MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 7 pdf

... 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 RC=−γ ... 2 RC=−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 ... Z 2 = U 2 cos θ 2 eiψ 2 ;ψ 2 = ωt −x sin θ 2 − z cos θ 2 cL(σzz) 2 =−iωU 2 cLλ +2G(cos θ 2 ) 2 eiψ 2 ;(σzx) 2 =+iωU 2 cLG sin 2 2 eiψ 2 ; (σzy) 2 ≡ 0[1.104]Now, the...

... as:x+dL /2 x−dL /2 (S)r ×ρ∂ 2 ξ∂t 2 −∂t1∂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 loadingZ(x) = ax0dxGS+ b [2. 34]a and b are two arbitrary real constants. 2. 2 .2. 3 ... [2. 23]The solution is:X(x) = adxES+ 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) =axES+ b [2. 25] 2. 2.1.3...

... GS∂Ys∂x 2 +∂Zs∂x 2 dx+1 2 L0GJT∂ψx∂x 2 +EIz∂ 2 Yb∂x 2  2 +EIy∂ 2 Zb∂x 2  2 dx[3 .2] The kinetic energy Eκis:Eκ=1 2 L0ρS{˙X 2 + (˙Ys+˙Yb) 2 + (˙Zs+˙Zb) 2 }dx+1 2 L0ρ(J˙ψ 2 x+ Iy˙ψ 2 y+ ... αEaa /2 −a /2 θ(z)dz [2. 86]˜N (x) = ES∂X∂x− αEaa /2 −a /2 θ(z) dz˜My(x) =(S)σxx(z)z dS =−EI∂ 2 Z∂x 2 − αEaa /2 −a /2 θ(z)z dz where I =a4 12 [2. 87] Figure 2. 37. Beam subjected to a transverse ... expression of the bending moment, we get:My=−2Eaz303χ + σ0ah 2  2 − z 2 0=−2Eaz303χ + Eaz0χh 2  2 − z 2 0=Eah 2 4z0−Eaz303χ [2. 101]It is...

... immediately that [3 .71 ] isequivalent to the system: ∂ 2 ∂x 2 EIy∂ 2 Z∂x 2 + ρS∂ 2 Z∂t 2 = 0; ∀x ∈[0, L]∂∂xEIy∂ 2 Z∂x 2 x0+−∂∂xEIy∂ 2 Z∂x 2 x0−= ... +EnInℓn0∂ 2 Zn∂x 2  2 dx=[XIZIZ′IXJZkZ′J]K11K 12 K13K14K15K16K 21 K 22 K 23 K 24 K 25 K 26 K31K 32 K33K34K35K36K41K 42 K43K44K45K46K51K 52 K53K54K55K56K61K 62 K63K64K65K66XIZIZ′IXJZJZ′Jwhere ... +ddxEId 2 dx 2 Zx=0= 0L 2 [Z(L)]=K 2 Z −ddxEId 2 dx 2 Zx=L= 0L3[Z(0)]=K3dZdx+ EId 2 Zdx 2 x=0= 0L4[Z(L)]=K4dZdx− EId 2 Zdx 2 x=L= 0where...

... −6γℓ 0 − 12 −6γℓ0 −6γℓ 4γℓ 2 06γℓ 2 ℓ 2 −10 0 1 0 00 − 12 6γℓ 0 12 6γℓ0 −6γℓ 2 ℓ 2 06γℓ 4γℓ 2 ⇒ K (2) = Kℓ10000 12 −6γℓ −6γℓ0 −6γℓ 12 2 ℓ 2 0 −6γℓ 2 ℓ 2 4γℓ 2 Then, ... 4γℓ 2 ⇒ K(1)= Kℓ4γℓ 2 06γℓ 2 ℓ 2 010 06γℓ 0 12 6γℓ 2 ℓ 2 06γℓ 4γℓ 2 U 2 W 2 ϕ 2 U3W3ϕ3T⇒U 2 W 2 ϕ 2 ϕ3TKℓ10 0−10 00 12 −6γℓ ... to be:K(1)α= Kℓ4γℓ 2 6γℓS 6γℓC 2 ℓ 2 6γℓS C 2 + 12 S 2 CS( 12 −1) 6γℓS6γℓC CS( 12 −1)S 2 + 12 C 2 6γℓC 2 ℓ 2 6γℓS 6γℓC 4γℓ 2  21 4 Structural elementsϕn(x) = sin(nπx/L)...

... or oddEhπ 2 4n 2 η+m 2 η 2 α 2 n,m+n 2 η +m 2 2ηβ 2 n,motherwiseEhπ 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 αn,mβn,mwhere ... system: 4EhM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 − m 2 ) 2 π 2 4n 2 η 2 +m 2 η 2 2−ω′n,m 2 ×αn,mβn,m=00 28 6 Structural elementsThe ... equations are:Eh1 − ν 2 ∂ 2 X∂x 2 + ν∂ 2 Y∂x∂y+ Gh∂ 2 X∂y 2 +∂ 2 Y∂x∂y=−αEh1 − ν∂θ∂xEh1 − ν 2 ∂ 2 Y∂y 2 + ν∂ 2 X∂x∂y+ Gh∂ 2 Y∂x 2 +∂ 2 X∂x∂y=−αEh1 −...

... Zεxx=−z∂ 2 Z∂x 2 +1 2 ∂Z∂x 2 +z 2 2∂ 2 Z∂x 2  2 +∂ 2 Z∂x∂y 2 εxy=−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εyy=−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 ; αn,m= βn,mout -of- phase modes:fn,m=c2Ln 2 +m 2 2−8n 2 m 2 π 2 (n 2 − m 2 ) 2 ; αn,m=−βn,mFigure 5. 17 shows the mode shapes (1, 1), (1, 2) and (2, 2) of a steel plate(though ... ax 2 + by 2 [6.81] 29 8 Structural elementsIn the particular case of a square plate, they are immediately obtained as:in-phase modes:fn,m=c2Ln 2 +m 2 2+8n 2 m 2 π 2 (n 2 − m 2 ) 2 ;...