TimeIntegration
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Setting the stage— By Ng Tin Yau (PhD) — 5/38
Pendulum Absorber
A pendulum absorber model is shown as follows:
In this model, the connecting rod is assumed to be rigid and all joints and contact points are smooth.
Setting the stage— By Ng Tin Yivalent First-Order System
Rewrite the second-order system as ¨ = M−1 (f − Cq ˙ − Kq) q and then put it in a matrix form ˙ q ¨ q Let A= 0 I − 1 −M K −M−1 C and b = 0 M−1 f (8) = M−1 (f ˙ q ˙ − Kq) − Cq = 0 I − 1 −M K −M−1 C q 0 + ˙ q M−1 f
Setting the stage— By Ng Tin Yau (PhD) — 6/38
Pendulum Absorber: FBD
The free-body diagram of the model is shown as follows:
The equation of motion in the x-direction for the block is given by m1 x ¨ = −kx + P sin θ + f (♣)
In the case where θ is small, then sin θ ≈ θ and cos θ ≈ 1. In addition if ˙2 sin θ ≈ 0, then we obtain a linearize system of the we also assume θ original model. m1 + m2 m2 L m2 m2 L x ¨ k 0 + ¨ 0 m2 g θ x θ = f (t) 0 (7)
k1 + k2 −k2 −k2 k2 + k3
In vector notation we have q= and M= m1 0 0 m2 C= c1 + c2 −c2 −c2 c2 + c3 K= k1 + k2 −k2 −k2 k2 + k3 x1 x2 and f = f1 f2
Thus, we arrive the standard form ¨ + Cq ˙ + Kq = f Mq
On the other hand, the force vector takes the form: ˙) = F (t, q, q ˙) F1 (t, x, x, ˙ θ, θ ˙) F2 (t, x, x, ˙ θ, θ = ˙2 sin θ f (t) − kx + m2 Lθ −m2 g sin θ
Table of Contents
1
Setting the stage Standard Form of Second-Order Linear Systems Classical Models Equivalent First-Order System & RK4 Numerical Time Integration The Central Difference Scheme Newmark Family Schemes
2
3
— By Ng Tin Yau (PhD) — 2/38
Equations of Motion
In many circumstances, engineers have to solve the following system of second-order differential equations: ¨ (t) + Cq ˙ (t) + Kq(t) = f (t) Mq (1)
Setting the stage— By Ng Tin Yau (PhD) — 8/38
(♥) (♦)
Governing Differential Equations
Now the two governing differential equations becomes ¨ cos θ = f − kx + m2 Lθ ˙2 sin θ (m1 + m2 )¨ x + m2 Lθ ¨ + (m2 cos θ)¨ m2 Lθ x = −m2 g sin θ In matrix form, m1 + m2 m2 L cos θ m2 cos θ m2 L x ¨ ¨ = θ ˙2 sin θ f − kx + m2 Lθ −m2 g sin θ (6) (4) (5)
Hence, the governing equations may be expressed in vector form ¨ = F (t, q, q ˙) Mq which is merely rewriting the Newton’s second law in terms of the generalized displacement q.
Equivalent First-Order System & RK4— By Ng Tin Yau (PhD) — 11/38
General Equation
For a general second-order system, we have ¨ = F (t, q, q ˙) Mq and then put it in a matrix form ˙ q ¨ q = ˙ q ˙) M−1 F (t, q, q
Direct Time Integration
Linear Systems
Ng Tin Yau (PhD)
Department of Mechanical Engineering The Hong Kong Polytechnic University
Nov 2015
— By Ng Tin Yau (PhD) — 1/38
Setting the stage— By Ng Tin Yau (PhD) — 9/38
Governing Differential Equations
Let q = [x θ]T . Then the mass matrix is given by M= m1 + m2 m2 L cos θ m2 cos θ m2 L
Setting the stage— By Ng Tin Yau (PhD) — 7/38
Pendulum Absorber: FBD
The acceleration components of the ball with respect to the body-attached coordinate frame are ˙2 − x m2 (Lθ ¨ sin θ) = P − m2 g cos θ ¨+ x m2 (Lθ ¨ cos θ) = −m2 g sin θ Now by combining eq. (♣) and (♥) to yield ˙2 − x m1 x ¨ = −kx + [m2 (Lθ ¨ sin θ) + m2 g cos θ] sin θ + f 2 2 ˙ ⇒ m1 x ¨ + m2 x ¨ sin θ − m2 Lθ sin θ + kx − (m2 g sin θ) cos θ = f On the other hand, using eq. (♦) we have ¨ cos θ + x −(m2 g sin θ) cos θ = m2 (Lθ ¨ cos2 θ) Thus, ¨ cos θ = f − kx + m2 Lθ ˙2 sin θ (m1 + m2 )¨ x + m2 Lθ
Equivalent First-Order System & RK4— By Ng Tin Yau (PhD) — 12/38
Linear Second-Order Equation
Example
¨ (t) + Cu ˙ (t) + Ku(t) = G(t) in Express the second-order system Mu ˙ (t) = Ax(t) + Bg(t). state-space form x Rewrite the second-order system as ¨ (t) = −M−1 Cu ˙ (t) − M−1 Ku(t) + M−1 G(t) u Define x(t) = and A= O I − 1 −M K −M−1 C B= I O O I g(t) = O M−1 G(t) u(t) ˙ (t) u
˙ and define z = [z1 z2 ]T and thus, we obtain an Let z1 = q and z2 = q equivalent first-order system: ˙ (t) = Az(t) + b(t) z ˙ 0 ]T . and z0 = [x0 x (9)
where M, C and K are the mass, damping and stiffness matrices. On the other hand, the vector f (t) is the external excitation to the system whereas the vector q(t) is the output response of the system. This equation is a direct consequence of modeling of many dynamic systems in engineering. On the other hand, suppose that we have the initial conditions: q(t = 0) = q0 ˙ (t = 0) = q ˙0 q Our goal is to provide a few classic methods to solve this equation numerically.
Pendulum Absorber
A pendulum absorber model is shown as follows:
In this model, the connecting rod is assumed to be rigid and all joints and contact points are smooth.
Setting the stage— By Ng Tin Yivalent First-Order System
Rewrite the second-order system as ¨ = M−1 (f − Cq ˙ − Kq) q and then put it in a matrix form ˙ q ¨ q Let A= 0 I − 1 −M K −M−1 C and b = 0 M−1 f (8) = M−1 (f ˙ q ˙ − Kq) − Cq = 0 I − 1 −M K −M−1 C q 0 + ˙ q M−1 f
Setting the stage— By Ng Tin Yau (PhD) — 6/38
Pendulum Absorber: FBD
The free-body diagram of the model is shown as follows:
The equation of motion in the x-direction for the block is given by m1 x ¨ = −kx + P sin θ + f (♣)
In the case where θ is small, then sin θ ≈ θ and cos θ ≈ 1. In addition if ˙2 sin θ ≈ 0, then we obtain a linearize system of the we also assume θ original model. m1 + m2 m2 L m2 m2 L x ¨ k 0 + ¨ 0 m2 g θ x θ = f (t) 0 (7)
k1 + k2 −k2 −k2 k2 + k3
In vector notation we have q= and M= m1 0 0 m2 C= c1 + c2 −c2 −c2 c2 + c3 K= k1 + k2 −k2 −k2 k2 + k3 x1 x2 and f = f1 f2
Thus, we arrive the standard form ¨ + Cq ˙ + Kq = f Mq
On the other hand, the force vector takes the form: ˙) = F (t, q, q ˙) F1 (t, x, x, ˙ θ, θ ˙) F2 (t, x, x, ˙ θ, θ = ˙2 sin θ f (t) − kx + m2 Lθ −m2 g sin θ
Table of Contents
1
Setting the stage Standard Form of Second-Order Linear Systems Classical Models Equivalent First-Order System & RK4 Numerical Time Integration The Central Difference Scheme Newmark Family Schemes
2
3
— By Ng Tin Yau (PhD) — 2/38
Equations of Motion
In many circumstances, engineers have to solve the following system of second-order differential equations: ¨ (t) + Cq ˙ (t) + Kq(t) = f (t) Mq (1)
Setting the stage— By Ng Tin Yau (PhD) — 8/38
(♥) (♦)
Governing Differential Equations
Now the two governing differential equations becomes ¨ cos θ = f − kx + m2 Lθ ˙2 sin θ (m1 + m2 )¨ x + m2 Lθ ¨ + (m2 cos θ)¨ m2 Lθ x = −m2 g sin θ In matrix form, m1 + m2 m2 L cos θ m2 cos θ m2 L x ¨ ¨ = θ ˙2 sin θ f − kx + m2 Lθ −m2 g sin θ (6) (4) (5)
Hence, the governing equations may be expressed in vector form ¨ = F (t, q, q ˙) Mq which is merely rewriting the Newton’s second law in terms of the generalized displacement q.
Equivalent First-Order System & RK4— By Ng Tin Yau (PhD) — 11/38
General Equation
For a general second-order system, we have ¨ = F (t, q, q ˙) Mq and then put it in a matrix form ˙ q ¨ q = ˙ q ˙) M−1 F (t, q, q
Direct Time Integration
Linear Systems
Ng Tin Yau (PhD)
Department of Mechanical Engineering The Hong Kong Polytechnic University
Nov 2015
— By Ng Tin Yau (PhD) — 1/38
Setting the stage— By Ng Tin Yau (PhD) — 9/38
Governing Differential Equations
Let q = [x θ]T . Then the mass matrix is given by M= m1 + m2 m2 L cos θ m2 cos θ m2 L
Setting the stage— By Ng Tin Yau (PhD) — 7/38
Pendulum Absorber: FBD
The acceleration components of the ball with respect to the body-attached coordinate frame are ˙2 − x m2 (Lθ ¨ sin θ) = P − m2 g cos θ ¨+ x m2 (Lθ ¨ cos θ) = −m2 g sin θ Now by combining eq. (♣) and (♥) to yield ˙2 − x m1 x ¨ = −kx + [m2 (Lθ ¨ sin θ) + m2 g cos θ] sin θ + f 2 2 ˙ ⇒ m1 x ¨ + m2 x ¨ sin θ − m2 Lθ sin θ + kx − (m2 g sin θ) cos θ = f On the other hand, using eq. (♦) we have ¨ cos θ + x −(m2 g sin θ) cos θ = m2 (Lθ ¨ cos2 θ) Thus, ¨ cos θ = f − kx + m2 Lθ ˙2 sin θ (m1 + m2 )¨ x + m2 Lθ
Equivalent First-Order System & RK4— By Ng Tin Yau (PhD) — 12/38
Linear Second-Order Equation
Example
¨ (t) + Cu ˙ (t) + Ku(t) = G(t) in Express the second-order system Mu ˙ (t) = Ax(t) + Bg(t). state-space form x Rewrite the second-order system as ¨ (t) = −M−1 Cu ˙ (t) − M−1 Ku(t) + M−1 G(t) u Define x(t) = and A= O I − 1 −M K −M−1 C B= I O O I g(t) = O M−1 G(t) u(t) ˙ (t) u
˙ and define z = [z1 z2 ]T and thus, we obtain an Let z1 = q and z2 = q equivalent first-order system: ˙ (t) = Az(t) + b(t) z ˙ 0 ]T . and z0 = [x0 x (9)
where M, C and K are the mass, damping and stiffness matrices. On the other hand, the vector f (t) is the external excitation to the system whereas the vector q(t) is the output response of the system. This equation is a direct consequence of modeling of many dynamic systems in engineering. On the other hand, suppose that we have the initial conditions: q(t = 0) = q0 ˙ (t = 0) = q ˙0 q Our goal is to provide a few classic methods to solve this equation numerically.
