Aircraft structures for engineering students - part 7 ppt

... 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 ... (10. 27) x 1.22 = 3.66mm 3 x 276 00 27 600 t;, = Hence 1 270 3.66 67, = - - - 3 47 Also 612 = 656 = 840, 623 = 78 3...

Ngày tải lên: 08/08/2014, 11:21

... determination of critical load for a flat plate 51 51 59 61 63 65 65 68 68 70 71 73 76 77 85 100 103 103 1 07 109 110 110 122 122 125 129 1 37 141 142 149 149 152 152 ... 2 E, sin’ e + E,+~/~ COS 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 (1.3...

Ngày tải lên: 08/08/2014, 11:21

... the cross-section of the bar. Rearranging and substituting for u and from Eqs (3.9) dw rzx dB dw r7 dB -A _- - - +-y, - ax G dz dy G dzX - (3.10) For a particular torsion ... M=-Pz Rz, hence -= 4 2 dR 2z between F and B P d3 dM d3 4 2 dR 2 M = - (L - z) - -Rz, hence - = - -2 and between B and...

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- ... m=ln=l EA:n[~4($+$)zsin - a b sin cos - -2 ( 1- v )- (sin - -qo EArn, m2n27r4 2mrx . 2nry a2P a b a a xx nz=l n=l The term mult...

Ngày tải lên: 08/08/2014, 11:21

... Flexural-torsional buckling of 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,,,, ... (6.84) into Eqs (6 .74 ), (6 .75 ) and (6.83), we have 7rZ 7rZ 7rz u = AI sin - , 8 = A3 sin - L L L (6.85) 1 (P-~)A~-PX~A~=O 2EIXX (P-9)A1+Pys...

Ngày tải lên: 08/08/2014, 11:21

... for vertical equilibrium L+P- w=o (8 .7) for horizontal equilibrium T-D=O and taking moments about the aircraft s centre of gravity in the plane of symmetry La - Db - Tc - Mo -PI ... x lO-”m/cycle, i.e. no crack growth. In the second region (1 0-8 -1 0-6 m/cycle) much of the crack life takes place and, for small ranges of AK, Eq. (8 .71 ) may be represe...

Ngày tải lên: 08/08/2014, 11:21

... (by symmetry) qb. 67 = qb;23 = (by qb.21 = -7 .22 X 1 0-4 (250 X 100) = -1 8.1N/mm qb.18 = -1 8.1 - 7. 22 x 1 0-4 (200 x 30) = -2 2.4N/mm qb: 87 = qb:21 = -1 8a1 N/mm (by Taking ... right-hand side of Eq. (ii) we have r=l 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(i...

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, ... 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 fuselages...

Ngày tải lên: 08/08/2014, 11:21

... integrating from the origin for s to any point s around the cross-section, we have d8 du dv dz dz dz W, - WO = 2AR,o - -( x - XO) - - (y -yo) (11. 67) where 240 = p~ ds. ... on the right-hand side of Eqs (1 1 .76 ) dUr - dkfr ~AR - - dz dz rR From Eq. (1 1 .74 ) d28 dz2 Mr = -ErR- so that Hence and (see Eq. (11.58...

Ngày tải lên: 08/08/2014, 11:21

... 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 Energy methods 6 8-1 09 108 16 5-9 complementary ... virtual forces 7 1-3 principle of virtual work 7 1-3 reciprocal theorem 68, 10 3 -7 strain energy 68, 69, 142, 143 total complementary e...

Ngày tải lên: 08/08/2014, 11:21