Aircraft Structures 3E Episode 1 pdf

... section beams Problems 10 Stress analysis of aircraft components 10 .1 Tapered beams 10 .2 Fuselages Contents vii 17 4 17 5 i77 18 0 18 8 19 7 19 7 209 211 21 1 220 223 225 232 233 ... (1. 38) max or E1 - Err ( = 2 {cf. Eqs (1. 14) and (1. 15)}. 1_ _IN_ 1. 14 Mi (1. 39) We now apply the arguments of Section 1. 13 to the Mohr circle of stress de...

Ngày tải lên: 13/08/2014, 16:21

... (1. 50) we have 1 + COSM 1 - cos 28 %=EI( 2 )+EII( 2 ) or E, = $ + + 4 - cII) COS 28 (1. 51) Similarly &b = $ ( 1 + EII) + $ (E1 - EII) cos 2(8 + a) (1. 52) ... (lSl), (1. 52) and (1. 53). in Eqs (1. 47). Thus The principal stresses are now obtained by substitution of and (1. 54) 1 E &I = - (Cq - vq1) and Solving Eqs...

Ngày tải lên: 13/08/2014, 16:21

... 10 9 - 2 3 4 5 6 1 8 1 16 15 14 13 12 11 10 - 216 .6 216 .6 216 .7 216 .1 216 .6 216 .6 216 .6 216 .6 216 .6 216 .6 216 .7 216 .6 216 .6 - - - 352.0 269.5 14 5.8 0 -14 5.8 ... 10 .1. 378 Stress analysis of aircraft components Table 10 .3 Skin panel Boom B, (mm’) Y, (m) qb (N/m) 12 23 34 45 56 67 78 89 1 16 16...

Ngày tải lên: 13/08/2014, 16:21

... 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 = ... 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 -...

Ngày tải lên: 13/08/2014, 16:21

... sine - 0 .16 0), = 0.30Tx/r, M4, = T(0.59sine - 0 .16 6) 1 Fig. P.4 .19 5 .1 Pure bending of thin plates 12 3 Fig. 5 .1 Plate subjected to pure bending. are given by 1 Z Ex=-, ... neutral plane. 11 4 Energy methods of structural analysis Fig. P.4 .10 P.4 .11 A bracket BAC is composed of a circular tube AB, whose second moment of area is 1. 51, an...

Ngày tải lên: 13/08/2014, 16:21

... thin plates 16 9 from which 42 EI EI PCR = - = 2.4 71 - 1 712 12 This value of critical load compares with the exact value (see Table 6 .1) of 7r2EI/ 412 = 2.467EI /12 ; the error, ... Eqs (6 .1 1) and (6 .12 ) (6 .12 ) (6 .13 ) Further, in a similar manner, from Eq. (6 .10 ) d2 $ (. y: dA + Et r yf dA) + e 2 (. y1 dA - Et J” 0 y2 d A) = - P...

Ngày tải lên: 13/08/2014, 16:21

... Problems 19 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 L Timoshenko, S. P. and Gere, J. M., Theory of Elastic Stability, 2nd edition, McGraw-Hill Book Company, New York, 19 61. Gerard, ... specified in 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 18 6 Struc...

Ngày tải lên: 13/08/2014, 16:21

... 17 6 17 9 18 0 18 1 18 2 18 3 18 4 18 5 18 6 18 7 18 8 18 9 19 0 19 1 19 2 19 3 19 4 19 5 19 6 19 7 1% 19 9 200 2 01 202 203 204 205 206 207 208 209 210 21 1 212 213 214 Cross-dam ... 16 0 Englne bay ventlng alr scoop 16 1 Cannon muzzle fairing 16 2 Lift augmentation retractable cross-dam 16 3 16 4 16 5 16 6 16 7 16 8 16 9 17 0 17 1...

Ngày tải lên: 13/08/2014, 16:21

... = 1. 5 aC~,T/aa = 2.0 Weight W = 16 00000N CM,O = - 0. 01 Mean chord E = 22.8 m At 18 300m p = 0 .11 6kg/m3 Am. P = 267 852N, AP = 36257N, AL = 2 71 9 31 N, n = 1. 19 ... against h from ESDU data sheets for aircraft having cloud warning radar and integrating gives 9000 dh 6000 dh - = 3.4 16 000 11 0 - = 14 , j3000 11 1) - = 303; F...

Ngày tải lên: 13/08/2014, 16:21

... e 1 loa (- ?s: + 58.7~~) dsl SY(& + 9a) = 11 52a3 0 We may replace sin 6’ by sin(O1 - 0,) = sin el cos O2 - cos O1 sin Q2 where sin O1 = 15 /17 , cosO2 = 8 /10 , ... Also B1(= B6) = 10 50mm2 i.e. 1. 5 x 600 ( KM)) 2+- 2.0 x 600 ( 2M) 2.5 x 300 2+- + (2 -1) + 6 6 6 B2 = 2 x 300 + from which B2(= Bs) = 17 91. 7~ 2 Finally...

Ngày tải lên: 13/08/2014, 16:21