Advanced Engineering Dynamics 2010 Part 3 ppt

Advanced Engineering Dynamics 2010 Part 3 ppt

Advanced Engineering Dynamics 2010 Part 3 ppt

... dxdt=O Carrying out the variation t2 +=L s,,.Lo[ p‘(&)6($) -T($)6($)pd2=O (3. 15) (3. 16) (3. 17) (3. 18) To keep the process as clear as possible we will consider the two terms separately. ... 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...

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Advanced Engineering Dynamics 2010 Part 1 pdf

Advanced Engineering Dynamics 2010 Part 1 pdf

... impulsive forces xi 1 1 1 2 3 5 5 6 11 12 13 13 14 15 17 17 18 19 21 21 23 25 27 29 31 33 35 39 41 43 Contents ix 6.12 6. 13 Waves in periodic structures 6.14 ... 55 58 59 61 64 65 67 72 75 76 80 83 85 85 85 88 90 93 100 1 03 106 107 107 109 1 I8 125 125 125 128 130 132 133 136 138 141 145 149 2 Ne...

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Advanced Engineering Dynamics 2010 Part 2 doc

Advanced Engineering Dynamics 2010 Part 2 doc

... Thus equation (1. 53) can be written Newton 's laws for a group of particles 15 1. 13 Newton's laws for a group of particles Consider a group of n particles, three of ... 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 j and F, is the resultant for...

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Advanced Engineering Dynamics 2010 Part 4 pot

Advanced Engineering Dynamics 2010 Part 4 pot

... 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 ... Figure 4.7 shows a plot of IJ 13 against 12/Z3 on which equation (4.67) plots as line ab. For a long slender rod with its axis along the 3 axis I3 = 0 and I, = I2 and is shown ... equati...

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Advanced Engineering Dynamics 2010 Part 5 pdf

Advanced Engineering Dynamics 2010 Part 5 pdf

... symmetric bodies will attract 4 31 4 g = Gp-rca - = Gp-rca 3 a2 3 each other as if each was a point mass concentrated at their respective centres. 5 .3 The two-body problem Figure ... (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...

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Advanced Engineering Dynamics 2010 Part 6 docx

Advanced Engineering Dynamics 2010 Part 6 docx

... e.1.e = l2d + m2z2 + n2z3 = (1 - n2)1, + n2z3 Finally equation (5.59) is Gm G [3( 1 - n2>1, + 3n24 - 2~, - z,] p= +- R 2 ~3 R 2 ~3 G (3n2 - I)(z, - 1,) (5.60) ... 1) into equation (5 .30 ) and rearranging gives i = + + 3 case or 1 r = 1 + ecose (5 .32 ) Ate = 0 (5 .33 ) (5 .34 ) y=y,=- 1 l+e 1 1-e and at 8 = x r = r...

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Advanced Engineering Dynamics 2010 Part 8 docx

Advanced Engineering Dynamics 2010 Part 8 docx

... h3’2 = - kV p 3 h and because as p + a, t + 1, V X,, = kVk = (6.52) (l/Z, + l/ZJ Thus X - = p3 XI, Introducing a non-dimensional time 2 13 1 13 - t=th v (6. 53) ... force/velocity). From equations (6.29), (6 .30 ) and (6 .3 1) we obtain 2v/c, - (1 - z,/Z,)g:-, (1 + ZJZ2) (6 .32 ) g'n = and (6 .33 ) ) clzl c2z2 ( (1 + Z,/...

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Advanced Engineering Dynamics 2010 Part 9 potx

Advanced Engineering Dynamics 2010 Part 9 potx

... exponen- tially, (6. 135 ) U" = U(- 1)" e" e-Kn Substituting equation (6. 135 ) into equation (6. 130 ) leads to o = 2J (dm) cosh (k'/2) (6. 136 ) for o > oca. ... motion is S(U,+~ - u,) - s(u, - n,-,) = mu, (6. 130 ) Let us assume (6. 13 1) Substituting equation (6. 13 1) into equation (6. 130 ) and dividing through by the common...

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Advanced Engineering Dynamics 2010 Part 11 pps

Advanced Engineering Dynamics 2010 Part 11 pps

... and a3 are constant so that the relevant A matrices are ce, o se, o oL41,= [ i [ -: ;,I ,[A12= [ %; 3& apos; ; a27 j CB, -S0, 0 a,C0, C8, -SO3 0 &e3 *[A 13 = ... scalar products. Thus (e’)(e’)T(p’) = (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’...

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