Advanced Engineering Dynamics 2010 Part 8 docx
... + 312 f3 = - e-’LL’L - - 2p [ -4p+ c [Le (6 .81 ) (6 .82 ) where B, = e-6p + e-4’ (1 - 8p) + e-’’’ (1 - 8p + 8p’) + 1 At the impact point, x = 0, the velocity ... but this time versus x. An increase of 21.r in the argument cor- kh = 2n (6 .88 ) or k = 2dh (6 .89 ) k is known as the wavenumber. Fig. 6.24 I 36 Impact and one-dimensio...
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Advanced Engineering Dynamics 2010 Part 6 docx
... of momentum relative to the centre of mass is L, = Api + Bqj + Crk (5 .84 ) (5 .85 ) (5 .86 ) 100 Dynamics of vehicles c = -1 1" dr = (d2 + r3/6)- 1" YK 241 ... - APdk (5 .87 ) The moment of forces about G is MG = Li + Mj + Nk Therefore (5 .88 ) L, = Ap + (C - B)qr M = Bq + (A - C)rq N = Ci + (B - A)pq (5 .89 ) (5...
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Advanced Engineering Dynamics 2010 Part 13 docx
... frames. From equation (9 .83 ) we get (9 .83 ) (9 .84 ) (9 .85 ) (9 .86 ) Thus from equation (9 .84 ) ux - v u,’ = (1 - vu,/cz) and from equations (9 .85 ) and (9 .86 ) uz u: = (1 - vu,/cz) ... x: (9 .88 ) (9 .89 ) (9.90) (9.9 1) If the length of the outward journey is L then when x = L, x’ = 0, so from equation t,“ and x{ are constants to be d...
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Advanced Engineering Dynamics 2010 Part 1 pdf
... Dispersive waves 55 55 55 58 59 61 64 65 67 72 75 76 80 83 85 85 85 88 90 93 100 103 106 107 107 109 1 I8 125 125 125 1 28 130 132 133 136 1 38 141 145 149 2 Newtonian ... 172 172 172 176 177 1 78 179 184 186 189 192 194 194 194 197 223 235 235 235 240 24 1 242 246 249 250 252 254 256 2 58 261 272 281 288...
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Advanced Engineering Dynamics 2010 Part 2 doc
... a group of particles 15 1.13 Newton's laws for a group of particles Consider a group of n particles, three of which are shown in Fig. 1.1 1, where the ith parti- cle has ... on the particle is the vector sum of the forces due to each other particle in the group and the resultant of the external forces. If & is the force on particle i due to particle ......
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Advanced Engineering Dynamics 2010 Part 3 ppt
... then the time integral will be integrated by Parts because 6u = 0 at t, and t2. The second term in equation (3. 18) is Integrating by parts gives (3.19) 50 Hamilton S principle ... termed To and is the kinetic energy of a single particle of mass m located at the centre of mass. The second term will be J 2 J Fig. 2 .8 Ha m i It o n’s Pr i nci p I e 3.1 ....
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Advanced Engineering Dynamics 2010 Part 4 pot
... l/,13. This is known as the moment of inertia ellipsoid. 4 .8 Torque-free motion of a rigid body From equation (4. 18) we have that if the torque Mo is zero then the moment of momentum ... (1. 48) we have that the linear momentum is the total mass times the velocity of the centre of mass. This is true whether the body is rigid or not, so equation (1.50) is valid (4...
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Advanced Engineering Dynamics 2010 Part 5 pdf
... (5.12) 78 Rigid body motion in three dimensions 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 ... possible values of 8 for the given initial conditions. It is seen that 8 oscillates between levels 0, and 8, whilst O3 is the value of 8 where8 = 0. 2Z, sin2 8 We sh...
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Advanced Engineering Dynamics 2010 Part 9 potx
... wavenumber K= Rk and the non-dimensional frequency (6. 182 ) (6. 183 ) (6. 184 ) 1 66 Impact and one-dimensional wave propagation Fig. 6 38 or U,l (PI = PAD2@) (6.161) The component equations ... along the axis of the wire. Thus (6.1 78) 0, = 0, e (6.179) 01- J up = GkeJ("'-k') (6. 180 ) p"0, = R (-Jk)0, (6. 181 ) J(0f - b) - 0&apos...
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Advanced Engineering Dynamics 2010 Part 11 pps
... (ef>(e)T@> (8. 8) Now 3. 3 il.,*f 3. kl (e‘)(e’)T - - ( ’:) ( 3 jl kl ) = [ jf. il jf.jl jl. kl ] k’ k’. i’ k1.j’ k’. k’ =[; 8 B] (8. 9) so that equation (8. 8) reads ... overall dynamics and not with the detail. This is a vast subject area of which dynamics is a substantial and vital part. 8. 2 Typical arrangements 8. 2.1 CARTESIAN CO-ORDI...
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