Aircraft structures for engineering students - part 10 pot

... method 7 9-8 5 for buckling of columns 16 5-9 for bending of thin plates 14 2-9 potential energy 70, 71, 7 3-6 , 14 4-9 , principle of stationary value of total Energy methods 6 8-1 09 108 16 5-9 complementary ... the form shown in Fig. 13 .10. Substituting the second natural frequency in Eq. (vii) we have - - ~2 1 - 16Xw2 - 1 - 16 x...

Ngày tải lên: 08/08/2014, 11:21

... the cross-section of the bar. Rearranging and substituting for u and from Eqs (3.9) dw rzx dB dw r7 dB -A _- - - +-y, - ax G dz dy G dzX - (3 .10) For a particular torsion ... M=-Pz Rz, hence -= 4 2 dR 2z between F and B P d3 dM d3 4 2 dR 2 M = - (L - z) - -Rz, hence - = - -2 and between B an...

Ngày tải lên: 08/08/2014, 11:21

... right-hand side of Eq. (ii) we have r=l qb:23 = qb,34 = -7 .22 X 1 0-4 (400 X 100 ) = -2 8.9N/mm qb,4j = -2 8.9 - 7.22 x io-4(ioo x 50) = -3 2.5N/m.m qb,56 = qb:34 = -2 8.9 ... qb.67 = qb;23 = (by qb.21 = -7 .22 X 1 0-4 (250 X 100 ) = -1 8.1N/mm qb.18 = -1 8.1 - 7.22 x 1 0-4 (200 x 30) = -2 2.4N/mm qb:87 =...

Ngày tải lên: 08/08/2014, 11:21

... Preface During my experience of teaching aircraft structures I have felt the need for a text- book written specifically for students of aeronautical engineering. Although there have been a ... = - [a, + a,, + az - 2v(ax + ay + 41 or In the case of a uniform hydrostatic pressure, a, = a,, = az = -p and 3(1 - 2~) E P e =- (1.49) The...

Ngày tải lên: 08/08/2014, 11:21

... 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 _- - when m=m' and . nry . n'ry sin-sin- ... and My are known. If either M, or My is zero then d2W - a2W d2W - d2W ax2 - 8Y2 ay2 - dX2 - -v- or - -v- and the plate has...

Ngày tải lên: 08/08/2014, 11:21

... Flexural-torsional buckling of thin-walled columns 185 0 P - ~EIJL~ -Pxs P - ~EI,,JL~ 0 PYS PYS - Pxs IOPIA - .rr2ET/L2 - GJ =O (6.86) d’v dz- d2 u EI, 7 = - PV EI,,,, ... 0 1 2 3 4 I5 I3 II 9- 7- 5, k - - - a/b (b) k 4 0- 36 - Clamped edges Simply supported 12345 a/b (C) Fig. 6.16 (a) Buck...

Ngày tải lên: 08/08/2014, 11:21

... AA. Thus 4.5 g mla =-3 g= 13.5kN Resolving forces parallel to the axis of the fuselage 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 ... 0.1, K,, = 1.45 for six specimens, K,, = 1.445 for 10 specimens, K,, = 1.44 for 20 speci- mens and for 100 specimens or more K,, = 1.43. For t...

Ngày tải lên: 08/08/2014, 11:21

... and -2 10 x lo6 = -0 .376~~ 5.58 x lo8 Yr 0z>r = Hence Pi = P3 = -P4 = -P6 = -0 .376 x 305 x 900 = -1 03212N and P2 = -Ps = -0 .376 x 305 x 1200 = -1 37616N ... - PZJ - SZ - Pr, 2- SZ (10. 4) (10. 5) Again we note that Sy2 in Eqs (10. 4) and (10. 5) is negative. Equation (10. 5) may be used to determine the sh...

Ngày tải lên: 08/08/2014, 11:21

... 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 Rearranging, ... = - - qt- 71t 7lt = - Et Glt Solving the first two of Eqs (10. 54), we obtain Also 71t = Gltylt I Equations (10. 55) may...

Ngày tải lên: 08/08/2014, 11:21

... dz & ;- T,(Z) = -+ - - dB dz TJ = GJ- (Eq. (9.60)) and d30 Tr = -Er - dz3 (Eq. (1 1.58)) so that Eq. (1 1.63) becomes d46' d2B Er - - GJ - - T~(~) d# ... matrix for a uniform beam 51 1 It is possible to write Eq. (12.44) in an alternative form such that the elements of [KJ are pure numbers. Thus 12 -6 -1 2...

Ngày tải lên: 08/08/2014, 11:21