Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 2 Part 7 pptx

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20=1.2+ + 1,2

)

fy Upper Front Flange 050 04

Upper Stringer 7 Upper Rear Flange No 2 & 6 (t 214 — 1 area of angles = 508

area of skin = 090 area stringer 1 area of angles = 508

area of web = 050 area skin 1 area of skin 090 » 648 area of web No.3,4& 5 (t = 050) area stringer 205 area skin 096 2 -04 upper skin~, -301 cf uw Ww Tw x 7 al we 3 oe X reference axis _— ~ ° £ _ —~ 3 Len § 9 X Centroidal Axis” * 2g 3] 3 3 3 5 sa 3 ale £3 a 14T“: z =8“ ổ Š › NI l2 10g %4 Đa, -032 lower skin

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Fig Al9.25 shows the cross section at

Station 20 divided into 14 longitudinal untts numbered 1 to 14, Since the external load con~ dition to be used places the top surface in compression, the skin will buckle and thus we use the effective width procedure to obtain the skin portion to act with each stringer Fig Al9.25 shows the effective skin which is used with each flange member to give the total area of mumbers (1) to (7) The skin on the bottom surface being in tension is all effective and Fig Al9.25 shows the skin area used with each bottom flange member

The next factor to decide is the stringer effectiveness as discussed and explained tn the previous example problem For the structure of Fig AlS.25 we will assume that the compressive failing stress of the stringers is the same 4s that for the corner members, thus we will have no correction factor to take care of the Situation of flange members having different ultimate strengths

Table Al9.2, columns 1 to 11, and the calculations below the table give the calculations for determining the section properties at Station 20, namely A, Iy, Ig and Ixg Table Al9.3 gives the same for wing section at

tation 47.5 The areas in column (2) are less Since sizes have changed between Stations 20 and 47.5

Calculation of longitudinal stress due to My and My bending moments: -

The design bending moments will be assumed and are as follows: - Station 20 #ế = 1,300,000 in.1b i ~ 285,000 in.1b ¥ " tation 47.5 = 1,000,000 in.1b = ~ 215,000 in.1b = Ñ 1

The moments about the y axes are not

needed in the bending stress analysis but are

needed in the shear analysis which will be made later

To solve equation (4)

K, and K, must be known he constants Ki,

For Station 20 from Table Al9.2, Iy = 230.3, Iz = 1030 and Iyg = - 50, whence ~ Ky = Ixz⁄/(x Iz - Ixg?) = -50/(230.3 x 10.30 - 50*) = -50/235500 = -.0002125 K, = 12/235500 = 1030/235500 = 004378 Ks = 1y/235500 = 230.3/235500 = 00098 A19 17 Substituting in equation (4) op = - [.00098 x - 2a5000 - (-.0002125 x 1500000) | x ~ [00478 x 1,300,000 - (~,0002126 x ~ 285000)] 2 | whence, Op) = 3.3 x - 5639 z- - (8) Column 12 of Table Al9.2 gives the results of this equation for values of x and z in columns 10 and 11 Multiplying these bending stresses by the stringer areds, the stringer loads are given in column 13 The sum of the loads in this column should be zero since total tension must equal total compression on a sec~ tion in bending Stresses at station 47.5: Iy = 187.4, Iz = 700, Iyg = -35.4 (Table A19.3) K, = 35.4/(187.4 x 700 - 35.4%) 5 -35.4/108950 = -.000324 Ka = 700/108950 = 00643 157.4/108750 = 001447 đụ = - | „001447 x -215000 ~ (~,000324 x 1,000,000) | x~ L(-oos4ø x 1,000,000 - (~.00824 x 215000) | 2 whence Oy, = -14.5 x -6360 2

Column 12 of Table A19.3 gives the results of this equation and column 13, the total stringer loads at station 47.5,

The stresses in column 12 of each table would be compared to the failing stress of the

flange members to obtain the margin of safety ANALYSIS FOR SHEAR STRESSES IN WEBS AND SKIN

The shear flow distribution will be calculated by using the change in axial load in the stringers between stations 20 and 47.5, a method commonly referred to as the AP method For explanation of this methed, refer to art Al5.16 of Chapter Al5

The shear flow in the y direction at a point n of the cell wall equals,

n

= AP

Yn = % ~ BO we 2 - eee (8)

where dg is a known value of shear flow at some point o and the second term 13 the

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TABLE Al9.2

SECTION PROPERTIES ABOUT CENTROIDAL X AND Z AXES Wing Section at Station 20 (Compression on upper surface) 1 | 3 4 5 6 7 a 9 10 LL 12 13 | 9 | Flange Area a az | aze2 x” axe f AKIZ 2 7P sØyA^ | 1 848 5.50 3.56 19.61 -gQ.10 ~0,06 9 -37880 24390 1 z | 1353] 5.90 | 2,08 | 12.30 5.65 1.99} 11.30 -39840 | -14070 | 3 | 1300} 5.85} 1.76 | 10.28 | 11.20 3.36 | 37.80 -39620 | -11890 j 4 | ,300| 5.55 | 1,66 | 9.29 | 16,85 5,05 | 85,00 -37950 | -11370 | s | 300] 3.08] 1.52 | 7.68 | 22.40 6.72 | 180,30 -35115 | -10510 ï 6 | 1353] 4.40 | 1.55 | 6.83 | 28.00 9.89 | 277.00 -31450 ' ~11090 7 | 630] 3.55 | 2:24 | 7.95 | 38.72 23.15 | 826.00 -26560 | -16700 | a | 692 | -8.40 | -5.81 | 48.80 | 35.72 24.70 | 881.00 40750 | 28230 9 | 270| -8.50 | -2.30 | 19.55 | 27.80 7.50 | 208.00 41280 | 11180 10 | 270 | ~8.50 | -2.30 | 19.55 | 22.60 6.10 | 138,00 41250 | 11170 ii | 160|-8.30 | -1.33 | 11.06 | 16.70 2.67] 44,80 40100 6430 12 | 27 | -8.00 | -2,16 | 17.30 | 10.90 3.94| 33.00 38400 | 10390 13 | 27 | -?.50 | -2.02 | 15.20 5.60 1,81 8.45 35570 9600 14 | 1768 | -6.50 | -3.00 | 32.50 | - 9.10 208) 0 29980 | 23020 £ |:.584 ~6.56 |238.0 96.44 | 2700 0.00 Za 1m - -1.176 nezal Ngtes:

See Fig, Al9.25 for Section at Station 20

#„ 96.44 5173 Reference axes X'X' and 2'Z' are assumed as shown

5.384 Properties are calculated with respect to these

Ig = 238 ~ 5.584 x 1.2762 = 20,3 int axes and transferred to the centroidal X and Z axes Tz ¬ 2700 - 9.584 x 17.37 „ 1030 in, Ing = -163.8 = 5.584 x-1.175 x 17.3 « -50 MT = 3.3 X - 5639 2 { 1 TABLE A19.3

SECTION PROPERTIES ABOUT CENTROIDAL X AND Z AXES

Wing Section at Station 47,5 (Compression on upper Surface) tị? 3 4 § 6 7 § 9 10 11 12 13 Flange A No A | oa az] age? 2 xt axt | ax:? 2 AM'Z! mm |P«epA 1 | 476 | s.€0 | 2.66] 14.95 - 0,05 9 = 0.27 -15,80 -39275 | -18700 2 | 318] 5.83 | 1.8¢] 10.80 1,40| - 6.18 8.16 -11,10 -40790 | -12890 3 66 | 5.75 | 1.53] 8.80 2.66] 28,60 15.30 - 5.50 -4025 | -10710 4 | 266 | 5.42 | 1.44] 1.81 4.15] 64.75 22.40 Ị 9,10 -38400/ -10200 | 5 | i266] 4:90 | 1:30! 6140 8.64} 120.00 27.60 | 3.70 ~35078 | - 9320 | 6 | 318) 4.307 1.37] 5.88 8.50} 226.50 36,60 | 11,20 -31360 | -10000 7 | 476 | 3.80 | 1.71] 6.17 18.10} 478,00 54.40 | 16,20 -37030Ì -12880 j a | 553 |-7.10 | -3.93 17.51| 555.00 | -124.50 16.20 41050 | 22770 + 9 | ,23E |-?.40 | =1.74 6.23] 165.00 | - 48.10 11,00 43090 | 10140 10 | 1225 | 7.40 | 1.74 j - 5,00] 107,00 | - 37.18 5.80 43180 | 10160 qi -0.90 28.30 | - 13.60 ~ 0.40 ¡ 41900 3240 ¡ 12 -1,63 21,70 | - 15.60 ~55.90 | 40085 9410 13 -1.47 4.75 | - 6.61 ~11.00 36060 8470 j 14 ~3,27 0 0.33 ~15, 60 30725 18600 ¡ £ ~2,80 | 129,1 71.3 11804 - 79.0 i g Ị Ê „ -2,80/4.61 © -0,61" i General Notes:

Ä „ 71.3/4.61 < 18,50" See Fig Al19.26 for section at Station 47.5

Ix = 189.1 - 4,61 x.612 „ 157.4 Reference axes X'X' and Z'Z' are assumed as shown

: #b s -14.5 X - 6380 7

Tz = 1804 - 4.61 x 15,5” « 700 lxz s -T9 - 4.61 x -.61 x 15.5 - 35.4

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AP equals the changes in stringer axial load

over a distance d in the y direction

Since the cell in our problem is closed the

value q, at any point is unknown We assume it

zero on wed 1-14 5y imagining that the web is

cut as shown in Fig Al$.27 Equation (6) thus

reduces to,

Columns 1 to 5 of Table A19.4 show the solution of equation (7) The shear flow values of q in column 5S are plotted on the cell wall in Fig Al9.27, remembering that dy = dy = qz rules giving direction of dy and qg refer to Chapter Al4, Art Al4.6 288.8 374.6 306.8 _ 245.3 414.2 tor 353,2 { aa} z c.g ' q wl ——— At M1 ox te 1 236.3 ` 280.0 —3ịaT '3545 Fig, A18, 27 TABLE A19.4

Caiculation of Flexural Shear Flow Assuming q 2 zero in Web Between Flange Members (1) and (14) (Shear values are average between stations 20 and 47 5) 1 2 3 4 5 8 7 fala P P \ qx | m qr Fane) ac Station| at station}- 3% | 2p | sq | mạ | at o 20 4T.5 -9|21.5| ìn a+ + |~24390w | -18700 | 206.8 2 |-14070 | -12090 | “a9.1/ 298-8) 48-8 3 | -11890 } -10710 | 42.9] 383-3) 38-8 4 =11370 ~10200 42.5 331 3 42, 8 3 | -10810 | = 9320 | 43.3] 331-8) 42-8 6 |-11080 | -1oo00 | 39.6] 378-5) vào 7 | 16700 } -12880 | 139.0| $52) 320 8 | 28240 | 22770 [a198.7] $59-2)197- 9 | 11180 | 10140 j= 37.8] 358-3) 47-8 10 | H170 | 10180 |- 36.1| BIT ae 11 5430 5240) 43.2] 280-0) 40-8 12 | 10390 9410 |+35.6| 201 5] sạng 13 9600 3470 |= 41.1 Tủn | sáng l4 | 29036 | 12600 | -sg0.3) 97) Đóng Total mq = 256060

For equilibrium of all the forces itn the plane of the cross-section My must squal zero for convenience we will select a moment y axis througn the c.g of the cross-section The moment of the shear flow q on any sheet element equal q times double the area of the triangle formed by joining the c.g with lines going to each end of the sheet element These double

À19.19 areas are referred as m values (See ?P1g A19.27), Column 6 of Table Alg.4 records these double areas which were obtained by use of 4 planimeter Column (7) gives the moment of each shear flow about the c.g and the total of this column gives the moment about the c.g of the complete shear flow system of Fig Al9.27 or a value of 256060 in.1b

The double areas (m) can be found approximately as follows:

The moment of the shear flow q on the web (2-3) about point O equals q times twice the area

(A, + A,) In most cases, the area A, can be neglected

By simple

geometry, the area

A, = 1/2 (X, 2, ~ X,2,) The moment of the shear flow q on wed (2-3) thus equals q (X, 2

~ X;¿ 2Z„ ) Since values of X and Z for all

flange points with reference to section (c.g.) are given in the Table A19.2, 1t is unnecessary to use the planimeter except for regions of sharp curvature

MOMENT OF EXTERNAL LOADS ABOUT c.g OF STATION 20

AS stated before the engineers in the

applied loads calculation group supply the shears and moments at various spanwise stations We will assume that these loads are: Vz = 12000

Ub., Vy =+2700 1b., My = - 390,000 in.lb The

location of the reference y axis used by the leads group will be assumed as located at point QO in Pig Al9.28 relative to cross-section at Station 20 (2 2 re cv Vz 120008 ~~] io T7 x St] m3 ƒ V„= 27008 * “x s0 nue 4b , My = 390000"# 33.3 —a Fig A19 28 Therefore moment of external loads about c.g 18, Mog, = 12000 x 35.5 - 2700 x 11.8 - 390000 = 41800 in.1b Moments Produced by Inclination of Flange Loads With Beam Section

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have components 1n the Z and X directions

Columns (4) and (7) of the Table Al9.5 give the

values of these in sảng components The slopes

dx/dy between stations 20 and 47.5 are found oy Scaling from Fig Al9.24 Fig Al9.29 shows

these induced in plane forces as found in Table Al9.5 TABLE A18 5

In Plane Moments About Section c.g Produced by in Plane Components of Flange Loads Station 20 1 2 3 4 3 § T § Pự= |MG,g, Pz= | Mog, = Flange |g ox az No ay p& |= Pxz se ye ay eo | (2 x) az} 1 -.046 1120 T480 ,025 +610 - 10600 2 0 8 9 -022 | -310 ~ 3610 3 9 0 Q 022 | -262 + 1600 4 9 9 9 „022 | -250 ~ 112 5 9 9 0 „022 | -23L 1180 6 9 9 0 025 | +28 2970 7 +021) - 350 -1660 025 | +417 1710 8 -021 584 -4300 „018 | -508 9350 9 9 g 9 G17 | -190 2000 10 0 9 0 „015 | -167 885 11 0 9 9 -815 | - 96 - 5T 18 9 9 0 „017 | ~LTT ~ 1130 13 9 0 8 Ø18 | ~173 = 2020 14 -.046 | ~1060 3840 „018 | -416 = 7260 z | 160 + 2224 NOTES:

Coiumn (2); P trom Table A19.2 Coiuma (5) and (8): Values of Z and X are found in

Columns 10, 11 of Table A19.2 610 417 5 231 Hog f 262 {4 250 180 508 Fig 419, 29 In plane forces produced by flange axial loads

The moments of these in plane components about the section c.g are given in Columns (5) and

{8} of Table AlS.S5 In general, these moments are not large

Total Moments of All Forces About Section o.¢ at Station 20:

Due to flanges = 7160 - 2224 =

(Ref Table Al9.5) 4936 in.lb Due to assumed static shear ?19w = 256060

in.lb (Ref Table 419.4)

Due to external loads = 41800 in.1d Then

256060 + the total unbalanced moment = 4636 + 418CO # 502798 tn.ì5,

For equilibrium, this must de balanced by a constant shear flow q,

hence

796 :

TELS 7 7 328 1b./1n,

(Note: 461.5 = total area of cell}

The shear stresses q, are listed in Column (8) of Tab1e A19.4,

The final or resultant shear ?1ow ay at any peint therefore equals

dr “q2

The resulting values are given in Column 9 of Table 419.4 Fig Al9.30 illustrates the results graphically

Final shear flow diagram

ues see Column 9 of Table Als.4

Fig A19.30 For val-

Having determined the shear flows, the shear stress on any panel would be q/t In checking the sheet for strength in shear and combined shear and tension, interaction rela~ tionships are necessary The strength design of sheet panels under combined stresses is covered in considerable detail in Volime II

A19,14 Bending and Shear Stress Analysis of 2-Cell Multipie Stringer Tapered Cantilever

Wing

A two-cell beam is also quite common in wing structural design A two-cell structure

in bending and torsion is statically indeterminate to the second degree since the shear flow at any one point in each cell is unknown However, due to continuity between cells the angular twist of each cell must be the same, which gives the additional equation necessary for solving a two-cell beam as compared to the single cell analysis

Example Problem

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Al9, 21

lations as was used in the previous single cell The first 7 columns of this table are the same problem, the bending and shear stresses will be as in Table Al9.4, since no stringers have been determined Zor the same structure 4s in the added to cell (1), and the shear q is assumed

rên

previous example except that the leading edge cell is considered effective, thus making 4 2-cell structure Since there are no spanwise stringers in the leading edge, very little skin on the compressive side will be effective On the tension side, the leading edge skin would be effective in resisting bending axial loads and thus the moment of inertia would be slightiy different from that found in example problem 1 Since this problem is only for the purpose of

illustrating the use of the equations, the leading edge skin will be neglected in computing the bending flexural stresses With this

assumption, the bending stresses and flange loads at stations 20 and 47.5 are the same as for the previous problem (See values in colum 12 and 13 of Tables AlS.2 and A19.3.)

Shear Flow Calculations: -

To compute the static shear flow, each cell is assumed cut at one point as shown in Fig AlS.31, and thus the shear flow is zero at points (a) and (b), Fig A19.32 Cell (1) b zero in cell (1)

To make the twist of each cell the same and also to make the summation of all torsional forces zero will require two unknown constant shear flows, q, in cell (1) and q, in cell (2) Thus two equations will be written, nameiy: 9,59, -+ - ee -e (8) Meg 7 9 cere (8) The twist 6 per unit length of a cell equals 1

Ome lal/t -+ +-+ (10) 2AG

The modulus of rigidity G will be assumed constant and thus will be omitted

Consider cell (1):

Columns 10 and 11) (Refer to Table Als.6,

Area of cell (1) = 83.5 sq in 9 2— [P44 €3 4, L/t «3 g, L4 Cau (2) iL = - ï T_T + - y Area = 83.5 Area = 461.5 sq in 9, 2x 83.5 9 1278 q, ~ Z30 q„ |; wHence 14 13 - 12 + ịị = 100 : 9 6, = 7,65 q„ ~ 1.278 Qạ- - = ~ ~—— -¬ (a) For cell (2): Table Al9.6 (Column 5) zives the value of

the static shear flow under these assumptions Area of cell (2) = 461.5 sq in

TABLE 419.6

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9= TEAS | 735050 - 230 q, + 2469 aa]

O, = 794 - 2495 gq, + 2.678 Gq, - = = - (d)

For continuity 9, must equal G,, hence equating (a) and (b):

7.889 q, - 4.086 q„ ~ 794 =0 (c)

For equilibrium the summaticn of all mom-

ents in the plane of the cross~section about the

section (c.g.) must be equal to zero, or

Š Hẹ,g, TÔ

The moment of the external loads about the section c.g is the same as in previous problem

=4

M external forces 41800 in.1b The induced moment due to the in plane components of the flange axial loads is likewise the same as in previous problem (see Table A19.5)

Maue to flange loads * 4956 in.1b The torsional moment due te the static

Shear flow from Column (7) of Table AlS.6 equals 256060 in.ib The torque due to the unknown constant shear flows of q, and q, 1s equal to twice the enclosed area of each cell times the shear flow in that cell, whence nN ) 5 2x83.5 q,+2x461.6 q, = 167 4, +923 a, due to q, and q, Therefore Me.g, = 167 4, + 923 q, + 502796 = 0 -~+- Solving equations (c} and (d), we obtain, q, = -62.8 Ib./in., aq, = -317 lb./1n These values are listed in columns (12) and (13) of Table A19.6 The final or resultant shear flow Gp on any Sheet panel equals the sum of q+ qd, + q, The results are shown in colum

(14) ot Table Al9.6 Fig Al9.22 shows the potted shear flow pattern Comparing this figure with Fig Al19.30 shows ths erfect of adding the leading edge cell to ‘the single cell of the previous problem Se a he ee ị ⁄ lễ sỉ i hi Fig.alg9.a2 Po, g s Ss id “4 SEG nto A19.15 Bending Strength of Thick Skin - Wing Section

Figs 1 and k of Fig A18,ð 11 approximate shapes for airfoils of

aircraft Such airfotls nave relatively low thickness ratios and since supersonic military aircraft have comparatively nigh wing loadings,

it {is necessary to go to thick skin in order to resist the wing bending moments efficiently The ultimate compressive stress of such structures can be made rather uniform and occurrin at stresses considerably above the yteld point of the material, Since structures must

the design loads without failura, it is

sary to be able to caiculate the ultimate bending resistance of such a wing section if the margin of safety is to be given for various

load conditions

The question of the ultimate bending

resistance of beam sections that fail at stresses beyond tne elastic stress range 1s treated in Article A13.10 and example problem 7 of Chapter Al3 and should ce studied again before pro~ ceeding with the following example problem,

Al9.16 Example Problem

To illustrate the procedure of Art AlZ,i0, a portion of a thick skin wing section as

illustrated in Fig Al9,.34 sill be considered, Fig Ag 34

For simplicity the section has been symmetrica about the x-x axis The material is alumim alley In this problem the material stress-— strain curves will be assumed the same in Doth tension and compression The problem is to determine the margin of safety for this team section when subjected to a design bending moment My = 1,850,000 in.1b

SOLUTICN:

Since it is desirable

formula op = MyZ/Ix, Ít is necessary to obtain a modified beam section to correct for the nonlinear stress-strain relationship since the give structure will fail under stresses in the inelastic zone The maximum compressive stress at surface of beam will be assumed at SOCCO psi

to use the bean

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mS

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES ation of crippling and column strength of the

stiffened skin, a subject treated later Curve (A) of Pig Al9.35a is a portion of the compressive stress-strain diagram of the aluminum alloy material from zero to 50,000 psi

Due to symmetry about the x-x axis, we need only to consider one half of the beam section We divide the upper hai? of the beam section into a horizontal strips, each 3/8 inch thick Each beam portion along these horizontal strips can be placed together to form the areas labeled (1) to (8) in Fig b of Fig A19.35 Since plane sections remain plane after bending

in both elastic and inelastic stress zones, Fig ¢ shows the beam section strain picture Fig A19.35 Fi Fig ¢ 8.2 ,01"/1n .010 + iB —- g 3 3.008 7 R 3 3.008 a " 5 004 1 @ 002 9 10 20 30 40 50 o.% 1000 (psi)

Table Al9.7 snows the calculations for obtaining the modified moment of inertia of the cross-section to use with the linear beam formula TABLE Al9.7 1 2 3 4 5 8 1 8 Area K= Arm Portion! A | Ớc | đc |ơc/GglAa=KAI 2 |AeZ” 1 |4.500|49200|46900] 1.049 | 4.71 | 2.8125/37 30 2 ]2.437|48000|40600] 1.182] 2.88 |2.4375117.14 3 [0.211] 46000] 34400| 1.340| 0.282 J2.0625] 1.20 4 |0.211|43200|28100] 1.539] 0.324 |1.6875] 0.92 5 J0.211139200|21840] t.790| 0.378 [2.3125] 0.65 6 J0.211|33200|15620| 2.121] 0.449 J0.9375| 0.38 7 |0.211120500) 9360| 3.1904 0.461 |0.5625J 0.15 8 J0.211| 7000| 3120] 2.240| 0.473 |0.1875| 0.02 ä 57.18 2 Iq = 115.52

Col (2) = Area of Strip

Col (3) Stress at midpoint of strip as read from Curve (A)

Col.(4) Stress at midpoint of strip as read from Curve (B) Col.(5) Nonlinear Correction Factor K = o,/a,!

Col.(6) Modified Area = Ag = KA

Col.(7) Arm from Neutral Axis to Midpoint of Strip

Col.(8) Modified section moment of inertia

Ald 23 The values in column (3) of Table Alg.7

represents the true compressive stress at the

midpoint of a strip area when the beam is

resisting its maximtm or failing bending ’moment The values in column (4) represent the com~ pressive stress at the midpoint of the strip

areas if the bending stress is linear and vary-

ing from zero at neutral axis ta 50000 psi at edge of beam section (Curve B of Fig Al9.35a)

To illustrate, consider strip area number (2) in Fig b of A19.35 Project a horizontal dashed line from midpoint of this strip until it intersects curves A and 3 at points (a) and (bd) respectively From these intersection points project downward to read values of 48000 and 40600 psi respectively

In using the linear beam formula, the stress intensity on strip (2) would be 40600 out actually it is 48000 The ratio between the two is given the symbol K Thus to modify the linear stress to make it equal to the nonlinear stress we increase the true strip areas by the factor K, giving the results of column (6) The modified moment of inertia (column 8) equals I, = 115.52 The design bending moment was 1,850,000 in.lb Consider point at midpoint of strip (1) ã = 2.8125 inches Oy = My Z/Ty = (1,850,000 x 2.8125)/115.52 45100 psi

This stress is based on the modified strip areas The true stress My strip 1 thus equals Koy = 1.049 x 45100 = 47400 The allowable stress at failure equals 49200 from column

(3) of Table Al9.7 Hence margin of safety = (49200/47400) - 1 = 04 or 4 percent

The margin of safety for points other on the beam section will likewise be 4 percent For example at midpoint of strip (3), Z = 2.0625 K=1.34

whence On, =[L (1,850,000 x 2.0825)/115.58 ]I.34 = 44400

whence margin of safety = (46000/44400) - 15 04

The moment of inertia without modifying the strip areas would come out to be Iy = 104.42

hence the stress at midpoint of strip (1) would

calculate to be ap = (1,850,000 x 2.8125}/104.42 = 49900 The allowabie strvss for linear stress yartation would be 46900 from column (4) of Table Hence margin of safety would be

ran

Hà

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(46900/42900) ~ 1 Z ~.O06 The elastic theory

thus gives a margin of sa?sty 1O percent less than the strength given when true stress-strain or non-linear relationship is used

If the same comparison was made for bending about Z axis of this same beam section the difference would de considerably more than 10 percent as more beam area 1s acting in the region of greater descrepancy between curves A and B,

A19.17 Application to Practical Wing Section

A practical wing section involves these

facts: - (1) The section is unsymmetrical; (2) external load planes change their direction under different flight conditions; (3) the material stress-strain curves are different in tension and compression in the inelastic ranges

Since the stress analyst must determine eritical margins of safety for many conditions,

it would be convenient to mve an interaction curve invoiving My and Mz bending moments which would cause failure of the wing section This

interaction curve could be obtained as follows:- (1) Choose a neutral axis direction and its location

(2) Assuming that plane sections remain plane, and taking the maximum strain as that causing failure of the compressive flange, use the stress-strain curve to determine the longitudinal stress and then the

internal load on each element of the cross- section A check on the location of the assumed neutral axis is that the total compression on cross-section must equal total tension Since the location was assumed or guessed, the neutral axis must be moved parallel to itsel? to another location and repeated until the above check is obtained

Find the internal resisting moment about the neutral axis and an axis normal to the neutral axes Resolve these moments into moments about x and z axes or My and Mg These resulting values of My and are bending moments which acting together will cause failure of the wing in bending (4) Repeat steps 1, 2 and 3 for several other

directions for a neutral axis which results will give additional combinations of

and Mz, moments to cause wing failure Thus an interaction curve involving values of My and Mz, which cause failure of wing in bending is obtained and thus the margin of safety for any design condition is readily

obtainable

Al19.18 Shear Lag Influences

In the beam theory, the assumption plane sections remain o

beam involving sheet and stringer this assumption means that the sheet pan:

have infinite shearing dity, whicn of course £8 not true as shearing stresses oroduce shearing strains The effect of sheet panel shear strains is te cause some stringers to resist less axial load than those calculated doy 5eam theory This decreased effectiveness of

stringers is referred to as "shear lag" effect, since some stringers terd to lag bacx from the

position they would take if plane sections re-

main plane after bending thas Ina

In general, the shear lag ef? stringer structures is not appreci for the following situations: - e she Sle excep mt 2 ai (1) Cutouts which cause one or more stringers to be discontinued (2) Large abrupt changes in external load applications

(3) Abrupt changes in stringer areas In Chapters A7 and A8, strains due to shearing stresses were considered in solving for distortions and stresses in structures involving sheet~stringer construction Even in these So-called rigorous methods, simplifying assumptions must be made as for example, shear

stress is constant over a particular sheet panel and estimates of the modulus of rigidity for sheet panels under a varying stata of buckling must be made The number of stringers and sheet panels in a normal wing is large, thus the structure 1s statically indeterminate to many degrees and solutions necessitate the use of high speed computors, Before such analyses can be made, the size and thickness of

each structural part must be known, thus rapid approximate methods of stress analysis are desirable in obtaining accurate preliminary sizes to use in the more rigorous elastic analysis

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 2P/3 <— 2P/3 2P/3~— Fig A19.36

The theoretical load in center stringer can be calculated by methods of Chapter A7 and À8, and the results would give the solid curve of Fig Al9.37 To simplify the solution, it is common practice to assume the load distribution in the center stringe to vary according to the dashed curve in Fig A19.37 which

indicates that in a distance 3b, the load 2P is equalized between the three stringers 2p/3|—— — - -—~ — | —— Theory N ¬ T~~" Approximate ‘PP! Load in Stringer (2) Fig À19.37

Al9.19 Application of Shear Lag Approximation to Wing with Cut-Out

Pig Al9.38 shows the top of a multiple stringer wing which includes a cut-out in the surface The stringers (5), (6) and (7) must be discontinued through the cut-out region 3 2 1 | i Root 3b ——l ca Fig A19.38

It is assumed that the effectiveness of these 3 interrupted stringers is given by the triangles in the figure At deam section 1-1 these stringers nave zero end load The stringer load is then assumed to increase linearly to full effectiveness when it intersects the sides of this triangle whose height

A19 25

equals 3b At beam section 2-2 stringers (5)

and (7) have become effective since they inter- sect triangle at points (a) on section 2-2 At section 3-3 point c, stringer (6) becomes fully effective

To handle shear lag effect in a practical wing problem another column would be inserted

in Table Al9.1 between columns (3) and (4) to take care of the shear lag effect The shear lag effectiveness factor which we will call R would equal the effectiveness obtained from a triangle such as illustrated in Pig A19.38

For example, the shear lag factor R at beam section 1-1 in Fig Al9.38 would be zero for stringers (5), (6) and (7) and one for all other stringers At beam section 2-2 stringers

(5) and (7) have a factor R = 1.0 dince they are fully effective at points (a) Stringer

(6) ts only SO percent effective since section 2-2 is halfway from section 1-1 to point (c), thus R = 0.5 for stringer (6) At beam section 3-3, stringer (6) becomes fully effective and thus R = 1.0 for all stringers The final modified stringer area (A) in column (4) of Table AlS.1 would then equal the true stringer area plus tts effective skin times the factors KR The procedure from this point would be the same as discussed before Thus shear lag ap— proximations can be handled quite easily by modifying the stringer areas Using these modified stringer areas, the true total loads

in the stringers are obtained The true stresses equal these loads divided by the true stringer area, not the modified area

A19.20 Approximate Shear Lag Effect in Beam Regions

where Large Concentrated Loads are Applied

Wing and fuselage structures are often required to resist large concentrated forces as for example power plant reactions, landing gear reactions, etc To illustrate, Fig A19.39 Tepresents a landing condition, with vertical

Trang 11

and flange members Fig (a) shows the bending moment diagram due to the landing gear reaction alone The internal resistance to this vending moment cannot be uniform on a beam section adjacent to section A-A because of the shear strain in the sheet panels or what is called Shear lag effect To approximate this stringer effectiveness, a shear lag triangle of lengtr 3b is assumed, and the same procedure as discussed in the previous article on cut-outs is used in finding the longitudinal stresses, It should be understood that the bending moments due to the distributed forces on the wing such as air loads and dead weight inertia loads are not included in the shear lag considerations, only the forces that are applied at concentrated points on the structure and must be distributed into the beam A side load on the gear or power plant would produce a localized couple plus an axial force besides a shear force as in Fig Al9.39 The resistance to this couple and axial force would likewise be based on the effectiveness triangle in Fig, Al9.39 A19, 21 Approximation of Shear Lag Effect for Sudden

Change in Stringer Area

Stringers of one size are often Spliced to stringers of smaller size thus creating 2 dis- continuity because of the sudden change in stringer area,

Pig Al9.40 shows the stringer arrangement in a typical sheet-stringer wing Stringer B is spliced at point indicated The stringer area Ag is decreased suddenly by splicing into a stringer with less area A

Top Surface of Wing r I 3b + —3b Fig A19 40

To approximate the shear lag effect, assume the area of stringer B at splice point to be the average area of the two sides or

(A, + A,)/2 This average area is then assumed to taper to A, and A, at a distance 3b from the splice point The shear lag effectiveness factor R will therefore be greater than 1.0 on the side toward the smaller stringer A, and less than one on the side toward the stringer with te greater area Ag, Since the average area was used for the splice point,

À18.22 Probiems

(1) Fig Al9.41 shows a cantilever, 3 stringer, Single cell wing It is Subjected to a

distributed airload of 2 lb./in.? average # Ls + ty —— A Tran Wing Tip = 20" 5 —H9" 21 0096 ————mỪ † /cai A 12" & + B Fig Al9 41

intensity acting upward in the z direction and 0.25 lo./in.? average intensity acting rearward in the x direction ‘The center of pressure for 2 forces 1s on the 25 percent of chord line measured from the leading edges edge and at mid-height of spar AB for the x air forces Assume the 3 stringers A, 8, C develop the entire resistance to ex~ ternal bending moments Find exial loads in stringers A, 8, C and the shear ?1ow in the 3 sheet panels of cell (1) at wing Stations located 50", 100" and 150" 2rom wing tip Consider structure to rear of cell (1) as only carrying airloads forward to cell (1) and not resisting wing torsion or bending

Fig A19 42

28"

Rear Spar |_ | Single >, prom gear zo"

Pin a Front gpart, 420 y

Fittings “*! al = = Te"

d “— tœ' 47 — 10" 4 67,20",

Trang 12

(3)

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES and Wy = 5 lb./in., acting to rear and

located at mid-deptn of wing Find re-

actions at points (a), (b) and (d) Find

axial loads on front and rear spars Find primary bending moments on front spar

Find shear flow on weds and walls Neglect

structure forward of front spar and rear-

ward of rear spar

Fig Al9.43 shows a portion of a single

Al9 27 Table A gives the stringer areas at sta~ tions © and 150 Assume stringers have linear variation in area between these two stations Use 30t as effective skin with compression stringers

Find axial loads in stringers at stations

150 and 150 and determine shear flow system at station 150

cell - multiple stringer cantilever wing (4) Same as problem (3) but add an internal The external air loads are: Web of 04 thickness connecting stringers

(3) and (8)

Wg = 100 1b./in acting upward and whose

center of pressure is along a y axis coin- (5) Same as problem (4) but add a leading edge ciding with stringer (3) cell with radius equal to one-nalf the

Trang 13

* North American Aviation FJ3 "Fury Jet’ View Shows Bottoat

Note Integral Construction of Skin and Stringers si PRED ER SS bóc

Trang 14

CHAPTER A20

INTRODUCTION TO FUSELAGE STRESS ANALYSIS

A20.1 General In general the purpose of an alir- plane is to transport a commercial payload or 4 military useful load The commercial payload

of a modern airliner may be 100 or more passengers and their baggage These passengers must be transported safely and comfortably For example, an airliner flies at high altitudes where temperatures may be far below zero and where the air density is such as not to sustain human life These facts mean that the body which carries the passengers must be heated, ventilated and pressurized to provide the

necessary safety Air travel must be acceptable to the passengers, thus the airplane body must shield the passengers from excessive noise and vibration, and furthermore efficient, restful and attractive furnishings must be provided to make travel enroute comfortable and enjoyable, The portion of the airplane which houses the

passengers on payload is referred to as the fuselage Fuselages vary greatly in size and configuration For example, the fuselage of 4 supersonic military airplane may nouse only one passenger, the pilot, the remainder of the fuselage interior space being used to house the power plant, to provide retracting space for landing gear, and to house the many mechanical and electronic installations which are necessary to fly the airplane and carry out the various operations for whicn the airplane was designed to accomplish Many groups of engineers with various backgrounds of training and experience are therefore concerned with the design of the fuselage The structures engineer plays a very important part because ne is responsible for the strength, rigidity and light weight of the fuselage structure

A20,2 Loads Basic Structure

The wing, being the lifting body is subjected to large distributed surface air forces, whereas the fuselage is subjected to relatively small surface air forces The fuselage is subjected to large concentrated forces such as the wing reactions, landing gear reactions, empen-

mage reactions, etc In addition the fuselage

houses many items of various sizes and weignts which therefore subject the fuselage to large

inertia forces In addition, because of high altitude flight, the fuselage must withstand internal pressures, and to nandle these internal pressures efficiently requires a circular cross- section or a combination of circular elements

The student snould refer to Chapter A4 for further discussion of loads on aircraft and also

to Chapter AS, where example calculations of fuselage shears and moments are presentec

Trang 15

A20 2 FUSELAGE STRESS ANALYSIS el PHOTO NO 2

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A20.3 Stress Analysis Methods

Section Effective Crass~-

It tS common practice to use the simplified beam theory in calculating the stresses in the skin and stringers of a fuselage structure If the fuselage 1s pressurized, the stresses in the skin due to this intermal pressure must be added to the stresses which resist the flight loads In wings the skin in the middle region of the airfoil is relatively flat and thus the skin is usually considered as made up of flat sheet panels In fuselages, however, the skin is curved and curved sheet panels have a higher eritical compressive buckling stress than flat panels of the same size and thickness In small airplanes, the radius of curvature of the fuselage skin is relatively small and thus the additional buckling strength due to this curvature may be appreciable A simple procedure of approximately including the effect of sheet

curvature will now be explained

Fig A20.1 illustrates a distributed stringer type of fuselage section Assume that external loads are applied which produce bending of the beam about the Y axis with compression on the upper portion of the cell compression side tension Š side wo 8 5 nu } 5 9 fl i j : k 8 Fig A20.1

Up to the point of buckling of the curved sheet between the skin stringers, all the material in the beam section can be considered fully effective and the bending stresses can be computed by the general flexure formula op = Mz/ly, where ly is the centroidal moment of inertia of

the entire section When a bending compression

stress is reached which causes the curved sheet between stringers to buckle, a re-arrangement takes place in the stress distribution on the section as a whole Theory as well as experi- mental results indicate that the ultimat> compressive strength of a curved sheet with edge stringers can be approximated by the following

two assumptions

A20.3 (1) A small width of sheet w, on each side of

the attachment line of skin to stringer ts considered as carrying the same compressive stress as the stringer, as was discussed in Chapter Al9 These effective sheet widths w, are shown as the blackened elements adjacent to the stringers on the compressive Side in Fig A20.1

The remainder of the curved sheet between stringers, namely, b-(w, + w,) carries a maximum compressive stress Sor F SEB t/r This value for dor is conservative £& is the modulus of elasticity of the skin material, t the skin thickness, and r the radius of curvature of the skin These curved sheet elements are shown by the natched skin lengths in Fig A20.1 Since the thin curved skin between the stringers normally buckles under a compressive stress far below the buckling strength of the stringers, the curved sheet is treated as an element with varying effective thickness which depends on the ratio of the curved sheet buckling stress đẹp to the bending stress 0, exist- ing at that point for bending of the fuselage section Hence the effective sheet thickness for the curved sheet panels can be written,

tạ = t (Øer/Cp)

or an effective area can be written

Ae = b't (op/Gy) - - eer where b' 1s the width of curved sheet between the effective sheet widths w,, w,, etc (See Fig a20.1}

To illustrate this approach in obtaining the effective cross-section of a fuselage section, an example problem will be presented The example problem will be broadened to some extent for the purpose of introducing the student to design procedure

A20.4 Example Problem

Let it be required to determine the stringer arrangement for the approximate elliptical

shaped fuselage section shown in Fig A20.2 The following data will be assumed: - Design bending moment about y axis = 160,000 in lb (producing compression on upper portion), Zee stringers, one inch deep and with an area equal to 0.12 sq in shall be used

The ultimate compressive strength of the zee stringer plus its effective skin and a length equal to fuselage frame spacing {s assumed to be 32000 psi The skin thickness ts 032 and all matertal 1s (2024) aluminum alloy with © =

10,300,000 psi The fuselage stringers are to

be symmetrical about section center lines

Trang 17

Height = 50" “3 Z stringers Width = 30” “hủ = 11 nàn Z=-3.38 5 - — ¿N.A Trial 2 —#[ TT ~T TT N.A Trial 1 14 Fig A20,2 Symmetrical about the Z axis ?1 233 Location of stringers 3 below Y'-¥' same as above Y’-Y" Solution:

The first thing to do is to determine approximately how many Z stringers will be required so that a section can be obtained to work with Since the internal resisting mom- ant must equal the external bending moment, one can guess at the internal resisting couple in terms of total compressive flange stress and an effective internal couple arn

Por elliptical and circular sections with distributed flange material, the approximate effective resisting arm of the internal couple ean be taken as 0.75 times the height a, and the average tension or compressive stress as 2/3 the maximum stress Thus equating the external bending moment to the internal resisting moment an approximate total area A, for the compressive side of the fuselage section can be

obtained

My = Ay (.67 Oy) (.75 nh}, whence,

Ag = „75 X 50 x 67 x 32000 460000 = 2.0 sq in Part of this total area 1s provided by the effective skin area The effective width to use with each rivet line equals w= Ct VE/S85¢- We will take C = 1.7 which ts a commonly used

value `

WF 1.7x 082710,500,C00/32000 = 975 in which equals a width of 975/.032 = 30.3 sheet thicknesses Since the bending stress decreases to zero as the neutral axis is approached, 2n¢ since the curved sheet between the Z stringers can carry loads up to its ouckling str th, 2 preliminary value ef effective width w = 40t will be assumed acting with each stringer total area of stringer plus effective skin equals 0.12 + 40 x 0327 = 0.16 sq i number of stringers requirec is there Q.16 = 12, Thus ti a = ®

Fig AZO.2 shows how the stringers were placed to give 12 stringers on the top nalr Since the skin on the lower half is in tension and therefore fully effective, the neutral axis will fall below the center line and thus the two stringers on the center line will be considered as part of the required 12 stringers A fuselage cross-section has now been obtained The desired final result is that the neximum compressive stress will be near but not over 32000 psi The procedure from this point is still a trial and error process since the effective sheet on the compressive side depands

tude of the compressive bending stress which in turn is influenced by the amount of sffective sheet and the buckling load carried by the curved sheet

Using the preliminary stringer arrangement of Fig A20.2, Tables A20.1 and 420.2 sive the calculation of the e*fective moment of inertia of the section about the horizontal neutral axis, Table A20.1 deals with the stringers and the effective sheet elements and Table A20.2 deals with the curved buckled sheet elements

In the trial No 1, the following assumptions are made:

(1) A width of 30 thicknesses of skin act

with 4ach stringer cn the upper or compressive side

(2) The area of the curved sheet between

the effective sheet widths as found in (1) is

modified to give an effective area by multiplying by a K factor of Gpr/Op, nheré Gor 18 the buckling compressive stress and dp is the bending stress at the center of the curved skeet element assuming 32000 at the extreme upper fiber of the beam section and zero at the hortzontal center line, witn linear variation in between these points

Trang 18

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A20,5 centerline axis of cell TABLE A20.1

TRIAL NO 1 TRIAL NO 2

1 2 3 4 § | 8 7 8 $ 10 1ì 12 Explanatory Notes for Table A20.1

ẩ " wey oes « —————

.* 1/838 se 53 TRIÁU NO L

tol area ä $ lšnm aul 32 † Col, 1 For numbering of stiff-

agl a Arm | az! ey, Loe cal 2% | az az? eners and sheet elements, see

majsq.ing) zt ụ 'ị cœ sả | oe Fig A20.2

gio} 15 -11200 | 1.66 acting separately, Tbe

212) 12 = 3680 | 2,88 entire skin between 213) 224 stiffeners is consider- 2i4l ,120 § a ed ag a unit S15] 224 3 § Col 3 All arms z' are measur- B16) 12 2 3 4.5 ed to hortzontal cen— 317/ ,180 a 5 terline axis ais} 1 Ễ gig) 16 4 8 TRIÁI NÓ, 2 #20| 12 » - Š z = distance to neutral

Z21| ,18 3 8 axis as found from re-

S#22| 12 2 5 sults of Triai No 1

ø2a| 088| a a 7 oy = 1600000 2/1470

4 @ + effective width based

SUM = 2.49 ~12,23| 714.3 on stress in Col 7 TABLE A20.2 TRIAL KO 1 TRIAL NO, 2 1 7 3 3 |4 5 6 7 | sto tiolu 12 | 13 14 15 | 16 7 17 | 18 3 - - = & về | g2 | ,xịu Ê aa Be "| tr 3 3 5 | Bel a | az jaze2i| js Sle 518 Ber) az | az? 38 ‘ g ™ M ị 24.58 125 | ah aa ws BS | wa | we ' Ti zee{eu-| sen »| 8ì 3 củ oe Saal - & = 4 bé = aia x b a a) em Ba" oe issn} 3 — 6 $ Ÿ | 1 | 2.27| 032 | 11 | -900o | -31900| ,282 | 020 | 24.9|0.50 |12.4| (28.28 |~30800 |.292 |2.25 | ,021 | 59|16.7 =| 3 | 4.04] (032 | 12 | -9000| -30300{ 296 | 038 | 23.7/0,90|21.3| ]27.08|-29500 | 208 | 3,99 | 039 11.06 |28.7 3 | 5 | 4.04|.032 | 24 | ~4510| -26000 | 173 | 022 | 20.3}0.45) 9.1! 23,68 -25800 | 174 |3.83| 022 | 0.53|12.3 | 7 | 4.04|.032 | 38 | -2600 | -20200| 128 | 016 |15.8]0.25 | 3.9| 19.18 |-20900 | 124 |3.80| 015 |9.29| 5,5 9 | 9 Ì 8.041.032 | 38 | -2600 | ~12900| Z01 | ,049 |10.1|0.40] 4,0| 13,48 |~14700 | ,177 |5.53| 031 |0.42 5.7 3 | 11 | 6.52|,032 | 38 | -2600| - 4100| ,63 | 13L | 3.2|0.42| 1.3j| 6.58 |- 7200 | 362 | 4.73} 055 | 0.36) 2.4 TƠTALS 286 2,92 |52.0| | tOTALS ,183 | 3.24|71,3

Explanatory Notes for Table A20.2 Explanatory Notes for Table A20.2

Trial No J Trial No, 2

Tol 1, 2, 3, 4 (see Fig A20,.) for weaning of terms) Gol 12 2% = distance to neutral axis Col, § # ø 10,300,000 for aluminum alloys as found in results of Trial Col 6 Cy varies as a straight line from 32000 at top of cell to zero at centerline Col, 13 ơy = 1600000 2/1470 Yo 1

Col 9 z! = distance from centroid of element to Col 14 based ou stress Oy of Col 13

Results of Trial

Considering results of both Tables and multiplying by 2 since only one half of cell was considered:

ga = total effective area = (2,49 + 266)2 = 5.51 sq in Gaz! = (412.29 + 2.92)2 = -18.62 Zw Saz'/Za = -18,62/5.51 = -3,38" Iya = 2(714,3 + 52) - 5.51 x 8ˆ 1470 in *

Results of Triai No 2

Total effective area = (2.627 + 183)2 = 5.62

Lazo (-2.86 + 3,24)2 = 0.76

Trang 19

Due to the symmetry of the section, Tables ma + A2Q.1 and A20.2 z!1v7e caiculaYyions for only one- Wa nó half of the material, thus the results are

multiplied by two General explanatory notes 4 : _ = + (Vz) 18.57 we 2, one b,u/n, ,

are given below each table, 9+ quán 19-87 0152 7z 1b./2n

The results of trial No 1 give a neutral axis 3.38" below the center line and = mement of

inertia of 1470 in.* In Trial No 2, the effective sheet widths are based on the moment of inertia of 1470, The resuits of trial No 2 give a moment of inertia of 1489 in.‘ with a neutral axis 135" above the first location If a third trial were used, making use of the 1489 moment of inertia, the change would be quite small since the effect of a small change in stress on the effective sheet width is negligible

The compressive stress on stringer No 2 using the resulting moment of inertia and neutral axis location, therefore becomes

Op = MyZ/ly = 160000 x 27.45/1489 = 29500 psi The allowable stress was 32000, hence the margin of safety is (32000/29500) - 1 = 08 or eight percent If a smaller margin of safety was desired some material would be eliminated and the calculations of Tables A20.1 and A20.2 would be repeated

Calculation of Shear Stress tn Skin at Neutral Axis

The equation for the shear flow q at some point on the skin is,

qs do - #2 Po TT {TT ~T~~ TT y

Due to symmetry of cross-section about the Z axis the shear flow đọ 1S zero at a point on the center line Z axis The summation of the term az between a point on the Z axis and the neutral axis is given in Table 420.3 The values of areas (a) and arms (z) are taken from Tables A20.1 and AZ0.2 TABLE A20.3 Element No | az qa | 021(28.88 - ,13) „ 0.59 @) | 152(27.58 ~ ,13) «w 4.17 (3) | ‹039(27.08 ~- 13) x 1.05 | 4) | +453(25.38 - 413) „ 3.86 | (s) ‹022(23.68 ~ ,13) = 0,52 (8) 156(21,58- 13) „ 3.44 roo) +0159 18 = 0.28 (8) «161(16.68 = 2.86 «ay 031(13.48 = O41 * day au | -173(10.28 | 0558.58 = 0.35 = Ls | The shear StreSS + = 4⁄E = 0152 V„/,082 = 415 Vz

The average shear stress on the sect

be tay = Vzg/eht = V2/2x50x 032 noe es

Thus for this shape of cross-secti stringer arrangement the maximum shear is 413/,.312 times the average

approximately 4/3 times as large

The procedure as given above is quite conservative relative to the true or actual margin of safety, because 4 linear variation of stress with strain has been assumed and failure of the section is assumed to occur when the most remote stringer reaches its ultimate compressive stress Actually in a static test of a fuselage to

destruction, the fuselage section as a whole will not collapse wren one stringer buckles, but will continue to take increasing load until other stringers have reached their ultimate strength Furthermore, in a typical fuselage Structure, stringers of various sizes, shaves and tnerefore different compressive strengths are used, and thus to obtain a better measure of the ultimate strength of a fuselage section, modifications in stress procedures are made to measure stringer effectiveness This subject was discussed in some detail in Arts 11 and 12

of Cnapter AlS, To illustrate stringer effectiveness in fuselage bending stress analysis, 4 simple example problem will be presented

A20.5 Ultimate Bending Strength of Fuselage Section Example Calculation

Fig A20.3 shows the cross-section or a circular fuselage, The Z stringers are arranged symmetrically with respect to the center line Z and X axes

Three sizes of Z stringers are used as tilustrated in Fig A20.4 and are labeled S., 8, and S, These symbols are used on Fig A20.3 to indicate where each type of stringer

Trang 20

ANALYSIS AND DESIGN OF .006 @ Axis, — — x Aseumed 2 Neutral Axis — „0088 Fig a - Strain Fig A20.3 Diagram 11 8 ¬ 9 _ F16 = rt r1 Tỳ" — Te *165 tL kgs 893 Hg GE Th

ee oT Stringer S, Stringer S, tg Loe We Stringer S, -p.03s TS

Area = 135 Area = 18 Area = 08

Fig A20.4

FLIGHT VEHICLE STRUCTURES A20.7

Since the location of the neutral axis ts unknown, a location will be assumed, namely, 7 inches below the center line axis as shown in Fig AZ0.3 The entire calculations for determining the effective moment of inertia cah best be done in Table form, as shown in Table AZ0.4, Due to symmetry about the Z axis only one-half of the structure need be considered since the results can be multiplied by two

Column (1) lists the stringer numbers relative to location and Column (2) according to types S,, S, and Sj Column (3) gives the

stringer area On the tension or lower side of

the section, the skin ts all effective and that

area of skin halfway to each adjacent stringer is assumed to act with stringers numbered 9 to

13 and this skin area ts recorded in Column (5)

On the compressive side the skin is only

partially effective The effective width w for

each stringer rivet line depends on the stringer

stress We will take the effective width w = 1.9t VE7ogp The effective area Ag will then equal wt These effective skin areas are recorded in Column (5) In solving this equation the stringer stress ogp has been taken as -36500 psi on stringer number (1), and then varying linearly to zero at the neutral axis as indi~ cated in Column (4) This assumption {s not true but accurate enough to obtain effective skin areas, To illustrate, consider stringer nưmber (1) The effective area Ag equals wt* =

1.9 x 032*(10,300,000/36500)4 = 032 sq in

Column (6) gives the sum of the stringer and effective skin areas or Ag + Ag

In this example problem, the effectiveness of the curved sheet panels between the sheet effective widths will be neglected since its in- fluence is small it could be included as illustrated in the previous example problem Column (7) lists the distances from the assumed neutral axis to the centroid of each stringer- Skin unit We now assume that plane sections remain plane or a linear strain variation Referring to Fig AZO0.5, it is noticed that when a unit strain of 006 is obtained in stringer S, type the compressive stress 1s 36500, which Tepresents its ultimate stress Stringer (1) is of S, type and is located fartherest from the neutral axis Sub Fig (a) of Fig A2Z0.3 shows the strain diagram with 006 at stringer

(1) and varying as a straight line to zero at

the neutral axis Column (8) of Table Az0.4 records the unit strain at each stringer cen- trold, The true stress at each stringer point due to these strain values is read from the curves on Fig A20.S and recorded in column (9) of the Table It should be noted that stringer

(3) although closer to the neutral axis than stringer (1) carries a higher stress than stringer (1) This 1s possible because when stringer (1) reaches tts naximum-stress, it bends dut continues to hold the same stress with

Trang 21

TABLE A20 4 H 2 3 4 5 6 + 8 9 10 11 12 13

No | Type Ag Sử ao Area (in) € o (2= 2°) 1 Sy 9.135 ~36500 „082 «167 35.7 | -.00600 | -36500 1.00 18T7 $.96 213 2 8: 9.135 -34700 - 034 188 33.8 | -.0057 -36500 1.05 178 6,00 203 3 Sa 0,180 -31000 - 036 216 30.3 | -.0052 -39100 1,28 272 8.25 250 4 Sy 6.135 ~236800 039 ,1174 26.0 | -.0048 ~36000 1,35 „2335 6.10 159 5 Ss 0 080 -20500 044 2124 20.1 | -.0034 ~31500 1.54 „181 3.85 TT § 85 0 086 -13400 -084 „134 13.7 | -.0023 -24000 1.79 ,240 3,30 45 + § 0 080 -7150 074 154 1.0 | -,0012 ~12500 1.75 270 1.89 13 8 8, 0, 080 0 0 „080 10,0 0 9 1,0 `,080 ị 9 9 9 8, 0 080 6130 +216 296 - 6,8 „ 0010 10000 1.83 483 i ~2, 88 17 10 Ss 0.080 12280 «216 - 296 -12.0 „0020 20500 1,67 494 | -5.91 1 11 8x 0.080 18800 318 „296 ~18.5 0028 30000 1.78 526 | -8 69 144 12 8s 0, 080 20400 „316 ,296 -20.0 0034 35000 | 1.T1 506 }-10, 10 202 13 8s 9.080 31700 216 ,296 -21.2 „0036 38000 1,75 „51T |-10.93 232 a 4.158 | -3.16 1626

increasing strain, but stringer (3) which has not reached its maximum strength of 39000 continues to take increasing load

Since we wish to use the beam formula oy = My2/Ix in computing stresses, we must modify the

stringer areas to give a linear stress variation since the formula is based on a linear stress variation The stringer modification factor K equals the ratio of the true stress in column

(9) of Table to linear stress value in column (4) or K = o/o* The results are recorded in column (10) The modified stringer areas are then equal to KA and are recorded in column

(11), Column (12) gives the first moment of the modified areas about the assumed neutral axis, giving a total value of -3.16

5

The distance rom the assumed neutral axis to the true neutral axis is thus,

Š = 3š KA2`/ZKA

= 73.16 _ #

* Tiss = - 9-76

The true N.A would fall about 70 inches below assumed position The effect on total

Sum of Column (13) would be negligible, thus Table A20.4 will not be revised

Column (13) gives the calculation of the effective moment of inertia with Z* being equal

to Z The effective moment of inertia is therefore twice the sum of Column (13) or 3252 Calculation of Ultimate Resisting Moment

The maximum stress at the most remote stringer which is number (1) is 36500 From the beam formula, „ My 2 oply/Z (36500 x 3252)/35.7 + 0.7 3,260,000 in.1b, #

This bending strength when compared to any design vending moment about the X axis would give the margin of safaty relative to bending strength If the moment of inertia nad been computed without regard to non-linear stress variation, or in other words, using K equal 1 for all stringers the neutral axis would have come out 4.9 inches below the centerline axis and the moment of inertia would have calculated to be 2362 in.* The resisting moment developed

would then be (36500 x 2382)/33.6 = 2,600,000

in.ibs Thus the true strength ts 25 percent greater than the strength for linear stress variation This result explains why such structures test overstrength 1f designed on linear stress variation basis

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES using the modified areas of Table A20.4, the

true stringer areas must be used to find the true stringer loads, which must be used in the Shear flow analysis

A20,6 Shear Flow Analysis for Fuselage Structures

The shear flow analysis can be made once the effective cross-sections of the fuselage are obtained The procedure is the same as was

111ustrated for wing structures in Chapter Als To illustrate, two example problems will be presented

Example Problem 1 Symmetrical Tapered Section Fig A20.6 shows a portion of a tapered circular shaped fuselage structure that might be representative of the rear portion of a fuselage for a small airplane Since this example is only for the purpose of illustrating Shear flow analysis, it will be assumed that the 16 stringers are the only effective mater-

fal In an actual stress analysis, the effective cross-section would have to be used as illustrated in previous articles A20.3 to A20.5

The problem will be to determine the

stringer stresses and the skin shear flow stress system at Station (0) under a given load system at Station (150) as shown in Fig A20.6

Solution No 1 - Solution by Considering Beam Properties

at Only One Section

If the change tn longitudinal Stringer or flange material is fairly uniform this method can be used with little error in the resulting shear flow stresses

Moment of inertia of section at station (0) about centroidal Y axis:

Area of all Stringers = 1" Frame spacing = 15" A20.9 Ty = (15* x 2x2) + (13.867 + 10.612 +6,742).1 x & = 180 in.*

Table AZ20.5 gives the necessary calculations for determining the flange bending stresses and the net total shear load to be taken by the cell skin Since the cell is tapered, the stringers have a Z component, thus the stringer axial loads nelp resist the external shear load summation of column (8) of Table AZ0.5 gives 333.4 lb for a summation for half the fuselage section

The

= 1233.2 1b

The results in this particular problem show that at station O the flange stringer system re- sists one third of the external shear load At station 150 the web system will resist the entire external shear load of 2000 lb since the load in the stringers is zero

In actual design the net web shear should be used since in many cases it will decrease the sheet thickness required one or more gauges Calculation of Flexural Shear Flow

Vz (web) paz 1333.2

q-=aq, — = do 7 = az

° Ty 9ˆ “186

3% 7 7,40 Š 4Z = = ———————~ (A)

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TABLE A20.5 rj) 2/3] 4 5 6 | 7] 8 9

Stitt - Op = |Px=Ơpa P22} Py=

ener] AZ™ 9| 18612] (3)(4) | GE | &% |p xd2l_p dy

No (psi) | lbs, dx| ax: 1 |15 |*.08|-25000|-1250 |-.0333) 0 |-41.6/ 0 2 | 13.86| 10|-23100|-2310 |-.0308.-.0127|-71.3 | -29.4 3 | 10.61) 10|-17700|-1770 |-.0236|-.0236| -41.7 | -41.7 4 | 5.14 ,10|- 9850|- 955 |-.0127|-.03081-12 1 | -29.5 5 | 0 | 10 0 0 9 0333| 0 ọ 6 |- 5.14 10| 9580| 955 | 01271-.0308|-12.1] 29.5 7 |-10.61l 10| 17700] 1770 | 0236|-.0236]-41.7| 41.7 8 |-13.86 10| 231001 2310 | 0308.01271-71.3| 29.4 9 |-15 |*.05} 25000| 1250 | 0333| 0 |-4i.6| 0 Shear taken by stringers = |-333.4 0 *1/2 of stringer one is assumed acting with each half of cell NOTES: Col 4 oy = -Mz/Ly = -2000 x 150 x 2/180 = -1667z

Col 5 Total x component of load in stringer member For practical purposes, it equals axial load in stringers since cosine of a small angie is practically one

Col 6 The slope of the stringers in the z and y directions can be calculated from the dimen-

sions of the two end sections and the length of

the cell (see Fig A20 6)

Col 8 The in plane components of the stringer axial &9 loads at station 0, start with stringer (1) 44,7 9- 7.40 x 05 x 15 = ~ 5.55 lb /in Gas = ~ 5-55 - 7.40 x 1 x 13.86 = ~ 15.80 Qa, = ~ 15.80 - 7.40 X 1 x 10.61 = ~ 23.66 G4 = - 23.66 - 7.40 x 1 x 5.74 = - 27,91

The torsional moment T about the centroid of the section at station (0) equals 5 x 2000 = 10000 in, 1b (clockwise when looking toward Station 150) Due to the symmetry of the sec~ tion at station 0, the in-plane components of the stringer loads produce zero moment about the section centroid

For equilibrium a constant internal shear flow q, 1s necessary to make 2 My = 0

qr = 729000

} 2k 2xx 15° = -7.06 ld./in Adding the torsional shear ?1ow q, to the flexural shear flow q, the following results are obtained: 4.4% -7.06 + 5.55 = -1.51 b./in Gas = 77.06 + 15.80 = 8.74 4;, ae = -7.,06 + 23,66 = 16.60 = ~7,06 + 27.91 20.85 ate a '

Fig A20.7 shows the results in graphical form, on the left side of the section the snears are of the same sign and therefore add tog Shear flow system #/in 16 °bE- Fig A20.7

Solution No 2 Shear Flow by Change in Stringer Loads

Between Adjacent Stations AP Method

The shear ?1ow will be calculated by considering the change in the ax‘a2 load in the longitudinal stringers between fuselage sections at stations (0) and (20)

Fig A20.8 shows the beam section at station 30, the stringer areas seing she same as at station ©, but the section as a whole is smaller due to the taper of the cell

area of each

stringer = 10

Fig 20.8 3 8

Table A20.6 gives the calculations for the flexural shear system The procedure is the same as illustrated for wing structures in Chapter A19,

Comparing the results of column 13 with the flexural shear flow as found by solution No we find the second solution gives a maximm shear flow of 28.95 Lb./in agair

27,91 for the first soluticn 7 st method

deals with the properties at only ome section

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AL ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A20 11 TABLE A20,6 1 2 3 4 5 6 7 8 9 10 11 12 13

Sta 0| Sta 30 Sta 0| Sta 30|Stringer Load AP Panel Flexural Shear

No a a Arm Arm | -1667Z) -1522Z) Py= | Py= |- (Col 8- Col 9}j Corr —0- =r APK

sq in | sq in, z z (psi.) | (psi.) |Ơpa | đpa 30 Factor 4 30 K 1b /in, 1 08 08 15.00| 14 00 ¡ -25000 | -21300 |-1250 | -1085 6.17 938 3.76 5 76 2 „10 „10 13.96| 12.93 | -23100 j ~19700 |=2310 | -1970 11.33 +935 10 60 18 36 3 10 -10 10.61] 9.80 | -17700 | -15050 |~1770 | -1505 8.83 -835 8.25 24.61 4 10 10 5.74 5.36] - 9550 |~ 8155 |~ 955 | - 815 4 65 +935 4.34 28.95 5 „10 „10 9 9 9- 9 0 9 8 9 9 28.95 § „10 V10 J- 3.74|- 5.36 8550 8155 955 815 - 4.65 -935 |- 4.34 24 61 7 „10 -10 |-10.61|- 9.90| 17700; 15050 | 1770 | 1505 - 8.83 835 |~- 8.25 = 16.36 8 10 -10 |-13,.88|-12.93| 23100| 18700 | 2310 | 1970 +11.33 +935 |-10.60 5 76 9 „08 Ø08 |~15.00]~14.0 25000) 21300 | 1250 | 1065 - 6,17 835 |- 5.76 NOTES: Col 6 Ớp = ~2000 x 150 z/180 = -1d87 z Col 7 oy = -2000 x 120 2/157.2 = -1532 z

Coil 10 Change in axial load in each stringer between stations 0 and 30 divided by distance between Stations This

result represents the average shear flow induced by the loading up of each stringer between stations 0 and 30 Col il The width of a skin panel at Station 0 is 5 88 inches and 5 § inches at Station 30, The shear flow on the edge

of the panels at Station 0 equals (5.5/5.88}) AP/30 (See Art A15.18 of Chapter Al5 for explanation) This

refinement is usually neglected and the average values as given in Col 10 are used which are conservative

Col 13 Due to symmetry of structure, the shear flow is zero on z axis Thus shear flow at any station equals the progressive summation of the shear flow values in Col 12

second method is recommended for practical analysis procedure

Since the section is symmetrical, there are no moments induced by the in-plane components of the stringer forces at station 0

The torsional shear flow forces are the same as in solution method No 1 and these are added to the values of column 13 of Table A20.6 and give a pattern similar to Fig 420.7

A20.7 Example Problem Tapered Circular Fuselage with Unsymmetrical Stringer Areas

Fuselage cross-sections are seldom 211 sym~ metrical relative to stringer and skin areas because the practical fuselage has cut-outs such as dcors, etc To tllustrate the unsymmetrical case a Simplified case will be presented

Fig AZO.9 shows 2

fuselage The stringer areas are such as to portion of a tapered

make the cross-sections unsymmetrical relative

Again for simplicity, we

to bending material

will assume the stringers are the only effective material In actual design practice the effectiveness of the skin and each stringer would have to be considered as explained in Articles A20.4 and 3

The problem will be to determine the

stringer stresses and the skin shear flow val-

ues at station (0) due to the given external loads of Pz = 4000 lb., Py = 1000 lb and Py = 1500 acting at station (150) as shown tn Fig A29.9

SOLUTION:

Since we choose to use the AP method in finding the shear flow system at station (0), we will find the stringer loads at two stations, namely, station (0) and station (30) The first

step is to find the moment of inertia of each fuselage section about centroidal z and y axes and the product of inertia about these axes

Table AZ0.7 (Columns 1 to 11) gives the calcu-

Trang 25

Fig A20.9 150" Sta tị ` ™ Piano Centerline ‡ Le 4s : lg i I (9 tl f 150 2 ‘h Section at Sta, 150 Section at Station 0 Sta 0

The skin stringers are located symmetrical2 centerline axes, however the stringer a the figure are not symmetrical with these axes

the stringers taper uniformly between the values as given for station O and i50 The cell would of course have interior transverse frames which are not shown on the rea sas given 1 % is assumed in this probl: figure TABLE A20 7 Section Properties at Sta 0 Total Stringer Loads at Sta 0 1 2 3 4 5 8 7 8 g 10 11 12 13 14

Ze = c= F/Ea [Pg =aGy- oc)

Stringer| Area} Arm | Arm | az! az'2 ay’ ay'2 azty' - Sp No, a „ y 2Ó |y'~ÿ = -1500/3.40| = (2) (12 + 13) a 80 | 10.5 |-12.00| 6.30) 68.201 -7.20| 86.50|=- 75.50[ 11.36 ~14 46 | 15080 -441 -9312 b -10| 18.98)- 8.48{ 1.90{ 36.00) -0.85 7,20|~ 16.10| 19.84 | -10 94 | -21960 -441 -2239 e 10 | 22.50 0 2.25{ 50 801-0 9 9 23.36Ì- 2.46 |~-22600 „441 -2204 d „10 | 18.98 8.481 1.90] 36.00) 0.85 7.20 16.10| 19.84 6 02 | -16742 -441 -1719 e +80] 10.5 12.00{ 8,40/ 88.10) 9.60/115.10] 100.80/ 11.36 9.54|- 7692 ~441 -6508 f „80 |-10.5 12.00|-8.40| 88.10 9 60 |115 10 |-100.80| ~ 9.65 9.54) 11958 -441 9216 £ 20 |-18.98 8.48|~3.80| 72.00| 1.70| 14.40|- 32.30| -18.13 8.02 18798 ~441 3673 h - 20 |~22 50 Ụ -4.50|101.60| Ö 0 Ụ -21.85)- 2.46| 18485 -441 3508 { „20 |-18.38|~ 8.48|-3.80| 72.00| 1.701 14, 40 32 30| -18.13|-10.94| 13610 -441 2834 ] „ 30 | =10 50) -12.00|-3.15] 33.10] -3,60| 43.25 37.80] + 9.85|-14 46 4592 -441 1246 Sum | 3.40 ~2,90| 643.9 8.40| 403.2 |- 37.70 ~1500

Reference Axes Z' and Y' are taken as the centerline axes (see Fig A20.9)

Location of centroid and transfer of properties to centroidal axes, 2 = -2,90/3.40 = - 855" ÿ = 8.40/3 40 = 2, 48" ly = 643.9 - 3 40 x 8557 = 641,4 Iz = 403.2 - 3.40 x 2.462 = 382.6 Izy = -37.7 - 3.40 x 2.46 x -.855 = -30.55 General Notes: Col 12 O} = 307 Oy - 936 1z

Col 14 Since the total tensile stresses equal to

total compressive stresses in bending,

the sum of Col 14 should equal the external applied normal load

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