Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 part 5 docx

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STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING

C4 28 IN TENSION, COMPRESSION, BENDING,

no restrictions on type of truss or

arrangement of members, however, the goal is the lightest truss Omit consideration of weight of any gusset plates at truss joints 1500 1400 1200 2B" te 28% oe 38" ole 28 —of ° Fig 4

Same as problem (6), but instead of a

cantilever truss use a simply supported truss with supports at points (A) and (B) Fig 5 shows a front beam and front lift

strut in an externally braced monoplane

The wing beam and lift strut are in the Same vertical plane The ultimate design

loads on the beam for the critical

conditions are w = 50 ib./in and w =

ibs per inch

down (a)

-30

Minus means load is acting

Design a streamline tube to act as

the lift strut Material {ts 2024-T5

aluminum alloy

Same as (a) but made from alloy steel Fry = 75000 Compare the weights of the two designs ae „" or — i Pa #/in ‘ PP Pp Ppp { Pin {b) Bmmmmnn; Fig 5

(9) Pig, 6 illustrates the strut and wire bracing for attaching float to fuselage of

a seaplane Determine the necessary sizes for the streamline struts AC and BD for the following load conditions

Condition 1 ¥V = -32000 lbs., H = - 8000 lbs Condition 2, V = ~ 8000 lbs.,

H = ~28000 lbs Material 2024-T3 aluminum alloy Tube size 2 - 065 round L = 44 in.,

C's 1.5 Material alloy steel Fry =

95000, welded at ends Design ultimate loads equal 22000 1b compression and 28000 lbs tension Find margin of safety

(11) Same as Problem 10 but heat treated to

Fey = 150000 after welding Use c™l (10) TORSION AND COMBINED LOADINGS, Thrust Ling _ Fig 6 (12) The ultimate design load if 20,000 lbs

compression L=30 in Use C=l Design the lightest round tube from the following materials and compare their weights

{a) Aluminum alloy 2024-T3

(b) Alloy steel Fty = 180,000 (c) Magnesium alloy Fey = 10,000 Same design load as in problem (10) but design the lightest streamline tube from 2024-T3 aluminum alloy material

A round tube is to carry an ultimate pure bending moment of 14000 in lbs Select the lightest tube size from the following materials and compare their weights

(a) Alloy steel Fry = 240,000, (b) 2024-73

aluminum alloy, (c) Magnesium alloy

Fey = 30000, (4) Titanium 6AL¬4V

alloy

A round tube 20 inches long is to carry

an ultimate torsional moment of 15000 in

Ib Select the lightest tube size from the following materials and compare their weights

(4) Alioy steel Fey = 180,000,

(b) Aluminum alloy 2024-T3, (c) Mag- nesium alloy Fry = 36000

Determine the lightest 2024-TS aluminun alloy round tube 10 inches long to carry a combined bending and torsional design load of 4500 and 3000 in.lbs respectively Same as Problem 16, but change material to alloy steel Fry = 95000

A 1-1/2 - 065 2024-T3 round tube 50 inches

long is used as a beam-column The distributed load on beam is 12 lb per inch and the axial load is 700 lbs What is the M.S under these loads

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CHAPTER C5

BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, BENDING AND UNDER COMBINED STRESS SYSTEMS

C5.1 Introduction

Chapter A1l8, Part 2, introduced the student to the theoretical approach to the problem of determining the buckling equation for flat sheet in compression with various edge or boundary conditions A similar theoretical approach has been made for other load systems, such as shear and bending, thus the buckling equations for flat sheet have been available for many years This chapter will summarize these equations and provide design charts for practical use in designing sheet and plate structures Most of the

material in this chapter is taken from (Ref 1),

NACA Technical Note 3781-Part I, "Buckling of

Plat Plates” by Gerard and Becker This report

is a comprehensive study and summary of practically all important theoretical and experimental work published before 1957 The report is especially useful to structural design engineers

C§.2 Equation for Elastic Buckling Strength of Flat Sheet in Compression

From Chapter Al8, the equation for the elastic instability of flat sheet in compression!

18,

nk, E

oy = ——— @* - (05.1)

12 (1-Uạ) Ô

Where kg = buckling coefficient which depends

on edge boundary conditions and

sheet aspect ratio (a/b)

E = modulus of elasticity

Vg = elastic Poisson’s ratio

bd = short dimension of plate or loaded

edge

t = sheet thickness

CS5.3 Buckling Coefficient ke

Fig C5.1 shows the change in buckled

shape as the boundary conditions are changed

on the unloaded edges from free to restrained In Fig (a) the sides are free, thus sheet acts as a column In Fig (bd) one side is restrained and the other side free, and such a restrained sheet is referred to as a flange

In Fig (c) both sides are restrained and this

restrained element ts referred to as a plate COLUMN BUCKLED FORM, Fig, (a) Fig {b) `

Fig C5,1 (Ref 1) Transition from column to piate a8 supports are added along unlaaded edges Note changes in

buckle configurations

Fig C5.2 gives curves for finding the puckling coefficient kK, for various boundary

or edge conditions and a/b ratio of the sheet

The letter C on edge means clamped or

fixed against rotation The letter F means a

free edge and SS means simply supported or hinged Fig C5.3 shows curves for k, for

various degrees of restraint (e) along the

sides of the sheet panel, where ¢ is the ratio of rotational rigidity of the plate edge stiffener to the rotational rigidity of the plate

Fig C5.4 shows curves for k, fora

flange that has one edge free and the other

with various degrees of edge restraint Fig CS.5 illustrates where the compressive stress varies linearly over the length of the sheet, a typical case being the sheet panels on the upper side of a cantilever wing under normal

flight condition

Fig CS.6 gives the ky factor for a long

sheet panel with two extremes of edge stiff-

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BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, C5 2 BENDING AND UNDER COMBINED STRESS SYSTEMS 6 “ 2 enmatas aL — sa | 4 ca F debe #[———- \ D7 | z|-—=—=+ \ , L TVE OF ALONE “6/47 m Ninh Le Fig C5.2 (Ref 1) Compressive-buckling coefficients for flat rectangular pilates sa 4 8ã» “a oN : ˆ ° z+ ————_- ` %

Fig CS.4 (Ref 1) Compressive-buckling-stress coefficient of flanges as a function of a/b for various amounts of edge rotational restraint LOADED EDGES C1 surety uy OQ ¢ & @ #6 2 MB 28 BB B 40 %

Fig C5.3 (Ret 1) Compressive-buckling-stress coefficient of plates as a function of a/b for varicus amounts of edge rotational restraint 7 ` : # # ! : A, AP Gs a5 + 4 422 We 2 beg eee $ 300 3 oe 2 2 200 “160 AST £ x 4 s œ a/b

Fig C5.5 (Ref 1} Average compressive-buckling-stress coefficient for rectangular flat plate of constant thickness

with linearly varying axiai load ke p17 E 2

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C5.3

which 1s a closed section and, therefore, a Substituting in Eq C5.1, relatively torsionally strong stiffener File

c5.6a gives the compression buckling Jan = mn? x 4.0% 10,700,000 (204? = 2480 pst coefficients ky for isosceles triangular cr 12 (1 - 3*) 5 x psi plates

This stress is below the proportional limit stress for the material, thus equation

I CS.1 applies and needs no plasticity correction i ca, CLAMPED C5.4 Equation for Inelastic Buckling Strength of Flat Sheet [ | (he 696) in Compression Ị | i |

TOASONACLY If the buckling or instability occurs at WEAN STRPENER a stress in the inelastic or plastic stress

range, then E and YV are not the same as for elastic buckling, thus a plasticity correction factor is required and equation C5.1 is

written,

NV ke BO

T Ser #S—————— ` ~~~~—~ (C5.2)

STIEFEWER 7 H T 12(1 - 1⁄2?)

Là | ra Where 7] is the plasticity reduction factor ,

2 200 250 p00 and equals dor plastic/og, elastic -

b/t *

Fig C5 8 (Ref 1} Compressive-buckling coefficient for long rectangular stiffened panels as a function of b/t and stiffener torsional rigidity CLaep TT + s

Fig, C5 6a (Ref 1) Uniform Compression Illustrative Problem Find the compressive

buckling stress for a sheet panel with (a) = 10

and b = 3 inches, thickness t = 04 and all edges are simply supported Material is 2024-T3 aluminum alloy Solution: b5 = 10,700,000 a⁄Ð = 10/5 "=2 corresponds to Case (c) in Fig C5.2 Ve 2 0.3, The boundary or edge condition Thus

using curve (c) for a/b = 2, we read ky = 4.0

The values of k, and Uy are always the

elastic values since the coefficient 7 contains

all changes in those terms resulting from

inelastic behavior

A tremendous amount of theoretical and “ experimental work has been done relative to

the value of the so-called plasticity cor-

rection factor Possibly the first values

used by design engineers were n = E,/E or <

N= Esec/E- Whatever the expression for ] it

must involve a measure of the stiffness of the

material in the inelastic stress range and

since the stress-strain relation In the plastic range is non-linear, a restrt must be made to

the stress-strain curve to obtain a plasticity correction factor This complication is

greatly simplified by using the Ramberg and Osgood equations for the stress-strain curve which involves 3 simple parameters (The

reader should refer to Chapter Bl for

information on the Ramberg-Osgood equations.)

Thus using the Ramber-Osgood parameters (Ref.1)

presents Figs C5.7 and CS.8 for finding the compressive buckling stress for flat sheet panels with various boundary conditions for both elastic and inelastic buckling or instability C5.5 Simple Problems to Mlustrate Use of Curves in Figs 5.7 and C5 8 Ser +4 là ~ The sketch shows a 3x9 Ess

inch sheet panel The sides ss 4 ga are simply supported The

material is aluminum alloy SA TTT

2024-73 The thickness is van

.094" = 10,700,000, H1 :

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Fig C$.7% Chart of Nondimensional Compressive Buckling Stress for Long

Hinged Flanges 7] = (Eg/E)(1 - Ve*)/(1 - VU) } | 2 ‡ ~ i! : 7 à 7 + + nivetr P.iB4.f | ; a ‘ | ; Ị i ; Tỉ Ƒ a | : + Sg TT wo oi i (22, $2 : I a i Ị s ⁄ † r + Ser, Lb | 7 2 | or I | i Foo /Fe.7 s " m——] i 2 ZS L Wa | ® d7 ‘ ! 4 | Ae Ƒ | : : PEL | : | | Ị : Ị | : j Ị nil i | Po poe ts | 2 @ + a @ + “ “ z8 “2ø at VEG, "

Fig C5,8 Chart of Nondimensional Conipressive Buckling Stress for Long Clamped Flanges and for Supported Plates with Edge Rotational Restraint

Ne (Be/2e{ +0.5 [1+ GBY/EQ)] `⁄2)ạ ~ Ve Va -V¥)

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Ve = 0.3 Pind the buckling stress oor

Solution: We use Fig C5.8 since it covers the boundary conditions.of our problem The parameter for bottom scale is, k,n”s ta < `“ (A) 12 (1-Uạ”) Øạ.„ For a/b = 9/3 = 3, we find kg from curve (c} of Fig C5.2 equals 4.0

The use of Fig C5.8 involves the use of o,,, and n the Ramberg-Osgood parameters Referring to Table Bl.1 of Chapter Bl, we find for 2024-T3 aluminum alloy that o,., = 39000

and the shape factor n = 11.5 — Tei phoosy

Substituting in (A):-

4,0 1° x 10,700,000 (2084) so)" #985 12 a - 3°) 39, 000 ~

From Fig C5.8 using 98 on bottom scale

and n = 11.5 curve, we read on left hand scale

that Gor/o,,, = 84.-

NN Then doy’ = 39000 x 84 = 32800 pst ~

If we neglected any plasticity effect, then! we would use equation C5.2 with ne =1,0, oy

n? x4 0x10,700, 000 cán" 12 - 3°)

Ser = = 58400 psi

Whereas the actual buckling stress was 32800,

or in this case the plasticity correction factor is 328/384 = 954

The sheet thickness used in this example

of 094 is relatively large If we change the

sheet thickness to 051 inches the results

would be practically no correction within the accuracy of reading the curves, and the buckling| Stress Ogr would calculate to be 11200 psi, which is below the proportional limit stress and thus no plasticity correction

C5.6 Cladding Reduction Factors

Aluminum alloy sheet is available with a thin covering of practically pure aluminum and

is widely used in aireraft structures Such

material is referred to as alclad or clad

aluminum alloy, The mechanical strength

properties of this clac material is considerably lower than the core material Since the clad is located at the extreme fibers of the

alclad sheet, it is located where the strains attain their highest value when buckling takes place Fig C5.9 shows make up of an aliclad Sheet and Fig C5.10 shows the stress-strain curves for cladding, core and alclad combina-

tions

C5.5

Thus a further correction must be made for

alclad sheets because of the lower strength

clad covering material Thus the buckling stress for alclad sheets can be written:

Fer = Nog

Cladding

Fig C5.9

Reference 1 gives simplified cladding reduction factors as summarized in Table (5.1

Thus the buckling stress for alclad sheets is determined for the primary strength properties as normally listed for such materials as illustrated in the two previous example problems The resulting oer is then reduced by use of equation C5.3, using values of 7 trom Table C5.1 Come Tnị MLD [arcane g |f€ g Fel on E

Fig, C5 10 (Ref, 1) Stress-strain Curves for Cladding, Core,

and Alclad Combinations Ø/Œ„are = Ì - Ý +01; 0 *ƠglØcore:

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adge rotational restraint

BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, BENDING AND UNDER COMBINED STRESS SYSTEMS

C5.6

BUCKLING UNDER SHEAR LOADS

C5.7 Buckling of Flat Rectangular Plates Under Shear Loads

The critical elastic shear buckling stress for flat plates with various boundary

conditions is given by the following equation: a „ r= T Ky E cr ad œ@ 12 (1 ¬ Uạ*)

Where (b) is always the shorter dimension of

the plate as all edges carry shear k, is the

shear buckling coefficient and is plotted as a

function of the plate aspect ratio a/b in Fig

CS.11 for simply supported edges and clamped

edges

If buckling occurs at a stress above the proportional limit stress, a plasticity correction must be included and equation C5.4

becomes

Test results compare favorably with the

results of equation (5.5 if 7], * Gg/G where G

18 the shear modulus and Gg the shear secant

modulus as obtained from a shear stress-strain

diagram for the material

A long rectangular plate subjected to ' pure shear produces internal compressive

stresses on planes at 45 degrees with the plate edges and thus these compressive stresses

cause the long panel to buckle in patterns at

an angle to the plate edges as illustrated in Fig C5.12, and the buckle patterns have a half

wave length of 1.25b

fae 1 25b —o

Fig C5 12 (Ref 7)

Fig C5.13 is a chart of non-dimensional shear buckling stress for panels with various

This chart fs

Similar to the chart in Figs C5.7 and C5.8 in

that the values go, „ and n must be known for the material before the chart can be used to

find the shear buckling stress

BUCKLING UNDER BENDING LOADS

C5.8 Buckling of Flat Plates Under Bending Loads

The equation for bending instability of flat plates in bending is the same as for

compression and shear except the buckling coefficient kp is different from k, or kg When

@ plate in bending buckles, it involves relatively short wave length buckles equal to 2/3 0 for long plates with simply supported edges (see Fig C5.14) Thus the smaller buckle patterns cause the buckling coeffictent Kp to be larger than Kg or Kg

Fig C5 14 (Ref 7) Bending Buckie Patterns For bending elastic buckling the equation is, n? 5 or = GE eh" ~ eee ee (c5.8) 12 (1 = Ug*) For bending inelastic buckling, Tị, TỶ 8 Scr = _> gb" zt (cS.7) 12 (1 - Ve")

Where Kp is the buckling coefficient and is obtained from Fig C5.15 for various a/b

ratios and edge restraint « against rotation

In the a/b ratio the lodded edge is (b),

The plasticity reduction factor can be obtained from Fig C5.3 using simply supported

edges

BUCKLING OF FLAT SHEETS UNDER COMBINED LOADS

The practical design case involving the

use of thin sheets usually involves a combined

load system, thus the calculation of the buckling strength of flat sheets under combined stress systems 1s necessary The

approach used involves the use of inter-action equations or curves (see Chapter Cl, Art

C1.15 for explanation of inter-action equations)

C5.9 Combined Bending and Longitudinal Compression

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C5.7 7 : T 2 Ị Ị 2 i Ì 3 Ị | Ÿ 6 „ + — 20, ' 2056 5 ’ T | 2 LA i 1 Ter/ 9.7 A or Fsqp/Fo.7 3 = WY ¬— AN à Àt ầ a) eb SN ; Z— £ S a FAY + — f Li | 2 B—— + ~ , i 2 f 2 3 4 OS ø a a 3 10 KT RE tạ — 120-Vg M.„ 8U Tà

Fig CS, 13 (Ref 1) Chart of Nondimensional Shear Buckling Stress for Panels With

Edge Rotational Restraint 7) = (Eg/E) (1 - Ve*/(1 - V4

ae Ver he Meas B.#vo KS “ ` À z a SYMMETRIC \ | MooE LAMPED EDGES 2 2 h, - 2 9 ⁄ 0 ANTISYMMETRIC SYMMETRIC 3 MODE MODE ; - é EDGES gol) õ 7 z 3 4 5 TF 2 ab %

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BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, C5.8 tudinal compression is, Rp*???đ+ Re 81/0 -ô + -

This equation was originally presented in Ref 2 and the interaction curve from plotting this equation is found in many of the

structures manuals of aerospace companies

Fig C5.15 is a plot of eq C5.8 It also shows curves for various margin of safety values Lt Fig C5.15 Combined Bending & Long Compression Rpts Real 1.0 0 1 3 3 4 5 6 ,7 8 9 1.0 L1

5.10 Combined Bending & Shear

The interaction equation for this com-

bined leading (Ref, 1 & 2) 1s,

(cS.10)

Fig C5.16 is a plet of equation C5.9 Curves showing various M.S values are also shown Rg is the stress ratio due to

torsional shear stress and Rgt is the stress ratio for transverse or flexural shear stress

CS.11 Combined Shear and Longitudinal Direct Stress (Tension or Compression } The interaction equation is (Ref 3,4) RL + RS “MS = lO «+-+ - (CS.11) + VRE aR) Rp 1,1 1.0 BENDING AND UNDER COMBINED STRESS SYSTEMS Fig C5 16 Combined Bending & Shear Rụ + Rạ=l .,1l 2 3 4 5 6 ,7 ,8 9 1,0 1,1 Rg, (Rg + Ret), Ret

Fig.-C5.17 ts a plot of equation CS.11, If the direct stress is tension, it is

included on the figure as negative compression using the compression allowable

CS.12 Combined Compression, Bending & Shear

From Ref 5, the conditions for buckling

are represented by the interaction curves of Fig C5.18 This figure tells whether the sheet will buckle or not but will not give the margin of safety

Rp:- if the value of the Re curve defined by Given the ratios Re, Rg and

the given value of Rp and Rg is greater

mumerically than the given value of Re, then the panel will buckle

1,0

Rp 9

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BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, BENDING AND UNDER COMBINED STRESS SYSTEMS C5 10

The margin of safety of elastically

buckled flat panels may be determined from Pig C5.19 The dashed lines indicate a typical application where Ry = 161, Rg = 23,

and Rp = 38 Point 1 is first determined for

the specific value of Rg and Rp The dashed diagonal line from the origin O through point

1, intersecting the related Re/Rg curve at

point 2, yields the allowable shear and bending stresses for the desired margin of safety

calculations (Note when R, is less than Rg

use the right half of the figure; in other cases use the left nalf)

C§.13 Mlustrative Problems

In general a structural component com-

posed of stiffened sheet panels will not fail when buckling of the sheet panels occurs since the stiffening units can usually continue to carry more loading before they fail However, there are many design situations which require

that initial buckling of sheet panels satisfy

certain design specifications For example, the top skin on a low wing passenger airplane should not buckle under accelerations due to

air gusts which occur in normal every day

flying thus preventing passengers from

observing wing skin buckling in normal flying

conditions Another example would be that no

buckling of fuselage skin panels should occur

while airplane is on ground with full load aboard in order to prevent public from

observing buckling of fuselage skin In many airplanes, fuel tanks are built integral with the wing or fuselage, thus to eliminate the chances of leakage developing, it is best to design that no buckling of sheet panels that

bound the fuel tanks occur in flying and

landing conditions In some cases aerodynamic

or rigidity requirements may dictate no buckling of sheet panels To insure that buckling will not occur under certain load

requirements, it 1s good practice to be conservative in selecting or calculating the boundary restraints of the sheet panels, Problem 1

Pig CS.20 shows a portion of a cantilever wing composed of sheet, stiffeners and ribs The problem is to determine whether skin panels

marked (A), (B) and (C) will buckle under the

various given load cases The sheet material

is aluminum alloy 20213

Load Case 1,

P, = 700 lb., P, = 0, P, 20

With only loads P, acting, the one cell stiffened cantilever beam is subjected to a

compressive axial load of 2 x 700 = 1400 lb

Since the P, loads are not acting through the

centroid of the cross-section, a bending moment

is produced about the x-x axis equal to 1400 x

3.7 = 5170 in lb = My, where 3.7 15 distance from load P, to x=x axis Area of Zee Pr Stringer = 18 Area of Corner Member = 0.25 sq in Fig C5 20

The sheet thicknesses, stiffener areas and

all necessary dimensions are shown on Fig

c5.20 The total cross-sectional area of beam

section including all skin and stringers is

3.73 sq in The moment of inertia about x-x

centroidal axis calculates to be 49.30 in.*

Since the beam section is symmetrical, the top panels A, B and C are subjected to the same stress under the P, load system

Compressive stress due to transferring

loads P, to centroid of beam cross-section is,

t, = 2P,/area = 1400/3.73 = 375 psi

Compressive stress due to constant bending moment of 5170 in lbs is,

f, = M,2/T,, = 5170 x 4,.233/49.30 = 444 psi Total T1 3 375 + 444 = 819 pst

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C5.11

and the rib flanges will be con- The Re + Rg = 431 + 3097 = 826 Since servatively assumed as simply Pris? the result is less than 1.0, no buckling

Supported (Fo, is same as dor) 4 SSF occurs

ssc

dss Canise 2

mk & SS fa=15 The M.S = -1

= & * {Ra + An^

“>2 TU Ge -

For "T217 tạ ss Re + V Rộ + 4Rã

(See Hq 05.1) môn pb a = rẽ 2 1= ,69

a/b of skin panel = 15/5 23 :

` Fig , ——_— : Load Case 3

From Fig C5.2 far Case Œ, g (a) `

we read kK = 4.0 P, = 700, Pa = 500, Py = 100 lb

Po, 202 fer 4.0 1057005000 (088)* ¡on nọ; The two loads P, produce bending and

12 (1 - 0.3") 5 P - fleXural shear on the beam,

Since Foor the buckling stress is less than the applied stress fg, the panels will

not buckle

M.S = (Fo, /f¢) -1 = (190/819) -1 = 1.32

Load Case 2

P, = 700 1b., Pa = 500, P, = 0-

The two loads P, acting in opposite directions produce a couple or a torsional moment of 500 x 16.5 = 8250 in lb on the beam structure, which means we have added 4 pure shear stress system to the compressive stress system of Case 1 loading

The shear stress in the top panels A, B and € 1s, fg = T/2at = 8250/2 x 138 x 035 = 854 pst — (Where A {s the cell inclosed area) ; The shear buckling stress is mk Eg Poor s &* - = (See Hq 65.4) 12 (1 - Uạ) T

a/p = 18/5 = 3 Prom Fig C5.11, for

hinged or simply supported edges, we read

Kg = 5.8

TẾ x 6.8 x 10,700,000

ne “Ser” 2 (1 - 3? (SE) = 2760 pst 2035 ,3_

The sheet panels are now loaded in

combined compression and shear so the inter-

action equation must be used From Art C5.12 the interaction equation is Ry + Rễ =1

Rẹ =ứ, đ 3 3 é ‡ b 819/1900 = 431

Rg = s ay œ 3 3 a 854/2760 = 309

The bending moment

produces a different end compressive stress on

the three sheet panels since the bending moment is not constant over the panel moment To simplify we will take average bending moment on the panel Mav) = 200 x §2.5 = 10500 in 1b f, dus to this bending = M,Z/Iy = 10500 x 4.255/49.3 = 903 psi Total f, = 903 + 819 = 1722 pai Ro = fo/Foa, = 1722/1900 = 906

The two loads P, produce a traverse shear

load V = 200 lb The flexural shear stress must be added to the torsional shear stress ag

found in Case 2 loading

Due to symmetry of beam section and Py, loading the shear flow q at midpoint of sheet panel (B) is zero We will thus start at this

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BUCKLING STRENGTH OF FLAT SHEET IN COMPRESSION, SHEAR, BENDING AND UNDER COMBINED STRESS SYSTEMS cS 12

The shear flow q on panel (A) varies from 4.20 to 7.20 or the average q = (4.2+7.2)/2 = 5.7 Thus the average shear stress is

5.7/,035 = 163 psi It is in the same

direction as the torsional shear flow and thus is additive

Total fg = 163 + 854 = 917 psi

Re = fs/fs„„ = 917/2760 = 332 Ro + RG = 1, Sudt.:- 906 + 332" = 1.016,

since the result is greater than 1.0, initial buckling has started The margin of safety 1S slightly negative and equals,

2

.905 + V 905” + 4x 332*

In this example problem, the panels were assumed simply supported, which is conservative

Reference to Fig C5.6 shows that k, could be

assumed higher as the panel is riveted to a

zee shaped stringer which has some torsional

resistance and thus panel is not free to

rotate at its boundaries

M.S = ~ l1 =~ O1

Panel (C) is less critical because the

flexural shear is acting opposite to the

torsional shear stress, thus fg total = 845 -

163 = 682 psi The Rg = 682/2760 = 246

Re + Rg = 905 + 246" = 966 Since the

result 1s less than 1.0, panel will not buckle Panel (B) carries a small shear flow, being Zero at center of panel and increasing

uniformly to 1.5 lb per inch at the edges, and flowing in opposite directions from the centerline Thus transverse shear will have

negligible effect Thus Rg = 9854/2760 = 309

Re + Rg = 905 + 3097 = 1.00, or panel (B) is on the verge of buckling under the

assumptions made in the solution

PROBLEMS

(1) A sheet panel is 3" x 9" x 051" in size

The 3" side 1s simply supported and the 9" side is free Determine the buckling load if the compressive load is applied

normal to the 3” sides Do so for 3

different materials, (1) aluminum alloy

7075-76, (2) magnesium HKS1A, (3) Titanium

T1-81

(2) In Problem (1) if all edges were simply

supported, what would be the buckling

load

(3) In Problem (1) if the 9" sides were

clamped or fixed and the

supported, what would be 3° sides simply

the buckling load

(4)

(8)

A sheet panel 5" x 12.5" x 051 has all edges simply supported The panel is subjected to combined compression and Shear loads which produce the following stressesi-

fo = 2400 psi, applied normal to 5" side fg = 2800 psi Will the sheet buckle under the given load system if made of

aluminum alloy 2024-T3 material What

is the margin of safety

Tf the material in problem (4) is changed to alloy steel Fry = 95000 psi, what would be the margin of safety If sheet was heat treated to Fry = 180,000, what would be the M.S

A 3" x 12" x 040" sheet panel {5 subjected to the following combined stresses

fe = 3000, fp = 10000, fg = 8000 The fy

and fp stresses are normal to the 3° side

If sides are simply supported, will panel buckle if made of 7075-T6 aluminum alloy What is M.S What will be the M.S if material is changed to Titanium T1-@Mn References:~ (19) (11)

NACA Tech Note 3781, 1957 ANC-5 Amendment 2 Aug 1946, NACA ARR.No L6A05

NACA ARR.No SKIS

ANC-5 Revision of lode

General References on Theory NACA Tech Note 3781

"Introduction to Structural Stability Theory" By Gerard Book published by McGraw-Hill

"Theory of Elastic Stability", dy Timoshenko, McGraw-Hill Co

"A Unified Theory Of Plastic Buckling

of Columns and Plates", NACA Report

898 By Stowell

"Plastic Buckling of Simply Supported Compressed Plates”, NACA T.N 1817

By Pride & Heimerl

Trang 14

CHAPTER Có

LOCAL BUCKLING STRESS FOR COMPOSITE SHAPES

C6.1 Introduction

Thin flat sheet is inefficient for carrying

compressive loads because the buckling stresses are relatively low However, this weakness or

fault can be greatly improved by forming the flat sheet into composite shapes such as angles, channels, zees, etc Most of the many composite shapes can also be made by the

extruding process Formed or extruded members

are widely used in Flight Vehicle Structures,

thus methods of calculating the compressive

strength of such members is necessary C8.2 Compressive Buckling Stress for Equal

Flanged Elements,

The simplest equal flanged member that can be formed ts the angle shape Other shapes with equal flanges are the T section and the cruciform section as shown in Pig Cé6.1

ce T-Section

These sections can be considered as a group of long flanges, as illustrated, for the angle Section in Fig C6.2 Since the flanges which make up the section are equal in size, each

flange will buckle at the same stress There- fore each flange cannot restrain the other and thus it can be assumed that each flange is simply supported along the flange junction as illustrated in Fig C6.2 Cruciform Section Fig C6.1 Đẹp Lợi pig SŠ E 4 ss g 34 ° É 5 ° 2 Ee ® Ề we = S988 Ễ fe b Eas i ra 4 ss fe 4d ss TF †TTTT mòn mô Fig C8.2

From Equation Œ5.1 of Chapter C5, the

buckling compressive stress for a long flange is, From Fig C5.2 of Chapter C5, ky = 43, then = 0:43 ne tye “T6 o.s (gì = 0-388 se ——= 06.1 Ser

If the buckling stresses are above the

proportional limit stress, use Pig C5.7 in ~“, Chapter C5 to take care of the plasticity effect /

For formed angles, the flange width b

extends to centerline of adjacent leg, but for

extruded angles, the width b extends to inside edge of the adjacent flange or leg

C8.3 Compressive Buckling Stress for Simple Flange-Web Elements

The most common flange-web structural shapes are channels, zees, and hat sections A

flange has one unloaded edge free, whereas a

web has no free unloaded edge and thus has an unknown restraint on the boundary between the web and the flange Fig C6.3 shows the break down of 4 Z section into two flange and one web Plate elements Speed Story Ey 8 ss 8 ss a 5 a g 2 Flange} o Web g |Flanga 2 5 3 fa 8 = fa ss ` §§ A ss FFF TF

Breakdown of Z into Flange and Web Elements

Fig C6.3

The buckling strength of the web and flange elements depends on the boundary restraint

between the two elements If this restraint which is unknown could be found in terms of a

known rotational restraint « as presented in Chapter C5, the buckling coefficients could be found from charts tn Chapter CS Having the C6.1

Trang 15

C86 2 LOCAL BUCKLING STRESS buckling stress for each element, the critical buckling stress will be the smaller of the two The buckling load based on the buckling stress is not the failing load as more load can be taken by the material in the corner regions before local failure or crippling takes place The subject of local crippling of formed and extruded shapes is covered in Chapter C7

Using the moment distribution method or a

step by step analysis procedure, several research studies have determined the restraint

factors between web and flange elements for

simple shapes like channels, Z, H, square tubes

and formulated design charts for such shapes (Refs 1 to S inclusive.)

C6.4 Design Charts for Local Buckling Stresses of Some Composite Web- Flange Shapes,

Figs C6.4 to C6.7 inclusive give charts for determining the local buckling stress of channel, Z, H, square tube and hat shaped

sections For formed sections, the width b

extends to centerline of adjacent element and

tor extruded sections the width b extends to

inside edge of adjacent element C8.5 Problems illustrating Use of Charts PROBLEM 1

~~ L-TR† 084

The Z section in Fig (a) FT

1s formed from alumimm alloy Flange

2024-13 sheet What compres— 1.5 Sive stress will start local

buckling of an element of the member Web Ì Fig (a) Solution:- Dy bự bự/Đy tw/te From Fig C6.4, we read Kw = 2.9 - neg a Sor * BS Tar ee Ser 2.9 n* x 10,700,000 ( +064 )* or 12(1-.37) Ss ‘Tage 1.5 ~ 064 = 1.436 0.75 - 032 = 0.718 = 0.718/1.436 = 50 064/.064 = 1.0 = 56100 psi |

This stress is above the proportional

limit stress of the material, thus a plasticity

correction must be made The buckling occurs

on the flange

From Table Bl.1 of Chapter Bl, we obtain for 2024-T3 aluminum alloy:- G,,, = 39000 and

ñ = 11.5

FOR COMPOSITE SHAPES

For the plasticity correction of shapes covered in Fig C6.4 (Ref 4), the plasticity correction for a flange free on one edze can be used with accuracy Thus we can use chart in Fig CS.7 of Chapter CS.to ccrrect for plasticity effects

The parameter for bottom scale of Fig

cS.7 ts equation (A) divided by ao.,, or 56100/39000 = 1.44 Using this value and the

n= 11.5 curve, we read from Fig C5.7 that Ger/Fo,7 = 1.02

Therefore the local buckling stress is Scr = 39000 x 1.02 = 39800 psi

PROBLEM 2

If the member tn Problem 1 is subjected

to a 300°F temperature for 2 hours duration,

what would be the local buckling stress From Table Bl.1 for this temperature condition, Se.r = 35700, Eo = 10,300,000, n= 15 kw n2 E (#m)* „ 2:8 n* x 10,300,000 1201~1e”](Øa.„) ‘Dw 1201-.3”) (35700) „Q64 432 _ (ng) z 1,51,

Using this value on bottom scale of Fig

C5.7 and n curve = 15, we read Øor/2as.z = 1.03,

Thus Ger = 1.03 x 35700 = 36800 psi PROBLEM 3

Same as Problem 1, but change material to

Titanium Ti~6Mm Sheet

From Table Bl.1 we obtain for this material:- Ey = 15,500,000, ơ„,„= 119500, n = 13.7,

2.9 n* x 15,500,000 , ,064 +

= Shee a *

Ger 12(1-.37) (Tage? = 01200 pst

This stress is near the proportional limit

stress so plasticity correction should be small

if any

Øẹy/G „,„ = 81200/119500 +68

From Fig C5.7, using n = 13.7 curve, we

read Ocr/d,,, = 68 Then dep = 119500 x 0.68 =

81300, thus no plasticity correction PROBLEM 4 The rectangular tube has the dimensions as shown in Fig (b) alloy 2014-T6 buckling stress

It is extruded from aluminun

Determine the local compressive

Trang 16

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES ⁄ C8.3

7 WES BUCKLES FIRST ⁄ | | |

|_ l \⁄ ?Ƒ—*» suoqzs zwsr

8 + : (Nước BUALES Finest | L : AK

Trang 17

c8.4 LOCAL BUCKLING STRESS FOR COMPOSITE SHAPES

Solution:- 3 TT In the design of rectangular tubes, the

+04 b L0 designer should select the tube thicknesses for b =# 1¬ 08 = 92 04 le both long and short sides so that buckling h b/h =2 Oe = 92/1.92 = 479 Tose 08 = 1.92 2 i = section for buckling strength occurs on both sides, thus giving the lightest

- AS pointed out previously in this chapter,

tp/En = 1.0 Fig ©) the load on the member which causes local

= buckling is not the failing or maximum load for From Fig C1.6, we read ky = 5.2 a a short length of the member This local fafling

or crippling stress is treated in the next

Oop = 222 Tee gor = 21900 psi | chapter Since the buckling stress may fall in

AS shown in Fig Cl.6, buckling occurs on the h Side of the tube The computed buckling stress 1s below the proportional limit stress, thus no plasticity correction

PROBLEM 5

Same as Problem 4 but change the thickness of the h side to 072, but leave the b side -04 in thickness Solution:- b = h = b⁄h tp⁄tn From Fig C6.6, Ky = 4.3 1 -~ 144 = 856 2 ~ 08 = 1.92 -856/1.92 = 446 04/.072 = 555 4.3 m* x 10,700,000 ,.072,a Sor = 120-.2) +)

This stress is above the proportional limit stress thus a plasticity correction is necessary

(Ref 4) gives no value for a plasticity correction but recommends the correction for a clamped long flange which ts a Slightly conservative correction This plasticity

correction should also be used for the hat shape as shown in Fig C6.7,

= $8600 psi

Thus Flg C5.8 of Chapter C5 can be used to correct for effect of plasticity

From Table Bl.1, for our material Øy„^ 53000 and n = 18.5

The value of the parameter for bottom scale of Fig CS.8 1s agr/o,,, = 58600/53000 =

1.10

From Fig C5.8, using n = 18.5 curve, we

read Ocr/o,,, = 91 Therefore oor = 53000 x

+91 = 48100

Thus by changing the long side of tube from 04 to +072, the buckling stress was tn- creased from 21900 to 48100 psi

the inelastic stress range, the buckles will not entirely disappear when load 1s removed Since limit or applied loads must be carried without permanent distortion, it 1s thus important to know when local buckling starts For those missile and space vehicles that carry no humans, the factor of safety on limit loads is

considerably less than for aircraft, thus the

spread between local buckling and local falling strength becomes important in design

C8.5 Buckling of Stiffened Flat Sheets Under Longitudinal Compression

In supersonic aircraft, it is important that the surface skin, particularly that on the wing, not buckle under flight conditions since

a buckled surface could effect the aerodynamic characteristics of the airflow around the wing,

thus it is important to know when the skin or its stiffening units inittally buckle in order to design so that suc Duckling will not occur under flight conditions

Gallaher and Boughan (Ref 6) and Boughan and Baab (Ref 7) determined the local buckling

coefficients for idealized web, Z and T stiffened plates The results of their studies are shown in Figures C6.8 to C6.12 and were taken fron

(Ref 4) The initial local buckling stress

for plate or stiffener is given by the equation:-

kg 1” 5B jtg,?

%r "TT 1) tố TT n nh re If the buckling stress is above the pro- \ portional limit stress of the material, correct '

for plasticity effect by using Fig C5.9 of Ỷ

Chapter C5, ⁄

Problem Illustrating Use of Charts ems trating Vso of Charts

Fig C shows a plate with idealized Z

Section stiffeners The material 1s 2024-73

Trang 18

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES BUCKLING OF SHIM AESTAAIMED AY STUFFENER : i REMI OF STEFENER | RESTHAMED BF SKM Ae RRR ROW + ` i § Oe Ss Wy Nao `» ở + + # a Of, Web stiffeners 0.5 <tw/tg <2

Fig C6 8 (Ref 7) Compressive-local-buckling coefficients for infinitely wide idealized stiffened flat plates, keE /ts\7 got (= oF 18 (Le Ve) \b = wo Œ 9 pods 2 Fig C6.11 (Ref 7) T-section stiffeners bị bự ty/ty = 1,0; > 10; — > 0,25, tự bg 3 z— L— aBuLu A 2 + * ra , sỡ 1 *% Fig C6.9 (Ref 6) Z-section stiffeners ty/tg = 0.50 and 0,79 & | % đi : % 4 ot | * + LỚN, † las 4, , \ z + | N 4 Ẳ ‘ 2 2— i 4 | INN 4 | 3 ty Ệ , * | I + se a ao a ^ a, Fig C6 10 (Ref 6) Z-section stiffeners tw/ty = 0.63 and 1.0, @ + 4 6 $4,

Fig C6 12 (Ref 7) T-section stiffeners,

tey/tt = 0.7; be/tt > 10; by/bg > 0 25,

C8.5

Trang 19

C6.6 LOCAL BUCKLING STRESS aluminum alloy Determine the initial buckling

stress under longitudinal compression bye i:5 2 375,

br 0.5 |

By Te 7 OSS, b= ZG

Using the above three values and referring to Fig C&.9, we read the buckling coerricient Kg to be 4.2 Substituting tn equation (B) = 4:2 n* x 10,700,000 ,.125,2 _ Ser = “TT a1 Lá = 39800 psi

This stress is no doubt above the pro-

portional limit stress so a check for plasticity

effect will be made For this effect we use Fig C5.8 of Chapter C5

For our 2024-T3 material, we find from Table 81.1 of Chapter Bl, that Ser/Go.7 = 39000 and the shape parameter n = 11.5

The bottom scale parameter on Fig C5.8 is equation (B) divided by Ø,,;„ thus it equals 39809/29000 = 1,02, Using this value and the n = 11.5 curve on Fig (5.8, we read on lert

Side scale that oor/a,,, = 86 Therefore the

duckling stress agp = 86 x 39000 = 33600 psi FOR COMPOSITE SHAPES (7) References

Lundquist, Stowell and Schuette: Principles of Moment Distribution Applied to Stability

of Structures Composed of Bars and Plates

NACA WRL-326, 1943

Kroll, Fisher and Heimerl: Charts for

Calculation of the Critical Stress for Local

Instability of Columns with I, Z, channel and Rectangular Tube Sections NACA WRL-429, 1945,

Kroll: Tables of Stiffness

Factor for Flat Rectangular Plates Under Compression: NACA WRL-398, 1943,

Becker: Handbook of Structural Stability Part If Buckling of Composite Elements NACA TN.3782, July 1957

Van Der Maas: Charts for the Calculation of the Critical Compressive Stress for Local Instability of Columns with Hat Sections Jour Aero Sci Vol 21, June 1954

Gallaher and Boughan: A Method for Cal- culating the Compressive Strength of Z Stiffened Panels that Develop Local In~ stability NACA TN.1482, 1947,

Boughan and Baab: Charts for Calculating

and Carry-over

Trang 20

CHAPTER C7

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION

SHEET EFFECTIVE WIDTHS

CT, 1 Introduction

Chapter C6 was concerned with the local buckling stress of composite sections when

loaded in compression Tests of short lengths

of sections composed of flange-plate elements

often show that after the section has buckled

locally, the unit still has the ability to carry a greater load before failure occurs In other words, the local buckling and local

failure loads are not the same For cases

where local buckling occurs at low stress, the crippling or failing stress will be higher When local buckling occurs at high stress such as 7 to 8 Foy, puckling and crippling stress are practically the same Fig C7.1 illustrates

the stress distribution on the cross-section

after local buckling has occurred but prior to local crippling or failure

Fig C7.1

AS the load on the section is increased,

the buckles on the flat vortions get larger but

most of the increasing load is transferred to the much stiffer comer regions until the stress

intensity reaches a high enough value to cause sufficient deformation to cause failure

A theoretical solution for the local

crippling stress for all types of shapes nas

not been developed as the boundary restraint between flange and plate elements is unknown

and also the manner in which the stress builds

up in the corner regions is not well understood

Consequently, the methods of solution are semi~ empirical in character, and the results of such

methods have been sufficiently proven by tests two methods of calculating crippling stresses will be presented in this chapter

C7.2 METHOD 1 THE ANGLE METHOD, or the Needham Method

This method which will be referred to as

the angle method or the Needham method was

COLUMN STRENGTH presented in (Ref 1) In this method the member section is divided into equal or unequal angles as illustrated in Fig C7.2 The

strength of these angle elements can be established by theory or tests The ultimate strength or failing strength can then be found by adding up the strengths of the angle elements that make up the composite section

Needham made a large number of tests on

angle and channel sections From a study of

these test resultsaas well as other published

test data on channels, Square and rectangular

tubes, etc., he arrived at the following equation for the crippling or failing stress of angle sections

Angle Angie 4 such _ fob ay unit | unit \_f units

a + i a)

s Fig, b igo

Basic Angle Basie Angle Basic’ Angle Unit Two Unit One Unit No

edges free edge free edge free

Fig C72

¬- (87.1)

crippling stress (psi)

compression yield stress (pst) Young’s modulus of elasticity in

compression (psi)

equivalent b/t of section = (a + b)/2Y

coefficient that depends on the

degree of edge support along the edges of contiguous angle units Specifically they are:-

Cg = 0.316 (two edges free)

Ce = 0.342 (one edge free)

Ca = 0.366 (no edge free)

ơ ~ or a

The crippling stress for angles, channels,

zees and rectangular tubes can be determined

directly from use of equation C7.1 The crippling load on an angle unit is then,

Pog = FosA

where A ts the area of the angle -

The crippling stress of other formed CT.1

Trang 21

CT2

structural shapes can be determined by dividing

the shape into a series of angle units and

computing the crippling loads for these

individual angle units by use of equations C7.1

and C7.2 The weighted crippling stress for

the entire section 1s obtained from the

following equation:-

Fre = 2 (crippling loads of ani les) _ (07.3)

cs 2 (area of angles

C?.3 Design Curves

Fig C7.3 gives curves for determining the crippling stress of angle units as per equation C7.1, and Fig C7.4 gives curves for determining the crippling loads for angle units Using these curves and equation C7.3, the erippling

stress of composite shapes other than angles,

channels, zees and rectangular tubes can readily

ve calculated

Illustrative problems, using this method,

will be given later and the results compared

with method 2 Crippling stresses for

composite sections should be limited to the

values given in Table C7.1 unless substantiated by test results

C†?.4 METHOD 2 For Crippling Stress Calculation

(The Gerard Method)

Introduction: References 2 and 3 gives the results of a very comprehensive study by Gerard on the subject of erippling stresses From a thorough study of published theoretical studies and most available test or experimental results, Gerard has developed and presented a more generalized or broader semi-empirical method of determining erippling stresses one sense it is generalization or broader application of the Needham method which was

presented as method 1 The student and

practicing structures engineer should refer to the above references for a complete discussion of how the resulting crippling stress equations

were obtained and how these check the extensive

test results In this short chapter we can only present the resulting equations, design curves for same and example problems in the use of the information, in the determination of crippling stresses

In

C7.5 Stresses and Displacements of Flat Plates After Buckling Under Conditions of Uniform

End Shortening

Fig C7.5 shows a picture of the resulting

stress distribution on flat plates after

buckling under conditions of uniform end Shortening as determined by Coan tn (Ref 4)

The Gerard method recognizes the effect of

distortion of the free unloaded edges upon the failing strength of the member section CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION ®)

Fig C7.5 Stresses and displacements of flat plates after buckling under conditions of uniform end shortening (reference 4), (a) Straight unloaded edges, (b) stress free unloaded edges free to warp in the plane of the plate

CT.6 The Gerard Equations for Crippling Stress The following equations are taken from

(Ref 3)

For sections with distorted unloaded edges as angles, tubes, V groove plates, multi-corner Sections and stiffened panels, the following crippling stress equation applies within Ÿ 10 percent limits:-

Fos/Foy = 0,86)

[(st⁄A)(/£ey)"⁄+]

For sections with straight unloaded edges such as plates, tee, cruciform and H sections,

the following equation for crippling stress

applies within = 5 percent limits

o.as

Fos/Foy = 0.87

7219199

[tet 7A) (B/Pey *| ~~ (C7.5)

For 2 corner sections, Z, J, and channel sections, the following equation applies within

10 percent limits

Fog/Foy = 8.2

fee YAM(B/FeyP/*]""" ~~ (07.8)

crippling stress for section (psi)

compressive yield stress (psi)

element thickness (inches}

section area (in.*)

= Young’s modulus of elasticity = number of flanges which compose the

composite section, plus the number of cuts necessary to divide the

Section into a series of flanges See Fig C7.6 for method of cutting compo~

Site sections to determine value of ¢

The cut-off or maximum crippling stress Pag for a composite section should be limited to the

Trang 22

0.09 EDGES FREE DONE EDGE FREE 008 NO EDGE FREE 0.07 006 NDISad ANV SISATWNY 0.05 004 003 0.02 001 40 Barb t et (Note: Fe, same as Fog) SHUNALINULS TISIHFA LHĐI141 AO

Fig C7.3 Dimensionless Crippling Stress vs b'/t (Ref 1)

(Note: Pc, same ad Pog) =

Fig C7.4 Dimensionless Crippling Load va by/t (Ref 1)

Trang 23

CT.4 angie Unloaded Edges Plate Tube 1 £——- Ì cuts-| - 1 1 Basic Section 1 Cut 4 Cuts gz2 2 Flanges 8 Flanges 32¢ 12=¢ Straight Unloaded Edges T-Section Cruciform H-Section + 4 — Cut Basic Section 0 Cuts 1 Cut g=3 4 Flanges 6 Flanges 4ag Tag

Fig C7.6 Method of cutting simple elements to determine g

values in Table C7.1 unless higher values can

be sidstantiated by test results The cut-off

values given in Table C7.1 are no doubt slightly conservative Design curves for equation C7.4, 5 and 6 are given in Figs C7.7, C7.8 and C7.9,

1 18 ,2 25,30 ,4

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION C7.7 Correction for Cladding

Since many formed sections are made from alclad sheets, the clad covering acts to reduce the value of buckling stress and thus 2

correction factor q must be used to taks care

of this reduction in strength This correction from (Ref 3) is,

ñ.= [1+5 (Øay⁄6er)f]/(1+32) - - - (07.7)

where,

Gg] = cladding yield stress

Ocr = buckling stress

£ = ratio of total cladding thickness to

total thickness f = 0.10 for alclad

2024-T3 and 08 for alclad 7075-T3 C7.8 Maximum Values for Crippling Stresses,

The cut-off or maximum crippling stress for

a composite section should be limited to the values in Table C7.1 unless test results are

obtained to substantiate the use of higher erippling stresses Table C7 1 Type of Section Max Fes Angles + 7 Foy

V Groove Plates Fey

Multi-Corner Sections, Including Tubes 8 Foy

Stitfened Panels Foy

Tee, Cruciform and H Sections 8 Foy 2 Corner Sections Zee, J, Channels 9 Fey 5.6 7.8.91.0 Fig C7.7

Trang 24

Fes Foy an 4 ANALYSIS AND DESIGN OF FLIGHT al 1B ,2 ,25 3 .4 5 6.T7.8.91.0 VEHICLE STRUCTURES Fig C7.8 Curve for Plates, Tees, Cruci Form and H Sections Fes = 0,67 [(gt YAME/Fey) */7]°°*° Foy { cy 1 See Table C7,1 for Cut-off Values l5 2 2.5 3 4 5 67 8919 Ty 1/2 avg) ⁄ - Fig CT.9

Trang 25

C7.8

CT.9 Restraint Produced by Lips and Bulbs

Quite often in formed sections, the ‘lange element which has a free edge is rather small in width as illustrated in Fig a Also for extruded sections, a bulb its often used as illustrated in Figs b The question then arises, is the lip or bulb sufficiently large enough to provide a simple support to the adjacent plate element Since the compressive buckling coefficient for a plate element is

Fig a Bulb

Bulb Lip Lip Buib Fig b

4,0 and 0.43 for a flange element, the use of a small lip or bulb can increase the value of the coefficient considerably above 0.43 and thus produce 4 more efficient load carrying element The problem of determining the dimensions of a lip or bulb to give at least a simply supported edge condition to the adjacent plate element has been investigated theoretically by Windenburg

(Ref 5) The results of his studies gives the following design criterion

sh Abe s _le

2.73 bạt det = 5

Where I; and Ar, are the moment of inertia and

area of the lip or bulb respectively (See Fig 07.10) dees th Tt Lip Đr (a) "Sy | LH Fig C7 10 From Fig C7.l0a for the lip, AL = bt, and IỊ = tbịŠ/5

In substituting these values in equation (C7.8), the dimensions of the lip are expressed

as

bị DL 5 Ot Đb

DLs

0.910 ® ""+* t

To determine by, and t, an additional re- quirement is specified, namely, that the buckling stress of the lip must be greater or equal to the buckling stress of the adjacent

Plate element

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION

C&, the compressive buckling Ve = 3 1s 0.388 for a flange From Chapter coefficient using element and 3.617 for a plate element There- fore, ~ s2? > + c2 s/È£\* ¡ 0.388 Gn = 3.617 BG) - - - (07.10; From equations ing relationship is C7.9 and C7.10, the follow~ obtained, ĐL „ be tự 5 0.528 Tỉ Fig C7.11 shows the results as a curve KỊP PROVIDES AT LẺAST SUMPLE SUPPORT FOR FLANGE 1° » SSD Lots an vã ole 20 46 $060 70 80 by +

Fig C7.11 Minimum lip dimensions required for flange to buckle as simply supported plate (Ref 5) In extruded sections, a circular bulb is

often used to stiffen a free edge as illustrated in Fig C7.10b The moment of inertia of the

bulb area about the centerline of the plate element is,

+

rae Be

As for the case of the lip, the buckling stress of the bulb must be greater or equal to the buckling stress of the adjacent plate element, which gives,

(Pt -1.82)° -0.574(2)* = 7.44 be/t ~ (C7,12)

Fig C7.12 shows design curve representing the above equation

Bo

“0 so

#š

Fig C7.12 Minimum buib dimensions required for buckle as simply supported plate (Ref 5)

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