Aircraft structures for engineering students - part 1 pps

... 10 3 10 3 10 7 10 9 11 0 11 0 12 2 12 2 12 5 12 9 13 7 14 1 14 2 14 9 14 9 15 2 15 2 15 6 16 0 16 2 16 5 16 9 17 3 17 4 1. 1 6 Experimental measurement of surface strains 3 1 Also i.e. ... Since Q, = 0, Eqs (1. 1 1) and (1. 12) reduce to Problems 33 Am. or = 10 0.2N/mm2, 0 = 24" 11 ' a 11 = -2 0.2 N/IIUII~, 6' = 11 4...

Ngày tải lên: 08/08/2014, 11:21

... 4.5 g mla =-3 g= 13 .5kN Resolving forces parallel to the axis of the fuselage N - T + mlacos 10 " - 4.5 sin 10 " = 0 N- 13 7 .1 + 13 .5~0 ~10 ~-4 .5sin1O0=O 1. e. 4.5 kN ... variation of 0 .1, K,, = 1. 45 for six specimens, K,, = 1. 445 for 10 specimens, K,, = 1. 44 for 20 speci- mens and for 10 0 specimens or more K,,...

Ngày tải lên: 08/08/2014, 11:21

... vj o 12 1~~ -6 1~ ~ o -1 21~ ~ -6 1~ ~ o -1 21~ ~ 61~ ~ o 12 1~~ 61~ ~ 00 000 0 0 -6 /L2 4/L 0 6/L2 2/L 00 000 0 0 -6 /L2 2/L 0 6/L2 4/L , (12 .45) We may deduce the transformation ... Thus 12 -6 -1 2 -6 -6 4 62 -1 2 6 12 6 -6 2 64 This form of Eq. (12 .44) is particularly useful in numerical calculations for an assem- blage o...

Ngày tải lên: 08/08/2014, 11:21

... Area @ F AB -R/2 -1 12 ~14 ~~ -R/2 -1 12 ~14 ~~ AB =I2 BC Ll2 CD LI2 DE LIZ BD LIZ EB LI2 A€ LI2 AB A R 1 RIA A R 1 RIA A -R -1 RIA A -R -1 RIA A R 1 RIA The ... 4000 20 000 2PB,f 13 213 0 0 16 013 0 0 0 - 16 0J2/3 0 FB 4000J2 -2 0000J2 d2pB.f 13 ~ 213 DC 4000 80 000 PB,f13 11 3 pDsf 1 32 013 320 PD,f 1 32 0...

Ngày tải lên: 08/08/2014, 11:21

... thin plates 12 3 Fig. 5 .1 Plate subjected to pure bending. are given by 1 Z Ex =-, Ey - Px PY Referring to Eqs (1. 47) we have 1 1 &, =-( cJx-vcry); E Ey =-( cry-Vcrx) E (5.2) ... q(x,y)sin-sin- dxdy . mrx m'rx nry n'ry sin - sin - dx dy = m=l 2 n=l T/:jIamsin-sin- a a b b - ab 4 - since a 2 _- -...

Ngày tải lên: 08/08/2014, 11:21

... u = AI sin - , 8 = A3 sin - L L L (6.85) 1 (P-~)A~-PX~A~=O 2EIXX (P-9)A1+PysA3=O 6 .1 2 Flexural-torsional buckling of thin-walled columns 18 3 or UB=U+(YS-YB)@ Similarly ... member 6 .1 2 Flexural-torsional buckling of thin-walled columns 18 5 0 P - ~EIJL~ -Pxs P - ~EI,,JL~ 0 PYS PYS - Pxs IOPIA - .rr2ET/L2 - GJ =O (...

Ngày tải lên: 08/08/2014, 11:21

... qb.67 = qb;23 = (by qb. 21 = -7 .22 X 1 0-4 (250 X 10 0) = -1 8.1N/mm qb .18 = -1 8 .1 - 7.22 x 1 0-4 (200 x 30) = -2 2.4N/mm qb:87 = qb: 21 = -1 8a1 N/mm (by Taking moments ... (N/mm2) 1 +660 640 278 x lo6 35.6 2 +600 600 216 x IO6 32.3 3 +420 600 10 6 x IO6 22.6 4 +228 600 31 x lo6 12 .3 5 + 25 620 0.4 x lo6...

Ngày tải lên: 08/08/2014, 11:21

... 0.09 1. 78 x 10 ' 345.6 7 .1 Cell 11 0 .14 9 0 .11 2 367 37 .15 18 .76 6.63 4.97 0.67 0.25 0 .12 0.07 435.6 3.09 x 10 ' 5. 51 x IO* 8.9 Cell I 11 0 .12 1 15 5 41. 10 2 .10 0.56 ... For cell I For cell I1 [-2 50qI + qII(250 + 725 + 233 + 725) - 233q 11~ ] (ii) de 1 dz - 2 x 355 OOOGREF _- For cell 11 1 (i...

Ngày tải lên: 08/08/2014, 11:21

... we substitute in Eq. (1 1. 28) for q from Eq. (1 1. 25) and for PB from Eq. (1 1. 26). Hence 1 #PA - Gt ( PA S,Z PA) - - - - - - - - - - - 2 dz2 dE 2B 2Bh A ... ~~ 12 " 10 84 1. 220 1 2L 216 0 1. 625 14 ,23 12 7 0. 915 34u 797 0. 915 34L 797 0. 915 I 10 8400 I1 202 500 I 11 528 000 An...

Ngày tải lên: 08/08/2014, 11:21

... the form shown in Fig. 13 .10 . Substituting the second natural frequency in Eq. (vii) we have - - ~2 1 - 16 Xw2 - 1 - 16 x (1/ 21) 15 - v1 15 Xw’ -= ~2 1 - 16 h2 1- 16 ... simplifies to (13 .15 ) 556 Elementary aeroelasticity For a non-trivial solution (1 - 16 d) -1 5Xw’ -( 1 - 6XJ) Expanding this determi...

Ngày tải lên: 08/08/2014, 11:21