ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C4.3
Table C4.1 Column Formulas for Round Steel Tubes
Short columns (#) Tranal- , Local
Material Ptụ; ksi|Fty, ksi|Fco, ksi bị Basic ;a ¡ Honal (£) Long columns (9) i failure Column formula (°) equation Uh 1025 see 55 36 36 36, 000~1.172(L'/0)” | C4.3 122 |286 x 10°/(L'/p)* (@® 4130 wee 98 75(9) 79.5 |79, 800-51 9(L"/p) *° c4.4 91 |286 x 10°/{L'/p)* (9) Heat-treated © alloy a steel | 128 | 103 113 |113,000-11.15(L1⁄9)2,.| C4.3 T3 |286x109/(L!⁄2)® | (3 Heat-treated alloy steel ] 150 | 132 145/145, 000-18 36(L'/p)* ) 4.3 63 286 x 10°/(L'/p)* ) Heat-treat: steel 180 | 163 179 179, 000-27 96(L'/p)7 1 C4.3 56/286 x 10°/(L'/p)* | 3 Equation C4, 1 may be used in the short column range if E 4 Not necessary to investigate for local instability when
is replaced by Et obtained from the combined stress- D/t 50
strain curves for the material, © This value is applicable when the material is furnished
DLt/o =L/ye: L'/p shall not exceed 150 without specific in condition N (MIL-T-6736) but the yield strength is
authority from the procuring or certificating agency reduced when normalized subsequent to welding to 60
© Transitional L/p is that above which columns are “long” ksi,
and below which they are short." These are approximate
values
are given in Tables C4.3 and (4.4 or inelastic instability of the column as a
: whole AS the slenderness ration L'/p gets
smaller, the Fg stress increases Now if the diameter of the tube is relatively large and the wall thickness relatively small or, in other words, if the diameter/thickness (D/t)
Table C4.3 Column Formulas for Magnesium-
Alloy Extruded Open Shapes?
GENERAL FORMULA
F _ KlFey)” ratio is large, failure will result by local
° (bạ) ertppling or crushing of the tube wall and this local failing stress is usually repre~ sented by the symbol Fog The values of Fog
in general have been determined by tests (see
(Stress values are in ksi)
‘Alloy K a ]m |Max Fe design charts for Fee versus D/t ratio)
MIAL Le eee , | 180 | 1/2 [1.0 | 0,90 Fey 4.9 Design Column Charts
AZ31B, AZ61A, AZ80A | 2,900 | 1/4 | 1.5 Fey,
AZ80A-T6, ZKGOA-TS | 3,300 | 1/4 [1.5 | 0.96 Fey In design, column strength charts are a
great time-saver as compared to substituting
in the various column equations, thus a number of column charts are presented in this chapter to facilitate the strength check of columns and the strength design of columns Fig c4.2 is a chart of L'/p versus Fe for heat treated round alloy steel tubing Fig C4.3 & Formulas given above are for members that do not fail by local buckling Table C4,4 Column Formula for AZ31B-H24 Magnesium-Alloy Sheet (1.05 Fey)®(L'⁄2 ) ®
Fo = 1.05 Fey 4n is a similar type of chart for aluminum alloy
round tubing Fig, C4.4 gives column charts Max F = F, for magnesium alloy materials A11 three
seen ey charts are taken from (Ref 1) Figs C4.5
and C4.6 represent a further simplication for
the design of steel and aluminum round tubing C4.7 Short Column Equations for Other Materials
C4.10 Section Properties of Round Tubing
For other metals for which short column equations are not available, the use of Euler’s equation, using the tangent modulus S¢ can be used (eq C4.2) Refer to Chapter C2 for information on how to construct column strength curves using this equation
4.8 Column Failure Due to Local Failure The equations as presented give the allowable stress due to failure by bending of
the column as a whole and the action is elastic Table C4.3 gives the section properties
of round tubing A tube 1s designated by giving its outside wall diameter (D) and its
wall thickness (t) Thus a 2-1/4 - 0S8 means
a tube with 2-1/4 inch outside diameter and a wall thickness in inches of 058, Since a tube
is symmetrical about any axis, the polar moment of inertia, which is needed in torsion problems, equal twice the rectangular moment of inertia as given in Table C4.3 For weight comparison, the weight of steel and aluminum tubing is
Trang 2STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADINGS,
Table C4.2 Column Formulas for Aluminum Alloys Alloy and Temper Product FF Short Transitional Long Columns Ha Columns
2014-T3, T4, T451 Sheet and Plate?; 2024-T3, T351, T36, T4, | Rolled Rod, Bar and T42 Shapes; Drawn Tube 5052—All Tempers 5083—All Tempers 5086—All Tempers 5454—-All Tempers All Products Fey(1+ ¥ Fey/1000) | Equation | 1.7327 y Z/F,, | Equation 5456—All Tempers C4.5 Cc4.8 6061-T4, T451, 74510, T4511
All Cast Alloys and | Sand and Permanent
Tempers Mold Castings
2024-T3510, T3531, T4, | Extrusions T2
2014-T6, T651
2024-T6, T81, T86, T851 | Sheet and Plate*;
7075-T6, T651 Rolled Rod, Bar and Fa(l+ VF 9/1333) | Equation | 1.346" E/? | Equatioa 7178-T6, T661 Shapes; Drawn c4.6 c4.8 Tube 6061-TS, T651 Sheet and Plate* 201©T6, Tesi, T8510, T6511, Tesz 20274-T6, 181, 18510, 2024-T8, TS, T8510, T8511, T8652 7075-T6, T6510, P8SIt, | Extrusiona, Forgings T082 7079-T6, T8510, T6511, T652 7178-T6, T6510, T8511 Rolled Rod, Bar and „(1+ VF.„/2000) | Equation | 1.2240-+/2/F., | Equation Shapes; : C4.7 C4.8 6061-T6, T651, T8510, | Drawn Tube; T6511 Extrusions X2020-T6, T651 Sheet and Plate* “Tocludes clad ss weil as bare sheet and piste
‘Transitional L’/y ls that above which the columns are “lone” and below which they sre “short”
Equation C4.8 may be used in the short column range If £7 is replaced by £; obtained from the compressive stress-strain curve for the material
Trang 3ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
given tn the last two columns of the table
C4.11 Some General Facts in Tubing Design 1 For a given area, the larger the tube
diameter, the greater the column strength if failure due to local crippling is not critical
The higher the D/t ratio of tube the lower the crippling or local failure strength If columns fall within the long colum category, the use of higher strength alloy steel or alimminum alloy will not increase strength of column since £ is practically * constant for all chrome-moly steel alloys
and likewise for all aluminum alloys Failure is due to elastic buckling of the column as a whole and is therefore a function only of I, L' and E
The column end restraint effects the neces- sary tube size Consult the design require ments of the Army, Navy, and C.A.A in this matter In general with welded steel tubular trusses a coefficient of C = 2 is permissible except for engine mount and Nacelle structures For trusses with riveted joints a value of not over 1.5 is generally permissible
The student should realize that practical limitations such as clearance requirements may determine the diameter of the tube
instead of strength-welght considerations Thus design can consist of checking the
tubes available under the given restrictions
C4 12 Effect of Welding of Steel Tubes Upon the Tension
and Column Strength,
Since welding effects the grain structure of the tube material adjacent to the weld, tests show the strength of the material adjacent to the weld is decreased as compared to the unwelded material If a tapered weld is used, the effect of the weld is decreased Table C4.4 shows the allowable stresses in tension to use when tension loads are carried
In short columns, the primary column failing stress may be greater than the local crippling strength of the tube adjacent to the weld at the end of the tube This local failing stress Jue to welding 1s referred to as the weld cut-off stress and the column compressive stress F, should not exceed this value This cut-off weld stress is shown by the horizontal lines in Fig C4.2 and C4.5
C4.5
Tabie C4 4
Tension Allowables Near Welds in Steel Tubing (X-4130) Normalized | Welded after HT [HT after |’ Type of Weld [Tube Welded |or Norm after Weld| Welding
*Tapered Welds
of 30° or Less( 947 Fry 90, 000 psi +90 Fry All others OAL Fig 80, 000 pst «80 Fry
“Note: Gussets or plate inserts considered 09 “taper'
with @
** For (X-4130) Special, comparable, values to the Fr,
equal to 90,000 and 80, 000, are stresses 94, 500 and
84, 100 psi, respectively
Ret Anc-5 + C4,13 Dlustrative Problems in Strength Checking
and Design of Round Steel Tubes as Columns
and Tension Members, PROBLEM 1
Tube size 1-1/2 - 058, Length L = 30 in End fixity coeffictent C = 1
Materlal:— Alloy steel, Fry = 95000 tube is welded at ends
Ultimate design loads are:- P = -14,500
lbs compression, and P = 18500 Ibs tension
Required the Margin of Safety (M.S.) The
Solution: The compressive (M.S.) will be determined first As the simplest solution,
we can use the column curves in Fig C4.5
Por a length of 20 and C = 1, from the upper right chart we projact upward to the inter- section with the 1-1/2 diameter tube and then horizontally to the left hand scale to read the column strength of 14800 lbs which we will call the allowable failing Pag
Pa 14800
> 14500
The tube strength could also be found by
using Fig C4.2 as follows:
L' =LVc =50V 1 = 30
L'/p = 30/.5102 = 58.7 is found from
Table C4,.3 as well as the tube area 0.2628
sq in Using 58.7 for L'/p on lower scale
and projecting upward to the Fry = 95000 curve, which is the lower curve, and then horizontally to left hand scale we read Fg = 56500 psi
M.S = -1= -1= @
Whence, Pa = FoA = 56500 x 2628 = 14850 1b
The solution obviously could be made by Substituting in the short column equation for steel having Fry = 95000, or
Trang 4C4.8 F, sAllowable Column Stress, ksi
Trang 6STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING
Trang 8
STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING C4.10 IN TENSION, COMPRESSION, BENDING,
Fg = 79500 - 51.9 (L'/p)***
79500 ~ 51.9 (58.7)*** = 56500 pst The short column equation applies since,
as shown in Table C4.1, the transitional L'/o 1s 91 and the value for our tube is 68.7 Tensile Strength
Since the tube is welded, the tube material adjacent to the weld is weakened The weld correction values are given in Table C4.4 We will assume a weld other than
tapered Let P, = allowable or failing tensile strength of tube Pg = Fty (weld factor) (area of tube) = 95000 x 841 x 2628 = 21000 lbs M.S = (Pg/p)-1 = (21000/18500) - 1 = 0.13, thus compression is critical PROBLEM 2 Case 1 Tube size 1-1/4 - 049, L = 40 tn e=l
Material: Alloy steel, Ft, = 95000 Find ultimate compressive load 1t will
carry
Solution: From Fig C4.5, Pg = 6000 lbs Case 2 If tube was heat treated to Fy, =
150,000, what compressive load would it carry
Solution: Fig C4.5 cannot be used since
Fry = 150,000, thus we will use Fig c4.2
L'=LA © = 40//1T = 4 From Table C4.5,
9 = 425 and area (A) = 1849
L'/p = 40/.425 = 94 From Fig (04.2, using the 150,000 curve, we find F, = 32500 Then Pg = FeA = 32500 x 1849 = 6000 1b Thus heat treating the tube from 95000 to 150,000 for Fey did not increase the column strength For a L'/p 2 94, it is a long column and failure is elastic and E is constant
The strength could also be calculated by Euler’s equation from Table C4.1
Fy = 286,000,000/(L'/p)*
= 286 ,000,000/(94)* = 32500 psi, the
same as previously calculated
Case 3, Same as Case 1, but assume tube 1s welded to several other tubes at its
end and that the end fixity developed
isc = 2
TORSION AND COMBINED LOADINGS
L' = LA ce = 40//2 = 28.4
In Fig C4.5 we can use L = 40 andc = 2 scale at bottom of chart or use c = 1 scale and L' = 28.4, Reading the chart we obtain Pa = 9200 lbs Thus the ¢ = 2 fixity
increased the strength of the tube from 6000 to 9200
Case 4 Same as Case 3 but heat treated to
Fry = 150,000 after welding
L'/p = 28.4/.425 = 66.8
From Fig C4.2 using 150,000 curve, we Tead Fy = 63000, whence
Pg = FoA = 63000 x 1849 = 11650 1b In this case heat treating produced additional strength, whereas in Case 2 it did not The reason for this is that failure
occurs in the inelastic stress range and heat
treating raises the material properties in the inelastic range The end fixity changed the column from a so-called long column to a short
column
The strength could be found also by sub- stituting in the short column equal for 150,000 steel as given in Table C4.1,
Fo = 145000 - 16.36 (L'/p)?
= 145000 ~ 18.36 (66.8)" = 63000 pst PROBLEM 3
Case l Tube size 2 - 065, L = 24, ¢ = 1.5 Material Fy = 95000 Welded at ends Ultimate design load = 25000 lbs What is M.S
LỤ “L/€ = 24//1.5 = 19.7 From Fig 04.5 for L = 19.7 one = 1 scale, we project upward to the 2 inch tube and note that it intersects the horizontal weld cut-off line which gives an allowable column load at left scale of Pg = 26700 lb Failure in this case is local crippling adjacent to welds at the tube ends Solution:
M.S = Pa/p = 26700/25000 - 1 = 07
Case 2 Assume tube is heat treated to Fry
= 125000 after welding What is tube strength
L'/p = 19.7/.6845 = 28.8
Using Fig C4.2 with L'/p = 28.8 and pro~ jecting up to 125000 curve, we again note that horizontal weld cut-off line is intersected
Trang 9ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES giving Fe = 95000, whence Py = 95000 x 3951
= 37500 lbs
If the tube had not been welded at ends the dashed part of the column curve could have been used, thus giving additional strength PROBLEM 4
Fig C4.7 shows a steel tubular engine mount structure for a 1050 H.P radial engine The ultimate design tension and compressive load in each member as determined from a stress analysis for the various flying and landing conditions are shown in ( ) adjacent to each member The true length L of each member is also shown Using chrome-moly steel tubes, Fey = 95000, select tube sizes for the given loads It is common practice to assume the column end fixity ¢ = 1 for engine mount members, since the mount is subjected to considerable vibration and the true rigidity given by the engine mount ring is difficult to accurately determine e Front View Side View Fig 4.7
Consider member (3) Ultimate design load = ~ 9250 Referring to the column charts of Fig C4.5, we find for C = 1 and L = 31.4 the following tube sizes for a strength near - 9250 Ibs 1-1/2 035, Pg = 9850 (weak) (Note-Pg = allowable load) 1-3/8 - 049, Pa = - 10350,weight=5.78 1b./100", M.S = (10350/9250) - 1 = 12 -10000, weight = 6.15, M.S = 10000/9250) - 1 = 08 1-1/4 ~ 058, Py =
Thus use 1-3/8 - 049 since it is the lightest as well as the strongest
Consider member (4), Load = - 5470, L = 30, eFl From Fig C4.5: 1~1/8~ 048, Pạ = -7100, wt.= 4.68, M.8.= ,20 1-1/4¬.085, Pa = -6500, wb.= 3.78, M.8.= ,18 1" ~,088, Pay = -6000, wt.= 4.86, M.S.= 10 C4, 11 The results show that 1-1/4 - 035 is the lightest Since there is danger in welding -035 thickness to the other heavier tube gauges particularly the engine mount ring which is usually relatively heavy for this size engine, a minizum tube thickness of 049 will be used, hence the 1-1/8 - 049 tube will be selected
Consider Member (2)
Design Loads 11650 tension and 4250 compression Since the tension load appears eritical, the tube will be designed for the tension load and then checked for the compressive load The Fru of the material equals 95000 psi Since the engine mount ina welded structure, the strength of the tube ad- Jacent to the end welds must be reduced to 841 x 95000 » 80000 psi (see Table C4.4)
Hence tube area required = 11650/80000 = 0.146 sq, in From Table C4.3, which gives the section properties of round tubes, we select the following sizes: 1L-.049, Area = 146, M.S = (.146/,.146) -1 = 0 1- 058, Area = 172, M.S = (.172/.146) -1 3.19 1-1/8 - 049, Area = 166, M.S = 14 1-3/8 - 035, Area = 147, M.S = 0 To obtain a reasonable margin of safety, the 1-1/8 - 049 will be selected
Many structural designers prefer to have large margins of safety on engine mount members as considerable trouble has been encountered in the failure or cracking of engine mount members
The strength of the 1-1/8 — 049 tube as a column for length = 31.4 and c = 1 equals -6700 lbs from Fig C4.5 which gives a margin of safety of (6700/4250) - 1 + 58 on the maximm compressive load The student should select a
tube size for member (1)
C4.14 Dlustrative Problems Using Aluminum Alloy and Magnesium Round Tubes as Columns and Tension Members
In general alloy steel round tubes must be heat treated to around 180,000 to 200,000 ultimate tension strength before they can com- pare favorably with aluminum round tubes on 4 material weight basis However, aluminum alloy as used for tubes cannot be welded satisfactorily and tims in a truss structure the end connections involving riveted and bolted connections add weight and design difficulties as compared to welded connections
Trang 10STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING
C4, 12 IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADINGS, PROBLEM 1 C4.15 Strength of Streamline Tubing
Case 1 Tube size 1 - 049 round, If a round tube Is exposed to the air-
L= 24, ¢ = 1, Material 2024-T3 stream, the air drag is about 15 times greater Find failing compressive load than if it were given a streamlined shape, thus
streamline tubes are used when the member is Solution: The column curves in Fig C4.6 are exposed to the airstream
slightly conservative because the equation used was Slightly different from the equation now specified in (Ref 1) Use L = 24, we read for 1-049 tube 4 failing load of 2800 1b As a second solution, we will use Pig C4.3 L's Le = 24/71 x24 L/P2 24/.3367 = 71.3 From Fig C4.3, we read Fo = 20000 Then Pg = Ped = 20000 x 1464 = 2930 1b The answer could be obtained by substi- tuting in equation C4.9, Fe = 50600 - 431 L'/p = 50600 - 431 (71.3) = 19900 Pg = 19900 x 1464 = 2920 lb
A column may also fail by local crushing or crippling of the tube wall, thus the crushing stress Fee should be determined to See if it is less than the primary bending
failing stress for the column
For our tube the diameter over thickness ratio D/t = 1.0/.049 = 20.40 Values of D/t are given in Table C4.3,
MU
Referring to the small chart in the upper right hand corner of Fig C4.3, we find for 4 D/t of 20.4 that Fog = 47500 psi Since this
stress {1s greater than the bending failing ~ column stress of 20,000, 1t 1s not critical
Case 2 Same as Case 1 but use ¢ change material to 6061-TS alloy L' eles AN 1-5 = 19.7, L'/p 1.5 and aluminum 19.7/.3567 $6.5 From Fig C4.3, Fe = 22500 Whence Pa = 22500 x 1464 = 3300 1b, Foc for D/t = 20.4 from Fig C4.3 2 38500 (not critical)
Case 3 Same as Case 2 but change material to magnesium alloy, Fey = 10,000
For L'/p = 58.5 and using lower curve on Pig C4.4, we read Fy = 7600
7600 x 1464 = 1110 lb Then Py = FyA =
Streamline tubes are drawn from round tubes In designating a streamline tube, the round tube from which it was made is used and then the fineness ratio is also given The fineness ratio is the ratio L/D, which dimen- sions are shown in Fig C4.8 The most common fineness ratio used is 2.5 to 1 Table c4.4 shows the section properties of streamline tubing having a fineness ratio of 2.5 to 1 Figs C4.9 and C4.10 give curves for finding the column failing stress F, and the local crushing stress Foc Fig C4.8 Bs 9441 R 19D Dz,87l4dd L 5D d = Equiv Round Dia fags f— Ll PROBLEM 1 Case l
A streamline tube made from a basic round tube of 2-1/2 - 065 size has a fineness ratio of 2.5 to l The length L is 30 in Take ¢ #1 Material is alloy steel Fey = 75000 Find the ultimate compressive load the member will carry
Solution: From Table C4.4 for 21/2 ~ 065
size we find the following section properties:-
Area (A) 4972, 9 (major axis) = 5137 in Then L' = LA/C = 30// T = 30, and L'/p = 30/.5137 = 58.5
D/t value for tube = 2,5/,065 = 38.5
From Fig C4.9 for L'/p = 58.5 and D/t =
38.5, we read Fy = 46500 psi For D/t = 38.5 and reading from small chart in upper right hand corner of Fig C4.9, we read Foo = 66500 Thus Fo is critical and Pg > 46500 x 4972 = 23000 lb
Case 2 Same as Case 1 but change material to 2024—-T6 aluminum alloy
For this material we use Fig C4.10 For L'/p = 58.5, we read Fe = 26000 psi For D/t = 38.5, we read Fog = 37500 (not
critical) Thus Pa = 26000 x 4972 = 12900 1b
.C4.16 Strength of Oval and Square Shaped Tubes in Compression
Tables C4.6 and C4.7 give the section properties for square and oval shaped tubes
respectively For the design of these shaped
Trang 11/
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C4.13
Streamline TABLE C4.4 SECTION PROPERTIES OF STREAMLINE TUBING (Fineness Ratio 2.5 to 1)
Equiv wall Axet w i Equiv Wall Axe
Roun cea & 2 Round! we t z , 1
GD fDec't |Gage | Maior | Minor | ™ per fe Minor | Minar | stnor | {S481 D0 [oage] Maice | Minor | 4° | peri | stor | Mace [Mayr | tiaor | Miaer | Miao 34 |.oxs] 20 [Lotte | 4286| 44786 | 3673 { 0017 | 0| 1446 | 0073 | 0140 | 3046 044 | 18 * “| 1079 | 36684 0032| 0103| 1435 | 0097 | 0187 | 2998 2M| 049) 18 0584 17 14489 | 3773 | 1283 | tors | 145|.5194| 2 4450 | 513 | 1186 { 2660) 3163| 2406| 4343 |1.0377 47644 2747 | LOS t<].0y5| 29 |1-1800| sooo} 094] 3140] 0028} 0122) 4753 | o5 | 0195 | 3572 oe] ie Bor | aie “Tat | 383 | Son So] 33a | Hones 0i9| 18 | 7 | “az | 4333| 0037 | 0149| d7) | 6156 | 0361 | 3323 095 13 T8 | 1440 | 1615| 2941| 3038| 7504| /4333|1.0144
098 | 17 1489 | 506i | 0042 | 01684 167L | 0182 | 030k | 3495 440) ‘agra | 3.090 | 2188| 3063| 4938| : : 4 #424| 3322| Lotst :
1 os 2 1.34834 S74 eed 37 oS a 2020 oe os bi 2% ong 8 3.7088) 1.5714 ae Late 1366 17533 11 sa 7848 | 1.1432
958) 17 ne : 3 340 [1.1398
cosa {iz | * "| A716 | [5835 | 006s | 0138| 1939 | 0278 | 0403 | 4033 ‘065 1864 | li6$| 2247| “sore | 2 THỊ 1118 1á |o| ‡o |l4070| «4436| t199 | 4074] 0065 | 616] 2292 | 0257 | 0329 | 4636 3 2364 | nee ams 5609 | Coat | 1241189 oa] ig | | Agee | | | Hới Tây | Gaon | lone | ae “20 3.371 | 2873| 3783| 3476 | 1.2427 | 6525 | 2.1196 065 | 16 2165 | 7359| 0103| 03201 2184 | 049 | 0569| 4923|| 3 | 058 azz | anes | 2493) 236 | đua | 3597 1.2056 4 07 | 2111| - 4311 + 1 |ðss{ 3o |t6837| 73 4336 | 45⁄2] 0088| 0446 2558 | 0355 | 0409 | 51, = T6 | dea | ose) sis | ties) ‘oe (rasn otis | ° j 84g | 5383| 0116| 0333| 2908 | 0462| os57 | -s107 095 1947 | 3229} 3768| 6103 | t3184| 4336 | 1.2332 098) 17 : 2172 | 7384| 0133 | 0372| 477 | 0560 | 0647 | 5077 -120 3.431 | 3923 | 4580| 6013 | 1.4189 | 7319 |1.1149 963 { l6 -2o | 8246| 0143 231 | 0618| 0714| 40331 136 4746 | 4834 | 3630 | 5879 | 2.0497 | 9869 |t.2L18 i} 095] 20 [3.9543] 7897| 143 | -s009 | øl18 | o300] 3838 | 0476 | 6349| 3683 || 314] 038 ag | tg | 7 - | 204i | 4999| 0157 | 400] 2777 | 0548 | 0680 | 4633 065 lợn | 2648Ì 380] 67 | 1.0617 | 4708 | 1.3511 2211 | ‘2661 } “3189 } 4747 | 1.1830 | 5248 |1.3437 958|17 | „ = | 3400 | -8138| 0181 | 0461| 2745 | 07% | G791 | 360% 08 + 3489 | 3071 1.4886 | “S604 | 1 3426 065 | 16 |” 2675 098 | ‘05o4] 2719 | 0833 | 0875 | S361 09 320 | 4150 | ⁄4470| 6639 | 1.6873 | 7449 |1.13%6 “120 6549 | 10887 | 9275 1.3304 1g |.035} 20 [2.0228} 8571| 161 | 5476 | 0154| 0359 0623 | 0997 | 6209 5 sate | 1 21707 [1317 09418 | 7 ` | 374] 2993| 0207| 0463| 304$ | 0848 | 0613 | 46144 1 68 [hi 913173 038 | 17 1627 | (8932 | 0239 | 0553| 204 | 09884 031 | 6131] | 394] 065 7286 | 1.4838 | 6108 |t.4%4L 063 | 16 .083 | 14 + | 3695 |1⁄256 | 036 | 0737| 2923 | d3 | 1303| 407 2934 | ‘9962 | 10262 | ‘osii| ‘2989 | -103 | 1053| 4108 “083 05 120 1% | 035 | 10 049 | 18 sas | 4746 | 3943| 0198 | 0426! 3363 | 0793 | 704) 6f4 ` | ‘2436 | (9248 | 0266 | 0573 3314 | 4083 4 36 958 | 17 2855 | 9707 | 0308 | 0448| 3283 | 1266 | A134 | 5659 | | 3% | 065] 16 7525 065 | 16 3186 | 1.083 | 0338 | 0715) 3356 | 1402 | 4243 | 66⁄4 083] 14 9582 083 | 14 402 | 1362 | 00 | 0883) 3192 | 1738 | 1545 | 6574 095) 3 1.9908 CN 4566 | 1.552 | 0453 | 09764 3149 | 1949 | 4732| 654 no] 138$ 134 |.035 | 20 [2.3600 | 1.0000] 4886 | 4401 | 0249 | 0498] 3633 | 0995 | 0830 | 7163 049 | 18 ¬ | 2618 | ‘8902 | 0336 | 0672| 3563 | 1363| 1123 | 7216 4 | 085) 16 la ‘ 8035 $8359] 2.2270) £025 | 1.6648 058 | 17 308) | 1948 | 0389 | 0778| 335L | 1394| 1312 | 7186 083] 14 1914 293 | 1.8161 | 1.0148 | 1.6604 063 | 16 344L | 1170 | 0438 | 0836| 332$ | 1765 | 1455 | 163 095) 13 11655 350 | 3.1917 | 1.1506 | 1.6549 983 | 14 1 4347 |1478 | 0521| 3032| 146L | 1192| 3809 | 7101 4939 | 1.65 0577 | A1 18 | 2463 | 2034 | 706L a 2 4 RSE : ae làm TT 4 4 I5 J9 | Le | để say bị mm | 288) Xe = |2247 [7926 | 5.9364] 2.1431 | 1.6239
1% |.035 | 20 [2.5285 | 1.074] 2033 | 4878 | 0308 | 0973) 3903 | 1227| 0943 | 7789 049 | 18 | 7 A “sai | 9596 | ‘onus | -0779) “sas | 4683 | 1256| 7 || #| 0634 16 [3730| 24081 | Beas 8895 | 2.6784) 5082) 1.7706
938 | 17 3341 |1136 | o4z | 004 35 | 185 | 4913 | 772 oa 4| „ | | Loses FT» Hư
063 | 16 3656 | 11257 | 0932 | 0993| 3744 | 2186 | 4683 | 7650 1 Đ| „| „ im 38428 | 1.3055 | Le “093 | 14 4673 | 11589 | 0680 | 4244| ‘3729 | 2719 | 2093 | 7684 aay - làn | sơn | 106i | 175
09 | 15 s2 |lâne | 0712 | 1346| 3687 | 3398 | 2770 | 2543 al Es 00 | 1ina|2.436s| 17314
> Be 1 | er | Gt? | Sie | ass| “tiga | 2a | 1 ng || ass 084 1€ |40681| 2.5725} 2.ast7 3367 | 4.0156 | 12894 | 1.8696
058 | t7 359 |1203 | o9L | loM| 4088 | 2462| 1793 | 3588 m3) - liêu Pu r1 r1hz
1065 | 16 38 | 1.345 | 0652 | “itary 4062 | 2568 | 3925 | 2217 vel it - |1 Tô | 1 2304| 1.3507 |1 44t 083 | 14 3999 | 1699 | 0793 | 1398| 3996 | 3313 | 1399 | 4136 ay fe Ò [Ea Soot | asso | 27498 | 1832 095 | 13 5685 |1933 | 0889 | 1456| 3954 | 374 | 2703 | 8113 L1) - liãm 4| 3001| 31t | L851 21⁄4 | 049 | 18 |3.01⁄4 3388 |1152 | 0735 | 1144| 4657 | 2946 | 1887 | 5325 058 2399 |1354 | 0854 | 1329| 3625 | 3440 | 2413 | 9294 065 083 2462 | L317 | 0944 | 1469| 4600 | 3835 | 2438 | #171 23680 |l93t | 116L | 1807| 49353 | 4794 | 3075 | 9223 “ 3432 |2186 | 4298 | 2020| +92 | 3409| 3479| 90 120 303G |1730 | 1336 | 2441] #462 | 6339 | 433 | | T T T Os@asic round tube digmeter i I ! , | i c «o| 4.9 $ _ 9 a | 2 r a 4 ôfas ọx Đ ore 4 3 ị xi š 205 20 40 60 86 00 8 | 0t 8 I ậ 3 Ỷ 3 a “ 20 4Q 60 8O 0Ó 20 Lip Fig C4.10 Streamline 2024-T3 Tubing F,-ALLOWABLE COLUMN STRESS IN THOUSANDS OF PSI F,¢- CRUSHING Fa, 28, 80 D/, of Basic Round Tube
NOTE: Higher values of allowable stress can be used in short column
range if substantiated by tests
Fig C4.9
ALLOWABLE COLUMN AND CRUSHING STRESSES CHROME MOLYBDENUM STREAMLINE TUBING ==
20 40
Fty= 75000 PSI
60 80 100 120 140
Trang 12STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING wail 09 Dect Oo 049 01 0g 035 9 os 049 058
C4 14 IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADINGS, TABLE C4.6 SECTION PROPERTIES OF STREAMLINE TUBING
‘Siew ae Wall l we \ 2
We, Ares per fe
Trang 13ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES tubes the primary column strength can be found
by using the curves in Figs C4.9 and C4.10 The crushing stress Fog for oval shaped tubes can conservatively be taken as that for streamline tubes as given in Figs C4.9 and 04.10 For square tubes the local crushing stress Foo can be taken as the crippling stress ofa flat plate For this stress refer to the chapter which covers the buckling and crippling stress of flat plates with various widths, thickness and boundary edge conditions
ULTIMATE BENDING STRENGTH OF ROUND TUBES
C4 17 Charts for Finding Modulus of Rupture Stress
Chapter C3 was concerned with the theory and methods of determining the ultimate yield and failing stress of a section in pure bending It was concerned with finding a fictitious stress Fp which, when substituted
in the well known beam formula M = Fpl/c, would
give the value of the bending moment which
would cause failure
The same procedure as was used in Chapter
C3 can be used to find the modulus of rupture stress (Fp) for round tubes However, since round tubes have been a standard and available structural member for many years, much testing has been done on tubes and as a result rather complete design curves are available for find- ing the modulus of rupture in bending (Fp) for round tubes Pigs C4.11 to C4.14 inclusive give curves for finding modulus of rupture Fp round tubes when fabricated from alloy steels, aluminum alloys, magnesium alloys and titanium alloy 270 260 250 Fhy 2160000 PSI 240 230 220 210 200 190 180 Fy BENDING MODULUS OF RUPTURE, KSt 160 = - O 5 10 1S 20 25 30 35 40 45 50 ost
Fig C4 14 Bending modulus of rupture for round
6Al-4¥ tubing (Titanium) Solution: Solution: C4.15 C4.18 Problems Involving Bending Strength of Tubes PROBLEM 1
A 1-1/4 ~ 058 round tube is used as a Simply supported beam with the supports at the ends The span or length of the beam is 24 inches It carries a uniform distributed load win pounds per inch Find the value of w to cause the tube to fail in bending if the tube is made from the following materials:- alloy steel Fry = 95000, 6061~Té6 aluminum alloy, and 6AL-4V titanium,
The maximum bending moment occurs
at the midpoint of the span and equals wL?/8 = Wx 24/8 =.72 W in lb
The section properties for the tube are D/t = 21.55, and I/y = 06187 obtained from Table C4.3
The beam equation involving the modulus of rupture Fy is M 2 Ful/y Substituting 72w for M, we obtain:
FI b w= aBy
Consider the alloy steel, Fry = 95000 From Fig C4.11 for D/t = 21.55, we read Fp = 117000 Then w = (117000 x 06187)/72 = 100.2 lb per inch Consider tube material as
6061-76 which has a Fry = 42000 pst
From Fig C4.12 for D/t
that FPo/FPtu = 1.06, Thus Fp
44500
21.55, we read 42000 x 1.06 = Then w = (44500 x 06187)/72 = 38.1 Tube material SAL-4V titanium
From Fig C4.14 for D/t = 21.55, we read Fp = 191600 Then w = (191600 x 06187)/72 = 118.7 lbs per in
PROBLEM 2
A beam simply supported at its ends has a span of 30 inches The ultimate design load consists of two equal loads of 2000 lbs each The beam is symmetrically loaded with each load located 12 inches from the ends,
Select the lightest round tube when made from the following material: (1) alloy steel
Fry = 220000, (2) 7075-T6 aluminum alloy
Compare the resulting tube weights The maximum bending moment ts constant and occurs between the load points
Trang 15
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
Since the allowable or failing bending stress is a function of D/t, and since we do not have a tube size, the design or solution procedure is by trial and error
Observation of the modulus of rupture curves show that as D/t increases Fp decreases This is due to the fact that failure in bending is a local failure and the thinner the wall and larger the diameter, the lower the buckling or crushing stress However, the larger the D/t value the greater the section modulus I/y of the tube, which means increasing bending resistance Thus we have two influences which act oppositely relative to effecting the bending strength
There are many ways of guessing a tube size for checking purposes In this example
problem we will assume two values for D/t and see what I/y would calculate to be The two values of D/t will be 45 and 25
Consider the material alloy steel Fru=220000:- For D/t = 45 from Fig.C4.11, Fp=232000 Then I/y = M/Fpy = 24000/232000 = 103
For D/t = 25, Fp = 266000
Then I/y = 24000/266000 = 089
Therefore we will refer to Table C4.3 and select tubes that have an I/y value near the
.089 to 103 range and then find their true pending strength Table (A) shows the selection and the necessary calculations, using Fig C4.11, Table A M.S = Tube Size | Vy |Area| D/t | Fy |M=FbLy (Ma/M)-1 1~7/8 - 035 | 09136 |.2023|53.6 |227000| 20800 ~0,13 1-3/4 -,049|.1083 |.2618/35.7 1248000; 26400 +0 10 1-5/8 -.049|.0928 | 2426/33 154250000 | 23200 =0, 04 The lightest available tube with a
positive margin of safety is 1-4/4 - 049 and its weight for a 30 inch length is 30 x 2618
x 0.283 = 2,22 los Ỷ
Consider the tube made from 7075-T6 aluminum alloy material which has a Fey = 77000 From Fig 04,12:- For D/t = 60, Fp/Fty = 84, thus Fp = 84 X 77000 = 64700 I/y = M/Fy = 24000/64700 = 37 For D/t = 30, Fp/Fty = 1.045, thus Pp = 1.045 x 77000 = 80500 I/y = 24000/80500 = 30 C4, 17 Therefore we will select a tube from Table C4.3 that has a I/y value in the region of 30 to 37
Try 2-3/4 - 058 I/y = 3233, D/t = 47.4 From Fig C4.12, Pp/Ftu = 0.90 Then Fp = 90 x 77000 = 69300 psi Then Ma = Fp I/y = 69300 X 3233 = 22400 This is less than the design pending moment of 24000 so this tube is weak Try 3-058 I/y = 3868, D/t = Fp/Fty = -085 > Fp = 885 x 77000 = Ma = Fy I/y = 68000 x 3868 = 26300 M.S = (26300/24000)-1 = 09 51.7 68000
A study of other tubes in Table C4.3 shows that no other tube would be lighter in weight
Tube weight = 30 x 5361 x 101 = 1.70 lbs., aS against 2.22 lps for the alloy steel heat treated to 220,000 Thus aluminum alloy tubes from a weight standpoint usually yleld
results better than most materials This
conclusion applies to only low temperatures, delow 250°F, as aluminum alloys lose strength rapidly for temperatures above 250° to 300°F The student should calculate the lightest titanium tube and the lightest magnesium tube
using Figs C4.14 and C4.13 respectively and
compare the weight results with the steel and
aluminum as found above
ULTIMATE TORSIONAL STRENGTH OF ROUND TUBES
C4.19 Torsional Modulus of Rupture
In Article A6.2 of Chapter A6, the torsion formula for circular sections, f, = Tr/J, was derived This equation assumes the maximum shear stress on the cross-section of a round bar or tube does not exceed the proportional
limit of the material, or the stress variation is linear as shown in Fig C4.15 and this situation exists under the flight vehicle limit loads Before a round bar made of ductile
T t a if
Fig C4.15 Fig C4 16 material fails in torsion, the Shear stresses fall in the inelastic or plastic range and
Trang 16
STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING C4 18
thus the internal shear stress distribution is similar to that indicated 1n Fig C4.16 The shear stress-strain curve is similar in shape to the tension stress-strain curve.and is equal to approximately 0.6 of the ordinates Thus to find the ultimate internal torsional resisting moment, we can divide the cross— section into a number of circular elements From the stress curve in Fig C4.16, the stress at the midpoints of the circular elements can be found Multiplying the area of the element by this stress times the distance of the element from the center axis gives the moment developed, and adding up the moment of the Shear force on all the circular elements gives the total allowable or ultimate internal resisting moment Ta
If this moment Tg is to be used in the torsion equation, we must replace the true stress curve in Fig C4.16 by a triangular stress curve with maximum value Fs, which will produce the same internal resisting moment as the true stress curve Thus
Tạ = Fạt d/r, or Pep = Tg r/J - - - - (04.10) Fst 1s called the torsional modulus of rupture stress It is a fictitious stress, being higher than the ultimate shear stress of the material Tubes subjected to torsion usually fail by inelastic or plastic instability or by elastic instability if the D/t ratio 1s large To obtain reliable values of Fst, resort is usually made to tests, and since the round tube or rod is the most efficient and most available shape, much testing has been done over the years thus design curves are Teadily available for round tubes of the most common flight vehicle materials
4.20 Torsional Modulus of Rupture Curves
Figs C4.17 to C4.24 inclusive present curves for finding the modulus of rupture stress Fs¢ for steel alloys from Fry = 95000 to 260,000 psi It should be noted that the torsional strength is influenced by the D/t and the L/D values of the tube Figs C4.25 to C4.30 inclusive give curves for the various aluminum alloys and Fig C4.31 gives informa- tion for one magnesium alloy All of these curves were taken from (Ref 1)
The curves are based on a theoretical investigation by Lee and Ades (see Ref 2), and have been found to be in good agreement with experimental results
C4.21 Problems Mlustrating Use of Torsional Modulus
of Rupture Curves
Problem 1
inches long A 1-5/8 ~ 065 round tube ts 16.25 What ultimate torsional moment
IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADINGS Soecitication ww-T- 625 Fru? 36 si fay Torsional Modulus of Rupture, asi Fig C4.31 Torsional modulus of rupture for magnesium-alloy round tubing
will it develop if made of the following materials: (1) alloy steel Fry = 95000,
(2) aluminum alloy 2024-13,
Solution: D/t =-25 J/r=2(t/y) =2x
0.11948 = 0.2390 L/D = 16.25/1.625 = 10 For alloy steel, Fey = 95000, we use Fig C4.17 where for D/t = 25 and L/D = 10, we read Fg¢/1000 = 47,3 Thus Fst = 47300 Ta = Fgt J/r = 47300 x 0.2390 = 12300 in.lb For 2024-T3 aluminum alloy, we refer to Fig C4.26 where we read Fst = 29000 Then Ta = Fst J/r = 6930 in lb 29000 x 0.2390 = Problem 2
A round tube 10 inches long is to carry an ultimate torsional moment of 7000 in lb Select the lightest tube if made from aluminum alloy 2024-T4 and alloy steel Feu = 200,000 and compare the resulting weight of each Solution: Since the modulus of rupture
depends on D/t and L/D and since the tube size is unknown, we will use the trial and check approach The design calculations can con- veniently be made in table form as follows:
Trang 17Fry= 150 kei F /1000 9 0 20 39 40 3 ry 70 s0 on
Fig C4.17 Torsional modulus of rupture - alloy steels heat treated to Fry = 95 ksi
9 10 20 » “ 0 “0 70 8 oA
Trang 18lÍ: i Fy/1000 F,y/1000 t0 9 4 20 9 "0 20 +® 3 0 o 70 a0 04
Fig C4,21 Torgional modulus of rupture -
alloy steels heat treated to Fiy = 200 ksi Fy" 240 kat i/o 4 20 0 10 20 3o 40 sô 60 70 a0 DA
Fig C4.23 Torsional modulus of rupture -
alloy steels heat treated to Fty = 240 ksi
fg/1000
fu»220 kai
10 20 30 +“ 50 “0
DA
Fig C4.22 Torsionai modulus of rupture -
alloy sleels heat treated to Fiy = 220 kai 70 tụ" 260 ksi 20 30 40 50 60 DA
Fig C4.24 Torsional modulus of rupture -
Trang 19ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C4, 21 Itt FORGING
En,"65 ki
FARM (Lied) 15 COMPUTID FOR F_ ~ 79,/088, eNO 0OE5 NÓT ALLOW FOR
TỒN POSUBILITY O# tEOUCEO STHNGTM Tl PARTING Manet
a a ” ” 2 2 “ ” ©
on
Fig C4.25 Torsional modulus of cupture - 2014-T6 aluminum alloy forging Fy tsi ° 10 ” „ « 2 ° 70 * on
Fig C4.27 Torsional modulus of rupture - 2024-74 aluminum alloy tubing
TOPS AOLLED AO F_u*77 kh
a 2 = * ° „ “ ”
on
Fig C4.28 Torsional modulus of rupture - 1075+T6 aluminum alloy rolled rod
sàn xangG F,yt64 ksi
on
Pig C4.26 Torsional modulus of rupture - 2024-73 aluminum alloy tubing,
anes rte
on
Fig C4.28 Torstonal modulus of rupture ~ 061-TS aluminum alloy tubing
MOTE: fet CURVE EEPRESERSTING MATERIA, #aINUEE (L/053 15 COMSUTED FOR E, + 45.038, AbD ODES HOT ALLOW FOR
1 POSSHUNITY OF ELOUCED STHEMG IH
" EY ” ” = a 7 -
on
Trang 21ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES 40" R w = 10#/in Ay TETEETIELTPETE ETP ETP et c Al ©1-1/4 - 049 Rd Tube 20 Fig C4, 33 BỊ wi? = 10 x 860 = 8600 40 _ +k—L_— 88.ẽ 7 „683 sec 21 cos 2 2) from Table A10.2 of Chapter A10 cos of 683 = .77597 hence —— = 1/.7787 = 1.29 cos L/2} Substituting in equation (A) “hax
From Fig C3.2, the modulus of rupture Fy for a 2024-T3 round tube which has D/t value of 25.5 equals 64000 psi Therefore the bending
strength M = Fpl/y = 64000 x 0534 = 3420 in.lbs = 9600 (1-1.29) = - 2490 in 1b
Stress ratio in bending Fp = ae = 728
Re + Rp = 191 + 728 = 917 which shows that member is not weak
-1lg9l+ V28) The margin of safety could be read đirectly from curves in ?1z C4.32,
The student should notice that the maximum bending moment of 2490 in lb is 24 percent greater than the primary moment which equals
wL?/8 = 10 x 40°/8 = 2000 The lateral de-
flection at the midpoint of the tube thus equals 490/400 = 1.22 inch
The secondary moments due to lateral de- flection do not vary linearly, so if design loads were increased the calculation of the maximum bending should be repeated instead of assuming that the moment would increase directly as the applied load to the beam
C4, 23 Combined Bending and Tension
The interaction equation for combined C4 23 bending and tension on round tubes, which is widely used, is Rp + Rg SL (C4.15) (Ref.2)
This equation is plotted in Fig C4.34,
The stress ratio Rt = ft/Fty The figure is
based on D/t = 10 and in general is conserva- tive for other D/t ratios ma
C4.24 Combined Bending and Torsion
The interaction equation for combined bending and torsion from (Ref 1) is,
Ro + Rep Zl - (C4.18)
Trang 22
STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADINGS C4, 24 for various values of Rp and Ret are also shown on the figure The expression for Margin of Safety is, 1 V Rb + Ret ILLUSTRATIVE PROBLEM 1 M.S = -1 (C4.17}
A 1-1/2 - 058 round steel tube (Fry =
125000) 1s 30 inches long It its subjected to an ultimate design bending moment of 10,000
in ibs and a torsional moment of 6000 in los Find the Margin of Safety
Solution: D/t = 25.85, I/y = 0912, L/D = 20 To find Fp, we refer to Fig C4.11 where for D/t = 25.85 and Fen = 125000 we read Fp = 148000 Then Mg = Pyl/y = 148000 x 0912 = 13500 in lb M 10900 Then Ry = 7 > Tas00 7 774
To find Fgt, we refer to Fig C4.18 where for D/t = 25.85 and L/D = 20 we read fsy⁄1000 to be 61, whence Fgz * 61000 =~ 7 ~ 6000 Ret =a 7 are? -o4 M8 =————- 1 V Rp + Rạp” a 1 = -12—4,-1 = 09 V7 + ee 917
thus the ultimate strength of the tube has 9 percent Margin of Safety under the combined loading
In a design problem which involves a trial and error procedure, using an equivelent
torsional moment Tg which will produce the same torsional stress as produced by the combined bending and torsional loads is quite useful in shortening the trial and error procedure
Let fo max) 7 Ter/2l -~~ - (c3.18)
Ts (max) 2180 equals ¥ thr {f£,/2)" ~ = - (C4.19)
Also f, = Tr/21, and fy = Mr/I Substituting these values in C4.18, =-EÍM/T+ HS Ís (max) = ml L+ (TAM) ] Equating C4.20 and C4.18 and solving for Tes (C4.20) Tg MV 1+(t/M)* -
Having the value of Tg, select tube sizes that will develop this torsional moment Ta as was done in Problem 2 of Art C4.21 These sizes are then checked for combined bending and torsion as illustrated above in the example problem
~.C4,24 Ultimate Strength in Combined Compression, Bending and Torsion
The interaction equation for combined compression, bending and torsion from (Ref 1) 18;
Rot + Rấp # (L-Re)”
Figs C4.36 and 37 show a plot of this equation The expression for the Margin of
Safety ts,
= 1
M.S = ———-1 z 3 -+ - (C4.25) Ro tv Rye * Roe
To illustrate the use of the interaction curves, let us assume the following values for the three stress ratios: -
Ry = 333, Rh = 20, Rez = -20
Then Rgt/R, = 20/.323 = 60 st/“e `
In Pig C4,36 locate point (a) at the
Intersection of Re = 333 and R} = 20 Since the intersection point (a) lies inside the Rgt/Re = 60 curve, we know that a positive
margin of safety exists A line is now drawn
through (0) and (a) and extended to an inter-
section with the Rsy/R„ = 60 curve at point
(b) Projecting vertically downward to Ro
Trang 23
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES C4, 25 Projecting horizontally from point (tr) to #* ses: Tin = 1.005
Rp scale, we read Rp = 323 Then M.S = * +
(.323/.200) - 1 = 62; which checks value L 1 1
previously found I7 the ratio of Rsv⁄/Re is sec 2” weet “EEET ˆ 1.87 (See Table
greater than one, we use the curves in Fig $ A10.2)
C4.37 and use the ratio Re/Rsy hence,
Substttating in equation C4.23, Max = 10*184.7 (1=1.87) = -1347 in 1p M.S = - 1= 62, as given +333 + ¥ 207 + 207 from use of curves 1.0 Ot ILLUSTRATIVE PROBLEM
A 1-1/8 ~ 049 round tube made from 2024-T3 aluminum alloy carries the ultimate design loads as shown in Fig C4.38 Find the
margin of safety under the combined loading << w= 10#/in, a |, gg 1-1/8 =,049 2024-73 Ra, Tube, J ea Fig C4 38 Solution:
The maximum bending moment due to symmetry will occur at midpoint of tube For a beam column carrying a uniform side load with no end moments, the maximum moment is given by the
following expression (See Chapter Al0, Table Al0.1)
M ‘max = wi (1 - see 23
1z /BE1 _ /10;200,0Q0 x 0240
P 1600 = V 154.7 = 12.45
The column strength for a 25 inch length and end fixity c = 1 can be read from Fig C4.6 and equals 3700 lb Then Re = P/P,g = 1600/3700 = 432
To find the ultimate bending strength, we refer to Fig C4.12, where for D/t 2 1.125/.049 = 22.95, we read Pp/Fry = 1.04 Then Fp = 1.04 x 62000 = 64500 Thus My = FpI/y = 64500 x 0427 = 2760 in 1b Therefore Rg = eee „487,
To find the ultimate torsional strength we refer to Fig C4.26 where for D/t = 22.95 and estimating location of L/D = 25/1.125 = 22.2 line, we read Fst = 27500 Then Ta = FstJ/r = 27500 x 0427 x2 = 2350 = 7 650 _ whence Ret = 7 * Gasq = 275 M.S ST —-1 Re + V¥ Rp* + Ret =) -432 + ¥ ,487% + 1276" “992 -1 = 01 C4, 2S Ultimate Strength in Combined Bending and Flexural Shear
The interaction curve from this type of
combined loading from (Ref 2) is,
Ro+ Rg zl + -+ (04.24)
M.§8 =— —-1 -~~ (4.25)
Vv RG + RS
The allowable flexural or transverse shear stress is taken as 1.2 times the allowable
torsional stress of the tube (Fst) ILLUSTRATIVE PROBLEM
Trang 24STRENGTH & DESIGN OF ROUND, STREAMLINE, OVAL AND SQUARE TUBING Cả 26 600 12.5 300 Diagram Bending Moment Diagram Fig C4 39
The critical section 1s adjacent to the midpoint of beam where shear and bending forces are largest
The ultimate bending strength can be found from Fig C4.26 where for D/t = 21.60 we read Fp/Fry = 1.07 Then Fy = 62000 x 1.07 = 66300 and therefore M, = FpI/y = 66300 x 06187 = 4100 Then Rp = M/M, = 3750/4100 = 915 — The allowable shear stress Fg can be taken as 1.2 Fez (Ref 2)
To find Fst, we refer to Fig C4.26 where for D/t = 21.6 and L/D = 20, we read Fst = 28400 Then Fg = 1.2 x 28400 = 34000
The maximum shear stress in a round tube
which occurs at the centerline axis is given by the equation, xáy —80_—_ fs Sửa (1 my nh) where V = vertical beam shear load A = tube area D = outside diameter d= inside diameter An approximate formula is
fg = ai, for D/t »10 error less than 14
Using the approximate equation, = 2X 300 †s = “Ẩn = 2760 pst = fa = 2760 „ Rs = 5g * gago0 ~ +082! Ms s— tL -15 + -1 V Rb + RS v 915" + 081” = 09 The effect of the shear stress is less than 1 percent In using Fg = 1.2 Fgz, 1f the result is greater than Fg, for the material, use Fs, for Fg
C4, 26 Ultimate Strength in Combined Compression, Bending, Flexural Shear and Torsion
The interaction equation for this combined
loading as presented in Ref 2 is,
we
IN TENSION, COMPRESSION, BENDING, TORSION AND COMBINED LOADWGS
Re + Rep +V Ry + RG FL -
Interaction curves for this equation are given in Fig C4.40, where Rg the stress ratio for flexural shear can be found as explained in the previous article (C4.26) ° a fSaen ee ena gC ° a n a ' gu ah + 2 2 , 2G 2.35 3 = fe hag tấn Se " 886 EE a OSE Bg + 2a ag pg 22 ——^;3 eS 3 gs Og 4.27 Ultimate Strength in Combined Tension and Torsion,
The interaction equation for this type or loading as presented in Ref 3 is,
RE +R, = 1 - (C4.27)
Where Ry = ft/Fry
M.S 2 a (04.28)
v RỆ + Rật
C4.26 Ultimate Strength in Combined Tension, Torsion and Internal Pressure p in psi
Ref 3 gives the following interaction equation,
Trang 25
ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES
RE + Rep + RO = 1
Where Rp = 2/ Prax » Where Paax =
where t = wall thickness and d = tube diameter The expression for Margin of Safety is, 1 V Fệ + Fấy + Rộ PROBLEMS Ñ) M.S = -1
Fig 1 shows a portion of a steel tubular fuselage of a small airplane The
critical tension and compression load is Shown adjacent to each truss member Assuming and end fixity coefficient c = 2 for all members, select tube sizes for all members of the truss The minimum size to be used is 3/4 ~ 035 The top and bottom longerons should be spliced at least once using telescoping sizes The material is
alloy steel Fy, = 95000 and truss is welded ! pot +— 3 —m— 1> — _U, -5340 Ú, -490 420 _U 21
For the cantilever welded steel tubular truss of Fig 2, select the lightest members for the truss loading as shown (2) 4000# 30 20 30" 30" 10006 2000# Fig 2
The top and bottom longerons should be continuous members Minimum size 3/4 +
-035 Use C = 2 for web members and c = 1.5 for longerons Material (chrome-moly steel Fy = 95000 psi} (3) (6) 4,27 If the truss of Fig 2 is heat treated to 180000 psi after welding, how much weight could be saved over the results obtained in problem (2)
Same as problem (2) but change material to 2024-T3 aluminum alloy round tubes Members to be fastened together by rivets and gusset plates For design of tension members assume that rivet or bolt holes
cutout 10% of tube area
Fig 3 shows a typical tubular engine mount structure The engine 1s supported at points A and B For design purposes assume that engine torque is reacted 60% at A and 40 at B Tube material is steel Feu = 95000 psi Use C = 1 for all
members Determine tube sizes for the following design conditions:
Condition I Vertical Load factor = 10 (down ) Thrust Load factor = 2
(forward) Engine Torque Load factor = 2 Condition II (Same as I except vertical
load factor is 5 up
General Data: xelznt of power plant in- Stallation = 440 lb = 400 1b Engine 120 at 2000 R.P.M Maximm engine thrust HP = The loads shown in Fig 4 are to be
transmitted to the support at the left, or in other words, 4 cantilever structure The problem is to design the lightest truss configuration using round tubes of alloy steel Fry = 95000 and welded to- gether at the truss Joints Use ¢ = 1.5 for end fixity of all members There are