Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 2 Part 8 pps

the necessary calcula- shear flow at station 0 Moment of Shear Flow about Intersection of Centerline Axes For equilibrium tn section at station 0}, the plane of the cross the summati

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Section Properties at Sta 30 Total Stringer Loads at Sta 30

the calculations Zor station (30) Mz = -1000 x 120+1500x2.11 = 116820 1n.1b,

Before the sending and shear stresses can Py = -1500 1b., Vz = 4000 lb., Vy = -1000

be calculated, the external bending moments ,

shears and normal forces at stations (0) and (30)

The bending moment about y neutral axis at Ớp (K,My - K My) y - (K,My -K\M,) 2

Substituting values from Tables A20.7 and

The shears at station (0) are ý = Py = A20.8:

4000 1b.-and Vy = Py = -1000 1b K, = -30.55/(641.4x 382.8 - 30.584) =

The normal load Py at station (0) referred = -30.55/244670 = ~.0001248

to centroid of section equals Py = -1500 lb

K, = 382.6/244670 = 00156

In a simtiar manner, the values at station

(ZO) are, (see Fig A20.10) K, = 641.4/244670 = ,00262

A20.14

Substituting K values in equation for dp:

By = = [00262 x-146310 ~ (~.0001248 x

611800) | yr [0018s x 612500 - (-.0001248 x -146310) | z

whence

Sp = 307.0 y -936.1 z (plus oy is tension) Station (30):

results of solving the equations for dy

Since an external load of 1500 lb is

img normal to the sections and through the

section centroids, an axial compressive stress

Øc is produced on the sections (See Columns

15) The total load P; in each stringer equals

the area of the stringer times the combined

bending and axial stresses (See column 14 of

each table)

act~

Calculation of Flexural Shear Flow q

Table A20.9 gives tions to determine the

based on the change 1n stringer loads between

stations (0) and (30) The correction of the

average Shear due to the taper in the skin

panels as was done in example problem (1},

Table AZO.6, column (11), 1s omitted in this

solution since it tends toward the conservative

Side Since the effective cross section is un-

symmetrical, the value of the flexural shear

flow q at any point is unknown thus a value for

q at some point is assumed In Table A20.9

the shear flow q in the web aj is assumed zero

Colum (5) gives the results at other points

under this assumption

the necessary calcula-

shear flow at station (0)

Moment of Shear Flow about Intersection of

Centerline Axes

For equilibrium tn

section at station (0}, the plane of the cross the summation of the

FUSELAGE STRESS ANALYSIS

mements the plane, of all internal and ex-

ternal ust De Zero Column (7) of Table A20.8 #1 moment of tne flexural shear

about ¢ point (See notes and Fig below Table ter exp tion.)

TABLE A20.9 SHEAR FLOW CALCULATIONS

1 2 3 4 5 8 7 8 9

ar =

8tringerlPx at|Px at mq jai

about O° (Fig a)

Moments Due to In Plane Components of Stringer Loads

Sines the stringers are not normal to

Section at station (0), the stringers nave

plane components which may produce a moment

about the intersection of the symuetrical axes which has been selected as 2 moment center

Table A2O0.10 gives the calculations for the in-

plane components and their moments about point

Oo

Moment of Externai Load System About Point (0°)

The 1000 1b load at station (150) acting

in the ¥ direction has a moment arm of 7" about

the point 0' of station (0)

Hence external moment = 1600 x 7 = 7000

in.15,

Therefore the total moment about the assumed moment center O' 5

OF STRINGER LOADS STA

Col (2) from Table A20.7

of stringers

i id divections Shear flow distribution

Fig A20.9

Col (5) {see Fig A2 ) where large concentrated loads are applied can

and (8) and y' from be determined by the procedure given in Articles Table A20 7 18 to 20 of Chapter Al9 A more rigorous

Fig b shows analysis can be made by the application of the

the Py and Pz components from i basic theory as given in Chapter Aé

Cols (4 5- (4) and (7) id (7) li x$0 154 The problem of shell stresses due to in-

Total moment about 0’ = Looking Toward Sta 150 ternal pressures is presented in Chapter Al6

-1834 + 1612 = ~22"# The strength design of the fuselage skin in-

volves a question of combined stresses The broad problem of the strength design of struc-

92670 due to shear flow q tural elements and their connections under all

-22 due to in plane components of stringers

7000 due to the external loads

enclosed area of cell)

This value of q is antered in colum 8 of Table A&0.9 The resulting shear flow in any

wed portion dp equals the algebraic sum of q

and q, (See Col 3, Table 420.9) Fig

A2O.11 shows the results in graonical form

A20.8 Discontinuities - Shear Lag - Pressurization

Stresses - Combined Stresses

A practical fuselage has many cut-outs

The approximate effect of these discontinuities

ag well as the shear lag effect at sections types of stress conditions is covered in Volume

Fig A20,12 shows the cross-section of a

circular fuselage All stringers have same

area, namely 0.12 sq in Skin thickness is C35 inches Stringers are l inch in depth

A20 16

All material is aluminum alloy §& = 10,500,000

psi The ultimate compressive strength of

stringer plus its effective skin is 35000 ø81.,

For effective sheet width use w >= 1.9t (E/ogp)®

For buckling strength of curved panels use

Øẹy = 3 Et/r Determine the ultimate bending

moment that the fuselage section will develop

for bending about horizontal neutral axis Use

linear stress distribution Follow procedure

as given {n example problem in Art AZO,4,

(2)

Fig A20.13 shows

4 spaces @7"=28" the cross-section of a

8, § 8s rectangular fuselage

-032 skin stringer locations

§ 5s 5 Three types of string-

dị ` $8 ers are used, namely,

3) ae oad $,,5,andS, Fig

8 $ ultimate compressive

oF (032 for each of the three

Se a nS Sa stringer types and

eS also the tension

Fig A20 13 stress-strain curve

of the material

Determine the ultimate bending resistance

of the fuselage section about the horizontal

neutral axis if the maximum unit compressive

strain is limited to cO08 Refer to Art A20.5

for method of solution

ADDITIONAL DATA Area stringer 5y

(3) Fig A20.15 shows a tapered circular

fuselage with 8 stringers The area of each

stringer is 0.1 sq.in Assume stringers develop

entire bending resistance Find the axial load

in stringers at statton (110) due to P„ and Py

loads at station (0) Also find shear flow

system at station 110 using AP method Use

properties at station (90) IN OBTAINING AVERAGE

A20,10 Secondary Stresses in Fuselage Stringers and Rings

The stresses that are found in the

stringers or longerons of a typical fuselage by use of the modified beam theory or by the more rigorous theory of Chapter AS, are referred to

as primary stresses Because of the necessity

of weight saving, most fuselage structures ars designed to permit skin buckling, which means

that shear loads in the skin are carried by

diagonal semi-tension field action This diagonal tension in the skin panels produces additional stresses in the stringers and also

in the fuselage rings These resulting stresses are referred to as secondary stresses and must

be properly added to the primary stresses in

the strength desim of the individual stringer

or ring Chapter Cll covers the subject of these secondary stresses due to diagonal semi-

tension field action tn skin panels It is suggested to the student that after studying Chapters A19 and A2O, that Chapters ClO and Cll

be referred to in order to obtain a complete stress picture for skin covered structures.

CHAPTER A21

LOADS AND STRESSES ON RIBS AND FRAMES

A21 Introduction For aerodynamic reasons the

wing contour in the chord direction must be

maintained : t apprectable distortion

n is quite thick, spanwise

ttached to the skin in order ding efficiency of the wing

ore to nold the skin-stringer wing surface

to contour shagé and also to Limit the length

of stringers tạ an affictent column compressive

Sỉ

e incrsase vns ber

a

strength, internal support or brace units are

required se structural units are referred

to as wing ribs The ribs also nave another

major purocse, namely, to act as a transfer or

distribution unit All the loads applied to

th2 wing are reactad at the wing sucporting

points, thus these applied loads must be trans-

farred into the wing cellular structure com-

posec 92? skin, stringers, spars, etc., and then

reacted at the wing support points The anrlied

loads may be only the distributed surface air-

leads which require relatively light internal

ribs to provide th carry thro’ or transfer

requirement, to rather rugzsd or hsavy ribs

which must absord and transmit large concen-

trated applied loads such as those from landing

gear reactions, power plant reactions and fuse-

lage reactions In between these two extremes

of applied load masnitudes are such loads as

reactions at Supporting points for ailerons,

flaps, leading sdge nigh lift units and the

many internal dea | weight loads such as fuel

and military armament and other installations,

Thus ribs can vary from a very light structure

which serves primarily as a former to a hsavy

structure which must receive and transfer loads

involving thousands of pounds

Since the airplane control surfaces (verti~

cal and horizontal stabilizer, etc.) are nothing

more than small size wings, internal ribs are

lixewise needed in these structures

The skin-~stringer construction which forms

the shell of the ?uselacs likewise needs in-

ternal forming units to nold ‘the fuselage

cross-section to contour shape, to limit the

column length of the stringers and to act as

transfer agents of internal and externally

applied loads Since a fuselace must usually

have clear internal space to house the vayload

such aS passengers in a cemmercial transoort,

these internal fuselage units which are usually

referred to as frames are of the open or ring

type Fuselage frames vary in size and strength

from very light former type to rugged heavy

types which must transfer large concentrated

loads into the fuselaze shell such as those

from landing gear reactions, wing reactions,

tail reactions, power clant reactions, etc

The dead weicht of all the payload and fixed

equipment inside the fuselace must be carried

to frames by other structure such as the fuselage floor system and then transmitted to the fuselage shell structure Since the dead weight must be multiplied by the design accel-

eration factors, these internal loads become quite large in magnitude

Another important purpose or action of ribs

and frames is to redistribute the shear at dis-

continuities and practical wings and fuselages contain many cut-outs and openings and thus

discontinuities in the basic structural layout

A21.2 Types of Wing Rib Construction

Figs A21.1 to 6 illustrate the comnon types of wing construction Fig 1 illustrates

a sheet metal channel fer a leading edge 3

7?

Skin

Ribs in 3 Spar Wing

Fig A2l.4 Fig A21.5

stringer, single spar, single cell wing

structure The rib is riveted, or spot-welded,

or glued to the skin along {t boundary Fig

2 shows the same leading edge cell but with

spanwise corrugations on the top skin and

stringers on the bottom On the top the rib

flange rests below the corrugations, whereas

the stringers on the bottom pass through cut-

outs in the rib Fig 3 illustrates the gen-

eral type of sheet metal rib that can de

quickly made by use of large presses and rubber

dies Figs 4 and 5 illustrate rib types for

middle portion of wing section The rib

flanges may rest below stringers or be notched

for allowing stringers to pass through Ribs

that are subjected to considerable torstonal

forces in the plane of the rib should have some

shear ties to the skin For ribs that rest

below “stringers this shear tie can be made by

a few sheet metal angle clips as illustrated in

Fig S Fig A21.7 shows an artist's drawing

of the wing structure of the Beechcraft Bonanza

commercial airplane It should be noticed that

various types and shapes of ribs and formers

are required in airplane design Photographs

A2l.1 to 3 illustrate typical rib construction

in various type aircraft, both large and small

Since ribs compose an appreciable part of the

wing structural weight, it is important that

they be made as light as safety permits and

also be ‘efficient relative to cost of fabrica-

tion and assembly Rib development and design

involves considerable static testing to verify

and assist the theoretical analysis and design

A21.3 Distribution of Concentrated Loads to Thin

In Art AZl.1 1t was brought out that ribs

were used to transmit external loads into the

wing cellular beam structure Concentrated

external loads must be distributed to the rid before the rib can transfer thse load to the

wing beam structure In other words, a con-

centrated load applied directly to-the edge oz

a thin sheet would cause sheet to buckle or

eripple under the localized stress Thus a structural element usually called a web sti??- ener or a web flange is fastened to the web and the concentrated load goes into the stiffener

which in turn transfers the load to the web

To set the load into the stiffener usually , quires an end fitting In general the distri- a buted air loads on the wing surface are usually

of such magnitude that the loads can be distri- buted to rib web by direct bearing of flange normal to edge of rib web without causing local Duckling, thus stiffeners are usually not

needed to transfer air pressures to wing ribs

load of 1000 lb is applied at point (A) in the

direction shown Another concentrated load of

1000 1b, is applied at point (E£) as shown

To distribute the load of 1000 1b at (A),

a horizontal stiffener (AB) and a vertical stiffener (CAD) are added as shown A fitting

would be required at (A) which would be attached

to both sti?feners The horizontal component

of the 1000 lb load which equals 800 1b is taken by the stiffener (AB) and the vertical component which equals 600 lb is taken 5y the

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Fig A21.7 General Structural Details of Wing for Beechcraft "Bonanza" Commercial Airplane

vertical stiffener CD The vertical load at £

would be transferred to stiffener EF through

fitting at E The problem is to find the shear

flows in the web panels, the stiffener loads

and the beam flange loads

om a wed pansl

The shear flows on web panels (1) and (2)

Nill be computed treating sach component of

the 1000 lb load as acting separately and the

results added to give the final shear 219W,

stiffeners CAD and AB and the external load at

In Fig Agzl.9 the shear flows q, and q,

41th the sense as show “Faking

moments about point E,

mM = 800x3~-1l2x10q, = 0, whence q, =

20 lb./in

iry = ¢00-20x10-10q, = 0, whence q, =

$0 1b./in

PHOTO A21.3 Rib Construction and Arrangement in High Speed, Swept Wing, Fighter Type of Aircraft

North American Aviation - Navy Fury - Jet Airplane

q, = 20 + 50 = 70 ` 1b./1n,

q 60 ~ SO = 10 1b./1n,

Fig A2l.11 shows the results Fig

AZ1.12 shows stiffener 4B as a free body, and

Fig A21.13 the axial load diagram on stiffener

AB, which comes directly from Fig A2l.11 by

starting at one end and adding the shear flows

Fig A2i.11 Fig A21.12

Fig A2l.14 shows a free body of the

vertical stiffener CAD, and Fig A2Z1.15 the

axial load diagram for the stiffener

gạt

ạm =%0,

—la Aj 30

gr de *HE00 Loony ¿ 630# (tension)

Fig A21.14 Fig A21.15

The shear lows q, and q, could of course

be determined using both components of forces

at (A) acting simultaneously For example,

consider free body in Fig A&l.15a

The shear flow q, in web panel (3) is ob- tained by considering stiffener EBF as a free body, see Fig A21.16

aFy = léq,+10x3-9x70

~1000 =0

whence, q, = 133,33 1b./in

agit 77 The shear flow q,

B 0 treating entire beam to

3" a.7* y right of section through

T E panel (3) For this free

Fig A21 16 2Fy = -600- 1000+ 12q,= 0

whence, q, = 133.23 Fig A21.17 shows diagram of axial load in stiffener EF as determined from Fig A21.16 by

starting at one end and adding up the forces to any section

After the web shear flows have been determined

Fig A217 the axial loads in the

beam flanges follow as

BL - 1570 1b the algebraic sum of the

shear flows Fig A21.18

E (tension) 1000 lb shows the shear flows

along each beam flange

as previously found The upper and lower oceam flange loads are indicated

by the diagrams adjacent to each flange

4700

+ 700 tb

Tension

3900# compression

In this example problem the applied extern-

al load at point (A) was acting in the plane of

the beam web, thus two stiffeners were suffi-~

client to take care of its two components Often

loads are applied which have three rectangular

components In this case, the structure should

be arranged so that line of action of applied

force acts at intersection of two webs as

tllustrated in Pig A21.19 where a load P is

applied at point (6) and its components Pz, Py

and Py, are distributed to the web panels by

using three stiffeners S,, S, and S, inter-

secting at (0)

In cases where a load must be applied

normal to the web panel, the stiffener must be

designed strong enough or transfer the load in

bending to adjacent webs

In this chapter, the webs are assumed to

resist pure shear along their boundaries In

most practical thin web structures, the webs

will buckle under the compressive stresses due

to shear stresses and thus produce tensile

fleld stresses in addition to the shear

stresses The subject of tension field beams

is discussed in detail in Volume II In gen-

eral the additional stresses due to tension

field action can be superimposed on those

found for the non-buckling case as explained

in this chapter

STRESSES IN WING RIBS

A21.4 Rib for Single Cell 2 Flange Beam

Fig A21.20 illustrates a rib in a 2-

flange single cell leading edge type of beam

Assume that the air-load on the trailing edge

portién (not shown in the figure) produces a

couple reaction P and a shear reaction R as

shown These loads are distributed to the cell

walls by the rib which ts rastened continuously

to the cell walls Let q = shear flow per

inch on rib perimeter which is necessary to

hold rib in equilibrium under the given loads

P and R

Taking mcments about some point such as (1)

of all forces in the plane of the rib:

mM, = -Ph + 24q 20

hence

@ = Ph/2a (A = enclosed area of cell)

ON RIBS AND FRAMES

Fig A21 20

With q known the shear and bending acment

at various sections along the rib can be deter~

mined For example, consider the section at 3-3

in Fig A21.20, Fig A21.21 shows a free body

of the portion forward of this section

The bending moment at section B-B equais:

Mg = 2qA, where A, is the area of the shaded portion

Let Fy equal the horizontai component of the flange load at this section

and the lower flange load equals Fp = Fy/cos 9„

The vertical shear on the rib web at 3-3

equals the vertical component of the shear flow

Q minus the vertical components of the flange

loads Hence

Web > 14> Fy tan 0, - Fy tan o,

zqa-2gk (tan 0, + tan @,)

Ilustrative Problem The rib in the leading edge portion of the wing as illustrated in Fig A21.22 will be

analyzed

A distributed external load as shown will

be assumed

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Solution:

The total air load aft of beam = $x40/2 =

160 lb The arm to its c.g location from the

baam equals 40/3 = 13.33" Hence the reactions

at the beam flange points due to the loads on

the trailing ecge vortion equals:

Let q be the constant flow reaction of the

cell skin on the rib perimeter which is neces-

sary to hold the rib in equilibrium under the

applied air loads

Take moments about some point such as the

lower flange (1)

IM = -213.2x%10+3x15*7.5+2x159.3 q

= 0

whence, q = 1232/278.6 = 4.42 1b./in,

With the applied forces on the rib known,

the shears and bending moments at various

sections as desired can be calculated For

example, consider a section B-B, 2.5" from the

leading edge Fig Ael.24

Bending moment at section B-B =

(b) and (c) An external load is applied at

point (a) whose components are 5000 and 3000 lbs as shown Additional reactions from a trailing edge rib are shown at points (b) and

(¢) A vertical stiffener ad is necessary to

distribute the load of §000 lb at (a) The following values will be determined: - (1) Rib web shear loads on each side of stiff-

ener ad

(2) Rib flange load at section ad

(3) Rib flange and web load at section just

to left of line be

SOLUTION It will be assumed that the 3 string-

ers develop the entire wing beam bending resist-

ance, thus the wing shear flow is constant be- tween the stringers The wing rid is riveted

to the wing skin and thus the edge forces on

the rib boundary will be assumed to be the same

as the shear flow distribution In other words,

the three shear f1oWS Gages Apa and dgp Hold the external loads in equilibrium The sense of these 3 unknown shear flows will be assumed as

shown in Fig A21.25,

To find dgag, take moments about point (b)

whence, Gp = 42.7 1b./in

With these supporting skin forces on the rib boundary, the rib is now in equilibrium

and thus the web shears and flange loads can

be determined Consider as a free body that

portion of the rib just to the left of the

stiffener ad centerline as shown in Fig 421.26

whence, dag = 74.6.1b./in

To find the shear in the web just to right

of stiffener ad, consider the free body formed

by cutting through the rib on each side of the

stiffener attachment line as shown in Fig

421.27 The forces as found above are shown

LOADS AND STRESSES ON RIBS AND FRAMES

To find web shear

At Jolnt (4) T' obviously equals 2158 lb, The

stiffener ad carries a compressive load of 5000

lb at its (a) end and decreases uniformly by

the amount equal to the two shear flows or

463 + 74.6 = 537.6 1b./in

take 2Fy = 0,

body,

= 0, whence, C!

The results obtained by considering Fig

A21.27 could also be obtained by treating the entire rib portion to the left of a section just to right of stiffener ad, as shown in Pig

A21.28,

To find rib flange load T’

about point (a) take moments

To find flange load C' take IF, = 0

The above values are the

The rib flange loads and

calculated for a section just web shear will be

to left of line