Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 part 11 ppt

If the rings have stresses in them due to loads other than diagonal tension 1.e, bulkhead type loads, see Chapter A21, then an interaction equation is Calculate a by the method of succes

ARG, =

tex

This formula {s simtlar to the one for the se 3| iên + sin2a (1-k) (1-1)

°9#factLve”" area of a Single upright in a plane

ved Deam system, (Art C11.20), where or use Fig C11.36 for bracketed expression

distance from c.g of ring to the skin

&

radius of gyration of the ring (cross~

Saction) about an axis parallel to the

skin

6

Equation (105) assumes that k, fg 4, t,

etc., are the same for the panels on each side

of the ring If not, then some average value

Should be used, or else fag must be written as

the sum of 2 diagonal tension affects (for each

Side panel) as was dona for the longeron in (S) above

The "“aaximum" stress, ÝRGMaX » im the ring

wiil be, as before, eq (59)

7, The next problem ts the determination of

the angle of diagonal tension, a There are

Several ways in which a can be evaluated Four

are as follows:-

3) When there is a simificant amount of

compressive strain, simply assume

a= 459, b) when there is no, or very little, axtal

strain present (only shear) use Fig

C1145 to obtain a

c) 1? sufficient tension stress (or

strain) 1s oresent to prevent duckling, from eq (78), then, of course, a = 90,

approximation as follows This is the

most tedious method and (a) and (b}

above will suffice in most cases, especially for preliminary design

f

(ty, from equation (104) )

é 286 {f., § RG from equation (103) ) Equation (106) must be solved by successive

approximation as was done in the stringer system

where Fog is the "natural" crippling

strength and RB 1S obtained from Fig

If the rings have stresses in them due

to loads other than diagonal tension (1.e, bulkhead type loads, see Chapter A21), then an interaction equation is Calculate a by the method of successive used

fy fag

gh + te 1.0

*cc Fag

where f, is the stress due to loads

other than diagonal tension effects

that no effective skin is used and that Prong 15 at most 34

e) The loads on rivets at splices, etc

are the same as for the stringer system, eq (96) and (97)

The procedure outlined in (1) to (8) above

can best be illustrated through the use of an

example problem

Fig C11.37 Example Problem,

Consider a longgron type fuselage structure

having a cross-section as shoym in Fig C11.47,

The details and section properties of the

longerons and rings are included in the figure

The properties of the structural cross-

section are given in Fig C1l.48 The

properties shown are for the case where the

bending moment causes the upper skin to be in

compression and the Lower in tension The upper

Sk1n thus buckles out early, as indicated

For this example problem assume that the

applied loads are,

Cross-Section a2 60 in? 2 =.365 in Ts 0252 in Detait "A"

IT 2.197 int Clad 7075-T6 Sheet Mul

Cross-Section and Details of Members

These are the loads at the middle of the

bay being checked

41.8" Effective Section in Geometry Bending - Skin Shown

Cross-Section Area = 2830 in.* "Dashed" is not Effective

Ty A, 1895 in * Fig C11 48

1 Using the geometry and properties of Fig

C11.48 and the engineering theory of bending,

we find the primary internal stresses to be,

ANALYSIS AND DESIGN OF F 2a Check top skin panel for four and k

Use properties of Fig (11.47 and the

applied loads to get Ro, eq (74) The skin

compression stresses will be "fictitious" since

the skin buckles out early, but will give the

proper constant, B, for interaction Assume

ÍCsEin at a point 2/3 up from upper longeron

to top surface as governing compressive

from buckling equations in Chapter ca

From Chapter C8, for our panel dimensions,

ky = 10 and kg = 12.83, thus

n® x10 x 10,300,000 (2928)* 2310 pst Foor *~ Te (1 - «3*)

4a k is then, from (3) and Fig C11.19, for

IGHT VEHICLE STRUCTURES

to serve as a criterion for buckling under -

combined shear and compression

2b The stresses at the top (upper longeron)

and bottom (lower longeron) of the side panels are -37,200 (comp.) and 14,800 (tension) respectively This gives an average "fictitious"

stress of -11,200 axial and = 26,000 psi bending

stress (equivalent) The actual reduced buckling

stress for the side panels could be obtained from an interaction for these axial and bending

stresses

Another way is to use an axial stress only,

"weighted" on the high side of the average,

arbitrarily, to account for the effect of the pending stress For this problem a compression stress value half-way between the averags,

~11,200, and the maximum, ~37,200, is used

- : Ki to cot a, Ks fe, cot ag eng = RG 2 Kixc® Lm = ~,00262

*L "tp 7 BAL + ,6 (1>k¿)Re "Dap BE + 5(1-Ka)Re ——_ 8 73 x 10"

.81 (4240) cot a eé.-€

2 ~37,200 ~ TET) s (2.610.105) 1£) { ^ tan? Ga # ee

TH.8)(.ce5) + 5 (1-.81)(.108) & = Egg +i tan” œ

_ «79 (11,780)}cot a 20.5)

35.4 1.085) + 5 (1-+.79)(.335)

~37 ,200 - 2970 cot a, - 9810 cot ad,

6a, The stress in the rings supporting the

upper skin is obtained from eq (105)

As discussed earlier, this assumes the upper

pariels on each side of the frame to have the

same shear stress

6b The stress in the rings supporting the

side panels is, similarly,

= ~79{11780) tan ds

RG St.0es) * 6 (1+ 78) -0235

a, and a, in the above equations refer to the

upper panels and side panels respectively

7 Angle of diagonal tension

The compression in the upper panel is so

large that a, can be reasonably taken as 45°

For the side panel the same is probably

true, in the upper portion, but this will be

checked using the method of successive

approximation and equations (106), (105), (104)

Thus, a, # 459 which is close to the 44° assumed

Let a, = 45°, This bears out the statement that any significant compression stress (or strain)

forces a towards 45°,

Had there been no significant average axial

strain (or stress) present a could be gotten from Fig C11.45 which 1s based on pure shear (no axial loads present) Using this for the above side panels, for example, we would proceed

45°, as frequently assumed for simplicity |

8 Now that a has been established as 45° for

the upper skin panels and Zor the upper portion

of the side skin panels, the stresses are

available (from (5) and (6) ) and strength checks can be made

a Longerons:

fp = -37,200; £ 5.7 - 2970(1.0) ~ 9810(1.0)

= - 12,780 Fog = $5,000

HP e Using Fig C11.38, and the

#00 + T00 - „801 = 1.0 (adaquate}

Thus the ring is not adequate Either

more area (thicuness) is required or a

closer ring spacing (or both) are needed

to lower fpq-

Skins:

The side panel has the highest stress and

will be, therefore, more critical from an

ultimate shear strength than the top

Caeck for Permanent Buckling

Usually there is the requirement that no

permanent sxin buckles shall occur at

limit load This is checked as follows:

The load transferred between the longeron and the ring can be calculated as

Pog = tag *Ara, » from (8b) and (6a)

27,900 x 0285

795 1b

This load is actually carried by the rivets attaching the outer ring flange to the longeron flange (next to the skin) and also

by the gusset action of the skin at the ring-longeron junction Two fasteners,

staggered if necessary, should be used to

attach the outer ring flange to the longeron flange (See discussion of joggles in

Chapter D3)

C11.38 Summary

From these examples involving both stringer and longeron type construction it can de sean that the effect of axial compression stresses (or strains) along with the shear stresses in the panels is two-fold:

a) It brings about, through interaction,

earlier buckling of the panels than would result from shear stresses alone

The result is, of course, a nigher value of k and, hence, of all the en-

Suing loads and stresses that are a

function of k

C11.48

bd) The angle of diagonal tension is

forced to approach 45°, more than

would result from shear stresses only

The effect of tension stresses 1s just the

opposite and can thus be conservatively”

ignored, or evaluated tf desired

From a time-saving standpoint, in pre-

‡minary design, an arbitrarily large value of

% and an angle of diagonal tension of 45° can

be assumed where significant axial compression

strains are present

The exact magnitude of the various diagonal

tension effects throughout a network of skin

panels defies simple evaluation from an

analytical standpoint This is particularly

true when both shear stresses and axial

stresses change from panel to panel as in most

practical structures and loadings No simple

analytic expressions are available But some

rational approach is necessary to complete the

design, or specimens for test programs, and the

approaches given in this chapter represent one

such procedure If margins are extremely

small, element tests for substantion are in

order The reader is encouraged to consult

the references for a more thorough understanding]

of the basic theory and its limitations,

particularly with regards to areas where sub~

stantiating test data is relatively meager

The stringer system is usually found, for

example, in fuselage structures where there

are relatively few large ‘eut-outs™ to disrupt

the stringer continuity This is more typical

of transport, bomber and other cargo carrying

aircraft The longeron type structure 1s more

efficient and suitable where a large number

"quick-access" panels and doors and other

"cut-outs" are necessary to service various

systems rapidly These would "chop-up" a

stringer system rather severely, making it

quite expensive from both a weight and manu-

facturing standpoint Therefore, longeron

systems are more usually found In fighter and

attack type aircraft, and in others with un-

usual features There are also, of course,

other factors influencing the choice of

structural arrangement

Some further notes concerning this general

subject are included in Chapter C3 This

includes “beef-up" of panels and axial members

bordering cut-outs or non-structural doors,

which is also related to "end-bay” effects

discussed in C11.24

——

* Tt is not conservative to ignore the reducing effect of tension

stresses on k if the compression stresses due to diagonal

tengion are being relied upon to reduce any primary load

tension stresses in stringers or longerons

DIAGONAL SEMI-TENSION FIELD DESIGN

C11.39 Problems for Part 2

1 In the example problem of Art 011.54, what M.S would exist if the ring were

made of 032 2024 (instead of 040)

(Assume a = 45°)

In the example problem of Art Œ11.324, how wide could the ring spacing, d, be made and still show a positive M.S for the ring {Assume a = 489.)

In the example problem of Art C11.34, how much additional torsion, T, could be applied before the ring would show 4

Assume the general section properties (1 and neutral axis location) remain the same

as In 011.3%

In the example problem of art C11.37

a) What (standard) gauge of 7075-TS sheet

aluminum would the ring have to be to show a minimum positive M.S

b) What ring spacing, d, would be required for the 032" ring to show a minimum

11.40 Problems for Part 1

(1) The beam as shown in Fig (A) is sub-

jected to a shear load of 8000 1b as shown

Determine the margin of safety for the given loading for the following units: {1} web,

(2) web flange rivets, (3) web stiffeners

NACA method

(2) Same as problem (1) but with web up- right spacing = 4”

Use

(3) Design a semi-tension field beam for

the beams and loading of Fig (B) Take wed

as 2024-T3 with minimum thickness being 020

Stiffener, | F—.064 Web se" >1) 7075 Clad

Web uprights and flange members to be 2014-TS

extrusions Rivets to be 2117-T3 Show cal~

culations for at least four sections along beam length Assume the beam braced laterally

by 025 skin on top and bottom

(4) Fig (C) shows the cross-section of a

single spar wing beam Strength check the

following units of the beam section (1) web, (2) web uprights, The upright spacing is 8

inches and the design shear load on section is

35000 1b web material is 7075 alclad

Diagonal Tension Part 1 Methods of Analysis N.A.C.A T.N 2661

(4) Kuan, Peterson, Levin:- A Summary of Diagonal Tension Part 2 Experimental Evidence N.A.C.A T.N 2662

Fig C11.36 Graph for Calculating Web Strain (Ref 3)

Fig C11.40 Correction for Allowable Ultimate Shear

Stress in Curved Webs

4 & 8 LO Fig Cli.4ld Angle of Pure Diagonal Tension

2024-T3 Aluminum Alloy Øuit = 62 ksi Rat i i Ree aL Dashed Line is Allowable Yield Stress RG st

LIfe bP LLL

APPROXIMATE VALUES OF THE ANGLE OF DIAGONAL TENSION a cà

FOR PANELS WITH h'> d (LONGERON SYSTEMS)

35 1) This curve is for panels loaded: in: shear:

- “" '“anly, no significant axial strains ~ a

2} The.curve is aa i ‘ion # the following ‹ uation

£4 cos “0 )(teps} ain®-o.+{

30 tee TTT womens nba Ait

Ay 7Ì Tư ườ T- Tỉ

25 _ 3] The.range ot =e values represented on.this curve ig-from: 02 to-1„ 8 ~

Valuea of.this parameter.outside this range (tor values: aPg- 20)

~- must be applied to’ the ‘equation above (Note 27 to: determine’

1997 3 4567991 2 3 4 567991 2 3 456

3 1) This curve good for flat or curved panels Conventional

panel stiffening members must be attached to edge 2) Curve does not apply to panels with stringers

3} Curve may be used for any material at any temperature provided proper values of Fson > Eo & Foy are used in the parameters

FSp

FScr

Fscr = Panel Shear Buckling Stress

Ee = Modulus of Elasticity in Compression Fey = Compression Yield Stress

4x Fsop 10°

Ee x Fey Fig C11 46