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Lagrange Approach Basilio Bona DAUIN Politecnico di Torino - - PowerPoint PPT Presentation

Lagrange Approach Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Lagrange Semester 1, 2016-17 1 / 50 Introduction A multibody system is considered as a system in which the dynamic equations derive from a


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Lagrange Approach

Basilio Bona

DAUIN – Politecnico di Torino

Semester 1, 2016-17

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 1 / 50

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Introduction

A multibody system is considered as a system in which the dynamic equations derive from a unifying principle. This principle is based on the fact that, in order to describe the motion of a system, it is sufficient to consider some scalar quantities. These quantities were in origin called vis viva and work function, today they are called kinetic energy and potential energy. Both are state functions, i.e., functions that map the value of the state vector into a scalar function. The concept of state will be defined later; for the moment we simply consider that the state corresponds to the two vectors of the generalized coordinates q(t) and of the generalized velocities ˙ q(t).

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 2 / 50

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This general principle is called the Principle of Least Action. Let us consider the space Q of the generalized coordinates q ∈ Q, as sketched in Figure for a two-dimensional space Q. A particle starts its motion at time t1 in Q1 = q(t1) and ends it motion at time t2 reaching the state Q2 = q(t2) (or vice-versa, since time can be reversed). Assume that the motion keeps constant the total energy, i.e., the sum E = K +P of the kinetic energy K and the potential energy P that the particle has at time t1.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 3 / 50

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Q1 and Q2 are connected by a continuous path (trajectory) called true trajectory, that is unknown, since it is what we want to compute as the result of the dynamical equation analysis. If we choose at random different trajectories, with the only condition that the two boundary points remain the same (perturbed trajectories), the chance to obtain exactly the true trajectory will be very small. What characterizes the true trajectory with respect to all possible other perturbed trajectories? Euler contributed to the solution of this problem, but Lagrange developed a complete theory, that was later extended by Hamilton.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 4 / 50

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The true trajectory is the one that minimizes the integral of the vis-viva (i.e., twice the kinetic energy) of the entire motion between Q1 and Q2. This integral is called action and has a constant and well defined value for each perturbed trajectory having constant total energy E (E depends only

  • n the initial state).

The least action principle states that the nature “chooses”, among the infinite number of trajectories starting in q(t1) and ending in q(t2), the trajectory that minimizes the definite integral S =

t2

t1

K ∗(q(t), ˙ q(t))dt

  • f a particular state function K ∗(q(t), ˙

q(t)). The integral between the initial time t1 and the final time t2 must obey to the boundary constraints in the two time instants.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 5 / 50

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The scalar quantity S is the integral of a function and is called a functional. A functional is a mapping between a function and a real number; the function shall be considered as a whole, i.e., not a single particular value; in this sense a functional is often the integral of the function. The minimization of a functional is based on a particular mathematical technique, called calculus of variations. The conditions that guarantee the minimization of S provide a set of differential equations that contain the first and second time derivatives of the qi(t); this set of equations completely describes the system dynamics.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 6 / 50

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These differential equations specify the evolution of a physical quantity as the result of infinitesimal increments of time or position; summing up this infinitesimal variations we obtain the physical variables at every instant, knowing only their initial value and possibly some initial derivative: we can say that the motion has a local representation. The action characterizes the motion dynamics requiring only the knowledge of the states at the initial and final times; every intermediate value of the variables can be determined by the minimization of the action, that is a global, rather than a local, measure. The Lagrange approach is based on the definition of two scalar quantities, namely the total kinetic co-energy K ∗ and the total potential energy P associated with the body. The reason for using the term co-energy instead of the term energy, will be clarified later.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 7 / 50

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Lagrangian approach

The Lagrange method allows to define a set of Lagrange equations, that have some advantages with respect to the vector equations provided by the Newton-Euler approach. The approach provides n second-order scalar differential equations, directly expressed in the generalized coordinates ˙ qi(t) e qi(t). If holonomic constraint are present, the constraint forces do not appear in the equations. The kinetic co-energies and the potential energies are independent of the reference frame used to represent the body motion. The kinetic co-energies and the potential energies are additive scalars: in a multi-body system the total energies/co-energies are the sum of each energy/co-energy component.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 8 / 50

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Linear and angular momenta

Linear momentum hL is the physical vector defined as the product of a body mass M for its linear velocity v (the center-of-mass velocity) hL(t) = Mv(t) In non-relativistic mechanics the mass M of a body is constant (except for some particular cases, as rockets consuming fuel, etc.) Angular momentum hA (also called moment of momentum or rotational momentum) is the physical vector defined as the product of a body rotational inertia Γ for its rotational velocity ω hA(t) = Γ(t)ω(t) While the mass of a body is usually constant, the inertia matrix (or inertia tensor) Γ may vary in time.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 9 / 50

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Kinetic energy and co-energy for single point-mass

The mechanical kinetic energy associated to a point-mass m is defined as the work necessary to increase the linear or angular momentum from 0 to h, i.e., K (h) =

h

0 dW

The infinitesimal work associated to the mass is given by dW = f ·dx+τ ·dα where the symbol · indicates the scalar product, and f is the resultant of the applied linear forces on the mass, dx is the infinitesimal linear displacement increment, τ is the resultant of the applied angular torques, and dα is the infinitesimal angular displacement increment. Moreover f = dhL dt τ = dhA dt

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 10 / 50

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The resulting infinitesimal work is therefore the sum of two terms dW = dWL +dWA = dhL dt ·dx+ dhA dt ·dα = v ·dhL +ω ·dhA and we can write K (h) =

h

0 v ·dhL +ω ·dhA

The kinetic energy is a scalar state function associated to the particle states (v,ω) and (hL,hA). Another state function associated to the point-mass, called mechanical kinetic co-energy, is defined as K ∗(v) =

v

0 hL ·dv +hL ·dω

As shown in Figure, between the mechanical energy and the co-energy a relation exists K ∗(v) = h·v −K (h) (for notational simplicity, only the linear velocity is considered)

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 11 / 50

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Figure: This relation is an example of the Legendre transformation

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 12 / 50

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In particular, if the mass particle is moving at a velocity significantly smaller that the speed of light c, i.e., it is not a relativistic mass, the relation is h = mv with m constant, and the two “energies” become K (h) =

h

1 mh·dh = 1 2mh·h = 1 2m h2 K ∗(v) =

v

0 mv ·dv = 1

2mv ·v = 1 2mv2 As one can see, in this case the kinetic energy and co-energy are the same since h2 = m2 v2 This does not happen for relativistic masses where m = m(v(t)).

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 13 / 50

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In an extended body composed by N masses mi the kinetic co-energy is the sum of the kinetic co-energy of each mass K ∗(v) = 1 2

N

i=1

mivi ·vi

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 14 / 50

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We consider the velocity v0

i = ˙

x0

i with respect to R0.

Each velocity i in R0 can be computed from the general relation ˙ x0

i (t) = ω0 01(t)×ρ0 i (t)+R0 1 ˙

x1

i (t)+ ˙

d

1(t) = ω0 01(t)×ρ0(t)+ ˙

d

1(t)

where the term R0

1 ˙

xi(t) is zero, since the point-masses are fixed with respect to the body-frame, i.e., ˙ x1

i (t) = 0.

Now we consider a purely translatory motion and then a purely rotational motion.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 15 / 50

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Translational motion

If the motion is purely translational all point masses have the same linear velocity v0 with respect to R0 ˙ x0

i (t) = ˙

d

1(t) ≡ v0(t) ∀i

then K ∗ = 1 2v0 ·v0

N

i=1

mi = 1 2mtot v0 ·v0 = 1 2mtot

  • v0

2 = 1 2(v0)T(mtotI)v0 where the mass mtot is the total body mass. The kinetic co-energy is equivalent to that of one particle with total mass mtot with the translational velocity v0. The total mass mtot can be ideally concentrated in the body center-of-mass C, whose position is xc xcmtot = ∑

i

ximi → xc = 1 mtot ∑

i

ximi

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 16 / 50

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Since the velocity is equal for all points of the body, v0 is also the velocity

  • f the center-of-mass C; if we use the symbol v0

c ≡ v0 for this velocity, we

can write K ∗ = 1 2mtot v0

c ·v0 c = 1

2mtot

  • v0

c

  • 2 = 1

2 (v0

c)T(mtotI)v0 c

that gives the usual rule: “the kinetic energy is half the product of the total mass for the total velocity squared.”

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 17 / 50

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Rotational motion

If the motion is purely rotational, then ˙ x0

i (t) = ω0 01(t)×xb i

i.e., vi = ω0 ×xi where ω0

01 ≡ ω0 is the total angular velocity and xi is the position of the

i-th mass in R0. Considering all masses K ∗ = 1 2

N

i=1

mi(ω0 ×xi)·(ω0 ×xi) since a·(b×c) = b·(c×a) and assuming a ≡ (ω0 ×xi), b ≡ ω0, c ≡ xi we obtain K ∗ = 1 2

N

i=1

miω0 ·xi ×(ω0 ×xi) = 1 2 ω0 ·

  • N

i=1

mixi ×(ω0 ×xi)

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 18 / 50

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The previous relation is equivalent to K ∗ = 1 2 ω0 ·

  • N

i=1

hi

  • = 1

2 ω0 ·h0 where h0 is the total angular momentum with respect to the origin O of the reference frame R0. Since h0 = Γ0ω0, we have K ∗ = 1 2 (ω0)TΓ0ω0 Notice the similarity: K ∗

trasl = 1

2 (v0

c)T(mtotI)v0 c

pure translation K ∗

rot = 1

2 (ω0)TΓ0ω0 pure rotation

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 19 / 50

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Total kinetic co-energy

Therefore, the total kinetic co-energy is K ∗ = K ∗

trasl +K ∗ rot = 1

2

  • (v0

c)T(mtotI)v0 c +(ω0)TΓ0ω0

This is a well known result, that can be expressed in words as: the total kinetic co-energy of a body is the sum of the translational kinetic co-energy of the center of mass plus the rotational kinetic co-energy around the center of mass. This relation is valid also for extended bodies.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 20 / 50

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Generalized coordinates

Considering the generalized coordinates q(t) and velocities ˙ q(t) and the Jacobians vc = JL(q)˙ q

  • r

ω = JA(q)˙ q we obtain K ∗(˙ q,q) = 1 2

  • ˙

qTJT

L (mI)JL ˙

q+ ˙ qTJT

AΓcJA ˙

q

  • and

K ∗(˙ q,q) = 1 2 ˙ qT JT

L (mI)JL +JT AΓcJA

  • ˙

q = 1 2 ˙ qTΓtot ˙ q where Γtot = JT

L (mI)JL +JT AΓcJA

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 21 / 50

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Potential energy

Potential energy is a form of energy that depends only on position; two types of position-related energies exist. One is due to the gravitational field, the other is the energy stored in the elastic components of the body, that accumulate energy under the effects of deformation. Since in our approach the considered bodied are rigid, the elastic parts are external to the bodies and are represented by ideal springs that connect various parts of the mechanical system.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 22 / 50

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A potential function P(x), is a scalar function that depends only on the position x = x y zT A force f is said to be conservative when it is the negative gradient of P(r) f(x) = −∇P(x) = − ∂P(x) ∂x ∂P(x) ∂y ∂P(x) ∂z T If a potential function exists, it is called potential energy of the system and it is unique apart from an additive constant. This implies that the effects on the body dynamics depend only from the potential energy variation, and not on its absolute value.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 23 / 50

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Gravitational energy

An example of conservative force field is the gravitational field around the

  • Earth. The potential field produces the so-called weight forces.

The potential energy due to a gravitational field and associated to a generic mass m is given by the following relation: Pg = −mg ·x0

c

where g is the local gravitational acceleration vector and x0

c is the body

center-of-mass position vector with respect to a plane going through the

  • rigin of R0 and orthogonal to g, that provides the conventional zero value
  • f potential energy (zero potential energy plane) as shown in Figure.
  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 24 / 50

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Figure: The potential energy due to the gravitational field.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 25 / 50

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Elastic energy

Another force field is related to potential energy, that due to elastic elements: these elements represent the abstract model of a proportional relation between displacement and force. If we assume a one-dimensional linear spring, the relation between the applied force f and the linear elongation e from the rest position of the spring is f = kee If we assume a one-dimensional torsional or torsion spring we can write a relation between the applied torque τ and the resulting angular deformation δ from the rest position of the spring τ = k′

ke and k′

e are the so-called elastic constants of the springs.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 26 / 50

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Figure: A one dimensional linear spring. The rest length of the spring is x0, and e is the extension/compression occurring when the force f is applied.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 27 / 50

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The potential energy is the integral of the virtual work performed by the spring deformation P(e) =

e

0 f ·de

  • r

P(δ) =

δ

0 τ ·dδ

We can define also in this case the potential co-energies, that are P∗(f) =

f

0 e·df

  • r

P∗(τ) =

τ

0 δ ·dτ

the relation between P∗(f) and P(e) is is given by the Legendre transformation P∗(f) = f ·e−P(e)

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 28 / 50

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A nonlinear spring: the relation between f(t) and e is nonlinear, but the relation f ·e = P(e)+P∗(f) holds.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 29 / 50

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When the elastic elements are linear with constant ke, the potential energy and co-energy are given by P(e) = 1 2eT(keI)e = 1 2ke e2 and P∗(f) = 1 2fT(keI)−1f = 1 2ke f2 When torsion springs with constant k′

e are considered, the potential energy

and co-energy are given by P(δ) = 1 2δ T(k′

eI)δ = 1

2k′

e δ2

and P∗(τ) = 1 2τT(k′

eI)−1τ = 1

2k′

e

τ2 In linear case, energies and co-energies are equal P(e) = P∗(f) and P(δ) = P∗(τ)

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 30 / 50

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SLIDE 31

If the elastic constants are different along the three directions, a more general relation applies P(e) = 1 2eTKee; P∗(f) = 1 2fTK−1

e f

and P(δ) = 1 2δ TK′

eδ;

P∗(τ) = 1 2τT(K′

e)−1τ

where Ke = diag(ke1,ke2,ke3) e K′

e = diag(k′ e1,k′ e2,k′ e3) are the elastic

constant matrices along the three dimensional axes.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 31 / 50

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Generalized forces in holonomic systems

The forces fi acting on the i-th mass can be classified according to three groups: N′′′ constraint forces fv

i due to constraint reactions.

N′′ conservative forces fc

i due to conservative fields.

N′ non conservative forces fnc

i .

The total force is the sum of these three types of forces f =

N′

i=1

fnc

i

+

N′′

i=1

fc

i + N′′′

i=1

fv

i

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 32 / 50

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Constraint forces

The virtual displacements δxi are always tangent to the constraints, while the constraint forces fv

i are always orthogonal to the constraints; from this

assumption it follows that fv

i ·δxi = 0

Therefore the work done by the constraint forces is zero (the forces “do not work”) δW v =

N′′′

i=1

fv

i ·δxi = 0.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 33 / 50

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Conservative forces

The N′′ conservative forces do a work that results δW c =

N′′

i=1

fc

i ·δxi = − N′′

i=1

∇Pi ·δxi = −

N′′

i=1

∇Pi ·

  • n

j=1

∂xi ∂qj δqj

  • =

n

j=1

  • N′′

i=1

−∇Pi · ∂xi ∂qj

  • F c

j

δqj This last expression highlights the so called generalized conservative forces F c

j

The virtual work can be expressed as a function of the generalized coordinates qj: δW c =

N′′

i=1

fc

i ·δxi = n

j=1

F c

j δqj = F c ·δq

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 34 / 50

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SLIDE 35

Non conservative forces

The N′ non conservative forces fnc

i

do a work equal to δW nc =

N′

i=1

fnc

i

·δxi =

N′

i=1

fnc

i

·

  • n

j=1

∂xi ∂qj δqj

  • =

n

j=1

  • N′

i=1

fnc

i

· ∂xi ∂qj

  • F nc

j

δqj This last expression highlights the so called generalized non conservative forces F nc

j

and allows to transform the virtual work from a function of the positions x to a function of the generalized coordinates qj: δW nc =

N′′

i=1

fnc

i

·δxi =

n

j=1

F nc

j δqj = F nc ·δq

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 35 / 50

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SLIDE 36

In conclusion, only two types of forces will do work in a system subject to holonomic constraints the j-th generalized conservative forces: F c

j (q) = − N′′

i=1

∇Pi · ∂xi ∂qj the j-th generalized non conservative forces: F nc

j (q) = N′

i=1

fnc

i

· ∂xi ∂qj The generalized force, being the result of a scalar product, will be itself a scalar quantity.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 36 / 50

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SLIDE 37

In case of torques acting on the system, the generalized forces due to them will give origin to the following generalized torques T c

j = N′′

i=1

−∇Pi · ∂αi ∂qj ; T nc

j

=

N′

i=1

τnc

i

· ∂αi ∂qj The symbol used to identify both the generalized forces and the generalized torques will be F.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 37 / 50

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Lagrange equations with holonomic constraints

In a multi-body system subject to holonomic constraints, the formulation

  • f the Lagrange equations may take different forms.

From the knowledge of the total co-energy of the system K ∗(q, ˙ q) =

N

ℓ=1

K ∗

ℓ (q, ˙

q)

  • ne derives the Lagrange equations: they are a set of n equations (one for

each generalized coordinates qi) defined as d dt ∂K ∗ ∂ ˙ qi

  • − ∂K ∗

∂qi = Fi i = 1,...,n where Fi = F c

i +F nc i

is the i-th generalized force, with a positive sign if applied by the external environment to the body, or a negative sign if applied by the body to the external environment.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 38 / 50

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SLIDE 39

Since the conservative forces due to the gravitational field are the negative

  • f the gradient of the potential, we can move to the left hand part of the

equation, resulting in d dt ∂K ∗ ∂ ˙ qi

  • − ∂K ∗

∂qi +

  • N′

k=1

∇Pk · ∂xk ∂qi

  • = F nc

i

The term inside the square bracket is equal to ∂P ∂qi ; therefore d dt ∂K ∗ ∂ ˙ qi

∂K ∗ ∂qi − ∂P ∂qi

  • = F nc

i

Moreover P does not depend on ˙ qi, so one can write ∂P ∂ ˙ qi = 0 and one obtains the most common form of the Lagrange equations d dt ∂K ∗ ∂ ˙ qi − ∂P ∂ ˙ qi

∂K ∗ ∂qi − ∂P ∂qi

  • = F nc

i

i = 1,...,n

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 39 / 50

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SLIDE 40

Lagrange state function

The Lagrange (state) function L is defined as the difference between the total kinetic co-energy K ∗ and the total potential energy P of he system L (q, ˙ q) = K ∗(q, ˙ q)−P(q) We can write n differential equations d dt ∂L (q, ˙ q) ∂ ˙ qi

  • − ∂L (q, ˙

q) ∂qi = F nc

i

(q) i = 1,...,n each one relative to the i-th generalized coordinate The term ∂L ∂ ˙ qi is the generalized momentum and is usually indicated by the symbol µi; the vector of generalized momenta is indicated by µ(q(t)).

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 40 / 50

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SLIDE 41

Dissipative and friction forces

The friction phenomena involve energy dissipated by the body as heat; they are due to complex interaction between solid/solid or solid/fluid surfaces; tribology is the science that studies the friction forces. If we keep the friction or other dissipative forces f fric

i

separate from the

  • ther non conservative forces, the Lagrange equation becomes:

d dt ∂L ∂ ˙ qi

  • − ∂L

∂qi = Fi −f fric

i

We can approximately describe the friction force f fric

i

as a nonlinear function of the relative velocity v between the two contact surfaces of the involved bodies. We can model the total friction force as in Figure and write f fric

total = f fric stiction +f fric coulomb +f fric viscous

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 41 / 50

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SLIDE 42
  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 42 / 50

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SLIDE 43

While stiction and Coulomb friction must be explicitly introduced as non-conservative forces, it is a common assumption to express the viscous dissipative phenomenon as the derivative of a dissipation function, also called Rayleigh function, given by: Di(˙ q) = 1 2 ˙ qT(βiI)˙ q = 1 2 βi ˙ q2 where the coefficient βi is the viscous friction coefficient, and ˙ q is the norm of the relative velocity between the moving body and the surface responsible of the viscous friction effect. This quadratic expression is NOT a dissipation “energy”, but only a conventional way to introduce it in the Lagrange equation, as follows d dt ∂L ∂ ˙ qi

  • − ∂L

∂qi + ∂Di ∂ ˙ qi = Fi i = 1,...,n Now the term Fi includes only the remaining non conservative forces

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 43 / 50

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SLIDE 44

Summary

Find the k = 1,...,N rigid bodies composing the system and compute the n degrees-of-freedom. If necessary, define the various body frames Rk Define a set of complete and independent generalized coordinates qi(t),i = 1,...,n ≤ 6N Compute the position vectors of each center-of mass xk(q(t)),k = 1,...,N Compute the linear velocity vectors of each center-of mass vc,k(q(t), ˙ q(t)) and the angular velocities vectors ωk(q(t), ˙ q(t)) of each body

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 44 / 50

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SLIDE 45

Energy computation

Compute the kinetic co-energy K ∗

k of each k-th rigid body with mass

mk and inertia matrix (wrt the center-of-mass) Γk as K ∗

k = 1

2mk

  • vc,k
  • 2 + 1

2ωT

k Γkωk

Set the local gravity acceleration vector g and represent it in R0 Localize the Ne elastic energy storage elements and model them with “ideal springs” with elastic constants kℓ,ℓ = 1,...,Ne Compute the gravitational potential energy Pg,k and the elastic energy Pe,ℓ as Pg,k = −mkgTpk Pe,ℓ = 1 2kℓ e2 where e is the elongation (positive or negative) of the spring.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 45 / 50

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SLIDE 46

Lagrangian function

Compute the total energies K ∗(q, ˙ q) =

N

k=1

K ∗

k

P(q) =

N

k=1

Pg,k +

Ne

ℓ=1

Pe,ℓ Compute the Lagrange function of the system L = K ∗ −P Compute the generalized forces Fi Write the n Lagrange equations d dt ∂L ∂ ˙ qi

  • − ∂L

∂qi = Fi i = 1,...,n i.e., d dt ∂K ∗ ∂ ˙ qi

  • − ∂K ∗

∂qi + ∂P ∂qi = Fi i = 1,...,n

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 46 / 50

slide-47
SLIDE 47

Lagrange equations

If there are linear dissipative elements, model them with a linear dashpot, having friction coefficient βi. Compute the dissipative function Di = 1 2βi

  • vf ,i
  • 2, where vf ,i is the

velocity associated to the friction producing element. Upgrade the Lagrange equations as follows d dt ∂K ∗ ∂ ˙ qi

  • − ∂K ∗

∂qi + ∂P ∂qi + ∂D ∂ ˙ qi = Fi i = 1,...,n When nonlinear elastic or friction elements are present, one should directly introduce the resulting nonlinear elastic or friction forces at the second term of the Lagrange equations.

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 47 / 50

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SLIDE 48

Characterization of the Lagrange equations

The Lagrange approach generates n differential equations d dt ∂L ∂ ˙ qi

  • − ∂L

∂qi = Fi i = 1,...,n Collecting the n equations in one vector equation one obtains d dt ∂L ∂ ˙ q

  • − ∂L

∂q + ∂D ∂ ˙ q = F If the equations are linear (or if we consider small perturbations around some equilibrium point), we will have a general formulation expressed as a second order differential vector equation A1¨ q(t)+A2 ˙ q(t)+A3q(t) = F

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 48 / 50

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SLIDE 49

The linear equations can be rewritten as M¨ q(t)+(D+G)˙ q(t)+(K+H)q(t) = F where M = MT is the mass or inertia matrix positive definite, symmetric D = DT is the viscous damping matrix symmetric G = −GT is the gyroscopic matrix skew-symmetric K = KT is the stiffness (elasticity) matrix symmetric H = −HT is the circulatory matrix (constrained damping) skew-symmetric

  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 49 / 50

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SLIDE 50

Lagrangian systems are holonomic systems where the forces are solely due to generalized potential functions P(q, ˙ q), Hamiltonian systems are those where the kinetic co-energy and the potential energy explicitly depend on time K ∗ = K ∗(q, ˙ q,t) and P = P(q, ˙ q,t) ... or, if you prefer the Wikipedia definition ... A Lagrangian system is a pair (Y ,L ) of a smooth fiber bundle Y → X and a Lagrangian density L which yields the Euler–Lagrange differential

  • perator acting on sections of Y → X.
  • B. Bona (DAUIN)

Lagrange Semester 1, 2016-17 50 / 50