Aircraft Structures 3E Episode 8 doc

... the life of the aircraft in terms of flights is Nflighr = l/l)total 8. 7.5 Crack propagation (8. 60) (8. 61) We have seen that the concept of fail-safe structures in aircraft construction ... and E(u,) from Eqs (8. 48) and (8. 50). From Eq. (8. 48) m sa = su,e = K, (1 + c/JG) from which where Su?, = kl Veu, and S;,, = kl VeuF Also Eq. (8. 50...

Ngày tải lên: 13/08/2014, 16:21

... in Eqs (12 .88 ) from Eqs (12 .82 ) gives E, =a2 Ey =a6 rxy = Q3 + Q5 or in matrix form {E}= [" 0 0 10 0 0 0 0 0 o,[q 1 001010 a5 a6 (12 .88 ) (12 .89 ) which is ... complete element we obtain From Eq. (12 .81 ) and by comparison with Eqs (12. 58) (12 .85 ) takes the form {fl} = [AI{d Hence (step 3) we obtain b (12 .85 ) d (12.59) we see tha...

Ngày tải lên: 13/08/2014, 16:21

... and notation 277, 2 78 287 -90 moment relationships 283 , 284 curved thin sections 288 , 289 Bending of thin plates, see Plates Bending of wings. see Stress analysis of aircraft components ... reciprocal theorem 68, 103-7 strain energy 68, 69, 142, 143 total complementary energy 68, 76-100, total potential energy 68, 71,73-6, 144-9, unit load method 68, 85 , 100-2, 342-4,...

Ngày tải lên: 13/08/2014, 16:21

... dan/d8 = 0. From Eq. (1 .8) 5 = -2a, cos 8 sin 8 + 20, sin 8 cos 8 + 2ryy cos 28 = 0 do Hence -(u.~ - u,,) sin 28 + 2TsJ cos 28 = 0 or (1.10) Two solutions, 8 and ... components Problems 8 Airworthiness and airframe loads 8. 1 8. 2 Load factor determination 8. 3 Aircraft inertia loads 8. 4 Symmetric manoeuvre loads 8. 5 8. 6 Gust loads...

Ngày tải lên: 13/08/2014, 16:21

... at the point, then if E,, &b and E, are the measured strains in the directions 8, (8 + a), (8 + a + p) to we have, from the general direct strain relationship of Eq. (1.3 1) ... directions have and E, = E~COS 2 8 + cII sin 2 8 since E, becomes become principal directions. Rewriting Eq. (1.50) we have 1 + COSM 1 - cos 28 %=EI( 2 )+EII( 2...

Ngày tải lên: 13/08/2014, 16:21

... vertical deflection of B is = 3.52mm 12 68 x lo6 AB:v = 180 0 x 200 000 and the horizontal movement of D is = 2.44~ 88 0 x lo6 = 180 0 x 200 000 The positive values of AB," ... pDsf 1 32013 320 PD,f 1 32013 320 1 480 13 240 BA 4000 60 000 2pB,f/3 213 pD,f 0 0 0 0 FC 4000 100 000 0 0 C = 12 68 C = 88 0 92 Energy methods of s...

Ngày tải lên: 13/08/2014, 16:21

... zero twisting moment and zero shear force by a single equivalent condition. Thomson and Tait ( 188 3) gave a physical explanation of how this reduction may be effected. They pointed out that the ... Equations (5.7) and (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 -...

Ngày tải lên: 13/08/2014, 16:21

... Substituting for u, v and 8 from Eqs (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 ... area Io of the column about the shear centre S. Thus Eq. (6 .81 ) becomes P d 28 (6 .82 ) Substituting for T(z) from Eq. (6 .82 ) in Eq. (11.64), the general equation for the tors...

Ngày tải lên: 13/08/2014, 16:21

... listed below A = 600mm2 Zxx = 1.17 x 106mm4 J = 80 0mm4 = 0.67 x 106mm4 I? = 2 488 x 106mm6 Zo = 5.32 x 106mm4 186 Structural instability t C 2.5mrn I - X 2.5mm ... particular aircraft. An outstanding example of such a development is the use of Hiduminium RR 58 as the basis for the main structural material, designated CMOO1, for Concorde. Hiduminium...

Ngày tải lên: 13/08/2014, 16:21

... retractable cross-dam 163 164 165 166 167 1 68 169 170 171 172 173 174 175 176 177 176 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 1% ... alleviation factor, F. Thus Eq. (8. 27) becomes (8. 33) PO ~E(~CL/~~)FUE n=l+ ,I It' Similar modifications are carried out on Eqs (8. 25), (8. 26), (8. 28) and...

Ngày tải lên: 13/08/2014, 16:21