Aircraft structures for engineering students - part 8 doc

... 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 fusela...

Ngày tải lên: 08/08/2014, 11:21

... (5 .8) define the deflected shape of the plate provided that M, and My are known. If either M, or My is zero then d2W - a2W d2W - d2W ax2 - 8Y2 ay2 - dX2 - -v- or - -v- ... 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 _- - whe...

Ngày tải lên: 08/08/2014, 11:21

... = - PV EI,,,, = -Pu d 48 P d 28 d24 ( A)G= El? GJ-Io- (6 .87 ) (6 .88 ) (6 .89 ) Equations (6 .87 ), (6 .88 ) and (6 .89 ), unlike Eqs (6.74), (6.75) and (6 .83 ), are uncoupled and ... 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 )...

Ngày tải lên: 08/08/2014, 11:21

... 8 and will attain a maximum or minimum value when dan/d8 = 0. From Eq. (1 .8) 5 = -2 a, cos 8 sin 8 + 20, sin 8 cos 8 + 2ryy cos 28 = 0 do Hence -( u.~ - u,,) sin 28 ... - ay sin 28 = 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...

Ngày tải lên: 08/08/2014, 11:21

... 80 0J3 30 50 - J3R/2 -J3/3/2 -2 000 + 20J3R 48. 2 CB 80 0 20 86 .6 + R/2 112 1732 + 10R 87 .6 -1 .8 BD 80 0J3 30 -J3R/2 -J3/3/2 20J3R CD 80 0 20 R 1 40R 2.1 AD 80 0 20 RI2 112 ... comparison with Eq. (3 .8) T = 2V0l 48 Two-dimensional problems in elasticity P b2 /8 I G r/r ~ -; g I L - - - - - - (a) (b) Fig....

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 ... COSY +fw - D = 0 (8. 13) and for pitching moment equilibrium about the aircraft& apos;s centre of gravity La - Db - Tc - Mo - PI = 0 (8. 14) Eq...

Ngày tải lên: 08/08/2014, 11:21

... 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 moments about ... 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...

Ngày tải lên: 08/08/2014, 11:21

... qb:27 = -1 .07 x qb,J6 = -1 .07 x x 388 0 x 230 = -9 5.5N/mm x 2 580 x 165 = -4 5.5N/mm qb$5 -4 5.5 - 1.07 x io-4 x 2 580 x (-1 65) = 0 qb.57 = 1.07 X qb; 38 = -1 .07 ... P3 = -P4 = -P6 = -0 .364 X 295 X 900 = -9 6642N and P2 -Ps = -0 .364 x 295 x 1200 = -1 28 856N At a section 1.9 m from the built-i...

Ngày tải lên: 08/08/2014, 11:21

... X for cos 8 an or, in abbreviated form - Fx,i = Fx,i cos 8 + Fy,i sin 8 = -Fx:i sin 8 + Fy,i cos 8 - Fx,j =FxJcos8+ FY;,,sin8 Fy,j =-Fx,jsin8+ Fy,jcos8 - ,u for ... dz & ;- T,(Z) = -+ - - dB dz TJ = GJ- (Eq. (9.60)) and d30 Tr = -Er - dz3 (Eq. (1 1. 58) ) so that Eq. (1 1.63) becomes d46' d2B Er...

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 1 08 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 energy 68,...

Ngày tải lên: 08/08/2014, 11:21