Mechanical Sytem Dynamics - Appunti del corso

MECHANICAL SYSTEM DYNAMICS

F = kΔl

F = c dΔl/dt

mẍ + cẋ + kx = F(t)

mẍ + cẋ + kx = 0

x(t) = X₀ e^λt

ẋ(t) = λ X₀ e^λt

ẍ(t) = λ² X₀ e^λt

mλ² X₀ e^λt + c λ X₀ e^λt + k X₀ e^λt = 0

(mλ² + cλ + k) X₀ e^λt = 0

λ² m + λc + k = 0

λ_1,2 = ^c/_2m ± √(^c/_2m)² - ^k/_m

ASSUME c = 0 → NO DAMPING

λ_1,2 = ± √^-k/_m = ± i √^k/_m = ± i ω₀

mẍ + kx = 0

X(t) → X_ω₀ e^iω₀t + X_ω₀ e^-iω₀t

ω₀ = √^k/_m [rad/s]

F₀ = ^ω₀/_2π [Hz]

X_(t) → X_ω₁ e^-iω_ot + X_ω₂ e^-iω_ot

x(t) = (a + ib) e^iω_t + (a - ib) e^-iω_t

= (a + ib) (cos ω_t + i sin ω_t) + (a - ib) (cos ω_t - i sin ω_t) = 2a cos ω_t - 2b sin ω_t

A cos ω_t + B sin ω_t

apply initial/border conditions

x(0) = X₀ ⇒ A cos ω_t + B sin ω_t → x(0) = A

ẋ(0) = V₀ ⇒ Aω₀ sin ω_t + Bω₀ cos ω_t → ẋ(0) = Bω₀

A = X₀

B = V₀ / ω₀

x(t) = X₀ cos ω_t + V₀ / ω₀ sin ω_t

if x(0) = X₀, ẋ(0) = 0 → x(t) = X₀ cos ω_t

continuous transfer of energy between kinetic and potential

x₀

t

T₀ = 2π / ω₀ OSCILLATION PERIOD

λ_1,2 = C / 2m ± √(C / 2m)² − k / m

assume C > 0, - (k / m) < 0

[C / 2m] [± ω₀] [C / 2m] [− ω₀²] < 0

[C / 2m] ω₀ > 0

C / 2m < ω₀

C / 2mω₀ < 1

2mω₀ CRITICAL DAMPING [N S / m]

C / C_crit DAMPING RATIO

if damping ratio is lower than 1

λ_1,2 = -C / 2m ± i √ (ω₀² − (C / 2m)²)

Eigenvector describes how the system oscillates at a given frequency (as a ratio between coordinates values)

δ(t) = C₁₁ X₀₁ e^iω₁t + C₂₁ X₀₂ e^iω₁t + C₁₂ X₀₁ e^iω₂t + C₂₂ X₀₂ e^iω₂t + ...

X(t) = | X₀₁ C₁₁ C₁₂ e^iω₁t + C₂₁ C₂₂ e^iω₁t + ...

X(t) = X₀₁ (A₁ cos ω₁t + B₁ sin ω₁t) + X₀₂ (A₂ cos ω₂t + B₂ sin ω₂t) + ...

X(t) = ∑_j=1^s X_0j (A_j cos ω_jt + B_j sin ω_jt)

Free motion = combination of eigenmodes - constants can be obtained by applying initial conditions

String element

∑F⃗ = m a⃗

-T sin α + T sin (α + dα) = m dx ¨W (vertical acceleration)

-Tα + T (α + dα) = m dx ¨W

T dα = m dx ¨W

α ≈ tan α ≈ δW/δx → δα ≈ δ²W/δx² dx

T δ²W/δx² = m δ²W/δt²

δ²W/δt² = (T/m) δ²W/δx² (valid only with no external forces)

kg/m = mass per unit length

mass = m . dx [kg/m]

(m/s²) / (kg/m) = speed²

δ²W/δt² = c² δ²W/δx² (wave speed propagation squared)

δ²W

δt² = α(x) β(t)

∫₀^W

0^W x⁴ = αʺ(x) β(t)

αʺ(x) β(t) EJ = α(x) β(t)

αʺ(x)

P(x) δ(x) m

β(t) P(t) = const. = -ω²

β̋(t) + ω² β(t) = 0

P(t) + ω² β(t) = 0

β(t): A cos t + B sin t

= C cos (ωt + ϕ)

αʺ(x) EJ m = -ω² α(x) = 0

αʺ(x) = m ω² α(x) = 0

αʺ(x) - γ² α(x) = 0

with γ² = mω² EJ

α(x) = c₁ e^γx

c₂ e^-γx

α(x) = α r -

αʺ_e - γ²

αʺ e^-γ - αe γ²

- e^-γ α

γ² = -λ²

λ² = γ² → λ_re = ±γ

α(x) = α_i3 e^iγx

+ α_l3 e^iγx

+ α_l4 e^-iγx

A cos γx + B sin γx

ch (γx) = e^γx e - e

2 hyperbolic cosine

sh (γx) = e^γx e - e

2 hyperbolic sine

sh (x) = e^γx

α(x) = A cos γx + B sin γ x + C

2 + D sh γx

- D 2 = -βαx

most general solution possible

α''(x) + ω_L² α(x) = 0

U(x,t) = A cos (ω_L/c x) + B sin (ω_L/c x)

α(x) = A cos (ω_L/c x) + B sin (ω_L/c x)

C is multiplied into the previous A and B

u(0,t) = 0

A = 0

u(L,t) = 0

B sin (ω_L/c) = 0

→ Trivial solution

ω_L/c = kπ k = 1,2,...

ω_k = kπ/L c = kπ/L √(E/ρ)

U_k(x,t) = (B_k sin (kπ/L x)) cos (ω_k t + ϕ_k)

k = 1 ω₁ = π/L √(E/ρ)

k = 2 ω₂ = 2π/L √(E/ρ)

When u_x(x,t) rises ↗ have TRACTION forces

↓ u_x(x,t) decreases ↘ have COMPRESSION forces

u(x,t) = (A cos (ω_L/c) + B sin (ω_L/c)) cos (ωt + ϕ)

No constraints

N(0) = 0

N(L) = 0

N = EA ∂u/∂x

∂u/∂x = (-ω_L/c A sin (ω_L/c) + ω_L/c B cos (ω_L/c x)) cos (ωt + ϕ)

N(0) = 0 -> B = 0

N(L) = 0

A =0 trivial solution

ω_L/c = kπ k = 1, 2, ...

→ ω_k = kπ/L c

U(x,t) = (A_k cos (ω_k/c x)) cos (ω_k t + ϕ_k )

k = 1

ω_k = π/L c

GENERAL SOLUTION

V. = 1/2 ∫0^L qⱼ_,x x qⱼ_,x^T EJ I dx

V_i = 1/2 ∫0^L q_,x^T K q_,x dx, [K] stiffness matrix

k_ij = ∫0^L sin iπx/L x sin jπx/L x EJ x (iπ/L)² dx

k₁ⱼ = 0, j ≠ 1

qⱼ, m_i

⋮

ϕ_k = sin kπx/L

Φ_k = cos kπ/L

Φ_k = -sin kπ/L

[A] ∫c^b ϕ_i(x)m(x)ϕ_j(x)dx

∫_c^b ϕ_i(x)m(x)ϕ_k(x)dx

m_ij = ∫₀^L Φ_i (m) m_i Φ_j(x) dx ⋯ condition

m_ij = 0 i ≠ j

m_ij ≠ 0 i = j pinned-pinned beam ∫₀^L sin iπ/L x m sin jπ/L

k_ij = ∫₀^L Φ_i(x) EJ Φ_j(x) dx

k_ij = 0 i ≠ j

k_ij ≠ 0 i = 0 pinned-pinned beam ∫_(iπ/L) ( sin iπ/L x ) x EJ dx x (iπ/L) EJ L/2

k₁ = (π/L)⁴ EJ L/2

k₂₂ = 16(π/2)⁴ EJ L/2

k₃ = (1/L)⁴ EJ L/2

h_i = 1/2 m L ω_i² x 1%

r. 2 m L ω_i h_i ⋯

(ω_i) q(→){[x]}[r_x]

(ω_i) q(→{[x]}[r_x]) q = 0

(ω_i)/m_i = √(in iπ/L)

EJ / L/2 m L = (2/L)⁴[EJ/m]