3—general criteria for analysis and design 350 3 350 3r 13

... ® 37 2 x 0.95 = lel 1b, Total forces in x direction on pilot equals 180 ~ 17 = 1 63 1b Total force in g direction = 36 0 + 180 - 161 = 37 9 lb Hence Resultant force R, equals v 37 9& #34 ; + 16a& #34 ; ... 18 10 a 7500 aasso M4180 7 Axial Load 130 00 Ib Diagram ~=Compression 133 40 Shear Diagram Ữ ZA -840 -7500 Ib Bending Moment Diagram Fig AS 13 ~ 120000 ~ 133 440 in Ib Example Problem 4 Fig AS.14 ... therefore equals 2 836 0 x sin 30 = 14180 1b, “ Fig AS. 13 shows the oleo strut as a free body with the reactions at A, D and E as calcu- lated Fig AS. 13 also shows the axial load, vertical shear and

Ngày tải lên: 07/08/2014, 10:20

... (5. 535 ° + 6.667*%) = 90,667 lạ =ZA X® = 1 (10.667%) + 2 (5 .33 3%) = 170.667 Tyg = 2A x2 = 1 xX 6.667 (- 5 ,33 3) + 1 (- 5 .33 3) (+ 5 .33 3) + 0.5 x 10.667 x 2.667 + 0.5 x 10.667 (- 5 .33 3) = - 21 .33 3 ... all forces both external and internal about point (0), IM, = 1000x2+400x3+171 23 = 2 032 3 in.1b (171 23 equals summation of column 43) Thus for equilibrium a negative moment of -2 032 3 Trang 22 ANALYSIS ... flow system of Fig A15.10 However it acts 13. 3 to right of shear center, hence it produces a clockwise moment of 100 x 13. 3 = 133 0 in.1b on the cell For equilibrium, this moment must be balanced

Ngày tải lên: 07/08/2014, 10:20

... 1.000 3. 000 0.125 0.250 0.500 1.000 2.000 38 .000 | Đnsis A B A B A B oO 30 32 30 30 29 34 36 34 35 34 18 19 16 15 14 26 28 23 24 21 12 13 10 10 10 20 21 19 22 18 22 22 22 32 | 23 23 23 23 23 43 ... 200 23. 20 x 10 5 45 15 117, 000 21.75 x 10 * 42.81 121, 500 20 30 x 10 + 40, 58 125, 800 18,85 x 10 ¢ 38 . 43 130 , 000 17.40 x 10 © 36 32 134 , 000 15.95 x 10 ¢ 34 ,25 137 , 700 14, 50 x 10 * 32 .22 ... properties: Pray ksi 38 39 36 36 38 2 24 16 16 2 14 14 11 11 14 19 19 : 19 45 45 50 55 55 60 23 + 23 (e/D =2.0) 32 32 & #34 ; 32 €, percent 8 9 7 7 6 E, 108 pst 6 .3 E., 108 pat - 83 G, 10t p8i

Ngày tải lên: 07/08/2014, 10:20

... 058 | 0. 935 | 00867) 530 00 30 75 2760 2 102 | 0.840 | 007 73] 52500 535 0 430 0 3 135 ; 0.75 00688 | 52100 7025 5040 4 153j 0.65 „00591 | 51500 T870 4820 5 „185 ¡ 0.55 00494 | 51000 8410 431 0 8 180; ... load P = 235 00 1b f, = 235 00 /.7854 = 29500 f2 = 235 00 /,4418 = 531 00 Portion 1 f /Fo., * 29900/72000 = 415 From Fig €2 .36 for n= 16.6, we read E,/E 21.0 Portion 2 f2/Fo., = 531 00/72000 = 738 From ... column curve for 7075-T6 Alum Alloy extruded material With L/P = S5, we read allowable stress Fo = 33 ,500 psi Therefore P= A = 33 ,500X0.7854 = 26 ,30 0 1b f, = 33 ,500 and fa = 26 ,30 0/.4418 59,500,

Ngày tải lên: 07/08/2014, 10:20

... Tô | 1 230 4| 1 .35 07 |1 44t 0 83 | 14 39 99 | 1699 | 07 93 | 139 8| 39 96 | 33 13 | 139 9 | 4 136 ay fe Ò [Ea Soot | asso | 27498 | 1 832 095 | 13 5685 |1 933 | 0889 | 1456| 39 54 | 37 4 | 27 03 | 81 13 L1) - ... | t3184| 433 6 | 1. 233 2 098) 17 : 2172 | 738 4| 0 133 | 037 2| 477 | 0560 | 0647 | 5077 -120 3. 431 | 39 23 | 4580| 60 13 | 1.4189 | 731 9 |1.1149 9 63 { l6 -2o | 8246| 01 43 231 | 0618| 0714| 4 033 1 136 ... 1469| 4600 | 38 35 | 2 438 | #171 236 80 |l93t | 116L | 1807| 4 935 3 | 4794 | 30 75 | 92 23 “ 34 32 |2186 | 4298 | 2020| +92 | 34 09| 34 79| 90 120 30 3G |1 730 | 133 6 | 2441] #462 | 633 9 | 433

Ngày tải lên: 07/08/2014, 10:20

... 1400 /3. 73 = 37 5 psi Compressive stress due to constant bending moment of 5170 in lbs is, f, = M,2/T,, = 5170 x 4,. 233 /49 .30 = 444 psi Total T1 3 375 + 444 = 819 pst Trang 12 ANALYSIS AND DESIGN ... 5.7/, 035 = 1 63 psi It is in the same direction as the torsional shear flow and thus is additive Total fg = 1 63 + 854 = 917 psi Re = fs/fs„„ = 917/2760 = 33 2 Ro + RG = 1, Sudt.:- 906 + 33 2& #34 ; ... factors between web and flange elements for simple shapes like channels, Z, H, square tubes and formulated design charts for such shapes (Refs 1 to S inclusive.) C6.4 Design Charts for Local Buckling

Ngày tải lên: 07/08/2014, 10:20

... ⁄% 3, 7te Trang 13 ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES whence, Dy/ty 0 .37 2 Ạ/ty The rivets are 3/ 32 diameter Brazier head type AN456, 2117- 13 material spaced at 0.75 inches 38 , ... load for each angle and add up the total ror the 6 angle units The crippling stress will Trang 3 ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Area of angle (1) = o3O9 =A Pog * 030 9 x 39 ,30 0 ... rivet type use C = 3 and for counter-sunk or dimpled rivets Trang 7 ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Figs C7.18 and C7,.19 show a plot of equation C7.22 for aluminum alloy materials

Ngày tải lên: 07/08/2014, 10:20

... 10,000 ~l©i Tìg C8 24 (Ref 16) The & #34 ;Á& #34 ; and & #34 ;B8& #34 ; curves are recommended for practical design Tne & #34 ;A& #34 ; level curve is recs for use for those structures where # re would ... Tngenieur Archiv Vol 3, 1 932 , (12) Gerard & Becker Handbook of Structural Stability Part IIL Buckling of Curved Plates and Shells NACA T.N 37 83, Aug 1957 ( 13) Hopkins and Brown The Effect ... C8$. 13 and Œ8.1Za give convenient design curves for finding the bending buckling Stress based on 99 percent probability and 90 BUCKLING STRENGTH OF MONOCOQUE CYLINDERS PRORAELITY CORE 35 00 4000

Ngày tải lên: 07/08/2014, 10:20

... 2400 433 2 130 0 8250 <a 9-625 00844 49400 929 5050 ị 8 p 00541 -38 000 497 32 90 & |e 9 .37 5 00822 4 930 0 828 5890 Ñ 8 .37 5 00559 | ~38 800 492 31 30 3 9,128 00800 49200 627 37 20 z 3. 4125 ... Frintore-| 10 93) 3, 532 ) 38 6] 1 .36 | negl 1 ,36 0 Ị 1/ 13 x ,051 [te ? ?35 10| 0 | 0 o |z 6.87 53 - 1 ,38 0 \ 1 .38 [Rewer avgiew | 1780| -3, 301|~.5aa/ l.e4 | 008 1,848 G T man ~,044 | -3, 0 63| 087|~ 208i ... ay al ? ?31 25 ‹00798 | .35 00 0*| 10940 99200 11.56 23] ,o1014 32 050 | '16280]- 4 i Ho ' ng 458090 37 3 ng b 11 .37 5 -00997 32 000 4870 $5450 ẩ | d| 31 250 200 734 |-45150 | 5640 47200 zl§ 32 ,225

Ngày tải lên: 07/08/2014, 10:20

... skin and spaced at 15& #34 ; ‘The rings are 1& #34 ; x 3& #34 ; x 1& #34 ; 24ST Z sections, 040& #34 ; in thickness (see Pig.Ccll. 43) An analysis will be made of stringer #3, skin panels 2 -3 and 3- 4, and ... value for a 2) Determine c, epg and egTR using this assumed value of a 3) Substitute ¢, epg and egpa into the formula for tan? a and get a & #34 ;new& #34 ; value for a 4) Repeat stens (1) to (3) as ... moment it must carry For example, consider the end stiffener of the beam of Fig C11. 13 in Art C11. 13 Using the data from Art (11.27, C11, 23 w=kqocota -69 (18 ,35 0 x 025) x 1.228 = 38 6 _1b./in This

Ngày tải lên: 07/08/2014, 10:20

... 4 130 and 8 630 Steel 150 KSI H T Curve 3- for All Aluminum and Steel Alloys a Ste Curve 4 - 4 130 and 8 630 Steel 180 KSI HT Curve 5 - 35 6-16 and AZ91C-T6 Sand Castings Curve 6 - 2024-T3 and 2024-T4 ... 1,975 110 2, 684 3, 189 218 3, 868 4, 937 37 3 5, 261 6, 8 83 592 6, B71 9, 104 884 8, 697 11, 5 63 1, 260 10, 738 14, 719 1, 730 15, 4 63 21, 8 73 2, 980 21, 046 29, 520 4, 740 27, 489 36 , 759 'ï, 010 ... 2014-T6 and 7075~T6 Die Forging (L) 4 130 and 8 630 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

Ngày tải lên: 07/08/2014, 10:20

... % 3 1 l$ % % % % Shear strength, Ib: 30 58, Fi =28 kei 99 208 36 8 33 6 | 802 | 1⁄40 | 2/290 | 3, 280 2117 23, Fy = 80 ksi 106 217 38 8 39 8 sez | 135 0 | 2460 | 3/ 510 2017-T31s, Fy =34 bat 30 17-T3, ... 51L | 98 3/ 16 0 025 cĩ TL] 87 1 /32 0 032 ¡ 108 | 139 1⁄4 0.040 137 | 189 ¡ 9? ?38 0,051 i 186 | 281 j 5/18 0.064 | 242 | 38 8 | 11? ?32 0.072 179 | 31 8 Ị 3/ 8 0.081 ¡ $20 Ì 4P cị 13/ 32 0.091 | 38 8 Ì 498 ... 204 37 0 | 36 2 eee 34 5 419 ae 538 594 401 515 610 614 811 481 557 a 669 902 562 6 23 788 761L 982 638 746 861 842 1,0 53 854 1,017 9 13 1,115 1,018 1 ,31 3 1,021 1 ,35 7 ¬ 1,574 & #34 ; 1,694 175 83 |

Ngày tải lên: 07/08/2014, 10:20

... consider Fig 93. 17 Here two tees, & #34 ;2& #34 ; and & #34 ;b& #34 ; carrying axial lead, P, are seen to be spliced togetaer by 2 pair of angles & #34 ;c& #34 ; Since the lower leg of & #34 ;Q& #34 ; 1s thicker ... Provided Fig D3 34 The recessed panel can continue to carry its shear load but there will be out-of-plane & #34 ;kick& #34 ; loads These are resisted by the framing members & #34 ;a& #34 ; and & #34 ;b& #34 ; as ... as shown in Fig C3.20, Fig D3 20 Before the cut-out was made the members shown by solid lines (flanges, stiffeners, weds) are present The members & #34 ;a& #34 ; and & #34 ;b& #34 ; are added to

Ngày tải lên: 07/08/2014, 10:20

... proponents of AM and XP have expressed themselves quite clearly and forcefully on the subject of agile modeling and programming, and, judging from the current bleak and stony landscape of systems ... 48(4), 2 83- 319 Erickson, J., & Siau, K (20 03, April) UML complexity In Proceedings of the Systems Analysis and Design Symposium, Miami, FL Erickson, J., & Siau, K (2004, December) Theoretical and ... programming and collective ownership, for example If that is the case for obvious and even for nonobvious relationships, then what effect upon the overall success of the project, and ultimately

Ngày tải lên: 07/08/2014, 11:22

... CHAPTER THREE Object-oriented analysis and design Object-oriented methods cover methods for design and methods for analysis Sometimes there is an overlap, and it is really an idealization ... SPACE AND TIME 15 Figure 2 .3 (a) New Zealand in absolute space (b) New Zealand in 19 53 time space (c) New Zealand in 1970 time space. (Reprinted by permission of Oxford Press. Source Gatrell 19 83, ... nodes and arcs, keep records of locations within the system, and handle numerous travel times at both nodes and arcs in the network. Although ‘ARC/INFO NETWORK can meet the requirements for standard

Ngày tải lên: 08/08/2014, 00:23

... overlay networks [34 ], hierarchical foreign agents [35 ], as well as recent Internet drafts: fast handovers for Mobile IPv6 [36 ] and low latency handovers in Mobile IPv4 [37 ] A handover mechanism... ... mechanism... 1998 [35 ] Caceres, R., and V Padmanabhan, “Fast and Scalable Handoffs for Wireless Networks, ” ACM Mobicom, 1996 [36 ] Dometry, G., (ed.), Fast Handovers for Mobile IPv6, Internet ... raised demands for QoS support. This is due to the variety of Internet applica - tions and the increased number of users, which have different demands for con - tent, type of information, and quality

Ngày tải lên: 14/08/2014, 14:20

... between the analysis ANALYSIS AND DESIGN OF COMBINED FOOTINGS AND MATS 33 6.2R-7 and performance (Bowles 1974). End-spring doubling for beams will give a minimal spring coupling effect. 3. 3.2 Estimating ... Requirements for Reinforced Concrete 8.2 -Cited references ACI Committee 33 6, 1966, Suggested Design Procedures for Combined Footings and Mats, ”... width ANALYSIS AND DESIGN ... ACI 33 6.2R-88 (Reapproved 2002) Suggested Analysis and Design Procedures for Combined Footings and Mats Reported by ACI Committee 33 6 Edward J. Ulrich Shyam N. Shukla

Ngày tải lên: 24/10/2014, 22:01

... Fig. 2.5 Fig. 3. 1 Fig. 3. 2 Fig. 3. 3 Fig. 3. 4 Fig. 3. 5 Fig. 3. 6 Fig. 3. 7 Fig. 3. 8 Fig. 3. 9 Fig. 3. 10 Fig. 3. 11 Fig. 3. 12 Fig. 3. 13 Fig. 3. 14 Fig. 3. 15 Fig. 3. 16 Fig. 3. 17 Fig. 3. 18 Model of ... 3. 3 Case 2: Rotating Disk Subjected to Stationary Load (RD-STL) 3. 4 Dynamic Response of Disks 3. 4.1 3. 4.2 3. 4 .3 In-plane Response Varying Speed Varying Load 8 13 14 16 16 21 21 24 25 27 33 ... R1 and R2 are radii of the outer arc 1-2 -3 and inner arc 4-5-6, respectively, η1, η2 and ? ?3 are the polar angles (in radian) of radial lines 1-6, 2-5 and 3- 4, respectively. The band width and

Ngày tải lên: 16/09/2015, 15:43

... parallel-connected converters under CLC scheme 16 18 19 20 21 24 3. 1 3. 2 3. 3 3. 4 3. 5 3. 6 3. 7 3. 8 3. 9 3. 10 3. 11 3. 12 3. 13 3.14 Basic UPEC cell . . . . . . . . . . . . . . Cout -Lin -PF relationship ... 1 R10x 1K R11 1K R 13 10 R15 10 R14 10 R16 10 +5V R20 56K +5V R 33 56K R9 56K 12 5 6 13 R36 56K R22 56K R29 56K +5V R25 56K +5V R40 56K C3 33 0pF R12 24K C4 33 0pF R17 24K + R10 18.6K U4A X1 MC 331 74-C - R21 56K R30 18.6K + U4B 14 7 +15V 1 U4D U4C X3 MC 331 74-A R 23 18.6K R32 56K X2 MC 331 74-D + - 3 2 - R28 56K R37 18.6K X4 MC 331 74-B + R39 56K 9 8 +5V +5V R19 10 D1 1N4148 D2 1N4148 R31 10 D5 1N4148 Vin Iin V+ V- D8 1N4148 D7 1N4148 D4 1N4148 D3 1N4148 R38 10 D6 1N4148 +5V +5V 14 11 X4x SKM150GB063D 6 105 106 [...]... ... /2 − Vs sin ωt (5.16) 84 Figure 5. 13: Hysteresis current control scheme 15 Current (A) 10 5 0 −5 −10 Tf −15 0 .32 0 .33 0 .34 0 .35 0 .36 0 .37 0 .38 0 .39 Time (sec) 0.4 Figure 5.14: Input

Ngày tải lên: 29/09/2015, 12:52

... HV 3s p11   3( 1)s HV s 3s   3( 1)p10 HV 3s p10   3( 1)p11 HV 3s p11  3( 1)p11 HV s p11   3( 1)d20 HV 3s 3d20   3( 1)d21 HV s 3d21   3( 1)d21 HV 3s 3d21   3( 1)d22 HV s 3d22 ... (II -3)   2(1)p11 HV 3s p11   2(1)p11 HV 3s p11   3( 1)s HV s 3s   3( 1)p10 HV 3s p10   3( 1)p11 HV 3s p11   3( 1)p11 HV 3s p11   3( 1)d20 HV s 3d20   3( 1)d21 HV 3s 3d21 ...   3( 1)s HVA1 3s   3( 1)p10 HVA1 p10   3( 1)p11 HVA1 p11   3( 1)p11 HVA1 p11   3( 1)d20 HVA1 3d20   3( 1)d21 HVA1 3d21    3( 1)d21 HVA1 3d21   3( 1)d22 HVA1 3d22   3( 1)d22

Ngày tải lên: 30/09/2015, 06:37