MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 7 pdf

... = 4, [4 .73 ] is written as follows:    2 L + ̟ 2 1 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 2 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 3 − ̟ 2 2 L 2 L 2 L 2 L 2 L + ̟ 2 4 − ̟ 2      q 1 q 2 q 3 q 4   =   0 0 0 0   by ... homogeneous system to be solved is:    2( γ L + ̟ 2 ) 2 L 2 L 2 L 2 L 2 L + π 2 − ̟ 2 2γ L 2 L 2 L 2 L...

Ngày tải lên: 13/08/2014, 05:22

... so (dr) 2 = g 2 α (dα) 2 + g 2 β (dβ) 2 [7 .29 ] The determination of (d n) 2 is obtained from the differentiation of n. d n = ∂ n ∂α dα + ∂ n ∂β dβ ⇒ ( d n ) 2 =  ∂ n ∂α  2 (dα) 2 +  ∂ ... n ∂α (dα) 2 + ∂r ∂β . ∂ n ∂β (dβ) 2 +  ∂r ∂α · ∂ n ∂β + ∂r ∂β · ∂ n ∂α  dα dβ [7. 31] With [7 .25 ] and [7 .26 ] it follows that: dr ·d n = g 2 α R α (...

Ngày tải lên: 13/08/2014, 05:22

... stresses 75 2. 1.6.1. Equilibrium of forces 75 2. 1.6 .2. Equilibrium of the moments 77 2. 2. Small elastic motion 78 2. 2.1. Longitudinal mode of deformation 78 2. 2.1.1. Local equilibrium 78 Solid ... 196 4 .2. 2.1. Longitudinal modes 196 4 .2. 2 .2. Torsion modes 20 0 4 .2. 2.3. Flexure (or bending) modes 20 0 4 .2. 2.4. Bending coupled with shear modes 20 5 4 .2. 3....

Ngày tải lên: 13/08/2014, 05:22

... that: λ +2G(cos θ) 2 G = λ +2G(1 −(sin θ) 2 ) G = λ +2G(1 −(γ sin β) 2 ) G = γ 2 (1 2( sin β) 2 ) = γ 2 cos 2 The system [1.1 12] is thus transformed into:  γ cos 2 −sin 2 sin 2 γcos 2  R C  =  −γ ... 2  R C  =  −γ cos 2 sin 2  [1.113] which has the non-trivial solution: R = sin 2 sin 2 −(γ cos 2 ) 2 sin 2 sin 2 +(γ cos 2 ) 2 ; C = 2 sin 2 cos...

Ngày tải lên: 13/08/2014, 05:22

... as:  x+dL /2 x−dL /2  (S) r ×  ρ ∂ 2  ξ ∂t 2 − ∂  t 1 ∂x  dS dx =  x+dL /2 x−dL /2  (S) (r ×  f (e) )dS dx +  M (e) (x; t)dL [2. 20] Using the relations [2. 1], [2. 8], and [2. 9], the following ... case of the longitudinal mode of deformation to the shear case. 2. 2 .2. 2 General solution without external loading Z(x) = a  x 0 dx GS + b [2. 34] a and b are two arbi...

Ngày tải lên: 13/08/2014, 05:22

... GS   ∂Y s ∂x  2 +  ∂Z s ∂x  2  dx + 1 2  L 0  GJ T  ∂ψ x ∂x  2 +  EI z  ∂ 2 Y b ∂x 2  2  +  EI y  ∂ 2 Z b ∂x 2  2  dx [3 .2] The kinetic energy E κ is: E κ = 1 2  L 0 ρS{ ˙ X 2 + ( ˙ Y s + ˙ Y b ) 2 + ( ˙ Z s + ˙ Z b ) 2 }dx + 1 2  L 0 ρ(J ˙ ψ 2 x + I y ˙ ψ 2 y + ... αEa  a /2 −a /2 θ(z)dz [2. 86] ˜ N (x) = ES ∂X ∂x − αEa  a /2 −...

Ngày tải lên: 13/08/2014, 05:22

... immediately that [3 .71 ] is equivalent to the system: ∂ 2 ∂x 2  EI y ∂ 2 Z ∂x 2  + ρS ∂ 2 Z ∂t 2 = 0; ∀x ∈[0, L] ∂ ∂x  EI y ∂ 2 Z ∂x 2      x 0 + − ∂ ∂x  EI y ∂ 2 Z ∂x 2      x 0 − = ... +E n I n  ℓ n 0  ∂ 2 Z n ∂x 2  2 dx =[X I Z I Z ′ I X J Z k Z ′ J ]        K 11 K 12 K 13 K 14 K 15 K 16 K 21 K 22 K 23 K 24 K 25 K 26 K 31 K...

Ngày tải lên: 13/08/2014, 05:22

... −6γℓ 0 − 12 −6γℓ 0 −6γℓ 4γℓ 2 06γℓ 2 ℓ 2 −10 0 1 0 0 0 − 12 6γℓ 0 12 6γℓ 0 −6γℓ 2 ℓ 2 06γℓ 4γℓ 2        ⇒ K (2) = K ℓ     1000 0 12 −6γℓ −6γℓ 0 −6γℓ 12 2 ℓ 2 0 −6γℓ 2 ℓ 2 4γℓ 2     Then, ... 4γℓ 2        ⇒ K (1) = K ℓ     4γℓ 2 06γℓ 2 ℓ 2 010 0 6γℓ 0 12 6γℓ 2 ℓ 2 06γℓ 4γℓ 2      U 2 W 2 ϕ 2 U 3 W 3 ϕ 3  T ⇒  U 2 W 2 ϕ...

Ngày tải lên: 13/08/2014, 05:22

... or odd Ehπ 2 4  n 2 η + m 2 η 2  α 2 n,m +  n 2 η + m 2 2η  β 2 n,m  otherwise Ehπ 2 4  n 2 η + m 2 η 2  α 2 n,m +  n 2 η + m 2 2η  β 2 n,m + 8n 2 m 2 π 2 (n 2 − m 2 ) 2 α n,m β n,m  where ... system: 4Eh M      π 2 4  n 2 + m 2 η 2 2  −  ω ′ n,m  2 2n 2 m 2 η 2 (n 2 − m 2 ) 2 2n 2 m 2 η 2 (n 2 −...

Ngày tải lên: 13/08/2014, 05:22

... Z ε xx =−z ∂ 2 Z ∂x 2 + 1 2  ∂Z ∂x  2 + z 2 2   ∂ 2 Z ∂x 2  2 +  ∂ 2 Z ∂x∂y  2  ε xy =−z ∂ 2 Z ∂x∂y + z 2 2 ∂ 2 Z ∂x∂y  ∂ 2 Z ∂x 2 + ∂ 2 Z ∂y 2  + 1 2  ∂Z ∂x ∂Z ∂y  ε yy =−z ∂ 2 Z ∂y 2 + 1 2  ∂Z ∂y  2 + z 2 2   ∂ 2 Z ∂y 2  2 +  ∂ 2 Z ∂x∂y  2  [6.89] In ... m 2 ) 2 ; α n,m = β n,m out -of- phase modes: f n,m = c 2L...

Ngày tải lên: 13/08/2014, 05:22