Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 part 10 pot

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C11, 18 DIAGONAL SEMI-TENSION FIELD DESIGN

In single uprigt ne outstanding legs ars| C11.23 Web Design

rallevad toa 2 extent by virtue of

c fact that the compre Ssiv3 stress jeer For design purposes, the peax value of the wisn distance from the web; the allowable nominal web shear stress within 4 bay is takan

stresses for single uprights ars therefore as,

somewhat naigher than those for double uprights

Because the forced crippling is of local lsnax Sify (L+ kU }(1+kC,) - - - ~ (62) nature, it is assumed to depend on the peak mm"

value _ of the upright stress rather than Refer to figs Cll.23 and Cl1.24 to

: determine values of factors C, and oz The C,

on the average value

The upright stress at which final collapse occurs 13 obtained by the following empirical method:~

(1) Compute the allowabls value of Punax for a perfectly, elastic upright material by

the formula:-

For 2024-T3 Aluminum Alioy:-

Fy221000k*/* (ty/t}*/9 (Por double uoriants’- Fy226000k2/* (ty/t)

$12)

f )*/3 (For single uprights)-(31b) for 7075-TS Aluminum Alloy:-

Fy226000K*/* (ty/t)*/* (Por double uprights )-(61

FysB2s00k*/9(ty/t}*/8(Por single uprigh Tay

ằ

(2) If Fy exceeds the proportional limit

for the upright material, use as allowable

value the stress corresponding to the compres- Sive strain F,j/E

Fy can also be obtained for various

materials from Fig C11.38 Natural Crippling failure

The term "natural crippling failure” ts used to denote 4 crtiopling failure resulting from a compressive stress uniformly distributed

over the cross-section of the upright By this definition it can occur only in double uprights

To avoid natural crippling fallure, che peak

Stress In the upright Fy in the upright

should De less than the cripoling stress of the

section for L/o —» 0 It appears that crippling

failure does not appear to be 2 controlling factor in actual destzns

nts General Elastic Instability of wed and Upr1z

Test experiance so far has not indicated se alastic instabiltty meed be von- ed design Apparently the web system

is safe against general slastic instability if the uprights are designed to tall by column

action or by forced cri2pli1g at a shear load not much lass than the shear strength of the

wed

term constitutes a correction factor to allow

for the angle a of the diagonal tension field differing from 45 degrees C, makes allowance

for the stress concentration due to the flexi-

bility of the beam flanges

Allowable Shear Stress Fs

The allowabla shear stress Pg 1s determined

by tests and denends on the value of the diagonal-tension factor k as well as on the details of the web to flange and web to upright fastenings Fim C11.25 gives empirical allowable curves for two aluminum alloys It should

be noted that these curves contain an allowance

for the rivet factor; inclusion of this factor in these curves is possidle because tests have

shown that the ultimate shear stress based on

the gross section (that is, without reduction of rivet holes) is almost constant within the

normal rang? of rivet factor (CR = 0.6)

For the allowable stress for other

materials refer to Fig C11.42 Permanent Buckling of Web

A check tor the development of permanent

Shear buckles can de made using Fig C11.46

In this figure Fs, g 18 the allowable web gross shear stress for no permanent buckles

The Air Force usually specifies no permanent

buckles at limit load

C11.24 Rivet Design, Wed to Flange

ets

The load per

flange rivets is inch run acting on web to

taken as,

eo Sa 414 «) (og

R * a + 0.414 kK) - ($3)

With doubla uprights the web to upright rivets must orovide sufficiant longitudinal shear strength to make the two uprights act as an integral unit until column fallure occurs The total shear strength (single shear strength

of all rivets) required in an upright is

an

2 co

R =

Trang 2

ANALYSIS AND DESIGN OF where

Feo = colmn yield strength of upright material (If Feo is expressed in ksi, Reotar will be in Kins.) tota

@ = static moment of cross-section of one upright about an axis in the

median plane of the web in.?

hì = width of outstanding leg of upright

ny/Lg # ratio obtainabl2 from formula (60) The rivets must also have sufficient

tenstle strength to prevent the buckled sheet from lifting off the stiffener The necessary

strengti is given by the criterion

Tensile strength per inch of rivets >

0.15 x t Fey

where Fp, 13 the tensile strength of the web

For Wed to Upright Rivets on Single Uprights The required tensile strength is given dy the tentative criterion,

inch of rivets >

Pensile strength per

.82 xt Fou

(The tensile strength of a rivet is defined as

the tensile load that causes any failure; if the sheet is thin failure will consist in the pulling of the rivet through the sheet.)

No criterion for shear strength of the

rivets on single uprights has been established;

the criterion for tensile strength is probably adequate to insure a satisfactory design

The vitch of the rivets on single uprights should be small enough to prevent inter-rivet

buckling of the web (or the upright leg if

thinner than the web), at a compressive stress equal to funy x The pitch should also be less

than 3⁄4 1n order to justify the assumption on edge support used in the determination of

Powne Ser

Upright to Flange Rivets

These rivets must carry the load existing between upright and beam flangs These loads

are,

a

Py =f, A, (for double uprights) - - (67)

Py = fy Au, (for single uprights) - - (68)

These formulas neglect the gusset effect {decrease of 2, towards the ends of the upright )

in order to be conservative Two fasteners

should de used to attach the upright to the

FLIGHT VEHICLE STRUCTURES

C11 19 flange, especially when the upright ts Joggled

(See Chapter D3.)

Cl11.25 Secondary Bending Moments in Flanges

The secondary moment in a flange, caused by the vertical component of the diagonal tension may be taken as,

where C, is 4 factor The

moment given by this

moment and exists at

given in Fig C11.24 formula is the maximum

the ends of the bay over the uprights If C, and k are near unity, the

moment: in the middle of the bay is half aa

large as that given by formula = (68) and of

opposite sign

C11 26 Shear Stiffness of Web

The theoretical effective shear modulus of

a wed Gg in partial diagonal tension is given in Figs Cl1l1.26a and Cl1.26b In Fig Cll.26a,

đỊpr is valid only in the elastic range The

correction factor for plasticity is given for 2024-T3 aluminum alloy in Fig C1l.26b

The effective modulus should be used in deflection calculations

Gg, for example, should be used in place of G

in the flexibllity coefficients for shear panels

in Art 47.10 of Chapter A7, if buckling skins or webs are present

C11.27 Example Problem Using NACA Method

(Method 2)

The beam used in example problem of Art C11.13 will be checked by the NACA Method

First check to see if given beam falls within the limitations of the NACA formulas

From Art C1ll.1€:-

tị

—È snould be zreaber than Q.6

For our beam t,, the leg thickness of the upright is 125” ang the wed thickness is 025", henca 125/.025 = 5, which is greater than 6

Also from art Cl1.16, the d/n value for

the beam should fall between 2 and 1.0 The d/h value for our beam {s 10/30 =

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C11 20

Calculation of the Critical Shear Stress (Fs ~)

Use will be made of Fig C1l.17,

de = = Se 2 400 = SEL ifener spacing 10 = Stiffener spacing (See Fig

t 025 Web thickness 11.16)

Qe _ 27.94

a =I F 2.79, See Fig C11.16 for Ags Using the above values, we find from Fig C11.17 that Fegp = 370 psi Calculation of the Loading Ratio t3/Fson From equation 55 Sw - 13500 - = het 29.45 x d25 7 16550 pst fs = where, Sw

hạ external shear load on web distance between flange centroids Equation 55 was used because the flanges will take very little of the shear load

Therefore the loading ratio,

£ § 18350

== = 49.6

Por 370

Calculation of Diagonal-Tension Factor k

With ts/Fsop = 49.6, we use Fig C11.19 to find value of k = -69, for zero curve since

Sheet is flat or R = 0

Calculation of Average Stress in Upright

ee Stress in Upright

Upright consists of one 1

extruded angle section x 1x 1/8, 2024 To obtain effective area, we use eq 58

Au 234

Au, 8 = tr 9ìa =a Cd t+ aS) 3025 a

where,

Au area of web upright or stiffener e distance from median plane of web

to centroid of single upright

® = centroidal radius of gyration of cross section of upright about axis parallel to web 2112 10x 025 „69, we obtain value of fy/ts = Using 71g C11.20, with A, /dt = Ue = 45 and k = „93 €ripplin:

DIAGONAL SEMI-TENSION FIELD DESIGN

Hence, fy, 2 18250 x 92 = 16900 psi, which

represents the average stress over the length of

the upright but applies only along the median plane of the web or along the line of rivets connecting upright to web

Calculation of Maximum Stress Upright eT ress Upright

a 10

Value on ?8 s0” «352

where, d upright spacing

hy = distance between centroids of up=

right-flange rivet connections

69, we find = 1.17 Hence, Using 352 as value of d/ay and k =

from Fig Cl1.21 that đun x⁄#u

fy 7 ~16900 x 1.17 = -19800 psi Calculation of Diagonal Tension Angle a

From Fig Cll.22 for values of k = «69 and †u⁄s = 16900/18350 = 92, we obtain tan a = .815 Calculation of Allowable Stresses for Upright

Check allowable stress for failure as a column:-

Since the design ts of the single upright type, we check the following two criteria:~

(1) Stress fy should be no greater than

the column yield stress for the upright material In the previous solution of this beam by Method 1, the crippling stress of the web upright was calculated to be -S2500 psi., which corresponds to the column yield stress Poo Since fy, (average) was -16900 psi,, the upright is far overstrength for this Particular strength check,

(2) The stress at the centroid of the up-

right should be no greater than the allowable

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Fy = 26000 k?/* (ty/t) */* Fy = 26000 x 69°/* (,125/.025)*/* = 34600 psi

The Fey for 2014-Té Ext = 53000 ¥

Correcting from Fey = 42000 for 2024 material by directly as the v of the yield stresses, which 1s approximately true, we obtain,

aja

Fy = l2aa0)2 sXx 31600 = 38800 psi M.S = (38800/19800) ~ 1 = 0.96

The stress is below the proportional limit

stress of the material so no correction is necessary

The stress Fy could also be found by use of curves in Fig C11.38

The web stiffener or upright is too much overstrength and should be re-designed

Check Web Design

The peak value of the nominal web shear

stress within a beam bay is taken as, t Smax = fg (l+kC,) (1+kc,) From Fig C11.23 using tan a = 815, we find 0, = 022 Using Fig Cll.24, we must find value of term 4$ wd = 0.7 dV + before the value of (Ig + Ip) hạ C, can be found 4 wd = 0.7 x 10 V „025 = 1.96 (.1075 + 0291) 29.45 hence from Fig C11.24, C, = 075 Substituting: ~ tenax = 18350 (1 + 69 x 022)(1 + 69 x 075) 19600 psi

Allowable Web Shear Stress

From Fig Cll.25a for kK = 69, the allow-

4ble 7g.ịị for 248T (Fry = 62000 psi.) equals

21000 psi for web without rivet washer and

23700 psi Zor web with rivet washer The

margin of safety for web with rivets without

washer is (21000/18600) - 1 = 07 For rivets

with washer the margin of safety would be

Cli 21 (23700/19600) ~ 1 3 21

Check of Rivets

Loads per inch of run web to flange rivets is given by the following equation R" == (1 + 0.414 k) a ~ 13500 = = Seep (1 + 0.414 x 89) = 610 1b, Load per rivet pitch of 3/4 inch = 75 x 610 = 456 lb

These values are practically the same as by

the first method of solution in Art C11.13

Single shear strength of 5/32 2117-TS rivet 2512 1b Bearing strength on 025 web = 486

1b (critical)

Strength per rivet pitch 22x 486

M.S = (972/456) - 1 = 1.12 = 972 lb Web to Upright Rivets

The required tensile strength ts given by the criterion, that the tensile strength of

rivets per inch of stiffener should be greater

than 0.22 to Fey which equais 0.22 x 025 x 62000 = 340 1b./inch

In our rivet problems we have specified no

web-stiffener rivets Thus rivets should be

specified that will develop 340 lb tension strength per inch of stiffener See Fig 11.374 for tension strength for some fastener- skin combinations Refer to further

discussion in latter part of Art Cl1.32 Upright to Flange Rivets

The web uprights are fastened to the flange both upper and lower by 1/4 dia AN steel bolt

The load on the upright at its ends is,

Py = fy Ay, 16900 x 112 = -1895 1b

Shear strength of 1/4 bolt = 3681

Bearing on vertical 3/32 leg of lower flange = 1.25 x 2340 = 2930 lb (erttical)

M.S = (2930/1895) - 1 = 54

Since a 1/i" dia rivet AL7ST, Fgy = 38 ksi

has a shear strength of 1970 lb it could be

used instead of the 1/4" bolt and the M.S would

be 04,

Check of Flange Strength

Section 50" from end Design external

we

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Cll, 22

bending moment = 50 x 13500 = 675000" Ib Diagonal tension factor = 69

Thus bending moment developed as a shear resistant beam = (1 - 69} 675000 = 209000” 1b

The remainder of the bending moment is developed as 2 Dura tension fisld beam, or 4 moment of 466000" 1b Bending Stresses Upper flange bending stresses, (extreme 209000 x 12.58 466000 x 10.94 270.5 ~ 210 fiver) fp = - = - 9700 - 24300 = - 34000 pst Lower Flange Bending Stresses_(extreme fider) 209000 x -17.4Z „ ~466000 x _-19,06 f th? - 270.5 210 = 13450 + 42400 = 55850 pst Average Axial Flange Loads Due to Bending

AS 2 Shear resistant beam

flange load equals average axial 1, MG -), rT ( te where M at tine (1 - k) = 675000 (1 - of web buckling .69) = 209000" 1b

hg = distance between flange centroids I # moment of inertia of entire beam

section about neutral axis

Iw = moment of inertia of web about beam N.A Substituting 209000 Flange load = 25.25

Average axial flange load due to bending with yeam in diagonal tension state, 4 2 486000 + + 15810 1b 29.45 Total Flange Average Axial Loads Due to Bending: ~15840 - 5510 = -21350 1b Po = 21350 lb Upper flange Fo = Lower flange Fy = Flange Axial Loads Due to fension Field Action = _ Sot F 2 cot a my

Spp = shear load carried by diagonal

tensicn field action = ks, where k= 69 DIAGONAL SEMI-TENSION FIELD DESIGN Hence Fp = - 282 ¥,18500 1,928 = - 5710 1b Total Flange Average Axial Loads Upper flange F_ = -21350 -5710 = -27060 1b Lower flange Fy = 21350-5710 = 15640 lb -27060 _ 575 — ~40000 pst 15640 _ 8 40300 psi Average stress upper flange

Average stress lower flange

Comparing these values with the values obtained by the first method of solution we find -44000 and 42450 respectively Thus the NACA method

decreases the load on the flanges *

Check of Fiange Stresses Due to Secondary Flange Bending Moments From Formula 65 % aqbk ts ta? cy From Fig 011.24, C, = 97 when wd = 1.96 Hence, M “3z 69 x 18350 x 025 x 107 x 97 = 2560 in.1lb This compares with 3210 in.1b by the first method of solution The flange stresses could

be found as in solution method 1

C11.28 General Conclusion

In general the NACA method gives higher margins of safety The NACA method is recom mended for actual design of semi-tension field

beams The NACA method has been extended to

cover curved webs and this subject is presented in Part 2 of this chapter

In the example problem the web was

relatively thin and the diagonal tension factor

k of 69 means that wrinkling is quite severe

as 69 percent of the snear load 1s carried by

diagonal tension In heavily loaded and rather

shallow depth beams, sucu as wing spars or wing

bulkheads subjected to large external loads,

the webs are much thicker and the k factor much less

`

The great saving in wed weight over that required for a non-buckling web design easily exceeds the weight increase in the flanges and the web uprights that diagonal tension field

action produces The rivet design in semi- , tension field design presents more detailed

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES and more complex than for a beam with non-

buckling web, C11.29 End Bay Effects

The previous discussion has been concerned

with the "interior" bays of a beam The

vertical stiffeners in these areas are subject, primarily, to only axial compression loads, as discussed The outer or "end bay” is a special case Since the diagonal tension effect

results in an inward pull on the end stiffener, it produces bending in it, as well as the usual compression axial load This action can be clearly seen in Fig Cll.7 Obviously, the end stiffener must be considerably heavier than

the others, or at least supported by additional

members to reduce the stresses due to bending Actually an end bay effect exists wherever a buckled panel ends and structural members along the edge of the panel must carry the pending due to the diagonal tension loads

Typical examples, in addition to the end stiffener discussed, would be the edge members

pordering a cut-out or a non-structural door

in a flat beam, or a curved fuselage or wing

panel Illustrations are shown in Fig C11.27

Flange End Flange Stiff

¬ Stiff — Members

xf v, 7 Provided

3 = = |2⁄4 ' =S to Frame!

Tens App Al Cut |e Cut-Out

Field-_ #1 |) toad Out |% Pull A b= Abella Ss 4 LLA \ (a) (b) Fig C11 27

The component of the rumning load per inch that produces bending in such edge members is given dy the formulas

w=KkKq tana

for edge members parallel to the neutral axis

(stringers) and

weak qcota

for members normal to the neutral axis

(stiffeners or rings) The longer the unsupported length of the edge member subjected to w, the greater wlll be the bending moment it

must carry

For example, consider the end stiffener of

the beam of Fig C11.13 in Art C11.13 Using

the data from Art (11.27,

C11, 23 w=kqocota

-69 (18,350 x 025) x 1.228 = 386 _1b./in

This is a severe loading for the end

stiffener to carry in bending in addition to its other loads A heavy member would be required

There are in general 3 ways of dealing with the edge member subjected to bending, the object being to keep the weight down

(1) Simply "beef-up" or strengthen the

edge member so {t can carry all of its loads (this 1s inefficient for long unsupported lengths)

(2) Increase the thickness of the end bay panel to either make it non-buckling or to reduce k and thereby the running load producing bending in the edge

member (this is usually inefficient

for large panels)

(3) Provide additional momber (stiffeners)

to support the edge member and thereby reduce its bending moment due to w

(this requires additional parts)

Actually a combination of these methods might be best

Consider method (3) above This will some~ times increase the local shear in the bay and should be considered Assume that 3 additional stiffeners are added (equally spaced in this casa) to support the end stiffener against bending and analyze the end bay internal loads resulting for the beam of Fig C11.13 and the analysis used for it in Art, C11.27 (NACA Method)

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C11, 24 DIAGONAL SEMI-TENSION FIELD DESIGN

Note that in Fig (a) there are two sets

of applied loads One is the basic applied shear load of 13,500 1b and the other is the component of the tension field load, w,

calculated as 388 lb./in earlier in this

article,

Pig (b) shows the shear flows in the end

bay panels, taken as constant in all the panels, due to the 13,500 1b load applied ‘The shear flows are shown as they act on the edge members,

Fig (¢) shows the shear flows in the

panels due to the applied load w Average shear} flows (in the center of each panel) are shown Actually the shear flow in the end bay will

29.45 _

2x io~ 572 1b./⁄1n to O 1b,⁄1n, at the center The values

of this variation at the center of each bay are shown for analysis purposes

vary from a maximum of q = 388 x

In Fig (@) the load systems of (b) and (c)

are added to obtain the final (preliminary) loads It can be seen here that the shear flow in the upper two panels of the end bay is Significantly increased over the nominal value

so 29,45 457 1b./in existing in the / other bays of the beam This means that the diagonal tension effects in this area will be

Greater (for the same web thickness) and must

be considered locally here in checking the upper flange, the end stiffener, the added support stiffeners, the rivets, the web, etc

Having determined the basic internal loads,

the members involving the end bay can de

checked for strength using the methods and data of Art Cll.14 to Cll.26 When, as itn the case of the upper added support stiffener, the shear flows are different in the adjacent bays, average values of q and k should be used in

checking the stiffener for strength Forma

57 assumes equal shears in adjacent bays In general, there is no simple analytical

way of calculating exact tension field load

variations when shear flows vary from panel to panel in a structural network ‘The procedure outlined above 1s but an elementary "approximation” that can be used for design purposes

If all margins are near zero, substantiating

element tests are in order

~

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Fig Cli 16 Graphs for Calculating Buckling Stress of Webs 9 200 ANALYSIS AND 400 10 o (@) Theoretical simply with 600 Ser? Fig C11 18 20 DESIGN OF FLIGHT coefficients supported VEHICLE STRUCTURES C11, 25 for oiglas ,9) Empirical rastrant coefficients, edges

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C11, 26 DIAGONAL SEMI-TENSION FIELD DESIGN ‘ 2 3 4 6 8 I0 20 30 42 60 80 ĐQ soo aco 60 !OQO tuy

Fig C11.19 Diagonai-tension factor k, (H h >d, replace by 2, if 4 (or 4) > 2, use 2.)

For Flat Sheet Use Zero Curve

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tana {max fu (8 ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11, 27 kzO rh] 3 PM oN ; Pe «P| = mm: ` ——-ở| 3 P— | ————' Em [mm F E——————== 5 2 - 2 4 6 8 10 a hụ

Fig C11.21 Ratio of Maximum Stress to Average Stress in Web Stiffener NOTE for use on curved webs:

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(a) 2024 Aluminum Alloy (b) Alelad 7075 Aluminum Alloy

Fig C11.25 Allowable Values of Nominal Web Shear Stress 10 3 9 | 2346 0 20 40 62 KO 200 600 @ fs Peer (a) Modulus Ratio for Elastic Web, 9 8 lề 24 fg, ksi “ae 32 sở

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11.29

PART 2 CURVED WEB SYSTEMS

(BY W F McCOMBS)

C11.30 Diagonai Tension in Curved Web Systems - Introduction

This type of structure has an important

place in the design of light metal structures,

The structural designer should have as good an

understanding of it as he must have for the

somevhat Simplsr plane web system Actually, most airframe shear web systems are curved rather than flat, the fuselage of a modern aircraft being the oustanding example To make 2 fuselage skin entirely non-buckling would require a very thick skin and/or a closely spaced suostructure supporting it This would

involve a considerable weizht penalty compared

so the buckling skin arrangement The typical

metal skin in a medern fighter or transport

airolane thus carries its limit and ultimate loads with a considerable degree of skin buckling In view-of this, the need for an understanding of diagonal tension effects in

curved web systems is obvious

11,31 General Discussion

Before getting into the details of design,

a Zeneral discussio of what happens ina

curved web system as the web buckles ts helpful Consider a semi-uonocoque structure with 4 Circular (or elliptical) cross-sectional shape as shown in Fiz C11.29 aoe Ring Attached ta Skin Skin SS Ring "Floating" Ring, Attached Only to Stringers Fig C11, 29 of a number of 2: 12 they are nun few in number,

The structures consists memoers (called "strlagers"

and "l2ngarone" if they are

4 to 8) which are supported oy frames or rings and covered with 4 skin The rings may be

attached to the skin and "notched" to let the striagers pass through, as in Fig (b), or they

may De located entirely under the stringers and not, therefore, attached to the skin In this

latter case they are called "floating" rings,

as in Fig (c) Sometimes both types of rings `

are present Comparing this structure to 2 plane web beam, the stringers correspond to the flanges, the rings correspond to the uprights and the skin corresponds to the web Thus, the stringers carry (or resist) axial loads The rings support the stringers and, if not of the "floating" type, also divide the skin panels

into shorter lengths The skin carries (or resists) shear loads

Now, assume the structure to be subjected to a pure torsion, T, as shown in Pig C11.29 Before the skin buckles, this torsion produces 8 shear in the skin panels given by the well-

known formula

a

a * By

Only the skin is loaded As in the case of the uprights of a plane web beam, the rings are not loaded There {s no load in the stringers

AS the torsion is increased, however, the skin shear stress eventually becomes larger than the critical buckling stress and the panels

buckle Any further increase in torsion must

now be carried as diagonal tension Five main things then occur as the torsion is increased above the buckling value, as illustrated in Fig C11.30

1, The skin panels buckle and flatten out between rings fastened to the skin, from their original curved shane This gives a polygonal cross-section (away from a ring) The angle of diagonal tension is less than that for a plane web beam, however, in the

range of 209 = 30°,

2 The stringers now feel an axial load, due to the pulling on the ends, (C11.30b), of

the structure by the buckled skin, Just

as in the case of the plane web beam 3, The stringers 2lso Zeel a normal loading

that tends to bend, or "bow", them inward

between supporting rings, as Shown in

Fig C1l.30c

bà The supnorting rings feel an inward loading

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C11 30 T (b) Generation of Axial Loads in the Stringers Net Loads Applied by the Skin to the

Sub-Structure (Stringers and Rings) (d) End View of Stringer Showing Loads & Supports PB Pag “RE Loads on Stringer and Support by Rings (Pq) PnG Loads on Ring Due to Supporting Stringers (Addi- (e) tional Loads also Present Due to Skin) Fig C11,30

stringsrs, coming entirely from (3) above,

and is thus concentrated at points This concentration produces not only compression!

but also internal bending moments in the floating rings This is shown in Fig J11.50e,

5 Any fasteners splicing skins together or

fastening the skins to the end rings feel

not only a shear panel type of loading dut also a normal loading, as in the case of a

plane sed Deam Also, the "folds" tn the skin due to the diagonal buckles ory on the rivets as they “attempt” to extend

across the rivet lines at the stringers ani rings

The important thing to realize here is that, although only a cure torsion has been apolied, considerable axial loads have deen

generated the stringers and rings And

cing moments have been induced in

even some t

fing2rs and in the rings or the "floating"

Now, assume that at the same time the

torsion load is being applied an increasing

compressive axial load, P, is being applied, simultaneously, as shown in Fig C11.21

Fig C11, 31

The following will now happen due to the presence of P,

1 The stringers will, of course, have to,

carry the compressive load, P, which will be divided among them There will be some "effective" skin to help

2 Less obvious, but very important, is the fact that the diagonal tension loads due to the torsion, T, will be considerably affected by the presence of the axial load, P, The larger P is, with respect to T, the greater will be its effect upon the diagonal tension effects This is as

follows

a) The skin panels will now buckle at a lower amount of applied torsion since

they are now also strained axially in

compression Actually there is a "com— bined” buckling consisting of compression and shear buckling This can be obtained from an interaction equation, discussed later

b} Since the critical shear buckling stress

is now lower the diagonal tension

factor, k, ts larger

c) All of the diagonal tension effects dependent upon K are increased These

include the axial loads induced tn the

stringers., the normal loads bending the stringers inward, the Loads induced in the rings and the loads felt by the fasteners

d) The angle of dctagonal tensio will be

larger, closer to 459,

Thus we see that the effect of compression is to

increase the loads due to diagonal tension

Now assume that instead of being compres— sion, the axial load, P, is tension, In this

case, the effects of diagonal tension, due to

T, are reduced

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 2 The skii panels can carry a larger shear

stress before buckling in Shear In fact, a relatively small amount of axial tension can prevent them from buckling at all, as

will be discussed later

The diagonal tension factor, k, will be smaller (or zero if no buckling occurs) 4, All diagonal tension effects dependent

upon k are reduced and the diagonal tension| angle is smaller

Fiaally, suppose that instead of an axial load, P, a bending moment is being applied Simultaneously with the torsion, T, as in Fig

C11.32

/ 7

T,

xỆ Fig C11.33

In this case, also, as the torsion is being increased from 4 small mount, M is also being

iacreased, (the ratio of M to T being constant, as was the case for axial load, >) Then,

from the usual bending theory, fs Mz, the 1

following will occur

1 The stringers (and skin) above the neutral

axis will feel compression loads, the

further away the greater the load The

upper skin panels will thus buckle earlier in combined compression and shear and produce the largest diagonal tension loads

on the stringers and rings

stant} in the rings and

1 this region

3 The skins near the neutral axis will feel

little or no strains due to bending and

will buckle about as in the case for the

pure torsion, T, producing equivalent

effects

In an actual airplane structure, a fuselage for example, the applied leading is more

complax Instead of simply an applied torsion,

there is also, usually, a vertical and perhaps

@ Sideward set of loads which produce shears

that vary from panel to panel And there may be not only a Dending moment, which changes,

along the fuselage, but also axial loads due to

11 31 landings, catapulting requirements, etc

Obviously all of this complicates the calculations and experience and judgment are of great help; dut the method of going about it

is fundamental and will now be discussed The N.A.C.A has conducted an extensive

program over the years with the object of determining a system for the design of structures having curved weds in diagonal tension The theory for this system, as in the case for the plane web beam, was given by

Wagner and others and modified as necessary

from the results of many tests The design method Is fully discussed in Ref (3) and the

substantiating test program in Ref (4) The

reader is encouraged to consult these refer- ences for a fuller presentation of the theoretical development and test results

AS mentioned earlier, curved web systems are of two general types One of these has an arrangement which results in the web, or skin, panels being longer in the axial direction, d, than in the circumferential direction, h This is typical of the stringer system in 4

fuselage, as shown in Fig C11.33

Fig, C11.33

That is, the geometry of the stringer spacing, h, and the ring spacing, d, is such that

ge

"Floating" rings, not being attached to the skin,

do not determine the spacing, 4

A second type of curved web structure may

be referred to as the longeron system Its main characteristic is that the skin panels are long in the circumferential direction This tyve of structure would, typically, in a fuselage, consist of a few axial members (a minimum of 3 but more usually 4 to 8 for a “fail safe" design) and a large number of closely spaced frames The frames would, in this system, be at about a 4" to 6" spacing as compared to a 15"=20" spac-

ing for a stringer system This gives

ead

as indicated in Fig Cl1.34

242 395

Trang 15

C11, 32 Longeron Ring ra Fig, C11 34

In the longeron system the frames are attached to the skia and longerons, there are no

"floating" rings

Most of the NACA data and design method ts for the stringer system The design method for the longeron system, also presented herein, has evolved from Ref (3) and also from the results of an investigation and test program at Chance-Vought Airerart Corp These latter checks are in some places different from the ones in the stringer system The main problem is the determination of the angle of diagonal tension, for either System Once this is known all of the diagonal tension stresses are known

‘The stringer system will be discussed first and tne longeron system will be discussed Secondly The basic approach is similar to that for plane web beams

C11 32 Analysis of Stringer Systems in Diagonal Tension Before the diagonal tension effects can

be calculated, the primary internal loads in

the structure, due to the applied loads, must

be determined This can be done as discussed

1n Chapter A20 The engineers theory of bending, can usually be used to determine the axial loads in the stringers, as in Art, A20.2-A20.5, The shear flows in the various skin panels can

be determined as in A20.6-A20.8 In the case of striager constructton a more accurate

determination 92 the skin shear flows may, if desired, be obtained if the skin panels are

considered flat, rather than curved, between

Stringers This is because the panels, after Duckling, actually flatten out, giving a

polygonal cross-sectional shape, except

imnediately adjacent to non-floating rings, as shown in 1g G11.20a, The diagonal tension effects can now be evaluated 1 DETERMINATION OF CRITICAL BUCKLING STRESS OF SKIN PANELS

The buckling strength of curved panels

under pure saear and compressive stresses is

covered in Chapter C9 The equation for

buckling Snear stress F Ser is,

(by?

F sop — HT ng B + 12 (1- Pa”) TẾ Txg E x

where simple support is usually conservatively assumed in determining Kg

When axial loads (and strains) are present,

as in practical structures subjected to bending as well as shear, the situation is more compli- cated As discussed in Chapter C3, the presence of compressive stresses together with shear Stresses causes the panel to buckle at a lowar value of shear than if no compression were present The presence of tension stresses, along with shear stresses, enables the panel to stand larger shear stresses before buckling occurs In fact, a relatively small amount of tension stress may prevent the panel from ever

buckling in shear

It ts important to determine the actual shear buckling stress when axial stresses, particularly compression, are present The Treason is that this affects the diagonal tension factor, k, and all of the ensuing stresses affected by k

2 First consider a shear panel subjected to 3 shear stress fg and a compression stress fo It has been demonstrated experimentally at the NACA, Ref (5) that.a curved panel, thusly loaded, buckles according to the interaction formula fs ft Fort G7 210 -<~ (71) Cer Ser

where Fog, and Fs op are the critical panel

duckling stresses for pure compression and pure

Shear respectively From Chapter C9, the ——

buckling stress for a curved sheet panel, in

end compression is given by the equation, P - H27 ke E bya Ser ~ 12 (1 ~ Uạ”) t Now for any part!eular panel, and F, Cer 7A Faop

Now, also for any given applied loading condition being checksd we Gan calculate

fo (= +) at the center of the panel (before buckling) and also fg (previously done) These

stresses will bear a constant ratio to each

other until buckling occurs, after which the

Trang 16

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

compression stress no longer increases Thus

we can write Ỷ a

t 7 B ¬==~=———~>~——-—-—~ (73)

or ff, = B fg

Substituting these formulas for Foor and fc back {nto the interaction formula (71) we obtain, fs tt (3+)? = 1.0 "Ser “Scr Solving this quadratic for fg we get fs = Pgor

where fg is the actual shear stress at which

the panel buckles due to the presence of compression stresses Calling this svress Ty and calling the expression in tue brackats

where Re is always less than 1.0 when compression stresses are presant

* Next, consider a shear panel subjected to

a shear stress, fs, and a tension stress, fr

For this case it has been experimentally demonstrated, Ref (6), that the interaction relationship defining buckling is

fs

oxy Scr

where Fg and F, cer are as previously discussed

The shear stress, fg, at which the panel will

buckle when the tension stress present is fr

can be solved for directly from equation (76), Sor a= 1.0+ ae - ee ee ee (77) "Sor Sor or, tt fs¿„ = (1.0 +o) Paap 7 TT (78) T Calling the expression in parenthesis in (78) Rp we get, tsor * C11 33 Since Rp is always greater than 1.0 when tension stresses are present, the actual shear buckling stress will always be greater than Poon the

buckling stress for pure shear loads only

Inspection of the term in parenthesis tn (78) making up Rr shows that as the tension stress

becomes several times larger than Foor the - value of faq, will become several times larger

than Fs Thus, if fy is large enough no

buckling will occur even when large shear

stresses are present

2 DIAGONAL TENSION FACTOR, k

The next step is the determination of the diagonal tension factor, k This is a function

Ỷ where, as điscussed above,

se

tsor = Papp Re OF fsoy = Fay Br

depending upon whether compression or tension

stresses are present

For a curved panel the formula for k, as determined from much test data, Ref (3) is the empirical relationship, f k = tanh | (.5+300 49) 10g,, —2-] - (80) Ra for R = radius of curvature with the auxiliary rules that a) it S>2 use only 2

b) if h=d replace poy ("1ongeron"

system) and, in this case, if ae 2 use only 2

Rather than calculate k from the formula, it can more easily be obtained from Fig C11.19 3 STRINGER LOADS, STRESSES, AND STRAINS

AS in the case of the plane web system, the total stringer load will consist of the primary axial loads, Pp , due to applied bending moments and/or axial loads, plus the dlagonal tension

induced loads, Pp,m,

P, stk Pp * Pap 7-7 tte

Pp is determined as in Chapter 420.2-a20.5 Pp.t, is determined, similarly to tne case for

the flanges of a plane wed beam, from the

"pulling" of the buckled skin on the and

frames

dT

Trang 17

C11, 34

Fig C11 38

AS shown in Fig C11.35, Pp,p, 18 the

diagonal tension load in a stringer bounded by panel “a” on one side and panel "b” on the other Let the width of panel (a) be hg and

the width of panel (b) be hy (For equally

spaced stringers, of course, h, # hy) Let the

shear flow in panel (a) be gg and that in panel

(b}) be qụ Then we can write

> Ba Ga By OOF Sy Ky My By OOF dị,

D.T 7 2 8

To keep the calculations Siu21er (as will be

aporeciated later) we can accept some in-

acouracy and use average values for the Aa + Bh da + đọ respective terms (3, a get 2 , 2 , gtc.) and Pout, 4 qa cota

remembering that this is an "“averas

which, for closely spaced stringers, 15

suffictant, especially for preliminary design

The total stringer load is then, from 80 and 92

Popg = Pp + x Qh cota

where ail terms are known except a

Tae stringer stress is obtained by

dividing tha terms on the right side by their

respective effective areas

= k ah cota

iste = ~~ eD.T

3) 4ey is the total effective area, stringer

and skin, used in determining the primary loads

2s in Tnap 20 If Pp is tension the area

is equal to ht, panel is fully effective DF Aa =

Ref (3) suggests a more accurate calculation

Rot depending upon whether axial compression or tension {ts oresent Again, «k and R, or Rr are also average values for the pansis™on each side of the stringer Thus we can write tera = Pp + k gh tot a Ảsmg † ÂøsswIN Agta * -Snt (1-K)Re 7 or cot , = Pp + k fg cota “STR A STR * “esxin ASTR+ 5(1-4)R, 9 nt ? - (85)

where all terms are known on the right hand

side of the equation except a

The total stringer strain is then

If fgpg is larger than the proportional limit stress, or in the neighborhood of the yield stress, Ey is not a constant and a stress strain dlagram should be used to read ¢ directly, using

fsra Again, fgpa, and hence egtp, cannot be determined until a is known (later)

There is also a secondary loading on the

stringer whieh tends to bend or "bow" it inward This is caused by the fact that the taut skins are pulled flat on each side of the stringer Thus as in Fig C11L.30 thare ts an inward component of the skin diagonal tension loading that pulis the stringer inward This loading

is not a simply distributed one; it is largest in the middle of the stringer and becomes smaller at the supports Ref (13) recommends that the effect of this loading be considered

as producing secondary bending moments in the stringer, taken as

Lo mm (37)

R = radius of curv Tais represents a "peak" moment at the middie and at the ring supports It will produce

tension on the inside of the stringer at the

middle and compression on the inside of the

stringer at the supports The recommended value of Meta is the result of many test

measurements, therefore it is of a semi-

empirical nature

4 STRESSES AND STRAINS IN RINGS

There are two types of rings, those attached

to the skin and those mot attached to the skin,

called "fleating rings," which support, and are

therefore loaded only dy, the stringers Rings 2

Trang 18

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11, 35

The stringers are actacned to these rings f = X ts pe (81)

locally by some shear clip arrangement The RG ARG

rings feel an inward acting loading which puts zr

them in “hoop compression" These loads come from the stringers and from the skin The Stringer being pulled inward by the skin, as described above and in Pig Cl11.30, in turn pushes inward on the Supporting rings The skin, not being flat at the ring, also pulls inward on the ring The result of all of this

is, essentially, according to Ref (3) a hoop

compression stress 1⁄2 the ring for the case of a cylinder under pure torsion, This is because tne loading is approximately equivalent to an evenly distributed tnward acting radial loading

For this case the radial loading can be taken

as, per incn along the ring, - fe dt-k tan a

Pre

Rang The axtal (hooo

will then be ) compression load in the ring Pog = DR =fg dt k tang - - -~— ~ (88) The axial compression stress in the ring will be ~ trq = P, 5G s——S kf, tana gg} Arg Ae, °RG+SKIN ~ 2G = ap * +3 (1-k)

Where ali terns on the right are known except a The axial strain in the ring will be

faa

£ RG Fro Sp eee eee ee (90)

where ipg and hence epg are unknown until a is determined

When loads other than pure torsion are applied fg and « will wary from panel to panel

and the noop compression stress will not be

nstant around the ring There will also be

Some varying secondary shear stresses in the

panel due to unequal "pulls" on each side of the panel, at she stringer, by the buexled skin the

When "floating" rings are used concentrated

inward acting radial loads are apelied by the

stringers This produces hoop compression and, since all the Loads ars concentrated, also

Some bending moments There is, of course, no

effective skin acting with these rings, The axial compression load is

Pop =f5 dt k tana

The axial compression stress ig then,

where all terms on the right hand side are known

except a,

The axial strain is given by

or can, of course, be gotten from a stress Strain diagram for the material,

The maxinun bending moment present ina floating ring is given by (93) kf, th? a tana te” R = radius of curv 12R

This occurs at the fdunetion with the stringer There is a secondary moment, half as large, midway between stringers {n the ring

S Se STRAINS IN THE SKIN PANELS ORIN PANELS

The strain in the skin panels is given in Ref (3) as

fs 2k

era

where u = Poisson’s Ratio = 32 for a1uminưn,

every term on the right hand side of the equation is known except a Fig C11.36 is of help in calculating e, giving the value of the bracketed tern

din ga * Sin 20 n-Wa+a)]

6 SEESRMINATION DETERMINATION OF a* OF a”

For the stringer system (4d >h) Ref (8) shows that a is related to the stringer strain,

the ring strain and the web, or skin panel,

strain by the formula Ere tan? ao z—————N.—— _ _ (95) € 1 hịa — °ạạ * Br &) R = radius of curv where € is a tension strain (+) and eRG and €9TR are entered as negative members for com—

pression, thus adding to e

a is determined by successive approximation using the three prior formulas for e, egp and

€pg and then checking with the formula (95) above That is,

———-_——_————_ _—_Ö

* For a faster estimate or for preliminary design,

be skipped for panels in compression and shear step 6 can

Trang 19

Cit, 36

1) Assume @ value for a

2) Determine c, epg and egTR using this assumed value of a

3) Substitute ¢, epg and egpa into the

formula for tan? a and get a "new"

value for a

4) Repeat stens (1) to (3) as many times

(say 5) as necessary to get a to a

"converged" value

If rings attached to the skin are being

used then epg in step (2) above is obtained

from (90), using (89) for fpq If "Ploating"

rings are used then egg 1s obtained from (92),

using (91) for fra

Once a is determined, the stringer stresses and ring stresses are known, having been used

in getting the strains egpr and eng for the

final check of a in (95) The bending moments in the stringers can then be calculated from

(87) and the bending moments in floating rings, if used, from (93)

7, LOADS ON THE RIVETS

The only remaining internal loads to be caloulated are those acting on the rivets These are of two types

a) There are the primary loads, in the plane of the skins which try to cause shear or bearing failures at the

riveted joints, as in any spliced skin b) There are also, "prying" forces on the

rivets which try to "pop off" the rivet heads or to pull the skin up and around|

the rivet heads This latter may

oceur, particularly, when counter- sunk or dimpled skins and flush head rivets are used and the rivet diameter

is too small or the rivet spacing is too large These are "secondary" Loads .i The primary rivet loads occur whenever the

skin {s spliced, which is usually, but not

necessarily, along a stringer or along a ring These loads would also be present if the skin

panel ended at stringer or ring, as at an

opening or "cut-out" and the panel had not

‘ean re-enforced by a doubler to prevent buckling

At a splice parallel to a stringer the

load per inch along the rivet line is due to the same effects as discussed for the plane

web beam it is

Load/inch = fst [rrx ( ——_ 1) —~ (96)

cos a

At @ splice (or opening) along a ring (a vertical splice} the loading is

Load/ineh = fg t [1+K qe u on

Note that in either case, if the panels involved are made non-buckling (on each side of the splice) k=0 and the load per inch is the same

as for a non-buckled web It is only around

a "cut-out" or opening that the panels are made

either non-buckling or made to buckle to a

lesser extent, and this is done to "relieve" the loading on the edge member rather than on

the rivets (see Art 011.29)

The second type of rivet loads, the prying loads, are not determinable by any formula Ref (3) recommends that an arbitrary eriteria be used as follows The tensile strength of the rivet-skin combination, Pp, should be such that

{it is as large as the number given oy

Pp per inch = 22 t Foy

where Foy is the ultimate tensile strength of the skin or web material being used

Pp is usually most critical for flush

attachments AS an aid in getting this, Figs Œ11.27a and C11.27b give information on the tensile strength of various rivet types and

sizes

C11 33 Allowable Stresses (and Interactions) 1 STRINGERS

Just as there are two types of basic loads (and stresses) in the stringers and rings (the primary ones and the ones due to diagonal tension effects) there are also two types of allowable stresses yor local failure An

interaction formula is thus used to predict adequate strength This its,for the stringer £ fst —P-+ MAX = 1,0 Feo Fgp where

fp = stringer stress due to the applied loads, this is the first term on the

right hand side of equation (85)

Foc® the allowable crippling stress for the

stringer, obtained as tn Chapter C7

Tem

- MAX

fsmuy 7 fST * Fen

fsm 1s the second term on the right

Trang 20

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES ter êm is a ratio obtained from Pig 3T

Cll.21 It 18 explained in the plane

wed beam discussion

Fgps the "forced crippling” (diagonal

tension effect) allowable stress for

the stringer is obtained from Fig C11.38, as in the case for the uprights in the plane web beam, or

Calculated from the formas (also

for Faq and taq) tị sr^ 26,200 k®⁄* Ga for 248T material tị Fgp= 32,500 k2/* (2STR)3/9 ror ;gep St t

material with The Same restrictions noted in Cll.22 (The total axial

stringer stress is f, + fgp at the area near the supporting rings, and

fp + tat ax in the middle between

supports.)

The stress in the stringer due to the bending moment, Mạng: of equation (87) can be calculated and added to fp in the proper

manner

Ref (3) also suggests the stringer be checked as a column, fully fixed at the

supporting rings and carrying the stress fone:

Some allowance should be made for beam column

action, due to the bending moment present in the middle of the stringer This can be done using some effective loading producing the

moments discussed or by carrying some extra margin of safety over and above the column

buckling stress,

2 RINGS

The rings nave allowable stresses similar in nature to the stringers The interaction equation is,

for the case of rings attached to the skin

fy) is the stress due to the ring carrying loacs other than the tension field ones (as a

bulkhead analysis would show) Faq is the allowable forced crippling stress for the ring, obtained in the same manner as that for the stiffener, (Fig C11.23),

AR aA b a th

fr RGMAX = fpg X —3— TRq , obtained in the same

manner as discussed for the stiffener It

C11 37 occurs in the ring midway between stiffeners,

Fog 1s the "normal" allowable crippling stress

for the ring If Fra => Foo then use Fog for

Fra

Floatt rings, not being subject to forced crippling by the skin, are checked in the usual manner for the stresses due to hoop compression

loads and the accompanying bending moment,

equations (17) and (19) The interaction

equation is,

fia tf,

ee (101)

ce

tpg 15 a constant stress between stringer

Junctions; it does not have a maximum peak as do the rings attached to skin

3 GENERAL INSTABILITY

A general instability check for the

stringers and skin can be made from the empirical criteria presented in Fig C11.39

This is obtained from test data and recommenda- tions of Ref (7) and Ref (3) The allowables

are based upon pure torsion tests The ade~ quacy of the structure is checked by

t

wom ilo

Ÿsrysr

It is suggested that a respectable margin of

safety be held (M.S = 15), The radii of

gyration in Fig C11.39 should be made assuming the full width of sheet to act with the stringer or ring respectively and that the sheet is flat because the criterion was obtained under these

assumptions

4 ALLOWABLE STRESSES IN THE SKIN (OR WEB) OES N INE SKIN (OR WEB) Ref (3) recommends that the allowable

stress in a web or skin be taken as, for non- flush attachments,

*

sau, 7 feqpp (68 +4) + (103)

where 4 = 3 tanh ẢRG „ al tanh AST 3 4 can, at ht

more easily, be read from Fig C11.40 Bart

is obtained from Fig Cll.4lb or €11.41e aftar |

obtaining appp from Fig Cll.4la Data from

tests by the Chance-Vought Aircraft Corp indicate that the allowable web stress can

Siaply be read from Fig Cl1.42 In etther case,

the allowables apply to fs, the gross stress in the panel The net shear stress between rivet

Trang 21

C11 38

C11.34 Example Problem

The roregoing explanation can be better explained or clarified through the presentation of an example problem

Assume that we have a fuselage with a

structural arrangement as in the example problem of A20.5 and Fig A20.3 and 420.4 of Chapter AZO Also, assume the moment of inertia and

neutral axis for a linear bending stress distri-

bution to apply These values are given on

page A20.8* Also let it be assumed that the

supporting rings are attachsd to the skin and

spaced at 15" ‘The rings are 1" x 3" x 1" 24ST

Z sections, 040" in thickness (see Pig.Ccll.43) An analysis will be made of stringer #3, skin panels 2-3 and 3-4, and the ring [t Is

assumed also that an axial skin splice occurs

along stringer #3 and a vertical splice along the ring to show a check of the fasteners

ẤN) ping "bY Ring "8" “án #, J«- 1525 “%, Me1,387,000"4 \ T= 4 saa(o#| - evo 11,700# v= 11,700# Fig C11.43

First the internal loads in the stringer and skin panels are determined The "average" loads at the middle of the bay will be used in this example to get the diagonal tension effects At the center of the bay, M = 1,475,000 in 1b T = 634,000 in 1b VY = 11,700 1b Max stringer stress occurs at Ring d:- „Ắ MZ _ 1,565,000 (22.9) ÝPwax a 2 —— Des = 21,600 psi 1,475,000 (32.9) _ = oe est Se ‡ Thy, 2202 20,400 psi Shear flows in skin panels:~ vQ Tt wg = T* , Where A consists of area of 26 triangles „ 31,790[.157(58.5) + 169(56.4)], 2

* A non-linear bending stress distribution can be used, but it also is affected by the diagonal tension compressive stresses

and involves considerable iteration Linear ones are often used DIAGONAL SEMI-TENSION FIELD DESIGN 634,000 2 ¡26 x x 7.2 61.6 + 113 + 175 1b./1n „ 11,790[.167(58.3) + 168(86,4)+.216(32.9)| Seg 2:02 + 115 = 96 + 113 = 209 1b./in

The diagonal tension effects will now be

calculated assuming average q’s, bending stresses, etc., for the panels 2-3 and 3-4

_ 175 + 209 _

3ạva-¿ 2 EA = 198 1b./1n

198 _ 192 _

fsạy, „ " TẾ” 5 ae 7 S888 PS!-

The average critical shear buckling stress will now be calculated using equation (75)

First, if mo skin buckling occurred the average compression stress in the two panels would be, approximately, the stress in the stringer between them, thus

Ícpane1s = fay = 20,400 psi then the constant, B, would be

fo _ 20,400 _ fs 6,000 3.40

The critical pure shear buckling stress equation from Chapter C9 is, a TỶ KÔ 5 ser 7 12 (1 - Ue") Prom Chapter C8, kg 2 16.8, - Ị 9 C.3-b = 12X16.8x 10,600,000 (,032)2 12 (1 - 3*) 7.2 3140 pst bya cr &) ¥ser

For pure compression the critical buckling stress equation from Chapter C9 13, 3 K.8 nr tên oF họa Fọc “1E {1 -Ua) th Re

Thus, A = Poor/F sor = 2980/3140 = 950

Since buckling stresses are below the pro-

portional limit stress of the material, no plasticity correction was necessary as is

usually the case in thin walled structures ` Thus stress ratio (A) could be obtained directly

4S ke/Kg‹

ì

Trang 22

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Next: - _°ð+ /()°®+ấ - A = Re = 5 = _ 5.40, /(8:401* 950 5 .050 = 126 =

and, from equation (75)

lScre 4 = Feop x Re = 3140 (,26) = 815 psi which is quite small, due to the presence of compression Next, teay Sen, = 7.86 815 now k can be determined: ‘td _ 300 (.032) ts 300 FS a (2.0)*= ,64 Fron Fig C11.19, k= 75

The expressions for stringer and ring stresses can now be written in terms of a: Substituting into equation (85) and using Ty for the first term on the right side,

ô75 (6000) cot _Â

20,400+ S SE) ga)? c5(0-:76) (86) EE

Toor

20,400 + 5570 cot a

this 1s the “average” stress in the stringer

Substituting into equation (89)

and using Apg = (1"+3"+1")(.040") = tor rings, -20 in.*, -75 (6000) tan a R8” — E6 TiS}(.caay * °8 (1¬ +78) £ 8,300 tan a

a will now be determined, by successive approximation using equations (94), (86), (92) and the above expressions for stresses in them, and also Fig C1l.36 for equation (94), Since there 1s so much compression involved, assume

a= 459, to start with From Fig C11.36 ek „ 718 95182 x 105 * 1086 x 10 1.83 (6000) _ = = 20,400 + 5570 (1.0) - 22so x 1075 srr © —_— {0.6 x 10 * since 2 > 2.0, use 2.0 (See page C1l 33) Bs Ỷ C11.39 _ 8,300 (1,0) „ _ faq Ð “Tồ.6 x 105 = TỐC X 10 Now golve for "new" a using equation (95) a (1036 + 2450)107¢ tan a= (1086 + 782)107* me + T1 (CO ) ,7.20:a ~ 0,822 tana = 907 a 42.29, which is less than the assumed 489, As a second trial assume a = 42.49, = 1.84 (from Fig C11.36) 8 e = 1.84 x 6000/10.6 x 10° = 141 x 107-*

Egnp = 20400 + $570 (cot a)/10.6 x 10° = 2505 x10-°

fag = 8,200 x tan a/10.6x10° = 752 x 107°

¬=

tan* a= (pi + 7eeyiore + ooeas St

whence a = 42.6° as against 42,.4° assumed The

accuracy of the theory does not warrant a closer

check, thus a wil] be taken as 42.4° |

The corresponding stringer and ring stresses

are as follows:

fone = 20,400 + 5570 (1,092) = 20,400 + 6090

These are not added for local strength checks

They are added for a column stability check

The term on the right is the average stress, forced crippling, due to the diagonal tension effects tog = 8,300 (.917) = 7,610 psi Also, from equation (87), _ £000(7.25) (.082) (15) (-75) (847) „ Moor’ 24 (0 299 1b./1n and t € th = Moon © 299 (S32) = 5,200 pst lsp

The rivet load/in along the axial splice

is, from eq (96),

Load/in.= g000( 032)] 1+ r6 -) =

243 1b./in

Trang 23

C11, 40

and, along the circumferential splice, from (97)

Load/in.= sooo(.g82)[¬ + 78 (a5) -1)] = Ø§2 15./1n

From the above stresses, tuy ratios,

and the allowables, checks for adequate strength) can be made as follows:- STRINGERS : - At the Junction with ring (b), fore = (21,600 + f,) + 6090 pst For adequate local strength, t.* ỨC tsp = mm — c st Sgt 2/3(-2*)x/4 Fg = 26;000K AE) 26,000(.78)5⁄4 ca = 24,900 pst (Also determinable from Fig 11.38) Thus, 21,600 + 5200 „ _Ê080_ = 39,500 24,900 ~ „92 (sufficient) At the center of the stiffener, ÝSTMax temax = ‘sr | Ton ) from Fig C4.21, tian ~1,% st Therefore, fgmwuy = 6090 (1.92) = 6210 psi then we get, using the interaction formula, 20,400 + 5200 2 ; t——= 6210_ 5900 = „90 (sufflc1 e nt ) Checking as a fixed end colum, _ fi (ee oe fez, vig = 412 -—5_: L/2p * Saray = 18-2

F COL == 39,500 (very short colimn range)

DIAGONAL SEMI-TENSION FIELD DESIGN thus

20,400 +-6090 _ 39,500 67 (sufficient)

RINGS

The rings attached to the skin are subject only to the forced crippling stresses, fpq.-

Midway between stringer junctions this

action 1S a maximum and for adequate local strength, ẨRGMaX z <x = 1.0 RG Using the data in Fig C1l.21, tr fRGwAy = tro ar ara 7,610 x 1.0 = 7,610 psi and thus #ì

Pena = 1810 x3 (sufficlent) Fra 23,100

The same compression stress {and lead, P = frq x Apg) exists at the stringer junction If

the ring is notched to let the stringer pass

through, as 1s usually done, the net section at the ring must be made capable of carrying this load, which is located at the centroid of the wnenotched ring cross-section Usually this means some "beef-up", locally, around the notched

section; sometimes incorporated into the clip attaching the stringer to the ring

Actually the rings, like the stringers, are

subject to the average of the shear stresses in the panels fore and aft of them These have

been assumed equal in this example SKIN

The allowable shear stress taken from Fig C11.42 is

Fg = 21,800

Using the method of Ref (3)

Paqiy, = 207800 (from Fig, Cll.dla and b)

A 38 (from Pig C11.40) Thus

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11, 42

either case shows there to be a large margin of safety for the actual shear stress

of fg = 6000 psi, (average value shown) (Panels nearer the neutral axis will, of

course, feel a larger shear stress.)

FASTENERS

Any fasteners used in this area must, of

course, be able to transfer the load/inch at

the splices as previously calculated This criterion might not design the spacing however They must also have a tension allowable, when

installed in 032 skin, of, equation (98),

Tens All.Load/Inch = 22 (.032) (62,000) = 435 1b

From Fig C11.37 it can be seen that 1/8" flush

head aluminum rivets in a dimpled 032 skin at 50” spacing should be adequate, or 1/8" prazier head rivets at a spacing of J5" ifa "flush" joint is not required The tension requirement is most severe in this case GENERAL INSTABILITY

A check against overall instability of the "network" of rings and stringers can be made using Fig Cl11.39 This is analagous to the column stability check for the uprights of a plane web beam In the case of curved webs, however, no "help" is given by the taut skin

in preventing instability

For this example problem the values used in calculating the shear stress in the skin at which “collapse” due to instability would occur are: = ,414 (Stringer S, of Fig Az0.4 in 0 st Chapter A2O) Pag = 819

(Both of these radii of gyration are gotten by including a full width of skin acting with the stringer, Wg =h, and with the ring, We = d, per Ref (3) Then (,414 x 819)7/% x10* (15 x 7.85)*/% (80)%⁄% (Psr PRG)”/° x10 *_ (an)*/? n4 = 2386 133 x 10° = 29 From Fig C11.39 Fs Eo INST | 3.0; hence Fsiygr go ~ 3x10.3x10 Ips > 30,900 psi and the margin of safety against collapse is, F, SINST -l= 30,900 „ M.S = Fe 6,000 = large

Other panels, of course, might have a larger

shear stress and a smaller allowable Fervor

and thus be more critical, but there is toc much stiffness available, as seen, to expect any instability troubles

LONGERON TYPE SYSTEM

C11.35

The longeron type of structural system is somewhat simpler from a total analysis stand-

point This is, of course, primarily because

there are fewer members carrying the axial loads and not as many shear panels with varying shear loads This type of structure may, or may not, be the most optimm arrangement from

a weight and manufacturing cost consideration for a particular airplane But this is not the

subject of this discussion since it depends upon an optimization study The methods of analysis presented here would, however, have 4a place in the calculations behind such a study Sone typical types of longeron structural systems cross-sections for a fuselage are shown in Fig C11.44

Fig (a) shows the minimum arrangement, ag

to number of longerons, since at least 3 axial load carrying members are necessary for

equilibrium when bending moments in more than

one plane are involved This arrangement, how~ ever, has a disadvantage in that it is not a

"tail safe" design This means that the failure of any one member will not leava a structure

capable of still carrying some arbitrary

percentage (usually 50% to 67%} of the design

ultimate loads

The system shown in Fig Cl1.44b is capable of doing this and is, therefore, the minimm

type acceptable from the “fail safe" standpoint

(4 longerons) More longerons may be used, as in Fig (c), occasionally, depending upon other factors of design and manufacturing

Trang 25

ii, 42

system for strength-weight efficiency, as

mentioned before The optimum spacing of the supporting rings, or frames, is determined

after many trial calculations have been made

involving different gauges and sizes and spacings of rings and thicknesses of skins,

As in the previous discussion of stringer construction, the rings support the skin, dividing it into smaller panels, lengthwise They also support the longerons, similarly to the uprights (stiffeners) in a plane web beam, against bending due to unequal tension field

"pulls" by the skin on each side of the longeron There are no "floating" rings in a longeron structural system

After buckling, the skin in this system has tension diagonals which, rather than being

"flat" between closely spaced stringers,

instead lie on a hyperboloid of revolution

That is, they flatten diagonally between the closely spaced rings This action is discussed

in Ref (8) and the reader should consult 1t

for the basic theory as presented by Wagner originally In this system the ring spacing

"d" 1s such that d < h; see Fig Cll.34

C11 36

The engineering procedure for calculating stresses and allowables for the longeron system

is somewhat similar to that used for the stringer system The reader will note the differences

1 First, at any bay being checked, determine

the primary internal load distributions in the longerons and shear panels due to the applied

loads This can be done as in Art C11.34

using the engineers theory of bending itn most

cases In other cases where judgment and experience and the mature of the structure

indicate it, the method of Chapter AS may be

used in determining the primary load distribution due to the applied loads,

2 Next determine the critical shear buckling stresses in the skin panels Since compression

stresses are nearly always also øresent in

practical situations, pure shear buckling does

not cccur Thus, as discussed in the case for

stringer design, some rational interaction must

be used to obtain a "reduced" shear buckling stress This can be done, for example, by

using some "average" compression stress in the panel, weighted toward the high side for con- servatism Thereby the interaction method of Article C1ll.32 can be used where

and (A) is determined for a curved panel of length "d" between rings and height (n) between

longerons measured along the circumference as 1n Pig C11.42 (B) ts the ratio of the con- pression stress to the shear stress, (fe/fs), for the particular loading condition being investigated The compression stress should be

calculated as if the panel being calculated had

not yet buckled Then, as in equation (75)

fsor * Re Foy

gives the reduced shear buckling stress, Tg

When tension strains, rather than compressive ones, are present with the shear we

have, as in equation (78)

Rp = 1.0 + ott TS io * OFS or

and then

fe.p = Br Fey, 26 in equation (79)

3 Next, the loading ratio, {se can be calculated using Tà as determined in (2) above,

4, Following this, the diagonal tension factor, k, can be obtained from Fig C1l.19 5 The total axial stress in the longeron can now be written as fh = ip + fon k, f, cot a, =f, - 2 D BAL nyt, + 25 (1~K,}Re Ke + cot ag ——————— ?———- - - - - a, (104) ht,” +25 (L=ka)Re

r "ứ primary longeron stress from Step (1)

(+) if tension and (~) ££ compression

= Longeron Area

= a a

and

kK, fg, cot a, Ro (or Rp), h, and t are as

previously defined, One set, subscript (1), is

for the panels above the longeron and the other,

(2) for the panel below the longeron

6 The average stress in the supporting ring

(or frame) due to diagonal tension effects is

given by the following formula {and note that

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