Introduction to Aerospace Engineering Lecture slides Challenge the future 1
Aircraft & spacecraft loads Translating loads to stresses Faculty of Aerospace Engineering 29-11-2011 Delft University of Technology Challenge the future
Learning objectives Student should be able to Calculate the stresses in Fuselage shell due to pressurization (case I) Fuselage shells due to applied torsional load (case II) Wing spars due to applied bending loads (case III) Aircraft loads 2 32
Fuselage structure Pressurization (case I) Revisit stresses in pressure vessel σ circ pr pr = ; σ = t long 2t R dϕ p And the Hooke s law for plane stress or biaxial stress condition ε ε circ long σ circ = E σ long = E σ ν E σ ν E long circ σ long F y ϕ p Aircraft loads 3 32
Fuselage structure Pressurization (case I) With σ circ =2σ long this is ε ε circ long σ circ ν = 1 E 2 σ circ 1 = ν E 2 For a metallic pressure vessel with ν 0.3 this means that ε circ =4.25 ε long Aircraft loads 4 32
Fuselage structure Pressurization (case I) For a sphere under pressure σ = σ = pr t 1 2 2 The strain in a sphere is equal in all directions ε sphere pr 1 = 1 2t E ( ν ) Aircraft loads 5 32
Fuselage structure Pressurization (case I) What ratio t sphere /t cylinder is needed to connect cylinder to sphere in a pressure vessel? Avoid discontinuities in ε circ to avoid these deformation mismatches Thus ε circ = ε sphere! Aircraft loads 6 32
Fuselage structure Pressurization (case I) Derivation ε cylinder Thus R R sphere cylinder = ε sphere ( ν ) 2t 1 sphere = 2 = 1 t cylinder ( 1 ν ) pr cylinder 1 ν pr sphere 1 1 = 1 t E 2 2t E cylinder sphere ( ν ) With R cylinder = R sphere this means t sphere 0.4 t cylinder Aircraft loads 7 32
Fuselage structure Pressurization (case I) Aircraft pressure cabin Role frames & stringers (5-10%) Other disturbances (doublers around cut-outs) Fracture of fuselage could be disastrous: exploding balloon No limit load (formal definition: once in the lifetime of the aircraft), limit load occurs every flight during pressurization of the cabin safety factor is 2 Aircraft loads 8 32
Fuselage structure Pressurization (case I) Aircraft pressure cabin Design pressure differential p d = p cabin - p h Cabin altitude: p cabin related to altitude of 2400-3000 m Safety factor j=2: p ult = 2 p d = 2(p cabin - p h ) Example Pressure differential is 45 kpa Aircraft altitude of h =9050 m cabin altitude of 2440 m Aircraft altitude of h =10350 m cabin altitude of 3050 m Aircraft loads 9 32
Fuselage structure Torsional loading (case II) The axial stresses in a tube under torsion Aircraft loads 10 32
Fuselage structure Torsional loading (case II) F 1 Function of cylinder against torsional loading Rectangular shape d 1 Deforms like d 2 F 2 M 1 Equilibrium if F 1 d 1 = F 2 d 2 = F n d n M 1 = M 2 = M T M 2 Aircraft loads 11 32
Fuselage structure Torsional loading (case II) Torsion causes shear stress & strain τ γ = G M T γ Equilibrium: τ outside = τ cross section M T τ Aircraft loads 12 32
Fuselage structure Torsional loading (case II) Torsion causes shear stress τ γ = G M T γ Equilibrium: τ outside = τ cross section M T Torsional moment τ π τπ 2 M = 2 r t r = 2 r t T circumference circle area circle τ Define q = τ t M T = 2qA Aircraft loads 13 32
Fuselage & wing structure Torsional loading (case II) Thin walled box, stringers not loaded, torsion moment MT Moment around an arbitrary point: q ds a = M T with a ds = 2 A All fractions together: M T = 2qA q = M T 2A Aircraft loads 14 32
Fuselage & wing structure Torsional loading (case II) For a thin walled closed box (enclosed area A) the shear flow is q = M T 2A enclosed area A All cross sections have the same shear-flow q A torsion box does not need to have a circular cross section enclosed area A enclosed area A enclosed area A Aircraft loads 15 32
Fuselage & wing structure Torsional loading (case II) The torsion box will not function at a cut-out A stiff frame with rigid corners around the cut-out is needed Aircraft loads 16 32
Wing structure Bending of wing spar Consider elementary spar Diamond shaped deformation is resisted by sheets Frame exercises a shear stress on the sheets As reaction: sheets exercise shear stress on the frame Assumption: bars of frame very stiff, no deformation Without sheets: Frame is not functioning Aircraft loads 17 32
Wing structure Bending of wing spar If frame functions, sheets in equilibrium High forces: pleat formation However: Structure still functions! Resistance to shear is the resistance to tension and compression under 45 degrees. Relationship: E, ν, G Aircraft loads 18 32
Wing structure Bending of wing spar Break down the elements book keeping F 1x Global Horizontal equilibrium: F - F 1x - F 2x = 0 Vertical equilibrium: F 1y - F 2y = 0 Moment equilibrium: F h + F 1y w = 0 Local Equilibrium in elements F 1y F 2x F 2y Aircraft loads 19 32
Wing structure Bending of wing spar Break down the elements book keeping For shear there is physically no difference but we need a sign convention _ + Aircraft loads 20 32
Wing structure Bending of wing spar In bending, the function of spar webs (shear) is essential! The webs have to be supported on the upper and lower side by caps in order to realize equilibrium webs transfer (external) transverse forces into shear-flows caps transfer shear-flows in normal forces Aircraft loads 21 32
Wing structure Bending of wing spar Global equilibrium (always check!) Horizontal: N N = 0 Vertical: U L 4 4 F + F + F q h = 1 2 3 3 0 Moment: 4 ( ) ( ) N h F l + l + l F l + l F l = 0 U 1 1 2 3 2 2 3 3 3 Aircraft loads 22 32
Wing structure Bending of wing spar Web plate 1 and upper and lower cap 1 Equilibrium of forces in cap 1 F q h = 1 1 1 N = q l = l U 1 1 1 2 1 N q l l L 1 1 1 2 0 F h F = = h q 1 is shear flow in web plate 1 Aircraft loads 23 32
Wing structure Bending of wing spar Web plate 2 and upper and lower cap 2 Equilibrium of forces in cap 2 F = q h q h = q h F 2 2 1 2 1 1 1 2 N = N + q l = l + l U U 2 2 1 2 3 2 1 1 2 N =... = l + l L 1 2 3 h F F + F h F F + F h h Aircraft loads 24 32
Wing structure Bending of wing spar Web plate 3 and upper and lower cap 3 Equilibrium of forces in cap 3 F = q h q h = q h F F 3 3 2 3 1 2 N = N + q l U 4 3 4 U 3 3 F F + F F + F + F 1 1 2 1 2 3 = l + l + l h 1 h 2 h 3 F F + F F + F + F 1 1 2 1 2 3 N =... = l + l + l L 1 2 3 h h h Aircraft loads 25 32
Wing structure Bending of wing spar Shear flow in the webs: Transverse force building up: q n = D h n D = F + F +... + F = F n 1 2 n n 1 n Aircraft loads 26 32
Wing structure Bending of wing spar Shear flow in the webs: Transverse force building up: Normal force in the caps at the location of stringers: q n = D h n D = F + F +... + F = F n 1 2 n n 1 n N m+ 1 1 m = D l h n= 1 n n Next slide What about normal forces in the caps between the stringers? Aircraft loads 27 32
Wing structure Bending of wing spar Equilibrium in A: N N q x D D = 0 ( ) 2 1 N = l x + l U U 2 U 3 x x h h 2 1 Equilibrium in B: ( ) N N q l x = 0 U x U 2 2 2 ( ) 2 1 N = l x + l U 2 1 x D h D h Normal force increases linearly from outboard to inboard! Aircraft loads 28 32
Wing structure Bending of wing spar The bending moment at location x in the spar is: U x x ( ) N h M = D l x + D l 2 2 1 1 We see that that for l 2 dm x dx = D 2 This is valid on every location x on the spar: dm x dx = D x Aircraft loads 29 32
Wing structure Bending of wing spar External forces F n Transverse shear forces known at every location D n q = D n n h n = 1 Normal forces in caps known at every location M m N = h M = D l n+ 1 n F n n n Aircraft loads 30 32
Wing structure Lift, engine and fuselage weight Shear stress in webs q D x x τ = = x t ht x Normal stress in spar caps N x σ = ± = ± x A x x M ha x x In a spar ha x is called the moment of resistance (W) Aircraft loads 31 32
Summary Loads to stresses Three cases studied Fuselage shell due to pressurization Fuselage shells due to applied torsional load Wing spars due to applied bending loads Aircraft loads 32 32