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

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

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

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

(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

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 allow- able 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

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

total B

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,

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,

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 =

which falls within the given range +333

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

Qe _ 27.94

a =I F 2.79, See Fig C11.16 for Ags

Using the above values, we find from Fig

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

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

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 =

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 up- right 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

colum stress for a slenderness ratio of

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,

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

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

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

typ œ

Cll, 22

bending moment = 50 x 13500 = 675000" Ib

Diagonal tension factor = 69

Thus bending moment developed as a shear re-

sistant beam = (1 - 69} 675000 = 209000” 1b

The remainder of the bending moment is

developed as 2 Dura tension fisld beam, or 4

AS 2 Shear resistant beam

flange load equals average axial

hg = distance between flange centroids

I # moment of inertia of entire beam

section about neutral axis

Iw = moment of inertia of web about beam

Average axial flange load due to bending with

yeam in diagonal tension state,

Spp = shear load carried by diagonal

tensicn field action = ks, where

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

Check of Fiange Stresses Due to Secondary Flange

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

design problems since the rivet loads are larger

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

3 = = |2⁄4 ' =S to Frame!

Field-_ #1 |) toad Out |%

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

for members normal to the neutral axis

(stiffeners or rings) The longer the un-

supported 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

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)

Pig C11.28a shows the end bay and the loads applied to it

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 "approxi- mation” that can be used for design purposes

If all margins are near zero, substantiating

element tests are in order

~

eee

Fig Cli 16 Graphs for Calculating

Buckling Stress of Webs

Fig C11.17 Buckling Stresses F,, er for Plates with Simply

Supported Edges E= 10,600 ksi (To left of dashed line, curves apply only to 24ST

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

Fig C11.20 Diagonal-tension analysis chart

Fig C11.21 Ratio of Maximum Stress to Average Stress in Web Stiffener

NOTE for use on curved webs:

For rings, read abscissa ast, for stringers, read abscissa a3,

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

Fig C11.25 Allowable Values of Nominal Web Shear Stress

(bo) Plasticity Correction for 24S-T3 Aluminum Alloy

Fig C11.26 Effective Shear Modulus of Diagonal-Tension Webs

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-

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

nich puts them in "hoop" compression For rings attached to the skin this loading is applied by the stringers and the skin, and

is thus "spread out" For floating riags this loading is applied only by the