Aircraft Structures 3E Episode 11 pptx

... 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

... idealizations of the same structure are required for different loading conditions. Example 9 .11 Part of a wing section is in the form of the two-cell box shown in Fig. 9.48(a) in which the ... about the point 2 we have or Spa sin e 1 loa (- ?s: + 58.7~~) dsl SY(& + 9a) = 115 2a3 0 We may replace sin 6’ by sin(O1 - 0,) = sin el cos O2 - cos O1 sin Q...

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... direct stress of Eqs (11. 19) is also zero since wo = 0 (see Example 9.7). The rate of twist is obtained by substituting for w from Eq. (11. 18) in Eq. (11. 11). Thus ] (11. 23) in which ... LcllllllrrrJrwr- I 11. 5 Constraint of open section beams Instances of open section beams occurring in isolation are infrequent in aircraft structures. The majority of wing s...

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... 11, Chapters 7 to 13, is retitled Aircraft structures, with Chapter 12, Airworthi- ness, now becoming Chapter 8, Airworthiness and airframe loads, since it is logical that loads on aircraft ... Broddle of British Aerospace for supplying photographs and drawings of aircraft structures. T.H.G. Megson 1989 Aircraft Structures for engineering students viii Contents 10...

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... determination of stresses in these regions requires, for some problems, separate calculation (see Chapter 11) . Fig. 2.3 Stress distributions illustrating St. Venant‘s principle. 26 Basic elasticity ... constant C may be chosen to fuKl the remaining requirement of compatibility. Thus, from Eqs (3 .11) and (i) 2C(-$+$) = -2Gx de or de hb2 dz (a2 + b2) C = -G- givi...

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... are ignored but may be calculated when required for particular cases. Of greater interest in aircraft structures is the calculation of the deflections produced by the large shear stresses experienced ... 0 CB 4000 80 000 PB,f/3 113 EB 4000 20 000 2PB,f 13 213 0 0 16013 0 0 0 - 160J2/3 0 FB 4000J2 -20000J2 d2pB.f 13 ~213 DC 4000 80 000 PB,f13 113 pDsf 1 32013...

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... has curvatures of the same sign. Further, if M, = My = M then from Eqs (5.5) and (5.6) 111 PX Py P _ Therefore, the deformed shape of the plate is spherical and of curvature ... of plate element; (b) radii of curvature of neutral plane. 114 Energy methods of structural analysis Fig. P.4.10 P.4 .11 A bracket BAC is composed of a circular tube AB, whose...

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... and Bsinpl= 0 For a non-trivial solution @e. v # 0) then sinpl=O or pl=n.rr wheren= 112 131 giving or Note that Eq. (6.3) cannot be solved for v no matter how many of the available ... remain straight. These parts enable the plate to resist higher loads; an important factor in aircraft design. At this stage we are not concerned with this post-buckling behaviour, but r...

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... control rizontal tail countering e aircraft s tendency to pitch in a vertical plane Aircraft weight Fig. 7.3 Principal aerodynamic forces on an aircraft during flight. The force on ... loads encountered in landing and taxiing subject the aircraft to concen- trated shock loads through the undercarriage system. The majority of aircraft have their main Undercarriage loc...

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... jettison 110 Formation lighting panel 111 Port roll control air valve 112 Port navigation light 113 Radar warning aerial 114 Port wing reaction control air duct 115 Fuel pumps 116 Fuel ... given type of aircraft will satisfy them. Thus, of the total number 8.3 Aircraft inertia loads 241 A \", , Wheel reaction R / Arrester hook Fig. 8.5 Forces on the a...

Ngày tải lên: 13/08/2014, 16:21