Aircraft Structures 1 2011 Part 10 ppt

... = + 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. 10 .7 'Open section'shear ... 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...

Ngày tải lên: 12/08/2014, 03:20

... 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 .14 If the ... Chapter 9. Ans. (a) 3.09 x 10 4N, (b) 1. 78 x 10 4N, 1. 19 x 10 4N. P.6.22 A simply supported beam has a span of 2.4m and carries a central concentrated load of 10 kN. T...

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

... 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

... 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) ... whence N = 12 4.6 kN Now resolving forces perpendicular to the axis of the fuselage S - rnlusin 10 " - 4.5~0s 10 " = 0 i.e. so that S - 13 .5 sin lo&q...

Ngày tải lên: 12/08/2014, 03:20

... 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 - qb;54 x 10 0 x 12 01 + 2 x 972OOqs,0 Substituting ... 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...

Ngày tải lên: 12/08/2014, 03:20

... 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-jointed fr...

Ngày tải lên: 12/08/2014, 03:20

... Example: you do a project with two other people everyone can do one-third of the research rather than 10 0% b. Example: each person read one chapter from the textbook and summarize it for the others. ... space in an apartment (but only one small room in a dormitory) c. There are not so many people around, so it is usually less noisy than a dorm d. You can feel more independent in an...

Ngày tải lên: 15/12/2013, 12:15

... a-SiN 1. 33 is given in Eq. (3), with constants for the equation listed in Table 1. () 2 2 1 22 1 1 A n λ λ λλ =+ − → ( λ in nm) (3) Parameter Value A 1 2.8939 λ 1 13 9.67 x 10 -3 ... Activities in 2 011 284 In Fig. 17 , 3-D ray tracing is used to calculate paths of light propagation for the randomly generated optical k-vectors over a 250 < λ 0 < 11 00...

Ngày tải lên: 19/06/2014, 21:20

... kg.m -3 12 37 12 37 12 40 12 37 Under flow density, kg.m -3 18 44 17 25 18 34 17 10 Over flow density, kg.m -3 11 30 11 14 11 51 114 4 M0 01 R u , (under flow volumetric fraction) 0 .15 0 .14 3 0 .13 04 0 .15 7 ... kg.m -3 13 00 13 00 12 99 13 00 Under flow density, kg.m -3 19 30 17 69 18 89 18 67 Over flow density, kg.m -3 11 88 11 82 12 03 11 89 M00...

Ngày tải lên: 21/06/2014, 18:20