ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES CT,7 C7.10 Ilustrative Problems ¡n Calculating Crippling Stresses Problem 1 equal leg angle shown in Fig a is aluminum alloy 2024-T3 Find the crippling stress for the The material Solution by Method 1 Material properties are:- Foy = 40000, E, = 10,700,000
For this method, we use Fig
C7.3 The parameter for bottom scale of Fig C7.3 is (a + b)/2t, where (a) and (b) are leg lengths measured to centerline of adjacent leg of angle For our case 4 = b = 1 ~ 025 = 0.978 Thus (a + b)2t = 1.95/0.1 = 19.5,
From Fig C7.3 using 19.5 on lower scale,
and the curve for two edges free, we read on the left hand scale that Foc/VFoy8 = 0330 Since we have only one angle, the crippling
stress Feg = Foo = 0330 x y40,000 x 10,700,000
= 21600 psi
Solution by Method 3 (Gerard Method}
For angle sections we use equation C7.4
A plot of this equation is given in Fig C7.7
The parameter for lower scale is,
Po
wan, where A = section area and g
equals the number of flanges plus cuts, or g = 2 for an angle section Substituting,
093, 40,000 1/2
Sy 08*) Tes700,000 > «2-188
Using this value in Fig C7.7, we read
Feg/Foey = 0.50 Therefore Fog = 0.50 x 40,000
= 30 , B60 psi
Problem 2 Same as Problem 1, but change
material to aluminum alloy 7075-T6 Foy = 87,000, Eq = 10,500,000 Solution Method 1 Fes = Fog = 0330¥67000 x 10,500,000 = 28,000 pst Solution by Method 2 ( „0983 3( 67,000 yl? 779,500, 000 2x 05 = 1.485 From Pig 07.7, Feg/Fey = 0.40, whence Fog = -40 x 67,000 = 26800 psi Problem 3 material to Titanium Ti-SMn Eo = 15,500,000 Same as Problem 1 but change Fey = 110,000, Solution Method i Fog = Feg = -O330V110,000 x 15,500,000 = 43000 pst Solution Method 2 „095 3( 119,000 33/2 „ 110,000._.+/⁄2 Sx 057) (TS, 800,066) 1.565 From Fig C7.7, Feg/Fey = 0.38, whence Pog = 0.38 xX 110,000 = 41,800 Problem 4
Find the crippling stress for the channel
section shown in Fig b if the material is
aluminum alloy 2024-T3, Fey = 40,000, Eg = 10,700,000 Solution Method 1 — t.50 ——| + As shown in Fig.c, 3)
the channel is composed 16 || -75 of 2 equal angle units Area =.137 +
(1) and (2) Since they Fig.b —=ll*05
are the same size, we — b—n
need only calculate the failing stress for one angle T a Angle + Dele Unit b 725 725 Unit a+b 725+ 75 = at 8T 14.5 Fig ¢
From Fig C7.3 for b'/t = 14.5, we read Fec/VfoyE = 045 (for one edge free)
`
Then Fog = Fog = 045 x
v40,O00 x 10,700,000 = 29400 psi
Solution Method 2
For @ channel section we use equation C7.6 which {sg plotted on Fig C7.9 The parameter
for bottom scale of Fig C7.9 is, Feyvx/s _„ ,;157.,_ 40,000 Barre =( s327v, 420,000 (x73 „ = “gã” ) 557706, 000" 8.50 A + From Fig C7.9 Fog/Fey = 65, whence, Fog 2 65 x 40,000 = 36,005 pst
Trang 2Cr.8
From Fig C7.9, Fes/Foy = 57, whence
Feg = 57 x 67,000 = 38,200
Problem 6
Find the crippling stress for the square tube as shown in Fig d Material is 2024-TS aluminum alloy Fey = 40,000, Eg = 10,700,000 Area = 373 Solution by Method 1 The square tube is Cut considered as made up
of 4 equal angles with no
edge free Fig d bi/t = (a+b)/2t = 1.95/0.1 = 19.5 From Fig C7.3, using upper curve, we obtain Pee/VfcyE = 0392 whence, Fog = Fog = 0392 x V40,000 x 10,700,000 = 25,600 psi Solution by Method 2 Area A = 373 g = number of cuts plus flanges or 4+ 8 = 12, For rectangular tubes we use equation C?7.4 or Fig C7.7 C290 —)1/*~ o.7sg À (Êeyj/ „ „375 at? EE? ta 08*” (6, 760,000 From Pig 07.7, Fes/Foy = 70 Therefore Fog = 40,000 x 70 = 28,000 pst Problem 7 Same as Problem 6, but change material to magnesium HK31A-0 Sheet, subjected to a temperature of 300°F for 1/2 hour Solution by Method 1 From Table Bl.1 of Chapter Bl, Foy = 11,100, Eg = 6,160,000, Fog = -0392V11,100 x 6,160,000 = 10,250 psi Solution by Method 2 soem on Metiod 2 A Foy,i/a _ 373 11,100 2/2 gt Gl = Gata 6,160,000’ = 528 From Fig 07.7, Fog/Foy = 96 Fog = 96 X 11,100 = 10,600 psi
From Table C7.1, the cut-off or maximum crippling for rectangular tubes is 8 Foy or
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION -8 x 11,100 = 8900 psi, Thus unless tests substantiate higher values, the crippling stress
Should be taken as 8900 psi
Problem 8, Same as Problem 6, but change material to stainless steel 17~7PH(TH1O50), Fey = 162,000, E_ = 29,000,000 Method 1 Fog = 0892V162,000x 29,000,000 = 85,000 pst Method 2, 373 162,000 1/2 _ Sex 087) SS 00,000) = 83 From Fig 07.7, Fog/Fey = 1595 Pog = 162,000 x 595 = 96,200 psi Problem 9 Sa — Area A = 255 Fig e
Determine the crippling stress for the formed section shown in Fig @ if material is aluminum alloy 2024-73 Fey = 40,000,
Ee = 10,700,000
Solution by Method 1 (Needham)
The section is divided into 6 angle units by the dashed lines in Fig e They are
numbered (1) to (3) since we have symmetry
The procedure will be to find the failing load for each angle and add up the total ror the 6 angle units The crippling stress will
Trang 3ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Area of angle (1) = o3O9 =A Pog * 0309 x 39,300 = 1215 Ib Angle unit (2) Area = 0459 (no edge free) b'/t = (a+bd)/2t = (.48+ 78)/.08 = 15.1 From Fig 07.3 Feo/VfoyE = 0475 Fog * ¥40,000 x 10,700,000 x 0475 = 31,100 psi Pee = 31,100 x 0459 = 1428 Ib Angle unit (3) Area = 0509 (no edge tree) bi /t = (.73 + ,606)/,08 = 16.7 FFom Fig Œ7.5, Fcc/VfoyE = 046 1046 x V40,000 x 10,700,000 = 30,200 pst Peg = «0509 x 50,200 = 1540 1b, Foo = Feg = 2Po,/area 2 (2x1215 +2 x 1428 + 2 x1540)/0,255 = 32,800 psi Solution by Method 2 (Gerard) AL foyys/s at™ G) g = number of flanges plus number of 212+5= 17 Substituting in the above tern, +255 40,000 \s/a _ œš 04*) (18,700,006) = 578 From Fig C7.7, Fog/Foy = 895, whence Fog = 40,000 x 895 = $5,600 pst
From Table C7.1, 1t {3 recommended for
multi-corner sections that Fy, maximum be
limited to 8 Fey unless tests can prove higher values
Fes 7 „8 x 40,000 = 32,000 psi Since this is less than the above calculated values,
it should be used Problem 10 Find the
crippling stress for the en
extruded bulb angle shown T >
in Fig f if material is +
2014-TS extruded aluminum (a) I bự
alloy th ) y
This particular bulb BI ng
section is taken from
Table A3.16 of Chapter A3 Fig f
c?.9
as Section No 2
Q.113 sq in The area from that table is
Solution by Method 2 (Gerard)
For this material Foy = 53,000, Ey = 10,700,000
The first question that arises is the bulb size sufficient to give an end stiffness to the (a) leg so that the bulb may be equivalent to the normal corner
In Fig f, bp = 0.78, hence be/; = 15 Referring to Fig C7.12, we observe that
for a br/t value of 15 we need a D/t ratio of
at least 3.8 The D/t value for our bulb
angle is (7/32)/.05 = 4.4, thus bulb has
sufficient stiffness to develop a corner next question that arises should the bulb angle still be classed as an angle section for which equation C7.4 applies or be classed as a channel or 2 corner section with the bulb acting as a short thick leg of the channel For this case, equation C7.6 would apply
The
The crippling stress will be calculated
by both equations
By equation C7.4 or Fig C7.7:-
If bulb is considered as a full corner
then g = 4 flanges plus 1 cut = 5 AL Feyys/a 2 gt? ED ( Prơm F1g C7.7, 3 88,000 x⁄4_ Sx 087) 10,700,000) “656 Fes/Foy = 82, hence Fog = 82 x SSO000 = 43500 By equation C7.6 or Fig C7.9, A Feyys/s | 112, 55,000 ji/s _ er Ge = Gos*) G95 560,000) °° * 7-70 From Pig C7.9, Fos/Foy = «7, hence Feg 3 »7 X 53,000 = 37,100
Possibly the best estimate of the crippling stress would be the average of the two above results or 40,300 psi
In Table C7.1, the so-called cut-off stress for angles is 7 Fey and channels 9 Foy If we use the average value or 8 Foy: it gives Fog = 8 x 53,000 = 42,400 as maximum permissible because of limited test
results on bulb angles
For the case where the bulb or lip does N not develop the stiffness necessary to
assume a full corner, then the bulb is only
Trang 4
CT 10
considered as an additional flange and the ¢
count would be four instead of 5, thus reducing
the crippling stress
EFFECTIVE SHEET WIDTHS CT 11 Introduction
The previous discussion in this chapter has dealt with the crippling stress of formed or extruded sections when acting alone, that is
not fastened to any other structure along its
length However, the major structure of
aerospace vehicles, such as the wing or body, involves a sheet covering which is strengthened
by attached or integral fabricated stiffeners
such as angles, zees, tees, etc Since the sheet and stiffeners must deform together, the
sheet will therefore carry compressive load and to neglect this load carrying capacity of the
sheet would be too conservative in aerospace structural design where weight saving is very
important
C7.12 Sheet Effective Widths
Fig C7.13a illustrates a continuous flat
thin plate fastened to stiffener and the entire unit is subjected to a uniform compressive load Up to the buckling strength of the sheet the compressive stress distribution 1s uniform over both stiffeners and sheet as in (Fig b) assum- ing same material for sheet and stiffeners As the load is increased the sheet buckles between the stiffeners and does not carry a greater stress than the buckling stress, However as the stiffeners are approached, the skin being stabilized by the stiffeners to which it is attached can take 2 higher stress and
immediately over the stiffeners the sheet can take the same stress as the ultimate strength of the stiffener, assuming that the sheet has a@ continuous connection to the stiffener Fig ¢ Shows the general stress distribution after the sheet has buckled This distribution depends om the degree of restraint provided by the stiffeners and the panel dimension
Various theoretical studies (Ref 6) have
been made to determine this stress distribution after buckling In general they lead to long and complicated equations To provide a simple basis for design purposes, an attempt has been made to find an effective width of sheet w which would be considered as taking a uniform stress (Fig d) which would give the same total sheet strength as the sheet under the true non-uniform stress distribution of Fig c
The question of sheet effective sheet has been considered by many individuals The names of VonKarman, Sechler, Timoshenko, Newell, Frankland, Margurre, Fischel, Gerard, and many more are closely associated with the present knowledge on effective sheet width
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION
Fig 0 Sheet stress distribution before buckling
CEPT EPP Pp yp yyy
Sheet stress distribution after buckling ins | Equivalent sheet effective width ty 1l + i] In" IT Fst | Fie tt] Hh 4 WT Had II “| — vi — Bà | | Le iT H FT g4 —+—— — gì il ul Lal ™ _ 5 uw 7 rev \ ale tl Rul Y ti Fig a gil St t Buy Fe Su 5 IL 7Ci Cee 77 7 77e vad? eligee Sheet-stiffener panel Fig C713
From Chapter C5, the buckling compressive stress of a sheet panel is,
Pee 2 Kone ya
or “2 (1 -¥%*) ‘d
If we assume that the stiffener to which
the sheet is attached provides a boundary
restraint equal to a simple support, then k, = 4.0, and if Poisson’s ratio V, 1s taken as
0.3, then equation C7.13 reduces to, For = 3.60E(t/b)*
The Von-Karman-Sechler method as first proposed consisted of solving equation 07.14
for a width (w) in place of (b), when Foy was
equal to the yield stress of the material since experiments had shown that the ultimate strength of a sheet simply supported at the edges was independent of the width of the sheet
Thus equation C7.14 changes to,
Foy = 3.60E(t/w)* , whence,
ws l.90t V8/Fey > - (07.15)
Since the crippling or local failing
stress of a stiffener can exceed the yield
} strength of the material, equation C7.15 was later changed by replacing Fey by the stress
in the stringer Ffop, thus giving,
Trang 5
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
W=1.90t VE/Fep - ee
Some early experiments by Newell indicated the constant 1.90 was too high and for light
stringers a value of 1.7 was more realistic,
thus 1.7 has been widely used in industry
If we assume the stiffness of the stiffener and its attachment to the sheet as developing a fixed or clamped edge condition for the
Sheet, then
For = 6.35E(t/b)* or w= 2.52t VE/Fgp
For general design purposes, it is felt
that 1.9 or equation C7.15 is appropriate for
determining the effective width w If stiffener is relatively light, use 1.7 C7.14 illustrates the effective width for
sheet-stiffener units which are fastened
together by a single attachment line for each flange of the stiffener Fig 2 = Rivet Lines One | Rivet Line A."Ă= Fig, C7.14 ew at wy
The crippling stress is determined for
the stiffener alone This stress is then used
in equation C7.15 to determine the effective widths w The total area then equals the
stiffener area plus the area of the effective sheet width w The radius of gyration should
include the effect of the effective skin area
Fig C7.15 illustrates the case where
stiffeners are fastened to sheet by two rows
of rivets on each stiffener flange In this ease, the rivet lines are so close together
that the effective width w for each rivet line
would overlap considerably A common practice
in industry for such cases its to use the effective width for one rivet line attachment
as per equation C7.16 to represent sheet width to go with each stiffener flange However,
in calculating the crippling stress of the
stiffener alone, the stiffener flange which 1s attached to sheet is considered as having 2 thickness equal to 3/4 the sum of the flange thickness plus the sheet thickness Staggered Rivet Rows CT 11
The effective sheet width as calculated by equation C7.16 assumes that no inter-rivet
buckling of the sheet occurs or, in other words, the rivet or spot welds are close enough together to prevent local buckling of the sheet between rivets when the sheet is carrying the stiffener
crippling stress The subject of inter-rivet
sheet duckling is discussed later in this
chapter
Pig C7.16 1llustrates a procedure to follow for determining the effective width w when sheet and stiffener are integral in manufacture Case l Case 2 tạ 3 tị « 2ts te 2 tg Tee Section tấn kế Fig C7.18
For Case 1, find the crippling stress for the tee section alone, assuming the vertical stem of the tee has both ends simply supported For value of t in equation C7.15, use
(tg + tr)/2 The effective stiffener area -
equals the area of the tee plus the area of the Sheet of width w
For Case 2, determine the crippling stress
for the I section acting alone Calculate w/2
from equation C7.15 to include as effective Sheet area The column properties should
include I section plus effective sheet
CT 12 Effective Width W, for Sheet with One Edge Free
In normal sheet-stiffener construction, the sheet usually ends on a stiffener and thus
we have a free edge condition for the sheet as
illustrated in Fig C7.16a The sheet ends at Fig C7 16a fed! 0 erat bw |
a distance b’ from the rivet line For a sheet
free on one edge, the buckling coefficient in equation C7.13 ts 0.43, thus equation C7.13
reduces to,
Trang 6C7.12
Then the total effective sheet width for
this end stiffener would thus equal w, + w/2 Cĩ.13 Effective Width When Sheet and Stiffener Have
Different Material Properties
In practical sheet-stiffener construction
it is common to use extruded stiffeners which
have different material strength properties in the inelastic stress range as compared to the Sheet to which the stiffener is attached For example, in Pig C7.17 the stiffener material could have the stress-strain curve represented
by curve (1) and the sheet to which it is attached by the curve (2) Now when the
stiffener is stressed to point (B), the sheet directly adjacent to the stiffener attachment line must undergo the same strain as the
stiffener and thus the stress in the sheet
will be that given at point (A) in Fig C7.17 This difference in stress will influence the effective width w Correction for this condition can be made in equation C7.17 by multiplying it by Fsy/Foep, which gives
w= 1.900(Fsw/Fer) (ge)
Where fst is the stiffener stress and fgy is the
sheet stress existing at the same strain as existing for the stiffener and obtained from a
stress-strain curve of the sheet material STRESS IN THOUSANDS P.S,1, a 9.002 004 006 008 010 012 STRAIN - IN./IN, Fig, C7.17
Por a rather complete study and comparison of the various effective widths theories as
compared, see article by Gerard (Ref 7)
Equation C7,.15 is in general conservative for higher b/t ratios,
Cï.14 Inter-Rivet Buckling Stress
The effective sheet area is considered to act monolithically with the stiffener How-
ever, if the rivets or spot welds that fasten
the shest to the stiffener are spaced too far
apart, the sheet will buckle between the rivets
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION before the crippling stress of the stiffener
is reached, which means the sheet is less
subjected compressive load Thus,
to save structural weight, str ra
Select rivet snacings that will »revent inte
rivet buckling of the shset In zeneral, the
Tivet spacing along the stiffeners in the upper surface of the wing will be closer to-
gether than on the bottom surface of ‘he wing since the design comsressive loads on the
top surface are considerably larger than those on the bottom surface
The following method is widely used oy
engineers concerned with aerospace structures
relative to calculating inter-rivet buckling
stresses It is assumed that the sheet
between adjacent rivets acts as a column with
fixed ends Ũ
The general column equation from Chapter C2 for stable cross-sections 1s,
Po = Cr*B&y/(L/p)* - (C7.19)
Where C 1s the end fixity coeffictent and varles from a value of 1 for a pin end support to 4 for a fixed end support
Tae effective column length L' = L/vC,
thus equation C7.19 can be written
sẽ 2/2121 dd
@ = N7E,/(L' /o)* (07.20)
Let p the rivet spacing be considered the column length L Assume a umit of sheet 1 inch wide and t its thickness Then moment of
inertia of cross section = 1 x t°/l2, and area
A=1xt=t Then radius of gyration 9 = 0.29t Then substituting in equation 27.20 to obtain the inter-rivet buckling stress Fir, a Pip? eee TT - TT TT ~ (c7.21) ( ẹ ⁄0.29t)* For clamped ends C = 4, thus ae ory lt eee ee 28 Fir ^ [070:5át) (67.28)
To vlot this equation, the tangent modulus
#y for the material must be Known, However, we
can use the various column curves fn Chapter ce
which show a plot of F, versus L'/p and in
equation C7.22 the term p/0.58t corresponds to
L'/o
The fixity coeffictent C = 4 can be used for flat head rivets For spot welds it should
be decreased to 3.5 For the Bragier rivet type
use C = 3 and for counter-sunk or dimpled rivets
Trang 7ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Figs C7.18 and C7,.19 show a plot of
equation C7.22 for aluminum alloy materials If the inter-rivet buckling stress cal- culates to be more than the crippling stress of the stiffener, then the effective sheet area
can be added to the stiffener area to obtain the total effective area This total effective
area times the stiffener crippling stress will give the crippling load for the total sheet-
stiffener unit
When the sheet between rivets buckles
before the crippling stress of the stiffener is reached, the sheet in the buckled state has the ability to approximately nold this stress as
the stiffener continues to take load until it
reaches the stiffener crippling stress This buckling sheet strength can be taken advantage or by reducing the effective sheet area Thus effective sheet width equals,
Noorrected * W(Fir/Fes) - - - - > = - (07.25)
The area of the corrected effective sheet is then added to the area of the stiffener
The crippling load then equals the crippling
stress of the stiffener times the total area The use of sheet effective widths in finding the moment of inertia of a wing or
fuselage cross-section is a widely used pro-
cedure in the analysis for bending stresses in
conventional wing and fuselage construction
Reference should be made to article A19.13 of
Chapter Al9 and article AZ0.3 of Chapter A20 for practical illustrations in the use of
effective widths
C7.15 Olustrative Problem Involving Effective Sheet,
Conventional airplane wing construction is illustrated in Fig C?7.20 The wing is covered with sheet, generally referred to as skin, and this skin is stiffened by attaching formed or extruded shapes referred to as skin stiffeners or skin stringers A typical wing
section involves one or several interior
straight webs and to tie these webs to the
skin, a stringer often referred to as the web
flangs member, is required to facilitate this connection Fig C7.2l is a detail of the flange member and the connection at point (1) in Fig C7.20
The stiffener or flange member is an extrusion of 7075-T6 aluminum alloy The skin and web sheets are 7075-T6 aluminum alloy The
skin is fastened to stiffener by two rows of 1/8 inch diameter rivets of the Brazier head
type, spaced 7/8 inch apart The web is
attached to stiffener by one row of 3/16
diameter rivets spaced 1 inch apart The C?.13 - —k 1/8 —* ee Fig C7 21 Detail of Point (1}
problem is to determine the crippling stress of
the stiffener, the effective skin area and the total compressive load that the unit can carry
at the failure point Since the stiffener 1s
braced laterally by the web and the skin, column
bending action is prevented and thus the
crippling strength is the true resulting strength of this corner member under longitudinal
compression (Additional stresses are produced
on these corner members if web buckles under
shear stresses and web diagonal tension forces are acting This subject is treated in the chapter on semi-tension field beams.) Solution: Area of stiffener = 0.24 sq For 7075-76 extrusion, Foy 10,500,000 in = 70,000, Ey =
The Gerard method will be used in calculating the crippling stress Equation C7.4, or the design curve of Fig C7.7, applies to this multi- corner shape The lower scale parameter for Fig C7.7 18,
s +/a „Q.24 70,000 sa3/2 _
se Š ) ŠxO.72” (10,500,068) 0.83
Using this value, we read from Fig C7.7 that Fes/Foy = 0.82 Thus Fog = 82 x 70,000 = 57,400 psi The g¢ value of 6 was determined as
shown in Fig (a)
Effective Sheet Widths: fof
Equation C?7.16 will be
used to determine the sheet
f
effective widths Flanges (f) = 5
For the skin t = 05 No of cuts 5
Materfal {s 7075-Té6é aluminum
alloy Ec = 10,500,000, Fig a
= 1.9t /Fog 2 1.9 % 0.5 ¥10,500,000/57 ,400
= 1.28 in Thus a piece of sheet 1.28/2 = 64 wide acts to
Trang 8Fup - INTER « EIYET BUCTILIEG STEESE - ksi C s3 G kề Gò BÓ Fig CT 18
Material Alum Alloy Bare Curve Designation Thickness Basta {BARE) 2024-T4 =, 250 2024-T3 2024-T36 2024-T6 2024-T81 2024-88 ~*, 250 T0T5-T6 016-.039 1075-T6 040-,240 DOr rr oo i f 5 E at a A é «1 Có CN Bề B Fig CT.18
Trang 9ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES of the dimensions in Fig C7.21 shows the
effective width with each skin rivet line would
overlap slightly thus we use only the width
between rivet line (see Fig b)
For the web t = 064 Since the web has
a free edge, the effective width calculation
will be in two steps + - 1:9 z 2 x 064 vÃ0,500,000/87,400 = 0.82 W, from equation C7.18 = 62t vB/Fos = 62 x +064 ¥10,500,000/57,400 = 0.55 inch Fig (>) shows 2 w/2 the effective sheet fered 815 ae 84 4 width as calculated u Total effective Sheet area = 1.28 x 05 + 875 x 05 + (0.55 + 0,82) 064 = 195 Total area = 195 + 0,24 = 435 Fig b The total
compressive load that entire unit ce carry
before failure is then equal to AF,., = 435 x
57,400 = 25,000 lbs
This result assumes that no later-rivet
buckling occurs under the stress of 57,400 pst
in the sheet between rivets
The skin rivets are Brazier head type spaced 7/8 inch apart or p= 7/8
As discussed under inter-rivet buckling, the end coefficient c for this type of rivet
should be less than 4 or assumed as 3
Fig C7.18 gives the inter-rivet buckling stress versus the p/t ratio This chart is based on a clamped end condition or C = 4, Since C = 3 will be used for the Brazier type rivet, we correct the p/t ratio by the ratio v4 / VS = 1.16
The corrected p/t = 1.16 x 875/.05 2 20.3
From Fig C7.18 using curve (8) which Is our material, we read Fir = 50,000 psi, thus
skin will not buckle between rivets as
crippling stress is 57,400 psi
The web rivets are of the flat head type and C = 4 can be used Spacing is 1 inch
Hence p/t = 1/,064 = 15.6 and from Fig C7.18,
curve 8, we read Fyp = 54,600, which is considerably more than the unit crippling
stress
In wing construction, the skin rivets are
C?, 15 usually of the flush surface type, either countersunk or dimpled If we make the skin rivets of the countersunk type, the end fixity coeffictant must be reduced to 1 to be safe Then the corrected p/t ratio to use with
Fig C7,.18 would be (V4 / VI )p/t = 2 x 875/.05 = 35 From Fig C7.18 and curve 8, Fir = 29,000 which is far below the calculated crippling
stress, thus the rivet spacing would have to be reduced Use 9/16 inch spacing corrected p/t = 2x 5625/.05 = 22 From Fig C7.18, Fir * 57,400 psi, which happens to be the crippling stress and therefore satisfactory C?.16 Failing Strength of Short Sheet-Stiffener Panels
in Compression
Gerard (Refs 2, 3) from a comprehensive
study of test results on short sheet-stiffener panels in compression, has shown that his equation C7.4, or Fig C7.7, can be used to give the local monolithic crippling stress for
Sheet panels stiffened by Z, Y and H at shaped
stiffeners The method of calculating the value of the g factor is illustrated in Fig C7.22, Fig C7.23 1s a photograph showing the erippling type of failure for a short panel involving the Z shape stiffener
C7.17 Failure by Inter-Rivet Buckling
Howland (Ref 8) assumed that the sheet
acts as a wide column which ts clamped at its ends and whose length 1s equal to the rivet spacing The inter-rivet buckling stress
equation 1s then, 4
= enn te
Pir EG) ote (07.24) ‘
The end fixity coefficient C is taken as 4 for flat head rivets and reduced for other types as previously explained for equation
c?.21
Ris the plasticity correction factor
i is the clad correction factor Vg is Poisson’s ratio (use 0.30) ts = sheat thickness, inches
p = rivet spacing or pitch in inches
For non-clad materials the curves of Fig c?’.24 can be used This figure is the same as Fig CS.8 of Chapter CS For the clad correction see Table C5.1 of Chapter C5
C7.18 Failure of Short Panels by Sheet Wrinkling
Trang 11ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES article This sheet buckling does not deform
the flange of the stiffener to which the sheet is attached However, if the rivet or spot weld spacing 1s such as to prevent inter-rivet
buckling of the sheet, then failure often occurs
by a larger wrinkling of the sheet as
tllustrated in Fiz C7.26 The larger wrinkle Snape subjects to flange of the stiffener to
which the sheet is attached to lateral forces and thus the stiffener flange often deforms
With the sheet wrinkle shape This deforming of the stiffener flange produces stresses on the stiffener wed, thus wrinkling failure is a
combination of sheet and stiffener failure
The action of the wrinkling sheet to deform the stiffener flange places tension loads on the rivets, thus rivet design enters into the
failing strength of sheet-stiffener panels under compression, Fig C7, 25 Inter-Rivet Buckling Fig C7 26 Wrinkling Failure
Several persons have studied this wrinkling
or forced crippling of riveted panels (Refs 9
and 10) A rather recent study was carried out
by Semonian and Peterson (Ref 11), which is
reviewed and simplified somewhat by Gerard in (Ref 2) The results as given in (Refs 11 and 2} used to calculate the wrinkling stress C1.19 Equation for Wrinkling Failing Stress Fy
From Ref 2 we obtain, _ ky TT l8
W_ 12 (1-1)
Py Gee sect tee “sya
ky, the wrinkling coefficient is obtained from Pig C7.27 This coefficient is a function
of the effective rivet offset 7 which is
obtained from Fig C7.23 Having determined Ky from Fig C7.27, equation C7.25 can be solved by use of Fig C7.24,
CT.20 Rivet Criterion for Wrinkling Failure
A criterion for the rivet pitch found from test data which results in a wrinkling mode failure 15,
8/Đg ~ 127/4 V122 =~ -
The lateral force required to make the stringer attachment flange conform to the
wrinkled shset, loads the rivet in tension An 7,17 approximate criterion for rivet strength from Re?, 2 18, O7 Dgp ,„ „* >a iw E== 2a (hy)* -+ - re Sse 4
The tensile strengta of the rivet Sy is
defined in terms of the shank area and it may
be associated with either shank failure or
pulling of the countersunk head of the rivet
through the sheet
For aluminum alloy 2117-T4 rivets whose tensile strength is s = 57 ksi, the criteria
are:-
$= 57 ksi., dạ/ta„y = 1.87
(27.28) -_ 199 160
se“ Te/tay ˆ tay)! de/tgy > 1.67
where tay, is the average of sheat and stiffener
thickness in inches The effective diameter dạ
is the diameter for a rivet made from 2117-T4 material
The effective diameter of a rivet of
another material is,
dg/d = (S,/3}*/4 ~ - (c7.29)
where Sr is the tensile strength of a rivet defined as maximum tensile load divided by
shank area in ksi units
CT.21 Problem 1 Mlustrating Calculation of
Short Panel Failing Strength
Trang 12C7.18 CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION Z2: L | a 2 I ^= | s af? 550 Curves for finding Fir as per eq CT, 24 and Fy as per eq C7 25 Fo - ⁄ Fr Fig C7 24 oo + a “2 “ae “ ag cn?®E ts kun®E ts og 1201-Ug> PO) F 12-UaFavrbg ⁄% 3, 7te
Trang 13ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
whence, Dy/ty 0.372
Ạ/ty
The rivets are 3/32 diameter Brazier head type AN456, 2117-13 material spaced at 0.75 inches 38, bp/dy 9.27, bọ/t„ = 6.56 0,982 1n." 2 inches = 2 x ,064 = 0.128 Total area of sheet and stiffener = 0.380 in.* Area of 2 stiffener Area of skin for bg
Crippling Stress of Stiffener Acting Along
Fos st)’
Since we have a 2 type of stiffener, equation C7.6 and Fig C7.9 applies The lower scale parameter in Fig C7.9 is, \ Ay Foy.1/s /0.252,, 40,000 s/s _ a) ¢ Ea) = Sa ) (T9500, 000" = 9.56 From Fig 07.9, Fos/Fo 9.6, Hence, Fog = 58 X 40,000 = 24,000 pki = Pos(st)* Crippling Stress of Panel Considered as a Monolithic Limit F es (M)"
Equation C7.4 or Fig C?.7 applies for monolithic failure of sheet-stiffener panels
The lower scale parameter on Fig C7.7 1s, cA ) ESL y2/2 = (0280 (40,000 ;x/2 gt 7,82 x 0647" ‘10,700,000 = 0.73 8 = 7.83 (see Fig C7.22b} From Fig C7.7, we read Fos/Foy = 725, hence Fog = 40,000 x 725 = 29,500 PBL = Fog(my-
Inter-Rivet Buckling Stress (Fry)
The rivet type is Brazier head and the spacing p is 3/4 inch
Equation C7.24 applies and Pig C7.24 is
used to solve the equation The lower scale parameter in Fig C7.24 is, Cre (Es)e 12 (1 - Ve) Fy, ° 8 Por Brazier nead rivet C = 3, Substituting: - Y= 3 #n? x 10,700,000 «064, eT = 1Đ - 22)(32,000) “Tg” = 5-48
The shape parameter n for our material is 11.5, Reference to Fig C7.24 for a value of 5.45 om bottom scale which is off the scale, we estimate the Fy p/P, a.? as above 1.1 Thus
C7.19 Pur = 39,000 x 1.1 = 43,000 This value is far about the stiffener or panel crippling stress as previously calculated so inter-rivet buckling is not at all critical
Failure by Sheet or Face Wrinkling (Fy) The wrinkling failure stress by equation To determine value of ky, we use curves in Figs C7.28 and C7.27 P/d = 75/.0937 = 8, bo/tw = 5.26 From Fig C7.28, we read f/ty = 6.5, whence f= 064 x 6.5 = 0.416 f/dy = 416/2.437 = 17 (dy/ty)/0s/tg = 38/(2.0/.064) = 1.21
From Fig C7.27, we read Ky = 4.4
To solve equation for Fy we use Pig C7.24 The lower scale parameter is, Kụ R? E cšs 12 (1 = Ve*){F,,,) ‘ds ys 4.4 mn x 10,700,000 (20642 For n= 11.5, we read from Fig 07.24 that Fw/Fo1> = «9, whence Fy = 9 x 39,000 = 35,100 pst Thus wrinkling failure is not critical as Fy is larger than Feg(m) and Fog (gp)
3.0) = 1.15
The results show that the crippling stress for the stiffener alone of 24,000 psi is the smallest value, or the stiffener is unstable as it fails first The entire panel unit will not reach its failing strength when stiffener stress is 24,000 because the skin wrinkling stress fy is higher An approximation suggested
in (Ref 2) is to assume stiffeners carry the same stress as the skin up to Fcs(sr) and beyond this the stiffener carries no additional load Thus the panel failing stress F(p) can be
calculated from the following equation m wes Fwsts * Posie) Ast “F Dt, * Ag s/s ST ~ 35,100 x 2x 064 + 24,000 X 0.252 _ = EER ar0 88 — 27,800 pst The total load carried by each stiffener plus its sheet {s 27,900 x 380 = 10,600 lbs
The failing strength of the riveted panel cannot exceed the monolithic panel failing stress
Fcs(M); Which was 29,000 psi for our panel or
kì > Œz
Trang 14C7 20 greater than the calculated failing stress of 27,800 psi neck of Rivet Strength From expression C7.26 Dibs < 1.27/42y +4 Se 3 375 = 1.27/(4.4)*/* = 603(satiefactory)
The criterion for required rivet strength to make the stiffener flange follow the
wrinkled sheet is from C7.27 0.7 bs PB a Se aap a a (Fw) 0.7 2) 0.755 an a Šr ” (T87700,000 (5/58) (3/85) (26,100) Sy > 13.7 psi From expression C7.28 8 = 87 KSi for dg/tay, = 1.67
For given panel de/tay, = 3/32/.064 = 1.46, thus rivets have plenty of tensile strength to produce a wrinkling failure
C7 22 General Design Limitations to Prevent Secondary Failure in Sheet-Stiffener Panels
Sheet stiffener units can be designed as
columns if the secondary forms of failure such
as inter-rivet buckling and face wrinkling are avoided The following design rules referring to Pig (A) will usually avoid these secondary weaknesses i bg Fig A (1) te,/tg 2 0.5 + - Promotes overall crippling
(2) 04< be, /by = 0.5 Rolling versus local buckling (3) Make by as small as vossible — Face wrinkling (4) p/og = 0.5 - - - - Inter-rivet buckling CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION p/dD= 8 } ) Tensile strength of attachment per inch in order to prevent Face wrinkling (5) p/D 2B -~ - (6 rivet or spot weld should be = 0.05 F y failure in wrinkling 5
(7) As a rough guide do not use bent up
Stringers if dg/ts ~ 30 in orcer to prevent fase wrinkling weaxness
CT.23 Y Stiffened Sheet Panels
A Y shape cannot be formed from sheet,
thus it must be extruded To make the ¥ shape
efficient, the various parts usually have a different thickness Furthermore, the extruded Material has different mechanical properties in the inelastic stress range as compared to rolled sheet that is used for the panel, thus these effects must be considered in calculating the crippling stress of the stiffener and the
complete panel unit
The effective thickness tụ of the stiffener
is determined by the following equation (Ref 3): ty FE dyty/2 dy
where by and ty refer to the lengtn and thick~
ness, respectively, of the cross-section elements
When the yield stress Foy of stiffener and Sheet are different and effective Foy can be
estimated as follows (Ref 3):
Foy = {Foytsn) * Foy(se) [(Ev/ts) - 3]} Ea/ts
- ee eee eee (C7 30) The monolithic crippling stress for the Sheet stiffener panel can be calculated from equation C7.4 or by the curve in Fig C7.7 Equation c7.4 is
Fos (tt)/Poy = 0-56 [lat VA) (E/Foy) */7]°°8*
The constant 0,56 in this equation applies
for Y and hat-stiffened panels when the ratio ty/ts = 1.0 For other ratios the correction
of this constant which is referred to as 8s
Trang 15ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C7,24 Example Problem Y Stiffened Panel
The compressive monolithic failing stress of a Y stiffened panel, as illustrated in Fig C?.22a, will be calculated by the Gerard method Fig C7.31 shows details of the panel unit
The stiffener is extruded from 2014-T6 aluminum
alloy for which Foy = 53000 and E, * 10,700,000 fne skin or panel Sheet 1s aluminum alloy
7075-T6 for which Fey > 67,000 and Ee = 10,500,000
eae OO
Fig C7.31 Stiffener area = 0.538 in.?
Sheet area = 4,21 x 064 = 270 in."
Total area (A) per stiffener unit = 0.808 in.”
Radius.of gyration of stiffener alone =1.123 in
Solution:-
Since the elements of the Y stiffener have
dirferent thicknesses, the effective ty by equation C7.29 1g needed 5 byt aby oy 1,111 x ,186 + 1.422 x 068+3.2x 064 + 1.188 x 064 1.111 + 1.422 + 3.2 + 1.188 or ty = 0763 inches
Since stiffener and sheet have different
material properties, an effective Foy from equation C7.30 will be calculated
r
Foy = {7,000 + 88000 IS \ / SEES = 54700
The curve in Fig C7.7 will be used toa
solve the equation for Fos(M)‹ The lower scale
parameter for Fig C7.7 modified by ty and Poy
18,
4 CS) tựa gt tg “Ee
The value of g from Fig C7.22a its 16.83
as the average value for a 6 stiffener panel
C7 21 unit Eg will be taken as 10,600,000 which is the average E for stiffener and Sheet Substi-
tuting in the above parameter:-
-808 64,700_\+/2 gag
Games aren ee) ‘TO R50 B00 18.83 x 0765x 064" *10,500,000 From Fig C7.7 we read
Fes/fey 2 76
Thus the monolithic crippling a failing stress Fogy = Foy X «76 = 64,700 x 75 = 48,200 psi
_ The curve as plotted in Fig C7.7 is fora t/tg ratio of 1.0 The ratio for our panel 1s
.0763/.064 = 1.19, The correction factor from Pig C7.30 is 1.03 Therefore Fes) = 1,08 x 48200 = 49,600 pst Load carried by one stiffener-sneet unit = 0.808 x 49600 = 40100 lb COLUMN STRENGTH CT.25 Column Curve for Members With Unstable Cross-Sections
Chapter C2 dealt with the column strength of members with stable cross-sections For example, if we took a round tube with relatively heavy wall thickness and tested various lengths
in compression to obtain the failing stress and then plotted these stress values Fy against the
slenderness ratio L'/o of the member, the test results would closely follow the curves ABFC in Fig C7.32 The type of failure would be elastic overall bending instability at stresses between points B-C and inelastic bending
instability at stresses for most of the range BA Euler’s equation as shown in Fig C7.22 can be used to determine the failing stress for both elastic and inelastic bending instability with the tangent modulus Ey, being used in the
inelastic stress range
Now suppose we test various lengths of a
member composed of the same material as used for obtaining curve ABC, but use a member with an
open cross-section, such as a channel, hat section, etc., with relatively small material thickness The test results for such members
would often follow a curve stmilar to DEFC in Fig C7.32 Thus it 1s obvious that Euler’s equation cannot be used in the range DEF, as the true failing stresses are far less than that given by the Culer column equation A short length of the member with L'/p less than 20 will fatl at practically the same stress, thus the failing stress for lengths up to L'/> = £0 will be practically the same and this stress has been given the name of crippling stress (Fes) and the
previous portion of this chapter has been con-
Trang 16C†?.22 e* (Lt /p)* \ Crippling Stress Region | | | | — Transition Region L}⁄Ð Fig C?.32
cerned with methods of calculating this local crippling or failing stress at point (F) the
elastic buckling of some part of the cross~ Section begins Between stress points F to BE the action for the member involves both overall elastic bending instability plus locul buckling
which becomes more extensive as the stress increases The portion EF of the column strength curve is often referred to as the
transition range At present no reliable theoretical theory has been developed for determining the failing stresses in this transition range, thus resort 1s made to semi- empirical methods which have been checked against test results and found to give reason- ably close results
C7, 26 Methods Used for Determining the Column
Failing Stress in the Transition Region
METHOD 1 JOHNSON-FULER EQUATION
Possibly the first method used in calcu- lating the column failing stress Fe in the transition range EF in Fig C7.32 was the well known Johnson-Euler equation which involves the
crippling stress The equation is, 2
Fo = Fog - Gea (L'/o)? -~ - (c7.32) where, Po = colum failing stress (psi)
Fog = crippling stress, assumed to occur at L'/p = 0, where L' = LA
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION
The equation gives a parabolic curve
starting from the crippling stress at L'/o = 0, and becomes tangent to the Euler curve at a stress value equal to one half the crippling stress Fig C7.33 shows a plot of equation C7.31 for aluminum alloy material for various values of the crippling stress which is the F, stress at L'/o = 0
This method is quite simple to use as the
only additional calculation required is the crippling stress of the column section which is obtained by methods previously explained in this chapter The other methods usually involve the buckling stress as well as the crippling stress Sinca the crippling stress 1s normally
constant below L'/o = 20, the assumption that
Fes is zero at L'/o 0 is slightly conservative
METHOD 2
The following method or slight variations of it appears in the structural design manuals of a number of aerospace comoanies The method or procedure for determining the column failing stresses in the so-called transition range
involves the use of the basic column curve for
stable cross-sections The procedure can best be explained by reference to Fig C7.34 A Fe F 19 For Lo flig, C7,34
The curve ABC is the Euler column curve for @ column with a stable cross-section and fora
given material It involves using the tangent
modulus Ey in the inelastic stress range The following steps are taken to determine the
column curve for the so-called transition range:-
Trang 17
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
{1) Locate point (0) on the basic colum
curve by drawing a horizontal line through
an Fo value equal to Foy, the yield stress
of the particular material being used (2) Draw a horizontal Line starting at point
D Point D is at an Fy, value equal to the (Fog) crippling stress for the colum
Section being considered Point (E) on
this line is determined by projecting vertically downward from point (0)
(4) Locate point (F) at a value of Fy = 9 For Where For is the buckling stress for the
cross-section Draw a horizontal line
through point (F) to intersect the column curve at point (G)
(4) Connect points E and G with a straight line The line HG then represents the column failing stress Fe for values of
L'/p between points EB and G
This method requires the use of the column curve for stable sections and the determination of the duckling stress and thus requires more calculations than Method 1 It also requires 23 graphical construction CT 23 JOHNSON & EULER COLUMN CURVES ALUMINUM ALLOY a Euler Curve Fo “wa Fon up)" 4n*E P Johnson Curve Fg 2 Fog ~ E = 10,200, 000 P 5.1L METHOD 5
Another method that is widely used aiso uses a parabolic curve to represent the colum strength in the transition range (see Ref 12)
The parabolic approximation has the follow- ing form:- m ly em == (c7.32) es cs § where
Fo 18 the column failing stress + Feg is the crippling stress
Fer is buckling stress for the colum
cross-section
Fg is the Euler column stress for the
particular column being considered as found from equation Pg = n®E/(L'/p)* The equation applies for Fa = Fer For
cases where For > Fo, where Foy, is the pro-
Trang 18
C?.24
CT.2T Example Probiems Involving the Finding
of the Column Strength of Columns With Unstable Cross-Sections m 2.0" = ch TT mm | aL eb material is aluminum
alloy sheet 7075-Té, Fig a
with Poy = 67,000 and E, = 10,500,000 If tha member fas a pinned end condition, what is the
column failing stress
PROBL=M 1
A rectangular tube
21 inches long has the
cross~section as shown in Fig (a) The
Solution The area of cross-section = 234 in.?
The least radius of gyration is 0.42 inches Before the column failing stress Fa for the 21 inch length can be found, the erippling stress Fes and the buckling stress For must be
determined
The crippling stress will be calculated by the Gerard Method The parameter for use with Pig C7.7 is,
mr |) AF ey )2/a_ 234 , 67.000 _r/a | Tex 06* 19,508,000) 9.975 `
Prom Fig C7.7, Fes/Fey = 57, hence Fog = 57 x 67,000 = 38,200 pst
The initial buckling stress Foy will be
determined by the theory of Chapter Cs, b^“1~- 08= 92, h=2- 08 = 1.96 b/n = 92/1.96 = 479, tụ/tụ = 1.0 From Fig C6.6 of Chapter cs, Xu = 5.2 " “er “12 (1 = Ue") th 5.2m" x10,500,000 , 04 Por Ð TẾT can) — tạp) = 22,800 pst Colưmn Strength by Method 1 (Johnson-Euler eq.) L's LAG 3 21/7 = 21 L'/p = 21/142 = 50 L = 21 inches The Johnson-Euler equation 1s, 2 ° Fos ta Fo * Fos - qa gp (L'/e)* , substituting, n = 38,200 ° 2 Fo = 38,200 4T“ 10,500,7000 (50)* = 29,400 psi This value could be read directly from Fig
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION
C7.33 for a value of L'/p OQ and a value of /2
Fo = Fog = 38,200 at L! uu ou
ta
Column Strength by Method
Tnis method requires a grapnica
construction which involves the column curve
for the given material and for a stable or
section
o5sS-
Fig C7.35 shows the column curve for our
material for a column with stadle cross-sect
It is identical to Fig 02.10 of Chapter 22 for the room temperature condition, except curve in Fig C7.35 has been draym to a sma’ vertical scale, Col Curve Stable Cross- Section F¢ (ksi) 9 10 20 30 40 50 60 70 80 L'/p
The graphical construction to obtain the column curve in the region between the erippling stress and the buckling stress ts as foliows:
At a column stress Fy = Foy = 67,000, draw a horigontal line to intersect ths basic column
curve at point (0) “Draw a horizontal line
from point (D) which equals the crippling stress of 38,200 psi Locate point (Z} by projecting vertically downward from point (0) Locate point (G) at an Fo stress equal to 9 Fop =
+9 ¥ 21,500 = 19,350 psi Connect points (BE)
and (G) by a straight line This line repre- Sents the column failing curve for the member cf our problem for L'/p values between points (š) and (9) Taking L'/o = 50, we project up~ ward to &G line and then horizontally to scale
at left to read Fy = 27,700 psi Column Strength by Method 3 —_—S EE tod
In this method equation 07.32 is used
Fes 3
=1~ (1 - Fer/fes)(Fer/e)
Trang 19ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Fg is the Suler column stress for column
with a stable cross-section To find Fr, we project upward from L'/p = 50 to the curve AOC in Fig C7.35 and then horizontally to read Fy = Fe = 41,700 psi Substituting in eq (A), Fa ws 38,200 1 - (1 - 21,500/38,200) (21,500/41,700) ple 2 Fo = 38,200 (1 - 0.225) = 29,500 pst - Example Problem 2 A formed Z section as shown in Fig (b} is used
as a column The length
L = 30", The member is braced in the x-x directioy
thus column bending failure
must occur about x-x axis
Material is aluminum alloy, Fey = 67,000, E_ = 10,500,000 Assume the end fixity co- efficient c = 1.5 Find the column failing stress Solution: Area of Section = 0.117 in.* Oy = 635 in
The initial buckling stress is needed in finding the column strength by = 1.5 + 04 21.46, d¢ = 0.75 - 02 = 0.73 by/Oy = 73/1.46 = 0.5, ty/te = 1.0 From Fig C6.4 of Chapter C6, we read Kw = 2.9 2.9 n® x10,500,000 , 04,2 Fa = ———————-z—— ——-— = or T~ 8U) ‘Frag? * 202900 pst
The crippling stress will be calculated by the Gerard Method using Fig C7.9, which applies for Z shapes The lower scale parameter is,
ì ` ye 67,000 4
Bơm Tö,500,000) “` Ý 15-6 1/3 -
rom Fig C7.9, we read,
Fog/Poy * 455, hence, Fog * 67,000 x 455 = 30,600 psi ` Column Strength by Method 1 L' = LA/e = 30 vĩ.5 = 24.5 L'/p = 24.5/,535 = 45.8 C7 25 Substituting in Johnson-fuler equation, ma 30,600 * ác 892 xoe 2 Fo = 30,600 -7— x 10,500,000 (45+8" =25,860 psi Column Strength by Method 2
On Fig C7.35 point £' is located at
Fo = Feg = 30,600 Point G' is located at
«9 X 20,900 = 18,800 psi The line £'G' is the column curve For L'/p = 45.3, we project up~ ward to line E'G' and then horizontally to left
scale to read Fy = 25,000 psi
Column Strength by Method 3
To find fg, use Fig C7.33 with L'/p = 45.8 and basic Buler curve, gives a value of Fg = 47,000 psi
Substituting in equation (A):—-
nao “ 1 ~ (1 - 20,900/30,600) (20,900/47 ,000)
Fg = 30,600 (1 - 141) = 26,300 pst C7.28 Column Strength of Stiffener
With Effective Sheet
Column members are often attached to thin
sheet by rivets or spot welding If the rivet spacing is such as to prevent tnter-rivet
buckling, then the sheet will assist the stiffener to carry a compressive load and to neglect the
sheet would be too conservative
If the attached sheet 1s relatively thin, that is, less than the stiffener thickness, the method of using the effective sheet width as a part of the column area is widely used by structural designers and will be used in this example solution
Example Problems For an example problem, we will assume that tha 2 stiffener in Problem 2 is one of
Several stiffeners riveted to a sheet of 025 thick- ness and of the same
material as the stiffener
Fig ¢
Solution
The rivet or spot weld spacing is made such as to prevent inter-rivet buckling Thus the column area will be as shown in Fig c, namely, the stiffener area plus the area of
Trang 20CT 26 CRIPPLING STRENGTH OF COMPOSITE SHAPES Al
stress {S a function of the radius of gyration,
the design procedure is of the trial and error category
First Trial
Assume the effective sheet width is based
on column strength of 2 stiffener acting alone The average column failing stress by the 3 methods in the Problem 2 solution was (25,860 + 25,000 + 26,300)/3 = 25,700 pst, The effective width equation to be used is, 1.9 t vE/Fep 1.9 x 025 x 10,500,000/25,700 = 965 in 0.965 x 025 = Effective sheet area 0242 Area of stiffener = Ay = 117 Total area = 1412 the the Adding will change White (Ref effective sheet to the stiffener radius of gyration Mr R J 13) has developed equation 07.33
which gives the variation in the radius of gyration in terms of known variables for any stiffener cross-section Since the failing stress of a column is directly proportional to the radius" of gyration squared, equation C7.33 can be equated to the ratio of the colum
stresses
a ~1#+ 2 GER ¬ e (C7.33)
ND SHEET-STIFFENER PANELS IN COMPRESSION where, O09 radius of gyration of stiffener
alone
9 = radius of gyration of sheet and stiffener
§ = distance from centerline of sheet to neutral axis of stiffener
t = shset thickness w = sheet effective width
Equation C7.33 has been put in curve form
as shown in Fig, 07.36, which will now be used to compute the radius of gyration p, for the
stiffener plus effective sneet 5 „75 + ,0125 = 7625 Po for stiffener alone „555 8/pọ ° 7625/.535 = 1.425 we +965 „2085 ——- wt/Ag = 965 x 025/.117 =
From Fig C7.36 for the above values of S/O, and wt/A,, we obtain
Trang 21ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
The revised effective width based on this
stiffener stress is,
w21.9 x 025 ¥10,500,000/24,500 = 98 in This value is only 02 inch more than the
value of 96 previously, thus the effect on the radius of gyration op will be negligible The column failing stress for the sheet stiffener combination is therefore 24,500 psi, and the compressive failing load would be 24,500 x
„1412 = 3460 lbs
~
C7.29 Sheet-Stiffener Panels With Relatively Heavy
Sheet Thickness „
For 2, Y and hat shaped stiffener-sheet
panels as previously discussed relative to local failing stress, use entire sheet and stiffener area in computing radius of gyration or, in other words, due not use effective sheet widths
but use entire sheet as effective PROBLEMS
Find the crippling stress for the angle
sections of Figs (a) and (b) when formed
from the following materials Use both
Needham and Gerard Methods (1) 2024-T3 aluminum alloy (2) 7075-T6 aluminum
alloy (3) 17-7PH(TH10S0) stainless steel (4) Same as (3) but subjected to an
elevated temperature of 700°F for 1/2 hour (1) duration Ff] ra + Fig b ver + 3 T6 + var 3 16 .072 + —1⁄8 —>‡ Tnạ4 1=
(2) Find the crippling stress for the two channel shapes of Pigs ¢ and d when formed of following materials (1) AISI
4130 steel, heat-treated to Fry =
180,000 (2) Same as (1) but subjected to
an elevated temperature of 500°F for 1/2 duration (3) Ti-8Mn Titanium (4) 7075-T6 aluminum alloy *— 1-3⁄4 ——| F—!-1⁄4—— Tí TRG 7/8 || 16 3⁄4|| TẾ 04 Fig ¢ dou Fig a he 064
(3) Find the crippling stress of the
rectangular tubes in Figs 6 and f when formed from following materials (1) 2024-13 aluminum alloy magnesium (2) HK314-0 (3) Ti-8Mn Titansium ¬ (8) (8) CT.27 (4) Same as (2) but subjected to 300°F for 1/2 hour duration k— 1~1⁄2 —| F— 3 — + lệ || Fg.e ¬le.04 LẺ 08 + +
Find the crippling stress for sections of Figs g and h when formed from 2024-T3 and 7075-T6 aluminum alloy material h— 1 4 3⁄44 3⁄9 05 T Fight 1/8 4 + |-1⁄4 ¬
Find the crippling stress for the extruded sections numbers 4, 8, 16, 22, 28 and 34 as given in Table A3.16 of Chapter A3 Material 2014-T3 aluminum alloy Fig 1 shows a corner member, in a stiffened wing section The skin is fastened to the stiffener by one row of 1/8 diameter rivets at 3/4 inch spacing The web is fastened to stiffener by 2 staggered rows of rivets, with rivet spac-
ing in each of 1-1/8 inch, and rivets ara
of the flat head type Material is 2024-TS alumimm alloy Also use 7075-T6 aluminum material
Find crippling stress for stiffener
Will inter-rivet buckling occur Find effective sheet widths and total failing load unit will carry .035 k— HN 316 te - 7.051 tả 1⁄2 f 1+, 416 Fig i Fe—7/16 eh te OS, hem Web aL
Determine the compressive failing strength of a short Z stiffened sheet panel unit such as that tllustrated in Fig C7.29 The
various dimensions are as follows Refer
to Fig C7.29 for meaning of symbols
tw = 072 bp = 1.0 bs = 2.50
Trang 22C7, 28 (8) {10) (11) (12) (1) (2) (4)
CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION
Stiffener and sheet material is 2024-T3 aluminum alloy Rivets are 1/@ diameter Brazier Head type 2117-TS material and
spaced 7/8 inch
Same as (7) but change material to 7075-T6 aluminum alloy
Same as (7) but
Titantun change material to Ti~8Mn In the Example Problem on a Y stiffened panel as given in Art C7.25, change sheet
thickness from 064 to O81 and calculate
the resulting panel strength
The hat stiffener in Fig g of Problem (4)
is one of several stiffeners riveted to skin of thickness 032 and of the same material as the stiffener If the length of the panel is 20 inches, what will be the column failing stress if end fixity ec #1.5 Also fore = 2.0 Use method involving effective sheet widths
Same as Problem 11 but use Z stiffener of
Fig h of Problem 4
References
Needham, R A.:~ The Ultimate Strength of Aluminum Alloy Formed Structural Shaves
in Compression Jour, Aero Sci Vol 21 April, 1954
Gerard, G.:- Handbook of Structural Stability Part 1V Fatlure of Plates and Composite Elements NACA TN 3784 August, 1957
Gerard, G.:= The Crippling Strength of Compression Elements Jour Aero Sci., Vol 25, Jan 1958,
Coan, J M.:- Large Deflection Theory for Plates with Small Initial Curvature Loaded
in Compression Jr Applied Mech., Vol 18, June 1951 (8) (4) (7) (8) (9) 9) (11) (12) (13)
Windenberg, D Proc Sth Inter-
national Congress for Applied Mechanics
pp 54-61, 1939
Poi
Experimental Study of Deformation and Effective Widths in Axially Loaded Sheet Stringer Panels NACA TN 684, 1949 Gerard, F.:- Effective Width of
Blastically Supported Flat Plates Aero Sei., Vol 13, Oct 1946
Jour
Howland, W.L.:- Sffect of Rivet Spacing on Stiffened Thin Sheet in Compression Jour Aero Sci., Vol 3, Oct 1936 Argyris, J # and Dunne, P.C.:- Structural Principles and Data
of Aero., 4th Edition, 1952 Handbook
Bizlaard, P D and Johnston, G S.:-
Compression Buckling of Plates Due to + Forced Crippling of Stiffeners Preprint No 408 S.M.F Fund Paper Inst of
Aero Sci Jan 1955,
Semenian, J W and Peterson, J P.i- An Analysis of the Stability and Ultimate Compressive Strength of Short Sheet- Stringer Panels with Special Reference to
the Influence of Riveted Connection
Between Sheet and Stringer NACA TN 24351, March, 1955
Gerard, G.:- Handbook of Structural Stability, Part V Compressive Strength
Trang 23
CHAPTER C8
BUCKLING STRENGTH OF MONOCOQUE CYLINDERS
This chapter presents information on the buckling strength of circular cylinders
under compressive, bending and torsional loads acting separately and in com= bination, without and with internal pressure Some information on the buckling strength of conical cylinders is presented
C8.1 Introduction
Before the advent of the high speed air- eraft and particularly the missile and space vehicle, the use of the unstiffened cylinder or monocoque type of structure was quite limited
However, the arrival of the space age has
caused the thin walled cylinder to become im- portant in the design of missiles and space
structures
The classical small deflection theory
which has proved adequate for determining the buckling strength of flat sheet structures as covered in Chapter Al8, has not proved adequate for determining the buckling strength of thin walled cylinders or curved sheet panels since
the theory gives results much too high when compared with the experimental or test results
A more recent large deflection theory has
shown closer agreement with experimental results, however, the theory requires a prior Knowledge of cylinder imperfections As a result of the theoretical limitations to date, ths strength design of such structures is based primarily on best fit curves for experimental or test results, using theoretical parameters and curves when they appear applicable
A missile structure in handling operations and flight maneuvers is subjected to tensile, compressive, bending and torsional load systems The cylinders may be pressurized or
unpressurized The structural design of such structures thus requires a knowledge of the
buckling strength under the various load systems, acting separately and in combination This chapter will be confined to giving test data and design curves for the buckling
strength of cylinders under the various types of stress systems
C8.2 Buckling of Monocoque Circular Cylinders Under Axial Compression
The term monocoque cylinder means a thin walled cylinder without longitudinal skin
stiffeners or transverse intermediate frames
attached to cylinder skin
Investigation by the author of the pro-
cedure used by a number of aerospace companies
-
for the strength design of monocoque structures indicated that the design curves given in
(Ref 1) were widely used It thus seems appropriate to give some design material from
(Ref 1)
Cylinders are usually given four classi-
fications relative to length:-
(1) Short Cylinders Short cylinders tend to behave as flat plate columns (cylinders with infinite radius) They develop buckles in a sinusiodal wave form similar to flat plate buckling The end fixity
has considerable influence
(2) Intermediate or Transition Length of
Cylinders Buckling involves a mixed
pattern combining sinusicdal and diamond shapes The effect of end fixity and cylinder length are only of nominal
importance
(3) Long Cylinders.- Such cylinders buckle in
a diamond Shaped pattern, and the length
and end conditions are not of much in-
portance
(4) Very Long Cylinders Such cylinders buckle by over-all column instability or act as a Euler type column
Fig C8.1 shows a photograph of the buckle failure of an intermediate length cylinder under axial compression
The name of Donnell is prominent in the development of a theory for the buckling strength
of cylinders From (Ref 2), Donnell’s eight- order differential equation can be used to
provide a small-deflection theoretical solution to the behavior of cylinders in the short to
long length range A solution of Donnell’s
Trang 24
BUCKLING STRENGTH OF
Fig C8.1 Type of Failure Axial Compression and No Internal Pressure (Ref 1.) a Fig C8 le Type of Buckling Failure Under Pure Bending Load No Internal Pressure (Ref 9.) MONOCOQUE CYLINDERS Fig C8 la Type of Failure Under Axial Load With Internal Pressure (Ref 1.)
Trang 25ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES L/A A= nalf wave length of buckles in cireumrerential direction The compressive buckling stress is given by; Ko n* 5 te For/n = Bessy mm c9.2
where, t = wall thickness, L is the cylinder length and V is Poisson’s ratio The term 7 is
the plasticity correction factor and equals 1.0
for elastic buckling
Minimization of Eq C3.1 with respect to
the parameter (m* + /°*)*/m*, gives the critical
buckling coefficient for long cylinders in the following form:-
Kg 3 (4/E/n*)Z = 0.7022 - 8.3
where 2 = (L/rE)/1-2? - c8.4 8ubstitution of Eq C8,3 into Eq C3.2 reduces to the well known classical equation
Cs(t/r)
where, Œ = 1//#(1- 3°) =
Tae buckling coefficient for simply supported end conditions in the transition
range can be determined by substituting the tmiting values of Ổ = 0 and m = 1 in aq C8.1
A similar solution for cylinders with fixed edges deviates from the solution for simply supported cylinders only in the flat plate and transition ranges
0.605 for Y= 20
Figs C8.8 to C8.5 inclusive taken from Reference 1, show the theoretical curve which
shows the buckling coefficient as a function of the geometrical parameter 2 Theoretically,
the short cylinder range would occur at Z = 0, The values of Ky of 1.12 and 4.12 for Z=1
correspond quite closely to the buckling coefficients of simply supported anda fixed ended plate colums The long cylinder behavior|
ig represented by 4 45 degree sloping straight
Line portion of the curve The curve con-
ngeting the short and long cylinder ranges ts
rred to as the transition range,
#1zs, 08.2 to $8.5 show the plot of
extensive test data and a $O per cent proba- bility curve derived by the author of (Ref 1)
oy 2 statistical approach Fig 8.6 from
( 1) shows a plot of much test data ona logarithmic chart Kg for as
10,000 A vest 7it curve and SO and 99 per-
cent probability derived curves are also seen as well as the tneoret curve Tor c = 605
C8.3
Observation shows that
are far above the test values Fig C&.7 (from Ref 1} is a set of de curves of Ky
versus Z2 for various r/t valies and for 90
percent probability
the theoretical results
C8,3 Additional Convenient Design Charts for Determining
Compressive Buckling Stress
Figs C8.8a and C8.8b are more convenient
design charts as the buckling stress Fo, can
be read from the chart, thus avoiding the calculations involved in using curves in Fig Câ.7 Fig C8.8a is for 99 percent probability and 95 percent confidence level and Fig C8.38b for 90 percent probability and 95 percent
confidence level The curves are based on tests
of steel and aluminum alloy cylinders only The
accuracy of these curves when used for other
materials has not been substantiated by tests
The 99 percent probability curves are recom-
mended as design allowables for structures whose failure would be highly critical The 90 per-
cent curves can be used for less critical structures
C8.4 Plasticity Correction
The plasticity correction for cylinders in the long range (L#/rt = 100) is given by the
curves in Fig C8.9 taken from (Ref 4) In
general most practical cylinders in aerospace structures will fall in the long cylinder range
Foor
Kee 2 oe vet Fe)
Fig C8.9 (Ret, 4) Nondimensional Buckling Chart for Amally Compressed Long Cireular
Crinders 7) + (Eq'8) [leven < e9/Q ŸV®]
When L?’rt » 100, Applicabie
C8.5 Buckling of Monocoque Circular Cylinders Under Axial
Load and Internal Pressure
eriments conducted many years azo
ly showed that the compressive Duckling
mgth of monocoque cylinders was increased ernal pressure was added to the closed
r Since weight saving is very important n, the use of pressurized