MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 1 ppt

... 307 Chapter 6. Plates: out -of- plane motion 311 6 .1. Kirchhoff–Love hypotheses 3 12 6 .1. 1. Local displacements 3 12 6 .1 .2. Local and global strains 313 6 .1 .2. 1. Local strains 313 6 .1 .2. 2. Global flexure ... vibration of straight beams 19 0 4 .2. 1. Travelling waves of simplified models 19 0 4 .2. 1. 1. Longitudinal waves 19 0 4 .2. 1 .2. Flexure waves 19 3 4...

Ngày tải lên: 13/08/2014, 05:22

... is: X − X 0 =  1 + κξ 0  1 − ξ 2 0  ξ 2( 1 + κξ 0 (1 − ξ 0 )) − ξ 2 2 ;0≤ ξ ≤ ξ 0 X + X 0 =  1 + κξ 0  1 − ξ 2 0  (ξ − 1) 2( 1 + κξ 0 (1 − ξ 0 )) + 1 − ξ 2 2 ; ξ 0 ≤ ξ ≤ 1 Limit cases: κ ... In terms of local quantities it reduces to: ρ ∂ 2  ξ ∂t 2 − ∂  t 1 ∂x =  f (e) (x, y, z; t) ρ ∂ 2 ξ j ∂t 2 − ∂σ xj ∂x = f (e) j (x, y, z; t); (j = 1,...

Ngày tải lên: 13/08/2014, 05:22

... B 1 =− 3A 1 4 EIZ ′′′ 1 (0.5) =−F 0 L 3 /2 ⇒ 6EIA 1 =−F 0 /2 ⇒ A 1 =− F 0 L 3 12 EI 11 6 Structural elements 2. 2.5 .2 Generalized mechanical impedances Besides the elastic supports, other kinds of ... =−Z ′ 1 (0.5) ⇒ Z ′ 2 (0.5) = Z ′ 1 (0.5) = 0 Calculation of the four constants is quite straightforward: Z 1 (0) = Z ′ 1 (0) = 0 ⇒ Z 1 (ξ) = A 1 ξ 3 + B...

Ngày tải lên: 13/08/2014, 05:22

... form as:   k 1 −k 1 1 −k 1 k 1 + k 2 0 10 0     X 1 X 2 λ 1   +   m 1 m 12 0 m 12 m 1 + m 2 0 000     ¨ X 1 ¨ X 2 ¨ λ 1   =   F 1 F 2 D 1   [3 .10 1] where k 2 and m 2 are the diagonal ... principle 17 3 first finite element, which is constrained by the condition X 1 = D(t) is written as: k 1 [X 1 , X 2 ]  1 1 11 ...

Ngày tải lên: 13/08/2014, 05:22

... Figure 1. 17. Substituting [1. 111 ] into [1. 109] and [1. 110 ] we arrive at the linear system: − (U 2 + U 1 ) ( λ +2G(cos θ ) 2 ) +γGV sin 2 = 0 (U 2 − U 1 ) sin 2 − γV cos 2 = 0 [1. 1 12 ] which ... 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...

Ngày tải lên: 13/08/2014, 05:22

... 0 1 0 0 0 − 12 6γℓ 0 12 6γℓ 0 −6γℓ 2 ℓ 2 06γℓ 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 2 ϕ 3  T K ℓ        10 ... 4γℓ 2      U 2 W 2 ϕ 2 U 3 W 3 ϕ 3  T ⇒  U 2 W 2 ϕ 2 ϕ 3  T K ℓ        10 0 10 0 0 12 −6γℓ 0 − 12 −6...

Ngày tải lên: 13/08/2014, 05:22

... energy: M 1 ˙ X 1 + M 2 ˙ X 2 = M 1 ˙ X ′ 1 + M 2 ˙ X ′ 2 1 2 M 1 ˙ X 2 1 + 1 2 M 2 ˙ X 2 2 = 1 2 M 1 ˙ X ′ 2 1 + 1 2 M 2 ˙ X ′ 2 2 [4 .10 0] where ˙ X 1 , ˙ X 2 and ˙ X ′ 1 , ˙ X ′ 2 denote the ... 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...

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 ... finite f (1, 1) = 783 Hz f (1 ,2) out -of- phase = 10...

Ngày tải lên: 13/08/2014, 05:22

... [6.68], D[[Z]]=p 0 ⇒  ∂ 2 ∂r 2 + 1 r ∂ ∂r + 1 r 2 ∂ 2 ∂θ 2  ∂ 2 Z ∂r 2 + 1 r ∂Z ∂r + 1 r 2 ∂ 2 Z ∂θ 2  = p 0 D [6. 12 4 ] The problem is independent of θ, so the corresponding derivatives are null and [6. 12 4 ] is ... out -of- plane motion 3 51 Then [6 .1 12 ] takes the form: χ rr =− ∂ 2 Z ∂r 2 ; χ θθ =−  1 r ∂ ∂θ  1 r ∂Z ∂θ  + 1 r ∂Z ∂r  ;...

Ngày tải lên: 13/08/2014, 05:22