Advanced Engineering Dynamics 2010 Part 4 pot
... - c [r]' [TI' (0) (4. 44) (4. 45) Also from Appendix 1 [r]" [TIX = (r) (ry - r2 [ 11 and therefore equation (4. 45) is (4. 46) 2 The term in the large parentheses ... Moment of inertia 63 (4. 32) (4. 33) (4. 34) If we premultiply equation (4. 32) by (~0~)~ and subtract equation (4. 33) premultiplied by (al)T the resulting scalar ... rotat...
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Advanced Engineering Dynamics 2010 Part 9 potx
... or in matrix form where p -1lR 0 OOP and d' p'ds (6. 140 ) (6. 141 ) (6. 142 ) (6. 143 ) (6. 144 ) (6. 145 ) A cross-section has a displacement u and a rotation 0 from its ... of wire and also to rigid body rotation, so the strain (4 = [TI1 (u) - [@IX (WIds = [TI1 (4 + [d9/WX (0) (6. 146 ) Now (ds) = (ds 0 0) and therefore 000...
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Advanced Engineering Dynamics 2010 Part 15 pot
... both symmetric. We now write equations (A2 .43 ), (A2 .40 ) and (A2 .42 ) in matrix form. Thus (6P)T[M]{(i') - (a)} = 0 (A2 .45 ) (A2 .46 ) (A2 .47 ) It is known that {[Xl[Yl}' = ... h2dq2 + - ah2 dq,dq,) ( 2 )( a93 ( a42 * )( a42 a3 - 4h3dq3 + 6 + -dq2 h3dq3 + -dq2dq3) - - ( a(g2) + p)dq2dq, a(@,) a42 Thus, since dA, = h,dq2h3...
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Advanced Engineering Dynamics 2010 Part 16 pot
... Proper 240 Dilation 240 Timoshenko Equation - Transverse Waves in Beams 159 Top, Sleeping 79 Transformation Matrix (4x4) 206 Twin Paradox 249 Tycho Brahe 1 Understeer 1 24 Vectors, Axial ... Hartenberg Representation 208 Dispersion 125, 149 Diagram 151 Dissipation function 27 Divergence 290 Doppler Effect Light 244 Sound 242 Drag I 12 Dyad 212 Dyadic 273 Eigenval...
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Advanced Engineering Dynamics 2010 Part 1 pdf
... 3 .4 3.5 Illustrative example Lagrange’s equations derived from Hamilton’s principle 46 46 47 49 51 52 4 Rigid Body Motion in Three Dimensions 4. 1 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 4. 8 ... 161 1 63 172 172 172 176 177 178 179 1 84 186 189 192 1 94 1 94 1 94 197 223 235 235 235 240 24 1 242 246 249 250 252 2 54 256 258 261 272...
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Advanced Engineering Dynamics 2010 Part 2 doc
... (XI X, . . . X*) (4 = [A1 (4) + (b) then we may write where and Hence we may write = 1 (4& gt;' [AIT [ml [AI (4) + (b)T[ml [AI (4) + [ml (b) (2. 24) Note that use has ... = i, we have for the ith particle ( 1 .44 ) C AJ + F, = mlrl I We now form the sum over all particles in the group c YAJ '2 Fl = c rnlrl (1 .45 ) I I I The firs...
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Advanced Engineering Dynamics 2010 Part 3 ppt
... integrating by parts, h 6x 1; - It:mi 6x dt - kx 6x dt = 0 4 (3.10) Rotating fiame of reference and velocity-dependent potentials 37 It is interesting to note that for a charged particle, ... hand the Lagrangian is E= 2 2 Fig. 3 .4 Non-holonomic systems 4 1 The modified form of Lagrange’s equation for angular motion yields (2 .46 ) (2 .47 ) In this equation...
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Advanced Engineering Dynamics 2010 Part 5 pdf
... sin (w) i i' = cos(y)i - sin(w)j (4. 73) (4. 74) (4. 75) Fig. 4. 17 Satellite motion 93 p=l n=-2 p=2 n= 1 p=3 n= 6 p =4 n= 13 Notice that the only negative value of ... Substituting equations (4. 86) and (4. 8.7) into (4. 88) gives 1 (P, - P,cos~)2 P: E = - Z,e2 + + - + mgh cos8 (4. 89) 2 21, sin2 0 213 Equation (4. 89) can...
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Advanced Engineering Dynamics 2010 Part 6 docx
... 96 Dynamics of vehicles 2E.9 + 2Kr - L*2 = 0 The values of r satisfying this equation are -2K f d(4K’ + 8Ei”) 4E’ For real roots 4d + 8Eio2 > 0 or r= (5 .40 ) K’ ... (5. 84) (5.85) (5.86) 100 Dynamics of vehicles c = -1 1" dr = (d2 + r3/6)- 1" YK 241 + r2) JK - - -[ JK 1" 7 1 tan(6/2) + 5tan3(e/2) ll (5. 54)...
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Advanced Engineering Dynamics 2010 Part 8 docx
... - - - = V - ka (l/Zl + l/Z2) (E4 (E42 (6 .46 ) Let h = k(l/Z, + l/&) so that equation (6 .46 ) becomes a + ha" = V (6 .47 ) If m = 1 then the interface behaves ... "0 - k, - 4 cs - or 0, = c,(k, - 43 ) + wo (6. 94) Now a typical wave in this region is u = U, COS (o,t - k,x) Substituting from equation (6. 94) gives u...
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