Aircraft structures for engineering students - part 2 pot

... F~FDD,fIm,,f AE 4000J2 -6 OOOOJ2 -2 J2pB,f/3 -2 J213 0 0 320 J2 0 EF 4000 -6 0 000 -2 pB,f/3 -2 13 0 0 160 0 FD 4000J2 -8 0000J2 -d2PB.f/3 - ~21 3 0 0 640J213 0 CB 4000 80 000 ... satisfied. PX ax = - [5y(x2 - z2) - ioy3 + 6dZy] 20 d3 ay = E( - 3yd2 - 2d3) 4d3 -P 40d3 Txy = - [5(3x2 - Z2)(y2 -...

Ngày tải lên: 08/08/2014, 11:21

... qb :23 = qb,34 = -7 .22 X 1 0-4 (400 X 100) = -2 8.9N/mm qb,4j = -2 8.9 - 7 .22 x io-4(ioo x 50) = -3 2. 5N/m.m qb,56 = qb:34 = -2 8.9 N/mm (by symmetry) qb.67 = qb ;23 = ... symmetry) qb.67 = qb ;23 = (by qb .21 = -7 .22 X 1 0-4 (25 0 X 100) = -1 8.1N/mm qb.18 = -1 8.1 - 7 .22 x 1 0-4 (20 0 x 30) = -2 2...

Ngày tải lên: 08/08/2014, 11:21

... plate 174 2 8-3 2 Fabrication of structural components 22 5-3 2 fuselages 23 2 integral construction 23 0 sandwich panels 23 0 ,23 1 sub-assembly 22 8, 22 9 Wings 22 8-3 0 Factors of safety - flight ... w2miS2i w2mn 6% I=O 2 w2ml ail w2m2si2 (w2miSii - 1) w m,S, I I 2 I w2mlsn1 Jm26n2 w miSni (w2wI,,~,,,, - 1) I (1...

Ngày tải lên: 08/08/2014, 11:21

... 2Tcy a, - ay sin28 = Jm.’ ‘Os 28 = JW and sin 2( 8 + n /2) = dv, - 2? xy cos 2( 8 + n /2) = 4U.Y - Uy) (0, - cy) + 4?xy JW Rewriting Eq. (1.8) as U. an =%(l+cos20) +-~ (1 ... Eq. (1.8) as U. an =%(l+cos20) +-~ (1 -cos28)+rx,sin28 ahd substituting for (sin28, cos 28 } and {sin2(8 + 7r /2) , cos 2( 8 + n /2) } in turn gives 2...

Ngày tải lên: 08/08/2014, 11:21

... M, and My are known. If either M, or My is zero then d2W - a2W d2W - d2W ax2 - 8Y2 ay2 - dX2 - -v- or - -v- and the plate has curvatures of opposite signs. The case of ... expressions for the greatest stresses due to bending at the centre and at the ends of the minor axis. 3PU - 3) 2~ t3 -+ - +- 32 Am. wO= (d a2b2 b4 f3pa...

Ngày tải lên: 08/08/2014, 11:21

... thin-walled columns 185 0 P - ~EIJL~ -Pxs P - ~EI,,JL~ 0 PYS PYS - Pxs IOPIA - .rr2ET/L2 - GJ =O (6.86) d’v dz- d2 u EI, 7 = - PV EI,,,, = -Pu d48 P d28 d24 ... becomes P d28 (6. 82) Substituting for T(z) from Eq. (6. 82) in Eq. (11.64), the general equation for the torsion of a thin-walled beam, we have d2v d2u dz dz2 -...

Ngày tải lên: 08/08/2014, 11:21

... 35000N Tailload P=nW-L=4.5~800 0-3 5000= l000N Drag D = ipV2SCD = i x 1 .22 3 x 6 02 x 14.5 x 0.0875 = 27 90N Forward inertia force fW = D (from Eq. (8.13)) = 27 90 N In Section ... civil, non-aerobatic aircraft for which nl = 2. 5, w = 24 00N/m2 and aCL/acw = 5.0/ rad. Taking F = 0.715 we have, from Eq. (8.33) n=l+ !j x 1 .22 3Vc x 5.0...

Ngày tải lên: 08/08/2014, 11:21

... load, we have sx = -4 121 12 + q23l23 (10.19) Now resolving vertically sy = q31 (h 12 + h23) - 412h 12 - q23h23 sxqO + sycO = -2 A12q 12 - 2A23q23 (10 .20 ) (10 .21 ) Finally, taking ... Cell area (mm’) ~~ 12, 56 1 023 1 .22 AI = 26 5000 23 127 4 1.63 AI1 = 21 3 000 34 22 00 2. 03 A111 = 413 000 483 400 2. 64 5 72 460 2. 64...

Ngày tải lên: 08/08/2014, 11:21

... substitute in Eq. (1 1 .28 ) for q from Eq. (1 1 .25 ) and for PB from Eq. (1 1 .26 ). Hence 1 #PA - Gt ( PA S,Z PA) - - - - - - - - - - - 2 dz2 dE 2B 2Bh A Rearranging, ... 1OOOmm 2 B2 = B3 = B6 = B7 = 600mm2 AFZS. ~1 = -1 90.7N/m2, 02 = -1 81.7N/mm2, 63 = -1 72. 8N/mm2, U4 = -1...

Ngày tải lên: 08/08/2014, 11:21

... Therefore, eliminating rows and columns corresponding to zero displacements from Eq. ( 12. 53), we obtain Fy ,2 = - W 27 /2L3 9/2L2 6/L' -3 /2L2 9/2L2 6/L 2/ L 6/L2 2/ L 4/L 0 -3 /2L2 ... vj o 121 ~~ -6 1~~ o -1 21 ~~ -6 1~~ o -1 21 ~~ 61~~ o 121 ~~ 61~~ 00 000 0 0 -6 /L2 4/L 0 6/L2 2/ L 00 000 0 0 -6 /L2 2/ L 0 6/L2 4/L...

Ngày tải lên: 08/08/2014, 11:21