Aircraft Structures 3E Episode 6 pps

... = - 76. 2mm. The remaining section properties are found by the methods specified in Example 6. 1 and are listed below A = 60 0mm2 Zxx = 1.17 x 106mm4 J = 800mm4 = 0 .67 x 106mm4 I? ... of Eqs (6. 98) or (6. 99) 2 Ot +'F W 2AF tan a CF = in which AF is the cross-sectional area of each flange. Also, from Eq. (6. 102) wb us = -tana ASd (6. 1 06...

Ngày tải lên: 13/08/2014, 16:21

... of Structures, John Wiley and Sons, Inc., New York, 19 56. Timoshenko, S. P. and Gere, J. M., Theory of Elastic Stability, McGraw-Hill Book Company, 1 960 . New York, 1 961 . ... I-_ eA(c) 2 61 - WM or W eA(c),2 = z61 (ii) where 6A(c),2 is the rotation at A due to W at C. Finally the rotation at A due to MA at A is, from Fig. 4. 26( a) and (c)...

Ngày tải lên: 13/08/2014, 16:21

... equal to the angle 64 in the strain diagram of Fig. 6. 7(c). Hence and Eq. (6. 9) becomes, from Eqs (6. 1 1) and (6. 12) (6. 12) (6. 13) Further, in a similar manner, from Eq. (6. 10) d2 $ (. ... EI.y.x - = w dz4 (6. 65) Also, the equation for the buckling of a pin-ended column about the Cx axis is (see Eq. (6. 1)) (6. 66) 6. 3 Effect of initial imperfections 1...

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... 161 Cannon muzzle fairing 162 Lift augmentation retractable cross-dam 163 164 165 166 167 168 169 170 171 172 173 174 175 1 76 177 1 76 179 180 181 182 183 184 185 1 86 ... fairing 61 Outrlgger leg doors 62 Starboard aileron 63 Aileron composite construction 64 Fuel jettison 65 Formation lighting panel 86 Roll control airvalve 67 Wing t...

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... ucos$ + usin$ (1 1 .65 ) Also, since shear strains are assumed to be negligible, Eq. (9. 26) becomes (1 1 .66 ) Substituting for vt in Eq. (1 1 .66 ) from Eq. (1 1 .65 ) and integrating from ... same for all walls. Ans. Fig. P.ll.l q12 = q 56 = 46. 6 N/mm, 932 = q54 = 1.4N/mm1 XR = -63 0.1 mm, 952 = 180.8 N/mm, q43 = 74.6N/mm7 YR = 0 (relative to mi...

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... 380 4 06 415 425 432 432 432 443 443 445 449 455 465 485 4 86 494 495 4 96 497 500 507 507 509 5 16 533 533 533 540 54 1 5 46 55 1 568 5 76 577 Index 582 6 Basic ... on internal faces AC and CB. Summation of forces in the x direction gives 26s - O. ,6, V - Ty.,6X + xi 6X6y = 0 which, by taking the limit as Sx approaches zero, beco...

Ngày tải lên: 13/08/2014, 16:21

... = 10- + \/( 360 + 290)2 + 65 02 which gives &I = 495 x Similarly, from Eq. (1. 36) EII = -425 x IOp6 From Eq. (1.37) 65 0 x 360 x lop6 + 290 x lop6 = tan20 = ... (Fig. 1.14). Axes OE and Oy are set up and the points Q1 ( 360 x lop6,: x 65 0 x and Q2 (-290 x lop6, - 4 x 65 0 x IOp6) located. The centre C of the circle is the intersec...

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... giving -ll&PL3 RL3 +- 768 EI 16EI dz= Substituting from Eqs (iv) and (v) into Eq. (iii) 11d3~~~ m3 RL A+IOAB +- 768 EI 16EI+4E ( ABA ) =O - 76 Energy methods of structural ... -2pB,f/3 -213 0 0 160 0 FD 4000J2 -80000J2 -d2PB.f/3 - ~213 0 0 64 0J213 0 CB 4000 80 000 PB,f/3 113 EB 4000 20 000 2PB,f 13 213 0 0 160 13 0 0 0 - 1...

Ngày tải lên: 13/08/2014, 16:21

... 105u;5. 6 Sqrn 10001~0 Substituting for and Skqm we have or D, = 16. 99 x lo2 ( - :)2 Jm (u;: 2u;5. 26 + u ;6. 26) due 2/10 ur u2 Uf from which or, in terms of the aircraft ... Weight W = 160 0000N CM,O = - 0.01 Mean chord E = 22.8 m At 18 300m p = 0.116kg/m3 Am. P = 267 852N, AP = 362 57N, AL = 271 931 N, n = 1.19 P.8.8 An...

Ngày tải lên: 13/08/2014, 16:21

... 2.0 x 60 0 ( ;;l) B1 = 300 + 3.0x400(2-l)+ 6 6 2+- which gives Also B1(= B6) = 1050mm2 i.e. 1.5 x 60 0 ( KM)) 2+- 2.0 x 60 0 ( 2M) 2.5 x 300 2+- + (2-1)+ 6 6 6 B2 = ... 3 and 4 the right- hand side is positive. Thus q12 = -6N/= q23 = -6 + q12 = -12N/mm and q34 = +6 + q23 = -6N/mm giving the same solution as before. Note that...

Ngày tải lên: 13/08/2014, 16:21