Aircraft structures for engineering students - part 4 docx

... supported 0 1 2 3 4 I5 I3 II 9- 7- 5, k - - - a/b (b) k 4 0- 36 - Clamped edges Simply supported 12 345 a/b (C) Fig. 6.16 (a) Buckling coefficients for flat plates in ... Flexural-torsional buckling of thin-walled columns 185 0 P - ~EIJL~ -Pxs P - ~EI,,JL~ 0 PYS PYS - Pxs IOPIA - .rr2ET/L2 - GJ =O (...

Ngày tải lên: 08/08/2014, 11:21

... q(x,y)sin-sin- dxdy . mrx m'rx nry n'ry sin - sin - dx dy = m=l 2 n=l T/:jIamsin-sin- a a b b - ab 4 - since a 2 _- - when m=m' and . nry . n'ry sin-sin- ... 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...

Ngày tải lên: 08/08/2014, 11:21

... = - [a, + a,, + az - 2v(ax + ay + 41 or In the case of a uniform hydrostatic pressure, a, = a,, = az = -p and 3(1 - 2~) E P e =- (1 .49 ) The constant E/3( 1 - ... e - E~ sin 9 cos 8 7= Substitution of E, and E~+~~~ from Eqs (1.31) and (1.30) yields -_ 7 - (E~-E~)sin2e-~,-os2e 2 2 2 (1. 34) vi Contents Reference...

Ngày tải lên: 08/08/2014, 11:21

... cantilever of Fig. 4. 15 Table 4. 4 (Tension positive) G3 Force (N) 64 000R/9 -7 00 3 OOOR -5 25 64 000R/9 -7 00 3 OOOR -5 25 AB 40 00 4R/3 41 3 CD 40 00 4R/3 41 3 BC 3000 R ... dM d3 4 2 dR 2 M = - (L - z) - -Rz, hence - = - -2 and between B and C P fi dM d3 4 2 dR 2 M =-( L-z) R(L-z), hence...

Ngày tải lên: 08/08/2014, 11:21

... 0.1, K,, = 1 .45 for six specimens, K,, = 1 .44 5 for 10 specimens, K,, = 1 .44 for 20 speci- mens and for 100 specimens or more K,, = 1 .43 . For typical S-N curves a scatter ... Thus 4. 5 g mla =-3 g= 13.5kN Resolving forces parallel to the axis of the fuselage N - T + mlacos 10" - 4. 5 sin 10" = 0 N- 137.1 + 13.5~0~10 ~...

Ngày tải lên: 08/08/2014, 11:21

... right-hand side of Eq. (ii) we have r=l qb:23 = qb, 34 = -7 .22 X 1 0 -4 (40 0 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 ... = 25mm of 4. 3 N/mm. In the wall 23 q23 = -6 9.0 x 1 0-~ 2 x 75ds2 - 34. 5 r i.e. q23 = -1 . 04, ~2 - 34. 5 Hence q23 varies linear...

Ngày tải lên: 08/08/2014, 11:21

... qb:27 = -1 .07 x qb,J6 = -1 .07 x x 3880 x 230 = -9 5.5N/mm x 2580 x 165 = -4 5.5N/mm qb$5 -4 5.5 - 1.07 x io -4 x 2580 x (-1 65) = 0 qb.57 = 1.07 X qb;38 = -1 .07 ... 4( 29.oq21 + 82.5932 + 123.7q43 + 145 .8q 54) = 100 x lo3 Substituting for 932, q43 and q 54 from the above we obtain 4( 381q21 + 18 740 .5) =...

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, ... 0 at z = 0. d8 - - dz-2a2bZG L(b+~)[l-(bt,+Utb) tb CoshpL bt, - atb ’ coshp(L - z) 10.5 Cut-outs in wings and fuselag...

Ngày tải lên: 08/08/2014, 11:21

... matrix for a uniform beam 51 1 It is possible to write Eq. (12 .44 ) in an alternative form such that the elements of [KJ are pure numbers. Thus 12 -6 -1 2 -6 -6 4 62 -1 2 6 12 6 -6 2 64 This form ... 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...

Ngày tải lên: 08/08/2014, 11:21

... shear and torsion 34 2 -4 multicell wings 40 4 ,40 5 unit load method 34 2 -4 Deflection of thin plates 12 2 -4 9 Diagonal tension, see Tension field beams Displacements in two-dimensional problems ... method 7 9-8 5 for buckling of columns 16 5-9 for bending of thin plates 14 2-9 potential energy 70, 71, 7 3-6 , 14 4- 9 , principle of stationary value of total...

Ngày tải lên: 08/08/2014, 11:21