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

Calculation of longitudinal stress due to My and My bending moments: - The design bending moments will be assumed and are as follows: - The moments about the y axes are not needed in t

)

Upper Stringer 7 Upper Rear Flange

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

No.3,4& 5 (t = 050) area stringer 205

tT Lower Surface Stringer

area of bulb angle = 11

area of angles = 508 area of skin +120 area of web

«892 Lower Front Flange

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 calcula-

tions 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: -

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 ~

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~

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 cal- culated 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,

TABLE Al9.2 SECTION PROPERTIES ABOUT CENTROIDAL X AND Z AXES Wing Section at Station 20 (Compression on upper surface)

See Fig, Al9.25 for Section at Station 20

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

Ig = 238 ~ 5.584 x 1.2762 = 20,3 int axes and transferred to the centroidal X and

SECTION PROPERTIES ABOUT CENTROIDAL X AND Z AXES

Wing Section at Station 47,5 (Compression on upper Surface)

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

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

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

Web Between Flange Members (1) and (14)

(Shear values are average between stations 20 and 47 5)

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 approxi- mately 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

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

In Plane Moments About Section c.g Produced by in Plane

Components of Flange 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

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

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 indeter- minate 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

To avoid repetition of similar type calecu-

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),

Thus two equations will be written, nameiy:

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

The modulus of rigidity G will be assumed con- stant 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

Table Al9.6 (Column 5) zives the value of

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

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

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

it {is necessary to go to thick skin in order to resist the wing bending moments efficiently

The ultimate compressive stress of such struc- tures 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 bend- ing 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,

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 non- linear 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

This value could de calculated from a consider-

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

obtaining the modified moment of inertia of the

cross-section to use with the linear beam

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)

The margin of safety for points other on the beam section will likewise be 4 percent For ex- ample 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

2

(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:-

(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

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 shear- ing 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: -

(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 in- volving 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

To illustrate the shear lag problem in its simplest state, consider the three stringer~

sheet panel unit of Fig Al9.36 The three stringers are supported rigidly at B and equal loads P are applied to the two edge stringers labeled (1) at point (A), The center stringer (2) has zero axial load at (A), but as end B is approactied, the sheet panels transfer some of the load P to the center stringer by shear stresses in the sheet At the support points 8 the transfer of load from side stringers to center stringer iS such as to make the load in all three stringers approximately equal or equal to 2P/3,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

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 distri-

bution 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

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

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 inter-

sects 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 re- quired 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 load The wing is a box beam with 7 stringers

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

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

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

(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

front spar depth Take skin thickness as

Wy = 6 1b./in, acting to rear and located 04 inches,

STA.0