Mechanical System Dynamics - Esercizi e appunti

1. equation of motion

2. eigenfrequency, damping ratio

3. regime solution

4. F_P F_T (transmitted force to the ground)

-kx + cẋ - Mẍ + mẍ + mω²y₀cosωt = 0

(M + m)ẍ + cẋ + kx = mω²y₀cosωt

2)

(M + m)ẍ + cẋ + kx = 0

x = X₀e^λt

[(M + m)λ² + cλ + k] X₀e^λt = 0

λ_1,2 = —_c/^{2(M + m)} ± √(—_c/^{2(M + m)})² — k/^{M + m}

undamped λ_1,2 = ± i √(k/^{M + m}) = ± iω₀

h = c/^{2(M + m)ω₀}

3)

h ≥ e^λt

• x(t) = x_s(t) + x_p(t)

• b = 0

• c = 0

x₀(t) = X₀e^λt

λ_1,2 = ± iω_n

• x_s(t) = X₀e^iω_nt + X₀e^—iω_nt
• = Acosω_nt + Bsinω_nt

x_s(t) = —A₀sinωt + Bcosωt

• 0 < h < 1

λ_1,2 = —c/^{2(M + m)} ± √(—c/^{2(M + m)})² — k/^{M + m}

= —hω₀ ± i√(ω₀² — (hω₀)²)

X_g(t) = e^-h₁ω₀t(X₀1 + X_ce^iω₁t)

+ e^-h₂ω₀t(A cos ω₁t + B sin ω₁t)

decrease over time damping effect

i h = i

λ_1,2 = -ω₀ ± i ω

X_g(t) = A e^-hω₀t + B e^ω₀1t

(A + B) e^-iω₁t

h > i :

λ_1,2 = -1/2Mrm ± √(1/2Mrm) k/Mrm

X₀(t) = X₀1 e^ω₁t + X₀₂ e^λ₂t

X_p(t) = Xt e^iω_tt

[-(M_rm) Ω_e + i c Ω + k] e^iω_tt = mΩ₂X_e e^iω_tt

= F₀ e^iω_tt

1

X_I = -Ω²(M_rm) + i Ω c + k = ^F₀

_G(Ω₀) F₀ _

4)

R(Ω) = kx + Cuż reaction forces

j = G(Ω) F₀ e^-iω_tt

(k + i C, Ω)

[m₁ 0] [x₁] = [3k - 2k] [x₁] F_e e^iωt

[0 m] [x₂] -2k 3k] [x₁] F_e e^iωt

Free vibration

[M] x + [R] x = 0

(-ω² [M] + [R] x = 0

det [ ω₂[M] + [R] x ] = Ø

(mω₂-K mω₂-sk) = Ø

ω₁² k

m

ω₂² sk

m

m₁x₁ + k₁x_i + 2x(k(x₁-x₂) = F₁ e^iωt

m₂x₂ + k₂x₂ = 2x(x₁-x₂) = F₂ e^iωt

section B

1. M_B = 0
2. We know M = EJ ^∂2w / _∂x²
3. W_z to 5 sin(θ) = 0 small angle sinθ' ≈ θ' ≈ tan θ = ∂w/∂x
4. W_z(x₂, t) = W_z(x₂, t)_{A = 0} same displacement at B

section C

string → still no constraint on rotation (can rotate even if the constraint is a clamp)

no displacement

6. A₂ cos(βL) + B₂ sin(βL) = 0

collect all BCs in matrix form

X = 0 TRIVIAL SOLUTION

det [H(ω_i)] = 0 we obtain ω_i (i = 1, 2, 3, … ∞ ) NATURAL FREQUENCIES

substitute ω_i back to the matrix [H(ω_i)] X' = 0

arbitrary choose A_i = 1 , use ω_i and rescale [H(ω_i)] X = 0

N(ω_i) = [H(ω_i)(2:6, 1)]

both expressions are ok

1. boundary conditions on section A
2. BC in section B

section B

T_xL + F₀ e^iωt - k h W₃|_x=L - [dW₃/dt]_{x=L, 0} - T_x|_x=L - m_e[dW₁/dt]_x=u = 0

Where T_xL = ES [d³W₁/dx³]_x=L e^iωt

3rd eq = iω[ A₂ cos βx + ... ] + D₂ sh βy_x e^iωt

[d²W₂/dt²] = -ω² [ ... ] e^iωt

F

• only appears in the sth BC (it is subjective, depends on the order of BCs)

X = [H(α)]⁻¹ F

H(α) X = F

2 possible conditions:

• b negligible (as previous cases)
• b not negligible (not ø)

w₁(x, t) = A_n cos βx + B_n sin (βx) + C_n ch βx + D_n sh (βx) cos (ωt + φ)

w₂(x₂, t) = (A_n cos βx₂ + B_n sin βx₂ C_n ch βx₂ + D₂ sh β₂) cos(ωt + φ)

Z(t) = Z_c cos(ωt + φ) only in function of time single dof

4 unknowns ⟶ 9 boundary conditions

D = ₁/₂ q̇^T [C]₀ q̇ - ₁/₂ q̇^T [C]_S q

= ₁/₂ q̇^T [C]_I q̇

[C]_S is diagonal ⟹ linear combination of [M] and [K]

[C]_I not diagonal ⟹ concentrated damping introduces extradiagonal terms

Virtual work

δL = F₀ e^int δW_e e^ext

Q = (δL/δq̇)^T

W_e = _δWe/^δq̇ - Φ₂^T δq

Q = (F₀ e^ext, Φ_n (0) )^T

= δ Φ_n (0) F₀ e^ext

Φ_n = Φ_n (0) F₀ e^ext

Math: a scalar transposed is the scalar itself

Lagrange Equation

d/dt (δT/δq̇)^T - (δT/δq)^T + (δU/δq)^T = Q

Math: if [A] symmetric [A]^T = [A]

d²x^T[A]x/dt² = ẋ^T[A]ṫ + ẋ^T[A]ẋ

= 2ẋ^T[A]

δT/δq̇ = [M] q̇

d/dt δT/δq̇ = [M] q̈

δT/δq = 0 ⟹ linear system

δU/δq = [C] q̇ ⟹ assuming [C] diagonal (neglecting extradiagonal terms)

δU/δq = [K] q

eom: [M] q̈ + [C] q̇ + [K] q = Q

mii q̈_i + Ci q̇_i + kii q = Qi ⟹ single DoF system

q_i(t) = W(x,t) = Φ(x) Q(t)

known vector corresponding to the nth mode

convert it back to physical coordinates

different frequencies between the distributed load and the columns of winds

W₁(x₁, t) = [A₁cos j₁x + B₁sin j₁x + C₁ch j₁x + D₁sh j₁x] cos (wt+ϕ)

W₂(x₂, t) = [A₂cos j₂x + B₂sin j₂x + C₂ch j₂x + D₂sh j₂x] cos (wt+ϕ)

γ'' = ^m/_EJ ω² same for both beams

section A

1. W₁|_x1=0 = 0
2. M_Av = 0
3. EJ ^∂W₁/_∂x²|_x=x1=0 = 0

section B

3. W₁|_x=L = W₂|_x=x2=0
4. ^∂W₁/_∂x|_x=L = ^∂W₂/_∂x|_x=x2=0
5. T_Bar T_Br - m_B^.(w²) (L)^∂W₂/_∂x|_x=x2=0 - kW₂|_x=x2=0 = 0
6. M_Br - M_Be - J_C ^∂²W₂/_∂t∂x|_x=x2=0 = 0

section C

7. W₂|_x2=L = 0
8. -M_Ce|_x2=L = 0
9. -EJ ^∂W₂/_∂x²|_x2=L = 0

[H(w)] ξ = 0

det [H(w)] = 0 → ω_n n = 1, 2, ... ∞ → ϕ_n

discretization of the system W_nm → Ω_nm Ω_max = 2 Ω

W₁(x₁, t) = [Φ_x(1) Q_i(t)]

W₂(x₂, t) = [Φ_x(2) Q_i(t)]

Φ =^Φ₁(1)/_Φ₁(2)^Φ₂(1)/_Φ₂(2)^Φ₃(1)/_Φ₃(2)... N_Mx1^Q₁/_Q₂/_QMx1