertical web height h, flange width b, subject to a vertical shear force V -centroid: b d h b i - symmetric about axis, principal axes of bending, H c 0 Q () s f s c V () c H (8.5) c th h h bh H E bt E t

22 8. Shearing of thin-walled beams 8.. Shear flow distributions in open sections Example 8. Shear flow distribution in a -channel h Ets Q ( s ) ( ) 0 Ehts f s c V V c c c H H H V (8.6) Because, f( s 0) 0 Q ( s ) Ets Because, h s f ( s 0) f( s b) h Ets f( s ) c V h s EV tev f( s ) c ts [ bhs ( hs )] c c H H hs tev c c H H (8.8) (8.7)

23 8. Shearing of thin-walled beams 8.. Shear flow distributions in open sections Example 8. Shear flow distribution in a -channel - upper and lower flange: linearly distributed, 0 at the edges - vertical web: varies in a quadratic manner, shear flow and the stress pointing upward - max. shear flow: mid-point of the vertical web

24 8. Shearing of thin-walled beams 8.. Shear flow distributions in open sections Example 8. Shear flow continuity conditions -wall joint : equilibrium of forces along the beam s axis -f+f=0, or f=f : The shear flow must be continuous at the junction J -wall joint : -f-f-f=0, or more generally fi 0 (8.9) sum of the shear flows converging to a joint must vanish. 4

25 8. Shearing of thin-walled beams 8..5 Shear center for open sections Problem is not precisely defined --- Whereas the magnitudes of the transverse shear forces are given, their lines of action are not specified. -> It is not possible to verify the torque equilibrium of the cross section. Definition of the shear center subjected to horizontal and vertical shear force V, V with lines of action passing through K, (xk, xk), no external torque applied, MK=0 5

26 8. Shearing of thin-walled beams 8..5 Shear center for open sections equipollence conditions Integration of the horizontal component of the shear flow over cross -section must equal the applied horizontal shear force dx ds f ds V will be satisfied since it simply corresponds to the definition of shear force dx V( x) f ds ds Integration of the vertical component of the shear flow over crosssection must equal the applied vertical shear force dx ds f ds V 6

27 8. Shearing of thin-walled beams 8..5 Shear center for open sections equipollence conditions Torque generated by the distributed shear flow is equivalent to the externally applied torque, about the same point. --- does require the line of action of the the applied shear forces about point K, the torque, M k (8.0) torque generated by the external forces w.r.t. point K = 0 frds k M 00V 0V k M k frkds 0 (8.9) 7

28 8. Shearing of thin-walled beams 8..5 Shear center for open sections --- point K cannot be an arbitrary point, its coords must satisfy the torque equipollence condition M (8.9) k frkds 0 Definition of the shear center location 8

29 8. Shearing of thin-walled beams 8..5 Shear center for open sections Alternative definition Perpendicular distance from an arbitrary point A to dx the line of action r r x x ds xa, x a : coord. of point A a O a a Subtracting this equation from Eq. (8.) dx dx rk ra ( xk xa) ( xk xa) ds ds Substituting into the torque equipollence condition, Eq. (8.9) dx dx fr ds ( x x ) f ds ( x x ) f ds ds ds a k a k a fr ds ( x x ) V ( x x ) V 0 a k a k a dx ds 9

30 8. Shearing of thin-walled beams 8..5 Shear center for open sections Torque generated about point A by the shear flow distribution M fr ds ( x x ) V ( x x ) V a a k a k a --- moment at A due to force and moment resultant at point K M M ( x x ) V ( x x ) V a k k a k a (8.40) M k 0 By Eq. (8.9) Eqs. (8.9), (8.40) --- Torque generated by the shear flow distribution associated with transverse shear force must vanish w.r.t. the shear center. 0

31 8. Shearing of thin-walled beams 8..5 Shear center for open sections Summary A beam bends without twisting if and only the transverse shear loads are applied at the shear center. If the transverse loads are not applied at the shear center, the beam will both bend and twist. If the cross-section features a plane of symmetry, the shear center must lie in that plane of symmetry.

32 8. Shearing of thin-walled beams 8..5 Shear center for open sections Example 8.6 Shear center for a -channel -axis i : axis of symmetry -> shear center lies at a point along this axis - It is necessary to evaluate the shear flow distribution by V, to determine the shear center location - Resultant force in each segment: by Eqs. (8.0) (8.) R EV b hb t f( s 0 ) ds c 4 H h h ( ) b hb t EV ( 0 ) c 4 H R f s ds V R f s ds R equipollence conditions R R 0 R V

33 8. Shearing of thin-walled beams 8..5 Shear center for open sections h h frk ds R R e R 0 hr hbt E b c R 4 H e h 6 b (8.4) Example 8.8 Shear center for a thin-walled right-angle section Figure 8. - Shear center in thin-walled right-angle section Lines of actions of two resultant of the shear flow distributions, R and R, will intersect at point K procedures no torque about this point must then be the shear center

34 8. Shearing of thin-walled beams 8..7 Shearing of closed sections c c c c s xh xh xh xh Same governing equation f () s c Et V 0 V ds H H still applies, but no boundary condition is readily available to integrate this equation. Exception: axis of symmetry If V acts in the plane of symmetry i i, (8.9) mirror image of shear flow distribution point A : joint equilibrium condition symmetry condition : f f 0 f f f f 0 shear flow vanishes at A and similarly B 4

35 8. Shearing of thin-walled beams 8..7 Shearing of closed sections st step : Beam is cut along its axis at an arbitrary point. auxiliary problem, shear flow distribution fo(s) s du nd step : fo(s) creates a shear strain infinitesimal axial strain G () s s 0 du sds ds ds f Gt (8.4) 5

36 8. Shearing of thin-walled beams 8..7 Shearing of closed sections rd step : total relative axial displacement at the cut u 0 f0 () s ds Gt 4 th step : fc is applied to eliminate the relative axial displacement, thereby returning the section to its original, closed state (fc : closing shear flow ) 6 total shear flow u t f c f () 0 s fc ds 0 (8.44) Gt displacement compatibility eqn. for the closed section f () 0 s ds Gt (8.45) ds Gt f s f s f s 0 c

37 8. Shearing of thin-walled beams 8..7 Shearing of closed sections Summary f0(s) for an auxiliary problem fc(s) by f s f s f s 0 c f c f () 0 s ds Gt ds Gt 7

38 8. Shearing of thin-walled beams 8..7 Shearing of closed sections Example 8.9 Shear flow distribution in a closed triangular section - shear flow distribution for open section: already computed in Example. 8.4 f f ( s ) 0 0 f 0 s 60 9t V s V ( s ) 60 t 7 5t t V s ( s ) 60 9t t V t (8.46) - constant closing shear flow : by Eq. (8.45) 8 f f ( s ) f ( s ) f ( s ) V 9t 5t 9t ds ds 0 ds 5t ds 0 Gt Gt Gt Gt 0Gt ds 08 (9t0t9 t) Gt Gt G f c V 0 V Gt t (8.47) G

39 8. Shearing of thin-walled beams 8..7 Shearing of closed sections Example 8.9 Shear flow distribution in a closed triangular section - final shear flow distribution f s f s f 0 c - Both shear flow in the auxiliary section and the closing shear flow are (+) when pointing along the local curvilinear variable 9

40 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections Procedure similar to that used for a single closed section must be developed. One cut per cell is required. Shear flow distribution in the resulting open sections is evaluated using the procedure in sec. 8.. f0(s), f0(s), f0(s) along,, losing shear flows are applied at each cut. : fc, fc Then, shear flow distribution : f,,, 0( s) fc f0( s) fc f ( s ) ( f ) along,,. 0 c f c 40

41 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections front cell : clockwise / aft cell : counterclockwise f0( s) fc f0( s) ( fc fc) ut ds ds 0 Gt Gt f0( s) fc f0( s) ( fc fc) ut ds ds 0 Gt Gt f0 () s ds fc ds fc ds Gt Gt Gt f0 () s ds f c ds fc ds Gt Gt Gt Extension to multi-cellular section with N closed cells Open section by N cut, one per cell: shear flow distribution in open section by the procedure in sec 8.. losing shear flows are applied at each cut and displacement compatibility conditions are imposed: N simultaneous equations. 4 Total shear flow distribution is found by adding the closing shear flow to that for the open section.

42 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections Example 8. Shear flow in thin-walled double-box section - multi-cellular, thin-walled, double-box section subjected to a vertical shear force, V - right cell wall thickness t, while the remaining three walls of the left cell wall thickness t - Due to symmetry, i : principal axis of bending -> - bending stiffness based on thin-wall assumption H c 0 tb tb c H E ( bt b t) b tb E - st step: transformed into an open section by cutting the two lower flanges 4

43 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections - shear flow distribution for open section 6V s 6V s V s f ( s ), f ( s ), f ( s ) bb b b b b V s s V s5 s5 f0( s), f0( s5) b b b b b b V s6 V s7 s7 f0( s6), f0( s7) b b b b b - nd step: closing shear flows, fc, fc, are added to the left and right cells - axial displacement compatibility at left cell f ( s ) f f ( s ) f f ( s ) f u ds ds ds Gt Gt Gt b f0( s7) fc fc b 7fc fc V ds G t Gt b b b b 0 c 0 c 0 c t

44 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections 44 - axial displacement compatibility at right cell f ( s ) f f ( s ) f f ( s ) f u ds ds ds Gt Gt Gt b f0( s7) fc fc b fc V ds 0 7 fc 0 G t Gt b V V - sol. of two simultaneous eqn.: fc 8, fc 6 69b 69b b b b 0 4 c 0 5 c 0 6 c t total shear flow in each segment of the section V s V s s f( s) 4 9, f( s) b b 69b b b V s 4V s4 f( s) 5 9, f( s) b b 69b b 4V s5 s5 4V s6 f( s5) 59 9, f( s6) b b b 69b b f s V 69b ( 7 ) s7 7 s b b (8.50)

45 8. Shearing of thin-walled beams 8..8 Shearing of multi-cellular sections - shear flows in the webs vary quadratically, while those in flanges linearly - Net resultant of the shear flows in the flanges must vanish because no shear forces is externally applied in the horizontal direction. - Resultant of the shear flows in the webs must equal the externally applied vertical shear force, V 45

46 8.4 The shear center 8.4 Shear enter hap. 6 Assumption that transverse loads are applied in such a way that the beam will bend without twisting More precise statement : the lines of action of all transverse loads pass through the shear center - If the sear forces are not applied at the shear center, the beam will undergo both bending and twisting 8.4. alculation of the shear center location Involves two linearly independent loading cases () [] [], unit shear force, no shear force along shear flow f [] () s V [] [] [] [] (), V, V 0 f ( s) i, V 0 [] 46 [] f () s dx dx, 0 ds ds - from Eq.(8.7), shear forces equipollent to [] [] [] [] V f ds V f ds c c (8.5)

47 8.4 The shear center K( x, x ) K K [] [] dx dx M f r ds f ( r x x ) ds K 0 K K c c ds ds r : distance from K to the tangent to contour, Eq. (8.) - shear center location : Eq (8.0) K -Rearranging dx dx ds ds [] [] [] xk f ds K 0 c x f ds f rds c c 0 similarly, [] x f rds K 0 c by Eq.(8.5) (8.5) [] x f rds K 0 c (8.5) 47

48 8.4 The shear center - alternate torque equipollence condition, Eq.(8.40) a a [] x x f rds K a c a (8.54) [] x x f rds K a c a ( x, x ) : coordinate of an arbitrary point A General procedure for determination of the shear center compute the x-s centroid and select a set of centroidal axes (sometimes convenient with principal centroidal axes) f f [] () s [] () s compute corresponding to compute corresponding to according to Sections 8.. or 8..7 V V, V 0 [] [] 0, V [] [] (8.55) 48 4 compute the coordinate of shear center using Eqs (8.5) and (8.5) or (8.54) and (8.55) - If the x-s exhibits a plane of symmetry, simplified i, i plane is a plane of symmetry, the s.c. must be located in that plane. x 0 K, Eq. (8.5) can be bypassed.

49 8.4 The shear center Example 8. Shear center of a trapezoidal section - closed trapezoidal section - shear flow distribution generated by a vertical shear force, V : sum of the shear flow distribution in the auxiliary open section and the f s f s f closing shear flow 0 c EV h h EV f ( s ) s h s, f ( s ) s h ( h h ) l, 0 c 0 c H l H EV h h h h EV f ( s ) s hs l, f ( s ) s h 0 c 0 4 c 4 H l H f c EV ( h h ) ( h h ) l ( h h ) lh H l h h c 6( ) (8.48) (8.49) 49 Fig Thin-walled trapezoidal section subjected to a vertical shear force, V

50 8.4 The shear center - location of the shear center: by Eq. (8.49) - Evaluation of integral - Due to the symmetry of the problem, - If h h, x by symmetry x f s f rds k o c o f s f s V, f f V, V o o c c x b h h h h l lh h h h k 4 l hh ll h h h h l h h k 0 x k 0 Fig Thin-walled trapezoidal section subjected to a vertical shaer force, V 50

51 8.5 Torsion of thin-walled beams 8.5 Torsion of thin-walled beams hap. 7 Saint-Venant s theory of torsion for x-s of arbitrary shape. solution of PDE is required to evaluate the warping or stress function. However, approximate solution can be obtained for thinwalled beams 8.5. Torsion of open section Sec. 7.4 Torsional behavior of beams with thin rectangular x-s Sec. 7.5 Thin-walled, open x-s of arbitrary shape, shear stresses are linearly distributed through the thickness, torsional stiffness ~ (wall thickness) (Eq. (7.6)), very limited torque carrying capability 5

52 8.5 Torsion of thin-walled beams 8.5. Torsion of closed section Fig thin-walled, closed x-s of arbitrary shape subjected to an applied torque, assumed to be in a state of uniform torsion, axial strain and stress components vanish n(s) =0 - local equilibrium eqn. for a differential element, Eq.(8.4) f s 0 (8.59) shear flow must remain constant along curve f () s f const. (8.60) - constant shear flow distribution generates a torque M M f() s r() s ds f r() s ds 0 0 c c M A (Eq. (8.56)) Af (A : enclosed area by ) (8.6) Shear Flow i M 0 c da O rds/ i 0 P r 0 ts () ds Fig Thin-walled tube of arbitrary cross-sectional shape s s Bredt-Batho formula 5

53 8.5 Torsion of thin-walled beams s - shear stress resulting from torque M M At( s) () s s (8.6) twist rate vs. applied torque simple energy argument - strain energy stored in a differential slice of the beam of length dx da r tds tds dx G - introducing shear stress distribution, Eq.(8.6) s s s c (8.6) c da - work done by the applied torque M ds dx 4 A c Gt( s) (8.64) d dw M d M dx M dx dx (8.65) 5 where twist rate d dx

54 8.5 Torsion of thin-walled beams - st law of thermodynamics dw da proportionality between and, torsional stiffness H 4A c 4A ds c Gt (8.66) (8.67) - arbitrary shaped closed x-s of const. wall thickness, homogeneous material H M H maximum thin-walled circular tube (maximize the numerator) x M 4GtA l ds Gt, l: Perimeter of (8.68) 54

55 8.5 Torsion of thin-walled beams Sign convention A : area enclosed by curve that defines the section s configuration r () s 0 A r ( s) ds 0 c : perpendicular distance from the origin, O, to the tangent to, its sign depends on the direction of the curvilinear variable, s A is (+) when s describes while leaving A to the left (-) in the opposite. f 0, A0 M Af 0 - s : clockwise direction,, f f A A M Af Af 0 55

56 8.5 Torsion of thin-walled beams 8.5. omparison of open and closed sections 56 losed section : shear stress is uniformly distributed through the thickness Open section : shear stress is linearly distributed Torsional stiffness (enclosed area) for closed section, Eq(8.67) Torsional stiffness (thickness) for open section, Eq(7.64) - Fig.8.5. ircular shape, thin-walled tube of mean radius R m open losed s Open s H GR t /, Eq(7.64) m closed tube tube, Eq(7.9) - Maximum shear stress subjected to t he same torque, R m t Rm Fig. 8.5 A thin-walled open tube and closed tube M t M G t G H R t open open max open m t H H H closed open GR t Rm t max M open max closed max Rm t m (8.69) MR M RG G H R t closed closed m max m closed m (8.70)

57 8.5 Torsion of thin-walled beams -Example : R m =0t H max that of closed section will be,00 times larger than that of the open section that of open section will be 60 times larger than that of the closed section closed section can carry a 60 times larger torque Torsion of combined open and closed sections x-s presenting a combination of open and closed curves (Fig. 8.5) - twist rate is identical for the trapezoidal box rectangular strips Torques they carry M M box box strip H H strip t w b t t w t h b Fig. 8.5 Thin-walled trapezoidal beam with overhangs 57

58 8.5 Torsion of thin-walled beams Torsional stiffness Total torque H box H 4 GtA / l strip Gwt / box M M M strip Eq.(8.68) Eq.(7.64) strip box H box wl t M H H box H ( b b ) h for thin-walled section, t, h torsional stiffness of the section is nearly equal to that of the closed trapezoidal box alone. H H box box box M H H M box H H box strip strip strip M H M box M H 58

59 8.5 Torsion of thin-walled beams Max. shear stress from Eqs. (8.6), (7.65) ratio box box M M At At max strip strip strip M H max box M 0t wt H strip max box l t b b h max the max. shear stress in the strip is far smaller than that in the trapezoidal box 59

60 8.5 Torsion of thin-walled beams Torsion of multi-cellular sections 4-cell, thin-walled x-s subjected to a torque (Fig. 8.5) - only uniform torsion exists, and hence the axial stress flow vanishes : Eq.(8.4) reduces to shear flow is constant Free-body diagrams of the portion of the section - Fig () axial stress flow=0, f f f s 0 - Fig () D - Fig () f f f f 0, F G B fi 0 (8.7) the sum of the shear flows going into a joint must vanish f A f B M E B F G A dx D Fig. 8.5 A thin-walled, multi-cellular section under torsion 60 f B f A f D B A dx () () D () f f f F E F G Fig Free-body diagrams of the thin-walled, multi-cellular section. f G B f B

61 8.5 Torsion of thin-walled beams Torsion of multi-cellular sections onst. shear flows are assumed to act in each cell of the section (Fig. 8.55) [] f f [] -f f E [] [] [] f [] f f f -f [4] f [4] [] [4] Fig Shear flows in each cell of a thin-walled, multi-cellular section Determination of the const. shear flow in each cell N N [ i] [ i] [ i] M M A f (8.7) i i onst. shear flows are assumed to act in each cell of the section Determination of the const. shear flow in each cell total torque = sum of the torques carried by each individual cell Bredt-Batho formula 6

62 8.5 Torsion of thin-walled beams compatibility condition twist rates of the various cells are identified. - Eq.(8.66) [] [] [ i] [ N] (8.7) [ i] [ i] [ i] [ i ] M ds A f ds [ i] [ i] [ i] [ i] 4( A ) Gt 4( A ) Gt A [ i ] [ i ] f Gt Eqs.(8.7), (8.7) N cells eqn.s for N cells shear flows [ i ] ds (8.74) 6

63 8.5 Torsion of thin-walled beams Example 8.7 Two-cell cross-section - Two-cell cross- section (Fig. 8.56): highly idealized airfoil structure - Eq. (8.7): total torque carried by the section is the sum of the torques carried in each cell N cell i i i M A f R f 6R f - Eq. (8.7): twist rates for the two cells are identical. twist rate for the front cell (8.75) [] f f f f ds R R f ( f f ) A Gt() s G R / t t GRt twist rate for the aft cell [] f f f R A Gt() s GR t t 6GRt ds R f 0 ( f f ) f 0 6

64 8.5 Torsion of thin-walled beams Example 8.7 Two-cell cross-section - Equating the two twist rate -> second eqn. for the shear flow f ( f f ) ( f f ) f which simplifies to f.04 f - This can be used to solve for f and f - largest contribution to the torsional stiffness comes from the outermost closed sections, which is the union of the frount and aft cells. The largest shear flow circulates in this outmost section, leaving the spar nearly unloaded torsional stiffness H M ( R.04 6 R ) f [] / ( GRt)[.04 / (.04 )] f.8 GR t

65 8.6 oupled bending-torsion problems 8.6 oupled bending-torsion problems 65 hap. 6 arbitrary x-s subjected to complex loading conditions important restrictions no torques transverse shear forces are assumed to be applied in such a way that the beam will bend without twisting Now can be removed Fig concentrated transverse load acting at the tip and it point of application, A, with coord. ( x, x ), a a distributed loads p( x), p ( x), p ( x) i i p ( x ) q ( x ) p ( x ) p ( x ) P q ( x ) q ( x ) K x x k Fig Beam under a complex loading condition P Q Q P Q P Point of application A( x, x ) i i a a P i A K Shear center (, ) k rosssection entroid ( x, x ) c c

66 8.6 oupled bending-torsion problems Solution procedure ompute location of the centroid, x (, x) c c * * * ompute orientation of the principal axes of bending i, i, i and the principal bending stiffness (sec. 6.6) ompute location of the shear center, ( x, x ) (sec. 8.4) 4 ompute torsional stiffness (chap. 7, or sec. 8.5.) 5 Solve the extensional problems Eqs. (6.), (6.) in principal centroidal axes of bending planes 7 ompute torsional problem d dx * B.. at root 0 d [ ( ) ( ) ( ) ( ) ( )] * * * * * * * * * * * * 4 H * g x x x p x x x p x a a dx * * * * * * * * * * a a dx (8.76) d H Q ( x x ) P ( x x ) at tip (8.77) : axis system defined by the principal centroidal axes of bending More convenient to recast the governing eqn. in a coord. system for which * axis is aligned with the axis of a beam i 66

67 8.6 oupled bending-torsion problems - Knowledge of centroid and shear center complete decoupling of a problem 4 independent problems axial problem bending problem torsional problem - If no torque and all transverse loads are applied at the s.c. R.H.S of Eq(8.77) =0, the beam does not twist ( x ) 0 ( x ) If not, the beam twists, rigid body rotation about the s.c. 67

68 8.6 oupled bending-torsion problems Example 8.8 Wing subjected to aerodynamic lift and moment - Wing coupled bending-torsion problem (Fig. 8.66) - principal axes of bending i and i : their origin at shear center -axis i : along the locus of the shear centers of all the cross-sections -> straight line called the elastic axis - aerodynamic loading : lift per unit span L A, applied at the aerodynamic center aerodynamic moment per unit span M A - differential eqn for bending in plane (i, i ) 68 d c du H LA dx dx (8.79) B: du du u 0 du at the root, 0 dx dx dx at the unloaded tip - governing eqn for torsion d d H MA ela dx dx B: at the root, d at the tip 0 0 dx e: distance from the aerodynamic center to the shear center

69 8.6 oupled bending-torsion problems Example 8.8 Wing subjected to aerodynamic lift and moment - typical transport aircraft: e = 5 5% chord - it is convenient to select the origin of the axes at the s.c., rather than at the centroid.: bending problem is decoupled from the axial problem. beam will rotate about the origin of the axes system. - The rotation Ф of the section is, in fact, the geometric angle of attack of the airfoil. - lift, L A, is a function of the angle of attack - aerodynamic problem: computation of the lift as a function of the angle of attack - elastic problem: computation of wing deflection and twist as a function of the applied loads - aeroelasticity: study of this interaction 69 Fig The wing bending torsion coupled problem

70 8.7 Warping of thin-walled beams under torsion Thin-walled beam subjected to an applied torque Shear stress generated Out-of-plane deformations, warping, in x-s : magnitude is typically small, but dramatic effect on the torsional behavior Particularly pronounced for non-uniform torsion of open sections ontrasts with Saint-Venant theory Twist rate varies along the span Uniform torsion, constant twist rate 70

71 8.7 Warping of thin-walled beams under torsion 8.7. Kinematic description Fig 8.70 : thin-walled beam subjected to a tip concentrated torque Q Displacement field Similar to that for Saint-Venant solution Each x-s is assumed to rotate like a rigid body about R ( center of twist, (x r, x r )) unknown yet u ( x, s) ( s) ( x ) Fig Thin-walled beam subjected to an applied torque. (8.80a) 7 Twist rate Unknown warping function u ( x, s) ( x x ) ( x ) r u ( x, s) ( x x ) ( x ) r (8.80b) (8.80c)

72 8.7 Warping of thin-walled beams under torsion Strain field u x u x d () s dx 0 u u x x 0 u u d ( x x ) r x x dx (8.8) u x 0 u u d ( x x ) r x x dx Non-uniform torsion is assumed axial strain 0 d dx 0 In-plane strain components =0 since rigid body rotation assumed Shear strain components partial derivatives of warping function and twist rate 7

73 8.7 Warping of thin-walled beams under torsion 8.7. Stress-strain relations Non-vanishing components of the stress E E () s d dx d G ( x xr ) G dx d G ( x xr ) G dx (8.8) Only non-vanishing shear stress component for thin-walled beams τ s dx s ds dx ds d dx d dx dx dx ( x xr) ( x xr) G dx ds dx ds ds ds { Total derivative of Ψ w.r.t. s { Distance from the twist center to the tangent to, Eq.(8.) 7 s d ds r r G for open and closed sections (8.8)

74 8.7 Warping of thin-walled beams under torsion 8.7. Warping of open sections Shear stress distribution in open-section linearly distributed across the wall thickness and 0 along the wall mid-line τ s = 0 along curve, Eq.(8.8) s d r G ds 0 (8.84) Warping function relation Purely geometric function, Γ(s) d dx dx r ro xr x r ds ds ds (8.85) Warping function d r ds () s () s x rx x rx c o (8.86) (8.87) 74

75 8.7 Warping of thin-walled beams under torsion 8.7. Warping of open sections Uniform torsion, d dx 0 axial strain/stress = 0 c and (x r, x r ) cannot be determined, simply represents a rigid body displacement field, does not affect the state of stress/strain Non-uniform torsion { varying applied torque constrained warping displacement at a boundary or at some point non-vanishing axial strain/stress although acted upon by a torque alone but, still N, M, M =0 Axial force N =0 Eq.(8.8a), (8.87) c tds 0 75

76 8.7 Warping of thin-walled beams under torsion 8.7. Warping of open sections Etds xr Extds xr Extds c Etds 0 Bending moment { 00 origin of the axes is selected to be at the centroid M { c S xtds 0 Etds { { Extds xr Ex tds xr Exxtds c Extds 0 S { (axial stiffness) (8.88) 76 (principal centroidal axes of bending) H H 0 0

77 8.7 Warping of thin-walled beams under torsion 8.7. Warping of open sections x Ex tds r H (8.89) M = 0 x Ex tds r H (8.90) 77

78 8.7 Warping of thin-walled beams under torsion Example 8.0 Warping of a -channel - -channel cross section subjected to a tip torque (Fig. 8.4) - axes in the figure: principal centroidal axes of bending i and i -axis i : axis of symmetry - st step : compute the purely geometric function, r o s d ro ds (8.86) : normal distance from the origin of the axes to the tangent of the curve Fig The warping function for a -channel 78

79 8.7 Warping of thin-walled beams under torsion - For the lower flange (s ) where, r h 0 / ( s ) hs /c applying boundary condition, ( s ) 0 at s 0 then, ( s ) hs / r d and r h / 0 0 ( s ) ds h( bd) ( s ) hs / h( b d)/ - nd step: evaluate the integration constants Et b h/ b h c ( s ) ds ( s ) ds ( s ) ds ( b d) 0 h / 0 S 79 - Final step: coord. of the twist center Et b h h/ b h x s ds s s ds s ds r c h H ( )( ) ( ) ( ) 0 / 0 hbt E d c 4 H

80 Et b h/ b x ( s )( bd s) ds ( s )( d) ds ( s )( sd) ds 0 r c 0 h/ 0 H hbt E d c 4 H The warping function then follows from eq.(8.87) as h h ( s ) s eb ; ( s ) es ; ( s ) s e where, e h b te H c /(4 ) - The location of shear center coincides that of the twist center.

81 8.7 Warping of thin-walled beams under torsion Warping of closed sections Shear stress distribution M s H At At d Eq. (8.8) s H r r ds G AGt () s constant through the wall thickness in closed section A = area enclosed by curve (8.6) governing equation for in closed sections (8.94) Process of integration of Eq. (8.94) similar to that for open section Purely geometric function Γ(s) d ds H r AGt o arbitrary B.. is used to integrate Eq.(8.95) (8.95) c and (x r, x r ) can be determined by the vanishing of F, M, M 8

82 8.7 Warping of thin-walled beams under torsion Warping of multi-cellular sections Section shear flow distribution f(s) due to applied torque f () s () s s () s Governing equation for the warping function d () s r ds Gt r t (8.97) Determination of the warping function --- exactly mirrors that for open and closed sections, except the following d () s r ds Gt o (8.98) 8

83 8.8 Equivalence of the shear and twist centers Shear center defined by torque equipollence condition, Eq.(8.9) enter of twist introduced for the analysis of thin-walled beams under torsion Eq.(8.5) Eq.(8.86) Integrating by parts similarly, by Eq.(8.58) x Et x x ds Ex tds x k r H H by Eq.(8.89) x k r k x f rds f o df xk dsf ds Equivalence of the shear and twist center for open sections. Equivalence also holds for closed sections direct consequence of Betti s reciprocity theorem. Eq.(0.7) d ds boundary 8

84 8.9 Non-uniform torsion Non-uniform torsion both shear and axial stresses generated by differential warping Markedly different behavior from that under uniform torsion Axial stress distribution uniform across the wall thickness axial flow n w t Although the axial stress does not vanish, the resulting axial force and bending moment do vanish local equilibrium equation, Eq.(8.4), is not necessarily satisfied For this local equilibrium to hold, a shear flow, f w, warping shear flow is generated to satisfy the local equilibrium n x w fw s 0 84

85 8.9 Non-uniform torsion Introducing Eq.(8.8a) for the case of open sections f s w d Et dx can be integrated by the procedure in Section 8. (8.99) - Simple -channel Fig. (8.75) Question of overall equilibrium the warping shear flow generate resultant transverse shear force? Eq.(8.7) Eq.(8.99) dx f V f ds x dsx f ds s w w w w boundary Integrating by parts d V Ex tds 0 w dx 0 since f w = 0 at the edge of the contour Similarly, V w 0 85

86 8.9 Non-uniform torsion Torque resultant about the shear center generated by the warping shear flow 86 Eq.(8.0) Integrating by parts Introducing Eq.(8.99) d Mwk fwrds k fw ds ds df M dsf ds w wk w boundary d Mwk Hw dx Total torque = that by the twist rate + that due to warping d Mk H Hw dx generated by shear stress distribution warping stiffness (8.00) (8.0) (8.0) (8.04) Additional contribution from the warping shear flow, =0 for uniform torsion Equilibrium equation for a differential element of the beam under torsional load d d d dx dx dx H Hw q H w E tds Eq.(7.5) (8.05)

87 8.9 Non-uniform torsion Example 8. Torsion of a cantilevered beam with free root warping - Uniform cantilevered beam of length L subjected to a tip torque, Q - Root condition: No twisting is allowed, but warping is free to occur -> attaching the beam s root to a diaphragm that prevents any root rotation, but does not constrain axial displacement - uniform properties along its length, Eq. (8.05) becomes - at the root : no twist occurs, - free warping at the root : axial stress must vanish, - at the tip: torque must equal the applied torque, - at the tip: axial stress must vanish once again, - Introduction of non-dimensional span-wise variable, - Governing eqn.: H d d 4 H w 4 dx dx 0 0 d 0 dx d Q H H d dx 0 dx x L d w dx '''' k 0 (8.06) 87

88 8.9 Non-uniform torsion - New B s: at the root, at the tip, k HL H w 0, 0 0, k QL Hw : ratio of the torsional stiffness to the warping stiffness (8.07) - General sol. of the governing differential eqn.: cosh k sinh k (8.08) 4 - Application of B s: QL H (8.08) -> identical to the uniform torsion solution d Q const dx L H w - torsional warping stiffness,, disappears from the solution. 88

89 8.9 Non-uniform torsion Example 8.4 Torsion of a cantilevered beam with constrained root warping - Same uniform cantilevered beam, but the root section is now solidly fixed to prevent any wapring at the root -> at this built-in end, no twisting occurs 0 d no axial displacement 0 - Governing eqn. is the same, Eq. (8.06). But B s are New B s: at the root, 0, 0 at the tip, 0, k QL Hw - General sol. is the same as Eq. (8.08) - Application of B s: QL sinh k sinh k H k cosh k uniform torsion dx influence of the non-uniform torsion induced by the root warping constraint 89

90 Fig Twist distribution for the closed rectangular section under non-uniform torsion. k =6.54 ( ), k =8.7 ( ), k =5.04 ( ), k =.5(O). Fig Twist distribution for the channel section under non-uniform torsion. k =.65 ( ), k =. ( ), k =0.808 ( ), k =0.404(O).

91 8.0 Structural idealization Actual thin-walled beam structures stringers added to increase the bending stiffness can be idealized by separating the axial and shear stress carrying components into distinct entities called stringers sheets Axial stress Shear stress assumed to be carried only in the stringers assumed to be carried only in the sheets (a) (b) Sheet Stringer 9 box beam, L shaped longitudinal members located away from the centroid much larger contribution to the bending stiffness sheet-stringer idealization considerably simplified analysis procedure for stress distribution

92 8.0 Structural idealization 8.0. Sheet-stringer approximation of a thin-walled beam Figure 8.80 no discrete stringers or with far smaller x-s area still possible to construct a sheet-stringer model Idealized structures Axial stresses are carried solely by the stringers Shear stresses are carried solely by the sheets 9

93 8.0 Structural idealization Approach to estimate the areas of the stringers Triangular equivalence method (sec. 6.8) guarantee the same bending stiffness and centroid location Linear distribution of axial stress, σ = σ [] +(σ [] - σ [] ) s/b - σ [] : stresses of point A - σ [] : stresses of point B - s : local position along the contour of width b the areas A, and A, of the stringers need to be determined. ) Axial stresses at A and B are the same as the actual constraints ) Force and moment equivalences are maintained 9 Force equivalence F s/ b tds bt A A b [] [] [] [] [] [] [] [] [] 0 Bending moment equivalence b bt M A s/ bstds b A 0 6 solution A [] [] [] [] [] [] [] bt [] [] [] 6, A bt [] [] [] 6 (8.0)

94 8.0 Structural idealization special cases Uniform axial stress σ [] = σ [] A [] =A [] =bt/ Pure bending σ [] = -σ [] A [] =A [] =bt/6 (8.) (8.) Different stress distributions are considered, equivalent idealized area need to be recomputed 8.0. Axial stress in the stringers The same approach as developed in hapter 6, axial stress σ [r] acting in the r-th stringer N H M H M H M H M S H H [ r] [ r] [ r] [ r] E x x Uniform stress is assumed in a small lumped case (8.) net axial force = A [r] σ [r] 94

95 8.0 Structural idealization 8.0. shear flow in the sheet components Local equilibrium condition, Eq.(8.4), since no axial stress Stringer equilibrium Figure 8.8 f s 0 f = const. (8.4) [ r] dx A fdx fdx 0 x f f f f f A [ r] [ r] x (8.5) ( / x ) dx axial equilibrium for the r-th stringer f f -Eq. (8.) (8.5) H V H V H V H V f E A x H H (8.6) H H H ( H ) [ r] [ r] [ r] [ r] 95

96 8.0 Structural idealization { general thin-walled x-s sheet-stringer idealization shear flow distribution is governed by a differential equation, Eq. (8.0) shear flow distribution is governed by a difference equation, Eq. (8.6) { Integration constant needs to be determined open section 0 at stress-free edge closed section Section 8..7 Shear flow resultants Figure 8.8 curved sheet carrying a constant shear flow f, and connecting stringers, shear force resultant [] [] V i f ds f dx f ( x x ) similarly, V f ( x x ) [] [] V V V f ( x x ) ( x x ) f L [] [] [] [] 96 direction parallel to the line connecting the stringers (8.8)

97 8.0 Structural idealization Moment resulting from the shear flow distribution w, r, t point O A M f r ds f r ds f da f Aˆ 0 o o ˆ A : area of the sector defined by the stringers (Fig. 8.8) Distance e of line of action from O ˆ f Aˆ e A V L (8.9) 97

98 8.0 Structural idealization Torsion of sheet-stringer sections Open section linear shear stress distribution through thickness, inefficient at carrying torsional loads If different thickness for individual sheets H H bt G H i sheets sheets Gbt i i i (8.0) 98

99 8.0 Structural idealization Example 8.5 Shear flow in a sheet-stringer -channel section - -channel section subjected to a shear load, V, and a bending moment, M - i : axis of symmetry, principal centroidal axes. - Under the bending moment, axial stress will be const. over the top flanges and bottom flanges, but will vary linearly in the web. - Use Eqs. (8.) and (8.) to evaluate the stringers. A bt, A bt 6 ht, 4 A bt 6 ht, A bt 99 - This idealization yields the same bending stiffness as that for the thin-walled section H Ebht Eth Ebht c 6 h b

100 8.0 Structural idealization - Equilibrium condition for stringer, Eq. (8.6), yields - Shear flow in the upper flange - Shear flow in the vertical web A f f 0 V h f f EA [] [] H 6 h/ b h V h V h/ bv V f f EA H h b h h b h h [] 6 / 6 / V 00 - Shear flow in the lower flange V f, f f 6 hb h 4 4 -Observation shear flow is const. in each sheet in contrast with the thin-wall solution (Fig. 8.5) Max. shear flow in the sheet-stringer idealization f max V h

101 8.0 Structural idealization Max. shear flow in the thin-wall solution f max 4bh V 6bh h Thus, sheet-stringer idealization underestimates the true shear flow and thus is not conservative. Sheet-stringer idealization exactly satisfy overall equilibrium requirements. Torque equipollence about an arbitrary point of the section yields the location of the shear center, K. This result exactly matches the location found using the thin-wall solution. 0

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