Table B2.2! Design Mechanical and Physical Properties of HK3tA Magnesium Alloy (Sheet, Plate and Sand Castings) Alloy 0.020 ằ nề HK31A Form co che co Sheet « and plate « Sand castings * Condition -0 -H24 -T6 Thickness, Ín 0.016- 0.261— | 0.601- | 1.001- 0.016- 0.126- 0.251- | 0.501- | 1.001- | 2.001- 0.250 0.500 1.000 3.000 0.125 0.250 0.500 1.000 2.000 38.000 | Đnsis A B A B A B oO 30 32 30 30 29 34 36 34 35 34 18 19 16 15 14 26 28 23 24 21 12 13 10 10 10 20 21 19 22 18 22 22 22 32| 23 23 23 23 23 43 46 49 52 49 5 49 61 54 57 60 57 58 57 24 25 34 35 33 36 BL pe eee 24 25 34 35 33 36 BL] Pe Pe pe ee “ ắ 122] 20 12 12 12 4 6 4 8 4 10 10 8 4 &, 10* pai 6.5 Et, 10° pai 6.5 24 0.0647 C, BTU/(ib.) (F) 0.25 (32° to 212° F) K, BTU/(br.} (ft) (P) ft 60.0 (at 68° F) «, 10-*in./in./EF 15 (68° to 392° F)
* Properties for sheet and platy are taken parallel {o the direction of rolling ‘Tranaverse propertles * Mochanical properties are based upon the guerantecd tensile properties from sepurately-cast
aro equal to or greater than the longitudluel properties
* Reference should be made ta the specific requirements of the procuring or certificating agency with regard (o the use of the above values in the design of castings
Trang 2B2.30 HKGLA MAGNESIUM ALLOY (SHE! § T NHperoire Eapasure up to 1000 Fr 8 & Per Cent Fy at Room Temperature bì § ° lO Ø0 XỐO 420 5G 600 700 áoo Temoarotura, F
Fig B2.90 Effect of temperature on the ultimate tensile strength (Fry) of HK31A-H24 magnesium allay ‘Strangin af temoercture Exposure up 1o 000 nr x Per Cent Fy, 0! Room Temparaiwe 8 ° oO 2 300 400 500 600 T00 800 Temperature, F
Fig B2.91 Effect of temperature on the tensile yield strength (Fry) of HK31A-H24 magnesium alloy 8 8 Siangfh ơt room renpeeenxei Espoeue 100 12 1000 be at Room Tempacature 8 & “ Par Cont F 3 ° 70 SỐ 400 9900 600 700 A0 Tempergrure, F
Fig 52.92 Effect of exposure at elevated temperatures on the room-temperature ultimate tensile strength (Fy,) of HK31A-H24 magnesium alloy
MECHANICAL AND PHYSICAL PROPERTIES OF METALLIC MATERIALS FOR FLIGHT VEHICLE STRUCTURES » PLATZ & SAND CASTINGS) (Cont.) 01 Room Temperature Per Cam £ 0 I0 20 SG 40 502 600 700 200 TemGerature, F
Fig B2.93 Effect of exposure at elevated temperatures on the room-temperature tensile yield strength (Fty) of HK31A-H24 magnesium alloy 8 Strength at tenperctune Expeaue up 10 1000 Ww 8 ề Per Cant Fe, at Room Temperature 9 &
° 200) Temoergture, F 390 400 34G &oOO 700 Sọo
Fig B2.94 Effect of temperature on the ultimate tensile strength (Fry) of HK31A-T6 magnesium alloy (sand casting) 100 § Strengin at temperonore 8 3% Excosue uo 10 1000 Ễ ge é ® ao gx é n > 4 1000 her Ề i ° 0 10 20 %G 4200 500 600 700 S0U Temoerature, F
Trang 3ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES B2 31
AZ61A, AZGZA, AZ8OA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS)
Table B2.22 Design Mechanical and Physical Properties of AZ61A7 Magnesium Alloy (Extrusions and Forgings) Allay AZ61A
Extruded bar, rod, and Extruded Extruded
Ÿorm solid shapes Hollow shapes Tubes Forging Condition -F Thickness,in <0.249 0.250-2.400 0.028-0.750 Basis Mechanical properties: Pray ksi 38 39 36 36 38 2 24 16 16 2 14 14 11 11 14 19 19 : 19 45 45 50 55 55 60 23 + 23 (e/D =2.0) 32 32 " 32 €, percent 8 9 7 7 6 E, 108 pst 6.3 E., 108 pat - 83 G, 10t p8i 24 Physical properties: w, tb/int 0.0647 C, Btu/(ib)(F) K, Bea/{(hr) fey (P)/ft] 46 (212° to 572°F) 0.25 (at 78°F)* a, lOCin/im/Foo 24 (65° to 212°F)
4 Properties or extruded bars, rods, shapes, tubes, and forgings are
Trang 4
B2.32_ MECHANICAL AND PHYSICAL PROPERTIES OF METALLIC MATERIALS FOR FLIGHT VEHICLE STRUCTURES
AZ61A, AZ6ZA, AZSOA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS) (Cont.}
Trang 5
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES B2 33
AZG1A, AZ6ZA, AZSOA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS) (Cont.)
Table B2.24 Material Specifications for AZ63A Magnesium Alloy Specification Type of product QQ-M-56 Sand castings Permanent-mold castings
TABLE 4.2.3.0(b) Design Mechanical and Physical Properties of AZ63A Magnesiums Alloy (Castings) Allay ÀZ63A
Sand and permanent-
Form, - mold castings Condition - -F | -T4 | -TS | -T6 Thickness (in.) BasisÈ Mechanical properties: Pu kết 10 1 18 10 "1 16 17 | 19 36 50 50 65 Fey, kat (e/D=1.8) 28 | 32 36 (e/D =2.0) 340 | 36 | | 45 #, percent 4 7 2 3 EB, 10¢ pi 6.5 E, 10¢ psi 6.5 G, 10* pat 24 AeTensila stresa-strain B*Comoreasiwe streas-stroin C= Tenwle tangent: modulus OsComoresave tengent- modula ° * a e a @ toán, QOOI infin Tangent Modulus, 0" pa
Fig B2.98 Typical stress-strain and tangent~modulus curves for AZ63A-T4 magnesium alloy (sand casting) at room temperature
ArTansile stress-strain QsCompressive stress-strain Ce Tensile tangent modulus OsComprensive tangent modulus
9 z ry ` ° 0
Swein, GOOI 1n Tangent Modulus, 10° pa
Trang 6Table B2.25 Design Mechanical and Physical Properties of 8Mn Titanium Alloy Alloy Form Condition Thickness, in 8Mn Sheet, plate, and strip Auncaled Basis A Mechanica! properties: Fry, ksi + 120 Tv 120 Fiy, ksi + t0 Tv 110 Bey, ksi L 120 T 110 Fy, kst 84 Porwy ket 170 130 #, percent 10 E, 106 psi 15.5 tk, 108 psi 16.0 G, We psi Physical properties: w, Ib/in.a 0171 €, Htuu/(b)(E) 8.118 (at 68°F) K, Btu/[(br) (fu) (F)/ft] 6.3 œ 10-*in,/in./E 4.8 (at 200°F) Strength at tempecoture Eaposure up to 1000 ty 00-800 Tamperaiure, F
Fig B2,98 Effect of temperature on the ulttmate tensile strength (Fty) of BMn annealed titanium
alloy Exposure up to 000 br Strength of temperature Per Cant Fry at Room Temperature Temperature, F
Fig B2.99 Effect of temperature on (he tensile
yield strength (Fty) of 8 Mn annealed titaniuin alloy Strength ot lemperctive Exposure up to 1000 he 200 400 s00 óc Temperature, F Fig B2.100 Effect of temperature on the compressive yleld strength (Puy) of 8 Mu annealed titanium alloy ‘Strength af temperature Exposure up 10 1000 he Per Cent Fy, ot Room Temperature Temperature, F
Fig B2,101 Effect of temperature on the ultimate shear strength (Fgy) of 8 Mn annealed tilanium alloy Exposure upto thr Per Cent Fat Room Temperature 200 S00 600.800.1000 Temperature, F
Trang 7
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 52 35
6A1-4V TITANIUM ALLOY (BAR & SHEET)
Table BZ.26 Design Mechanical and Physical Properties of 6Al-4V Titanium Alloy Alloy 6Al4V Form Bar Sheet Condition Annealed Thickness,in 215 <0.187 Đass A A Mechanical properties: Fry, ksi 130 130 130 120 120 120 126 126 " 126 80 76 Porw ksi (e/D=1.5) 196 191 (e/D=2.0) 248 244 Pony, kat (/D=1.5) 174 163 (e/D=2.0) 205 198 ¢, percent 10 10 5, 10% pai + 16.0 15.4 16.4 18.4 18.0 16.9 6.2 Physical properties: œ, lb/in.! 0.160 €, Btu/(b)(Œ) sẻ 0.135 (at 68°F) Ä, Bu/[(hr)Œ9)/R] 3.8 (at 63°F) s, 10““in./in./E 4.6 (at 200°F) T ROOM TEMPERATURE ~ 9 “ mẽ n0 Temparatare, €
Fig B2 104 Effect of temperature on Fry, Fry and E of 6Al-4V annealed titanium alloy (sheet and bar)
Strengin ot temperature roesure we ms † M Fy a} Boom Temperature Per Cont Temgercnee, F
Fig B2.105 Effect of temperature on the compressive yield strength (Fey) of 6Al-4V annealed titanium alloy (sheet and bar) Gt Room Temperature 8 Per Cont Fy, 3 Temperature, F
Fig Ba 106 Effect of temperature on the ultimate bearing strength (Fpry} of 6AI-4V annealed titanium alloy (sheet and bar)
Longuwvanat Strength at temperature Espoture vụ tø ‡ he Per Cot Fy, af Room Temperatuce Temperature, F
Fig B2.107 Effect of temperature on the ultimate shear strength (Fsu) of 6Al-4V annealed titanium alloy (sheet
Trang 8B2 36
INCONEL X NICKEL ALLOY (SHEET) Table B2.27 Design Mechanical and Physical Properties
of Inconel X Nickel Alloy Tnconei X Sheet Precipitation heat-treated Mechanical properties: Pry, kat b 183 T 155 Pry kei 100 100 103 105 108 (e/D =2.0) 286 Pry kat (e/D =1.8) - (e/D =2.0) 186 20 31.0 31.0 Physical properties: o, Ib/int 0.304 C, Btu/(b)(F) 0.109 K, Beu/{ hr) (ftv) tt] a, 10-*in./in./F 8.7 (80° to 212°F) 6.4 (100° to 200°F) Pe Cant € ed Eg ot Roan Nepeiee 8 0 ee Terowenee,
Fig B2,112 Effect of temperature on the tensile modulus (E) of Inconel X nickel alloy
Fig B2.113 Effect of temperature on the ultimate bearing strength (Fpru) of precipitation heat treated Inconel X nickel alloy MECHANICAL AND PHYSICAL PROPERTIES OF METALLIC MATERIALS FOR FLIGHT VEHICLE STRUCTURES Pw Cand Fu, at Room Tengevelee gS) a aa 8m Tenoeene,£
Fig, B2.108 Effect of temperature on the ultimate tensile strength (Ftu) of precipitation heat treated Inconel X nickei alloy Pee Cont Fy, of Roam Reoeee
Fig B2.109 Effect of temperature on the tensile yield strength (Fry) of precipitation heat treated Inconel X nickel alloy Fee Cane Fo Boom
Fig B2.110 Effect of temperature on the compressive yield strength (Fey) of precipitation heat treated Inconel X nickel alloy Pe Cont Ff Foam Yepedue
Fig, B2.111 Effect of temperature on
the ultimate shear strength (Fgy) of
precipitation heat treated Inconel X nickel alloy,
Trang 9
PART C
PRACTICAL STRENGTH ANALYSIS & DESIGN OF STRUCTURAL COMPONENTS
CHAPTER C1
COMBINED STRESSES THEORY OF YIELD AND ULTIMATE FAILURE
1,1 Uniform Stress Condition
Aircraft structures are subjected to many
types of external loadings These loads often
cause axial, bending and shearing stresses
acting simultaneously If structures are to be
designed satisfactorily, combined stress re- lationships must be known Although in practi-
cal structures uniform stress distribution is not common, still sufficient accuracy for
design practice is provided by using the stress relationships based on uniform stress assump-
tions In deriving these stress relationships,
the Greek letter signa (a) will represent a
stress intensity normal to the surface and thus
a tensile or compressive stress and the Greek letter tau (t) will represent a stress intensity parallel to the surface and thus 4 shearing stress 1,2 Shearing Stresses on Planes at Right Angles, 21 Fig C11
Fig Cl.1 shows a circular solid shaft subjected to a torsional moment The portion
{A) of the shaft exerts a shearing stress Ty 0n section (1-1) and portion (B) exerts a resist-
ing shearing stress tz on section (2-2) Fig Cl.2 illustrates a differential cube cut fram
shaft between sections (1-1) and (2-2) For
equilibrium a resisting couple must exist on
top and bottom face of cube Taking moments
about lower left edge of cube: Ty dxdy (dz) - Tz dzdy (dx) = 0
hence, Ty = Tz (1)
Thus if a shearing unit stress occurs on one plane at a point in 4 bedy, a shearing unit stress cf same intensity exists on planes at right angles to the first plane
Cl.3 Simple Shear Produces Tensile and Compressive Stresses,
Fig Cl.3 shows an elementary dlocx of
unit dimensions subjected to pure shearing T1 T1 Pad af | ST, +1 P Fig C13 stresses
Fig Cl.4 shows a free body after the block has been cut along a diagonal section
For equilibrium the sum of the forces along the x~x axis equals zero
ar, = - 20) + 2 (x1 cos 45%) = 0
9 9
hence, ơ = 8 (+ 1 cos 4s ) cos 45° _ ~~~ (8) Therefore when a point in a body is sub- jected to pure shear stresses of intensity 1+;
normal stresses of the same intensity as the
shear stresses are produced on a plane at 45 with the shearing planes
C1.4 Principal Stresses
For a body subjected to any combination of stresses 3 mutually perpendicular planes can be
found on which the shear stresses are zero The
normal stresses on these planes of zero shear stress are referred to as principal stresses C1.5 Shearing Stresses Resulting From Principal
Stresses
In Fig C1.5 the differential block 1s subjected to tensile principal stresses Gy and
Gg and zero principal stress o, The block is
cut along a diagonal section giving the free
body of Fig Cl.6 The stresses on the diagonal
section have been resolved into stress compon- ents parallel and normal to the section as shown
For equilibrium the summation of the stresses along the axes (1-1) and (2-2) must equal zero
Gy dudy - oy dzdy cos © - og dydx sin @ = 0,
Trang 10C1.2 [—~Øx dzdy pe § quay = sin @, whence Im = Oy cos*9 + og sin*?Q -
The normal stress cy at a point is always less than the maximum principal stress oy or Oz at the point TPae = 0 — + dudy - cy dzdy sin 6 + og dydx cos 9 = 0 + S23v dydx _ But quay = c08 @ and -—^ quay sin 9 1
hence, T= gg sin 6 cos @ - oy cos 9 sin @
or, 7: log - Ox) sin 9 cos 9
(1/2) (og - oy) sin 2 0, where og is maximum principal stress and oy is minimum
principal stress
or, +
Since sin 2 9 1s maximum when @ = 45°,
tH (OW od, /2 - - - ee (3)
Stated in words, the maximum value of the shear- ing unit stress at a point in a stressed body ig one-half the algebraic differences of the maximum and minimum principal unit stresses
COMBINED STRESSES, THEORY OF YIELD AND ULTIMATE FAILURE C1.6 Combined Stress Equations
Fig Cl.7 shows a differential block sub-
jected to normal stresses on two planes at right
angles to each other and with shearing forces on
the same planes The maximum normal and shear-
ing unit stresses will be determined
Pig Cl1.8 shows a free body dlagram of a portion cut by a diagonal plane at angle 6 as shown AD dydx Fig C1,8
For equilibrium the sum of the forces in the z and x directions must equal zero, IFy = 0 On dudy cos 9 + v dudy sin 9 = oy dady - Txzdxdy ZQOe -+-+ - 2 eee (4) 7 aF, = 0 on dudy sin 9 - t dudy cos @ - og dydx - #xz3zdy FO - ee - ee ee ee ee ee (8) By dividing each equation by du and noting that dzdy dudy dydx „ dudy
= cos 9 and = gin ©, we obtain:
Trang 11ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES (a + oz) sin 9 - (T- Tyg) cos 9=O - - -(7)
The maximum normal stress gy will be maximum when @ equals such angle @' as to make t = zero
Thus 1f T= 0 amd @ = 9’ tn equations (6) and
(7), we obtain,
" oO
(Gn - Gx) cos @' - Tyz Sin ©@' " a (on - Gz) sin @' - ty, cos @!
In equations (8) and (9) oy represents he
principal stress Dividing one equation by
another to eliminate 9',
On Tez - Sx = _*X2 im - Op whence,
On* ~ (Ox - 0g) 1 + OZ BZ %_*, oF
Ox + Oy tA (E22) at 2) + tu = - (10)
In equation (10), tensile normal stress is
plus and compression minus, For maximum oy use
plus sign before radical and minus sign for minimum oy
Ơn =
To find the plane of the principal stress- es, the value of 6' may be solved for from
equations (8) and (9), which gives:
9' is measured from the plane of the largest normal stress oy or gz The direction of rotation of 9' from this plane is best de- termined by inspection Thus if only the shearing stresses +xz were acting, the maximum principal stress would be one of the 45° planes, the particular 45° plane being easily deter~ mined by inspection of the sense of the shear
stresses Furthermore if only the largest normal stress were acting it would be the maxi~ mum principal stress and 9' would equal zero Thus if both o and tT act, the plane of the principal stress will be between the plane on
which o acts and the 45° plane As stated before o refers to either oy or o, whichever
is the largest
Maximum Value of Shearing Stress (Tmạy,)
The maximum value of t from equation (3)
equals,
“max Ð (Ơn(max,) ~ nian ))/?
Substituting the maximum and minimum values of Gy from (10) tn (12), we obtain maximum shear-
ing stress as follows: Sx Sn e+ Tez" ~ ¬ (15) Tmax = 2 C1.7 Mohr's Circle for Determination of Principal Stresses
It 1s sometimes convenient to solve
graphically for the principal stresses and the maximum shear stress Mohr's circle furnishes
a graphical solution (Fig Cl.9a}) In the
Mohr method, two rectangular axes x and 2 are chosen to represent the normal and shearing
stresses respectively Taking point 0 as the
origin lay off to scale the normal stresses oy
and gz equal to OB and OA respectively If ten~
gion, they are laid off to right of point O and
to the left if compression From B the shear
stress Tyg 1s laid off parallel to Og and with
the sense of the Shear stress on the face DC of
Fig Cl.9b, thus locating point C With point
E the midpoint of AB as the center and with radius EC describe a circle cutting OB at F and
G AD will equal BC and will represent the
shear on face AB of Fib b It can be proven that OF and 0G are the principal stresses Cmax, and Ggin respectively and EC 1s the maximum
shear stress Tmgy, The principal stresses occur
on planes that are parallel to CF and ca (See
Figs c and d) The maximum shear stress occurs
on two sections parallel to CH and CI where HEI
1s perpendicular to OB If oy should equal zero
then O would coincide with a 2 # !n(max) Ble aif Boy = (b) Oz oz L () a se Đn(min) (4) %E - 2 On+5 ox iets _ tx] ? 2 oz Ox 5 (a) Flg C1.9 C1.8 Components of Stress From Principal Stresses by Mohr's Circle,
In certain problems the principal stresses may be known as in Fig Cl.9 and it is desired
to find the stress components on other planes
designated by angle 9 In Fig Cl.11 the axes
x and 2 represent the normal and shear stresses
respectively The principal stresses are laid
off to scale on ox giving points D and § respec~
tively Construct a circle with A the midpoint
Trang 12C1.4
to 26 It can be proven that OB represents the
normal stress on the plane defc of Fig Cl.10, and CB represents the shear stress + on this plane Ox | Oz, Ox are principal Fig C1.10 stresses Gy, on plane (defc) z #fủg C1.11 C1.9 Example Problems Example Problem 1
The maximum normal and shear stresses will be determined for the block loaded as shown in
Fig Cl.12
The graphical solution making use of Mohr's circle is shown in Fig C1.13 From reference axes x and 2 thru point 0, the given normal stresses ¢x = 10000 is laid off to scale on ox
and toward the right giving point B From B the
shear stress Tyz = 5000 1s laid off parallel to
oz to locate point C With = the midpoint of
OB as the center of the circle and with radius EC a circle is drawn which cuts the Ox axis at
# and G The maximum and minimum princtpal
stresses are then equal to oF and oG which
equals 12070 and -2070 respectively The maxi-
Mum shear stress equals EC or 7070
Algebraic Solution: From eq (10), (5% — " + Ty" COMBINED STRESSES THEORY OF YIELD AND ULTIMATE FAILURE “ạyz = 6000 y x ei ox = 19000 | | —-3 x = 10000 Ị " x về Ftg C1.12 Txz = ô000 z “xz = 5000 x = x | Fig, C1 13 | [ra ox= 10000 ¬ Ởn(min) % “TC OSn(max)=1207 =-2070 7 2 Substituting values, 5 ^/ (9) + 50007 = 5000 5000 + 7070 = 12070 psi = 5000 ~ 7070 =-2070 psi ont Looe +90 + + 7070 hence, on (max ) = Sn(mn, ) (1/2) (On mare ) Tmax ~Onmin,)) (Ref -Ba.22) (1/2) (12070 ~ (-2070}) = 7070 psi max can also be computed by equation (13), whence, z imax, 22 (22009 + 3) + 50007 = 2 7070 pst tan se" = 2 Tua 2 2X 5000 2, 5y 9z 10000 - 0 hence, Q® = 22.5, Example Problem 2
The maximum normal and shear stresses will
Trang 13ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Flg Cl.15 shows the graphical solution
using Mehr's circle From point 0, ox = 10000
and dz = -20000 are laid off equal to OB and OA respectively." Tyg equal to 12000 is laid off
parallel to 0Z at A locating C With = the
midpoint of AB as the center of a circle of radius EC a circle 1s drawn which cuts the ox
axis at PF and D The maximum normal and shear
stresses are indicated on the figure Zz Fig C1.15 tr x L Oz = 20000 ——-«0y=10000 Ơn(max)^ ^24200—†~Øn(mịnƑ 1430 Algebraic Solution 5 ơn = #47 9z + (z2) + typ? z = 10000 ; 20000 + (¿29 -222)) + l2000* = ~ 5000 = 19200 nence, On max.) = -5000 ~ 19200 = -24200 pst 'n(mtn,) = ~S000 + 19200 = 14200 2 x 12000 = “max, = + 19200, tan 20! 1000 = (-20000) ~ * 8 eo" = 18° - 50"
C1.10 Triaxial or Three Dimensionai Stresses
For bodies which are stressed in three
directions, the state of stress can be defined completely by the six stress components as
illustrated in Fig Cl.16 Using the same
procedure as was carried out for a two-dimen- sional stress system, it can be shown that there are three principal stresses o,, o, and os,
whose values are the three roots of o in the
following cubic equation C1,5 o = (ay + dy + Og)07 + (ayoy + Gye Ỳ z ¥ 2 + YG Zz 2 a —Yyz” “Yxy }ø~ (đxØvy9z + 9YyzYxzYxy ~ xX yz" - ØyYxz” ~ ØzYxy” = 0 y Fig C1 16
Fig C1.17 shows the principal stress system which replaces the system of Fig Cl.16
It can be shown that the maximum shear stress
max, 18 one of the following values etl ymax = 75 {o1 ~ Øa) + or’ max ~ i+ |e pole or Y max, =
The planes on which these shear stresses act are indicated by the dashed lines in Fig
Cl.18, namely, adhe, ddge and dcef The
largest of the shear stresses in equations (15) depends on the magnitude and signs of the principal stresses, remembering that tension is plus and compression 1S minus when making the substitution in equations (15) 1s Lk % đa Lo & % os Fig C1.18 Fig C1.17 Cl.11 Principal Strains
The strains under combined stresses are usually expressed as strains in the direction
of the principal stresses Consider a case of
simple tension as illustrated in Fig Cl.19, The stress o, causes a lengthening unit strain é€ in the direction of the stress o,, anda
Shortening unit strain e«’ in a direction at
right angles to the StTreSS Ơi,
The ratio of e’ toe is called Poisson's
ratio and is usually given the symbol » Thus,
h=e'/e
Trang 14
C1.6 COMBINED STRESSES THEORY OF YIELD AND ULTIMATE FAILURE
a to The strain energy can be expressed in
2 weer TT terms of stress by substituting values of « in
it ! terms of o from equations (17) into equation
| ' Øa ~—]| (19), which g1ves,
efi thet 1
Fg j7? Os ws U =zc Ol + O8+ 03 - 2u(cida + Gedy + gar)
` TT Fig c1.19 Ƒ ig ee 20 (20)
Since e¢ = o,/£, we obtain, For a two dimensional stress system,
og, = 0 and equation (20) becomes e'2pa/E - (18)
1 a = ẹ
= - + o2) + 2
Now consider the cubical element in Fig U = ay (ol - 24 oa, + 98) (21)
C1.20 subjected to the three principal stresses
O,, O, and o,, all being tension The total
unit strain e, in the direction of stress o,
will be expressed Obviously, a, tends to
stretch the element in the direction of o, whereas stresses go, and o, tend to shorten the element in the direction of a,, hence, - oe » whence a, =B - Ko, + 0y) and similarly for e, and ¢,, Ea = B - Ho, +o) fs = 3 - Ho +a.)
For a two-dimensional stress system, that is, stresses acting in one plane, Øạ = 0 and the principal strains become,
es =H (% ~ po)
tte l-u,) $F - (18)
ts = Elo, +g)
Equations 17 and 16 give the strains when
all the principal stresses are tensile stresses |
For compressive principal stresses use a minus sign when substituting the principal stresses
in the equations
Ci, 12 Elastic Strain Energy
The strain energy in the elastic range for the unit cube in Fig Cl.20 when subjected ta combined stresses 1s equal to the work done by the three gradually applied principal stresses
A, Og and a, These stresses produce strains
equal to e,, 6, and e, and thus the work done
per unit volume equais the strain energy Thus
if U equals the strain energy, we obtain, U «Sm + Saxe + Sả» wwe eee (19)
Factor of Safety
C1.13 Structural Design Philosophy Limit and Ultimate Loads Factors of Safety Margin of Safety The basic philosophy governing the structural design of a flight vehicle 1s to develop an adequate light weight structure that will permit the vehicle to accomplish the operations or missions that were established
as design requirements The job of a
commercial airliner is to carry passengers and
cargo from place to place at the lowest cost To carry out this job a certain amount of flight and ground maneuvering is required and
the loads due to these maneuvers must be
carried safely and efficiently by the structure A military fighter airplane must be maneuvered
in flight far more severely to accomplish its
desired Job as compared to the commercial air-
liner, thus the flight acceleration factors for the military fighter airplane will be considerably higher than that of the airliner In other words, every type of flight vehicle will undergo a different loan environment,
which may be repeated frequently or infre-
quently during the life of the vehicle The
load environment may involve many factors such
as flight maneuvering loads, air gust loads, take off and landing loads, repeated loads, high and low temperature conditions, etc
Limit Loads Limit loads are the calculated
maximum loads which may be subjected to the flight vehicle in carrying out the job it 1s designed to accomplish during its life time
of use The term limit was no doubt chosen
because every flight vehicle is limited relative to the extent of its operations A flight vehicle could easily be designed for loads greater than the limit loads, but such extra strength which is not necessary for safety would only increase the weight of the structure and decrease the commercial or military payload or in general be detrimental to the design
Factor of safety can be
defined as the ratio considered in structural
Trang 15ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Yield Factor of Safety This term ts defined
as the ratio of the yield strength of the
structure to the limit load
Ultimate Factor of Safety This term is
defined as the ratio of the ultimate strength
of the structure to the limit load
Yield Load This term is defined as the limit
load multiplied by the yield factor of safety
Ultimate Load This term can be defined as th
limit load multiplied by the ultimate factor
of safaty This resulting load is often
referred to by engineers as the design load, xhlich 1s misleading because the flight vehicle structure must be designed to satisfy doth yield and ultimate failure and either one may
be critical
Yield Margin of Safety This term usually
expressed in percent represents the additional yield strength of the structure over that strength required to carry the limit loads
Yield Strength -
Yield Margin of Safety = Limit Load 1
Ultimate Margin of Safety This term usually
expressed in percent represents the additional ultimate strength of the structure over that strength required to carry the ultimate loads
Ultimate Strength
Ultimate Load
Ultimate Margin of Safety =
C1,14 Required Strength of Flight Structures Under Limit Loads:-
The flight vehicle structure shall be designed to have sufficient strength to carry
simultaneously the limit loads and other
accompanying environmental phenomena for each
design condition without undergoing excessive
elastic or plastic deformation Since most
materials have no definite yield stress, it 18
common practice to use the unit stress where a 002 inches per inch permanent set exists as the yleld strength of the material, and in general this yleld strength stress can be used
as the maximm stress under the limit loads
unless definitely otherwise specified
Under Ultimate Loads:-
The flight vehicle structure shall be designed to withstand simultaneously the ultimate loads and other accompanying environ-
mental phenomena without failure In general
no factor of safety 1s applied to the environ- mental phenomena but only to the limit loads
Failure of a Structure:-
This term in general refers to a state or
condition of the structure which renders it
@
1
CLT incapable of performing its required function Failure may be due to rupture or collapse or
due to excessive deflection or distortion
C1.15 Determination of the Ultimate Strength of a Structural Member Under a Combined Load System Stress Ratio-Interaction Curve Method
Since the structural designer of flight vehicles must insure that the ultimate loads
can be carried by the structure without failure, it is necessary that reliable methods be used to determine the ultimate strength of a
structure Structural theory as developed to
date is in general sufficiently developed to accurately determine the ultimate strength of a structural member under a single type of loading, such as axial tension or compression,
pure bending or pure torsion However, many
of the members which compose the structure of a flight vehicle are subjected simultaneously to various combinations of axial, bending and
torsional load systems and thus a method must
be available to determine the ultimate strength of a structure under combined load systems A strictly theoretical approach appears too
difficult for solution since failure may be due to overall elastic or inelastic buckling,
or the local elastic or inelastic instability The most satisfactory method developed to date is the so-called stress ratio, inter- action curve method, originally developed and
presented by Shanley In this method the
stress conditions on the structure are repre-
sented by stress ratios, which can be con-
sidered as non-dimentional coefficients denoting the fraction of the allowable stress
or strength for the member which can be
developed under the given conditions of com- bined loading
For a single simple stress, the stress
ratio can be expressed as, R = stress ratio “‡
where f 1s the applled stress and F the
allowable stress The margin of safety in
terms of the stress ratio R can be written,
Load ratios can be used instead of stress ratios and is often more convenient
For example for axial loading,
Trang 16C18
For pure bending,
R=M/M, , where M = applied bending
moment and M, the allowaole bending moment
For pure torsion, R= T/T
moment and Tq the allowable torsional moment » Where T is applied torsional
For combined loadings the general conditions for failure are expressed by Shanley as follows:-
RỆ + RỄ + RZ + ———— = 1.0 ~ ¬ - -(24)
In this above expression, R,, R, and Ry could refer to compression, bending and shear and the exponents x, y, and z give the
relationship for combined stresses The
equation states that the failure of 4
structural member under a combined loading will result only when the sum of the stress ratios is equal to or greater than 1.0
For some of the simpler combined load systems, the exponents of the stress ratios in equation (24) can be determined by the various well known theories of yield and failure that have been developed However, in many cases of combined loading and for particular types of structures the exponents in equation (24) must be determined by making actual failure tests of combined load systems
Since the stress ratio method was pre- sented by Shanley many years ago, much testing
has been done and as a result reliable inter-
action equations with known exponents have been obtained for many types of structural
members under the various combined load
systems Ina number of the following chapters,
the interaction equations which apply will be used in determining the ultimate strength design of structural members
C1, 16 Determination of Yield Strength of a Structural Member Under a Combined Lead System AS explained in Art C1.14, the flight vehicle structure must carry the limit loads without ylelding, which in general means the yleld strength of the material cannot be exceeded when the structure is subjected to
the limit loads In some parts of a flight
vehicle structure involving compact unit or Pressure vessels, biaxial or triaxial stress conditions are often produced and 1t 1S
necessary to determine whether any yielding will occur under such combined stress action
when carrying the limit loads For cases where no elastic instability occurs, the
following well known theories of failure have
been developed
COMBINED STRESSES, THEORY OF YIELD AND ULTIMATE FAILURE
i Maximum Principal Stress Theory
2, Maximum Shearing Stress Theory
3 Maximum Strain Theory
4 Total Strain Energy Theory
5 Strain Energy of Distortion Theory
6 Octahedral Shear Stress Theory
The reader may review the explanation and
derivation of these 6 theories by referring
to such books as listed at the end of this
chapter
Test results indicate that the yield strength at a point in a stressed structure is more accurately defined by theories 5 and
6 followed in turn by theory 2 Since
theories 5 and 6 give the same result, they might be considered as the same general theory
In this chapter we will only give the resulting equations as derived by theory 6, since
theories 5 and 6 appear to be the theories used in flight vehicle structural design
C1.17 The Octahedral Shear Stress Theory
Since this theory gives the same results as the well known energy of distortion method
it is often referred to as the Equivalent
Stress Theory The octahedral shear stress
theory may be stated as follows:~ In elastic
action at any point in a body under combined stress action begins only when the octahedral shearing stress becomes équal to 0.47 fe,
where fg is the tensile elastic strength of the material as determined from a standard tension test Since the elastic tensile
strength {s somewhat indefinite, it 1s common
practice to use the engineering yield strength
Fey: In this theory it 1s assumed that the
tensile and compressive yield strengths are
the same, -
Figs Cl.21 and Cl.22 illustrate the conditions of equilibrium involving the octa-
hedral shear stress In Fig Cl.2l1, the cube
is subjected to the 3 principal stresses as shown A tetrahedron 1s cut from the cube
Trang 17ANALYSIS AND DESIGN OF F
principal axes, while the normal to the fourth
sids makes equal angles with the principal
axes The octahedral shear and normal stresses
are the resulting stresses on the fourth side The equation for the value of the normal octahedral stress is,
foot = (fa + fo + f2)
The equation for the octahedral shear
stress is,
Segoe TEV Farha) # (fafa) (Panta)
Now the octahedral shear stress 1s 0.47 of the normal stress
- -(26)
Let Ÿ be the effective axial stress in uniaxial tension or compression which results in the given octahedral shear stress
3 ft * fScct /0.47 2S VE ‘Sect
Therefore multiplying Eq (26) by
3//2 we obtain for a condition of principal triaxial stresses,
Bye Vint ete tet) - ~~ -(28)
Let F equal the allowable tensile or
compressive stress If the yield strength is
being determined then " Margin of Safety M.S.=>-1 - - -(29) nw For a biaxial stress system taking f, =0, we odtain, Fevtie+ti-tf, - (30)
It 1s often more convenient to use the x, y
and z component of stresses instead of the
principal stresses Fig €1.23 111ustrates
the various component stresses tz fsx Eszy f Sxz ty; f IS nh | ; Sxy J om Fig Cl 23
IGHT VEHICLE STRUCTURES C19
For a triaxial stress system, 1 : z =3 tx-fa) +(ta-ty) +(fy~†x) lon - ee eee ~(#1) For a biaxial stress system, fy, ts fs, 20 va ‡# = V/V tt tế 4 ta a -— -í ft fx + f2 ¬ Íyf„ + Say (32) Cl.18 Example Problem 1
A cylindrical stiffened thin sheet fuse-
lage is fabricated from 2024 aluminum alloy
sheet which has a tensile yleld stress
Pry = 40000, Find the yield margin of safety
under the following limit load conditions
(1) A limit bending moment produces 2 bending
stress of 37000 psi (tension) at top
point of fuselage section The flexural
shear stress is zero at this point Same as condition (1) but pressurization of fuselage produces a circumferential tension stress of 8600 psi and a longi- tudinal tension stress of 4300 psi Same as condition (2) but a yawing maneuver of airplane produces a limit torsional shearing stress of 8000 psi in fuselage skin
SOLUTION: Condition (1)
This is a uniaxial stress condition for point being considered
Pry = 40000
ft
Yield M.S = = 37000 ~ 1= 08 SOLUTION: Condition (2)
There are no flexural shear stresses at
the fuselage point being considered Since
no torsion is being applied to fuselage no torsional
Shear stresses exist The |
stress system at the point being considered is thus a
diaxial stress system and
Trang 18€C1.10 COMBINED STRESSES THEORY OF YIELD AND ULTIMATE FAILURE, whence f = 37700 psi of 40000 M.S = 77 Ls 37700 - - 12 06 - SOLUTION: Condition (3) Since a torsional shear stress has now te
been added, the new } „-/szx”2000
stress is still two :
dimentional, however †x„——| tt,
the given tension N
stresses are not } fay 8000
principal stresses fz
due to the addition of the torsional shear stress
fx=41300 psi fz=8600 psi fg =800C pst
Instead of finding the principal stresses and using Eq (30), we will use the fy and f,
stresses and use Eq (32) aVtg+t?- t,t, + ot2 = V 41500% 86005 41300 x 8600 + 3 x 6000" 40000 40200 — rà H.8 = ? = 40200 psi - 1 -.01 Thus yield is indicated since M.S is negative, Example Problem 2
A cylindrical pressure vessel 1s 100
inches in diameter and 1 inch thick The vessel is made of steel with Fry = 42000 psi
Determine the internal pressure that will produce yielding
SOLUTION: This applied stress system is
biaxial with no flexural or torsional shear
Let: p equal internal pressure fy
t = wall thickness = 1 in ‡ @ = diameter = 100" | — —~(f f, = circumferential stress due * to pressure p | = 24 = BA t,= ot and f, aE From Eq 30 BeVrti+ri- eye,
The vessel wall ts to be stressed to the
yield stress of 42000, thus f = 42000 Whence TT (oa (4200) = ("+ (EE) - (GS Solving, p = 970 psi FROBLENS (1) The combined stress loading at a point in a structure is as follows:- f2 = -1000, fy = -2500, fg = 2000 Determine
the magnitude and direction of the principal stresses Determine the maximum shearing stress Solve both analytically and graphically
(2) Same as Problem 1, but change fz ta 4000
and fy to +3000 and fg to 2500
(3) A solid circular shaft 1s subjected to a
limit bending moment of 122000 inch
pounds and a torsional moment of 250,000
inch pounds If diameter is 4 inches
and the yield tensile stress 1s 42,000, what is yield Margin of Safety
(4) A thin walled cylinder of diameter 6
inches is subjected to an axial tensile load of 15,000 pounds, and a torsional
moment of 12,C00 inch pounds What
should be the wall thickness if the permissible yield stress 1s 30,000 psi
(5) A closed end cylindrical vessel is 15
inches in diameter and a wall thickness
of 0.25 inches The vessel is subjected to an internal pressure of 10,000 psi,
and a tensile load of 22,000 pounds If
the yield tensile stress of the material is 75,000 p 1, what torsional moment can be added without causing yield
REFERENCES: =
Nadal, Theory of Flow and Fracture of Solids
Timoshenko, Strength of Materials
Freudenthal, The Inelastic Behavior of
Engineering Materials and Structures
Marin, J., Engineering Materials,
Trang 19
CHAPTER C2
STRENGTH OF COLUMNS WITH STABLE CROSS-SECTIONS
C2.1 Methods of Column Failure Column Equations In Chapter A18, the theory of the elastic and tnelastic instability of the column was
presented The equations from Chapter A18
for a pin end support condition are:- For elastic primary failure, Where Fo = fatlure = P/A stress compressive unit str°ss at Young’s modulus tangent modulus column length crs " radius of gyration of cross~ section
Fig C2.1 shows a typical plot of Fy versus Lf If the column dimensions are such as to cause it to fail in range CD in Fig C2.1, the primary failure is due to elastic
instability and equation (1) holds This range of L/o values is ofter referred to by
engineers as the long columi range AB c D to ¬ | Local Crippling Limit Fe u/p Fig C2.1
The range BC represents the range of Lp
values where failure is due to inelastic in-
stability of the column as a whole and equation
(2) applies This range BC is often re.erred
to as the short column range C2
The range AB in Fig C2.1 is for a range of Lf values of below 20 to 25, and repre- sents a range where failure is due to plastic
crushing of the colum In other words, the
column is too short to buckle or bow under
end load but crushes under the high stresses This column range of stresses ts usually referred to as the block compression strength
A column, however, may fail by local buckling or crippling due to distortion of the column cross-section in its own plane The horizontal dashed line in Fig C2.1 represents the condition where the primary column strength is limited by the local weak-
ness This line moves up or down according
to the value of the local weakness The determination of the column strength when
failure is due to local weakness is covered in another chapter
C2.2 Column End Restraint Fixity Coeificients Column Effective Length
The column strength is influenced by the end support restraint against rotation and by any lateral supports between the column ends The letter c is commonly used to indicate the
end fixity coefficient, and c = 1.0 for zero end restraint against rotation, which can be
produced mechanically by a pin or ball and
socket end support fitting Thus including
the end restraint effect equations (1) and (2) can be written, a PF, = 0% 5 » Fo = (Lp )*
Let L' = effective length of the column which equals the length between inflection points of the deflected column under load Then L' = L/Ve Thus equation (3} can be written as, TÊE pr, = Bt (Li /p)4 Fos te -~+- (5) (L'/p)*
Trang 20Fig C2.2 shows the deflected colum
curve under the load P for various end and
lateral support conditions The effective
lengths L' and the end fixity coefficients
are also listed ¬ ma TH | [gre yy eo TOLet! Ls \usa Ì mÀ | L/4 i ¬ —w—— mì — f Czl⁄4 h Cz2.08 of Fig C2.2
C2,3 Design Column Curves for Various Materials For routine design purposes it is con- venient to have column curves of allowable failing column stress Fo versus the effective
slenderness ratio 1'/p In equation (5) we
will assume values of Fe, then find the tangent modulus Ey corresponding to this
stress and then solve for the term L'/o
Table C2.1 shows the calculations for
17.7 PR (TH1060) stainless steel sheet at
room temperature The results are then plotted
in Fig C2.7 to give the column strength curve Similar data was calculated for the material under certain exposure time to different elevated temperatures and the results are also
plotted in Fig c2.7 Figs C2.3 to C2.15
give column curves for other materials under various temperature conditions Use of these curves will be made in example problems later
in this chapter The horizontal dashed line ig the compressive yield stress Values
above these cut-off lines should be sub- stantiated by tests C2.4 Tangent Modulus E;, from Ramberg-Osgood Equation The basic Ramberg-Osgood relationship for E, is given as follows: (See Ref 1) Be + >
Por eBa gies 7
E, = tangent modulus of elasticity 5 modulus of elasticity For definition of other terms see Article B1.12 of Chapter B1, STRENGTH OF COLUMNS WITH STABLE CROSS-SECTIONS TABLE C2.1 17-7 PH(TH1050) Stainiess Steel Sheet Pry = 180, 000 = ‘6 Fay = 160, 000 B= 29 x 10 Fo Et L'/p =nv Ei/ Fe | 20, 000 29 x 10° 119, 56 30, 000 29 «x 10° 97 62 40, 000 29 «x 108 34.54 50, 000 29 x10 T5 82 80, 000 29 x10 ° 69.03 70, 000 29 «x 10° 63.91 80, 000 29 x10 59.78 89, 600 29 x10* 56, 49 96, 000 27.55 x 10° 53.19 102, 000 28 LÔ x 10 # 50 22 107, 400 24.65 x 10° 47 5T 112, 200 23.20 x 10 5 45 15 117, 000 21.75 x 10 * 42.81 121, 500 20 30 x 10 + 40, 58 125, 800 18,85 x 10 ¢ 38.43 130, 000 17.40 x 10 © 36 32 134, 000 15.95 x 10 ¢ 34,25 137, 700 14, 50 x 10 * 32.22 147, 000 11.60 x 10¢ 27 89 158, 100 8,70 x 10 ¢ 23, 29 167, 600 5,80 x 10° 18, 47 173, 800 2.90 x 104 12 82
This equation is plotted in Fig 02.16 Por a given material, n,.F,,, and © must be
Known Then assuming values of F, we can
find corresponding values of E/E from Fig
C2.16 For values of 2, F4,, and n refer to
Table Bl.1 in Chapter Bl
€2,5 Non-Dimensional Coiu.in Curves
Quite useful non-dimentional column curves have been derived by Cogzone and Melcon (See Rez 3)
The Euler column equation 1s P = nđ8Â/(L'/o)*, which can be written,
The problem therefore resolves itself
into obtaining and expression for E,/F trom
the non-dimensional relationship Tos do this
multiply both sides of equation (7) by Fo i/F
and equate to B*
{ee 1
— B/E = rs F 23
ro 78 ee)
Fig C2.17 shows a plot of this equation as taken from Ref 3, and shows F/F,., versus B for various values of n
- = (8)
Trang 23
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C2.5 + ˆ " — fae ee =— T5 RT Fig C2.11 - an - Fig, C3.14
TO1S-T6 Alum, Alloy Extrusions os eT ~ TOS-74 Ales, Alley Clad Sheet
Trang 24C2.6 STRENGTH OF COLUMNS WITH STABLE CROSS-SECTION Fig C2140 M0 —- R.T, + T0T8-T€ Aktte.- Alloy-Hand Forging 14.6.0 I _—~ - T1-6Aš-4V Titantum Alioy Flg, C2.15