aircraft structures 3e episode 10 pdf

... Table 10. 1 0 0 0 @ 6x,/6r Boom Pz, (kN) 6y,lSz -100 0.1 1 2 3 4 5 6 -133 -100 100 133 100 0 -0.1 -0.1 0 0.1 -0.05 -0.05 -0.05 0.05 0.05 0.05 Py,, a (kN) (kN) P, (kN) -10 0 10 -10 0 IO ... Pqr = ai,rBr (10. 8) where B, is the cross-sectional area of the rth boom From Fig 10. 4(b) (10. 9) Further, from Fig 10. 4(c) or, substituting for Py,, from Eq (10. 9) (10. 10) The axial load ... Eqs (10. 13) for P.,,r and P,,,r from Eqs (10. 10) and (10. 9) we have sx, P:,r - s = sxw + x : r=l sz 1 sy = sy:w + pz,r I sY ' r= 1 6z (10. 14) Hence (10. 15) 368 Stress analysis of aircraft

Ngày tải lên: 13/08/2014, 16:21

... Application to deflection problems 83 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 ... in the rapid approximate solution of problems for which exact solutions do not exist Also, many structures which are statically indeterminate, that is they cannot be analysed by the application ... this chapter with the solution of deflection problems and the analysis of statically indeterminate structures We shall also include some methods restricted to the solution of linear systems, viz

Ngày tải lên: 13/08/2014, 16:21

... in all the members when P is 10 000 N All bars are of the same material ( E = 70000N/mm2) and have a cross-sectional area P.4.10 The plane frame ABCD of Fig P.4.10 consists of three straight ... 1996 Trang 51 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, John ... Trang 1106 Energy methods of structural analysis Example 4.10 An elastic member is pinned to a drawing board at its ends A and B

Ngày tải lên: 13/08/2014, 16:21

... 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 ... the aircraft of Example 8.2 4 /10m 8.3 Aircraft inertia loads 243 The horizontal and vertical inertia forces ma, and ma, act at the CG, as shown in Fig 8 .7; pn is the mass of the aircraft ... 8.4 we determined aircraft loads corresponding to... acceleration and aircraft response were based on the assumptions that the aircraft s flight is undisturbed while the aircraft passes

Ngày tải lên: 13/08/2014, 16:21

... beam 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 5 0 + 2 x 200 x 2 x 100 The section is symmetrical about C y so that Ixy = 0 and since ... much less than primary warping and is usually ignored in the thin-walled sections common to aircraft structures Equation (9.66) may be rewritten in the form or, in terms of the applied torque ... 1 0 - 4r 2 x 75&3 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 Hence q34 = -69.0

Ngày tải lên: 13/08/2014, 16:21

... Fig 10.22 (10.29) If the moment centre is chosen to coincide with the point of intersection of the lines of action of S, and S,, Eq (10.29) becomes (10.30) The wing section of Example 10.6 ... Fig 10.27 Evaluating 6 for each wall and substituting in Eq (10.28) gives for cell I Fig 10.27 qb (Wmrn) distribution in beam section of Example 10.9 (view along z axis towards C) Trang 104.6 ... then Jds/t* Trang 14288 51.38 5.20 0.93 0.09 1.78 x 10' 8.9 Cell I11 0.121 155 41.10 2.10 0.56 0.03 198.8 0.64 x 10' 4.1 - * l * l - * 10.3.7 Method of successive approximations - shear

Ngày tải lên: 13/08/2014, 16:21

... from Eqs (10.45) and (10.46) E~E~A = EfEfAf + Em€lAm so that Writing Af/A = vf and A,/A = v,, Eq. (10.47) becomes El = VfEf + VmEm (10.47) (10.48) Equation (10.48) is ... Substituting in Eq. (10.50) or, from Eq. (10.48) Now substituting for urn in the first two of Eqs (10.51) ut1 Yt -=- Et Et or (10.51) (10.52) 428 Stress analysis of aircraft components ... Eqs (10.54), we obtain Also 71t = Gltylt I Equations (10.55) may be written in matrix form as (10.54) (10.55) (10.56) in which c21 = c22 = c33 = Gt (= CI2, see Eq. (10.52))

Ngày tải lên: 13/08/2014, 16:21

... for continuum structures In the previous sections we have discussed the matrix method of solution of structures composed of elements connected only at nodal points For skeletal structures consist- ... in Fig 12.10 The stiffness matrices for the beam-elements 1-2 and 2-3 are obtained from Eq (12.44); thus Fig 12.9 Idealization of beams into beam-elements Trang 9- L O Lb i Fig 12.10 Assemblage ... 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 the

Ngày tải lên: 13/08/2014, 16:21

... 444 Aircrafi Structures for Engineering Students provides a comprehensive self- contained course in aircraft structures. Starting with the structural mechanics of aircraft this book ... warping function 60 Torsion of wings, see Stress analysis of aircraft components Total complementary energy 68, 76-100, 108 of a beam subjected to a temperature of an end loaded cantilever ... 497-500 stiffness method 494 Maxwell, J.C. 103 Membrane analogy 61-3 Miner and Palmgren’s linear cumulative Model analysis of a fixed beam 106, 107 Modulus of elasticity 24 Modulus of

Ngày tải lên: 13/08/2014, 16:21

... distribution (Pm) lo00 lo00 loo0 loo0 lo00 loo0 825/25/150 8251501125 3001401660 1050 1 390 91 0 1360 92 0 96 0 1050 1070 93 0 30 5 60 1 140 60 5t 1 It Water vapour (glrn’ 24 hr) 2 0.75 0.25 2 ... Plastics 0 295 50 100 200 150 Flow ratlo Fig 4.42 Claming pressures for different cavity geometries (typical values for easy flow materials) prudent to increase this value by 10-20% due to the

Ngày tải lên: 13/08/2014, 09:20

... aircraft required 120 110 100 200 500 Number of aircraft produced 100 0 Fig 8.36 Aircraft recurrent cost versus production run shared by the increased number of aircraft produced the ... factored distance 100 Weight (lb×10 –3 ) 110 MTOW 120 3000 4000 6000 8000 Unfactored hourly distance (ft) Fig. 8.35 Landing distance versus aircraft weight “chap08” — 2003/3/10 — page 259 — #58 ... developments for the aircraft This may colour future decisions on the layout and capabilities of the aircraft Most existing aircraft have... 1-56347 -105 -1 4 Raymer, D P., Aircraft Design:

Ngày tải lên: 13/08/2014, 15:21

... mass (each)= (π · 100 · 1 · 8) 2767/(1000 · 1000) = 7 kg Add 10 kg(22 lb) for fairing and support structure and add a contingency of 25 per cent: Total brace mass (both)= 2 · (7 + 10) · 1.25 = 42 ... required for aircraftperformance so should be adequate to meet the aircraft service needs.9.7.7 Initial aircraft layout The previous sections have set out the geometrical requirements for the aircraft ... (MF/MTO)} For the HALE aircraft there are some difficulties that arise from the definitions of aircraft systems to be included in the aircraft empty mass ratio Many of the systems on the aircraft are directly

Ngày tải lên: 13/08/2014, 15:21

... depth to which the system is able todiagnose faults, the other codes are listed below  Codes 101 or 102: keypad faults in the Master Display Unit  Codes 201–208: communication faults between ... (altitude) aˆaddd.dˆaamˆaamˆaacle 13 character field ˆa aaˆaaˆad.d if altitude < 10 mˆa aaˆadd.d if altitude < 100 mˆa Trang 482 Electronic Navigation SystemsAs an example, the first line of ... waves in free space is 3 × 108ms–1, then the distance travelled by a pulse may be measured in terms of the time taken to travel that distance, i.e if a pulse took 1000 µs to travel a certain

Ngày tải lên: 13/08/2014, 16:21

... is noted The rudder angle is then reduced to 10°starboard and the new steady turn rate noted This is repeated forangles of 5°S, 5°P, 10°P, 15°P, 10°P and so on The resulting steady rates of turn ... certain path relative to the ground in conditions Trang 9Figure 10.10 Vertical axis rudder (a) Construction (b) OperationTrang 10of wind and tide Other vessels demanding good positional control ... rudder movement reducing the torque required of thesteering gear Trang 7Figure 10.7 Semi-balanced rudderFigure 10.8 Flap rudder In semi-balanced and unbalanced rudders the fixed structure ahead

Ngày tải lên: 13/08/2014, 16:22

... known as negative offset (Fig. 10.11). With negative offset the Fig. 10.10 Swivel pin inclination positive offset Fig. 10.11 Swivel pin inclination negative offset Fig. 10.12 Directional stability ... wheels. 10.2.4 Transverse double wishbone suspension (Figs 10.20, 10.21 and 10.22) If lines are drawn through the upper and lower wishbone arms and extended until they meet either inwards (Fig. 10.20) ... 10.2.5 Parallel trailing double arm and vertical pillar strut suspension (Figs 10.23 and 10.24) In both examples of parallel double trailing arm (Fig. 10.23) and vertical pillar strut (Fig. 10.24)

Ngày tải lên: 21/07/2014, 17:20

... Scale: Motor Gear 10 in. 10 mils Desired off-line vertical motor shaft position Desired off-line vertical gear shaft position Shaft to coupling spool method East Top view Scale: 10 in. 10 mils Desired ... stem will move outward as it rotates to the bottom of the pump shaft producing Side view Scale: 10 in. 10 mils Observed amount of proximity probe gap change from OL2R conditions Vertical probe gap ... readings will be when aligning your machinery tocompensate for OL2R movement: Side view Scale: 10 in. 10 mils Observed amount of proximity probe gap change from OL2R conditions Vertical probe gap

Ngày tải lên: 05/08/2014, 11:20

... that is saturated with water vapor at ambient temperature and 25 to 100 psig is used. At 25 psig and 100°F this gas is saturated with 1,500 Ib/MMscf of water vapor. At ... Dehydrating 50 MMscfd at 1,000 psig and 100°F Tower Tower Diameter Internals (inch) A. Tray Bubble Cap B. Structural Packing (Figures 6-9, B 1-300 B 1-100 Flexipac#l Flexipac #2 C. Random ... "U," can be approximated as 10 to 12 Btu/hr-ft 2 -°F for glycol/glycol exchangers, 45 Btu/hr-ft 2 -°F for the gas/ glycol exchanger, and 100 Btu/hr-ft 2 -°F for the reflux

Ngày tải lên: 06/08/2014, 02:20

... flat steel (R  10000 kg/m3, C  500 J/kg•K, k  100 W/m•K) sheet emerges from a furnace at 10 cm/s and 800nC At distances of 10 meach, there are three rolling dies; see Figure 1.10(d) The initial ... you expect to obtain for the different systems (a) Food-freezing plant to chill vegetables to –10nC by circulating chilled air past the vegetables (b) A shell and tube heat exchanger, with hot ... (c) A system consisting of pumps and pipe network to transport water from ground level to a tank 100 m high (d) A vapor compression system for cooling a cold storage room (e) Flow equipment such

Ngày tải lên: 06/08/2014, 15:20

... between several ten nm and 10µm, is possible using IBS devices.For a reasonable range of annealing times up to about 106s, a diffusivity rangebetween 10−23m2s−1 and 10−16m2s−1 can be examined ... extended diffusion anneals and large enough diffusivities, D > 10−15m2s−1 , lathe sectioning can be used Diffusivities D > 10 −17m2s−1 are accessible via microtome sectioning In cases where ... Isotopically Controlled Heterostructures The use of enriched stable isotopes combined with modern epitaxial growthtechniques enables the preparation of isotopically controlled heterostructures.Either chemical

Ngày tải lên: 06/08/2014, 15:21