Aircraft Structures Beams Torsion & Section Idealization

Transcription

1 Universit of Liège Aerospace & Mechanical Engineering Aircraft Structures Beams Torsion & Section Idealiation Ludovic Noels omputational & Multiscale Mechanics of Materials M3 hemin des hevreuils 1, B4000 Liège

2 Balance of bod B Momenta balance Linear Angular Boundar conditions Neumann Dirichlet Elasticit T Small deformations with linear elastic, homogeneous & isotropic material n b (Small) Strain tensor, or Hooke s law, or with Inverse law l = K - 2m/3 2m with Aircraft Structures - Beam - Torsion & Section Idealiation 2

3 General expression for unsmmetrical beams Stress Pure bending: linear elasticit summar q With a urvature M xx In the principal axes I = 0 Euler-Bernoulli equation in the principal axis for x in [0 L] Bs f(x) T M xx u =0 x du /dx =0 M>0 L Similar equations for u Aircraft Structures - Beam - Torsion & Section Idealiation 3

4 Beam shearing: linear elasticit summar General relationships f(x) T M xx u =0 x du /dx =0 M>0 L Two problems considered L Thick smmetrical section h Shear stresses are small compared to bending stresses if h/l << 1 Thin-walled (unsmmetrical) sections Shear stresses are not small compared to bending stresses Deflection mainl results from bending stresses h 2 cases Open thin-walled sections» Shear = shearing through the shear center + torque losed thin-walled sections» Twist due to shear has the same expression as torsion t L Aircraft Structures - Beam - Torsion & Section Idealiation 4

5 Beam shearing: linear elasticit summar Shearing of smmetrical thick-section beams Stress With b() t h h t Accurate onl if h > b Energeticall consistent averaged shear strain t A * T + x T dx with g g dx x Shear center on smmetr axes T g max Timoshenko equations dx & q On [0 L]: q g x Aircraft Structures - Beam - Torsion & Section Idealiation 5

6 Beam shearing: linear elasticit summar Shearing of open thin-walled section beams Shear flow In the principal axes T S q T s T T x T Shear center S On smmetr axes At walls intersection T Determined b momentum balance Shear loads correspond to Shear loads passing through the shear center & Torque h t t S b t Aircraft Structures - Beam - Torsion & Section Idealiation 6

7 Beam shearing: linear elasticit summar Shearing of closed thin-walled section beams T Shear flow T Open part (for anticlockwise of q, s) T q p s onstant twist part T The completel around integrals are related to the closed part of the section, but if there are open parts, their contributions have been taken in q o (s) q p T da h ds s T Aircraft Structures - Beam - Torsion & Section Idealiation 7

8 Beam shearing: linear elasticit summar Shearing of closed thin-walled section beams Warping around twist center R T T T With u x (0)=0 for smmetrical section if origin on q p da h ds s Shear center S the smmetr axis T ompute q for shear passing thought S Use S With point S=T q p ds s Aircraft Structures - Beam - Torsion & Section Idealiation 8

9 Torsion of closed thin-walled section beams General relationships We have seen s s + s s s ds s s xx q + s q ds s q + x q dx xx + x s xx dx x ds dx q q s s M x If the section is closed Bredt assumption for closed sections: Stresses are constant on t, and if there is onl a constant torque applied then s s = s xx = 0 onstant shear flow (not shear stress) x M x Aircraft Structures - Beam - Torsion & Section Idealiation 9

10 Torsion of closed thin-walled section beams Torque As q due to torsion is constant M x Displacements It has been established that q p da h ds s So in linear elasticit n But for pure torsion q is constant du x x ds Remark mt is not constant along s but it is assumed constant along x s dx du s As s xx = s s = Aircraft Structures - Beam - Torsion & Section Idealiation 10

11 Torsion of closed thin-walled section beams Displacements (2) It has been established that for a twist around the twist center R u s As R p R p q Y for all values of s (so all value of Y ) The onl possible solution is, & So displacement fields related to torsion are linear with x Aircraft Structures - Beam - Torsion & Section Idealiation 11

12 Torsion of closed thin-walled section beams Rate of twist Use Relation developed for shearing, but with q due to torsion constant on s Torque expression R p R p q u s Y Twist constant with x Torsion rigidit Torsion second moment of area for constant m : Aircraft Structures - Beam - Torsion & Section Idealiation 12

13 Torsion of closed thin-walled section beams Warping Use Relation developed for shearing, but with q due to torsion constant on s R p R p q u s Y Swept from twist center R Torque expression Warp displacement Aircraft Structures - Beam - Torsion & Section Idealiation 13

14 Torsion of closed thin-walled section beams Twist & Warping under pure torsion Twist Warp Deformation Plane surfaces are no longer plane It has been assumed the keep the same projected shape + linear rotation Longitudinal strains are equal to ero All sections possess identical warping Longitudinal generators keep the same length although subjected to axial displacement Aircraft Structures - Beam - Torsion & Section Idealiation 14

15 Torsion of closed thin-walled section beams Zero warping under pure torsion Warp Zero warping condition requires for all values of s R p R p q u s Y As right member is constant the condition of ero warping is constant with respect to s Solutions at constant shear modulus ircular pipe of constant thickness Triangular section of constant t (p R is the radius of the inscribed circle which origin coincides with the twist center) Rectangular section with t h b = t b h R Aircraft Structures - Beam - Torsion & Section Idealiation 15

16 Example Torsion of closed thin-walled section beams Doubl smmetrical rectangular closed section B A onstant shear modulus t h M x Twist rate? Warping distribution? t b b D h Aircraft Structures - Beam - Torsion & Section Idealiation 16

17 Twist rate Torsion of closed thin-walled section beams As the section is doubl smmetrical, the twist B A center is also the section centroid t h M x Twist rate t b b h D For a beam of length L and constant section Torsion rigidit Aircraft Structures - Beam - Torsion & Section Idealiation 17

18 Torsion of closed thin-walled section beams Warping As the section is doubl smmetrical, the twist center is also the section centroid Warping It can be set up to 0 at point E B smmetr it will be equal to ero wherever t h B t b A M x s E h b D a smmetr axis intercept the wall & On part EA & Aircraft Structures - Beam - Torsion & Section Idealiation 18

19 Warping (2) Torsion of closed thin-walled section beams On part EA B A t h t b b M x s E D h So using smmetr and as distribution is linear Zero warping if b t h = h t b x M x Aircraft Structures - Beam - Torsion & Section Idealiation 19

20 Torsion of thick section Torsion of a thick section The problem can be solved explicitl b M x recourse to a stress function Hdrodnamic analog Shear stresses have the same expression than the velocit in a rotational flow in a box of same section x M x t M x t M x Aircraft Structures - Beam - Torsion & Section Idealiation 20

21 Torsion of thick section Torsion of a thick circular section Exact solution of the problem B smmetr there is no warping sections remain plane t M x In linear elasticit r Shear stresses Torque Torsion rigidit At constant shear modulus (required for smmetr): For circular cross sections (onl) I p =I T Maximum shear stress Aircraft Structures - Beam - Torsion & Section Idealiation 21

22 Torsion of thick section Torsion of a rectangular section Exact solution of the problem with stress function Assumptions Linear elasticit onstant shear modulus Maximum stress at mid position of larger edge t max M x h Torsion rigidit (constant m) b Approximation for h>>b h/b a /3 b /3 & Aircraft Structures - Beam - Torsion & Section Idealiation 22

23 Torsion of thick section Torsion of a rectangular section (2) Warping As For a rigid rotation (first order approximation) t q h b For a thin rectangular section & Doubl smmetrical section Aircraft Structures - Beam - Torsion & Section Idealiation 23

24 Torsion of open thin-walled section beams Rectangle approximation of open thin-walled section beams Thin rectangle & For constant shear modulus t M x h Warping Thin curved section t If t << curvature an approximate solution is t M x t s t n Open section composed of thin rectangles Same approximation t 3 l 2 t 2 l 3 t 1 l 1 t Aircraft Structures - Beam - Torsion & Section Idealiation 24

25 Torsion of open thin-walled section beams Warping n Warping around s-axis Thin rectangle Here i are not equal to 0 Part around s-axis du x s ds dx du s x Warping of the s-line (n=0) We found If R is the twist center As t xs (n=0) = 0 R p R t q M x t s t n u s Eventuall s-axis warp (usuall the larger) t Aircraft Structures - Beam - Torsion & Section Idealiation 25

26 Torsion of open thin-walled section beams Example t f = 1.5 mm U open section onstant shear modulus (25 GPa) Torque of 10 N. m Maximum shear stress? Warping distribution? M x S h = 50 mm t w = 2.5 mm t f = 1.5 mm b = 25 mm Aircraft Structures - Beam - Torsion & Section Idealiation 26

27 Torsion of open thin-walled section beams Maximum shear stress t f = 1.5 mm Torsion second moment of area M x Twist rate S h = 50 mm t w = 2.5 mm t f = 1.5 mm Maximum shear stress reached in web b = 25 mm Aircraft Structures - Beam - Torsion & Section Idealiation 27

28 Torsion of open thin-walled section beams Twist center t f = 1.5 mm Zero-warping point T Free ends so the shear center S corresponds to twist center R See lecture on structural discontinuities B smmetr, lies on O axis M x S h = 50 mm t w = 2.5 mm Appl Shear T to obtained S Shear flow for smmetrical section t f = 1.5 mm b = 25 mm With Aircraft Structures - Beam - Torsion & Section Idealiation 28

29 Torsion of open thin-walled section beams Twist center (2) t f = 1.5 mm Shear flow for smmetrical section (2) T On lower flange M x S O h = 50 mm t w = 2.5 mm t f = 1.5 mm Momentum due to shear flow Zero web contribution around O Top and lower flanges have the same contribution b = 25 mm s Moment balance Be carefull: clockwise orientation of q, s Aircraft Structures - Beam - Torsion & Section Idealiation 29

30 Torsion of open thin-walled section beams Warping of s-axis Origin in O as b smmetr u x (O )=0 On O A branch Area swept is positive M x S A Rp O A s B h = 50 mm At point A On AB branch Area swept is negative A b = 25 mm A Rp B M x S O s h = 50 mm b = 25 mm Aircraft Structures - Beam - Torsion & Section Idealiation 30

31 Torsion of open thin-walled section beams Warping of s-axis (2) A B Origin in O as b smmetr u x (O )=0 (2) A Rp On AB branch Area swept is negative M x S O s h = 50 mm b = 25 mm At point B Branches for <0 obtained b smmetr Aircraft Structures - Beam - Torsion & Section Idealiation 31

32 Warping of s-axis (3) On O A branch Torsion of open thin-walled section beams On AB branch Branches for <0 obtained b smmetr S M x x Aircraft Structures - Beam - Torsion & Section Idealiation 32

33 ombined open and closed section beams Wing section near an undercarriage ba Bending There was no assumption on section shape Use same formula Shearing Shear center has to be evaluated for the complete section Shearing results into a shear load passing through this center & a torque Shear flow has different expression in open & closed parts of the section Torsion Rigidit of open section can be neglected most of the time But stress in open section can be high Aircraft Structures - Beam - Torsion & Section Idealiation 33

34 h b = 0.2 m h f = 0.1 m ombined open and closed section beams Example Simpl smmetrical section onstant thickness Shear stress? b f = 0.1 m O T = 100 kn t = 2 mm b b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 34

35 h b = 0.2 m h f = 0.1 m ombined open and closed section beams entroid B smmetr, on O axis? b f = 0.1 m O T = 100 kn t = 2 mm b b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 35

36 h b = 0.2 m h f = 0.1 m ombined open and closed section beams Second moment of area As = m b f = 0.1 m O T = 100 kn t = 2 mm b b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 36

37 ombined open and closed section beams h b = 0.2 m h f = 0.1 m Shear flow As I x = 0 & as shear center on b f = 0.1 m with At A & H shear stress has to be ero If origin on A, q(0) = 0 q B A s O F G H Branch AB orresponds to an open section D T = 100 kn I t = 2 mm E b b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 37

38 ombined open and closed section beams h b = 0.2 m h f = 0.1 m Shear flow (2) Branch B Branches FG & GH b f = 0.1 m B smmetr q B A s O F G H D T = 100 kn I b b = 0.2 m E t = 2 mm Aircraft Structures - Beam - Torsion & Section Idealiation 38

39 ombined open and closed section beams h b = 0.2 m Shear flow (3) losed part: With & Let us fix the origin at O B smmetr q(0) = 0 (if not the formula would have required anticlockwise s, q) q = q o (s) Branch O F q B A O s F G H D T = 100 kn I b b = 0.2 m E t = 2 mm Aircraft Structures - Beam - Torsion & Section Idealiation 39

40 ombined open and closed section beams h b = 0.2 m Shear flow (4) Branch FE Shear flux should be conserved at point F q B A O s F s G H Shear flux on branch D T = 100 kn I E b b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 40

41 h b = 0.2 m ombined open and closed section beams Shear flow (5) Branch EI Other branches b smmetr q B A O s F s G H D T = 100 kn I E Aircraft Structures - Beam - Torsion & Section Idealiation 41

42 h b = 0.2 m ombined open and closed section beams Shear flow (6) Remark, if smmetr had not been used, shear stress at O should be computed (but require anticlockwise s and q for these signs of T & T ) With p O F = p O & q O F = -q O & B O s F G ds O F = ds O q A s H etc D T = 100 kn I E Aircraft Structures - Beam - Torsion & Section Idealiation 42

43 Example losed nose cell 0.02 m 2 area 0.9 m outer length Open ba onstant shear modulus m = 25 GPa Torque 10 kn. m Twist rate? Shear stress? ombined open and closed section beams t c = 1.5 mm A c = 0.02 m 2 l = 0.9 m t c = 1.5 mm t b = 2 mm M x b b = 0.6 m t b = 2 mm h = 0.3 m Aircraft Structures - Beam - Torsion & Section Idealiation 43

44 Twist rate As an approximation the 2 torsion rigidities are added ell Ba losed section with constant m Open section with constant m ombined open and closed section beams t c = 1.5 mm A c = 0.02 m 2 l = 0.9 m t c = 1.5 mm t b = 2 mm M x b b = 0.6 m t b = 2 mm h = 0.3 m Twist rate Aircraft Structures - Beam - Torsion & Section Idealiation 44

45 Shear stress ell losed section ( ) Ba Open section ( ) ombined open and closed section beams t c = 1.5 mm A c = 0.02 m 2 l = 0.9 m t t t c = 1.5 mm t b = 2 mm M x b b = 0.6 m t b = 2 mm h = 0.3 m Aircraft Structures - Beam - Torsion & Section Idealiation 45

46 Structural idealiation Example 2-spar wing (one cell) Stringers to stiffen thin skins Angle section form spar flanges Design stages onceptual Define the plane configuration Span, airfoil profile, weights, Analses should be fast and simple Formula, statistics, Preliminar design Starting point: conceptual design Define more variables Number of stringers, stringer area, Analses should remain fast and simple Use beam idealiation» See toda FE model of thin structures» See next lectures Detailed design All details should be considered (rivets, ) Most accurate analses (3D, non-linear, FE) Aircraft Structures - Beam - Torsion & Section Idealiation 46

47 Wing section idealiation Principle of idealiation Booms Stringers, spar flanges, Have small sections compared to airfoil Direct stress due to wing bending is almost constant in each of these The are replaced b concentrated area called booms Booms Have their centroid on the skin Are carring most direct stress due to beam bending Skin Skin is essentiall carring shear stress It can be assumed That skin is carring onl shear stress If direct stress carring capacit of skin is reported to booms b appropriate modification of their area Aircraft Structures - Beam - Torsion & Section Idealiation 47

48 Wing section idealiation Panel idealiation Skin panel Thickness t D, width b arring direct stress linearl distributed Replaced b Skin without thickness 2 booms of area A 1 and A 2 Booms area depending on loading s xx 1 x s xx 2 b t D Moment around boom 2 b Total axial loading A 1 A 2 s xx 1 x s xx Aircraft Structures - Beam - Torsion & Section Idealiation 48

49 Wing section idealiation h l = 0.4 m h l = 0.4 m Example 2-cell box wing section Simpl smmetrical Angle section of 300 mm 2 t a = 2 mm t b = 1.5 mm t l = 3 mm t m = 2.5 mm t r = 2 mm h r = 0.2 m Idealiation of this section to resist to bending moment? Bending moment along -axis 6 direct-stress carring booms Shear-stress-onl carring skin panels l l a = 0.6 m b = 0.6 m A 1 A 2 A 3 t a = 2 mm t l = 3 mm t m = 2.5 mm t r = 2 mm l a = 0.6 m t b = 1.5 mm A 6 A 5 A 4 l b = 0.6 m h r = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 49

50 Wing section idealiation h l = 0.4 m Booms area Bending moment Along -axis Stress proportional to stress distribution is linear on each section edge ontributions Flange(s) area Reported skin parts Use formula for linear distribution A 1 A 2 A 3 t a = 2 mm l a = 0.6 m t b = 1.5 mm t l = 3 mm t m = 2.5 mm t r = 2 mm A 6 A 5 A 4 l b = 0.6 m h r = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 50

51 Section idealiation consequences onsequence on bending Idealiation depends on the loading case Booms area are dependent on the loading case Direct stress due to bending is carried b booms onl For bending the axial load is equal to ero But direct stress depends on the distance from neutral axis It can be concluded that for open or closed sections, the position of the neutral axis, and thus the second moments of area Refer to the direct stress carring area onl Depend on the loading case onl Aircraft Structures - Beam - Torsion & Section Idealiation 51

52 Section idealiation consequences Example Idealied fuselage section Simpl smmetrical Direct stress carring booms Shear stress carring skin panels Subjected to a bending moment M = 100 kn. m Stress in each boom? A 1 = 640 mm 2 A 2 = 600 mm 2 A 3 = 600 mm 2 A 4 = 600 mm 2 A 5 = 620 mm 2 A 6 = 640 mm 2 A 7 = 640 mm 2 A 8 = 850 mm 2 A 9 = 640 mm 2 M O 1 = 1.2 m 2 = 1.14 m 3 = m 4 = m 5 = m 6 = m 7 = m 8 = m Aircraft Structures - Beam - Torsion & Section Idealiation 52

53 Section idealiation consequences entroid Of idealied section A 1 = 640 mm 2 A 2 = 600 mm 2 A 3 = 600 mm 2 1 = 1.2 m 2 = 1.14 m 3 = m A 4 = 600 mm 2 A 5 = 620 mm 2 A 6 = 640 mm 2 A 7 = 640 mm 2 A 8 = 850 mm 2 A 9 = 640 mm 2 M O 4 = m 5 = m 6 = m 7 = m 8 = m Aircraft Structures - Beam - Torsion & Section Idealiation 53

54 Section idealiation consequences Second moment of area Of idealied section A 1 = 640 mm 2 A 2 = 600 mm 2 A 3 = 600 mm 2 1 = 1.2 m 2 = 1.14 m 3 = m A 4 = 600 mm 2 A 5 = 620 mm 2 A 6 = 640 mm 2 A 7 = 640 mm 2 A 8 = 850 mm 2 A 9 = 640 mm 2 M O 4 = m 5 = m 6 = m 7 = m 8 = m Aircraft Structures - Beam - Torsion & Section Idealiation 54

55 Section idealiation consequences Stress distribution Stress assumed constant in each boom As we are in the principal axes A 1 = 640 mm 2 A 2 = 600 mm 2 A 3 = 600 mm 2 A 4 = 600 mm 2 A 5 = 620 mm 2 A 6 = 640 mm 2 A 7 = 640 mm 2 A 8 = 850 mm 2 A 9 = 640 mm 2 M O 1 = 1.2 m 2 = 1.14 m 3 = m 4 = m 5 = m 6 = m 7 = m 8 = m Aircraft Structures - Beam - Torsion & Section Idealiation 55

56 Section idealiation consequences onsequence on open-thin-walled section shearing lassical formula Results from integration of balance equation With So consequences are s s + s s s ds s q + s q ds s q + x q dx xx + x s xx dx x ds dx q q Terms & should account for the direct stress-carring parts onl (which is not the case of shear-carring-onl skin panels) Expression of the shear flux should be modified to account for discontinuities encountered between booms and shear-carring-onl skin panels s xx s s Aircraft Structures - Beam - Torsion & Section Idealiation 56

57 Section idealiation consequences onsequence on open-thin-walled section shearing (2) Equilibrium of a boom of an idealied section T Lecture on beam shearing Direct stress reads x T With & Eventuall x dx s xx q i+1 (no sum on i) q i+1 q i+1 q i q i dx s xx + x s xx q i Aircraft Structures - Beam - Torsion & Section Idealiation 57

58 Section idealiation consequences onsequence on open-thin-walled section shearing (3) Shear flow T x T dx Aircraft Structures - Beam - Torsion & Section Idealiation 58

59 Section idealiation consequences h = 0.4 m Example Idealied U shape Booms of 300 mm 2 - area each Booms are carring all the direct stress Skin panels are carring all the shear flow Shear load passes through the shear center Shear flow? T = 4.8 kn S A 3 A 4 A 2 A 1 b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 59

60 Section idealiation consequences h = 0.4 m Shear flow Simple smmetr principal axes T = 4.8 kn A 3 A 4 Onl booms are carring direct stress S Second moment of area A 2 q A 1 b = 0.2 m Shear flow Aircraft Structures - Beam - Torsion & Section Idealiation 60

61 Section idealiation consequences omparison with uniform U section We are actuall capturing the average value in each branch t T A 3 A 4 T q S A 2 q h A 1 S h O q t t b b q s Aircraft Structures - Beam - Torsion & Section Idealiation 61

62 Section idealiation consequences onsequence on closed-thin-walled section shearing lassical formula With And for anticlockwise q and s So consequences are the same as for open section Aircraft Structures - Beam - Torsion & Section Idealiation 62

63 Section idealiation consequences h m = 0.2m Example Idealied wing section Simpl smmetrical Booms are carring all the direct stress Skin panels are carring all the shear flow Shear load passes through booms 3 & 6 Shear flow? T = 10 kn A A 2 = 250 mm 2 3 = 400 mm 2 A 4 = 100 mm 2 h l = 0.1 m O A 5 = A 4 A 6 = A A 3 7 = A 2 A 1 = 200 mm 2 h r = 0.06 m A 8 = A 1 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 63

64 Section idealiation consequences h m = 0.2m Open part of shear flow Smmetrical section Shear center & centroid on axis I x = 0 (we are in the principal axes) Onl booms are carring direct stress Second moment of area T = 10 kn A A 2 = 250 mm 2 3 = 400 mm 2 A 4 = 100 mm 2 h l = 0.1 m O A 5 = A 4 A 6 = A A 3 7 = A 2 A 1 = 200 mm 2 h r = 0.06 m A 8 = A 1 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 64

65 h m = 0.2m Open part of shear flow (2) hoose (arbitraril) the origin between boom 2 and 3 Section idealiation consequences T = 10 kn A A 2 = 250 mm 2 3 = 400 mm 2 A 4 = 100 mm 2 s 0 h l = 0.1 m O A 5 = A 4 A 1 = 200 mm 2 h r = 0.06 m A 8 = A 1 A 6 = A 3 A 7 = A 2 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 65

66 Section idealiation consequences h m = 0.2m Open part of shear flow (3) hoose (arbitraril) the origin between boom 2 and 3 (2) -q o =28.9 kn. m q o =18 kn. m h l = 0.1 m -q o =32.5 kn. m -1 O h r = 0.06 m q o =22.4 kn. m q o =18 kn. m -q -1 o =28.9 kn. m -1 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 66

67 Section idealiation consequences h m = 0.2m onstant part of shear flow (anticlockwise s, q) If origin is chosen at point O With & -q o =28.9 kn. m q o =18 kn. m h l = 0.1 m -q o =32.5 kn. m -1 O h r = 0.06 m q o =22.4 kn. m q o =18 kn. m -q -1 o =28.9 kn. m -1 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 67

68 Section idealiation consequences h m = 0.2m onstant part of shear flow (2) -q o =28.9 kn. m q o =18 kn. m h l = 0.1 m -q o =32.5 kn. m -1 O h r = 0.06 m q o =22.4 kn. m q o =18 kn. m -q -1 o =28.9 kn. m -1 b l = 0.12 m b m = 0.24 m b r = 0.24 m Aircraft Structures - Beam - Torsion & Section Idealiation 68

69 Section idealiation consequences h m = 0.2m Total shear flow -q o =28.9 kn. m q o =18 kn. m h l = 0.1 m -q o =32.5 kn. m -1 O h r = 0.06 m q o =22.4 kn. m q o =18 kn. m -q -1 o =28.9 kn. m -1 b l = 0.12 m b m = 0.24 m b r = 0.24 m -q=34.2 kn. m -1 -q=5.3 kn. m q=37.8 kn. m -1 O 5 -q=34.2 kn. m q=5.3 kn. m -1 q=12.7 kn. m -1 1 q=17.1 kn. m -1 8 q=12.7 kn. m Aircraft Structures - Beam - Torsion & Section Idealiation 69

70 Section idealiation consequences onsequence on torsion If no axial constraint Torsion analsis does not involve axial stress So torsion is unaffected b the structural idealiation Aircraft Structures - Beam - Torsion & Section Idealiation 70

71 Exercise: Structural idealiation Box section Arrangement of Direct stress carring booms positioned at the four corners and Panels which are assumed to carr onl shear stresses onstant shear modulus Shear centre? 10 mm 10 mm Angles 60 x 50 x 10 mm Angles 50 x 40 x 8 mm 8 mm 300 mm 10 mm 500 mm Aircraft Structures - Beam - Torsion & Section Idealiation 71

72 Lecture notes References Aircraft Structures for engineering students, T. H. G. Megson, Butterworth- Heinemann, An imprint of Elsevier Science, 2003, ISBN Other references Books Mécanique des matériaux,. Massonet & S. escotto, De boek Université, 1994, ISBN X Aircraft Structures - Beam - Torsion & Section Idealiation 72

73 Exercise: Structural idealiation As shear center lies on O b smmetr we consider T Z Section is required to resist bending moments in a vertical plane Direct stress at an point is directl proportional to the distance from the horiontal axis of smmetr, i.e. axis The distribution of direct stress in all the panels will be linear so that we can use the relation below 1 T 2 T b 300 mm O S A 1 A mm 3 s xx 1 x s xx 2 In addition to contributions from adjacent panels, booms areas include the existing spar flanges Aircraft Structures - Beam - Torsion & Section Idealiation 73

74 Exercise: Structural idealiation Booms area 10 mm 1 T 2 10 mm Angles 60 x 50 x 10 mm Angles 10 mm 50 x 40 x 8 mm 500 mm 8 mm 300 mm 300 mm O 4 T S 500 mm 3 B smmetr A 3 = A 2 = 3540 mm 2 A 4 = A 1 = 4000 mm Aircraft Structures - Beam - Torsion & Section Idealiation 74

75 Exercise: Structural idealiation Shear flow Booms area 1 T 2 A 3 = A 2 = 3540 mm 2 A 4 = A 1 = 4000 mm 2 B smmetr I = mm O T S mm 3 As onl booms resist direct stress Aircraft Structures - Beam - Torsion & Section Idealiation 75

76 Exercise: Structural idealiation Open shear flow 1 T s 2 T 300 mm O S > mm Aircraft Structures - Beam - Torsion & Section Idealiation 76

77 Exercise: Structural idealiation onstant shear flow Load through the shear center no torsion 1 T s x 10-3 T 300 mm O T S > x 10-3 T mm 3 With and Aircraft Structures - Beam - Torsion & Section Idealiation 77

78 Exercise: Structural idealiation Total shear flow 1 T s x 10-3 T 300 mm O T S > x 10-3 T mm 3 1 T x 10-3 T 2 T x 10-3 T 300 mm O S > x 10-3 T mm x 10-3 T Aircraft Structures - Beam - Torsion & Section Idealiation 78

79 Exercise: Structural idealiation Shear center Moment around O Due to shear flow Should be balanced b the external loads 1 T x 10-3 T 2 T x 10-3 T 300 mm O S > x 10-3 T mm x 10-3 T Aircraft Structures - Beam - Torsion & Section Idealiation 79

80 Annex 1: Deflection of open and closed section beams Twist due to torsion As torsion analsis remains valid for idealied section, one could use the twist rate losed section M x Open section q p da h ds s In general How can we compute deflection for other loading cases? Aircraft Structures - Beam - Torsion & Section Idealiation 80

81 Annex 1: Deflection of open and closed section beams Smmetrical bending For pure bending we found Therefore the virtual work reads x M a M Let us assume smmetrical axis, M = 0 & pure bending (M constant) onsider a unit applied moment, and s (1) the corresponding stress distribution The energeticall conjugated displacement (angle for bending) can be found b integrating the strain distribution multiplied b the unit-loading stress distribution Aircraft Structures - Beam - Torsion & Section Idealiation 81

82 Annex 1: Deflection of open and closed section beams Virtual displacement Expression for pure bending M x Dq = -u,x (L) M In linear elasticit the general formula of virtual displacements reads s (1) is the stress distribution corresponding to a (unit) load P (1) D P is The energeticall conjugated displacement to P In the direction of P (1) orresponds to the strain distribution e Aircraft Structures - Beam - Torsion & Section Idealiation 82

83 Annex 1: Deflection of open and closed section beams Smmetrical bending due to extremit loading Example smmetrical axis, M = 0 & bending due to extremit load u =0 du /dx =0 M>0 L T x ase of a semi-cantilever beam Eventuall s (1) is the stress distribution corresponding to a (unit) load T (1) Du is the energeticall conjugated displacement to T in the direction of T (1) that corresponds to the strain distribution e Aircraft Structures - Beam - Torsion & Section Idealiation 83

84 Annex 1: Deflection of open and closed section beams General pure bending If neutral axis is a-inclined With It has been shown that x M a M q a Eventuall, as M is constant with x M xx M M x Dq = -u,x (L) M Dq = u,x (L) x M Aircraft Structures - Beam - Torsion & Section Idealiation 84

85 Annex 1: Deflection of open and closed section beams General bending due to extremit loading Bending moment depends on x T Integration b parts Du x T Du x Semi-cantilever beam Aircraft Structures - Beam - Torsion & Section Idealiation 85

86 Annex 1: Deflection of open and closed section beams General bending due to extremit loading (2) Virtual displacement method With s (1) due to the (unit) moments M (1) resulting from the unit extremit loading With D P u displacement in the direction of the unit extremit loading and corresponding to the strain distribution T T Du x Du x Aircraft Structures - Beam - Torsion & Section Idealiation 86

87 Annex 1: Deflection of open and closed section beams General bending due to extremit loading (3) Virtual displacement method (2) After developments, and if D P u is the displacement in the direction of T (1) = 1 In the principal axes I = Aircraft Structures - Beam - Torsion & Section Idealiation 87

88 Annex 1: Deflection of open and closed section beams Shearing Internal energ variation Variation of the work of external forces T g g dx x Defining the average deformation of a section See use of A for thick beams Vectorial value g max Applied shear loading Aircraft Structures - Beam - Torsion & Section Idealiation 88

89 Annex 1: Deflection of open and closed section beams Shearing (2) Virtual work With D T u the average deflection of the section in the direction of the applied unit shear load With q (1) the shear flux distribution resulting from this applied unit shear load With q the shear flux distribution corresponding to the deflection D T u {q(0)} meaning onl for closed sections Aircraft Structures - Beam - Torsion & Section Idealiation 89

90 h = 0.4 m Annex 1: Deflection of open and closed section beams Example Idealied U shape Booms of 300-mm 2 - area each Booms are carring all the direct stress Skin panels are carring all the shear flow Actual skin thickness is 1 mm Beam length of 2 m Shear load passes through the shear center at one beam extremit Other extremit is clamped Material properties E = 70 GPa m = 30 GPa Deflection? T = 4.8 kn S A 2 u =0 du /dx =0 M>0 A 3 A 4 L b = 0.2 m A 1 T x Aircraft Structures - Beam - Torsion & Section Idealiation 90

91 Annex 1: Deflection of open and closed section beams h = 0.4 m Shear flow (alread solved) Simple smmetr principal axes T = 4.8 kn A 3 A 4 Onl booms are carring direct stress S Second moment of area A 2 q A 1 Shear flow b = 0.2 m Aircraft Structures - Beam - Torsion & Section Idealiation 91

92 Annex 1: Deflection of open and closed section beams h = 0.4 m Unit shear flow Same argumentation as before but with T = 1 N T A 3 A 4 S A 2 q A 1 b = 0.2 m Displacement due to shearing Aircraft Structures - Beam - Torsion & Section Idealiation 92

93 Annex 1: Deflection of open and closed section beams Bending Moment due to extremit load Deflection due to extremit load In the principal axes u =0 du /dx =0 M>0 L T x Total deflection No torsion as shear load passes through the shear center Aircraft Structures - Beam - Torsion & Section Idealiation 93