MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 14 potx

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

... branches, such that: ̟ 2 (1)n =  2 1 +  2 2 + √  2 ; ̟ 2 (2) n =  2 1 +  2 2 − √  2 ;  =   2 1 −  2 2  2 + 2 4 c [8. 72] In Figure 8.11, the natural frequencies of the out -of- plane modes are ... η 2 n 4 + n 2 2 ( 1 + ν ) ;  2 2 =  1 2 + n 2 2 ( 1 + ν )  ;  2 c = n 2  1 + η 2  2 [8.70] which leads to the following modal system:...

Ngày tải lên: 13/08/2014, 05:22

... 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. Rayleigh’s quotient 20 7 4 .2. 3.1. Bending of ... external loads 318 6 .2. 2. Boundary conditions 319 6 .2. 2.1. Kirchhoff effective shear forces and corner forces 319 6 .2. 2 .2. Elastic boundary conditions 322 6 .2....

Ngày tải lên: 13/08/2014, 05:22

... 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 2 sin 2 sin 2 +(γ cos 2 ) 2 [1. 114] These ... 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:...

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 + ... expression of the bending moment, we get: M y =−  2Eaz 3 0 3 χ + σ...

Ngày tải lên: 13/08/2014, 05:22

... 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 − = F (e) z (t) [3. 72] example 3.–Beam ... +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...

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

... 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 2 L + 4π 2 − ̟ 2 2γ L...

Ngày tải lên: 13/08/2014, 05:22