Aircraft Structures 3E Episode 4 pps

... = 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 ) ... series *6pa4b2 &6pb4d 9(3b4 + 2a2b2 + 3d) ’ = t-(3b4 ’ + 2a2b2 + 3a4) mm nry w= 2 2Amnsin- U sin m=l n=l m 5 nrrr;l 4Wsin- sin U Ans. A,,,,, = r4Dab[(m2/d)...

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... are listed below A = 600mm2 Zxx = 1.17 x 106mm4 J = 800mm4 = 0.67 x 106mm4 I? = 248 8 x 106mm6 Zo = 5.32 x 106mm4 186 Structural instability t C 2.5mrn I - X ... = 45 ". In practice, both flanges and stiffeners deform so that a is somewhat less than 45 ", usually of the order of 40 " and, in the type of beam common to aircraft...

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... airbrake 40 Alrbrake hydraulic Jack 41 Formation lighting strip 42 Avionics bay access door, port and starboard 43 Avionics equipment racks 44 Fuselage frame and stringer construction 45 ... centrellne pylon 141 Zero scarf forward (fan air) nozzle 142 Ventral gun pack (two) 143 Aden 25-mm cannon 144 Engine drain mast 145 Hydraulic system ground connectors 146 Forw...

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... twist. Problems 48 7 Am. 912 = q45 = 44 .1 N/=, q51 = 80.2N/mm, q23 = q 34 = 42 .9N/m, qx = 37.4N/m, XR = 79.5mm, YR = 0 (referred to mid-point of web 24) P.11.3 A singly ... 43 1 = (1502 .4 - 18 94. 7~0 ~4 - 2102.1 sinq5)/R N/mm 1 Fig. P.11.3 P.11 .4 A uniform, four-boom beam, built-in at one end, has the rectangular cross-section shown in Fig...

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... 13 .4 Introduction to ‘flutter’ References Problems 380 40 6 41 5 42 5 43 2 43 2 43 2 44 3 44 3 44 5 44 9 45 5 46 5 48 5 48 6 49 4 49 5 49 6 49 7 500 507 507 509 516 533 533 533 540 ... analysis of aircraft components 10.1 Tapered beams 10.2 Fuselages Contents vii 1 74 175 i77 180 188 197 197 209 211 21 1 220 223 225 232 233 233 235 23...

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... possible forms of the stress function to those satisfying the biharmonic equation a 44 a 44 a4q5 ax4 ax2ay2 ay4 -+2- +-=O The final form of the stress function is then determined by the boundary ... -ey, v= ex Referring to Eqs (1.20) and (1 .46 ) (3.9) 28 Basic elasticity (-290 x 1 04, Y Q, (360 x 1 04, ix 650 x 1 04) Fig. 1. 14 Mohr's circle of...

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... cantilever of Fig. 4. 15 Table 4. 4 (Tension positive) G3 Force (N) 64 000R/9 -700 3 OOOR -525 64 000R/9 -700 3 OOOR -525 AB 40 00 4R/3 41 3 CD 40 00 4R/3 41 3 BC 3000 R ... 160J2/3 0 FB 40 00J2 -20000J2 d2pB.f 13 ~213 DC 40 00 80 000 PB,f13 113 pDsf 1 32013 320 PD,f 1 32013 320 1 48 013 240 BA 40 00 60 000 2pB,f/3 213 pD,f 0 0 0 0...

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... 1.19 P.8.8 An aircraft of all up weight 145 000 N has wings of area 50 m2 and mean chord 2.5m. For the whole aircraft CD = 0.021 + O.O41C;, for the wings dCL/da = 4. 8, for the tailplane ... Y Y 4 N (a) (b) Fig. 9 .4 Determination of neutral axis position and direct stress due to bending. 3 04 Open and closed, thin-walled beams The last two terms in Eq....

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... q3, in the wall 34, must be equal to -138.5 - 1 04 = - 242 .5N/mm. Hence q 34 = -69.0 x 10- 2(75 - ~4) ds4 - 242 .5 4r which gives q 34 = -1. 04~ ~ + 69.0 x - 242 .5 (vi) Examination ... that q 34 has a maximum value of -281.7 N/mm at s4 = 75mm; also q4 = -172.5N/mm. Finally, the distribution of shear flow in the wall 94 is given by q 94 =...

Ngày tải lên: 13/08/2014, 16:21