mathematics and mechanics of solids-2012-lidström-209-42
Post on 14-Apr-2018
214 Views
Preview:
TRANSCRIPT
-
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
1/35
http://mms.sagepub.com/Mathematics and Mechanics of Solids
http://mms.sagepub.com/content/17/3/209The online version of this article can be found at:
DOI: 10.1177/1081286511407111
2012 17: 209 originally published online 9 August 2011Mathematics and Mechanics of SolidsP Lidstrm
On the equations of motion in constrained multibody dynamics
Published by:
http://www.sagepublications.com
can be found at:Mathematics and Mechanics of SolidsAdditional services and information for
http://mms.sagepub.com/cgi/alertsEmail Alerts:
http://mms.sagepub.com/subscriptionsSubscriptions:
http://www.sagepub.com/journalsReprints.navReprints:
http://www.sagepub.com/journalsPermissions.navPermissions:
http://mms.sagepub.com/content/17/3/209.refs.htmlCitations:
What is This?
- Aug 9, 2011OnlineFirst Version of Record
- May 7, 2012Version of Record>>
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/content/17/3/209http://mms.sagepub.com/content/17/3/209http://www.sagepublications.com/http://mms.sagepub.com/cgi/alertshttp://mms.sagepub.com/cgi/alertshttp://mms.sagepub.com/subscriptionshttp://mms.sagepub.com/subscriptionshttp://mms.sagepub.com/subscriptionshttp://www.sagepub.com/journalsReprints.navhttp://www.sagepub.com/journalsReprints.navhttp://www.sagepub.com/journalsPermissions.navhttp://mms.sagepub.com/content/17/3/209.refs.htmlhttp://mms.sagepub.com/content/17/3/209.refs.htmlhttp://online.sagepub.com/site/sphelp/vorhelp.xhtmlhttp://online.sagepub.com/site/sphelp/vorhelp.xhtmlhttp://online.sagepub.com/site/sphelp/vorhelp.xhtmlhttp://mms.sagepub.com/content/early/2011/07/31/1081286511407111.full.pdfhttp://mms.sagepub.com/content/early/2011/07/31/1081286511407111.full.pdfhttp://mms.sagepub.com/content/17/3/209.full.pdfhttp://mms.sagepub.com/content/17/3/209.full.pdfhttp://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://online.sagepub.com/site/sphelp/vorhelp.xhtmlhttp://mms.sagepub.com/content/early/2011/07/31/1081286511407111.full.pdfhttp://mms.sagepub.com/content/17/3/209.full.pdfhttp://mms.sagepub.com/content/17/3/209.refs.htmlhttp://www.sagepub.com/journalsPermissions.navhttp://www.sagepub.com/journalsReprints.navhttp://mms.sagepub.com/subscriptionshttp://mms.sagepub.com/cgi/alertshttp://www.sagepublications.com/http://mms.sagepub.com/content/17/3/209http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
2/35
Article
On the equations of motion in
constrained multibody dynamics
Mathematics and Mechanics of Solids
17(3): 209242
The Author(s) 2011
Reprints and permission:
sagepub.co.uk/journalsPermissions.nav
DOI: 10.1177/1081286511407111
mms.sagepub.com
P Lidstrm
Division of Mechanics, Lund University, Sweden
Received 11 November 2010; accepted 18 March 2011
AbstractThe equations of motion for a constrained multibody system are derived from a continuum mechanical point of view.This will allow for the presence of rigid, as well as deformable, parts in the multibody system together with kinematicalconstraints. The approach leads to the classical LagrangedAlembert equations of motion under constraint conditions.The generalized forces appearing in the equations of motion are given as expressions involving contact, internal and bodyforces in the sense of continuum mechanics. A detailed analysis of physical constraint conditions and their implicationfor the equations of motion is presented. A precise distinction is made between constraints on the motion on the onehand and resistance to the motion on the other. Transformation properties covariance and invariance under changesof configuration coordinates are elucidated. The elimination and calculation of the so-called Lagrangian multipliers isdiscussed and some useful reformulations of the equations of motion are presented. Finally a Power theorem for theconstrained multibody system is proved.
Keywords
constraints, equations of motion, multibody dynamics
1. Introduction
A central issue in multibody dynamics is the formulation of geometrical or kinematical constraints representing
conditions on the relative motions between parts of the mechanical system and the associated formulation of
the equations of motion. In general, the kinematical constraint will lead to a system of first-order ordinary
differential equations:
g0 +gq = 0m1
where g0 = g0(t, q) Rm1 andg = g(t, q) Rmn are the constraint matrices andq Rn the configurationcoordinates for the multibody. A formulation of the equations of motion for the constrained mechanical system
may, for instance, be obtained by invoking the LagrangedAlembert equations of motion:
d
dt
T
q
T
q Q = 01n
Corresponding author:
P Lidstrm, Division of Mechanics, Lund University, P.O. Box 118, S-22100 Lund, SwedenEmail: per.lidstrom@mek.lth.se
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
3/35
210 Mathematics and Mechanics of Solids 17(3)
where T = T(t, q, q) is the kinetic energy of the multibody and Q = Q(t, q, q) R1n is the so-called sum ofthe generalized forces acting on the multibody due to internal, contact and body forces. One part of Q is the
force Qkc = Qkc(t, q, q) R1n due to the kinematical constraints. The so-called non-holonomic principlestates that
Qkc = Tg
where the multipliers = (t) Rm1 represent components of constraint forces and moments. A justifica-tion of the non-holonomic principal may be based on the principle of LagrangedAlembert, cf. Bloch et al. [1],
where a discussion, from an historical point of view, on the derivation of the equations of motion is presented.
In Lidstrm [2], the equations of motion for a multibody system, containing rigid as well as elastic parts,
were derived using kinematical concepts and dynamic laws taken from continuum mechanics. Constraints on
multibody systems were, however, not considered. It is the purpose of this paper to include constraints in
the general discussion. The interaction between rigid parts is discussed in terms of kinematical constraints and
the associated constraint forces. Interactions are expressed in coordinate systems compatible or non-compatible
with the constraints. The main issues of this paper are the derivation of the equations of motion for a constrained
multibody system and the elimination and calculations of the constraint forces.
Classical and modern developments concerning constraint enforcement in multibody systems and their
numerical implementation are reviewed by Lalusa and Bauchau [3, 4]. A good reference concerning modernmathematical developments, with applications in control theory, is Bloch et al. [1].
The paper is organized as follows. In Section 2 the basic notations and definitions used throughout the
paper are reviewed. In Section 3 the multibody system is introduced. In Section 4 the interaction between
rigid parts is analysed in terms of contact forces and kinematical constraints. A precise distinction is made
between constraints on the motion on the one hand and resistance on the motion on the other. Constraints on
the motion are maintained by reactions, that is, constraint forces and moments, which are a priori unknown.
They may be restricted by inequalities as, for instance, in the case of contact forces at a unilateral contact
between parts. Forces and moments responsible for resistance on the motion are prescribed constitutively.
The compatibility of the system of generalized coordinates with the imposed kinematical constraints and the
implications for the equations of motion is discussed. In Section 5 the equations of motion under prescribed
bilateral kinematical constraints are derived using the principal of virtual power. This leads to the classical
LagrangedAlembert equations of motion involving multipliers. The multipliers will, as is well known and as
will be further illuminated by the discussion presented in this paper, represent components of constraint forces
and moments. The Lagrangian multipliers correspond precisely to those components of forces and moments
that are responsible for the maintenance of the kinematical constraints. Forces and moments responsible for
resistance on the motion have to be specified separately and then added to the equations of motion. In Section 6
a further analysis of the equations of motion is undertaken. This involves the elimination and calculation of the
Lagrangian multipliers and a reformulation of the equations of motion that decreases the order of the ordinary
differential equations. Finally, in Section 7, a Power theorem for the constrained multibody system is proved.
2. Notation
In this paperR denotes the set of real numbers and Rn the set of n-tuples of real numbers. The set of n-dimensional, real column vectors is denoted by Rn1 and the null vector in Rn1 is written 0n1. R
mn denotes
the set of real matrices of order m n with the null matrix written 0mn. If A Rmn then AT Rnm is the
transpose of A. The rank of a matrix A Rmn is written rank(A). IfA is a square matrix then det(A) denotesthe determinant ofA and if det(A) = 0 then A1 denotes its inverse. Inn denotes the identity matrix in R
nn. If
A Rmn then kernel(A) =
u Rn1 |Au = 0m1
andrange(A) =
v Rm1 | v = Au , u Rn1
.
Let E denote a three-dimensional Euclidean point space with the corresponding translation vector space
V , that is, dim(V) = 3. Points in E are denoted by x,y, . . . ,X, Y, . . . and vectors in V by a, b, . . . ,u, v, . . ..
The space of all second-order tensors A on V , that is, linear mappings V V, is denoted End(V). The setof all rotations on V is denoted SO(V)=
A End(V)
ATA = AAT = 1, detA = 1
. The space of all skew-
symmetric tensors is denoted by Skew(V) = A End(V) AT = A. If A Skew(V) then ax(A) V
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
4/35
Lidstrm 211
denotes its axial vector, that is, Au = ax(A) u, u V. Given a V, then the tensora 1 Skew(V) isdefined by (a 1)u = a u, u V. IfA End(V) then a A = (a 1)A.
3. Preliminaries
This introductory section follows closely that of Lidstrm [2]. For more information on details and motivations
for the present approach the reader is referred to that paper.
Consider a multibody B consisting of N rigid or elastic bodies: B, = 1, . . . ,N. We write B =N
=1B.
Let the regular region Bt in Edenote the present (spatial) placementofB, at time t. The function representing
the transplacement of the part B from its reference placement B0 to its present placement Bt is given by the
mapping : B0 Bt , defined by
x = (X, t), (X, t) B0 R (1)
The transplacement (1) is assumed to be a twice-continuously differentiable mapping. Basic kinematical quan-
tities, such as material velocity x, deformation gradientFandGreen-St.Venant strain tensorEetc., belonging
to part B , will be super-indexed by according to x = t , F = X , andE =1
2 (FTF 1). We
assume that int(Bt )
int(Bt )) = , = , reflecting the impenetrability of matter. We also assume, as a
matter of convenience, that B0B
0 = . The transplacement of the multibody (, t) : B0 Bt is defined
by
x = (X, t) = (X, t), X B0 (2)
where B0 =
N=1
B0 andBt =
N=1
Bt . Note that the transplacement, (2), is a twice-continuously differentiable
mapping. We assume that it is possible to introduce configuration coordinates:
q = (q1, q2, . . . , qn) Rn (3)
where is an open set, so that the set of possible placements (configurations) of the parts of the multibody
may (locally) be represented by the mappings
x = q (t, q1, q2, . . . , qn;X) = q (t, q;X), (t, q;X) R B
0 (4)
where q : R B0 E is assumed to be a twice-continuously differentiable mapping in all its arguments
(t, q;X). Then (k, l = 0, 1, . . . , n, q0def= t),
2 q
qlqk=
2 q
qkqland
2 q
Xqk=
2 q
qkX.
The motion of the multibody is locally given by a function q = q(t), t T = ]t1, t2[ R and a mapping
x = q(t, q(t);X)def= q (t, q(t);X), t T, X B
0 (5)
The motion of a rigid partB is given by a function q = q(t), t T = ]t1, t2[ R and the mapping
x = q (t, q;X) = x
A(t, q) +R (t, q)(X XA ), X B
0 (6)
where A B is an arbitrarily selected material point in B with reference place XA and present place x
A =xA(t, q) =
q (t, q;X
A ). R
= R(t, q) SO(V) is the rotation tensorforB. The velocity field of the rigid partB may be written
x = x (t, q, q;x) = xA(t, q, q) + (t, q, q) (x xA), x B
t
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
5/35
212 Mathematics and Mechanics of Solids 17(3)
where
xA =
nk=0
xA,kqk and =
nk=0
;kqk
and
x
A,k= x
A,k(t, q)
def=
xA
qkand
;k=
;k(t, q)
def= axR
qkRT , k = 0,1, . . . , n (7)
The vectors xA,k and;k are called the partial velocity vector and the partial angular velocity vectors, respec-
tively. In analogy with definition (7)1 we introduce x
A;k
def=
xA qk
. It follows that, cf. Lidstrm [2], Corollary 4.1,
xA;k = x
A,k and;k =
qk, k = 1, . . . , n.
The kinetic energy for the multibody Bmay be written, cf. Lidstrm [2], Theorem 6.1:
T = T(t, q, q) =1
2(M0 +M1q + q
TM2q)
where M0, M1 andM2 are the mass matrices of the multibody. The equations of motion for the multibody read
qTM2 = Qcif + Qi + Qc + Qb (8)
where
Qcif =
qT
M2
q
1
2
M2
q
tq + qT
M2
t+ skew
M1
q
+
1
2
M1
t
M0
q
(9)
is the complementary inertia force. For a demonstration of (9) cf. Lidstrm [2], where the matrix notation is
explained.
Qi =
N
=1Q,i, Qc =
N
=1Q,c andQb =
N
=1Q,b (10)
are the generalized internal force, the generalized contact force and the generalized body force, respectively,
acting on the multibody. Furthermore
Q,ik =
B
0
P Fq
qkdv(X), Q
,ck =
B
0
q
qk t0 da(X), Q
,bk =
B
0
q
qk b0 dv(X) (11)
Here Fq = Fq (t, q,X) denotes the deformation gradient, t
0 = P
n0 the traction vector on the boundary
surface of part B where P = P (X, t) is the first PiolaKirchhoff stress tensor, n0 = n0 (X, t) the external
unit normal on B0 andb = b (X, t) is the (mass-) specific body force. If the elastic parts are constituted by
St. VenantKirchhoff material, then the generalized internal force may be written, cf. Lidstrm [2]:
Qi
=
Ve,q
q (12)
where
Ve =
N=1
B
0
1
2 (trE)2 + |E |2
dv(X)
(13)
The generalized contact force may be written
Qc = Qec + Qic (14)
where Qec represents the contact force from the exterior ofB, while Qic represents the contact force from all
internal contacts between parts ofB.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
6/35
Lidstrm 213
Assume thatB andB are in geometrical contact over the surface St , at time t. We introduce the following
vector fields defined on St :
xk(t, q,x) = q (t, q(t); (
q )
1(t, q;x))
qk, x
k(t, q,x) = q (t, q(t); (
q )
1(t, q;x))
qk, k = 0,1, . . . , n
where ( q )1(t, q; ) is the inverse of q (t, q; ). The generalized force, Q
k , from B on B is then given by
Q
k =
S
t
xk t da(x), = , k = 1, . . . , n (15)
where the contact force t is the traction vector acting on B from B . The corresponding generalized force,
Q
k , from B on B is then given by
Q
k =
S
t
x
k t da(x), = , k = 1, . . . , n (16)
The generalized force Qic, from the internal contacts between all parts of the multibody, may now be written as
Qick =
N, = 1 <
I
k (17)
where
I
k = Q
k + Q
k , k = 1, . . . , n (18)
is called the mechanical interaction between parts B andB . It follows that
Ik =
St
x,k tda(x), k = 1, . . . , n (19)
where x,k = x
k x
k andt = t(x, t) is the contact traction from B on B, cf. Lidstrm [2], Corollary 7.1.
4. The interaction between rigid parts
In Lidstrm [2], a general discussion of the interaction between parts, involving rigid as well as flexible parts,
was undertaken. Here we focus on rigid parts. Constraint conditions are often employed in the description of
the interaction between rigid parts in a multibody system.
Characteristic of multibody systems is the presence ofjoints that impose constraints on the relative motion
between rigid parts of the system. Many joints, in practical applications, may be modelled in terms of so-called
simple joints; examples are the cylindrical, prismatic, screw, revolute, sphericalandplanarjoints, cf. Geradin
and Cardona [5]. These are joints where the mechanical interaction is transmitted by a surface contact. The
relative motion of coincident material points on the contact surface between the two parts B andB are then
similar all over the contact surface. Other types of joints may involve line, as well as point, contacts.
Let St denote the contact surface at time t, let O be an arbitrary point in space and let C be a material
point in B, instantaneously in contact with the material point C B . The contact point in space C, is then
placed at
xC = xC = q (t, q;XC ) = xC =
q (t, q;XC ), XC S
0,t, XC S
0,t
and
xC = x,sysO +
(xC xO) and x
C= x
,sysO +
(xC xO)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
7/35
214 Mathematics and Mechanics of Solids 17(3)
B
B B B
Figure 1. Prismatic and revolute joint.
The relative velocity of the two material points is then given by
x,,sysC = x
,,sysO +
, (xC xO) (20)
where ,def= is the relative angular velocity. The impenetrability condition, n x
,,sysC 0 (cf.
condition (63) in Lidstrm [2]), may be expressed according to
n x,,sys
O + n , (xC xO) 0 (21)
We now give two examples of consequences emanating from condition (21). In the case of a prismatic jointthe
two parts are constrained to translate relative to one another along an axis with direction e = e(t), e e = 1, see
Figure 1. This constraint is then represented by the conditions x,,sys
C = e x
C , x
C R and, = 0. From
this it follows that x,,sysO = e x
C for all points O. The requirement of impenetrability (21) then reads
n x,,sys
O = n e x
C 0 (22)
an inequality that has to be satisfied for all x
C R and this is possible if and only if
n e = 0 (23)
that is, the contact surface normal n has to be perpendicular to the direction of relative translation e.
If the parts are connected by a revolute joint, then they are constrained to rotate relative to one another
around an axis (O, e), e = e(t), e e = 1 where O is a point on this axis, that is, a moving point in E with
place xO = xO(t). This constraint is represented by x,,sysO = 0 and
, = e, R and then, according
to (20), x,,sys
C = e (xC xO), xC S
t . The requirement of impenetrability reads
n e (xC xO) 0, xC St
which has to be satisfied for all R and this is possible if and only if
n e (xC xO) = 0, xC St (24)
The conditions (23) and (24) will put restrictions on the geometry of the contact surface. In order to analyse
this we introduce cylindrical coordinates (r, ,z) at O, with (O, e) defining the z-axis, that is, x = xO + err+ ez,
where er = er(t, ) is a unit vector ander e = 0. The contact surface St is assumed to be locally represented
by a smooth function r = r(t, ,z) > 0, that is
St =
xC E|xC = xC(,z, t)
def=xO(t) + er(t, )r(t, ,z) + e(t)z, 1 < < 2, z1 < z < z2
Proposition 4.1 Condition (23) is equivalent to S
t being a prismatic surface, that is, r = r(t, ). Condition
(24) is equivalent to St being a surface of revolution, with (O, e) as its axis, that is, r= r(t,z).
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
8/35
Lidstrm 215
Proof: One obtains xC
=
er
r+ er
r
and
xC
z= er
r
z+ e
and consequently
xC
xC
z =er
err
r
z +er
e r+ er e
r
= e r r
z + err+ er e
r
and then, taking H = xC
xCz
= ( r rz
)2 + r2 + ( r
)2 > 0,
n e =1
H
e r
r
z+ err+ er e
r
e =
1
Hr
r
z= 0
r
z= 0 r = r(, t)
proving the first part of the proposition. For the second part we consider
n e =1
He r r
z+ err+ er e
r
e =1
Her e r er r
and then
n e (xC xO) = n e (xC xO) =
1
H
er e r er
r
(err+ ez)
= 1
Hr
r
= 0
r
= 0 r = r(z, t)
and this proves the second part of the proposition.
Remark 4.1: Since St is part of a rigid body we may of course conclude that r does not depend on time, that
is, r= r() andr= r(z), respectively.
Rigid parts B andB are said to be rigidly connected, at time t T, if for some point O; x,,sysO (t) = 0and, (t, q(t)) = 0, meaning that the parts, at time t, move as one rigid body.
Proposition 4.2 If the rigid parts B andB are kinematically connected at time t T at three contact points,
not lying on the same straight line, then x,,sysO;k = 0 and
,;k = 0, k = 0,1, . . . , n and the parts are thus rigidly
connected at time t T.
Proof: We have, fork = 0,1, . . . , n, xk = x,sysO;k +
;k (x xO) and then
x,k = x
k x
k = x,sys
O;k x,sys
O;k + (;k
;k) (x xO) = x,,sys
O;k + ,;k (x xO) (25)
Let C,D andEdenote the contact points. Then (cf. Lidstrm [2], Lemma 7.2)
x,,sys
O;k + ,;k (xC xO) = 0, x
,,sys
O;k + ,;k (xD xO) = 0
x,,sys
O;k + ,;k (xE xO) = 0
and from this it follows that ,;k (xD xC) = 0 and
,;k (xExC) = 0. However, since (xD xC)(xExC)
= 0 it may be concluded that ,;k = 0 and subsequently x
,,sys
O;k = 0. The parts are then rigidly connected since
x,,sysO =
nk=0
x,O;kq
k = 0 and, =
nk=0
,;k q
k = 0.
The interaction between rigid parts in mechanical contact is given by the following proposition.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
9/35
216 Mathematics and Mechanics of Solids 17(3)
Proposition 4.3 IfB andB are in mechanical contact over the surface St , at time t T, then
I
k = x,,sys
O;k +
,;k
O , k = 0, 1, . . . , n
where
= St
tda(x) and
O = St
(x xO) tda(x)
are the force and moment sums of the contact forces acting from B on B.
Proof: From (19) and (25) one obtains
I
k =
S
t
x,k t
da(x) =
S
t
(x,,sysO;k +
,;k (x xO)) t
da(x)
= x,,sys
O;k
S
t
t da(x) + ,;k
S
t
(x xO) tda(x)
which proves the proposition.
Remark 4.2: Consider two rigid parts B andB , which are in mechanical contact. Let the contact interaction
be represented by the force sum f and the moment sum MO of the contact forces acting on part B from part
B . The reaction from B on B is given by f andM
O , respectively, where f = f andM
O = M
O .
The contribution to the generalized force, Qick , from this contact may be written as
Q
k = x,sysO;k f
+ ;k M
O , k = 1, . . . , n
where
f =
S
t
tda(x) andM
O =
S
t
(x xO) tda(x)
and then
I
k = x,,sysO;k f
+ ,;k M
O , k = 0,1, . . . , n (26)
A more systematic analysis of the constraint conditions will now follow. Assume that there is a unit direc-tion, c = c(t), such that c x
,,sys
O = 0, that is, the component, in direction c, of the relative system velocity atO is equal to zero. This expresses a bilateral kinematical constraint on the motion of the multibody equivalent
ton
k=0
g
k (t, q)qk = 0 (27)
where g
k (t, q) = c(t) x,,sysO;k (t, q), k = 0,1, . . . , n. Similarly
n
k=0h
k (t, q)qk = 0 (28)
where h
k (t, q) = c(t) ,;k (t, q), k = 0,1, . . . , n, expresses the kinematical constraint that the relative angular
velocity has zero component in the direction defined by c.
As an example, assume that the moving point O E, with place xO = xO(t), is kinematically connected to
both B andB , that is, xO = x,sysO = x
,sysO , t T. Note that O need not be a material point in any of the
two parts. Then we have the constraint
x,,sysO = 0, t T (29)
This is equivalent to the conditions
n
k=0g
ik (t, q)qk = 0, t T, i = 1, 2, 3 (30)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
10/35
Lidstrm 217
BB
Figure 2. Point contact.
where g
ik (t, q) = ci(t) x,O;k(t, q) andc = (c1 c2 c3) is a set of basis vectors in V . However, (29) may also be
written
0 =d
dt
(xA(t, q) +R(t, q)(XO X
A ) x
B(t, q) R (t, q)(X
O X
B )) (31)
where XO = X
A + RT(t, q)(xO(t) x
A(t, q)) andX
O = X
B + RT(t, q)(xO(t) x
B(t, q)) are the places corre-
sponding to O in the reference placements andA andB are material points in B andB , respectively. Equation
(31) is equivalent to the condition
C(t, q) = 0, t T, q (32)
where
C(t, q) = xA(t, q) x
B(t, q) +R(t, q)(XO X
A ) R
(t, q)(X
O X
B ) a
= q (t, q;XO) q (t, q;XO) a
where a is a constant vector. Thus the kinematical constraints (30) are equivalent to the geometrical constraint(32).
As a contrast to this we consider two rigid parts B andB with a single contact point C on their boundary
surfaces. Assume that the boundary surfaces have a common tangent plane at the contact point. The relative
angular velocity may be written as
, = np + r ,
r n
= 0 (33)
where p is the pivoting component, r the rolling composant of the relative angular velocity and n
a
normal vector to (see Figure 2). The impenetrability condition, (21), may be expressed according to
n x,,sysC = n
x,,sysO + n
r (xC xO) 0 (34)
a condition that obviously does not involve the pivoting component of the relative angular velocity. The
constraint
x,,sys
C = 0 (35)
represents a situation where B is rolling and pivotingon B without slipping.
By introducing an orthonormal set of basis vectors c = (c1 c2 c3) in V where c3 = n the constraint
condition (34) is equivalent to the three scalar constraints ci x,,sys
C = 0, i = 1, 2, 3. This condition may bewritten as
n
k=0g
ik qk = 0, i = 1, 2, 3 (36)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
11/35
218 Mathematics and Mechanics of Solids 17(3)
where g
ik (t, q) = ci x,,sys
C;k . Note that the impenetrability condition (34) is equivalent to
nk=0
g
3k qk 0 (37)
Since the point C is rigidly connected neither toB
nor toB
it is, in this case, not possible to obtain anequivalent geometric constraint of the type (32). However, the third condition in (36):
c3 x,,sys
C =
nk=0
g
3k qk = 0, t T (38)
representing zero normal relative velocity between the parts, may be shown to be a equivalent to a condition of
the type
C(t, q(t)) = 0, t T (39)
where C: T R is a continuously differentiable scalar valued function. Equation (39) is said to represent
a holonomic constraint on the multibody ifC(t, q)
q= 01n. Now, if C(t, q0) = 0 and if for some 1
l n,C(t, q0)
ql= 0 then, in a neighbourhood of q0 , equation (39) may, as a consequence of the
implicit function theorem, be solved for ql expressed as a function of t, q1, . . . , ql1, ql+1, . . . , qn, that is, ql =f(t, q1, q2, . . . , ql1, ql+1, . . . , qn). The number of configuration coordinates in the equations of motion may then
be reduced by one. This is done by eliminating ql in the kinetic energy, T = T(t, q, q) and in the generalized
forces, Q = Q(t, q, q). This elimination procedure may however lead to equations of motion not optimal from anumerical point of view. From (39) it follows that
d
dtC(t, q(t)) =
nk=0
gkqk = 0 (40)
where gk = gk(t, q) = C(t, q)qk
, k = 0,1, . . . , n and a holonomic constraint, according to (39), thus implies
the kinematical constraint (40), relating the generalized velocity components qk.
A kinematical constraint according to (27) is said to be holonomic if it is equivalent to a condition on the
form (40), that is, precisely if there exists real valued functions
= (t, q) = 0 andG= G(t, q) (41)
such that
gk(t, q) = (t, q)G(t, q)
qk, k = 0,1, . . . , n (42)
This then implies, since = 0,n
k=0
gkqk =
nk=0
G
qkq
k = G= 0 G = 0 Cdef= G c = 0 where
c R is a constant. The function = (t, q) is called an integrating factor to the kinematical constraint. If itis not possible to find functions andG, satisfying (41) and (42), then the kinematical condition (27) is said
to be non-holonomic. There are general integrability conditions giving requirements for conditions of the type
(27) to represent a holonomic constraint. A necessary and sufficient condition is contained in the Frobenius
integrability theorem, see the Appendix. It is well known that the conditions
nk=0
g
ik qk = 0, i = 1, 2 in (36)
are in general non-holonomic.
We now turn to the statement and proof of the holonomic character of condition (38).
Proposition 4.4 The constraint (38) is holonomic.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
12/35
Lidstrm 219
Proof: Let C be the contact point with place xC Bt B
t and let NC be a neighbourhood in E of xC.
Consider the sub-surface st = Bt NC of the material boundary surface B
t . Let it be (locally) represented
by the equation h(t, q,x) = 0, that is, x st h(t, q,x) = 0 where the mapping h : T N R
is a continuously differentiable function. The corresponding sub-surface st = B
t NC is assumed to be
represented by the equation h (t, q,x) = 0. Obviously h (t, q,xC) = h (t, q,xC) = 0 and we may assume that
xh
(t, q,x) = 0, x NC andxh
(t, q,x) = 0, x NC. Note that with
h(t, q,x) =h (t, q,x)
H(t, q,x)where H(t, q,x) = |xh
(t, q,x)|
one obtains
x St xh(t, q,x) =
1
Hxh
h xH
H2=
xh
H
xh (t, q,x) = 1Thus, h andh may be chosen so that x st |xh
| = 1 andx st
xh = 1 and, since the surfaceshave a common tangent plane with normal vectorc3 at xC:
c3 = xh (t, q,xC) = xh
(t, q,xC) (43)
We haveh(t, q, (X, t)) = 0, X s0 (44)
where s0 is the referential placement of st , that is, s
t =
(s0 , t). Differentiating (44), with respect to time,
one obtains
th (t, q, (X, t)) + qh
(t, q, (X, t))q + xh(t, q, (X, t)) x(X, t) = 0, X s0
and by taking X = XC ,xC = q (t, q;XC) one obtains, using (43) and leaving out the arguments (t, q,xC):
th + qh
q + xh x (XC , t) = th
+ qh q + c3 x
(XC , t) = 0, XC s0
Similarly th + qh
q + c3 x (XC , t) = 0, XC s
0 and from this it follows
0 = c3 x,,sysC = (th + qh q (th + qh q)) = (th, (t, q) + qh, (t, q)q) =
d
dth, (t, q(t))
where h, (t, q) = h(t, q,xC) h (t, q,xC). This proves the proposition.
Constraint conditions, such as (27) and (28), will in general have to be accounted for as additional differen-
tial equations along with Lagranges equations. If, however, q-coordinates are suitably chosen, these equations
may be identically satisfied. A q-coordinate system is said to be compatible with the constraints (27) and (28)
ifn
k=0g
k (t, q)qk = 0 and
n
k=0h
k (t, q)qk = 0, (t, q) T , q Rn
that is, if the constraint conditions are identically satisfied for all (t, q) T , q Rn.
Proposition 4.5 The q-coordinate system is compatible with the kinematical constraints (27) and (28) if and
only if
g
k (t, q) = 0, h
k (t, q) = 0, (t, q) T , k = 0,1, . . . , n.
Proof: For constraint (27), if
nk=0
g
k (t, q)qk = 0, (t, q) T , q Rn1 then, by taking q = 0n1, we
conclude that g
0 (t, q) = 0, (t, q) T . Thus
n
k=1g
k (t, q)qk = 0, (t, q) T , q Rn1. Then, by
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
13/35
220 Mathematics and Mechanics of Solids 17(3)
taking qi = 1, 1 i n, qk = 0 k = i, we conclude that g
i (t, q) = 0, (t, q) R . The correspondingproof for constraint (28) is identical.
Given a direction, defined by the unit vectorc, we may decompose the interaction force and moment sums
according to
f = cf +f
andM
A = cM
O +M
O,
where
f = c f,f
= (c f) c and M
O = c M
O ,M
O, = (c M
O ) c
Note that c f
= 0 andc M
O, = 0.
We now have the following proposition.
Proposition 4.6 For a q-coordinate system compatible with the constraints given by (27) and (28)
I
k = x,,sysO;k f
+ ,;k M
O,, k = 0,1, . . . , n.
Proof: From (26)
Ik = x,,sysO;k f + ,;k MO = x,,sysO;k (cf +f ) + ,;k (cMO +MO,)
= x,,sys
O;k cf +
,;k cM
O + x,,sys
O;k f
+ ,;k M
O, = gk(t, q)f + hk(t, q)M
+ x,,sys
O;k f
+ ,;k M
O, = x,,sys
A;k f
+ ,;k M
O,
where we have used Proposition 4.5
A constraint condition is a geometrical or kinematical condition of the type given in (27). A constraint mech-
anism is a physical arrangement that, by giving rise to necessary constraint forces and moments (reactions),
guarantees that a certain constraint condition is maintained. It should be noted that a specific constraint con-
dition may be realized using different kinds of constraint mechanisms. The constraint mechanism maintaining
the constraints (27) and (28) is equilibratedif, for instance, f
= 0 andM
O, = 0, which means that the force
f
and moment M
O acting on B
from B
are both parallel to c, that is
f = cf andM
O = cM
O (45)
where f andM
O are scalar components. The mechanical interaction between B andB is said to be equili-
brated ifI
k = 0, k = 1, . . . , n for some system of configuration coordinates q = (q1, q2, . . . , qn) Rn,
cf. Lidstrm [2].
Corollary 4.1 For constraint conditions (27) and (28) compatible with the q-coordinates and with a constraint
mechanism satisfying (45) the interaction is equilibrated.
Proof: This is a direct consequence of Proposition 4.6 and (45).
We now look at a situation where the q-coordinates are not necessarily compatible with the constraints. Weintroduce an orthonormal set of basis vectors c = (c1 c2 c3), ci = ci(t) and define the functions
gik(t, q) = ci(t) x,,sys
A;k (t, q), hik(t, q) = ci(t) ,;k (t, q), i = 1,2,3, k = 1, . . . , n
We do not here make any specific assumption about the orthonormal basis c. It may, in a specific situation, be
rigidly attached to one of the parts B orB .
Proposition 4.7 The interaction may be written as
I
k =
3
i=1(g
ik f
i + h
ik M
i ), k = 0, 1, . . . , n
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
14/35
Lidstrm 221
where f
i = ci f and M
i = ci M
O , i = 1, 2, 3 are components of the constraint forces and momentsrelative to the basis c.
Proof: From I
k = x,,sys
O;k f +
,;k M
O , x,,sys
O;k =
3i=1
g
ik ci and,;k =
3i=1
h
ik ci one obtains
I
k =
3
i=1
g
ik ci
f +
3
i=1
h
ik ci
M
O
=
3i=1
(g
ik ci f + h
ik ci M
O ) =
3i=1
(g
ik f
i + h
ik M
i )
and this proves the proposition.
Remark 4.3: A coordinate system not compatible with the constraints may usually be obtained by performing
an efficient coordinate change q = q(t, q) with n > n.
The net powerof the mechanical interaction, Pnet = P
net(t, q, q), expendedover the contact surface between
parts B andB is defined by Pnet = P
+P , cf. Lidstrm [2], where
P =
S
t
x tda(x),P =
S
t
x t da(x), , = 1, . . . ,N, =
In Lidstrm [2], Proposition 7.3, it is demonstrated that
Pnet =
nk=0
I
k qk (46)
Proposition 4.8 The net powerP
net
is given by
Pnet =
3i=1
n
k=0
g
ik qk
f
i +
n
k=0
h
ik qk
M
i
Proof: Equation (46) in combination with Proposition 7 gives
Pnet =
nk=0
I
k qk =
nk=0
3
i=1
(g
ik f
i + h
ik M
i )
qk =
3i=1
n
k=0
g
ik qk
f
i +
n
k=0
h
ik qk
M
i
It is clear from Proposition 4.5 that if constraint forces and moments are wanted as a result of the analysis ofthe multibody dynamics, then a system of configuration coordinates non-compatible with the constraint should
be used. As an example we may consider the ideal revolute joint, see Figure 1. We assume that the axis of
revolution is (O, c3). Then the kinematical constraints are
x,,sysO = 0, c1
, = 0, c2 , = 0 (47)
This is equivalent to (we partially drop indices , whenever this is possible without endangering the
intelligibility)n
k=0gikq
k = 0, i = 1,2, 3 and
n
k=0hikq
k = 0, i = 1, 2 (48)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
15/35
222 Mathematics and Mechanics of Solids 17(3)
B
B
Figure 3. Screw joint.
Then from (47) Pnet = (
nk=0
h3kqk)M3 = 0, (t, q) T , q R
n1, since the joint is assumed to be ideal.
Thus, c3 , =
nk=0
h3kqk = 0 M3 = 0 and then
Ik =
3i=1
gikfi + h1kM1 + h2kM2
where we here have five a priori unknown reaction components: f1, f2, f3,M1,M2. If the configuration coor-
dinates are non-compatible with the constraints (48) then gik = gik(t, q) and hik = hik(t, q) will not all be
identically zero and constraint forces will appear in the interaction I
k . Note that in the general case five
unknowns f1,f2,f3,M1,M2 will appear in the equations of motion by the addition of the interaction. This is in
addition to the n unknown q-coordinates. The additional unknowns are balanced by the addition of the five
differential equations (48).
Next consider the ideal screw joint (see Figure 3). We assume that the axis of the screw is ( O, c3). The
kinematical constraints are
x
,,sys
O = c3
,
2 L,
,
= c3
,
,
,
R
(49)
where L is the lead of the screw.
Conditions (49) are equivalent to
nk=0
gikqk = 0, i = 1, 2,
nk=0
hikqk = 0, i = 1, 2 (50)
nk=0
g3kqk =
,
2L,
nk=0
h3kqk = ,
For an ideal screw joint it follows from Proposition 8 and (50)1,2 that
Pnet =
n
k=0
g3kqk
f3 +
n
k=0
h3kqk
M3 = 0, (t, q) T , q R
n1
which, according to (50)3,4, is equivalent to,
2Lf3 +
,M3 = 0. Then, if , = 0, this is equivalent to
L
2f3 +M3 = 0 and
I
k =
3
i=1(gikfi + hikMi) = g1kf1 + g2kf2 +
g3k h3k
L
2
f3 + h1kM1 + h2kM2
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
16/35
Lidstrm 223
and we have five a priori unknown reaction components: f1, f2, f3,M1 andM2. Note that by eliminating , in
(50) one gets five constraint equations.
Next we consider a revolute joint with friction present. We assume that the axis of revolution is (O, c3). The
kinematical constraints are
x,,sysO = 0, c1
, = 0, c2 , = 0
c3
, = 0 if |M3
| M3,c
(51)
where M3,c = M3,c(f1, f2, f3,M1,M2) is a given function determined by the properties of the contact surfaces ofthe revolute joint. In this case we have six a priori unknown reaction force components: f1, f2, f3,M1,M2,M3and
I
k =
3i=1
(gikfi + hikMi), k = 1, . . . , n (52)
If, however, it turns out that |M3| > M3,c then one has to assume that the constraint c3 , = 0 is no longer
valid and one may then take
M3 = M3,r(,3 ;f1, f2, f3,M1,M2) (53)
as a constitutive assumption for the resistance to the motion, where
,
3 = c3
,
. The condition on the netpower is then given by
Pnet = (
nk=0
h3kqk)M3 =
,3 M3 0, (t, q) T , q R
n1
and the constitutive function M3,r must then satisfy
M3,r(,3 ;f1,f2,f3,M1,M2) =
< 0 if
,3 > 0
> 0 if ,3 < 0
One way to accomplish this is to assume that
M3,r(,3 ;f1,f2,f3,M1,M2) < 0 if
,3 > 0
M3,r(,3 ;f1,f2,f3,M1,M2) = M3,r(
,3 ;f1,f2,f3,M1,M2)
In this case there are five a priori unknown reaction components: f1,f2,f3,M1,M2 and
I
k =
3i=1
gikf+ h1kM1 + h2kM2 + h3kM3,r
Now consider the case of a geometrical point contact C between two rigid parts B andB with the constraint
condition given by
ci x,,sysC =
nk=0
g
ik qk = 0, i = 1, 2 (54)
where c = (c1 c2 c3) is an orthonormal basis set with c3 = n. If there is no rolling or pivoting resistance,
that is, ifM
C = 0, we have the interaction I
k =
3i=1
gikfi. In this case there are three a priori unknown reaction
force components: f1, f2 and f3. For a unilateral constraint of Coulomb type we have the following additional
conditions
f3 0, f2
1 +f2
2 sta |f3| (55)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
17/35
224 Mathematics and Mechanics of Solids 17(3)
where sta 0 is the coefficient of static friction. The net power of the mechanical interaction may be writtenas
Pnet =
3i=1
((
nk=1
gikqk)fi + (
nk=1
hikqk)Mi) = (
nk=1
g3kqk)f3 (56)
where we have used the fact that M1 = M2 = M3 = 0 and the constraint conditions (54).
From (37), (55)1 and (56) it follows that Pnet 0. However, according to Lidstrm [2], Equation (82),
Pnet 0 and then it follows that P
net = 0. From this we conclude that
f3 < 0
nk=1
g3kqk = 0 and
nk=1
g3kqk < 0 f3 = 0
Thus if the parts are mechanically in contact then they are kinematically connected and if they are in geometrical
contact and separating then they are not in mechanical contact. If it turns out that
f21 +f
22 > sta |f3|, then
one has to assume that the constraint (54) is no longer valid and that x,C, = x
,C c3(c3 x
,C ) = 0. Following
Coulomb, one may then take
f =kin x,C,x,C, + c3
f3where kin 0 is the coefficient of kinetic friction.
5. The equations of motion involving kinematical constraints
The distinction between constraints on the motion on the one hand and resistance to the motion on the other is
fundamental in multibody dynamics. Constraints on the motion are maintained by reactions, that is, constraint
forces and moments that are a priori unknown. They may be restricted by inequalities as, for instance, in the
case of contact forces at a unilateral contact between parts. Forces and moments responsible for the resistance
to the motion have to be prescribedconstitutively.
Consider a multibody with a motion described by (5) and with the following m, (1 m n) prescribed
bilateral kinematical constraints:
g0 +
nk=1
gkqk = 0, = 1, . . . , m (57)
wheregk = gk(t, q), = 1, . . . , m, k = 0,1, . . . , n. As was demonstrated in the previous section, constraintsinvolving rigid parts are of this type. A holonomic constraint according to (39) is assumed to be represented by
its kinematical counterpart, that is:
C
t+
n
k=1C
qkqk = 0
It should be noticed that, within the description adopted here, using a finite number of configuration coordinates,
these constraints will also appear in connection with some constraints connecting rigid and elastic parts. If the
parts B andB are geometrically connected over the surface St during the time interval T
q (t, q(t);XC ) = q (t, q(t);XC ), XC S
0,t,XC S
0,t, t T
where S0,t andS
0,t are the surfaces in contact in their reference placements, then the corresponding kinematical
constraint is written as
n
k=0( q,k(t, q(t);XC )
q,k(t, q(t);XC ))qk(t) = 0, XC S
0,t,XC S
0,t, t T (58)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
18/35
Lidstrm 225
Now, if a discretization in the sense of the finite element method is employed and a number of so-called nodal
points are introduced on the contact surfaces:
Z1C , . . . , ZnnodeC
S0,t and
Z1C , . . . , Z
nnodeC
S
0,t
where nnode is the number of nodes, then (58) is replaced by a finite number of constraints according to
n
k=0
( q,k(t, q(t); Z
C
)
q,k(t, q(t); Z
C
))qk(t) = 0, = 1, . . . , nnode
Remark 5.1: In (3)(5) it is assumed that (t, q, q) T Rn. The constraint (57) implies that the generalized
velocity q at (t, q) T is required to belong to a hyper-plane in Rn. The discussion in this paper is limited tobilateral constraints. The introduction of unilateral constraints will increase the complexity of the mathematical
model, since then constraints may become inactive and active as a result of a dynamics which thus has to
involve impacts. A discussion of the equations of motion under unilateral constraints is presented by Ltstedt
[6] and Ballard [7].
If the constraint equations (57) are independent in the sense that the constraint matrix, g = (gk) Rm(n+1),
satisfies rankg = m then g is said to have full rank. The constraint matrix may be written g =
g0 g
Rm(n+1) where
g0 =
g10...gm0
Rm1 andg = g11 g1n... . . . ...
gm1 gmn
Rmnand then (57) obtains the compact format
gq = g0 +gq = 0m1 (59)
The constraint (59) is said to be homogeneous (in the generalized velocities) ifg0 = g0(t, q) = 0m1, (t, q) T. The constraint conditions in (59) may be given an equivalent formulation using other constraint matrices.
Let A = A(t, q) Rmm be a non-singular matrix and define 0 = Ag0, = Ag. Then =
0
has full
rank if and only if g = g0 g has full rank. Furthermore 0 + q = 0m1 g0 +gq = 0m1.Remark 5.2: Consider two rigid parts B andB , which are constrained to have the same velocity at three
contact points A, B and C, not lying on a straight line, that is, x,,sys
A = x,,sys
B = x,,sys
C = 0. Then, if D is
a fourth contact point, the constraint x,,sys
D = 0 will, according to Proposition 4.2, not be independent of theprevious ones. This kind of redundancy, giving rise to a constraint matrix not having full rank, may usually be
eliminated from multibody systems. The presence of redundant constraints in multibody systems gives rise to
mathematical complication. A treatment of this case is presented by Ltstedt [6].
The equations of motion for the multibody now read
qTM2 = Q
cif + Qi + Qc + Qb
g0 +gq = 0m1(60)
where Qcif andQi are defined by (9) and (12), respectively. The generalized contact force is given by the sum
of the internal and external contact forces, that is:
Qc = Qic + Qec (61)
where Qec represents the contact force from the exterior of B. The generalized internal contact force is,
according to Proposition 4.7, given by
Qick =
N, = 1 <
3i=1
(g
ik f
ik + h
ik M
ik ), k = 1, . . . , n (62)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
19/35
226 Mathematics and Mechanics of Solids 17(3)
This may be written
Qick = Qkc + Qrick , k = 1, . . . , n (63)
where
Qkck =
m
=1F k, k = 1, . . . , n (64)
is the generalized contact force due to the kinematical constraints, F , = 1, . . . , m are the componentsof forces and moments responsible for the maintenance of the constraints (57) and Rmn is the
constraint matrix extracted from (62). Thus Qric represents that part of Qic not containing the constraint
forces. The generalized force Qric is called the residual internal contact force. With the matrix notation
F =
F1 F2 . . . FmT
Rm1 (64) may be written as
Qkc = FT (65)
and the equations of motion then read
qTM2 = Q
cif + Qi +FT + Qic + Qec + Qb
g0 +gq = 0m1(66)
The linear space of virtual velocity fields on B is given by
Wq =
w : B0 V| w = wq(t, q, w;X) =
nk=1
q (t, q;X)
qkwk, X B0 , w
k R
cf. Lidstrm [2].
We introduce the following space of constrained virtual velocity fields:
Wcq =w : B0 V| w Wq, X B
0 , gw = 0m1, w =
w1 . . . wn
T
Rn1
(67)
This is a linear subspace ofWq.
Lemma 5.1 If = Agwhere A = A(t, q) Rmm is non-singular then
Qicw = Qricw, w Rn
satisfying gw = 0m1
Proof: Ifw Rn and gw = 0m1 then, according to (63) and (65):
Qicw = (FT + Qric)w = (FTAg+ Qric)w = Qricw
We now retake the derivation of Lagranges equations presented in Lidstrm [2]. In the application of
the principal of virtual power the space of virtual velocities, Wcq , satisfying the constraint conditions, is nowadopted. The equations obtained do not contain anything new compared to (66). In fact these sets of equations
will turn out to be identical if the constraint matrix has full rank. The derivation will, however, lead to an
automatic inclusion of the appropriate reaction forces. Remarkably enough, these reactions are now obtained
without an explicit analysis of the interactions. The proof of the following theorem originates with dAlembert
[8] and Lagrange [9] and versions of it have appeared in, for instance, Hertz [10], Pars [11], Lanczos [12] and
Neimark and Fufaev [13].
Theorem 5.1 (LagrangedAlembert Equations of Motion)
qTM2 = Q
sum
g0 +gq = 0m1(68)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
20/35
Lidstrm 227
The sum of generalized forces on the multibody, Qsum, is given by
Qsum = Qres + (69)
where
Qres = Qcif + Qi + Qec + Qric + Qb (70)
is called the residual generalized force. Furthermore
k =
m=1
gk, k = 1, . . . , n (71)
where 1, . . . , m is a set of real variables, i = i(t).
Proof: Starting from the principle of virtual power one obtains the following condition on the generalized forces
Qa, Qi, Qc andQb, cf. Lidstrm [2]:
w = 0, w Rn
where =Qa Qi Qic Qec Qb. Let =Qa Qi Qric Qec Qb then from Lemma 5.1 it follows that
w = w (72)
for all w Rn1 satisfying gw = 0m1. Let 1, . . . , m be a set of real variables. Then, from (72), it follows that
0 = w Tgw = ( Tg)w =
nk=1
(k
m=1
gk)wk (73)
if gw = 0m1. Now, since 1 m n, n m of the components in w may be arbitrarily and independently
chosen. If 1 m < n let us take these to be wm+1, . . . , wn. We then consider the equations
k =
m
=1
gk, k = 1, . . . , m (74)
This is a linear system of m equations in the m unknowns 1, . . . , m. Since the kinematical constraints are
assumed to be independent the sub-matrix g = (gk), , k = 1, . . . , m is non-singular. Equations (74) will thenhave a unique solution = (t, q, q, q), = 1, . . . , m. This inserted into (73) gives the equation
nk=m+1
(k
m=1
gk)wk = 0 (75)
However now, since wm+1, . . . , wn may be independently chosen and (75) is satisfied for all wm+1, . . . , wn, we
have to conclude that
k
m
=1
gk = 0, k = m + 1, . . . , n (76)
These conditions, together with those in (74), may be summarized as
k
m=1
gk = 0, k = 1, . . . , n (77)
Ifm = n then (74) reads
k =
n=1
gk, k = 1, . . . , n
and this proves the theorem.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
21/35
228 Mathematics and Mechanics of Solids 17(3)
The n + m ordinary differential equations in (68) contain n + m unknowns: q1, . . . , qn, 1, . . . , m consideredas functions of time. The functions = (t) are usually calledLagrangian multipliers and the relation (71)
may be written as
= Tg (78)
In the following proposition it is demonstrated that the Lagrangian multipliers are equal to the components of
forces and moments responsible for the maintenance of the constraints (57). From (66) 1 and (68)1 it followsthat
= Qkc (79)
Proposition 5.1 If = Agthen = ATF.
Proof: Using (65), (78) and (79) it follows that FT = Tg (ATF )Tg = 01n = ATF since
rank(g) = m.
The physical interpretation of the multipliers, as components of constraint forces and moments, is thus,
according to the previous proposition, closely related to the actual mathematical representation of the physical
constraint condition. This fact is clearly underlined by the discussion given in Section 4. In many practical
situations it might be that A = diag 1 . . . n , where k = 1 or + 1, k = 1, . . . , n corresponding to apossible change in sign gk gk for some of the values: = 1, . . . , m. Then, for the corresponding
,one has = F .
Remark 5.3: Note that for a holonomic constraint, according to (39), the corresponding constraint force
component is equal to C
qk.
Remark 5.4: The constraint conditions given by (59) are affine in the generalized velocity components. As was
demonstrated in Section 4, these types of constraints naturally arise from assumptions concerning the physical
character of the mechanical interaction between the parts of a multibody. Theoretically there is a possibility of
considering more general constraints of the type g(t, q, q) = 0m1 where g : T Rn Rm1 is some
given non-linear function of q. However, these so-called non-linear constraints do not seem to play a major
role in the modelling of multibody systems, although there may be some applications to the control theory ofsuch systems. Furthermore, in the non-linear case, the principle of virtual power cannot be used to set up the
equations of motion. In the perspective of its fundamental status this seems to be a major drawback. There is a
vast literature dealing with non-linear constraints, see for instance Neimark and Fufaev [13], Papastavridis [14]
and Udwaida and Kalaba [15]. In these papers the principle of virtual power is replaced by some other principle
that does not seem to be equally well founded. In Weber [16] the problem is discussed using the canonical
formalism and this approach appears to avoid some of the obscurities.
The equations of motion may be written asqTM2 Q
res Tg = 01ng0 +gq = 0m1
(80)
The initial conditionsq(0) = q0, q(0) = q0 (81)
have to satisfy the constraint conditions, that is:
g0(0, q0) +g(0, q0) q0 = 0m1 (82)
Remark 5.5: If we write
Qres = Qcif + Qcon + Qnon (83)
where Qcon is a conservative force with potential V = V(t, q), that is:
Qcon = V
q(84)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
22/35
Lidstrm 229
then (80) is conventionally written
d
dt
L
q
L
q Qnon Tg = 01n
g0 +gq = 0m1
(85)
where L = L(t, q, q) = T(t, q, q) V(t, q) is the Lagrangian andQnon is the non-conservative generalized forcecontaining the non-conservative parts of Qres. If the specific body force is given by the acceleration of gravity
b(X, t) = g, where gis a constant vector, then V = V(t, q) will contain the sum
N=1
(Ve + V
g ), where Ve is the
elastic energy according to (13) and Vg = (xc xO) gm
is the potential energy in the gravitational field,
where xc is the place of the centre of mass of part B in its present placement:
xc = xc (t, q) = xO +
1
m
B
0
(q(t, q;X) xO)0 (X) dv(X))
Note that ifB is rigid then one may take Ve = 0.
For a conservative multibody system, that is, Qnon
= 01n, equation (85)1 may be derived using the so-calledLagrangedAlembert principle. Given a mechanical system defined by the Lagrangian L = L(t, q, q) and theconstraint conditions g0 +gq = 0m1, where rank(g) = m . Let t1, t2 R, t1 < t2 be two fixed points in time
and let q1, q2 Rn1 be two fixed points in q-space. Introduce the space of trajectories in q-space with fixed
end-points and satisfying the constraints, that is:
U =
q C1([t1, t2]) q(t1) = q1, q(t2) = q2, g0(t, q(t)) +g(t, q(t)) q(t) = 0m1, t [t1, t2]
and, forq U, the space of variations:
W(q) =
w C1([t1, t2])
w(t1) = w(t2) = 0n1, g(t, q(t))w(t) = 0m1, t [t1, t2]
Note that W(q) is a linear subspace of C
1
([t1, t2]). The mapping I : U R, defined by
I(q) =
t2t1
L(t, q, q)dt
is calledthe action of the mechanical system. Then (see Rund [17])
I(q + w) I(q) = I(q, w) +R(q, w)
where the variation of the action, I, is given by
I(q, w) = t2
t1
L
q(t, q(t), q(t))
d
dt
L
q(t, q(t), q(t))w(t)dt, q U, w W(q)
and
R(q, w)|w|
0, |w| 0.
Theorem 5.2 (LagrangedAlembert principle). The trajectory q U satisfies
I(q, w) = 0, w W(q)
if and only if there is a continuous function : [t1, t2] Rm1 such that
d
dt
L
q(t, q(t), q(t))
L
q(t, q(t), q(t)) = (t)Tg(t, q(t)) (86)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
23/35
230 Mathematics and Mechanics of Solids 17(3)
For the proof of this theorem we need the following lemma.
Lemma 5.2 Let a : [t1, t2] R1n andb : [t1, t2] R
mn be continuous functions.
Assume that rank(b(t)) = m, t [t1, t2] and that
t2t1
a(t)w(t)dt = 0, w : [t1, t2] R1n
such that b(t)w(t) = 0m1, t [t1, t2] (87)
Then there exists a continuous function : [t1, t2] Rm1 such that a(t) + (t)Tb(t) = 01n, t [t1, t2].
Proof: We prove this lemma by contradiction. Assume that there is a point t0 [t1, t2] such that
a(t0) + (t0)Tb(t0) = 01n
Then there is a vector k0 Rn1 such that b(t0)k0 = 0m1, but a(t0)k0 = 0. Let us assume that a(t0)k0 > 0.
Introduce the matrix B(t) = b(t)b(t)T Rmm. Since rank(b) = m it follows that B is symmetric and positivedefinite and thus non-singular. Define
w(t) = k0 b(t)TB(t)1b(t)k0
This is a continuous function equal to k0 at t0 and
b(t)w(t) = b(t)k0 b(t)b(t)TB(t)1b(t)k0 = b(t)k0 b(t)k0 = 0m1
The same holds for (t)w(t), where : [t1, t2] R is a continuous function. Since a(t0)k0 > 0 anda and w
are continuous, we have a(t)w(t) > 0 for all t close to t0. If we choose (t) 0 such that (t) = 1 when t isclose to t0 anda(t)w(t) > 0 where (t) > 0 we have
t2t1
a(t)(t)w(t)dt > 0
This is a contradiction, so for all t [t1, t2] there is a vector (t) such that a(t) + (t)Tb(t) = 01n. That is
continuous follows from
a(t)b(t)T = (t)Tb(t)b(t)T = (t)TB(t) (t) = B(t)1b(t)a(t)T
This proves the lemma.
Proof of Theorem 5.2: Put
a(t) =L
q
(t, q(t), q(t)) d
dt
L
q
(t, q(t), q(t)) andb(t) = g(t, q(t))
then from I(q, w) = 0, w W(q) it follows that (87) is satisfied and consequently (86) is valid. On the otherhand, from (86) it follows that
I(q, w) =
t2t1
L
q(t, q(t), q(t))
d
dt
L
q(t, q(t), q(t))
w(t)dt =
t2t1
(t)Tg(t, q(t))w(t)dt = 0
w W(q) and this proves the theorem.
Remark 5.6: For a conservative mechanical system Theorem 5.2 demonstrates that the LagrangedAlembert
principle produces equations of motion equal to those derived by the principle of virtual power. Note, however,
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
24/35
Lidstrm 231
that the incorporation of non-conservative forces into the LagrangedAlembert principle is not straightfor-
ward. These forces usually have to be added afterwards. This seems to be a major drawback for the Lagrange
dAlembert principle as compared to the principle of virtual power. The formulation of a LagrangedAlembert
principle in this context is non-trivial and will usually result in equations of motion different from the ones
derived from the principle of virtual work. See Bloch et al. [1] and Rund [17] for a discussion on the history
and validity of various variational principles.
A change of configuration coordinates:
q = (q1, q2, . . . , qn) Rn q = (q1, q
2, . . . , q
n
) Rn
is given by a continuously differentiable mapping q : T
q = q(t, q), (t, q) T (88)
Obviously, the following relationship between the generalized velocities, q andq, holds:
q =q
t+
q
qq (89)
Then we have for the constraint condition (59):
0m1 = g0 +gq = g0 +g(q
t+
q
qq) = g
0 +g
q (90)
where
g0(t, q) = g0(t, q(t, q)) +g(t, q(t, q)) q(t, q)
t Rm1 (91)
g(t, q) = g(t, q(t, q)) q(t, q)
q Rmn
The change of configuration coordinates (88) is said to be efficientif rank( q
q) = min(n, n), cf. Lidstrm [2].
Proposition 5.2 If a specific q-coordinate system is compatible with the constraint (27) then any q-coordinatesystem related to q by (88) will be compatible with the constraints (27). On the other hand if the q-coordinate
system is compatible with the constraint, n n, and if the coordinate change (88) is efficient then theq-coordinate system is compatible with the constraint.
Proof: From g0 = 0m1 andg = 0mn it follows, from (91), that g0 = 0m1 andg
= 0mn . This proves thefirst part. The second part follows from
g(t, q)q(t, q)
q= g(t, q) = 0mn g(t, q) = 0mn
since rank( q
q) = n.
Proposition 5.3 Let the coordinate change (88) be efficient with n n and let the constraint matrix g havefull rank, then g has full rank.
Proof: Ifu Rm1 then uTg = 01n uTg
q
q= 01n u
Tg = 01n u = 0m1.
In the next proposition we demonstrate that the Lagrangian multipliers are invariant under efficient
coordinate changes.
Proposition 5.4 Let the coordinate change (88) be efficient with n n and let the constraint matrix g havefull rank. Then the Lagrangian multipliers are invariant under this change of configuration coordinates, that is:
(t, q, q) = (t, q(t, q),q(t, q)
t+
q(t, q)
qq)
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
25/35
232 Mathematics and Mechanics of Solids 17(3)
Proof: We have the transformation properties Qic = Qic qq
, Qric = Qric qq
. Using (62) one may then conclude
that
Qkc = Qic Qric = (Qic Qric) q
q= Qkc
q
q Tg = Tg
q
q(92)
Thus, according to (92)2, and the assumption that g has full rank:
(Tg Tg) q
q= 01n (
T T)g = 01n = 0m1
and this proves the proposition.
The set of linearly independent kinematical constraints in (57) is said to be holonomic if there are functions
: T R, , = 1, . . . , m with the matrix (t, q) = ( (t, q)) Rmm being non-singular and
functions G : T R, = 1, . . . , m such that
n
k=0 gk(t, q)qk =
m
=1 (t, q)G(t, q, q), = 1, . . . , m, (t, q) T , q Rn (93)
where G(t, q, q) =
nk=0
G(t, q)
qkqk; otherwise they are non-holonomic. The following result is well known
Theorem 5.3 Take a set of linearly independent holonomic kinematical constraints, according to (57). Then
there exists an efficient change of configuration coordinates to a q-coordinate system compatible with these
constraints.
Proof: If the constraints in (57) are holonomic then according to (93)
n
k=0
gk
qk = 0 G = 0 G = c , = 1, . . . , m
where c are constants. From (93) it follows that
gk(t, q) =
m=1
(t, q)G(t, q)
qk, = 1, . . . , m, k = 1, . . . , n, (t, q) T
This implies that rank
G(t, q)
q
= m where G(t, q) =
G1(t, q) c1 . . . Gm(t, q) cm
Rm. Introduce
the mapping F : R1+n T r s Rm defined by F(t, q) = G(t, q) where q = (r,s), r r
R
nm
, s s
Rm
is a re-arrangement of the q-coordinates so that the matrixG
s
Rmm
is non-
singular. From the implicit function theorem it follows that there is a differentiable mapping f : R1+nm T r Rm such that F(t, (r,f(t, r))) = 0. Put n = n m and introduce the coordinate transformationq = q(t, q), q
r Rn
defined by r = q, s = f(t, q). It then follows that the constraint conditionsG = c, = 1, . . . , m are identically satisfiedq
. This coordinate change is efficient since
q(t, q)
q=
Inn
( Fs
)1(q,f(q))Fr
(q,f(q))
Rnn
and consequently rank
q(t, q)
q = n = n m = min(n, n).
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
26/35
Lidstrm 233
Remark 5.7: A necessary and sufficient condition for the constraint (57) to be holonomic is given by the
Frobenius integrability theorem, see the Appendix.
Remark 5.8: If the constraint conditions in (57) are organized in such a way that the first h; 0 h m
constraints are holonomic and the following m h constraints are non-holonomic, then h is an invariant underregular coordinate changes.
Remark 5.9: Since it is not possible to choose a q-coordinate system that is compatible with a non-holonomic
kinematical constraint, the corresponding reaction components will always have to appear in the mechanical
interaction.
Remark 5.10: The q-coordinate system in Theorem 5.3 is called a minimal coordinate system for the multibody.
The numbern = n m is called the number of degrees of freedom for the multibody.
6. Elimination and calculation of the constraint reactions
First consider the case of holonomic constraints and a change of configuration coordinates, q = q(t, q),
where the starred configuration coordinates q are compatible with the constraints. Then, according toProposition 4.5
g0 = 0m1 and g = 0mn (94)
From Lidstrm [2], Proposition 6.3 and Lemma 6.1, it follows that
M2 = ( q
q)TM2
q
qandQi + Qc + Qb = (Qi + Qc + Qb)
q
q(95)
Proposition 6.1. For a coordinate change, leading to configuration coordinates q that are compatible with the
constraints, the equations of motion may be written as
qTM2 Q
res = 01n (96)
where Qres = Qcif + Qi + Qc + Qb.
Proof: By introducing the coordinate transformation q = q(t, q) into (80) and using (95), the equations ofmotion in the starred coordinates read
qT M2 Q
res Tg = 01n
g0 +g q = 0m1
which, according to (94), is equivalent to (96).
Remark 6.1: Note that if the coordinate change is efficient, n n and M2 is non-singular then M2 is non-
singular, cf. Lidstrm [2].
Remark 6.2: It follows that Qres = Qresq
q qT
d
dt
q
q
TM2
q
q.
In the differential equation (96), the multipliers have been eliminated due to a special choice of coordinates.
The initial conditions, accompanying (96), are given by
q(0) = q0, q(0) = q0 (97)
where q0 andq0 are determined from the equations q0 = q(t, q0) andq0 = q(0, q0)
t+
q(0, q0)
qq0 andq0
andq0 satisfy (82).
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
27/35
234 Mathematics and Mechanics of Solids 17(3)
Below an explicit expression for the multipliers = (t, q, q) will be given. With this expression, and ifq = q(t) is solved for in (96), (97), we may calculate the constraint forces and moments from
= = (t, q(t, q(t)), q(t, q(t))
t+
q(t, q(t))
qq(t))
We will now look at a method where the multipliers are eliminated from the differential equations but never-theless retained as a result of the procedure. A reformulation of the equations of motion (96) and (97) may be
found using the following scheme: Firstly, take the time derivative of the constraint conditions (57), obtaining
nk=l
gvkqk = (t, q, q), = 1, . . . , m (98)
Then, solve forq in (80)1 as a function oft, q, q and, that is:
qk = k(t, q, q, ), k = 1, . . . , n (99)
This is always possible if the q-coordinate system is regular. Note that the dependence of qk on , in (99), is
linear. By substituting (99) into (98) one obtains a system of linear equations in , which can be solved if the
constraint equations are supposed to be independent. We then get
= (t, q, q) (100)
Inserting this expression for into (80)1 one obtains a set of differential equations:
qk = fk(t, q, q)
Solving these equations for q, under the given initial conditions (81), results in the motion q = q(t) and thisinserted into (100) will give the Lagrangian multipliers = (t). The method sketched above may always, at
least in principle, and under the assumptions made, be used to eliminate the Lagrangian multipliers. We now
formalize the discussion above starting with the following lemma.
Lemma 6.1 For the function = (t, q, q) in (98) we have
(t, q, q) = gv0
t
nl=1
gv0
qlql
nk=1
gvk
tqk
nk,l=1
gvk
qlqkql =
nk,l=0
gvk
qlqkql (101)
Proof: The constraint conditions (57) imply ( = 1, . . . , m)
nk=0
(gkqk +gkq
k) = 0 =
nk=0
gkqk =
nk=0
gkqk (102)
However, gk =
n
l=0
gk
ql
ql and this inserted into (102) proves the lemma.
Remark 6.3: In matrix format, (101) may be written as
= g0 gq = g0
t
g0
qq
g
tq
g
qq
q =
g0
qq
g
qq
q =
g
qq
q
where =
1 2 . . . mT
Rm1. For the matrix notation used here we refer to Lidstrm [2],Appendix A.1.
Lemma 6.2 IfM2 is non-singular and ifghas full rank then the matrix = g M12 g
T Rmm is symmetric and
non-singular. Furthermore, is invariant under a regular coordinate change, that is, (t, q) = (t, q(t, q)).
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
28/35
Lidstrm 235
Proof: The symmetry is obvious and the non-singularity follows from the following argument. If u Rm thenuT u = uTg M12 g
T u = (gT u)TM12 gT u > 0 gT u = 0 u = 0m1 since the constraint matrix g has full
rank andM12 is positive definite. The invariance of is demonstrated by
= gM12 gT = g
q
q q
q
T
M2 q
q
1
g qq
T
= g q
q
q
q
1M12
q
q
T q
q
TgT = gM12 g
T =
and this proves the lemma.
We may now formulate the main theorem of this section. In what follows we assume that det M2 = 0 and
rank(g) = m.
Theorem 6.1 (The initial value problem; reformulation A). The initial value problem (80)(82) is equivalent to
M2 q Qres Tc +g
T1 (g
qq) q = 0n1
qk
(0) = qk0, q
k
(0) = qk0
g0(0, q0) +
nk=1
gk(0, q0)qk0 = 0, = 1, . . . , m
(103)
where
Qresc = QresRT,R = Inn P and P= g
T1g M12 Rnn
Qresc is called the constrained residual generalized force. The matrices R and P are projections, that is, R2 =
P2 = Inn andP = 0nn. Furthermore, range(gT) is a linear subspace of range(P). The Lagrangian multipliers
are given by
=
1g M12 Q
resT + 1
g
qq
q
(104)
Remark 6.4: A similar result, but of less generality, has appeared in Hemami and Weimer [18] and Ltstedt [6].
Proof: From (80)1 it follows, after transposing M2 q QresT gT = 0n1. Since M2 is non-singular this is
equivalent to q = M12 QresT +M12 g
T. However, according to (98), one then obtains = gq = g M12 QresT +
g M12 gT = g M12 Q
resT + and, since according to Lemma 6.2 the matrix is non-singular, we may solvefor the Lagrangian multipliers:
= 1g M12 QresT + 1 = 1g M12 Q
resT 1 ( g
qq) q
and then eliminate the Lagrangian multipliers from the equations of motion thus obtaining
M2 q (Inn gT1g M12 )Q
resT +gT1 ( gq
q) q = 0n1
orM2 q RQresT +gT1
g
qq
q = 0n1 where
R = Inn P,P= gT1g M12 R
nn (105)
P is a projection since P2 = gT1g M12 gT1g M12 = g
T11g M12 = gT1g M12 = P where we
have used Lemma 6.2, and then R is a projection as well. Furthermore, if u range(gT) then u = gTv, v R
m1 and
Pu = PgTv = gT1g M12 gTv = gT1v = gTv = u
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
29/35
236 Mathematics and Mechanics of Solids 17(3)
and consequently, u range(P). Since rank(gT) = m 1 it is concluded that P = 0nn. Now, to prove theconverse, we just have to show that ifq = q(t) satisfies (103) then (80)2 is fulfilled. From (103)1 it follows that
gq = gM12 Qres Tc +gM
12 g
T1 = gM12 RQresT + .
From (105) one finds that gT1g M12 R = PR = 0nn and this implies g M12 R = 0mn, since rank(g) = m.
Consequently
gq = gq + gq + g0 = 0m1 ddt
(g0 +gq) = 0m1 g0 +gq = 0m1
where we have invoked (103)3. This proves the theorem.
Remark 6.5: It follows from standard theorems on ordinary differential equations that for sufficient regularity
of the participating functions in (103)1 and the assumption that M2 is non-singular and g has full rank, the
initial value problem defined by (103) has a unique solution in a neighbourhood of t = 0, cf. Coddington and
Levinson [19].
Remark 6.6: A direct proof of the invariance of the Lagrangian multipliers, under a regular coordinate change,
is obtained by considering (104) and using the fact that 1g M12 (Qi + Qc + Qb)T and1g M12 Q
cifT + 1are invariant.
The use of so-calledlinear coordinates will result in a constant mass matrix, that is, a matrix not dependingon t andq. This will greatly simplify the equations of motion.
Corollary 6.1 IfM2 is a constant matrix, then by performing a regular change of coordinates according to
q = M1/22 z (106)
the equations of motion may be written as
z+ QresTM,c +M1/22 g
T1(g0,z
t
g0,zz
z gzt
M1/22 z (
gzz
z)M1/22 z) = 0n1
zk(0) = zk0, zk(0) = zk0
g0,z(0,z0) +gz(0,z0)z= 0m1
where gz(t,z) = g(t,M1/22 z)M
1/22 , Q
resM,c = Q
resc M
1/22 = Q
resM1/22
T, = Inn and = M1/22 PM
1/22 .
Furthermore, in this case
Qcif = qTskew(M1
q)
1
2(
M1
t (
M0
q)T)
Proof: From (106) it follows, since M2 is a constant matrix, that q = M1/22 zand this inserted into (103)1 gives
M2 q + QresTc +g
T1 (g0
t
g0
qq
g
tq (
g
qq)q) = M
1/22 z+ Q
resTc +
gT1(g0,z
t
g0,z
zz
gz
tM
1/2
2
z (gz
zz)M
1/2
2
z) = 0n1
and then after pre-multiplying this equation with M1/22 one obtains
z+M1/22 Q
resTc +M
1/22 g
T1(g0,z
t
g0,z
zz
gz
tM
1/22 z (
gz
zz)M
1/22 z) = 0n1
where gz(t,z) = g(t,M1/22 z)M
1/22 . Now
M1/22 Q
resTc = M
1/22 (Q
resRT)T = M1/22 RQ
resT = M1/22 (Inn P)M
1/22 M
1/22 Q
resT
= (Inn )M1/22 Q
resT = (QresM1/22
T)T = QresTM,c
The constraint conditions (80)2 readg0 +g M1/22 z= 0m1 and this proves the corollary.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
30/35
Lidstrm 237
The following lemma will be useful in the forthcoming discussion.
Lemma 6.3 Let 1 m < n and L Rnm with rank(L) = m. Then L(LTL)1LT Rnn is an orthogonalprojection and range(L(LTL)1LT) = range(L).
Proof: See Lancaster and Timetsky [20] or Lidstrm and Olsson [21].
Proposition 6.2 = M1/22 PM1/22 is an orthogonal projection and range() = range(M
1/22 g
T).
Proof: By taking L = M1/22 g
T it follows that rank(L) = m and = LTL. Consequently
= M1/22 PM
1/22 = M
1/22 g
T1g M12 M1/22 = M
1/22 g
T1g M1/22 = L(L
TL)1LT
and then the proposition follows by invoking Lemma 6.3.
Proposition 6.3 RT is a projection onto kernel(g) andM2RT = RM2 = RM2R
T.
Proof: We have
RT = Inn M12 g
T1g = M12 (Inn gT1g M12 )M2 = M
12 R
TM2 (107)
Theng RT = gg M12 g
T1g = g 1g = 0mn (108)
and thus RT is a projection into kernel(g). Furthermore, from Proposition 6.2
rank(RT) = rank(R) = n rank(P) = n rank() = n rank(gT) =
n rank(g) = dim(kernel(g)) (109)
and thus RT is a projection onto kernel(g). Finally, from (107), it follows that M2RT = M2M
12 RM2 = RM2
and this proves the proposition.
Remark 6.7: Since (RTu)TM2v = uTRM2v = u
TM2RTv, the matrix RT represents an orthogonal projection with
respect to the scalar product (u, v)M2 = uTM2v on R
n1.
By post-multiplying equation (80)1 with RT we may eliminate the multipliers, since then
qTM2RT QresRT TgRT = qTRM2R
T QresRT = 01n
where we have used Proposition 6.3. We shall now take this a bit further by assuming that the generalized
velocity may be represented by
q = RTr+s (110)
where r Rn1 is considered as a new variable ands = s(t, q) Rn is a suitably selected function. According
to Proposition 6.3, RTr kernel(g). Consequently gq + g0 = g(RTr+ s) +g0 = gs + g0. Since 1 m < n
and rank(g) = m we may find a function s = s(t, q) Rn1 such that g(t, q)s(t, q) + g0(t, q) = 0m1 andthe constraint condition is then identically satisfied. Note that the function s is not uniquely defined, since if
g s +g0 = 0m1 then g(s + u) +g0 = 0m1, u kernel(g). Thus q = RTr+s = RTr +s where
r = r u ands = s + u, u kernel(g) (111)
Now, from (110), it follows that q = RTr+RTr+ s and this inserted into (80)1 gives
( RTr+RTr+ s)TM2 Qres Tg = rTRM2 + r
T RM2 + sTM2 Q
res Tg = 01n
Post-multiplying with RT, one obtains
rTRM2RT + rT RM2R
T + sTM2RT QresRT = 01n
where we have used Proposition 6.3. We may summarize this in the following theorem.
by Sudhanwa Kulkarni on December 1, 2012mms.sagepub.comDownloaded from
http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/http://mms.sagepub.com/ -
7/27/2019 Mathematics and Mechanics of Solids-2012-Lidstrm-209-42
31/35
238 Mathematics and Mechanics of Solids 17(3)
Theorem 6.2 (The initial value problem; reformulation B). The equations of motion (80) imply thatrTRM2R
T + rT RM2RT + sTM2R
T QresRT = 01nq RTrs = 0n1
(112)
where g s + g0 = 0m1. The appropriate initial conditions are q(0) = q0, r(0) = r0, where q0 and r0 have to
satisfy q0 = R(t, q0)Tr0 +s(t, q0) andq0 andq0 have to fulfil (82). Equation (112) is a system of 2 n first-orderordinary differential equations in the 2n unknowns q and r. Once this system has been solved one may use
(104) to calculate the multipliers. Note that the equations (112) are invariant under the transformation (111).
Remark 6.8: The mass matrix RM2RT, in equation (112), is not positive definite since dim(kernel(RT)) = m >
0. This may eventually cause problems when solving (112) numerically.
As a remedy for the problem noticed in the previous remark let us once again consider the projection RT.
Since, according to (109), rank(RT) = n m > 0, it is possible to construct a matrix S Rn(nm) from RTbyselecting n m columns from RT to build up S so that rank(S) = n m. Obviously, g S = 0mm. Let a R
nm
and
top related