Aircraft Structures 3E Episode 1 pdf
... critical load for a flat plate 76 77 85 100 103 103 107 109 110 110 122 122 125 129 137 141 142 149 149 152 152 156 160 162 165 169 173 174 Contents vii 6.9 6 .10 6.11 6.12 6.13 Local instability ... of open and closed section beams 9 .10 Deflection of open and closed section beams Problems 276 276 10 Stress analysis of aircraft components 10. 1 Tapered beams 10. 2 Fuselages 29 295 300 307 316 ... 327 331 342 345 362 62 374 viii Contents 10. 3 10. 4 10. 5 10. 6 Wings Fuselage frames and wing ribs Cut-outs in wings and fuselages Laminated composite structures Reference Further reading Problems...
Ngày tải lên: 13/08/2014, 16:21
Aircraft Structures 3E Episode 2 pdf
... Basic elasticity Y Q, (360 x 104 , i x 650 x 104 ) (-290x 04, Fig 1.14 Mohr's circle of strain for Example 1.3 Now substituting in Eq (1.35) for E,, &I E, and -yrY + = 10- \/(360 + 290)2 + 6502 which ... (1.57) cII = - 10. 9N / m Since Q, = 0, Eqs (1.1 1) and (1.12) reduce to 32 Basic elasticity and (ii) respectively Adding Eqs (i) and (ii) we obtain 01 + a11 = a, Thus ax = 80.9 - 10. 9 = 70N/mm2 ... their directions at a point on the surface of the shaft A ~ S aI= ~ / m m ~ ,e = 3104 3' aII = - ~ / m ~ ,e = 1 2104 3' P.1.7 An element of an elastic body is subjected to a three-dimensional stress...
Ngày tải lên: 13/08/2014, 16:21
Aircraft Structures 3E Episode 3 potx
... (mm) 100 200 300 400 500 600 700 800 -0.3 -1.4 -2.5 -1.9 2.3 4.8 10. 6 What will be the angular rotation of the beam at the prop due to a 30N load applied 200mm from the wall, together with a 10N ... 000 -20000J2 100 000 -2J2pB,f/3 (N) -2pB,f/3 -d2PB.f/3 PB,f13 PB,f/3 2pB,f/3 2PB,f 13 d2pB.f 13 FD,f -2J213 -213 -~ 113 113 213 213 ~213 0 0 @ aF&f /apBqf pDsf PD,f pD,f 0 (N) @ x 106 aFDsf laPD.f ... problems 83 w/un it length Pf -1 AT t L I Fig 4 .10 Deflection of a uniformly loaded cantilever by the method of complementary energy length (see Fig 4 .10) First we apply a fictitious load Pf at the...
Ngày tải lên: 13/08/2014, 16:21
Aircraft Structures 3E Episode 4 pps
... beam due to uniform temperature I% Fig 4.28 Bending of beam due to linear temperature gradient 107 108 Energy methods of structural analysis z (1 + at) I t R (b) (a) Fig 4.29 (a) Linear temperature ... London, 1996 1 10 Energy methods of structural analysis Argyris, J H and Kelsey, S., Energy Theorems and Structural Analysis, Butterworths, London, 1960 Hoff, N J., The Analysis of Structures, John ... when P is 10 000 N All bars are of the same material ( E = 70000N/mm2) and have a cross-sectional area of 300mm2 Ans P = 2947N: F12 = 2481.6N(T), F23 = 1861.2N(T), F34 = 2481.6N(T), = 3102 .ON(C)...
Ngày tải lên: 13/08/2014, 16:21
Aircraft Structures 3E Episode 5 ppsx
... Elasticity, McGraw-Hill Book Company, New York, 1953 P.5.1 A plate 10mm thick is subjected to bending moments M , equal to 10N m/mm and M y equal to N m/mm Calculate the maximum direct stresses ... point occurs at a slenderness ratio of approximately 100 , as shown in Fig 6.5 t 900 c N E z 600- v a b" 300 Yield stress - - - - - - I 0' I 10 I I 200 300 (l/d Fig 6.5 Critical stress-slenderness ... from Fig 6.16 that k is very nearly constant for a / b > This fact is particularly useful in aircraft structures where longitudinal stiffeners are used to divide the skin into narrow panels (having...
Ngày tải lên: 13/08/2014, 16:21
Aircraft Structures 3E Episode 6 pps
... specified in Example 6.1 and are listed below A = 600mm2 Zxx = 1.17 x 106 mm4 Zo = 5.32 x 106 mm4 J = 800mm4 = 0.67 x 106 mm4 I? = 2488 x 106 mm6 188 Structural instability From Eqs (6.90) P ~ ~ 4.63~ io5 ... described in Chapter Ans (a) 3.09 x 104 N, (b) 1.78 x 104 N, 1.19 x 104 N P A simply supported beam has a span of 2.4m and carries a central concentrated load of 10kN The flanges of the beam each ... +0.0113As), 238910Nmm Part II Aircraft Structures Principles of stressed skin construction With the present chapter we begin the purely aeronautical section of the book, where we consider structures...
Ngày tải lên: 13/08/2014, 16:21
... critical load for a flat plate 76 77 85 100 103 103 107 109 110 110 122 122 125 129 137 141 142 149 149 152 152 156 160 162 165 169 173 174 Contents vii 6.9 6 .10 6.11 6.12 6.13 Local instability ... of open and closed section beams 9 .10 Deflection of open and closed section beams Problems 276 276 10 Stress analysis of aircraft components 10. 1 Tapered beams 10. 2 Fuselages 29 295 300 307 316 ... 327 331 342 345 362 62 374 viii Contents 10. 3 10. 4 10. 5 10. 6 Wings Fuselage frames and wing ribs Cut-outs in wings and fuselages Laminated composite structures Reference Further reading Problems...
Ngày tải lên: 13/02/2014, 20:20
... can be used to make synchronisation urgent without causing a time-actionlock (see Sections 9.2 .10 and 9.4) The hiding operator turns observable into internal actions, which are, as we just mentioned, ... Consider the following scenario Source1 sends its first packet to Place1, at time t = Later, say at t = 10, Source2 sends its first packet to Place2 By the time that Source2 attempts to send the second ... particular, those which result from synchronisation) are urgent on their upper bounds (Section 9.2 .10) 12.3 Time-actionlocks (i) x>=1 b! (x>=2) (ii) x>=1, y>=3 b ((x>=2 | y>=4), x>=1,y>=3) y>=3...
Ngày tải lên: 12/08/2014, 07:22
MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 10 pdf
... (y))2 dy [6 .105 ] Plates: out-of-plane motion 343 the functional of kinetic energy is readily found to be: (κ) En,m = ρhLµ1 [6 .106 ] As an example we consider a strip of steel L = 10 m, l = m, ... from the symmetric form: ℓ (p) En,m = L ∂φnm ∂x (0) Mnm : χnm + Fxx dy [6 .104 ] dx By using the relations [6.66] and [6 .102 ] it is found that: (p) En,m = DL nπ L (0) + µ1 − 2ν Fxx L nπ L nπ L ... En,m = (κ) En,m Kn,m Mn,m [6 .103 ] where the functional of potential and kinetic energies are calculated by using the postulated mode shapes Since the trial functions [6 .102 ] not comply with the boundary...
Ngày tải lên: 13/08/2014, 05:22