MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 4 pptx

... Z(x) = ax 4 + bx 3 + cx 2 Loading: 24 a = f 0 EI → a = f 0 24 EI Free end at x = L Z ′′′ (L) = 24 ax +6b = 0 ⇒ b =− f 0 L 6EI Z ′′ (L) = f 0 L 2 2EI − f 0 L 2 EI + 2c = 0 → c = f 0 L 2 4EI then: Z(ξ) ... 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...

Ngày tải lên: 13/08/2014, 05:22

... +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 32 K 33 K 34 K 35 K 36 K 41 K 42 K 43 K 44 K 45 K 46 K 51 K 52 K 53 K 54 K 55 K 56 K 61 K 62 K 63 K 64 K 65 K 66               X I Z I Z ′ I X J Z J Z ′ J        where ... by...

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 ... attached concentrated mass 20 7 4 .2. 3 .2. Beam on elastic foundation 20 9 4 .2. 4. Finite element approximation 21 0 4 .2. 4. 1. Longitudinal...

Ngày tải lên: 13/08/2014, 05:22

... get: X 2 = U 2 sin θ 2 e iψ 2 ; Z 2 = U 2 cos θ 2 e iψ 2 ; ψ 2 = ω  t − x sin θ 2 − z cos θ 2 c L  (σ zz ) 2 = −iωU 2 c L  λ +2G(cos θ 2 ) 2  e iψ 2 ; (σ zx ) 2 = +iωU 2 c L G sin 2 2 e iψ 2 ; ... 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 =...

Ngày tải lên: 13/08/2014, 05:22

... [2. 23] The solution is: X(x) = a  dx ES + b [2. 24] where a and b are two arbitrary constants. If the beam is homogeneous and of constant cross-section, [2. 24] gives: X(x) = ax ES + b [2. 25] 2. 2.1.3 ... 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 arbitrary r...

Ngày tải lên: 13/08/2014, 05:22

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

Ngày tải lên: 13/08/2014, 05:22

... instance N = 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...

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...

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 ... symmetries of the problem. The equilibrium equations are: D...

Ngày tải lên: 13/08/2014, 05:22

... ςdn ⇒ (d  P) 2 = (d r) 2 + (dς ) 2 + ς 2 (d n) 2 + 2 dr ·d n [7 .28 ] dr is in the tangent plane so (dr) 2 = g 2 α (dα) 2 + g 2 β (dβ) 2 [7 .29 ] The determination of (d n) 2 is obtained ... [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. 1 24 ] The problem is indepen...

Ngày tải lên: 13/08/2014, 05:22