Aircraft Structures 3E Episode 5 ppsx

... (5. 44) and for the complete in-plane loading system we have, from Eqs (5. 42), (5. 43) and (5. 44), a potential energy of v = -'rr 200 [ ( g)2 + Ny( $)2 + ax ay (5. 45) ... unloaded edge free I I 5 one simply supported 0 1 2 3 4 I5 I3 II 9- 7- 5, k - - - a/b (b) k 40- 36 - Clamped edges Simply supported 123 45 a/b (C) Fig. 6.16...

Ngày tải lên: 13/08/2014, 16:21

... 380 406 4 15 4 25 432 432 432 443 443 4 45 449 455 4 65 4 85 486 494 4 95 496 497 50 0 50 7 50 7 50 9 51 6 53 3 53 3 53 3 54 0 54 1 54 6 55 1 56 8 57 6 57 7 Index 58 2 6 Basic ... analysis of aircraft components 10.1 Tapered beams 10.2 Fuselages Contents vii 174 1 75 i77 180 188 197 197 209 211 21 1 220 223 2 25 232 233 233...

Ngày tải lên: 13/08/2014, 16:21

... by substitution of and (1 .54 ) 1 E &I = - (Cq - vq1) and Solving Eqs (1 .54 ) and (1 .55 ) gives E 01 = - - 9 (&I + %I) (1 .55 ) (1 .56 ) 50 Two-dimensional problems ... 57 -x Fig. 3 .5 Lines of shear stress. Consider now the line of constant q5 in Fig. 3 .5. If s is the distance measured along this line from some arbitrary point then dq5dy...

Ngày tải lên: 13/08/2014, 16:21

... 413 CD 4000 4R/3 413 BC 3000 R 1 DA 3000 R 1 AC 50 00 -5R/3 -51 3 1 25 000R/9 8 75 DB 50 00 -5R/3 -51 3 1 25 000R/9 8 75 C = 48 OOOR 4.6 Application to deflection problems ... (Fig. 4. 25( b)), is - 1. 05 - 0.6 = - 1. 65 mm from which the angular rotation of the beam at B, OB, is given by D = -1.4mm. 1 1. 65 OB = tan- - = tan-&a...

Ngày tải lên: 13/08/2014, 16:21

... = 0 and from Eq. (5. 14) @w/axay = 0, so that Eq. (5. 37) simplifies to 5. 6 Energy method for the bending of thin plates 1 45 a' T z (b) Fig. 5. 1 5 (a) In-plane loads ... forces X Y Fig. 5. 1 1 Equivalent vertical force system. 5. 2 Plates subjected to bending and twisting 1 25 Fig. 5. 3 Anticlastic bending Equations (5. 7) and (5. 8) def...

Ngày tải lên: 13/08/2014, 16:21

... compression elements, J. Aeron. Sci. 25( 1), 37 -52 Jan. 1 958 . Kuhn, P., Stresses in Aircraft and Shell Structures, McGraw-Hill Book Company, New York, 1 956 . P.6.1 The system shown in Fig. ... torsion I = 50 0 mm; b = 25. 0 mm, t = 2 .5 mm, E = 70 000 N/mm2, EIG = 2.6 AYE. OCR = 282 N/m2. Fig. P.6.19 7.1 Materials of aircraft construction...

Ngày tải lên: 13/08/2014, 16:21

... non-aerobatic aircraft for which nl = 2 .5, w = 2400N/m2 and aCL/acw = 5. 0/ rad. Taking F = 0.7 15 we have, from Eq. (8.33) n=l+ !j x 1.223Vc x 5. 0 x 0.7 15 x 15. 25 2400 giving ... 150 Rolls-Royce Pegasus 11 Mk 1 05 vectored thrust turbofan 151 Formation llghting strips 152 Engine oil tank 159 Bleed air soill duct 154 Air conditioning i...

Ngày tải lên: 13/08/2014, 16:21

... r(ue) from Eq. (8 .51 ) 3.23 1 05~ ;5. ’6 E(ue) = 1000/~0 Equation (8 .54 ) then becomes ) dtle D, = - 1; ( g)2 ( Su,e 7 SA:-) ’ ( -3.23 x 5. 26 x 105u ;5. ’6 Sqrn 10001~0 ... 36 257 N, AL = 271 931 N, n = 1.19 P.8.8 An aircraft of all up weight 1 45 000 N has wings of area 50 m2 and mean chord 2.5m. For the whole aircraft CD = 0.021 + O....

Ngày tải lên: 13/08/2014, 16:21

... Eq. (9 .52 ) 4A2G 4 x 200002 x 250 00 (GJ)cl = = (900 + 300)/1 .5 which gives (GJ),, = 50 00 x lo7 Nmm2 The torsional rigidity of the open portion is found using Eq. (9 .59 ), thus ... x 4 x 8.04~1 x 250 00 x 316.7 i.e. the warping distribution is linear in 02 and ~2 = -0.01 x 25 = -0.25mm In the wall 21 AR = 4 x 8.04 x 25 -4 x 25s2...

Ngày tải lên: 13/08/2014, 16:21

... aircraft components Table 10.2 1 2, 16 3, 15 4, 14 5, 13 6, 12 7, 11 8, 10 9 381.0 352 .0 269 .5 1 45. 8 0 -1 45. 8 -269 .5 - 352 .0 -381.0 302.4 279.4 213.9 1 15. 7 0 -1 15. 7 ... 269 .5 1 45. 8 0 -1 45. 8 -269 .5 - 352 .0 381.0 352 .0 269 .5 1 45. 8 0 -1 45. 8 -269 .5 - 352 .0 0 -30.3 -53 .5 -66.0 -66.0 -53 .5 -30.3 0 -32.8...

Ngày tải lên: 13/08/2014, 16:21