wing stress check notes lw-001 · 2020. 9. 16. · 57-003 r1 flight envelope 6. mmpds-03 metallic...

SPORT CAMPER Report N°: LW‐001

Technical Report Rev: ‐3

Model: LoCamp Date: 01/11/12

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WING STRESS CHECK NOTES

LW-001

Prepared by: Camillo PEDETTI Edoardo PORRINI

Verified by :




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Table of contents 1. References ...................................................................................................................................... 3 2. List of Abbreviations ..................................................................................................................... 4 3. General ........................................................................................................................................... 7 4. Wing geometrical data ................................................................................................................... 9 5. Wing loads definition .................................................................................................................. 10 6. Front spar sizing .......................................................................................................................... 22 7. Aft Spar sizing ............................................................................................................................. 52 8. Center frame fwd spar analysis .................................................................................................... 74 9. Center Frame Aft Spar Analysis .................................................................................................. 84 10. FEM Loading ............................................................................................................................. 85 11. Ribs analysis ............................................................................................................................ 114 12. Flap and Aileron Hinge Analysis ............................................................................................ 127 13. Wing Tanks Check ................................................................................................................. 135 14. Wing drag analysis .................................................................................................................. 139

TABLE of REVISIONS

Rev. Description Author Date

0 Original issue See cover Sept. 2011

1 Additional analysis added C. Pedetti 01/02/12

2 Comments from authority introduced C. Pedetti Aug 2012

3 Revised buckling analysis of spars web, additional analysis added as per authority request

C. Pedetti 15/11/12




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1. References

1. F 2245-04-06 Standard Specifications for Design and Performance of a Light Sport Airplane 2. EASA CS-VLA-2009 3. 57-001 R0: Aerodynamics Characteristics Calculation (partially applicable) 4. 57-002 R0 - Aerodynamics Characteristics 4416 profile 5. 57-003 R1 Flight envelope 6. MMPDS-03 Metallic Material Properties Development and Standardization 7. Bruhn: Analysis and Design of Flight Vehicle Structures 8. SAWE Weight Engineers Handbook




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2. ListofAbbreviationsDimensions

Length l, L mm Height h mm Diameter D, d mm Thickness t mm Radius r mm Pitch p mm Width W, w mm Eccentricity, edge distance e mm Area, cross section A mm^2 Constant K, k - Celsius degrees °C F degrees °F Angle degrees

Subscripts

Maximum max Minimum min VM Von Mises Total tot Tension, Torsion t Compression c Shear s, sh Bearing b, br Ultimate u Ultimate Level U.L. Yield y Limit Level L.L. Critical cr Allowable all Resultant res Average avg

Materials properties

Ultimate tensile strength tu-Ftu MPa Tensile Yield strength ty-Fty MPa Compressive Yield strength cy-Fcy MPa Ultimate Shear strength su-Fsu MPa Ultimate bearing strength bru-Fbru MPa Bearing Yield strength bry-Fbry MPa Young modulus –tension E MPa Young modulus –compression Ec MPa Shear modulus G MPa Poisson ratio nu,, - Elongation , delta % Strain allowable (composite) _t, _c, _sh




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Sections properties

Area A mm^2 Moment of inertia I mm^4 Polar moment of inertia Ip mm^4 Torsion constant J mm^4 Radius of gyration mm Static moment Q mm^3

Abbreviations

dia., d, D diameter MS Margin of Safety RF Reserve Factor FQF Fatigue Quality Factor Pr, Pr Preload TT Tightening Torque FEM Finite Element Model 2D Two dimensional 3D Three dimensional config. Configuration Proto Prototype I/F Interface O/B OutBoard TBD To Be Defined TBC To Be Confirmed TBW To Be Written K,k Kilo (1.0*e+3) Ff fitting factor A/C Aircraft ID,id identification N Newton m Meter mm Millimeter Pa Pascal (N/m^2) MPa Mega Pascal (N/mm^2) CSK Countersunk bck buckling R.T. Room Temperature Lbs Pounds Psi Pound per square inch (Ksi =10^3psi) Cm Centimetres GFC Fiber Composite P.T. Pull Through S.F. factor of safety Rt tensile/axial ratio Rs shear ratio Rb bending ratio ext external LC Load Case HI High




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VM Von Mises fwd forward Eigenvalue FI Failure Index SPC Single Point Constraint CoG Centre of Gravity i.a.w. in accordance with

Loads

Force P, F N, kN Moment M Nmm, Nm Bending Moment BM Nmm, Nm Axial load P, F N, kN Shear Load Q, S N, kN Torque T Nm Pressure p, dp N/mm^2, N/m^2

Stress and strain

Direct stress MPa Shear stress MPa Principal stress pr MPa Von Mises stress VM MPa Strain Shear strain

Composite notations

Longitudinal modulus E11 MPa Transverse modulus E22 MPa In plane shear modulus G12 MPa Longitudinal tensile strength F1t MPa Longitudinal compressive strength F1c MPa Transverse tensile strength F2t MPa Transverse compressive strength F2c MPa Unidirectional UD Coefficient of thermal expansion longitudinal _11 mm/mm°C Coefficient of thermal expansion transversal _22 mm/mm°C Interlaminar shear strength ILSS MPa




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3. General This report is dealing with the wing preliminary sizing for aircraft LoCamp, according to F 2245-04-06. Standard Specifications for Design and Performance of a Light Sport Airplane. The wing preliminary sizing is based on some assumptions:

To be conservative the wing weight is not subtracted from the MTOW of 600 Kg. The total wing load is spread on the two spars as function of their position with respect to the wing

center of pressure at 3 degrees of AOA . Chapter 6 to 9 is preliminary sizing focused mainly to front and rear spars and the front and rear fuselage truss sections connecting the wing to the fuselage. Chapter 10 is focused on skin, both upper and lower, to check against buckling behavior. Chapter 11 is focused on ribs, to check instability behaviour.

3.1. CommonlyUsedFormulas Sign conventions generally accepted in their use are quantities associated with tension action (loads, stresses, strains, etc.) are usually considered as positive, and quantities associated with compressive action are considered as negative. When compressive action is of primary interest, it is sometimes convenient to identify associated properties with a positive sign.

3.1.1. SimpleUnitStresses ft = P / A (tension) fc = P / A (compression) fb = My / I = M / Z (bending) fs = S / A (average direct shear stress) fx = SQ / Ib (longitudinal or transverse shear stress) fx = Ty / Ip (shear stress in round tubes due to torsion) fs = (T/2At) (shear stress due to torsion in thin-walled structures of closed section. Note that A is the area enclosed by the median line of the section.) fA = BfH; fT = BfL (axial and tangential stresses, where B = biaxial ratio) fA = fc + fb (compression and bending)

fsmax [fs (fn ) ] (compression, bending, and torsion)

/




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fn max = fn/2 + fs max

3.1.2. Deflections(axial) e = δ / L (unit deformation or strain) E = f/e (This equation applied when E is obtained from the same tests in which f and e are measured.) δ = eL = (f / E)L = PL / (AE) (This equation applies when the deflection is to be calculated using a known value of E.)

3.2. Materialproperties Ref. ALCOA data-sheet




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4. Winggeometricaldata

Wing airfoil: NACA 4416 Wing surface area: 14.91 m2 (ref. 1) Wing span: 9.1 m (ref. 1) Aspect ratio: 4.8 (ref. 1) Taper ratio: 1 (ref. 1) Wing aerodynamic center location (% MAC): XAC = 0.244m MTOW = 600 Kg nZ = +4, -2g




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5. Wingloadsdefinition Using the above data, the following spreadsheet provides for 16 wing stations the shear and the bending moment that subsequently will be shared between the two spars as function of their relative location with respect the wing center of pressure.

Figure 5-1: Wing assy, view from top

Figure 5-2: Wing assy, view from bottom




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Figure 5-3: Wing unit load and shear




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Figure 5-4: Wing bending moment

5.1. Loadsdefinitiononspars The assumption made is that the shear and bending moment distribution obtained on the previous pages will be shared between the two spars as function of their location between wing center of pressure. To share the shear and moment, a beam on two supports is considered where the two supports are the spars while the loads are applied at the wing center of pressure for an AOA of 3 deg. In the following pages the pressure distribution every 3 degrees from 0 degree to 15 degree is provided, then the assumption made is that at AOA of 3 degree the max Nz is reached.




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Figure 5-5: Pressure distribution at 0 deg. AOA




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16





17





18





19





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Figure 5-11: Spars and C.P. location from leading edge

591

FrontsparRearspar

265 819

1594




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Beam on two supports:

Rla Rlp

P

326 493

P =1200 Kg (L.L.) max shear Equilibrium equations: P-RLA – RLP = 0 (shear equilibrium) -P*326 + RLP * 819 = 0 (moment equilibrium) Then: 1200 = RLA + RLP -1200*326 = - RLP * 819 then: RLP = 1200*326/819 = 478 Kg RLA = 1200 – 478 = 722 Kg The ratio between the fwd spar load and the aft spar load is: Kspar = 722/1200 = 0.6 Then it is considered that the front spar carry 60% of the total wing loads, while the remaining 40% loads will be carried out by the rear spar.




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6. Frontsparsizing The shear and moment applied on front spar are obtained from the load reported on §5 multiplying shear and moment by 0,6.

RIB WING STA WING LOAD SHEAR BENDINGNo. mm Kg/mm Kg Kgmm

0 0 0,263736264 720 16380001 370 0,263736264 661 13824322 443 0,263736264 650 13345673 885 0,263736264 580 10627694 1010 0,263736264 560 9915115 1135 0,263736264 540 9227256 1260 0,263736264 521 8564127 1589 0,263736264 469 6936948 1918 0,263736264 416 5481049 2247 0,263736264 364 419642

10 2576 0,263736264 312 30830811 2905 0,263736264 260 21410312 3234 0,263736264 208 13702613 3563 0,263736264 156 7707714 3892 0,263736264 104 3425615 4221 0,263736264 52 856416 4550 0,263736264 0 0

FWD SPAR 60%OF TOTAL WING LOAD

The above shear and bending values have been rounded to have no decimal places per the following criteria: e.g.: 720,4 720 720,5 721 720,6 721




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R H upr joint = 1382432/194 = -7126 Kg (L.L.) Using bending moment at STA 370 R H lwr joint = -R H upr joint = 7126 Kg (L.L.) R V upr joint = 661/2 = 330,5 Kg (L.L.) R V lwr joint = R V upr joint = 330,5 Kg (L.L.) The upper joint is made by 2 bolts type AN8 with ref. (7) the allowable shear is: Ps all = 14730 lbf = 6682 Kg The bolts works in double shear then:

RH upr

RH lwr

RV upr

RV lwr




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Ps app = (7126/2)/2 = 1782 Kg (L.L.) horizontal shear Ps app = (330,5/2)/2 = 83 Kg (L.L.) vertical shear Ps app RES = (17822+832)0,5 = 1784 Kg (L.L.) shear resultant M.S. = (6682/(1784*1,5*1,15))-1 = 1,17 (U.L.) max nz load case

6.1. LugAnalysis The spar fitting is now checked as a lug (inboard bolt hole only). The material considered for the spar cap and spar web is: Aluminum 6082-T6 From EN-AW-6082 that is more conservative with respect the ALCOA data-sheet properties provided on paragraph 3.2, the mechanical properties of this aluminum are:

Ftu = 290 MPa Fty = 250 MPa Fsu = Ftu * 0.577 = 290 * 0.577 = 167 MPa E = 10%

Lug geometry:

Lug applied loads: FH = 7126/2 = 3563 Kg (L.L.) 52430 N (U.L.) FV = 330,5/2 = 165,25 Kg (L.L.) 2432 N (U.L.)

Fv

Fh




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LUG DATA & ALLOWABLE

W, (mm) = 35 DWG No.: fwd spar D, (mm) = 12 Material: 6082-T6 t, (mm) = 22 Ftu, (MPa) = 290

e, (mm) = 17.5 Ftu; min, (MPa) = 290 c, (mm) = 0 Fty, (MPa) = 250

TENSION ALLOWABLE BEARING ALLOWABLE

At , (mm2) = 506 Abr,(mm2) = 264 Ktu = .8186401 Pbru, (N) = 94656

Ptu, (N) = 120127.3

TRANSVERSE LOAD ALLOWABLE A1, (mm2) = 291.6711 Aav, (mm2) = 277.5309

A2, (mm2) = 253 BETA DEG. = 0 A3, (mm2) = 253 Ktru = .9913455

A4, (mm2) = 291.6711 Ptru, (N) = 75897.41

Paxial = 52430 N (U.L.) Ptransv = 2432 N (U.L.)

Ra = 52430/94656 = 0.55 0.64 with F.F.

Rtr = 2432/75897.41 = 0.03 0.036 with F.F.

M.S. = (1/(Ra1.6 + Rtr

1.6)0.625)-1 ---------------------------------------------------------------

M.S. =------ WITH F.F. ----------> 0.55 (U.L.) Nz max ---------------------------------------------------------------

Bushing Yield Bearing Allowable Pbry, N = 0

LUG Yield Allowable Py, N = 71971.69 The outboard fitting hole is checked as a net section including the bolt load plus the by-pass load. For the bearing see the above allowable calculated for the inboard lug hole. The total load at the outboard hole net section is: PTOTAL = 52430 x 2 = 104860 N (U.L.) The net section area is: Anet = (35-12)*22 = 506 mm2 The max net stress is: ft = 104860/506 = 207 MPa (U.L.)

M.S. = (290/207)-1 = 0.40 (U.L) max nz




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6.2. Fwdsparbucklinganalysis

Figure 6-12: Web panels dimensions and numbering





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The buckling coefficients and allowable are obtained following the methods reported on ref. 7.

Non-buckling beam webs analysis

k t fsh D h fsc C C' b H/t C/t

mm MPa mm mm MPa mm mm mm

0,73 1 42,2 64 194 84,4 100,5 82 164,5 194 100,5

q all = 88,5 MPa

The panel No. 31 that buckles at ultimate load has been checked as fully simply supported that is conservative. Because permanent buckling and buckling at limit load does not occur, it is considered that a load redistribution on spar caps will help at ultimate load for the web buckling.




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6.3. Sparmainsectionsanalysis To be conservative all the sections analysis between STA Y370 and STA Y 1589 are done considering the single web thickness of 1 mm instead of double web thk. =2 mm.

Figure 6-16: Spar sections




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Figure 6-17: Spar section




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Figure 6-18: Spar sections




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6.3.1. CheckatwingSTAY370




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Fz 0 N Axial

Sx 0 N Shear_X dir.

Sy 9915 N Shear_Y dir.

Mx -20736480 Nmm Moment_X dir.

My 0 Nmm Moment_Y dir.

Material :

Allowable Loads Ftu 290 MpaFty 250 Mpa

Stress calculated using global axes (specify global or principal axes)

Axial Shear Bending Von Mises M.S.Point MPa MPa MPa MPa (U.L.)

1 0,00 5,84 158,48 158,80 0,832 0,00 5,84 110,04 110,50 1,623 0,00 5,84 110,04 110,50 1,624 0,00 5,84 -110,04 110,50 1,625 0,00 5,84 -110,04 110,50 1,626 0,00 5,84 -158,48 158,80 0,837 0,00 5,84 -158,48 158,80 0,838 0,00 5,84 -110,04 110,50 1,629 0,00 5,84 -110,04 110,50 1,62

10 0,00 5,84 110,04 110,50 1,6211 0,00 5,84 110,04 110,50 1,6212 0,00 5,84 158,48 158,80 0,83

Applied Loads on C.G.

Al 6082-T6

Section Stress




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Figure 18- Considered section for crippling analysis

********************************************** * SECTION INERTIAL PROPERTIES CALCULATION *

**********************************************

Fwd spar sta Y370 SECTION NR. 1

SECTION DATA

1 4.00 1.00 24.00 0.50 12.00 0.00 2 4.00 24.00 1.00 12.50 24.00 0.00 3 4.00 1.00 10.00 24.00 19.00 0.00

# TYPE DIM 1 DIM 2 A Y AY AY2 IOX

1 4. 1.00 24.00 0.2400E+02 0.1200E+02 0.2880E+03 0.3456E+04 0.1152E+04 2 4. 24.00 1.00 0.2400E+02 0.2400E+02 0.5760E+03 0.1382E+05 0.2000E+01 3 4. 1.00 10.00 0.1000E+02 0.1900E+02 0.1900E+03 0.3610E+04 0.8333E+02

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# X AX AX2 IOY AXY IOXY

1 0.5000E+00 0.1200E+02 0.6000E+01 0.2000E+01 0.1440E+03 0.0000E+00 2 0.1250E+02 0.3000E+03 0.3750E+04 0.1152E+04 0.7200E+04 0.0000E+00 3 0.2400E+02 0.2400E+03 0.5760E+04 0.8333E+00 0.4560E+04 0.0000E+00

SECTION AREA A = 58.00 mm2 CENTROID POSITION X = 9.52 mm Y = 18.17 mm MOMENTS OF INERTIA IX = 0.297361E+04 mm4 IY = 0.541732E+04 mm4 PRODUCT OF INERTIA IXY = 0.187283E+04 mm4 TORSIONAL INERTIA <THIN WALLED OPEN SECTION> 0.193333E+02 mm4 RADIUS OF GYRATION X = 7.16 mm Y = 9.66 mm PRINCIPAL INERTIA AXES ROTATION (CCW) = 28.44 DEGREES PRINCIPAL MOMENTS OF INERTIA IXP = 0.195930E+04 mm4 IYP = 0.643162E+04 mm4 RADIUS OF GYRATION XP = 5.81 mm Y = 10.53 mm

CRIPPLING ALLOWABLES:

EL B T FCC FCC(Y) FCC(U) 0/1

1 24.00 1.00 109.86 109.86 109.86 1 2 24.00 1.00 241.51 241.51 241.51 0 3 10.00 1.00 221.71 221.71 221.71 1

MATERIAL: CRIPPLING GENERAL SOL. - FORMED FCY = 250.00 N/MM2 FCU = 290.00 N/MM2

E = 70000. N/mm2 FCC = 183.62 N/mm2

FCC<Y>= 183.62 N/mm2 FCC= 183.62 N/mm2

M.S. = (Fcc/fc)-1 = (183.6/158.5)-1 = 0.16 (U.L.) L.C. Nz max




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Max Nz

U.L. (specify L.L. or U.L.)

Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 6,50 167,12 167,50 0,732 0,00 6,50 116,03 116,58 1,493 0,00 6,50 116,03 116,58 1,494 0,00 6,50 -116,03 116,58 1,495 0,00 6,50 -116,03 116,58 1,496 0,00 6,50 -167,12 167,50 0,737 0,00 6,50 -167,12 167,50 0,738 0,00 6,50 -116,03 116,58 1,499 0,00 6,50 -116,03 116,58 1,49

10 0,00 6,50 116,03 116,58 1,4911 0,00 6,50 116,03 116,58 1,4912 0,00 6,50 167,12 167,50 0,73

Load Case Id. :

Analysis performed at :


Al 6082-T6

Section Stress

With ref. page 33 the spar web flange has always the same dimensions therefore the crippling allowable is again: Fcc = 183.6 MPa





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39

Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 7,20 176,84 177,28 0,642 0,00 7,20 122,79 123,42 1,353 0,00 7,20 122,79 123,42 1,354 0,00 7,20 -122,79 123,42 1,355 0,00 7,20 -122,79 123,42 1,356 0,00 7,20 -176,84 177,28 0,647 0,00 7,20 -176,84 177,28 0,648 0,00 7,20 -122,79 123,42 1,359 0,00 7,20 -122,79 123,42 1,35

10 0,00 7,20 122,79 123,42 1,3511 0,00 7,20 122,79 123,42 1,3512 0,00 7,20 176,84 177,28 0,64



Al 6082-T6

Section Stress

Load Case Id. :


M.S. = (Fcc/fc)-1 = (183.6/176,84)-1 = 0.04(U.L.) L.C. Nz max




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41






42


This section shall be checked because the web reinforcing ends at this station, therefore the spar caps at the level of the web reinforcing are cutted at this station too. The analysis is done with the hypothesis that the first layer of caps strip of 1 mm is not effective. The shear and bending of STA1918 are taken.




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Max Nz

L.L. (specify L.L. or U.L.)

Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :



Axial Shear Bending Von Mises M.S.Point MPa MPa MPa MPa (L.L.)

1 0,00 7,26 134,62 135,21 0,852 0,00 7,26 93,47 94,31 1,653 0,00 7,26 93,47 94,31 1,654 0,00 7,26 -93,47 94,31 1,655 0,00 7,26 -93,47 94,31 1,656 0,00 7,26 -134,62 135,21 0,857 0,00 7,26 -134,62 135,21 0,858 0,00 7,26 -93,47 94,31 1,659 0,00 7,26 -93,47 94,31 1,65

10 0,00 7,26 93,47 94,31 1,6511 0,00 7,26 93,47 94,31 1,6512 0,00 7,26 134,62 135,21 0,85


Al 6083-T6

Section Stress

Load Case Id. :



M.S. = (Fcc/fc)-1 = (183.6/134,6)-1 = 0.36 (U.L.) L.C. Nz max




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The above M.S. provide evidence that the change in thickness of the spar web in correspondence of wing STA 1939 and, as consequence, the cut of the spar caps leaving the total thickness of the spar caps equal to 10 mm instead of 11 mm does not produce problems to the spar structural behavior.





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Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 8,08 137,35 138,07 1,102 0,00 8,08 99,27 100,25 1,893 0,00 8,08 99,27 100,25 1,894 0,00 8,08 -73,75 75,06 2,865 0,00 8,08 -73,75 75,06 2,866 0,00 8,08 -111,83 112,70 1,577 0,00 8,08 -111,83 112,70 1,578 0,00 8,08 -73,75 75,06 2,869 0,00 8,08 -73,75 75,06 2,86

10 0,00 8,08 99,27 100,25 1,8911 0,00 8,08 99,27 100,25 1,8912 0,00 8,08 137,35 138,07 1,10

Load Case Id. :



Al 6082-T6

Section Stress






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Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 7,80 69,62 70,91 2,532 0,00 7,80 -69,62 70,91 2,533 0,00 7,80 -69,62 70,91 2,534 0,00 7,80 -63,67 65,08 2,845 0,00 7,80 -63,67 65,08 2,846 0,00 7,80 -69,02 70,33 2,557 0,00 7,80 -69,02 70,33 2,558 0,00 7,80 69,02 70,33 2,559 0,00 7,80 69,02 70,33 2,55

10 0,00 7,80 63,67 65,08 2,8411 0,00 7,80 63,67 65,08 2,8412 0,00 7,80 69,62 70,91 2,53

Load Case Id. :



Al 6082-T6

Section Stress


M.S. = (Fcc/fc)-1 = (183.6/69,6)-1 = 1.64 (U.L.) L.C. Nz max




48

6.4. Junctionsanalysis

6.4.1. Fwdsparfittingarea

The most external strips of both upper and lower caps are made of sheets of 6082-T6 aluminum alloy. To get the load associated to the most external strip the total load on upper and lower caps is spread on each metal strip as function of the thickness. Both upper and lower caps are made of a series of strips 35 mm wide with different thickness from 1 to 2 mm.




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Figure 6-19: Spar cap thickness at wing root

Figure 6-20: Spar cap strips thk.

Layer 1

Layer

Layer 3

Layer 4

Layer 5

Layer 6

Layer 7




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Total load to be transferred in longitudinal direction: Plong = 104860/2 = 52430 N (U.L.) L.C.: Nz max. The transverse load is judged negligible. This load is the same for upper and lower spar caps, the load direction change only. The K factor to which the total load shall be multiplied to obtain the single strip load is obtained as follow for each strip: Ki strip = tstrip/ total thk Then: Layer 1 load = Plong * K1 strip = 52430* (2/22) = 4766 N (U.L.) (this load is valid for layer 1, 3,4, 5) Layer 2 load = Plong * K2 strip = 52430*(1.5/22) = 3575 N (U.L.) Layer 6, 7 load = Plong * K6 strip = 52430*(1 /22) = 2383 N (U.L.) The layer 1 is joined to the other layers through No. 3 bolts ¼” DIA type AN4. With ref. 7 the AN4 bolt allowable are: Pshear all. = 3682 lbs = 16384 N Ptens all = 4080 lbs = 1815 N The bearing allowable is obtained considering conservatively that the material Fbru is equal to Ftu, then: Fbru = Ftu = 290 MPa Pbru = Fbru*D*t = 290*6.35*2 = 3683 N Pbru tot = 3683*3 = 11049 N (3 bolts effective) Pshear tot = 16384*3 = 49152 N (3 bolts effective) M.S. = (11049/(4766*1.15)-1 = 1.01 (U.L.) L.C. Nz max Looking to the type and number used to join together the cap strips it is possible to state that the junctions are “passed by inspection”. The conclusion of the front spar sizing shows that the only modifications required are:

1. New definition of spar cap strips thickness 2. Additional sheet on web between STA Y370 and STA Y1907 Here below the updated drawing details are provided:




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Figure 6-21: Spar web thickness modification

The modification consists of an additional sheet of 6082-T6 thk = 1 mm to be added between STA Y 370 and STA Y1907.

Figure 6-22: Spar web stiffeners added -detail

The added web sheet incorporates the upper and lower cap strips.




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7. AftSparsizing

RIB WING STA WING LOAD SHEAR BENDINGNo. mm Kg/mm Kg Kgmm

0 0 0,263736264 480 10920001 370 0,263736264 441 9216212 443 0,263736264 433 8897123 885 0,263736264 387 7085134 1010 0,263736264 373 6610075 1135 0,263736264 360 6151506 1260 0,263736264 347 5709427 1589 0,263736264 312 4624638 1918 0,263736264 278 3654039 2247 0,263736264 243 279761

10 2576 0,263736264 208 20553911 2905 0,263736264 174 14273512 3234 0,263736264 139 9135113 3563 0,263736264 104 5138514 3892 0,263736264 69 2283815 4221 0,263736264 35 570916 4550 0,263736264 0 0

AFT SPAR 40%OF TOTAL WING LOAD




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R H upr joint = 921621/125 = -7373 Kg (L.L.) Using bending moment at STA Y370 R H lwr joint = -R H upr joint = 7373 Kg (L.L.) R V upr joint = 441/2 = 220.5 Kg (L.L.) R V lwr joint = R V upr joint = 220.5 Kg (L.L.) The upper joint is made by 1 bolts type AN8 with ref. (7) the allowable shear is: Ps all = 14730 lbf = 6682 Kg The bolts works in double shear then: Ps app = (7373)/2 = 3686.5 Kg (L.L.) horizontal shear Ps app = (220.5)/2 = 110.25 Kg (L.L.) vertical shear Ps app RES = (3686.52+110.252)0,5 = 3688 Kg (L.L.) shear resultant M.S. = (6682/(3688*1,5*1,15))-1 = 0.05 (U.L.) max nz load case




54

7.1. LugAnalysis The spar fitting is now checked as a lug. The material considered for the spar cap and spar web is: Aluminum 6082-T6 From EN-AW-6082 that is more conservative with respect the ALCOA data-sheet properties provided on paragraph 3.2, the mechanical properties of this aluminum are:

Ftu = 290 MPa Fty = 250 MPa Fsu = Ftu * 0.577 = 290 * 0.577 = 167 MPa E = 10%

Lug geometry:

Lug applied loads: FH = 3686.5 Kg (L.L.) 55298 N (U.L.) FV = 110.25 Kg (L.L.) 1654 N (U.L.)




55

LUG DATA & ALLOWABLE W, (mm) = 35 DWG No.: fwd spar D, (mm) = 12 Material: 6082-T6 t, (mm) = 22 Ftu, (MPa) = 290 e, (mm) = 17.5 Ftu; min, (MPa) = 290 c, (mm) = 0 Fty, (MPa) = 250 TENSION ALLOWABLE BEARING ALLOWABLE At , (mm2) = 506 Abr,(mm2) = 264 Ktu = .8186401 Pbru, (N) = 94656 Ptu, (N) = 120127.3 TRANSVERSE LOAD ALLOWABLE A1, (mm2) = 291.6711 Aav, (mm2) = 277.5309 A2, (mm2) = 253 BETA DEG. = 0 A3, (mm2) = 253 Ktru = .9913455 A4, (mm2) = 291.6711 Ptru, (N) = 75897.41 Paxial = 55298 N (U.L.) Ptransv = 1654 N (U.L.) Ra = (55298/94656)*1.15 = 0.667 Rtr = (1654/75897.41)*1.15 = 0.025 M.S. = (1/(Ra

1.6 + Rtr1.6)0.625)-1

--------------------------------------------------------------- M.S. =------ WITH F.F. ----------> 0.49 (U.L..) L.C.:Nz max --------------------------------------------------------------- Bushing Yield Bearing Allowable Pbry, N = 0 LUG Yield Allowable Py, N = 71971.69 The outboard fitting hole is checked as a net section including the bolt load plus the by-pass load. For the bearing see the above allowable calculated for the inboard lug hole. The total load at the outboard hole net section is: PTOTAL = 55298 x 2 = 110596 N (U.L.) The net section area is: Anet = (35-12)*22 = 506 mm2 The max net stress is: ft = 110596/506 = 219 MPa (U.L.)

M.S. = (290/219)-1 = 0.32 (U.L) max nz




56

7.2. Sparmainsectionsanalysis Following the same criteria used for the fwd spar, the aft spar main sections are checked starting from the available drawing data.

7.2.1. CheckstressatY370




57

Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 4,06 177,42 177,56 0,632 0,00 4,06 99,80 100,04 1,903 0,00 4,06 99,80 100,04 1,904 0,00 4,06 -99,80 100,04 1,905 0,00 4,06 -99,80 100,04 1,906 0,00 4,06 -177,42 177,56 0,637 0,00 4,06 -177,42 177,56 0,638 0,00 4,06 -99,80 100,04 1,909 0,00 4,06 -99,80 100,04 1,90

10 0,00 4,06 99,80 100,04 1,9011 0,00 4,06 99,80 100,04 1,9012 0,00 4,06 177,42 177,56 0,63


Al 6082-T6

Section Stress

Load Case Id. :





58

Crippling analysis

Figure 7-23: Considered section for crippling analysis

********************************************** * SECTION INERTIAL PROPERTIES CALCULATION *

**********************************************

Aft spar sta Y370 SECTION NR. 1

SECTION DATA

1 4.00 1.00 24.00 0.50 12.00 0.00 2 4.00 24.00 1.00 12.50 24.00 0.00 3 4.00 1.00 10.00 24.00 19.00 0.00

59

# TYPE DIM 1 DIM 2 A Y AY AY2 IOX

1 4. 1.00 24.00 0.2400E+02 0.1200E+02 0.2880E+03 0.3456E+04 0.1152E+04 2 4. 24.00 1.00 0.2400E+02 0.2400E+02 0.5760E+03 0.1382E+05 0.2000E+01 3 4. 1.00 10.00 0.1000E+02 0.1900E+02 0.1900E+03 0.3610E+04 0.8333E+02

# X AX AX2 IOY AXY IOXY

1 0.5000E+00 0.1200E+02 0.6000E+01 0.2000E+01 0.1440E+03 0.0000E+00 2 0.1250E+02 0.3000E+03 0.3750E+04 0.1152E+04 0.7200E+04 0.0000E+00 3 0.2400E+02 0.2400E+03 0.5760E+04 0.8333E+00 0.4560E+04 0.0000E+00

SECTION AREA A = 58.00 mm2 CENTROID POSITION X = 9.52 mm Y = 18.17 mm MOMENTS OF INERTIA IX = 0.297361E+04 mm4 IY = 0.541732E+04 mm4 PRODUCT OF INERTIA IXY = 0.187283E+04 mm4 TORSIONAL INERTIA <THIN WALLED OPEN SECTION> 0.193333E+02 mm4 RADIUS OF GYRATION X = 7.16 mm Y = 9.66 mm PRINCIPAL INERTIA AXES ROTATION (CCW) = 28.44 DEGREES PRINCIPAL MOMENTS OF INERTIA IXP = 0.195930E+04 mm4 IYP = 0.643162E+04 mm4 RADIUS OF GYRATION XP = 5.81 mm Y = 10.53 mm

CRIPPLING ALLOWABLES:

EL B T FCC FCC(Y) FCC(U) 0/1

1 24.00 1.00 109.86 109.86 109.86 1 2 24.00 1.00 241.51 241.51 241.51 0 3 10.00 1.00 221.71 221.71 221.71 1

MATERIAL: CRIPPLING GENERAL SOL. - FORMED FCY = 250.00 N/mm2 FCU = 290.00 N/mm2

E = 70000. N/mm2 FCC = 183.62 N/mm2 183.62 N/mm2




60


61

Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 6,62 179,21 179,58 0,612 0,00 6,62 124,43 124,96 1,323 0,00 6,62 124,43 124,96 1,324 0,00 6,62 -124,43 124,96 1,325 0,00 6,62 -124,43 124,96 1,326 0,00 6,62 -179,21 179,58 0,617 0,00 6,62 -179,21 179,58 0,618 0,00 6,62 -124,43 124,96 1,329 0,00 6,62 -124,43 124,96 1,32

10 0,00 6,62 124,43 124,96 1,3211 0,00 6,62 124,43 124,96 1,3212 0,00 6,62 179,21 179,58 0,61


Al 6082-T6

Section Stress

Load Case Id. :


MATERIAL: CRIPPLING GENERAL SOL. - FORMED (see page. 33) FCY = 250.00 N/mm2 FCU = 290.00 N/mm2

E = 70000. N/mm2 FCC = 183.62 N/mm2

FCC<Y>= 183.62 N/mm2 FCC= 183.62 N/mm2




62


63




1 0,00 6,73 169,81 170,21 0,702 0,00 6,73 166,30 166,71 0,743 0,00 6,73 166,30 166,71 0,744 0,00 6,73 115,49 116,07 1,505 0,00 6,73 115,49 116,07 1,506 0,00 6,73 -115,36 115,95 1,507 0,00 6,73 -115,36 115,95 1,508 0,00 6,73 -166,17 166,58 0,749 0,00 6,73 -166,17 166,58 0,74

10 0,00 6,73 -169,50 169,90 0,7111 0,00 6,73 -174,00 174,39 0,6612 0,00 6,73 -157,84 158,27 0,8313 0,00 6,73 -157,84 158,27 0,8314 0,00 6,73 -172,36 172,75 0,6815 0,00 6,73 -168,22 168,62 0,7216 0,00 6,73 -166,17 166,58 0,7417 0,00 6,73 -166,17 166,58 0,7418 0,00 6,73 -115,36 115,95 1,5019 0,00 6,73 -115,36 115,95 1,5020 0,00 6,73 115,49 116,07 1,5021 0,00 6,73 115,49 116,07 1,5022 0,00 6,73 166,30 166,71 0,7423 0,00 6,73 166,30 166,71 0,7424 0,00 6,73 168,46 168,87 0,7225 0,00 6,73 170,82 171,21 0,6926 0,00 6,73 156,30 156,73 0,8527 0,00 6,73 156,30 156,73 0,8528 0,00 6,73 172,37 172,76 0,68

Section Stress

MATERIAL: CRIPPLING GENERAL SOL. - FORMED (see page. 32) FCY = 250.00 N/mm2 FCU = 290.00 N/mm2

E = 70000. N/mm2 FCC = 183.62 N/mm2

FCC<Y>= 183.62 N/mm2 FCC= 183.62 N/mm2




64

7.2.4. CheckstressatYW1589andY2576 Being the change of spar cap thickness so close to the STA Y1589, the section analysis is done considering the dimensions of section between STA Y1589 and Y2576.

65



1 0,00 8,04 174,26 174,81 0,662 0,00 8,04 170,65 171,22 0,693 0,00 8,04 170,65 171,22 0,694 0,00 8,04 118,52 119,33 1,435 0,00 8,04 118,52 119,33 1,436 0,00 8,04 -118,33 119,14 1,437 0,00 8,04 -118,33 119,14 1,438 0,00 8,04 -170,46 171,03 0,709 0,00 8,04 -170,46 171,03 0,70

10 0,00 8,04 -173,87 174,43 0,6611 0,00 8,04 -178,49 179,03 0,6212 0,00 8,04 -161,91 162,51 0,7813 0,00 8,04 -161,91 162,51 0,7814 0,00 8,04 -176,81 177,35 0,6415 0,00 8,04 -172,56 173,12 0,6816 0,00 8,04 -170,46 171,03 0,7017 0,00 8,04 -170,46 171,03 0,7018 0,00 8,04 -118,33 119,14 1,4319 0,00 8,04 -118,33 119,14 1,4320 0,00 8,04 118,52 119,33 1,4321 0,00 8,04 118,52 119,33 1,4322 0,00 8,04 170,65 171,22 0,6923 0,00 8,04 170,65 171,22 0,6924 0,00 8,04 172,87 173,43 0,6725 0,00 8,04 175,29 175,84 0,6526 0,00 8,04 160,39 160,99 0,8027 0,00 8,04 160,39 160,99 0,8028 0,00 8,04 176,88 177,43 0,63

Section Stress

MATERIAL: CRIPPLING GENERAL SOL. - FORMED (see page 32) FCY = 250.00 N/mm2 FCU = 290.00 N/mm2

E = 70000. N/mm2 FCC = 183.62 N/mm2

FCC<Y>= 183.62 N/mm2 FCC= 183.62 N/mm2 M.S. = (Fcc/fc)-1 = (183.6/178.5)-1 = 0.03 (U.L.) L.C. Nz max The above M.S. covers both STA Y1589 and Y2576




66


To be conservative the 1mm thk strips, one per side, on top and bottom portion of the spar web close to the flanges have been neglected in the inertia calculation.

67

Max Nz


Fz 0 N Axial

Sx 0 N Shear_X dir.




Material :




1 0,00 6,67 77,00 77,86 2,212 0,00 6,67 -77,00 77,86 2,213 0,00 6,67 -77,00 77,86 2,214 0,00 6,67 -67,83 68,81 2,635 0,00 6,67 -67,83 68,81 2,636 0,00 6,67 -76,08 76,95 2,257 0,00 6,67 -76,08 76,95 2,258 0,00 6,67 76,08 76,95 2,259 0,00 6,67 76,08 76,95 2,25

10 0,00 6,67 67,83 68,81 2,6311 0,00 6,67 67,83 68,81 2,6312 0,00 6,67 77,00 77,86 2,21


Al 6082-T6

Section Stress

Load Case Id. :


MATERIAL: CRIPPLING GENERAL SOL. - FORMED (see page 33) FCY = 250.00 N/mm2 FCU = 290.00 N/mm2

E = 70000. N/mm2 FCC = 183.62 N/mm2

FCC<Y>= 183.62 N/mm2 FCC= 183.62 N/mm2 M.S. = (Fcc/fc)-1 = (183.6/76)-1 = 1.41 (U.L.) L.C. Nz max




68

7.3. Bucklinganalysisofaftsparweb The buckling coefficients as well as the allowable are obtained following the methods reported on ref. 7.

Figure 7-24: Panels dimensions and numbering




69






70


The non buckling type panels are checked following the method reported in ref. 7 page C10.17: qall = K*t*(fsh*(1-(D/h)2+fsc(D/h)0.5)*(C’/b) where: K = 0.85 -0.0006 (h/t) = 0.85-0.0006(125/1) = 0.775 D = 62 mm b = 160 mm (min) , 169 mm (max) h = 125 mm C = b-D =169-62 = 107 C’ = C-2B = 107- 2(4) = 99 (B from table C10.6 of ref.5 )

Non-buckling beam webs analysis

k t fsh D h fsc C C' b H/t C/t

mm MPa mm mm MPa mm mm mm

0,73 1 6 62 62 84,4 107 99 169 125 107

q all = 192,3 MPa NOTE: To recover the buckling negative margins on the spar webs panels close to the spar root the panel dimensions “a” shall be reduced as follow (red rows) installing a series of panel breakers (“L” fastened to the spar web).




71

The red angles shall be added as panel breaker in the middle of the existing panels to prevent shear buckling of the spar web.




72

7.4Junctionsanalysis The spar caps are made of a series of metal strips of 6082-T6, each strip is joined to the other by bolts or solid rivets in function of the span location. The analysis of the junctions is focused at the root where the max loads are applied.

As already done for the fwd spar the load distribution on each metal strip is based on the strip thickness. The 4 bolts above shown on the left side of the fitting hole 12 mm DIA are AN4 type (1/4” DIA).

Figure 7-28: Spar cap - strips thicknesses

Layer 1 Layer 2

Layer 3

Layer 4Layer 5

Layer 6




73

The max spar cap load at root is equal to Rh lwr or upr = 110595 N (U.L. ref. page 55), the vertical load is considered negligible (3308 N ult.). This load is spread on each cap strip as function of its thickness. Then being the total cap thk = 22 mm: The K factor to which the total load shall be multiplied to obtain the single strip load is obtained as follow for each strip: Ki strip = tstrip/ total thk Then: Layer 1 load = Plong * K1 strip = 110595* (5.5/22) = 27649 N (U.L.) (this load is valid for layer 1) Layer 2 load = Plong * K2 strip = 110595*(2/22) = 10054 N (U.L.) (this load is valid for layer 3) Layer 6, 7 load = Plong * K6 strip = 110595*(1 /22) = 5027 N (U.L.) (this load is valid for layers 2, 4, 5, 6). The layer 1 is joined to the other layers through No. 4 bolts ¼” DIA type AN4. With ref. 7 the AN4 bolt allowable are: Pshear all. = 3682 lbs = 16384 N Ptens all = 4080 lbs = 18155 N The bearing allowable is obtained considering conservatively that the material Fbru is equal to Ftu, then: Fbru = Ftu = 290 MPa Pbru = Fbru*D*t = 290*6.35*5.5 = 10128 N Pbru tot = 10128*4 = 40512 N (4 bolts effective) Pshear tot = 16384*4 = 65536 N (4 bolts effective) M.S. = (40512/(27649*1.15)-1 = 0.27 (U.L.) L.C. Nz max Looking to the type and number used to join together the cap strips it is possible to state that the junctions are “passed by inspection” because the remaining strips are much less loaded.




74

8. Centerframefwdsparanalysis

Figure 8-29: Fwd spar – center frame

The center frame is composed by 2 sheet metal frames 8 mm thickness with local reinforcing of 4 mm in correspondence of the lugs made of 2024-T3 aluminum. These 2 sheets are bolted together interposing square tubes 20x20 of steel 4130 N. The load condition used for the center frame analysis is the Nz max as done for the front and rear spar check. The applied loads are taken from table on page 22. The center frame analysis is done for strength an stability in the most loaded sections starting from the lugs upper and lower, then the frame members and the junctions between the sheet metal frame and the square steel tubes that work as spacer.




75

8.1. Materialsallowable




76




77

8.2. Luganalysis Lower Lug geometry:

Figure 8-30: Center frame lug geometry

Lug thickness = 8 + 4 = 12 mm Lug material: 2024-T3 sheet Ftu = 63*0.703*9.81 = 434.5 MPa Fty = 42*0.703*9.81 = 290 MPa Applied loads (ref. page 24): Faxial = 7126/2 = 3563 Kg (L.L.) 53445 N (U.L.) Ftransv = 330,5/2 = 165.25 Kg (L.L.) 2479 N (U.L.) Being the center frame lug a double shear lug the above forces shall be divided by 2 checking the single lug of 12 mm thk., then: Faxial = 53445/2 = 26722.5 N (U.L.) Ftransv = 2479/2 = 1239.5 N (U.L.)




78

LUG DATA & ALLOWABLE W, (mm) = 40 DWG No.:

D, (mm) = 12 Material: 2024-T3 t, (mm) = 10.9 Ftu, (MPa) = 434.5

e, (mm) = 25 Ftu; min, (MPa) = 434.5 c, (mm) = 5 Fty, (MPa) = 290

TENSION ALLOWABLE BEARING ALLOWABLE

At , (mm2) = 305.2 Abr,(mm2) = 130.8 Ktu = .7822833 Pbru, (N) = 64501

Ptu, (N) = 103738.1

TRANSVERSE LOAD ALLOWABLE A1, (mm2) = 171.7598 Aav, (mm2) = 173.0602

A2, (mm2) = 152.6 BETA DEG. = 0 A3, (mm2) = 207.1 Ktru = 1.110692

A4, (mm2) = 171.7598 Ptru, (N) = 63123.5

--------------------------------------------------------------- M.S = -------(U.L.)-------> 1.402329

M.S. =------ WITH F.F. ----------> 1.088982 L.C.: Nx max ---------------------------------------------------------------

LUG Yield Allowable Py, N = 46372.05

The inner hole is checked considering the bearing load plus the by-pass load at the net section. Anet = 12*2*40-(12*2*12) = 672 mm2 ft = 53445*2/672 = 159 MPa (U.L.) M.S. = (434.5/159)-1 = High (U.L.)




79

8.3. LugReinforcingjunctionanalysis Being the lug thickness composed by a main sheet of 8 mm (2024-T3) plus a local reinforcing done by a sheet of 4 mm (always in 2024-T3), the load associated to the reinforcing is based on is stiffness.

Figure 8-31: Center frame wall thickness

Then: Ptot unit = (7126/2)*9.81 *1.5 = 52430/10.9 = 4810 N (U.L.) Nz max (double shear lugs) Pthk 4 mm = 4810 * 4 = 19240 N The above load shall be transferred back to the base sheet of 8 mm thk through No. 4 bolts AN4 (see plot on next page) on the lower position and No. 5 bolts AN4 on the upper position. The upper position junction is checked also for inter-rivet buckling.




80

Figure 8-32: Center frame junctions geometry

Papp bolt = 19240/4 = 4810 N (U.L.) Junction allowable

Bolt shear allowable With ref. 7 the AN4 bolt allowable are: Pshear all. = 3682 lbs = 16384 N Ptens all = 4080 lbs = 18155 N

Bearing allowable Fbru = 129*0.703*9.81 = 890 MPa Pbru = Fbru*D*t = 890*6.35*4 = 22606 N M.S. = (16384/(4810*1.15))-1 = 1.96 (U.L.) L.C. Nz max On the upper reinforcing sheet he inter rivet buckling is checked due to the big distance between the second wing spar bolt junction and the first reinforcing sheet junction to the main plate. The Fir = Fcy = 39 Ksi = 269 N fc = (19240/(40*4)) = 120.3 MPa (U.L.)

M.S. = (269/120.3)-1 = High (U.L.) inter-rivet buckling




81

8.4. Frameanalysis The center frame is analyzed using a simple FE model. The plot below shows the model with the applied load for the L.C. Nz max and the axial and bending moment on each frame element. It is possible to see that the main loaded elements are the upper and lower crossing beams while the vertical and oblique elements are negligible loaded.

The upper and lower longitudinal beam elements have the same cross section, but while the upper element are loaded in compression therefore the stability check is provided, the lower elements are loaded in tension therefore the net section strength is checked.




82

8.4.1. Uppercrossbeamanalysis

Figure 8-33: Center frame sections geometry

The maximum compression load applied on upper cross beam is: Pcomp max = 7126 Kg (L.L.) = 106890 N (U.L.) L.C. : Nz max Column analysis Column length: 370 mm Column inertia (min.) : being the section composed by 2 sheet (8 mm thk) of aluminum and in between a square tube 20x20x1.24 made of 4130 N the equivalent section all in aluminum is considered.

Figure 8-34: Upr cross beam – equivalent section dimensions




83

Section properties: A = (40*36)-(12.8*32.8) = 1020 mm2

Jmax = (403*36/12)-(32.83*12.8/12) = 154359 mm4

Jmin = (363*40/12)-(12.83*32.8/12) = 149788 mm4 Pcr =PI2* EI/(L’2) = 3.142*75171*149788/3702 = 811753 N

M.S. = (Pcr/Papp)-1 = (811753/106890)-1 =High L.C.: Nz max

8.4.2. Lwrcrossbeamanalysis The lower cross beam is checked for tension considering the net section without the contribution of the square tube in steel: Anet = (40*8*2)-(6.35*8) = 538 mm2 ft = 106890/538 = 199 MPa (U.L.)

M.S. (Ftu/ft)-1 = (434.5/199)-1 = 1.18 (U.L. L.C.: Nz max)




84

9. CenterFrameAftSparAnalysis

Being the type of design, the materials and the thicknesses of each parts the same of those already checked for the center frame fwd spar, the analysis of this part is considered covered by those done for the fwd spar frame.




85

10. FEMLoading Based on the assumptions made on previous paragraphs the FE Model has been loaded in correspondence of front and rear spar as shown in the next plot.

Figure 10-35: FEM Loading

The couple of forces span wise applied at wing end are needed to obtain the correct shear and bending moment at the wing root. In the next page two plot representing the theoretical wing loading compared to the FEM loading are provided to show the consistency of the FEM loading.




86




87

The following plots shows the leading edge panels definitions, each panel side contour is associated to the front spar, ribs and leading edge locations.

Figure 10-36: Wing upper skin




88

Figure 10-37: Wing lower skin

Figure 10-38: Wing panels definition (No. of CQUAD 4 from rib to rib)




89

Figure 10-39: Upper skin CQUAD 4 ID at the root area




90

Figure 10-40: Lower skin CQUAD 4 ID at the root area




91

The following plots shows the FEM rib against the real rib.

Figure 10-41: FEM rib

Figure 10-42: Real rib




92

Figure 10-43: Wing ribs definition (Type Bar2)

Figure 10-44: Front and center ribs Bar2 ID at the root area




93

Figure 10-45: Wing max deflection for nz max




94

Figure 10-46: Wing max deflection for nz max

10.1. Skinbucklinganalysis The skin buckling analysis is done using the standard method provided in ref. (6). All the panels are considered simply supported. The skin panels are in aluminum 60.61-T6 t= 0.5 mm, therefore from ref. (5) the mechanical properties of this material are used. Being the skin thickness constant overall, the buckling analysis is done only for 7 panels (upper and lower) starting from the wing root. The buckling margins of the remaining panels are higher than those reported in the following pages. For the upper skin mainly loaded in shear and compression, the assumption made is that the skin buckle for shear and for a part of the high compression stress coming from FEM. The FEM linear analysis could not spread correctly the compression exceeding the panel buckling capability, then the delta compression load exceeding the minimum amount that combined with the applied shear provide M.B. = 0.0 is loaded on the upper spar cap. In case the panel buckle with the applied shear and the small amount of remaining compression a post-buckling analysis is performed. In the following pages the plots of the nodal shear, compression/tension limit stress on the first 7 panels (both upper and lower) is provided, then in a tabular form the panels buckling allowable, the applied stress and the buckling margins are provided.




95

Figure 10-47: Panels number




96




97




98




99

Figure 10-48: Panel 2 upr shear stress (L.L.) nz max L.C.

Figure 10-49: Panel 2 upr compression stress (L.L.) nz max L.C.




100






101






102






103





104






105





106

Figure 10-60: Wing top view, max Principal nz max L.C.




107

Figure 10-61: Wing top view, max Principal nz max L.C. (root area detail)




108

Figure 10-62: Wing top view, max Shear nz max L.C. (root area detail)




109

Figure 10-63: Wing bottom view, max Shear nz max L.C. (root area detail)




110

Figure 10-64: Wing bottom view, max Principal nz max L.C. (root area detail)

The following table provide the buckling margins of upper skin panel in the root area:

Table 10-1: Buckling margins of upper skin

In the following page a comparison between the upper cap loads obtained for the front spar on the report LW-001 rev. 1 is provided showing that the spar analysis done in the report LW001 rev. 1 is conservative. The plot shows the front spar upper cap loads at the wing root obtained with the by hand analysis and with the FEM analysis (corrected because the FEN is analyzed with a linear solution).




111

Figure 10-65: Front spar upr cap loads comparison

The lower panels that are mainly loaded in shear and tension are checked for shear buckling without the help of tensile stress.

fs Fscr Rs M.B.

(U.L.)

MPa

1 15,79 34,1 0,46 1,16

2 9,06 21,6 0,42 1,38

3 11,97 22,8 0,53 0,90

4 10,62 22,8 0,47 1,15

5 8,61 20,1 0,43 1,33

6 4,59 20,1 0,23 3,38

7 10,21 20,1 0,51 0,97

(U.L.)

Upper skin panels

Panel IDMPa




112

Each panel is known in term of applied stress defining side by side the stress distribution. In the following the panel sides definition is provided:

As an example the stresses on panel 1 upper are provided:

Side 1 Side 2 Side 3 Side 4 Side 1 Side 3

1,45 35,70 4,17 3,16 26,00 5,49

2,38 8,61 21,60 3,28 9,90 7,59

2,30 4,17 22,15 2,38 ‐5,22 ‐7,80

0,49 16,45 ‐28,85 ‐14,59

1,31 4,82 ‐16,16 ‐15,76

0,98 2,24 ‐12,23 ‐12,85

2,63 3,80 1,42 ‐14,10

17,58 3,72 8,67 ‐38,58

50,64 3,16 ‐14,87 ‐45,60

35,70 ‐90,80 ‐72,29

115,46 48,48 82,11 8,82 ‐122,14 ‐208,49

Shear

15,29 ‐24,80

Compression




113

10.2. Skinbuklinganalysisconclusions The skin analysis shows substantially a good behavior, but the following modifications are suggested to avoid local stress concentrations in the root area:

An additional suggestion is to change the skin thickness up to WS1000 from t=0,5 mm to t= 0,6 mm.

10.3. Skinjunctionanalysis Considering the blind rivets Avex 1604 da 3.2 mm DIA with shear allowable equal to 700 N, and bearing allowable of 984 N the safety margin on the max shear stress area equal to a 16 MPa(U.L) panel 1 is: M.S. = (700/(300*1.15))-1 = 1.02 (U.L.) max nz load case




114

11. Ribsanalysis The ribs analysis is done using the standard method provided in ref. (6). All the ribs are considered with fixed ends and distributed axial load. The ribs are in aluminum 60.82-T6 t= 0.5 mm, therefore from ref. (5) the mechanical properties of this material are used. The ribs are divided in three main groups, single and double thickness and front. Being the ribs geometric properties constant overall, for each group the rib analysis is done only for the most loaded (traction and compression) elements. The margins of the remaining ribs elements are higher than those reported in the following pages. In the following pages the plots of the elements load, compression/tension is provided, then in a tabular form the rib allowable, the applied load and the compression/tension and crippling margins are provided.

Single thickness ribs

-1500

-1000

-500

0

500

1000

1500

2000

20000 20050 20100 20150 20200 20250

Elements number

N

Figure 11-66: Single thickness elements load




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Double thickness ribs

-1500

-1000

-500

0

500

1000

1500

15000 15010 15020 15030 15040 15050 15060 15070 15080

Elements number

N

Figure 11-67: Double thickness elements load




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Leading edge ribs

-800

-600

-400

-200

0

200

400

600

800

1000

1200

25000 25050 25100 25150 25200 25250 25300 25350

Elements number

N

Figure 11-68: Leading edge elements load

Group Elements number Load (N) Elements number Load (N)Single thickness 20185 1353.5 20216 1495.2Double thickness 15050 1140.3 15027 996.6

Leading edge 25014 698 25000 961.7

Tension Compression

Table 11-2: Most loaded elements




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Figure 11-69: Leading edge elements

Figure 11-70: Double and single thickness elements

Elements 25014

Elements 25014

Elements 20186

Elements 15050

Elements 20216

Elements 15027 Elements 15008




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The following table provide the tension and compression margins of most loaded ribs (the section area include the skin contribution):

Element ID Section Area ft Ftu M.S.

U.L.mm^2 Mpa Mpa

20185 71.5 28.4 295 5.915050 149 11.5 295 16.125014 51 20.5 295 8.620216 29.5 76.0 295 1.615027 64 23.4 295 7.425000 32.5 44.4 295 3.4

Table 11-3: Tension and compression safety margins

The following table provide the crippling margins of the most loaded ribs:

Elements ID Fcy E Fcc/(FcyE)^0,5 Fcc Section area Load fc (U.L.) M.S.Mpa Mpa Mpa mm^2 N Mpa

20216 250 70000 see example 1 95 29.5 1495.2 76.0 0.2515027 250 70000 see example 2 96.1 64 996.6 23.4 3.1125000 250 70000 see example 3 45.2 32.5 961.7 44.4 0.02

Table 11-4: Crippling safety margins

These are the example for definition of Fcc: Example n.1

Section area

2mm

57mm

5mm

0.5mm thickness

11mm hole diameter

Fcc = (A1*Fcc1 + A2*Fcc2 + A3*Fcc3) / (A1 + A2 + A3)

Skin contribution




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Fcc1 = (Fcy*E)^0,5*c1 Fcc2 = (Fcy*E)^0,5*c2 Fcc3 = (Fcy*E)^0,5*c3

c1 = 0.09

c2 = 0.0165 c3 = 0.053

Fcc = (376.5 + 1587 + 554.3) / (23 + 1 + 2.5) = 95 Mpa

Example n.2

Section area

2,5mm

56mm

2,5mm

0.5mm thickness

Fcc = (A1*Fcc1 + A2*Fcc2 + A3*Fcc3) / (A1 + A2 + A3)

Fcc1 = (Fcy*E)^0,5*c1 Fcc2 = (Fcy*E)^0,5*c2 Fcc3 = (Fcy*E)^0,5*c3

c1 = 0.09

c2 = 0.017 c3 = 0.09

Fcc = (470.6 + 1991 + 470.6) / (28 + 1.25 + 1.25) = 96.1 Mpa

Example n.3

Skin contribution




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Section area

5mm

60mm 0.5mm thickness

Fcc = (A1*Fcc1 + A2*Fcc2) / (A1 + A2)

Fcc1 = (Fcy*E)^0,5*c1 Fcc2 = (Fcy*E)^0,5*c2

c1 = 0.053

c2 = 0.0073

Fcc = (554.25 + 915) / (2.5 + 30) = 45.2 Mpa

11.1. Aileronandflapmaneuver Two additional antisymmetric load conditions are considered derived from aileron and flap maneuvers. The loads are defined from FAR23 table A6 (the same for CS-VLA):

W = 600 Kg (1323 lb) MTOW S = 156 ft^2 (wing surface)

n = 4 nW/S = 34

average aileron load = 16 lb/ft^2 (Table A6 FAR23)

average flap load = 22 lb/ft^2 (Table A6 FAR23) Saileron = 5.9 ft^2 Sflap = 7.8 ft^2




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Aileron load = 42.5 Kg

Flap load = 78 Kg

The aileron load is added to the wing load defined in §5 divided by 2 (nz = 2g) see figure 11.71, the flap load is added to the wing load divided by 4 (nz = 1g) see figure 11.72.

Figure 11-71: Aileron load condition




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Figure 11-72: Flap load condition

The elements load results are reported in the following figures:

Figure 11-73: Aileron load condition, single thickness ribs




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Figure 11-74: Aileron load condition, double thickness ribs

Figure 11-75: Aileron load condition, leading edge ribs




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Figure 11-76: Flap load condition, single thickness ribs

Figure 11-77: Flap load condition, double thickness ribs




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Figure 11-78: Flap load condition, leading edge ribs

Tables 11.5 and 11.6 report the most loaded ribs for the two additional maneuver:

Table 11-5: Aileron load condition, most loaded elements

Table 11-6: Flap load condition, most loaded elements




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11.2. Ribsanalysisconclusions The ribs analysis performed show that the load condition defined in §5 (max Nz) is the most conservative (compare table 11.1, 11.4 and 11.5). For flap maneuver condition only, the most compressed elements, for double thickness rib is n.15008 but the rib area is bigger and the load lower respect the n.15027 element (see figure 11.12).

Figure 11-79: Elements 15008 and 15027 position

The ribs analysis shows substantially a good behavior, an improvement is cold bond the back to back double thickness ribs.

Element n.15027

Element n.15008




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12. FlapandAileronHingeAnalysis

The hinges are made of formed 6082-T6 sheet. The wing rear spar side per DWG SC5720021101-L.

Figure 12-1 : Wing and Aileron / Flap hinge




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Figure 12-2: Exploded viwe of aileron / flap hinge

Figure 12-3: Exploded view of aileron / flap hinge -movable side




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Ref, F2245_06 Appendix the flap and aileron loads are obtained as follow: Legend: n= load factor 4g W = MTOW (lbs) 600 Kg = 132.3 lbs S = wing surface (ft2) 156 ft2 nW/S = 34 Then: � Waileron = 16 lb/ft2 � Wflap = 22 lb/ft2 The biggest load is for the flap then with ref. F2245_06 TAB. X1.1 we get: Flap up n1 = 4g Flap down nf = 0.5 n1 Flap surface: 7.8 ft2 Wflap = 7.8 x 22 = 171.6 lb = 78 Kg Aileron surface: 5.86 ft2 Waileron = 5.86 x 16 = 94 lb = 42.5 Kg From above being the hinge drawing the same for flap and aileron it is possible to state that checking the hinge for the flap load the aileron is covered too.




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Figure 12-4: Wing attached spar caps

The hinge detail drawing is shown below:




131




132




133

Conservative assumption: � Flap down 60° � Load for flap up considered for flap down (arms max) Pv = 39 cos 30 = 34 Kg Ph = 39 sin 30 = 19.5 Kg Rh = ((34 x 154.24)/125) + ((19.5 x 88.6)/125) = 56 Kg Rv = 34/2 = 17 Kg




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The reactions are made by two bolt 4.8 mm DIA then: Bolt tension = 56/2 = 28 Kg (L.L.) Bolt shear = 17/2 =8.5 Kg (L.L.) Bolts “passed by inspection”. Bearing check on hinge holes: Rres = (282 + 8.52)0.5 = 29 Kg (L.L.) fbr = 29 x 1.5 x 1.15 / (4.8 x 1.5) = 6.9 Kg/mm2 (U.L.) The material Fbru is obtained multiplying the Ftu by 1.5 then: Fbru = 310 x 1.5 = 465 MPa M.S. = (465 / 69) – 1 = High (U.L. F.F. included) max flap load Angle type fitting check e = 8 mm b = 10 mm

h = 57 mm

Pall = 300 lbs = 136 Kg




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M.S. = (136 / 28 ) – 1 = High (L.L.) max flap load Lug analysis

The hinge lug is hexagonal to prevent the special bushing rotation, but for the analysis purpose it is considered circular 10 mm DIA.

Lug geometrical data: D = 10 mm W = 28 mm e = 14 mm t = 1.5 mm Ftu = 310 MPa Fty = 260 MPa Lug applied loads: Paxial = 0 N; Ptransverse = 19.5 Kg (L.L.) Lug allowable: Ptu = 6265 N; Pbru = 506 N; Ptru = 2559 N The load is considered, conservatively, fully transverse, then:

M.S. = (2559 /(195 x 1.5 x 1.15)) – 1 = High (U.L. F.F. included)

The flap down is covered by the analyses in the previous pages because the inertia factor to be considered for the load distribution calculation is 2 g instead of 4g.

Looking to the applied loads and the geometry of the hinge at the flap/aileron side it is possible to state that the hinge at the flap/aileron side is “passed by inspection”.

13. WingTanksCheck

The wing tanks are 2 per wing and each is 32 litres. The tanks are supported in three stations by three beams hinged-hinged to the front and rear spars through supports bolted to the spars caps. The total weight per wing is considered equal to 32 x 2 = 64 Kg, this weight include the tank structure plus the fuel. Each of the 3 beams is loaded in 2 points by a vertical shear (down) of: Pshear = 64 / 3 / 2 = 10.65 Kg (L.L.)




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Figure 13-1: Tanks support drawing

All the tank supports, those attached to the spars caps and the crossing beams are thk = 1.5 mm and are formed sheet of aluminum alloy 6082-T6. The analyis done is for a simply supported beam with two concetrated loads as follow:




137

Ra -10.65 -10.65 + Rb = 0 10.65*146 + 10.65*(255+143) – Rb* (362+255+143) = 0 Rb = 10.65*146 + 10.65*(255+143) / (362+255+143) = 7.6 Kg Ra = 13,7 Kg The beams are “hat” section, (see dwg for dimensions) and the geometrical properties are: 1. Area = 180 mm2 2. xc,g, = 21 mm 3. yc.g. = 20.5 4. Jxx = 29161 mm4 5. Jyy = 30602 mm4 The max deflection is 0.76 mm The max bending moment is 27358 Nmm The max shear is 137 N




138

fb = 27358*1,5*20,5/29161 = 29 MPa (U.L.) fs = 137/(36,5*1,5*2) = 1,25 MPa Also considering a max nz = 4 g therefore doubling the above calculated stresses we cna state: Looking the applied stress and the material allowable (Ftu = 310 MPa) it is possible to state that the beam are considered “Passed by inspection”. The low reaction forces at beams constraints and the beams support material and design allow to state that as for the supports at spars we can state that are “Passed by inspection”.




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14. WingdraganalysisThe wing drag effects on wing structure is done in the following hypothesis: Drag = ½ ρ V^2*S*Cd where: � ½ ρ =0.125 � V = VD = 68.3 m/s ref. Rep. 57-003 Rev. 1 pag. 4 � S = wing wet area = 14.5 m2 � Cd = 0.008 ref. Profilo NACA4416 � Drag = 52 Kg




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Geometrical properties of the spar cap: A = 627 mm2 J1 = 78101 mm4 J2 = 25283 mm4 Applied stresses are: fb = 16700*11*1.5/(2*25283) = +- 5.5 Kg/mm2 The max stress previously calculated on the front spa cap in the same location is 158.48 MPa (U.L.) ftot = 158.48 + 55 = 213.5 MPa M.S. = (Ftu / ftot) – 1 = (290 / 213.5) – 1 = 0.36 (U.L.) nz max + DRAGmax The rear spar is covered by the above analysis.

wing stress check notes lw-001 · 2020. 9. 16. · 57-003 r1 flight envelope 6. mmpds-03 metallic...

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