Aircraft Structures 3E Episode 16 potx

... section beams Problems 10 Stress analysis of aircraft components 10 .1 Tapered beams 10 .2 Fuselages Contents vii 17 4 17 5 i77 18 0 18 8 19 7 19 7 209 211 21 1 220 223 225 232 233 ... (1. 38) max or E1 - Err ( = 2 {cf. Eqs (1. 14) and (1. 15)}. 1_ _IN_ 1. 14 Mi (1. 39) We now apply the arguments of Section 1. 13 to the Mohr circle of stress de...

Ngày tải lên: 13/08/2014, 16:21

... satisfied. PX ax = - [5y(x2 - z2) - ioy3 + 6dZy] 20 d3 ay = E( - 3yd2 - 2d3) 4d3 -P 40d3 Txy = - [5(3x2 - Z2)(y2 - d2) - 5y4 + 6y2d2 - $1 The boundary stress ... dO/dz X 2Ty T (a2+b2) dw - 2Tx T (a2 +h2) - - - dw dx 7rab3G+G m3b3 y7 % =-E 7ra3b3 - or (vii) Integrating both of Eqs (vii) T(b2 - a2) 7ra3 b3 G T(b2 - a2) nu3 b3...

Ngày tải lên: 13/08/2014, 16:21

... 2 13 0 0 160 13 0 0 0 - 160J2 /3 0 FB 4000J2 -20000J2 d2pB.f 13 ~2 13 DC 4000 80 000 PB,f 13 1 13 pDsf 1 32 0 13 320 PD,f 1 32 0 13 320 1 480 13 240 BA 4000 60 000 2pB,f /3 2 13 ... -2J2pB,f /3 -2J2 13 0 0 32 0J2 0 EF 4000 -60 000 -2pB,f /3 -2 13 0 0 160 0 FD 4000J2 -80000J2 -d2PB.f /3 - ~2 13 0 0 640J2 13 0 CB 4000 80...

Ngày tải lên: 13/08/2014, 16:21

... = dx 2 8x (5 .41 ) The potential energy V, of the N, loading follows from Eqs (5 .40 ) and (5 .41 ), thus V, = - 1 r N, (g ) dxdy 200 Similarly (5 .42 ) (5 .43 ) The potential ... deflection from Eq. (4. 35) and the bending deflection from Eqs (4. 27); thus (4. 36) Example 4. 17 Determine the deflection of the tip of the cantilever in Fig. 4. 30 with th...

Ngày tải lên: 13/08/2014, 16:21

... (5. 44) and for the complete in-plane loading system we have, from Eqs (5. 42), (5. 43) and (5. 44), a potential energy of v = -'rr 200 [ ( g)2 + Ny( $)2 + ax ay (5. 45) ... unloaded edge free I I 5 one simply supported 0 1 2 3 4 I5 I3 II 9- 7- 5, k - - - a/b (b) k 40- 36 - Clamped edges Simply supported 123 45 a/b (C) Fig. 6.16...

Ngày tải lên: 13/08/2014, 16:21

... = - 76. 2mm. The remaining section properties are found by the methods specified in Example 6. 1 and are listed below A = 60 0mm2 Zxx = 1.17 x 106mm4 J = 800mm4 = 0 .67 x 106mm4 I? ... of Eqs (6. 98) or (6. 99) 2 Ot +'F W 2AF tan a CF = in which AF is the cross-sectional area of each flange. Also, from Eq. (6. 102) wb us = -tana ASd (6. 1 06...

Ngày tải lên: 13/08/2014, 16:21

... 163 164 165 166 1 67 168 169 170 171 172 173 174 175 176 177 176 179 180 181 182 183 184 185 186 1 87 188 189 190 191 192 193 194 195 196 1 97 1% 199 200 201 ... hydraulic actuator 74 Starboard outrigger wheel 75 BL755 6OO-lb ( 272 -kg) cluster bomb (CBU) 76 Intermediate pylon 77 Reaction control air ducting 78 Alleron control ro...

Ngày tải lên: 13/08/2014, 16:21

... 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

... carrying area. 9. 9.1 Bending of open and closed section beams The analysis presented in Section 9. 1 applies and the direct stress distribution is given by any of Eqs (9. 6), (9. 7) or (9. 9), depending ... in Sections 9. 5 and 9. 6 apply. 9. 9.5 Alternative method for the calculation of shear flow distribution Equation (9. 73) may be rewritten in the form apr q2-41 =-&a...

Ngày tải lên: 13/08/2014, 16:21

... Py,r (10. 13) r=l r= 1 Substituting in Eqs (10. 13) for P.,,r and P,,,r from Eqs (10. 10) and (10. 9) we have sx = s.x:w + P:,r - 1 sy = sy:w + pz,r I' sY (10. 14) ... P.9.18 10. 1 Tapered beams 371 33.2 77.5 110. 7 Izll 6 33.2 77.5 4 i Fig. 10. 7 'Open section'shear flow (Wmm) distribution in beam section of Example 1...

Ngày tải lên: 13/08/2014, 16:21

... 233q11~] (ii) de 1 dz - 2 x 355 OOOGREF _- For cell 111 (iii) 1 [-233q11+ q 111( 736 i- 233 + 736 + 368)] d6’ -= dz 2 x 161 OOOGREF In addition, from Eq. (10.22) 11. 3 ... Example 10.7 61 = 1542 + 250 = 1792 611 = 250 + 725 + 233 + 725 = 1933 6111 = 736 + 233 + 736 + 368 = 2073 61,~ = 250 611, 111 = 233 Hence, from Eqs...

Ngày tải lên: 13/08/2014, 16:21

... - ' mm mm P.10 .12 A singly symmetric wing section consists of two closed cells and one open cell (see Fig. P.10 .12) . The webs 25, 34 and the walls 12, 56 are straight, while ... distance of lOOmm from and parallel to side 12. The modulus of rigidity G is constant throughout the section. Wall 12 34 23 Length (mm) 375 125 500 11.2 Built-in end of a c...

Ngày tải lên: 13/08/2014, 16:21

... anticlockwise. 00 mm2 1.0 mm Fig. P.11.4 Matrix methods of structura I analysis Actual aircraft structures consist of numerous components generally arranged in an irregular manner. These ... idealizations and their effect on structural analysis have been presented in Chapter 9 for aircraft structures. Outside the realms of aeronautical engineering the represen- ta...

Ngày tải lên: 13/08/2014, 16:21

... uniform beam Our discussion so far has been restricted to structures comprising members capable of resisting axial loads only. Many structures, however, consist of beam assemblies in which ... aCL/aa is the wing lift curve slope. Rearranging gives or (13.2) Elementary aeroelasticity Aircraft structures, being extremely flexible, are prone to distortion under load. When...

Ngày tải lên: 13/08/2014, 16:21

... section beams 307 -16 Bredt-Batho theory 307-9 condition for zero warping 315, 316 displacements 309-15 Neuber beams 316 shear flow distribution 307 warping distribution 309 -16 aircraft components ... 340 Aircrafi Structures for Engineering Students provides a comprehensive self- contained course in aircraft structures. Starting with the structural mechanics of...

Ngày tải lên: 13/08/2014, 16:21