Aircraft Structures Design Example
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1 University of Liège Aerospace & Mechanical Engineering Aircraft Structures Design Example Ludovic Noels Computational & Multiscale Mechanics of Materials CM3 Chemin des Chevreuils 1, B4000 Liège
2 Goal From Specifications Preliminary design Design the rear fuselage of a two-seat trainer/semi-aerobatic aircraft Stringers and Frames Skin thickness Aircraft Structures: Design Example 2
3 Goal (2) Specifications and preliminary design Permit to calculate forces acting on the airplane Fuselage has to resist the induced loading Step 1: Forces acting on the airplane are Aerodynamic forces ( flight envelope) Thrust Self-weight Step 2: The rear fuselage consists into Stringers Frames Skin Rivets All these elements must be calculated to resist the stresses induced by the forces (bending moments, torques and shearing) y Wing lift Inertia Drag z Pitching moment Weight x Thrust Tail load Tail load Aircraft Structures: Design Example 3
4 University of Liège Aerospace & Mechanical Engineering Step 1 Loading acting on the rear fuselage Aerodynamic forces and self-weight
5 Specifications: flight envelope Required flight envelope n 1 = 6.28 V D = m/s V C = 0.8V D = m/s n 2 = 0.75 n 1 = 4.71 n 3 = 0.5 n 1 = 3.14 Representative values for an aerobatic aircraft? Further requirements Additional pitching acceleration allowed at any point of envelope Weight W in [kn] For asymmetric flight: angle of yaw allowed at any point of envelope The angle of yaw increases the overall pitching moment coefficient of the aircraft by / degree of yaw Aircraft Structures: Design Example 5
6 Data - Aircraft Fully loaded aircraft Weight W = kn Moment of inertia about center of gravity G: I q = kg. m 2 Positions of G and of the body drag centers known (see figure) Body drag coefficients C D,B (engine on) = C D,B (engine off) = Engine Maximum horse power: 905 hp Propeller efficiency: 90 % Aircraft Structures: Design Example 6
7 Geometry/Aerodynamics Span b=14.07 m Gross area S = m 2 Data - Wing Aerodynamic mean chord c = 2.82 m Lift and drag coefficients See picture Pitching moment C M = C L Angle of incidence Between fuselage axis and root chord -1.5 Additional pitching moment coefficient Aircraft Structures: Design Example 7
8 Data - Tailplane Geometry/Aerodynamics Span b t = 6.55 m Gross area S t = 8.59 m 2 Aerodynamic centre P Yaw asymmetry of the slipstream asymmetric load on the tailplane Resulting torque M is the mach number & y is in degree C Section CC M tail Loading induced by yaw on the tailplane C Aircraft Structures: Design Example 8
9 Data - Fin Geometry/Aerodynamics Height h F = 1.65 m Area S F = 1.80 m 2 Aspect-ratio A F = h F2 /S F = 1.5 Lift-curve slope a 1 With A the aspect ratio of an equivalent wing: A = (2h F ) 2 /2S F =2 h F2 /S F =2A F = 3 a 1 = 3.3 In yawed flight of angle y Fin has also an incidence of y Aerodynamic loading 3.7 m 1.13m Where y is in rad Centre of pressure of the fin 1.13 m above the axis of the fuselage 3.7 m aft section AA Aircraft Structures: Design Example 9
10 Initial calculation Flight envelope Values n 1 = 6.28 V D = m/s V C = 0.8V D = m/s n 2 = 0.75 n 1 = 4.71 n 3 = 0.5 n 1 = 3.14 V S V SA (stalling speed on point A)? Assumption: Lift from wing only C L,max From Aircraft Structures: Design Example 10
11 Balancing out calculations Loads are calculated for various critical points of the flight envelope Case A Point A y Wing lift z Pitching moment x Thrust Tail load Tail load Engine on Inertia Case A* Point A Engine off Drag Weight Case C Point C Engine off Case D 1 Point D 1 Engine off V S Case D 2 Point D 2 Engine off Aircraft Structures: Design Example 11
12 Balancing out calculations Case A point A/engine on Data (point A) n 1 = 6.28 V = V sa = 96.7 m/s C L,max = 1.38 α 0 L,max=18 C D,W = V S C D,W C L,max Aircraft Structures: Design Example 12
13 Balancing out calculations Case A point A/engine on 0.06 m 0.12 m From measures on drawing Or math, see next slide V A s Fuselage axis T L 1.07 m 6.28 m P V s A G M D B D W P nw Aircraft Structures: Design Example 13
14 Balancing out calculations Case A point A/engine on 0.06 m 0.12 m Math L V A s l a d T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 14
15 Balancing out calculations Case A point A/engine on 0.06 m 0.12 m Methodology Forces that can be directly calculated Trust: T from engine nw: n & W are known Drag (body B, wings W) from V and drag coefficients Pitching moment M from the pitching moment coefficients and the angle of yaw y Pitching moment acceleration Force equilibrium, and moment equilibrium 2 Equations involving P, L Find other forces acting on the rear fuselage Tailplane torque, fin load, fin load torque, total torque T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 15
16 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Trust of the engine Data Maximum horse power 905 Propeller efficiency η = 90 % 1 [hp] = 746 [W] Thrust T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 16
17 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Weight Data Fully loaded weight W = kn Loaded weight T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 17
18 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Drag Data C D,W = C D,B = V = 96.7 m/s S = m 2 Wing drag Body drag T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 18
19 Balancing out calculations Case A point A/engine on Pitching moment Data n 1 = 6.28 V D = m/s V = 96.7 m/s Fully loaded weight W = kn Pitching moment coefficient Maximum for the maximum yaw angle allowed during maneuver Maximum angle of yaw allowed Pitching moment coefficient Wing Maximum pitching acceleration allowed Wing/fuselage incidence Pitching of aircraft due to yaw (in ) Aircraft Structures: Design Example 19
20 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Pitching moment (2) Data V = 96.7 m/s S = m 2 MAC: c = 2.82 m Pitching moment coefficient C M = Pitching moment T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 20
21 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Moments about G Vertical equilibrium T L 1.07 m 6.28 m P V s A M G D W D B P nw Aircraft Structures: Design Example 21
22 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m 2 equations, 2 unknowns P = [N] L = [N] Remark: for other cases, a is not known Requires iterations on a in order to determine C L Should also be done here as V S was computed using C L of wing (10% error) T L 1.07 m 6.28 m P 18 V s A G M D B D W P nw Aircraft Structures: Design Example 22
23 Balancing out calculations Case A point A/engine on Tailplane torque Due to asymmetric slipstream (yaw) Data b t = 6.55 m S t = 8.59 m 2 M tail 3.7 m Fin load Due to yaw Data S F = 1.80 m 2 a 1 = m Aircraft Structures: Design Example 23
24 Balancing out calculations Case A point A/engine on 0.04 m 0.12 m Total torque (rear fuselage) 3.7 m 1.13m M tail T L 1.07 m 6.28 m P V s A M G D W D B M tail P F fin nw Aircraft Structures: Design Example 24
25 Balancing out calculations - End Summary Other cases follow the same method Case n [-] a [ ] P [N] F fin [N] M fus [Nm] A A C V S D D P a-1.5 F fin 1.13 m a-1.5 M tail a V Aircraft Structures: Design Example 25
26 Fuselage Stringers Frames Skin Frames Fuselage loads Preliminary choices Un-pressurized fuselage frames will not support significant loads Frames are required to maintain the fuselage shape nominal in size Combination of stringer and skin will resist self-weight and aerodynamic loads Shear forces Bending moments Torques Aircraft Structures: Design Example 26
27 Geometrical data Fuselage loads Preliminary choices Circular cross-section Simple to fabricate Simple to design Will meet the design requirements Possible arrangement 24 stringers arranged symmetrically spaced at Section AA: ~168 mm Section BB: ~96 mm All stringers have the same cross-sectional area Aircraft Structures: Design Example 27
28 Fuselage loads Data Material: Aluminum alloy Stingers & skin 0.1 % proof stress: [N/mm 2 ] = [MPa] Shear strength: [N/mm 2 ] = [MPa] Self-weight: Assumptions Fuselage weight is from 4.8 to 8.0 % of the total weight Tailplane/fin assembly is from 1.2 to 2.5 % of the total weight Half of the fuselage weight is aft of the section AA The weight distribution varies proportionally to the skin surface area Aircraft Structures: Design Example 28
29 Ø 1.28 m Fuselage loads Data Ø 0.73 m Ø 0.1 m Assumptions on geometry Rear fuselage is uniformly tapered A C B D A B C 2.13 m 2.44 m 4.57 m D CC is a section midway AA and BB. The center of gravity of the tailplane/fin assembly has been estimated to be 4.06 m from the section AA on a line parallel to the fuselage centre line Aircraft Structures: Design Example 29
30 Fuselage loads Data Ø 1.28 m Ø 0.73 m Ø 0.1 m Data Weight of rear fuselage: 1198 [N] Skin area of rear fuselage: 9.91 [m 2 ] A C B D Self-weight / m of the fuselage A B C 2.13 m 2.44 m D 4.57 m Aircraft Structures: Design Example 30
31 Fuselage loads Shear force and bending moment due to self-weight Self-weight induces Shear forces : SF Bending moments : BM SF and BM are calculated by equilibrium ( MNT ) 4.06 m 4.57 m 2.13 m 2.44 m m A C B D [N/m] [N/m] [N/m] W tailplane+fin =674 [N] Resultant = W rear fuselage = 1198 [N] 38.0 [N/m] Aircraft Structures: Design Example 31
32 Fuselage loads Shear force and bending moment due to self-weight Transform self-weight into A triangular repartition: q 1 (x) And a constant linear force: q m 4.57 m q 2 =38 [N/m] A q 1 (x) B 0 [N/m] D [N/m] W tailplane+fin =674 [N] Aircraft Structures: Design Example 32
33 Fuselage loads Shear force and bending moment due to self-weight 0.06 m 0.12 m Effect of load factor T L 1.07 m 6.28 m P V s A M G D W D B M tail P F fin The self weight is multiplied by the load factor n Forces are not applied in the cross section plane Forces are (a-1.5 ) inclined with this section Will be multiplied by cos (a-1.5 ) later Rigorously, an axial loading should also be considered Distance along fuselage axis is multiplied by cos (a-1.5 ) When computing bending moment We do not compute the fuselage compression Should be done and risk of buckling avoided nw Aircraft Structures: Design Example 33
34 Fuselage loads Shear force and bending moment due to self-weight Reactions at section AA SF A 4.57 m 4.06 m Q 2 (resultant) q 2 =38 [N/m] BM A A B D q 1 (x) Q 1 (resultant) W tailplane+fin =674 [N] Aircraft Structures: Design Example 34
35 Fuselage loads Shear force and bending moment due to self-weight Reactions at section CC SF C m 4.57 m 4.06 m q 2 =38 [N/m] A BM C C B D q 1 (x) Q 1 Q 2 W tailplane+fin =674 [N] Aircraft Structures: Design Example 35
36 Fuselage loads Shear force and bending moment due to self-weight Reactions at section BB 4.57 m 4.06 m 2.13 m SF B 2.44 m q 2 =38 [N/m] A BM B B D q 1 (x) Q 2 Q 1 W tailplane+fin =674 [N] Aircraft Structures: Design Example 36
37 Total shear forces, bending moments and torque Resultant forces in each section Example section AA M x M y T y M z 3.7 m T z P a-1.5 F fin 1.13 m SF A a-1.5 BM A x M tail a V nw Aircraft Structures: Design Example 37 z
38 Total shear forces, bending moments and torque Resultant forces in each section (2) Case n [-] a [ ] P [N] F fin [N] M fus [Nm] A A M x M y T y C D D M z Example case A, section AA SF A = 1872 n [N] & BM A = 4693 n cos(a-1.5 ) [Nm] T z Aircraft Structures: Design Example 38
39 Total shear forces, bending moments and torque Table of sections loading SF A = 1872 n [N] & BM A = 4693 n cos(a-1.5 ) [Nm] SF C = 1409 n [N] & BM C = 2959 n cos(a-1.5 ) [Nm] SF B = 1059 n [N] & BM B = 1651 n cos(a-1.5 ) [Nm] Sect Case T y [N] T z [N] M y [Nm] M z [Nm] M x [Nm] AA A A C D D CC A A C D D BB A A C D D Aircraft Structures: Design Example 39
40 University of Liège Aerospace & Mechanical Engineering Step 2 Design of the rear fuselage Stringers, frames, skin and rivets
41 Fuselage design calculation Material and approach 2 types of design are possible Elastic design Allowed stress = 0.1% proof stress / s s : safety factor = 1.5 Ultimate load design Ultimate stresses Actual load multiplied by an ultimate load factor For linear systems : both designs give the same results We use the elastic design as 0.1 proof stress is known Aircraft Structures: Design Example 41
42 Stringers section: Data 0.25D 0.353D 0.433D 0.483D 0.5D Frames Un-pressurized fuselage frames will not support significant loads Frames are required to maintain the fuselage shape nominal in size Circular cross-section 24 stringers arranged symmetrically and spaced at around Section AA l ~168 mm Section BB l ~96 mm All stringers have the same cross-sectional area l y D D y z z Aircraft Structures: Design Example 42
43 0.25D 0.353D 0.433D 0.483D 0.5D Stringers section Direct stress Induced by M z and M y Obtained from (as I yz = 0) Unknown B: the area of the stringers in a section B should be chosen such that Second moments of area y D D z Aircraft Structures: Design Example 43
44 Stringers section Values of M x and M y Stress Sect Case T y [N] T z [N] M y [Nm] M z [Nm] M x [Nm] AA A A Worst case Case D1 (dive) M z and M y have same sign y and z of opposite sign C D D CC A A C D D D BB A M z z M y y A C D D Aircraft Structures: Design Example 44
45 Stringers section Calculate Bσ xx Sect. AA CC BB Data D 1.28 m D 1.01 m D 0.73 m M y 42 kn. m M y 29 kn. m M y 16 kn. m For each Section M & D change Stringer y & z change Determine minimal value of stringers area B such that Stringers n -y [m] M z 55 kn. m M z 39 kn. m M z 23 kn. m z [m] Bs xx [kn] -y [m] z [m] Bs xx [kn] -y [m] z [m] Bs xx [kn] Max. Bs xx [kn] Min. B [mm 2 ] Aircraft Structures: Design Example 45
46 Stringers and frames From calculation B min (AA) > B min (CC) > B min (CC) Lighter stringer between CC and BB type A stringers : B > 57.7 mm² type B stringers : B > 51.4 mm² Aircraft Structures: Design Example 46
47 Stringers and frames Stringer type Z-section Type A stringer: B = 58.1 > 57.7 mm² Type B stringer: B = 51.9 > 51.4 mm² Aircraft Structures: Design Example 47
48 Frames Frames Non load bearing Must be of sufficient size to be connected with stringers (AA/BB/CC sections) Use of brackets Must be of sufficient size to allow to be cut So stringers can pass trough them Aircraft Structures: Design Example 48
49 Skin thickness Skin must resist shear flow due to Shear loads T y, T z & Torque M x Calculate the shear flow D M x At each boom there is a discontinuity T y y 10 T z z As stringers have constant area in one section and as I yy = I zz = 3BD Aircraft Structures: Design Example 49
50 Skin thickness Shear flow 0.25D 0.353D 0.433D 0.483D 0.5D Shear flow due to T Z Following equations D q D T z y 7 6 z By symmetry (no torque) q 6 7 = - q Aircraft Structures: Design Example 50
51 Skin thickness Shear flow Shear flow due to T Z (2) Final form of qd/t z D 7 6 z T z q y Aircraft Structures: Design Example 51
52 Skin thickness Shear flow 0.25D 0.353D 0.433D 0.483D 0.5D Shear flow due to T y Following equations D q D T y y But we also have 7 6 z By symmetry (no torque) q = - q Aircraft Structures: Design Example 52
53 Skin thickness Shear flow Shear flow due to T y (2) Final form of qd/t y D 7 6 z T y q y Aircraft Structures: Design Example 53
54 Skin thickness Shear flow caused by torque 0.25D 0.353D 0.433D 0.483D 0.5D Shear flow due to torque D q D M x y 7 6 z Aircraft Structures: Design Example 54
55 Skin thickness Shear flow caused by torque Maximum shear flow As T z >0, T y < 0 & M x < 0 Maximum shear flow is in a skin panel between stringers 12 and 18 Example: D 7 qd/t z 6 z T z y D 7 6 z T y qd/t y y Aircraft Structures: Design Example 55
56 Maximum shear flow (2) Equations Skin Thickness Maximum shear flow Aircraft Structures: Design Example 56
57 Skin Thickness Maximum shear flow Maximum shear flow (3) Critical case: D1 Sect Case T y [N] T z [N] M y [Nm] M z [Nm] M x [Nm] AA A A C D D CC A A C D D BB A A C D D Aircraft Structures: Design Example 57
58 Skin Thickness Maximum shear flow Computation See Table Minimum skin thickness From Sect. AA CC BB Data D 1.28 m D 1.01 m D 0.73 m T y -15 kn T y -15 kn T y -15 kn T z 15 kn T z 12 kn T z 12 kn M x -41 kn. m M x -40 kn. m M x -40 kn. m n q [kn. m -1 ] q [kn. m -1 ] q [kn. m -1 ] Panels But must support rivets 1mm skin thickness is chosen Max. q [kn. m -1 ] Min. t [mm] Aircraft Structures: Design Example 58
59 Rivets size Skin/stringers 0.25D 0.353D 0.433D 0.483D 0.5D What are the forces acting on the rivet? At a stringer, we found for T z This corresponds to The shear flow balanced by All the rivets linking the skin to the stringer Therefore the shear load per unit stringer length acting on the rivets fixing the skin to the stringer i is R i D 7 q D T z z y Maximum between stringers 6 and 12 T y <0 & T z >0 y <0 & z >0 R Critical case is still D1 Remark: the torque does not lead to a discontinuity in the shear flow Aircraft Structures: Design Example 59
60 Rivets size Skin/stringers Results Maximal load: 4.91 kn per unit stringer length Sect. AA CC BB Data D 1.28 m D 1.01 m D 0.73 m T y -15 kn T y -15 kn T y -15 kn T z 15 kn T z 12 kn T z 12 kn n -y [m] z [m] -R [kn/m] -y [m] z [m] -R [kn/m] -y [m] z [m] -R [kn/m] Stringers Max. R [kn/m] Aircraft Structures: Design Example 60
61 Rivets size Skin/stringers Rivets Maximal load: Rivets 4.91 kn per unit stringer length For 2.5 mm diameter countersunk rivets with skin thickness 1.0 mm Allowable load in shear: 668 N The number of rivets/m is given by R This corresponds to a rivet pitch of 125 mm Too large: does not ensure structural rigidity We choose 25 mm rivet pitch: 40 rivets/m Aircraft Structures: Design Example 61
62 Rivets size Frames/skin What are the forces acting between frames & stringers? These forces correspond to the direct stresses in the stringers Already calculated Maximal forces Section AA: 9 kn Section CC: 8 kn Section BB: 6.4 kn Sect. AA CC BB Data D 1.28 m D 1.01 m D 0.73 m n -y [m] M y 42 kn. m M y 29 kn. m M y 16 kn. m M z 55 kn. m M z 39 kn. m M z 23 kn. m z [m] Bs xx [kn] -y [m] z [m] Bs xx [kn] -y [m] z [m] Bs xx [kn] Stringers Max. Bs xx [kn] Min. B [mm 2 ] Aircraft Structures: Design Example 62
63 Rivets size Frames/skin Number of rivets between the skin and frames Section AA Maximal forces: 9 kn This force is resisted by the rivets between the skin and frame Distance available for one stringer l = m Resistance of one rivet: 668 N l y Section CC z Section BB Distance between the rivets: 10 mm for all frames Aircraft Structures: Design Example 63
64 University of Liège Aerospace & Mechanical Engineering Fuselage design - End Plans and Figures
65 Design of the rear fuselage Aircraft Structures: Design Example 65
66 Rear fuselage: details (A14A) 10 mm Aircraft Structures: Design Example 66
67 Rear fuselage: details (A14B) 10 mm Aircraft Structures: Design Example 67
68 Rear fuselage: details (A14C) 10 mm Aircraft Structures: Design Example 68
69 Rear fuselage: details (A14D) 10 mm Aircraft Structures: Design Example 69
70 Rear fuselage: details (A14E) 10 mm Aircraft Structures: Design Example 70
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