Aircraft Structures 1 2011 Part 16 pps

... shear strain 16 , 18 , 19 strain gauge rosette 29 strains on inclined planes 21, 22 Strain 16 -32 586 Index Materials of aircraft construction 21 1-20 aluminium alloys 214 -16 composite ... 1- 20 aluminium alloys 214 -16 composite materials 2 18 -20 glass 218 history 211 -14 plastics 217 , 218 steel 216 , 217 titanium 217 analysis of pin-joint...

Ngày tải lên: 12/08/2014, 03:20

... 000 -2pB,f/3 - 213 0 0 16 0 0 FD 4000J2 -80000J2 -d2PB.f/3 - ~ 213 0 0 640J 213 0 CB 4000 80 000 PB,f/3 11 3 EB 4000 20 000 2PB,f 13 213 0 0 16 013 0 0 0 - 16 0J2/3 0 FB ... 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...

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

... (6 .11 7) For horizontal equilibrium (FT - FB) COS /3 - gttd COS' = 0 (6 .11 8) Taking moments about B wz - FTd COS @ + $gttd2 Cos2 a = 0 (6 .1 19) Solving Eqs (6 .11 7), ... (6 .11 7), (6 .11 8) and (6 .11 9) for q, FT and FB at = td 2w sin 2a (1 -$tan@) 1 tan@ FT=-[z+?( W )] d cos @ 1 - -tan@ FB=-[z-F( W )] d cos @ E...

Ngày tải lên: 12/08/2014, 03:20

... 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 giving...

Ngày tải lên: 12/08/2014, 03:20

... (ii) and (iii) in Eq. (9.69) we have 2Ak = [ 61 25 x 1. 15 ~1 dsl + 2( 312 .5 - 4.02s2)2.5ds2 200 1 + jy 2( 1 1 1. 5 + 12 .54 1. 5 ds3 Evaluation of Eq. (iv) gives 2Ak = 424mm2 ... 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 T...

Ngày tải lên: 12/08/2014, 03:20

... 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 = 965 = 912 13 = q13 14 = 66.0 ... = + p',r + pI,r )1& apos;2 or, alternatively (10 .9) (10 .10 ) (10 .11 ) (10 .12 ) 10 .1 Tapered beams 3 71 33.2 77.5 11 0.7 Izll 6 33.2 77.5 4 i Fig....

Ngày tải lên: 12/08/2014, 03:20

... signed int16 and extending it to int32, throwing OverflowException one overflow. conv.ovf.i2.un 83 Converts the unsigned been worth one signal of the evaluation stack to signed int16 and ... conv.i1 67 Converts the been worth one signal of the evaluation stack to int8, then extends (pads) it to int32. conv.i2 68 Converts the been worth one signal of the evaluation stack to int...

Ngày tải lên: 03/07/2014, 17:20

... 5350: Part C9: 19 78 and ASTM D 316 7-76 ( 81) BS 5350: Part C10: 19 79 and BS Part C14: 19 79 BS 5350: Part C 11: 19 79 and ASTM BS 5350: Part C12: 19 79 and ASTM D BS 53.50: Part ... practice 16 /89 2.6 2.5 2.4 2.3 2.2 2 .1 2.0 E 2 1. 9 2 1. 8 2 e E 1. 7 o 1. 6 1c > c. ._ r" 1. 5 Ql +4 2 1. 4 1. 3 1. 2 1. 1 1...

Ngày tải lên: 11/08/2014, 15:20

... 0.003 0.0 21 0 .12 1 0.033 0.003 0. 015 0. 018 0. 414 0.855 0.846 0. 211 2. 611 (10 0%) 0.057 1. 512 0.0 31 0.672 1. 989 0.250 0.352 0.045 9.450 0. 513 42.03 62.55 1. 260 12 0. 71 (4623%) ... Pm 14 5.4 13 1.9 13 0.7 1. 0 0 9.50 4 3.50 7.83 2 2.87 0.80 4 0.30 1. 17 4 0.03 16 .35 4 5.80 5.27 4 2.63 16 8.9 15 7.8 13 7.3 1. 13 6.6 18 .4...

Ngày tải lên: 12/08/2014, 02:22