ANALYSIS AND DESIGN OF FLIG HT VEHICLE STRUCTURES C11 43 The angle, a, must satisfy the equation, K fs tan « ‘rq * ARG at „5 (Ì ~ Kỳ ỐC (208) from Ref (3), and this {s different from the case for stringers, ee here, as in equation (58), tan” q = TT T————¬ - (108) Đo £ - Bạn tt ®) tan* a ARG, = 8 1+ a where, as before, from equation (94) tex
This formula {s simtlar to the one for the se 3| iên + sin2a (1-k) (1-1)
°9#factLve”" area of a Single upright in a plane
ved Deam system, (Art C11.20), where or use Fig C11.36 for bracketed expression distance from c.g of ring to the skin
it
&
radius of gyration of the ring (cross~
Saction) about an axis parallel to the
skin
6
Equation (105) assumes that k, fg 4, t, etc., are the same for the panels on each side of the ring If not, then some average value Should be used, or else fag must be written as
the sum of 2 diagonal tension affects (for each
Side panel) as was dona for the longeron in (S) above The "“aaximum" stress, ÝRGMaX » im the ring
wiil be, as before, eq (59) = PRGA *ROyax = Tag (Tz) where Ỷ ts obtained from Fig Cll.21 " RG
7, The next problem ts the determination of the angle of diagonal tension, a There are Several ways in which a can be evaluated Four
are as follows:-
3) When there is a simificant amount of compressive strain, simply assume
a= 459,
b) when there is no, or very little, axtal strain present (only shear) use Fig
C1145 to obtain a
c) 1? sufficient tension stress (or
strain) 1s oresent to prevent duckling, from eq (78), then, of course, a = 90,
a
approximation as follows This is the
most tedious method and (a) and (b}
above will suffice in most cases, especially for preliminary design
f
(ty, from equation (104) )
é 286 {f., § RG from equation (103) )
Equation (106) must be solved by successive
approximation as was done in the stringer system
discussion
8 Strength checks can then be made as follows, using stress ratios
2) Longerons:
t đạm
eh pe F 1.0
ec L
where Fog is the "natural" crippling
strength and RB 1S obtained from Fig
011.38
Frequently, tn the case of thicker longerons, the following is used as a
strength check:
‡#_ + f tr D.T < 1.0
ec
That is, the primary and diagonal tension effects are Simply added and
checked against Foo
9) Rings:
f
ae F1.0 “RG
If the rings have stresses in them due to loads other than diagonal tension (1.e, bulkhead type loads, see Chapter A21), then an interaction equation is Calculate a by the method of successive used
fy fag
gh + te 1.0
Trang 2
C1144 DIAGONAL SEMI-TENSION FIELD DESIGN
where f, is the stress due to loads
other than diagonal tension effects ce) Skins; Strength check: t i "s 1.0 where Fy is obtained from Fig Cll.42 Permanent suckling: Usually no perm buckling is allowed at limit load ŸStrw†p < qe Fis 3p.b, where Fs > is obtained from Fig C11.46 °° a) General Instability: £ § §š = 1.0 Fsiner where F SINST is obtained from Fig C11.39 It 1s recommended that a margin of safety of 15 to 20 be maintained here Note in Fig C11.39
that no effective skin is used and that Prong 15 at most 34
e) The loads on rivets at splices, etc
are the same as for the stringer system, eq (96) and (97)
The procedure outlined in (1) to (8) above
can best be illustrated through the use of an
example problem
Fig C11.37 Example Problem,
Consider a longgron type fuselage structure
having a cross-section as shoym in Fig C11.47, The details and section properties of the
longerons and rings are included in the figure The properties of the structural cross-
section are given in Fig C1l.48 The
properties shown are for the case where the bending moment causes the upper skin to be in compression and the Lower in tension The upper
Sk1n thus buckles out early, as indicated
For this example problem assume that the
applied loads are, M = 2,135,000 in.1lb (compression in upper skin) r= T->| + 50h Ring ay Cross Section Splice A = 0787 in? Plate tonferon
Cross-Section a2 60 in? 2 =.365 in Ts 0252 in Detait "A"
IT 2.197 int Clad 7075-T6 Sheet Mul 2 = 580 in, oy 7075-T6 Extruded Mtl sÀ : Fey=10, 000 psi Detail "A" Bg =10.5x 105 ‘Skin, , 025" 7075-T6 SY” Longeron Cross-Section and Details of Members Fig C11.47 = 600,000 in lb (reversible) = 18,000 1b, (produces comp in upper skin) T V
These are the loads at the middle of the
bay being checked
41.8" Effective Section in Geometry Bending - Skin Shown
Cross-Section Area = 2830 in.* "Dashed" is not Effective
Ty A, 1895 in * Fig C11 48
1 Using the geometry and properties of Fig C11.48 and the engineering theory of bending,
Trang 3
ANALYSIS AND DESIGN OF F 2a Check top skin panel for four and k
Use properties of Fig (11.47 and the applied loads to get Ro, eq (74) The skin compression stresses will be "fictitious" since the skin buckles out early, but will give the proper constant, B, for interaction Assume ÍCsEin at a point 2/3 up from upper longeron
to top surface as governing compressive buckling _ Mã _ -8,135,000(27.7) „ fesetn FT 7 1896 n1 teen.) Tt „600,000 „ Ís * Dất ” EEGÓ) (V005) ~ S40 pst Tnen, from aq (73), (h) and R = > compression bucxliag 5
from buckling equations in Chapter ca
From Chapter C8, for our panel dimensions, ky = 10 and kg = 12.83, thus a e n® x10 x 10,300,000 (2928)* 2310 pst Foor *~ Te (1 - «3*) 20 a ye Facp = * i = oe ee = 2980 pst hence, from eq (72) _ Feor _ 2310 A Fei, 2980 74 Then, from eq (74) BR, [= 7.36 wa ttm Ry 5 and fsop 7 Re Papp = 106 (2980) = 515 psi fs 3a The Loading ratio, =—-, is fs er Ÿs 4840 =z F 15.54 Tsp 318
4a k is then, from (3) and Fig C11.19, for
IGHT VEHICLE STRUCTURES
11, 45
OO th | 300 (.025) (yee os Rd 30 ares
k= 82
Next, repeat steps 2a-4a for the side panels This involves an arbitrary interaction to serve as a criterion for buckling under -
combined shear and compression
2b The stresses at the top (upper longeron)
and bottom (lower longeron) of the side panels are -37,200 (comp.) and 14,800 (tension) respectively This gives an average "fictitious" stress of -11,200 axial and = 26,000 psi bending
stress (equivalent) The actual reduced buckling
stress for the side panels could be obtained from an interaction for these axial and bending
stresses
Another way is to use an axial stress only,
"weighted" on the high side of the average,
arbitrarily, to account for the effect of the pending stress For this problem a compression stress value half-way between the averags, ~11,200, and the maximum, ~37,200, is used Thus ft = 711,200 = 37,200 5 = 24,200 psi then, since fg = 11,780, to _ 24,200 _ B =7 “11/780 7 2-0 and, as before, Fe As 774 (2—F} Ser Renes 2.05 2.05.2 -= + ¢ ) +4 Re = ze - 774 = (335 and fsạy = Rọ Fs„„ = 555 (2980) = 998 pst 3b The loading ratio is then 77 † s 11,780 86 7 11.8 4b From Fig C11.19, using 11.8 frem (8b) k= 79
3 The upper longeron stress can then be gotten from eq (104);
Trang 4
C11.46 DIAGONAL SEMI-TENSION FIELD DESIGN ỉ =
- : Ki to cot a, Ks fe, cot ag eng = RG 2 Kixc® Lm = ~,00262
*L "tp 7 BAL + ,6 (1>k¿)Re "Dap BE + 5(1-Ka)Re ——_ 8 73 x 10"
Bits © Rata then, .81 (4240) cot a eé.-€ 2 ~37,200 ~ TET) s (2.610.105) 1£) { ^ tan? Ga # ee TH.8)(.ce5) + 5 (1-.81)(.108) & = Egg +i tan” œ _ «79 (11,780)}cot a 20.5) 35.4 1.085) + 5 (1-+.79)(.335)
~37 ,200 - 2970 cot a, - 9810 cot ad, 6a, The stress in the rings supporting the upper skin is obtained from eq (105) First, Ag 0787 Mpg = — Ss = re > 0288 22 75a 1+ Ñ 1+ CA Then, fea = — 01 (eto) an as 10,600 tan a 7 + ,5 (1~,81) 5(.025)
As discussed earlier, this assumes the upper pariels on each side of the frame to have the same shear stress
6b The stress in the rings supporting the
side panels is, similarly, = ~79{11780) tan ds RG St.0es) * 6 (1+ 78) -0235
f = 27,900 tan da
a, and a, in the above equations refer to the upper panels and side panels respectively
7 Angle of diagonal tension
The compression in the upper panel is so large that a, can be reasonably taken as 45°
For the side panel the same is probably
true, in the upper portion, but this will be checked using the method of successive
approximation and equations (106), (105), (104) and (94) Assume a = 449 (and u = 45°) Fig C11.36 and equations above, From = is _ 11,780 \ c = 1,88 >) = 1.38 pie “3x 108" 00215 c, = aH = -.00561 - 00028(1.19) ~ 00095 (1.19) = -.00507 = -00215 - (-.00507) = 1.00 „00216 - (—.00262) +; tạ2)*(.708)
Thus, a, # 459 which is close to the 44° assumed
Let a, = 45°, This bears out the statement that any significant compression stress (or strain)
forces a towards 45°,
Had there been no significant average axial
strain (or stress) present a could be gotten from Fig C11.45 which 1s based on pure shear
(no axial loads present) Using this for the above side panels, for example, we would proceed as follows: EB Aya - 10.3K10" (See z, R= “11,780 Bo) = 23-5 RG 2 ALL A 6 ae 7 +228 3 he = Ges (ose) ~ “8? Then from Fig C11.45, using the above para- meters, a = 36° (for no compression)
Tus the.effect of compression, as previously calculated, is significant in foreing a towards
45°, as frequently assumed for simplicity | 8 Now that a has been established as 45° for
the upper skin panels and Zor the upper portion
of the side skin panels, the stresses are
Trang 5
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11, 47 Rings:
The ring under tne side panel skin will be most critical (fs is larger): tag = 27,900 (1.0) = 27,900 TRGMAx = 87,900 (1.15) = 30,500 (Using Fig G11.21) t ˆ⁄4 củ = 7 Then, using Fig C1l1.38, For R = 30" N = 15,200 Fag = Nxc = 19,200x1.187 = 22,800 ~ FRG - 22,800 M.S === -1 * 30-500 “RG ~1=~ ,25
Thus the ring is not adequate Either
more area (thicuness) is required or a closer ring spacing (or both) are needed
to lower fpq-
Skins:
The side panel has the highest stress and will be, therefore, more critical from an
ultimate shear strength than the top panel, fs = 11,780 pst From Fig Cll.42, for k = 79 and Foy = 72,000, Four = 25,000 psi M.S = (25,000/11,780) + 1 = 1.13
Caeck for Permanent Buckling
Usually there is the requirement that no
permanent sxin buckles shall occur at
limit load This is checked as follows: = fs 11,780 | tsitait Tis 7 ios 7 79840 pst Using Pig 011.46 „ 5 * Feop Eg X Poy % 10" 10.4x10°x52,000 4 (2,980) x10" yg ˆ 8 = 8 = 2.8, Fs, , = 2.8(2980) = 3,340 pst FSop P.B 1g MeSig g 2 PB Peay pe = BOs ogg 7,840 e General Instability Using Fig C11.39, with © ona ® ,554 and pg = 565 (Using no effective skin, per the figure) 7/4 +39 (Sp) : .39(,565)/° (dn)*73 RsZ+ x1 0*x= (Sx52.4)°/% (40)3/+ x 10 # 11,25 From Fig C11.39 Fetnst = 3.0xEx10-5 = 30,900 psi Thus *stnet 30,900 ease i ee i M.S ts 11,780 1= 1.62 (large) f, Rivets
The rivet requirements can be checked in the same manner as was done for the stringer type structure along any skin splices and for the tension field "prying” forces The load transferred between the longeron and the ring can be calculated as
Pog = tag *Ara, » from (8b) and (6a)
27,900 x 0285 795 1b
This load is actually carried by the rivets attaching the outer ring flange to the longeron flange (next to the skin) and also by the gusset action of the skin at the ring-longeron junction Two fasteners,
staggered if necessary, should be used to
attach the outer ring flange to the longeron flange (See discussion of joggles in
Chapter D3)
C11.38 Summary
From these examples involving both stringer and longeron type construction it can de sean that the effect of axial compression stresses (or strains) along with the shear stresses in the panels is two-fold:
a) It brings about, through interaction,
earlier buckling of the panels than would result from shear stresses alone The result is, of course, a nigher value of k and, hence, of all the en-
Suing loads and stresses that are a
Trang 6
C11.48
bd) The angle of diagonal tension is forced to approach 45°, more than would result from shear stresses only The effect of tension stresses 1s just the opposite and can thus be conservatively”
ignored, or evaluated tf desired
From a time-saving standpoint, in pre-
‡minary design, an arbitrarily large value of % and an angle of diagonal tension of 45° can be assumed where significant axial compression
strains are present
The exact magnitude of the various diagonal tension effects throughout a network of skin panels defies simple evaluation from an analytical standpoint This is particularly
true when both shear stresses and axial
stresses change from panel to panel as in most
practical structures and loadings No simple
analytic expressions are available But some rational approach is necessary to complete the
design, or specimens for test programs, and the
approaches given in this chapter represent one such procedure If margins are extremely small, element tests for substantion are in order The reader is encouraged to consult the references for a more thorough understanding] of the basic theory and its limitations,
particularly with regards to areas where sub~
stantiating test data is relatively meager The stringer system is usually found, for example, in fuselage structures where there are relatively few large ‘eut-outs™ to disrupt the stringer continuity This is more typical of transport, bomber and other cargo carrying
aircraft The longeron type structure 1s more
efficient and suitable where a large number "quick-access" panels and doors and other
"cut-outs" are necessary to service various
systems rapidly These would "chop-up" a
stringer system rather severely, making it
quite expensive from both a weight and manu- facturing standpoint Therefore, longeron
systems are more usually found In fighter and attack type aircraft, and in others with un- usual features There are also, of course,
other factors influencing the choice of structural arrangement
Some further notes concerning this general
subject are included in Chapter C3 This includes “beef-up" of panels and axial members bordering cut-outs or non-structural doors, which is also related to "end-bay” effects
discussed in C11.24
——
* Tt is not conservative to ignore the reducing effect of tension
stresses on k if the compression stresses due to diagonal tengion are being relied upon to reduce any primary load tension stresses in stringers or longerons
DIAGONAL SEMI-TENSION FIELD DESIGN C11.39 Problems for Part 2
1 In the example problem of Art 011.54, what M.S would exist if the ring were
made of 032 2024 (instead of 040) (Assume a = 45°)
In the example problem of Art Œ11.324, how wide could the ring spacing, d, be made
and still show a positive M.S for the ring {Assume a = 489.)
In the example problem of Art C11.34, how much additional torsion, T, could be applied before the ring would show 4 M.g = 0, (Assume a = 45°.) Repeat example problem Art C11.34 using applied loads M T v 2,600,000 1n./1b 9 40,000 1b
Assume the general section properties (1 and neutral axis location) remain the same as In 011.3%
In the example problem of art C11.37
a) What (standard) gauge of 7075-TS sheet
aluminum would the ring have to be to show a minimum positive M.S
b) What ring spacing, d, would be required for the 032" ring to show a minimum
positive M.S (M.S = 0)
Repeat the example problem of Art C11.37 using as applied loads
= 2,400,000 in./Ib 0
sa 28,000 lb
{Assume section properties are not changed.) 7Ó what ts required to eliminate any negative
margins of safety in Problem (6) 11.40 Problems for Part 1
(1) The beam as shown in Fig (A) is sub-
jected to a shear load of 8000 1b as shown Determine the margin of safety for the given
loading for the following units: {1} web,
(2) web flange rivets, (3) web stiffeners
NACA method
(2) Same as problem (1) but with web up- right spacing = 4”
Use
(3) Design a semi-tension field beam for
Trang 7ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C11.49 3xd x # Stiffener One Side gr 04 Web 2024-T3 ++ [+ + + + | t + + tx All ets LIST Fig (A) 80, 000# 703/1n Design Loading Lat mm “avs + 1x1x3/32 Dia @ 3/4” 16 1x1x3/32 Vy Tye j Dia @ 1-1/4" 40#/in + mm 2 sai TÌ| _ TH |_, 1-1/2 3 —_— 2 Tổ ~ — 4 web Stiffener, | F—.064 Web se" >1) 7075 Clad Rave jy
Web uprights and flange members to be 2014-TS
extrusions Rivets to be 2117-T3 Show cal~
culations for at least four sections along beam length Assume the beam braced laterally
by 025 skin on top and bottom
(4) Fig (C) shows the cross-section of a
single spar wing beam Strength check the
following units of the beam section (1) web, (2) web uprights, The upright spacing is 8
inches and the design shear load on section is
35000 1b web material is 7075 alclad
REFERENCES
(1) H Wagner, "Structures of Thin Sheet Metal, Their Design and Construction", N.A.C.A
Memo 490
(2) H Wagner, "Flat Sheet Girder With Thin
Metal web."
Part I - N.A.C.A Technical Memo 604 Part II -N.A.C.A Technical Memo 605 Part III - N.A.C.A Technical Memo 606 (4) Kunn, Peterson, Levin:- A Summary of
Diagonal Tension Part 1 Methods of Analysis N.A.C.A T.N 2661
Trang 81,80 DIAGONAL SEMI-TENSION FIELD DESIGN eB 2k fg sin 2đ H= 0.32 + sin 20 (1~k)(1«¿) oy Rivet head (AN430) head (AN455] heod (AN442) =1” 20 25 30 35 40 45
a,deg bead (AN456)
Fig C11.36 Graph for Calculating Web Strain (Ref 3) 2 Tense sirengity_ Fensile strength of rivet shonk _" ” : oO gi 2 4 ma ivet “diameter 5
Fig C11 37a Tensile Strength of Four Types of AL7S-T3 Aluminum-Alloy Rivet in 245-T3 Aluminum-Alloy Sheet (Ref 3) Approximate £028 36 50 —— Sacre ty Nese 6, degrees 2 „ 100 Zia 82 Le s 5 „ ‘af rivet shank th _of rivet -Sheet_thickness, Rivet ~ diameter
Fig C11 37b Tensile Strength of NACA Machine-Countersunk Flush Rivets
Trang 10C11, 52 25 30 * fea » ksi I 9 Tlg C11.415 DIAGONAL SEMI-TENSION FIELD DESIGN 9 5 Og L6 20 i
Fig C11.40 Correction for Allowable Ultimate Shear Stress in Curved Webs 35 30) pay : 4345 38 25 ksi 30 25 20 20 1 30 35 40 0 2 4 6 8 10 2.67 k Rp = th ; Ry = ott Fig Cll.4le Alclad 7075-T6 Aluminum Alloy Oyj, = 72 ksi R “ng” Ast Fig Cll 4la tỄ Ape 489 t+ Re 454o 35 30 z5 20 3 3s 40 45 sO Spor See
4 & 8 LO Fig Cli.4ld Angle of Pure Diagonal Tension 2024-T3 Aluminum Alloy Øuit = 62 ksi Rat i i Ree aL
Trang 12
C11 54 DIAGONAL SEMI-TENSION FIELD DESIGN
LIfe bP LLL
APPROXIMATE VALUES OF THE ANGLE OF DIAGONAL TENSION a cà FOR PANELS WITH h'> d (LONGERON SYSTEMS) 45 40} NOTES: ae -
35 1) This curve is for panels loaded: in: shear:
- “" '“anly, no significant axial strains ~ a 2} The.curve is aa i ‘ion # the following ‹ uation
£4 cos “0 )(teps} ain®-o.+{
30 tee TTT womens nba Ait
rain? 6 [2 Grew) oe? a + ee ta a —
_ ¬ nhỉ Ang/dE+e20-k.: So: -
Ay 7Ì Tư ườ T- Tỉ
25 _ 3] The.range ot =e values represented on.this curve ig-from: 02 to-1„ 8 ~ Valuea of.this parameter.outside this range (tor values: aPg- 20) ~- must be applied to’ the ‘equation above (Note 27 to: determine’ Q ~ Degrees (The Angle of Diagonal Tension) 1997 3 4567991 2 3 4 567991 2 3 456 % OTES:
3 1) This curve good for flat or curved panels Conventional panel stiffening members must be attached to edge 2) Curve does not apply to panels with stringers
3} Curve may be used for any material at any temperature provided proper values of Fson > Eo & Foy are used in the parameters
FSp FScr
Fscr = Panel Shear Buckling Stress
Ee = Modulus of Elasticity in Compression Fey = Compression Yield Stress
“OL * 1.0 § 10 50 ,500 1000
4x Fsop 10°
Trang 13
CHAPTER Di
FITTINGS AND CONNECTIONS BOLTED AND RIVETED
Di.1 Introduction
The ideal flight vehicle structure would be the single complete unit of the same matertal involving one manufacturing operation Un-
fortunately the present day types of materials
and their method of working dictates a composite structures Furthermore general requirements of
repair, maintenance and stowage dictate a structure of several main units held to other
units by main or primary fittings or con- nections, with each unit incorporating many
primary and secondary connections involving
fittings, bolts, rivets, welding, etc No doubt main or primary fittings involve amore
weight and cost per unit volume than any other part of the aerospace structure, and therefore
fitting and joint design plays an important
Part in aerospace structural design
Di.2 Economy in Fitting Design
Por structural economy the structural destgner in the initial layout of the flight
vehicle should strive to usa a minimum number of fittings particularly those fittings con- necting units which carry large loads Thus
in a wing structure splicing the main beam flanges or introducing fittings near the centerline of the airplane are far more costly than splices or fittings placed farther out-
board where member sizes and loads are con-
Siderably smaller Avoid changas in direction
of heavy members such-as wing beams and fuse~
lage longerons as these involve heavy fittings If joints are necessary in continuous beams place them near points of inflection in order that the bending moments to be transferred
through the joint be kept of small magnitude
In column design with and fittings avoid intro-
ducing eccentricities on the beam and on the other hand make use of the fitting to increase
column end fixity thus compensating some of the weight increase due to fitting weight by saving in the weight of the beam
For economy of fabrication, the structural designer should have a good knowledge of shop
processes and operations he cost of fitting
fabrication and assembly varies greatly with the
type of fitting, shape and the required toler- ances Poor layout of major fitting arrangement
may require very expensive tools and jigs for shop fabrication and assembly
Fittings lixewise add considerably to the
cost of inspection and rejections of costly fittings because of faulty workmanship or materials are quite frequent » thus adding greatly to the unit cost
DI.3 Fitting Design Loads Minimum Margins of Safety,
As discussed in other chapters, limit loads are the maximum loads, which a flight vehicle may be subjected to during its 1life- time, when carrying out the required ground and flight operations The limit loads must be carried by the structure without exceeding the
yield stress of the material used in the structure The ultimate or desizn loads are
the limit loads multiplied by 4 factor of safety, usually 1.5 for aircraft and less for
missiles The structure must have suffictant
strength to carry the ultimate or design loads
without failure
The normal flight vehicle structure tn- volves many parts which are joined together by various types of connections In general, an
additional blanket factor of safety is required
for design of these connections This blanket
factor of safety {s normally 1.15 to 1.20 The
stress analysis of most connections or fittings is more complicated than for the primary
structural members due to such factors as com-
bined stresses, stress concentration, bolt-hole tightness, etc., thus an additional factor of safety 1s necessary to give a similar degree of strength reliability for connections as provided
in the strength design of the members being connected
D1.4 Special or Higher Factors of Safety
A blanket factor of safety for all types of
fittings or load conditions is not logical The
manner in which a load is applied to a joint often involves a dynamic or shock load, as for example Joints or fittings in landing gear Single pin connections often undergo rotation
or movement between adjacent parts, thus pro-
ducing faster wearing away of material in oper- ation Repeated loads often present a fatigue problem In an airplane or missile there are
certain main fittings which, if they failed,
would definitely cause the loss of the vehicle
Thus the design fitting requirements of the
military and civil aviation agencies involve
many special or higher factors of safety This
is so particularly in design involving castings
DI.L
Trang 14
DI.2 FITTINGS AND CONNECTIONS BOLTED AND RIVETED
D1i.5 Aircraft Boits
The aircraft bolt is used primarily to
transfer relatively large shear or tension
loads from one structural member to another Fig Dl.1 shows three standard aircraft bolts
in common use There ars other types but they
will not be presented in this limited chapter on connections Hexagon Head Clevis C) {a gang © 5| — TH Ptg DI.1
The hexagon head bolt is an Amny-Navy standard bolt made from SAE 2330-3.5 percent nickel steel, heat treated to an ultimate tensile strength of 125,000 psi The bolt head is of sufficient size as to develop the full tensile strength of the bolt
The internal-wrenching bolt is a high
strength steel bolt usually heat treated from
160,000 to 180,000 psi It is especially suitable for main splice fittings because of its high strength and the relatively small space required for the bolt head
The clevis bolt is referred to as a shear
bolt because its nead is not designed to develop the full tensile strength of the bolt The clevis bolt 1s usually used when a group
or cluster of bolts is required to transfer a
load by shear loads on the bolts The smaller bolt heads thus save weight and also provide greater bolt head clearances The clevis bolt W111 develop about one half the tensile strength of the standard AN hexagon head bolt
The three bolts are also made from aluminum
alloy for diameters over 1/4 inch In many fitting designs weignt can be saved by using aluminum alloy bolts
General Rules in Using Bolts
Bolt threads should not be placed tn
bearing or shear The length of the bolt shank Should be such that not more than one thread axtends below fitting surface, which can be done by the use of washers
Bolts less than 3/8 diameter should not be
For steel bolts, 3/16
used in major fittings
inch diameter should be the smallest size to be
used in any fitting
Bolts connecting parts having relative motion or stress reversal should have close tolerances to decrease shock loads
For bolts connecting members having rela- tive motion a lubricator should be incorporated
in the surrounding parts of the fitting; the fitting should not be drilled to provide lubri-
cation
Bolts should be used in double or multiple
shear if possible in order to increase strength efficiency in bolt shear and to decrease bending tendency on bolt
D1.6 Aircraft Nuts
Fig D1.2 illustrates four standard steel aircraft nuts Nut material is more ductile
than bolt material, thus when the nut is tightened the threads will deflect to seat on
the bolt threads
The Castle nut is probably the most common aircraft nut It develops the full rated
strength of the bolts The nut has slots milled in it so that nuts can be prevented from turning
The Shear nut is only one half as thick as the Castle nut, and thus has only threads enough to develop one half of bolt tensile strength
It 1s used with a clevis bolt which has a screw
driver slot to limit the torsion in tightening
the nut The nut is also castellated for cotter pin lock COTTER = #¬ ui CASTLE NUT SHEARNUT PLAIN NUT SELF LOCKING AN - 310 AN - 320 AN - 315 AC ~ 365 Fig D1.2
The Plain nut which is very seldom used in present design is used for permanent locations and is locked by "peening” or riveting the end
of the bolt over the nut, an operation which
destroys the bolt protective coating or finish Aluminum Alloy Nuts Aluminum alloy nuts
are not uséd on bolts designed for tension
Sometimes alumimum alloy nuts are used on steel bolts on land planes to save weight provided the bolts are cadmium plated If the bolt is used in places where the nut is repeatedly removed,
neither bolt or nut should be aluminum alloy because of the danger of injuring the relatively
Trang 15ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Self Locking Nuts Self locking nuts are
widely used in aircraft industry and there are
a number of types on the market Fig D1.2
shows one type of self locking nut involving 4
fiver insert when the bolt reaches the fiber
collar it tends to push the fiber up because the hole in the collar is smaller than the bolt, and is not threaded The fiber ring thus
sets up a heavy downward pressure on the oolt automatically throwing the load carrying sides
of the nut and bolt threads into positive contact Thus all play in the threads is
eliminated and a friction is set up between
every bolt and nut thread in constant This
constant pressure which 1s maintained on threads, provides the friction which prevents
nut from moving under vibration The use of the self locking nut reduces the assembly costs
as it eltminates the bothersome cotter pin which takes extra operations of the mechanic
and is very difficult to install on the nut
in the many Joints and corners of an airplane In another type of self locking nut, the locking force is provided by the spring action of the
upper part of a specially designed nut D1.7 Bolt Shear, Tension & Bending Strengths
Table Dl.1 gives the section properties and the ultimate shear, tension and bending strengths for AN Standard Steel bolts at room
temperature Fig D1.3 shows the correction
to be applied to the strength values in Table
D1l.1 when bolts are subjected to elevated
temperatures Table D1l.2 gives the tensile
and double shear strengths of Steel Internal
Wrenching bolts Table Di.2a gives ultimate
shear, tension and bending strengths for aluminum alloy bolts The value of Fp, the
modulus of rupture, was’ determined by the
method given in Chapter C3 Table D1 1 Ultimate Shear, Tensile & Bending Strengths of AN Steel Bolts (Fry = 125,000, Fg, = 75,000, Fy = 180, 000}
Tarea of Ultimate single Ultimate
solid Moment of| shear strength | Ultimare tensile} Bending
Size of pin| section, | inertia of ae full strength | Moment [Moe bolt | in?” |solid, in.* | diameter, 1b | (in thread), 1b | in, lbs | 0.180 | 02835 | 000840 2, 128 2,210 121 | ưa 3⁄16 - | 07869 | 0004882 -.04808 | , 0001818 3, 880 5, T50 4, 080 8, $00 276 338 3⁄8 18 „0008710 1001787 11,250 8, 280 10, 100 18, 800 1, 480 932 1⁄3 3/16 -004814 003088 14, T00 18, T00 23, 800 18, 500 22210 3, 140 5/8 1007482 23, 000 30, 100 4320 3⁄4 01558 33, 180 44, 000 1, 450 | 7⁄8 102878 45, 050 80, 000 11, 880 La 104908 58, 900 80, 700 17,670 of Re = ‘ 0 100 200 300 400 300 600 DI.3 Table Di 2
Ultimate Shear and Tensile Strengths of Steel Internal Wrenching Boits (Eyy = 160, 000)
uit Double ult Double
Size | Tensile Shear Size } Tensile Shear
Dia Strength | Strength Dia Strength | Strength lbs ibs lbs, lbs 1⁄4 8, 180 9, 300 5/8 43, 600 58, 300 5/18 9, 820 14, 600 3⁄4 83, 200 83, 900 3/8 15, 200 21, 000 1/8 86,100 | 114, 200 7/16 20, 600 28, 600 1,0 114,000 | 149,200 1⁄2 27, 400 37, 300 1-1/8 | 144,000 | 188, 900 9/18 34, 800 47, 200 1-1/4 | 186,000 | 233, 200 Table D1 2a
Ultimate Shear, Tensile and Bending Strengths
of 2024 Aluminum Alloy Bolts SHEAR TENSION BENDING Fgụ-35, 000 Tt-62, 000 F-72, 000 992 1,088 48 1,717 1,975 110 2, 684 3, 189 218 3, 868 4,937 373 5, 261 6, 883 592 6, B71 9, 104 884 8, 697 11, 563 1, 260 10, 738 14, 719 1,730 15, 463 21, 873 2, 980 21, 046 29, 520 4, 740 27, 489 36, 759 'ï, 010
D1,8 Bolts in Combined Shear and Tension
When bolts are subjected to both shear and
tension loads, the resulting strength is given
by the following interaction equation:- (Ref 1, 2} 3 a = + Ge 2#l++-+ +-+ - (01.1) where, shear load tension load
shear allowable load from Table D1.1
tension allowable load from Table D1l.1
ged
x
unin
ie
Figs Dl.4 and D1.5 is a plot of equation
for the various AN steel bolt sizes The curves
are not applicable where shear nuts are used
The curves are based on the results of combined
load tests with nuts fingertight D1 8a Bushings,
It is customary to provide bushings in the lugs of single bolt or pin fittings subjected
to reversal of stress or to slight rotation
Thus if wear and tear taxes place a new bushing
can be inserted in the lug fitting Steel
Trang 16Dl4 FITTINGS AND CONNECTIONS BOLTED AND RIVETED M.S 30 Interaction Formula và y2 Triển 25 Where: x = Shear Load y = Tension Load 15 ‘TENSION LOAD IN THOUSANDS OF POUNDS s # Ỹ q 1Í
Ps, allow or allow ¡ allow P low
a = ANC-5 Shear Allowable
20 b = ANC-5 Tension Allowable x
1§ 20 PS tow
SHEAR LOAD IN THOUSANDS OF POUNDS
Fig D1.4 Combined Shear and Tension on AN Steel
Bolts (Fry = 126, 000, Fay = 75, 000) 140 130 120 = = 3 80 T0 60 50 40 TENSION LOAD IN THOUSANDS OF POUNDS 30 20 10 20 30 40 Where: 50 SHEAR LOAD N THOUSANDE OF POUNDS x=Shear Load yaTension Load
a=ANC-5 Shear Allowables baANC-5 Tension Allow ables Interaction Formula ra raction J §0 TỤ LUNG 25 100 Fig, D1.5 Combined Shear and Tension on AN Steel Bolts
single bolt fitting lugs to increase the allow~ able bearing stress on the lug since the
bushing increases the bearing diameter 1/8 inch since bushings are usually 1/16 inch in wall thickness, If bushings are not used on single bolt connections sufficient edge distance should be provided to ream hole for next 3124
polt in case of excessive wear of the unbushed hole If considerable rotation occurs a
lubricator should be provided for a plain bushing or an o1l-impregnated bushing should
be used
D1.9 Singie Bolt Fitting
Possibly the simplest method of joining two members together is the use of 4 Bingle bolt or pin connection Such a joint can trans- mit relatively large loads and yet the joint is
easily and quickly disconnected
Fig D1.6 illustrates the four general methods of connecting two members by a single
bolt First the connection 1s made symmetrical
about the centerline of the load on the joint Thus {s Fig Dl.éa the load P on the male part of the fitting divides equally and symmetrically to the two female plates or units which make up part of the fitting unit If the male and female parts of the connection are to be tied together by a single bolt it is evident that
the connecting plates will be weakened due to the bolt hole unless extra material is added at the bolt hole section
Pig Dl.éa shows a fitting consisting of three rectangular plates of uniform section throushout fastened together by a single bolt Obviously the weak section for the plates in
tension would be a section through the center-
line of the hole if this section is strong
enough to carry the given loads, then the re-
maining part of the fitting members are considerably over strength To avoid this
over-strength which means extra weight of fitting units, a single bolt fitting unit is often made
like one of the examples indicated in Fig b, ¢ and d of Fig D1.6
In Fig b, the plates are made constant thickness but increased in width in the vicinity
of the hole section In Fig ¢c, the width of the plates is kept constant but the thickness
of the plates are increased in the vicinity of
the nole section, and in Fig d, both width and
thickness of plates are changed
DI, 10 Methods of Failure of Single Bolt Fitting and the
Allowable Failing Loads
As the load on a fitting is transferred
from one side of the fitting to the other, in- ternal stresses are produced which tend to cause the fitting to fail itn several different ways
wR
¬
Trang 17ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES D5 Đụ _ 7 SP a= ye i = l2 si 7 P ta Be Fig.b Fig a Fig D1.7 Fig D1.6
Failure by Bolt Shear,
In Fig D1.7 the bushing is not continuous Đetween the plates, but each of the three
plates have separate bushings As the pull P 1s placed on the fitting it tends to shear the
bolt at sections (1~1) and (2-2), (Fig Dl.7a)
Fig D1.7> illustrates the forces or pressures on the bolt and the fatlure which can take place if the stresses are sufficient
Let Py represent the maximum or ultimate
load on the fitting This force Py must be
resisted by the shear strength of the bolt at the two sections (1-1) and (2-2} Hence,
Purport shear) SPeyeAs2 5 - oe (2)
where Fg, = ultimate shearing stress for bolt material
A = cross-sectional area of bolt
Failure by Bolt Bending
saan oie Sonding
The main concern regarding bolt bending
is that under the limit loads any bending deflection of the bolt be not permanent as such
deformation would makes removal of the bolt dizficult The subject of bolt bending
stresses is discussed in a later article
Failure of Lug Portion of Fitting
es orton of Ritting.—
The lug portion of the fitting refers to
that portion of the fitting that involves the
hole for the single bolt that connects the
male and female parts of the fitting unit The
simplified assumptions regarding failing action
and the resulting equations which follow have
been Widely used for quick approximate check
of the lug strength The procedure which
follows will be referred to as Method 1 a Pu ~— aC Shr — ( 3) flg D1.8 b, = Fig D1.9 Ago ay @3)¬1 C9 -Œn Fig a ` Fig b Fig ¢ Py G + O Fig D1 10 Fig a Fig b METHOD 1 OF BOLT & LUG STRENGTH ANALYSIS a EAGT ANALYSIS Failure in Tension
Fig D1.6 indicates how a fitting plate can pull apart due to tension stresses on a section through the centerline of the bolt hole Both the male and female parts of the fitting must transfer the load past the centerline of the hole, thus both parts must be considered in the design of a fitting quating the allowable load P, to the ultimate resisting tensile stresses at points (a) and (b) Fig D1.8, we obtain,
Pu(tension) = Ftu (2R- D)t
where Pry = ultimate tensile strength of plate material
Equation (2) assumes that the tensile stress on
the cross-section is uniform This ts not true as the flow of stress around the hole causes a stress concentration To take care of this stress concentration requires a margin of safety
of 25 percent
Failure by Shear Tear Out sSintre ey shear Tear Out
Fig Dl.4 {llustrates the manner itn which
failure can occur by the shearing tear out of 2
Trang 18DLE FITTINGS AND CONNECTIONS BOLTED AND RIVETED tre lo on the Stress portion ( wrt tte ee (5) 5q: ing whera tane we 40 A margin of sa ant should be maintained If ta jected to
T——>—————~ (3) infrequent rotati out with load involving no shoc require a
margin of safety I? shock or
aring strength ¢ iS -
weaning Strengoh of ta vipration with i tation is present,
rea reguirs a margin y of 150 percent
it Shock is consider: vin such structures
a 1 ° n
on practice to take the Bs landing a enne » Roteving, mooring to the edge distance BEB OIE
a nole times the a ton _ afeaty 2 ven
š 2 tines to Since thera General Comment on Margins of Safety for Luzs are two shear area: - ya ha, mis is slightly con- ` 4
In general it is good desizn cractice to server ve because | at ona sneer the ts design lugs conservatively as the weight of
arger “i peraits one 8 ose we area lugs 1s small relative to their importance in
along line 1-1 which ts limited by the 40 insuring the safety cf the flight vehicle In-
degree line as shown in Fig Dl.Se Fe tin nae NA TA te Thu
2 by Bearing of Bust
„ D1.10a, the pull P causes the bolt t9 press against bushing wall which
sses against the plate wall If th
S high enough the plate material ad-
mt to the hole will start to crush and 2low
thus allowing the bolt and oushing to move
which results in the elongated hole as
{llustrated in Fig 5 Zquating the load Py to
She ultimate bearing strength on tha bearing
surfaces we can write, Py = Ppp Dt - - where Fhr = allowable bearing stress 9 = diameter of bushing t = plate thickness Failure by 32 ụ
4 bushing th ate hole and thus it is ti fit
fitting dbeit is re ren
therefore a certain t atween the 50
and bushing inside di necessary in order to insert anc r If fitting ts Sudjected to reve f 11 Sl2p In
the fitting tends he £*Et
182 the fitting zat
rotation takss cl tends ii ar bet: - itis ustoma z ween ing since : @ ins 1 2
accuracies in manufacture are difficult to
control It {8 good design practice to previde
suffictent matertal to permit drilling for a
Dushing {f bushing is not used in original
design, if castings are used as fittings, much nisher factors of safety on the limit loads
are specified because of the low ductibility 97
the material in castings
DI.11 Method 2
Loading Lug Strength Analysis Under Axial
Due to a comprehensive study and test
program by Cozzone, Melcon and Hobbit (Refs 3 and 4), the procedure as ziven in Method 1 is
somewhat modified The important difference is that curves derived from test results give
the stress concentration factor to use for tension on the net secticn and the shear out
failure as assumed in Methed 1 has deen re-
placed by a combined shsar-out bearing failure Fig S1,11 snews the lug-pin combinations and types of failure as taxen from Rat 3 Bushing Gy x 3 —¬ — a b @Œ: Ì_ Tension Fallure wen Ty — v0 5S) )- 2 đ 1 ơ
Shear Bearing Failure— |
c Hoop Tension Failure——
a
Trang 19ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES The methods of failure and the methods of
lug strength analysis are as foliows:
Pu = Ke Fey Ag
Because of strass concentration, the
Stress on the net cross-section cannot be taken as uniform The ultimate allowable tension load Py for lug equals,
where Ky is the stress concentration factor as found from Fiz Dl.12 and Table 01.3 Fey = ultimate tensile strength of the material and
Ay = net tension area,
notes: Fa St
1 Am seattionn M.S of 0, 15 ever and shone any coming or (itting factor ta appropriate 1 Valuns ace toe room cemperatuey use only,
ot
Fig D112 Lug Design Data
Tension Efficiency Factors for
Axially Loaded Lugs (Ref 3, 4)
Table D1 3
Curve Nomenciature for Axial Loading for Fig D1 12 L, LT and ST Indicate Grain in Direction F in Sketch L = Longitudinal LT - Long Transverse ST - Short Transverse (Normal) MATERIALS Curve 1 - 2014-T6 and 7075~T6 Die Forging (L) 4130 and 8630 Steel 2014-T6 and 7075-T6 Plate = 0.5 (L, LT) 7075-76 Bar and Extrusion (L)
2014-T6 Hand Forged Billet = 144 in.? (L) Curve 2 - 2014-T6 and 7075-T6 Plate > 0.5 in, = 4,0 in (L, LT)
7075-T6 Extrusion (LT, ST)
2014-16 Hand Forged Billet = 144 in.” (L) 2014-T6 Hand Forged Billet = 36 in.? (LT) 2014-T6 and 7075-T8 Die Forgings (LT) Curve 3 - 2024-T4, 2024-T2 Extrusion (L, LT, ST) Curve 4 - 2014-T6 and 7075-T6 Plate = i in (L, LT) 2024-T4 Bar (L, LT)
2024-T3, 2024-14 Plate (L, LT)
Curve 5 - 2014-T6 Hand Forged Billet > 36 ¡n,* (LT)
Curve 6 - Aluminum Alloy Plate, Bar, Hand Forged Billet and Dte Forging (ST) NOTE: For Die Forgings ST Direction Exists Only at Parting Plane
T078-T6 Bar (T)
Curve 7 - AZ91C-T6 Mag Alloy Sand Casting 356-T6 Aluminum Alloy Casting
EEE 2s erent zth
Failure dus to shear out and cearing are closely related and are covered by a single
calculation based on emptrical curves, The ultimate or failing load in this shear-bearing type of failure ts:-
Poru = Kory Fey App tote eee ee (7)
The values of Kbru, the shear-bearing
efficiency factor, is given by curves in Fig D1.15 “ & For shear-bearing yield strength the equation ts, Pory = Kory Ftu App - > 7 + +e (8) Le de ==—— hee eres he = cơ are A" io catet 19 08 ded fe at "lay khổ (xe im epee ot E Fave eee La fe ne Le neat ne ge ne se at bee ne ce p
fig D1.13 Lug Design Data Shear- Bearing Efficiency Factors for
Axially Loaded Lugs (Ref 3, 4)
Trang 20D1.8
Py* Kory Abr Fy
FOR ALL MATERIALS 2.0 Y 0.6 and higher ¢/D values 5 10 15 20 25 30 3.5 4.0 e/D Fig D1.14 (Ref 5)
Values of Shear-Bearing Factors of Lugs
Bolt or Pin Bending
The subject of bolt bending strength is
treated in art Dl.14
D1, 12 Lug Strength Analysis Under Transverse Loading
Cases arise where the lug of a fitting unit is subjected to only a transverse load
Melcon and Hobbit in (Ref 4) express the ultimate transverse or falling load by a single
equation: -
Pru = Keun ddr Fen - tcp ttt (10)
Similarly the yield strength of lue ts,
Pry # key Apr Pty o- - ttt (11)
The efficiency failing and yield co- efficients Xey and Ky are given by the curves
in fig Dl.15 The curve nomenclature for the
curves in Fiz D1l.15 1s given in Table D1.4 in using Fig D1.15, a value called Aay 18
needed, the value of which is shown in the
equation shown on Fig D1.15
D1.13 Lug Strength Analysis Under Oblique Loads Fitting lugs are often subjected to adlique loads Ref 4 gives the following anoroacn to this loading case
Resolve the applied load into axial and transverse components Then use the following interaction equation;~- Rg? + Rept"? z1 FITTINGS AND CONNECTIONS 1.8 1,6 xavng the $ T graia dle rectign im the plane of the tog Aay Apr me
Fig D115 Lug Design Data
Tension Efficiency Factors for Transversely
Loaded Lugs (Ref 3, 4)
(See Table Di 4 for Curve Nomenciature)
Table D1 4 (To he Used with Fig D1 15)
Curve Nomenciature for Transverse Loading
Curve 1 ~ 4130 and 8630 Steel thru 125 KSI H T Curve 2 - 4130 and 8630 Steel 150 KSI H T
Curve 3- for All Aluminum and Steel Alloys a Ste Curve 4 - 4130 and 8630 Steel 180 KSI HT
Curve 5 - 356-16 and AZ91C-T6 Sand Castings Curve 6 - 2024-T3 and 2024-T4 Plate = 0.5 in Curve 7 - 220-T4 Sand Casting
Curve 8 - 2014-T6 and 1075-T6 Plate = 0.5 in 9 - 2024-T3 and 2024-T4 Plate > 0.5 in also
2024-T4 Bar
Curve 10 - Approximate Cantilever Strength for All
Aluminum and Steel AHoys If Kty is Below
this Curve a Separate Calculation as a
Cantilever Beam is Warranted
Curve 11 - 2014-T6 and 7075-T6 Plate = 0.5 in = 1.0 in
707§-TS Extrusions
2014-T6 Hand Forged Billet = 36 in *
4 2014-T6 and 7075-T8 Die Forgings
| Curve 12 - 2024-16 Plate, 2024-T4 & 2024-T42 Extrusions
{
L
Curve
Curve 13 - 2014-T8 and 7075-T6 Plate ~ 1 in
Curve 14 - 2014-T6 Hand Forged Billet > 36 in.”
or margin of safety is, L
ˆ On Ra T1
Trang 21
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES values obtained for equations (6)
or (7),
Rer = transverse component of applied
ultimate load divided by the values of Pty in equation (10)
Di.14 Bolt Bending Strength
In general static tests of single bolt fittings will not show a failure due to bolt
bending failure However, {t is important that
sufficient bending strength be provided to
prevent permanent bending deformation of the
fitting bolt under the limit loads so that
bolts can be readily removed in maintenance operations Furthermore, bolt bending weakness can cause peaking up a non-uniform bearing
pressure on the fitting lugs thus influencing
the lug tension and shear strength The
unknown factor in belt bending 1s the true
value of the bending moment on the bolt because
the moment arm to the resultant bearing forces is unknown An approximate method (Ref 4) for determining the arm (b).to use in calcu- lating the bending moment on bolt is given tn Fig D1.16, which gives b = Sti.+ 25ta + ¢ + where g 1s clearance or gap between lugs The resulting bending moment is considered to be
conservative (See Ref 4 for other refine-
ments relative to determining moment am b.) t + bos Load on Pin 2 ai + ——— ; eT E i} b + ›‡r= Sqr T t 2 2 — 2 P = 4 + ? 2 ;— tt = =<] = |i peer Beg OH Fig Di 16 DI 15 Dlustrative Problems
PROBLEM 1 An engine mount fitting ts attached
to the nacelle fitting by a 5/6 diameter AN
steel bolt The ultimate or design loads on the bolt are tension = 16700 lbs and shear =
5800 lbs Due to vibration a 25 percent
margin of safety will be required
Solution:- This is a problem of combined
tension and Shear on @ bolt Use is mada of
curves in Fig Dl.4 which are interaction
curves for combined tension and shear On Fig Đ1.4 at a value on lower scale equal to the shear load of 16700 lb., labeled point (x) on the figure, we erect a vertical dashed line as
Dg shown On the vertical scale we locate point
(y) at tension load of 5800 Llbs., and draw
horizontal dashed line to give point (a) A
straight line through points (0) and (a) is
drawn and extended to intersect the curve for
a 8/8 AN bolt at point (b) Projecting down~ ward from point (b) to lower scale, we obtain ŸS (ã11ow) = 22400 lb and projecting horizontally we obtain Pt(a11ow) = 7700 1b
Then M.S = PS (211ow)%) - 1-3 (22400/ 16700) ~ 1 = 34 or M.S = (PE(itaw)) -1
= (7700/5800) - 1 = 34,
The specified required M.S was 25, thus bolt strength ts satisfactory
PROBLEM 2
Fig D1.17 shows a single pin fitting The lug material is AISI Steel, heat treated to Pry = 125000 psi The bolt is AN steel, Fry, 2 125000 The bushing is steel with Fẹụ = 125000 The fitting is subjscted to an ultimate tension load of 15650 lb The fitting will be strength checked for the design load The check will be made by both Methods 1 and 2
1/2" Dia AN Steel Bolt Dy 16" Thickness Steel is Bushing De 1-3/16 — Pal, 15x 15650 = 180004 Fig D1, 17 SOLUTION BY METHOD 1
A fitting factor of safety of 1.15 will be
used which is standard practice for military
airplanes
Design Fitting Load = 1.15 x 15650 = 18000 1b
Check of Bolt Shear Strength
Bolt ts in double shear From Table D1.1, single shear strength of 1/2 inch diameter AN steel bolt is 14700 lb
Trang 22moment am y bolt is, 0156 = 218 tachas| 1/44 tnen} = 5 x 18000 x 218 @ = 160000 psi of inertia) From Table D1.1, Fy for 10, 000, Steel dolts is Therefore M.S = (180000/160000) -1 = 12
Tais lug is more critical than lug 8
Bince thickness of lug B is more than one-nal? a2? lug A «of Tension Through Bolt Hole P¿ = 125000 x (1.1875 - 625) 375 = 36400 1b
To take care of stress concentration,
Method 1 says maintain a M.S of 25, thus lug ensile str = 82000 x (.5937 - 3125}2 x .375 = 45300 Ib t § value i 6000 Ib I t under tne design fitting load of ater shear { h 18 2srm1ssible %9 take 2 slightly ee 7ig D1.9G6), The
to the 40 degree line would
out area to give 4 positive
This method of calculating nh is conservative as will be ¢ result in Method 2 solution, same ultimate be critical area is less boa margin of safety 1s FITTINGS AND CONNECTIONS BOLTED AND RIVETED
ar Strength and Bolt Bending
2 calculated in the same manner as đ 1 and thus the calculations wiil not ted
Tension Net Section
Pa = Ke Feu Ac
Ky is the tension efficiency factor to take
care of stress concentration due to the hole and
is determined from Fig, 01.12 Table D1.3 says
to use curve number 1 for all steels To use Fig D1.12 requires the ratio W/D = (1.1875/.625)
= 1.9, Then from Fig D1.12 we read Kt = 98, whence, Py = 98 x 125000 x (1.1875 - 625).375 = 28500 lb M.S = (28500/18000) ~ 1 = 58
Fig Dl.12 says that a M.S of 15 ts appropriate over that of all required fitting factors of safety, thus our M.S of 58 provides
more than this additional M.S of 15,
Shear Bearing Strength
Poru * Xoru Feu App
Kory 1s the shear~bearing efficiency factor and is obtained from Fig 01.13, D/t e⁄0 .625/.376 = 1.565 +5837/.625 = 95 From Fig D1.13, we read Kypy = 80 Ppry = 60 x 125000 x 625 x 375 = 23400 1b M.S (23400/18000} - 1 = 30
The reader should note shear out strength
by Method 2 is considerably larger than by Method 1 Fig DL.13 says a 15 M.S 1s appro-~
priate, thus our 0.30 ts satisfactory
Bushing Yield
Pory =
with Fey
Pary = 1.65 x 112000 x ,5CO x 375 = 59000 195,
Since 211 margins of safety are vositive, the strangtn of fitting unit is satisfactory
Trang 23ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
nole slightly back of the center of the
out strength This
he “yas | thickness
2t bending moment and
£ 7/18 diameter bolt The A lug iden- tical to lug 4 in Problem 2 is subjected toa transverse load Poy, woich ing load Find the load Pey SCLUTION: Frem equation (10) the failing load is - Pu
Pru = Key Apr ren
Tae failing coefficient Kry is determined
by use of curves in Fig D1.13 The lower
scale paramater for Fig Di.15 18 Aay/Apr»
were
For the meaning of these (A) areas, refer to sketch of lug in Fig D1.15 Simple
calculations give the following values for these areas A, 2A, 3 014, Ag 2 Ag = -105 whence, Agy = = 126 is used for psi Then
Fig D1.18 snows a steel forked fitting bolted to a doudle channel $ de from “Œ7S-T6 aluminvm alloy The hinge pin ts 6/3
atameter AISI Steel, neat treated to Fry =
150000 Steel fitting 1s also 150000 steel
A strength check of the fitting will be carried out Di 11 AIST Steel f Heat-Treated ! £ t gil ~i—! ami rhs m̧ 18 E1 TI§ etd G4 | (150000 Fry) ‡ -# | pe | f / 4-8 pia an Bolts
7075 AL ALLOY Section Hinge pin = 5/8 Dia,
Area of 2 Channels = 70 sq in Bushing = 1/16 wail (steel) - “ bs lo Fig DI 18 SOLUTION:
Design fitting load = + 33000 tension and
-25000 lbs compression These loads include
the fitting tactor of safety of 1.15
Let Py equal ultimate strength of the
fitting unit in the various failing modes
Oheck Shear Strengtn of Hinge Pin
The hinge pin diameter is 5/6 and material
is AISI Steel, neat treated to Fry = 150000 Fsy 3 95000 Pin is in double shear
Py = Feu Ag = 95000 x 3068 x 2 = 58400 1b M.S = (58300/33000) - 1 = 77
Sheck Bending Strength of Hinge Pin
Trang 24
ĐI 12
Py = 150000 + 146000 (1.7 + 1) = 288000 pst M.S = (258000/226000) - 1 = 14
Check Shear-Bearing Tear Out of Lug
Poru = Xora Feu Aor D/t = 75/.625 = 1.2, e/D = 6875/.75 = 92 From Fig D1.13, Koru = -72 Poru = 72 x 190000 x 75 x 625 = 51500 lb M.S = (51500/33000)-1 = 56 (margin desirable at least 0.15) Check Tensile Strength of Lug Section Through Pin Hole Py = Ke Feu At Ke obtained from Fig D1.12 using curve 1 W/D = 1.125/.75 = 1.5 From Fig D1.12, Ky = 99 P, = 99 x 160000 x (1.125 ~ 75).625 = 34800 M.S = (34800/33000) - 1 = 06 It would be desirable to nave a M.S of „15, Check Tensile Strength of Section 2-2 of Steel ritting
From Fig D1.18, four 5/16 diameter AN
Steel bolts are used to attach fitting to channel members Since bolts are the same
size, it will be assumed that each of the 4 bolts transfers 1/4 of the total fitting load
ig D1.19 shows the load in the steel fitting
and the channel members
by section 2-2, The total load passes (6500 Fy ony osTRIBUTION 4 ON STEEL HTTING Fig D1 19 Net tension area = (1.125 - 3125).375 = 308 FITTINGS AND CONNECTIONS
BOLTED AND RIVETED
A stress concentration factor K, from Fig
DL.12 will be used to be conservative w/D = (1.125/.2125) = 3.6 Fig DL.12 gives Ky = -91 Py = 91 x 150000 x 305 = 41700 ib M.S = (41700/83000) - 1 = 26 Check Section of Steel Fitting at Section 3-3 Load from Fig D1.19 = 16500 1b Net Section = (1.125 - 3125).25 = 204 sq in Use same Ky as before Py = 91 x 180000 x 204 = 27900 1b M.S = (27900/16500) - 1 +69
Check Shear Strength of 4 - 5/16 Dia Boits
Bolts in double shear
Py 2 4x 2 x 5750 = 46000 lb M.S = (46000/33000} - 1 = 39
Check Bearing of Bolts on Steel Fitting Section 33 or 4-4 ts critical as bearing area is less Allowable bearing stress for 150000 steel is 218000 Py = For Apr = 218000 x 3125 x 25 = 17000 Ib M.S = (17000/8250) + 1 = 1.06
Bearing of Bolts on 7075-T6 Channels
Fpr for 7075 aluminum alloy = 106000 Py = 106000 x 3125 x 25 = 8300 M.S = (8500/8250) - 1 = 01 Check Tensile Strength of Channels at Section 4-4 Load on Section = 33000 Net area = 70 - 25 x 3125 = 622 Feu for 7075 material = 78000 Py = 78000 x 622 = 48500
M.S = (48500/33000)-1 = 47 (which will take
care of any stress concentration
Check Shear Out of Channels Behind Bolt B,
Trang 25
ANALYSIS AND DESIGN OF FPUIGHT VEHICLE STRUCTURES D1 13
Pụ = Psụ Ag`= 43000 x 234 = 10100 1b Therefore we can write,
M.S = (10100/8250) - 1 = 23 (figured con- pee @SL) eee ee N servatively by Method 1) n 3 Ps TTT Me Shear Out of Steel Fitting Behind Bolt B, Psu = 95000 Shear out area Ag = (4375 - 1565)2 x 25 = 141 Py = 141 x 95000 = 13400 M.S (13400/8250) - 1 = 62 Bushing Yield Strength
ry = 1.85 Foy Apr- Fey * 113000 for 125000
steel
Pory = 1.85 x 113000 x 825 x 625 = 81700 lb
Load = 33000, not critical
D1.16 Bolt Loads for Multiple Bolt Fitting
Bolts Sizes Different Concentric Loading
In designing or strength checking a
multiple bolt fitting, the question arises as to what proportion of the total fitting load
does each bolt transfer This distribution could be affected by many things such as boit
fit or bolt tightness in the hole; bearing deformation or elongation of the nolt nole;
shear deformation of the bolt or pin; tension or compressive axial deformation of the fitting members and the member being connected, and a number of other minor influences
Since aircraft materials have a consider-
able degree of ductility, if the fitting is
properly designed, the loads on the bolts when
the loads on the fitting approach their maximum
value, will tend to be in proportion to the shear strength of the bolt That is, if the
combined shear strength of the bolts is the eritical strength, the yielding of the fitting material in bearing, shear and tension will tend to equalize the load on the bolts in
proportion to their shear strengths For stresses below the elastic limit of the fitting plates the bolt loac distribution no doubt is
more closely proportional to the bearing area
of each bolt Since the primary interest is
failing strength, the bolt load distribution
in proportion to the bolt shear strengths 1s usually assumed
Leti-
P # design load on fitting
Ps = allowable shear strength of any bolt
Py 3 load on bolt an
Psn 2 allowable shear strength of bolt n
Example Problem of Bolt Load Distribution
Fig C1.20 shows a multiple bolt fitting unit subjected to a concentric load of 100000
lbs Determine the load transferred by each bolt - = [2Sae0 PSI She Te tie 2/8 40 35-15 Fe I2Sseo i Fig D1 20 Bolt Dia, Double Shear Strength = Ps ũ x x B Distribution of loads to each bolt: Py = P eh = 100000 x SSB = ss200 1 Pp = 100,000 x 37274/117340 = 31800 1b Po = 100,000 x 22844/117340 = 19200 1b Pa = 100,000 x 11502/117840 = 9800 1b Total load on the four bolts adds up to 100,000 1b, DI.17' Muitiple Riveted or Boited Joints Subjected to Eccentric Loads
Pig D1.21 shows a plate fitting attached to another member by means of four bolts or rivets (The circles represent the bolts) The
fitting plate is subjected to the loads Py and
Py acting as shown Let it be required to find the resultant loads on the bolt group due to the
given loads
The centroid of resistance for the dolt
group will then corresrond to the centroid of
the bolt areas Fig Dl.22 shows the fitting
unit with the force Py and Py replaced by an
equivalent force system at the bolt group
centroid point (0} This equivalent force
syStem will be:-
Py = -200 13., Py = 1000 lb