Aircraft Structures 3E Episode 10 pdf

... 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 10....

Ngày tải lên: 13/08/2014, 16:21

... photographs and drawings of aircraft structures. T.H.G. Megson 1989 Aircraft Structures for engineering students viii Contents 10. 3 Wings 10. 4 Fuselage frames and wing ribs 10. 5 Cut-outs in wings ... on the analysis of open and closed section beams Problems 10 Stress analysis of aircraft components 10. 1 Tapered beams 10. 2 Fuselages Contents vii 174 175 i7...

Ngày tải lên: 13/08/2014, 16:21

... dw rzx dB dw r7 dB -A_- - - +-y, - ax G dz dy G dzX - (3 .10) For a particular torsion problem Eqs (3 .10) enable the warping displacement w of the originally plane cross-section ... at a point on the surface of the shaft. A~S. aI = 121.4~/mm~, e = 3104 3' aII = -46.4~/m~, e = 1 2104 3' P.1.7 An element of an elastic body is subjected to a th...

Ngày tải lên: 13/08/2014, 16:21

... displacements y and r= 1 4 .10 The reciprocal theorem 103 xw-T * m- 1 m-u pe r p os i tG- An extremely useful principle employed in the analysis of linearly elastic structures is that of ... point load of 40 N is applied to its end. Dist. (mm) 0 100 200 300 400 500 600 700 800 Def. (mm) 0 -0.3 -1.4 -2.5 -1.9 0 2.3 4.8 10. 6 What will be the angular rotation...

Ngày tải lên: 13/08/2014, 16:21

... 1 10 Energy methods of structural analysis Argyris, J. H. and Kelsey, S., Energy Theorems and Structural Analysis, Butterworths, London, Hoff, N. J., The Analysis of Structures, ... P.4.13 shows a plan view of two beams, AB 9150mm long and DE 6100 mm long. The simply supported beam AB carries a vertical load of 100 000N applied at F, a distance one-third of the s...

Ngày tải lên: 13/08/2014, 16:21

... For mild steel this point occurs at a slenderness ratio of approximately 100 , as shown in Fig. 6.5. t I I I * 100 200 300 (l/d 0' Fig. 6.5 Critical stress-slenderness ratio ... 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 rather...

Ngày tải lên: 13/08/2014, 16:21

... (6 .102 ) wb us = -tana ASd (6 .106 ) (6 .107 ) where As is the cross-sectional area of a stiffener. Substitution of at from Eq. (6.95) and oF and crs from Eqs (6 .106 ) and (6 .107 ) ... 6.1 and are listed below A = 600mm2 Zxx = 1.17 x 106 mm4 J = 800mm4 = 0.67 x 106 mm4 I? = 2488 x 106 mm6 Zo = 5.32 x 106 mm4 186 Structural instability t C...

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... tank 100 Port wing Integral fuel tank 101 Flapvane 102 Port slotted flap, lowered 103 Outrigger wheel fairing 104 Port outrigger wheel 105 Torque scissor links 106 Port aileron 107 ... N - T + mlacos 10& quot; - 4.5 sin 10& quot; = 0 N- 137.1 + 13.5~0 ~10~ -4.5sin1O0=O 1.e. 4.5 kN Fig. 8.6 Shear and axial loads at the section AA of the aircraft of Exa...

Ngày tải lên: 13/08/2014, 16:21

... from Eq. (8.51) 3.23 105 ~;5.’6 E(ue) = 100 0/~0 Equation (8.54) then becomes ) dtle D, = - 1; ( g)2 ( Su,e 7 SA:-) ’ ( -3.23 x 5.26 x 105 u;5.’6 Sqrn 100 01~0 Substituting ... 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...

Ngày tải lên: 13/08/2014, 16:21

... -69.0 x 10- 2 x 75&3 4r and q3 = - 104 N/mm in the wall 03. It follows that for equilibrium of shear flows at 3, q3, in the wall 34, must be equal to -138.5 - 104 = -242.5N/mm. ... section. Taking moments of area about this upper surface (4 x 100 x 2+4 x 200 x 2)y= 2 x 100 x 2 x 50+2 x 200 x 2 x 100 + 200 x 2 x 200 which gives J = 75 mm....

Ngày tải lên: 13/08/2014, 16:21