Aircraft Structures 1 2011 Part 7 potx

... 8000 i x 1. 223 x 602 x 14 .5 N = 1. 113 L ipV2S cL= From Fig. 8 .10 (a), a = 13 .75 " and CM,cG = 0. 075 . The tail arm I, from Fig. 8 .10 (b), is 1 =4 .18 cos(a-2)+0.31sin(a-2) ... into Eq. (8 .12 ) we have =nW or dividing through by 4pV2S We now obtain a more accurate value for CL from Eq. (iv) 1. 35 4 .12 3 CL = 1. 113 x 0. 075 = 1. 088 gi...

Ngày tải lên: 12/08/2014, 03:20

... conditions for this particular case are v = 0 at z = 0 and 1. Thus A = 0 and Bsinpl= 0 For a non-trivial solution @e. v # 0) then sinpl=O or pl=n.rr wheren= 11 213 1 giving or ... 03 4 2 $PCRTnZA: U + V = -En A, - - n =1 41 n =1 413 (6. 51) Assigning a stationary value to the total potential energy of Eq. (6. 51) with respect to each coeffic...

Ngày tải lên: 12/08/2014, 03:20

... d2pB.f 13 ~ 213 DC 4000 80 000 PB,f13 11 3 pDsf 1 32 013 320 PD,f 1 32 013 320 1 48 013 240 BA 4000 60 000 2pB,f/3 213 pD,f 0 0 0 0 FC 4000 10 0 000 0 0 C = 12 68 C = ... CB 800 20 86.6 + R/2 11 2 17 32 + 10 R 87. 6 -1. 8 BD 800J3 30 -J3R/2 -J3/3/2 20J3R CD 800 20 R 1 40R 2 .1 AD 800 20 RI2 11 2 IOR 1 .o C = -268 + 12 9.2R...

Ngày tải lên: 12/08/2014, 03:20

... 7. 70 [1 + 0 .75 (b/d)'] The relationship between the diagonal tension factor and buckling stress ratio is b and d having their usual significance. T/TCR 5 7 9 11 13 15 k 0. 37 ... Example 6 .1 and are listed below A = 600mm2 Zxx = 1. 17 x 10 6mm4 J = 800mm4 = 0. 67 x 10 6mm4 I? = 2488 x 10 6mm6 Zo = 5.32 x 10 6mm4 Problems 203 Fig. P.6...

Ngày tải lên: 12/08/2014, 03:20

... qb;23 = (by qb. 21 = -7. 22 X 10 -4(250 X 10 0) = -18 .1N/mm qb .18 = -18 .1 - 7. 22 x 10 -4(200 x 30) = -22.4N/mm qb: 87 = qb: 21 = -18 a1 N/mm (by Taking moments about the ... (9 .76 ) and (9 .77 ) we have, from Eq. (9.38) 0 = [qb, 81 X 60 x 480 + 2qbT12(240 x 10 0 + 70 x 240) + 2qb,23 x 240 x 10 0 -2qb.43 x 12 0 x 10 0 -...

Ngày tải lên: 12/08/2014, 03:20

... 3 -3 2 -13 3 0 -0.05 0 6 .7 - 17 7. 3 0 0.3 0 0 3 -10 0 -0 .1 -0.05 10 5 -10 1.3 0.6 0.3 -3 3 4 10 0 -0 .1 0.05 -10 5 10 1.3 0.6 0.3 -3 3 5 13 3 0 0.05 0 6 .7 17 7. 3 0 0.3 ... Table 10 .3 49 8 = 49 IO = ql6 1 = q2 1 q32 = qS7 ‘?loll = q1 516 = 30.3 + q 21 q43 = q76 = 411 12 = q1 415 = 53.5 + 9 21 q54 =...

Ngày tải lên: 12/08/2014, 03:20

... 214 -16 composite materials 2 18 -20 glass 218 history 211 -14 plastics 2 17 , 218 steel 216 , 2 17 titanium 2 17 analysis of pin-jointed frameworks 500 -7 analysis of space frames 5 07- 9 ... buckling 17 3 instability of stiffened panels 17 5 -7 interrivet buckling 17 7 Kirchhoff, G.R. 13 3 maximum values of stress 13 7 neutral plane 12 2 plastic...

Ngày tải lên: 12/08/2014, 03:20

... Silhouette-basedHumanActivityRecognitionUsingIndependent ComponentAnalysis,LinearDiscriminantAnalysisandHiddenMarkovModel 259 1 1 1 1 1 1 ( ) ( ) ( ) ( , ) ( ) ( ) ( ) t ij j t t t q q t ij j t t i j i a b O j i j i a b O j               (13 ) 1 1 1 1 1 1 ( ) ( ) ( ) ( , ) ( ) ... NewDevelopmentsinBiomedicalEngineering258 1 1 1 1 1 1 (...

Ngày tải lên: 21/06/2014, 19:20

... every particle i do 9: compute weight using p(Z(k +1) |X v (k +1) ) 10 : end for 11 : re-sample the particles 12 : if current observed feature exists in the map then 13 : for every particle i do 14 : ... p(Z(k +1) |X v (k +1) 8: end for 9: calculate robot pose & landmark position from particles & associated weight 10 : re-sample the particles 11 : until robot stop navigati...

Ngày tải lên: 11/08/2014, 16:22

... and ventilation 17 7 Figure 9 .10 Single-duct system Warmer air ducts Fresh air _£E Cooler air ducts 1 2 3 4 "i ' ,fc 12 Recircuiated air U13 =13 ; " ;13 1 Mixing box 2 Fan 3 ... valve 6 Humidifier 7 Pre-heater 8 Zone heaters 9 Re heater 10 Plenums 11 Warmer air plenums 12 Cooler air plenums 13 Pre insulated spiro ducting 14 Air t...

Ngày tải lên: 12/08/2014, 03:20