MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 3 ppt

... 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 real constants. 2. 2 .2. 3 ... 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....

Ngày tải lên: 13/08/2014, 05:22

... analysis 3 32 6 .3. 1. Natural modes of vibration 3 32 6 .3. 1.1. Flexure equation of a plate prestressed in its own plane 3 32 6 .3. 1 .2. Natural modes of vibration and buckling load 33 5 6 .3. 1 .3. Modal ... Deterministic excitations 22 0 4 .3 .2. 1. Separable space and time excitation 22 0 4 .3 .2. 2. Non-separable space and time excitation 22 1 4 .3. 3. Truncatio...

Ngày tải lên: 13/08/2014, 05:22

... expression of the bending moment, we get: M y =−  2Eaz 3 0 3 χ + σ 0 a   h 2  2 − z 2 0  =−  2Eaz 3 0 3 χ + Eaz 0 χ   h 2  2 − z 2 0  =  Eah 2 4 z 0 − Eaz 3 0 3  χ [2. 101] It is ... results are quoted below: Ellipse: J T = πa 3 b 3 a 2 + b 2 Square: J T = 0,1406 a 4 Rectangle: J T = γa 3 b 3 a 2 + b 2 with a/b 24 8∞ γ 0 .28 6 0 .29...

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 ... −ρ...

Ngày tải lên: 13/08/2014, 05:22

... 2  R C  =  −γ cos 2 sin 2  [1.1 13] 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

... 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 ... 0 0 12 −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    ...

Ngày tải lên: 13/08/2014, 05:22

... = 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 ... conditions of linear momentum and kinetic energy: M 1 ˙ X 1 + M 2 ˙ X 2 = M 1 ˙ X ′ 1 + M 2 ˙ X ′ 2 1 2 M 1 ˙ X 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 ... ν ∂ 2 Z ∂x 2      y=±ℓ /2 = 0; ∂ 2 Z ∂x∂y    ...

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 ... differentiation of n. d n = ∂ n ∂α dα + ∂ n ∂β dβ ⇒ ( d n ) 2 =  ∂ n ∂α  2 (dα) 2 +  ∂ n ∂β  2 (dβ) 2 [7 .30 ] or d n ·d n = g 2 α R 2 α (dα)...

Ngày tải lên: 13/08/2014, 05:22