Section I.2
Lagrange’s Equations - Linearized EOM’s
In the preceding section of the notes, we developed the Lagrangian formulation of the
EOM’s for N-DOF systems. This formulation allows use to develop the EOM’s in a
systematic way and also allows us to exclude the contribution of forces that do no virtual
work on the system.
These EOM’s are often nonlinear, with the nonlinearities resulting from geometric effects
in the kinematics and/or from nonlinear material behavior. Our goal in this course is to
study small oscillations of systems. For small oscillations, we can generally use a
“linearized” form of the EOM’s. This linearization process can be done on an exampleby-example basis, as you have likely done earlier in your introductory vibrations courses.
Here we will develop a systematic approach for linearizing the EOM’s. In the process, we
will also discover some properties of the system matrices that will help us later on in the
course in understanding the qualitative nature of the linearized response.
Explicit Representation of Lagrange’s Equations
For a system of n particles, we have written the kinetic energy as:
n
n
1
1
T = " mi r˙i •˙ri = " mi vi2
2 i=1
2 i=1
Furthermore, the velocity terms above was also written explicitly in terms of the
generalized coordinates, their time derivatives and time as (as a result of the chain rule of
! differentiation):
N
v i = r˙i = #
" ri
"q j
j=1
q˙ j +
" ri
"t
N
=#
" ri
"q
k=1 k
q˙ k +
" ri
"t
Substituting the velocity expression into that of kinetic energy gives:
!
$ N "r
n
" ri ' $ N " ri
" ri '
1
i
&
)
&
)
˙
˙
T = # mi #
qj +
• #
qk +
2 i=1 &% j=1"q j
"t )( &% k=1"qk
"t )(
Expanding this dot product:
!
I.2-1
(1)
*
n
, " ri
1
T = # mi +
2 i=1 , "t
*
n
, " ri
1
= # mi +
2 i=1 , "t
-
.
" ri $ N " ri ' $ N " ri ' " ri $ N " ri ' $ N " ri ',
•
+
•&#
q˙ k ) + & #
q˙ j ) •
+ &#
q˙ j ) • & #
q˙ k )/
"t "t &% k=1"qk )( &% j=1"q j )( "t &% j=1"q j )( &% k=1"qk )(,
0
" ri
.
" ri $ N " ri ' $ N " ri ' $ N " ri ',
&
)
&
)
&
)
•
+ 2 •&#
q˙ k + #
q˙ j • #
q˙ k /
"t
"t % k=1"qk )( &% j=1"q j )( &% k=1"qk )(,
0
" ri
$ N "r ' 1 n $ N "r ' $ N "r '
i
i
i
• && #
q˙ k )) + # mi & #
q˙ j ) • && #
q˙ k ))
&
)
% k=1"qk ( 2 i=1 % j=1"q j ( % k=1"qk (
n
" ri " ri n
" ri
1
= # mi
•
+ # mi
2 i=1 "t "t i=1 "t
=
n
N $ n
N N $ n
1
" ri " ri
" ri " ri '
1
" ri " ri '
&
)
&
˙
m
•
+
m
•
q
+
m
# i "t "t #&# i "t "q ) k 2 # #&# i "q • "q ))q˙ jq˙k
2 i=1
k(
j
k(
k=1% i=1
j=1 k=1% i=1
=
1
"r "r
1
mi i • i + # hk q˙ k + # # m jk q˙ j q˙ k
#
2 i=1
"t "t k=1
2 j=1k=1
n
N
N N
or
!
T = T0 + T1 + T2
where
n
!
T0 =
1
"r "r
mi i • i
#
2 i=1
"t " t
;
INDEPENDENT of q˙ ' s
;
LINEAR in q˙ ' s
;
QUADRATIC in q˙ ' s
N
T1 = # hk q˙ k
(2)
k=1
N N
T2 =
1
# # m jk q˙ jq˙k
2 j=1k=1
n
!
hk = # mi
i=1
" ri " ri
•
"t "qk
n
!
!
m jk = # mi
i=1
;
INDEPENDENT of q˙ ' s
" ri " ri
•
= mkj
"q j "qk
;
INDEPENDENT of q˙ ' s
What we have accomplished here is to separate out the terms in the kinetic energy
expression to expose the explicit appearance of the q̇ terms. Specifically: T0 is
I.2-2
(3)
(4)
independent of the q̇ terms, T1 is linear in the q̇ terms and T2 is quadratic in the q̇
terms. In these expressions, we see that the sets of coefficients hj and mjk are
independent of the q̇ terms.
Furthermore, since the “j” and “k” indices on the right hand side of equation (4) can be
interchanged (the order of two vectors in a dot product is not important), the m jk
coefficients above are “symmetric”; that is, m jk = mkj . This will be an important result
later on in observing that Lagrange’s equations will always produce a symmetric mass
matrix.
!
!
I.2-3
Let us now use this representation of the kinetic energy to produce a more explicit form
of Lagrange’s equations.
(a)
Taking the partial derivative of T with respect to q p gives:
N
N N
"m jk
"T "T0
"h
1
=
+ # k q˙ k + # #
q˙ j q˙ k
"q p "q p k=1"q p
2 j=1k=1 "q!p
!(b)
Taking the partial derivative of T with respect to q˙ p gives:
N
N N
$ " q˙
"T "T0
" q˙
1
" q˙ '
j
=
+ # hk k + # # m jk &&
q˙ k + q˙ j k ))
"q˙ p "q˙ p k=1 "q˙ p 2 j=1k=1 ! % "q˙ p
"q˙ p (
N
= 0 + # hk
k=1
N N
N N
" q˙ j
" q˙ k 1
1
" q˙
+ # # m jk
q˙ k + # # m jk q˙ j k
"q˙ p 2 j=1k=1
"q˙ p
2 j=1k=1
"q˙ p
N
N
N
N
1
1
= h p + # m pk q˙ k + # m jp q˙ j
2 k=1
2 j=1
1
1
= h p + # m pk q˙ k + # m pj q˙ j
2 k=1
2 j=1
;
since m jp = m pj
N
= h p + # m pk q˙ k
k=1
!(c)
Differentiating "T /"q˙ p with respect to time gives:
N #
&
dm pk
d # "T & dh p
%%
(( =
+ )%
q˙ k + m pk q˙˙k (
dt!$ "q˙ p ' dt k=1$ dt
'
!
and using the chain rule (and recalling that the hj and mjk are functions only of
the q’s and t):
dh p
dt
N
=#
k=1
"h p
"qk
q˙ k +
"h p
"t
I.2-4
!
N
dm pk
=#
dt
j=1
"m pk
"q j
q˙ j +
"m pk
"t
gives:
!
#
&
N
"h p
"h p N %# N "m pk
"m pk &
d # "T &
(q˙ k + m pk q˙˙k (
%%
(( = )
q˙ k +
+ ) %)
q˙ j +
%
%
(
˙
dt $ "q p ' k=1 "qk
"t k=1 $ j=1 "q j
"t ('
$
'
!
Pulling together (a)-(c) and using Lagrange’s equations:
d # "T & "T
"R " V
%%
(( )
+
+
= Qp
dt $ "q˙ p ' "q p "q˙ p "q p
*
## N
&
"m pk
"m pk &
%
%
(
+ "q q˙k + "t + +%% + "q q˙ j + "t (q˙k + m pk q˙˙k ((
j
k=1 k
k=1$$ j=1
'
'
N
"h p
"h p
N
)
N
N N
"m jk
"T0
"h
1
" R "V
) + k q˙ k ) + +
q˙ j q˙ k +
+
= Qp
"q p k=1"q p
2 j=1k=1 "q p
"q˙ p "q p
*
N #
"h p "h k &
"R " (V ) T )
˙
˙
%
m
q
+
+ pk k +% "q ) "q ((q˙k + "q˙ + "q 0
k
p'
p
p
k=1
k=1$
N N #
"m pk 1 "m jk &
"h p N "m pk
(q˙ j q˙ k +
+ + +%%
)
++
q˙ k = Q p
"q j 2 "q p ('
"t k=1 "t
j=1 k=1$
N
or
!N
N
N N
N
#h
#m
#R #U
" m pk q˙˙k + " g pk q˙k + #q˙ + #q + " " N jkp q˙ jq˙k + #tp + " #tpk q˙k = Q p
p
p
k=1
k=1
j=1 k=1
k=1
(5)
where
!
g pk =
!
"h p
"qk
#
$ "h
"h p '
"h k
)) = #gkp ; SKEW # SYMMETRIC coefficients (6)
= #&& k #
"q p
% "q p "qk (
I.2-5
N jkp =
"m pk
"q j
#
1 "m jk
2 "q p
(7)
(8)
U = V " T0 = "dynamic potential"
!
!
Remarks
a) Equation (5) represents the most general form of Lagrange’s equations for a system
of particles (we will later extend these to planar motion of rigid bodies). This form
of the equations shows the explicit form of the resulting EOM’s.
b) For all systems of interest to us in the course, we will be able to separate the
generalized forces Q p in terms of components that are independent and dependent
on time. That is we can write: Q p = Pp + f p ( t ) where Pp are not explicitly
functions of time.
c) A “rheonomic” system is one on which time-dependent position constraints exist
!
are imposed. Otherwise, the system is known as “scleronomic”. We will deal with
both rheonomic and !
scleronomic systems in !
this course.
d) For many rheonomic systems, the coefficients m pk and g pk will NOT be explicit
functions of time. This will be the case for all systems of interest to us in this
course. As a result, our EOM’s will be of the form:
N
N
" m pk q˙˙k + " g pk q˙k +
k=1
k=1
N !
N
#R #!U
+
+ " " N jkp q˙ j q˙ k = Pp + f p ( t )
#q˙ p #q p
j=1 k=1
(5a)
e) Note that for a scleronomic system, the position vectors are NOT explicit functions
of time. As a result, equation (1) reduces to:
!
N
"r
˙ri = # i q˙ j
"q j
j=1
!
!
since " ri /"t for a scleronomic system. As a result, T0 = T1 = 0 (see equations (2)
and (3)). Consequently, the general form of EOM’s for scleronomic systems
reduces to:
!
N N
#R #U
" m pk q˙˙k + #q˙ + #q + " " N jkp q˙ jq˙k = Pp + f p (t)
p
p
k=1
j=1 k=1
N
!
I.2-6
(5b)
Linearized form of Lagrange’s Equations
Review of Taylor series expansions
In many cases, one is interested in investigating small amplitude oscillations in systems
resulting from initial conditions and external excitation. The oscillations will represent
small motion about an equilibrium state. Depending what is initially chosen as the
generalized coordinates in the model, the size of the numbers that represent the response
may not actually be small, but the deviation from the equilibrium solution will be small.
(We will reconsider this distinction later on in an example.) As a result of the
assumption, we will systematically transform the solution from our set of original
generalized coordinates, qj , to a set of perturbation coordinates zj and linearize the
EOM’s in a Taylor series expansion about the equilibrium state, q0j , where
z j = q j " q j0
!
!
Recall that the Taylor series expansion of a function g = g(q1,q2 ,K,qN ) about the point
q 0 = g(q01,q02 ,K,q0N ) is given by:
( )
N
"g
"q
k=1 k
1
( )
q
N
= g q0 + #
zk +
q
N
" 2g
k
j=1 k=1
0
N
"g
"q
k=1 k
N
(qk $ q0k ) + !# #
2
" q "q
g(q) = g q 0 + #
0
j q
(qk $ q0k )(q j $ q0 j ) + K
0
N
1
" 2g
##
2 j=1 k=1"qk"q j
zk z j + K
q
0
Example
For a function g as a function of a single coordinate q, the Taylor series expansion
! about q = 0 reduces to:
dg
d2 g
d3 g
2+ 1
gq = g0 +
q+1
q
q3 + …
dq 0
2 dq2 0
3! dq3 0
!
=
!
r=0
r
1 d g qr
r! dqr 0
For g(q) = sin(q), we have
g q = 0 + (1) q + 1 (0) q2 + 1 (-1) q3 + 1 (0) q4 + 1 (1) q5 + …
2
3!
4!
5!
!
=
!
r=0
(-1)r
q2r+1
(2r+1)!
which is the well-known power series expansion for the sine function.
I.2-7
Equilibrium solutions
The equilibrium solution q 0 is the solution of the EOM’s with q˙˙ = q˙ = 0 and with f(t) = 0;
that is, from equation (5a):
"U
"q p
( )
!
!
= Pp q 0
q
(6)
0
Once the equilibrium solution is found from above, we are now in a position to linearize
the EOM’s about this state. In the linearization, we will drop all terms, which are
! nonlinear in the coordinates z j = q j " q j0 . The following is a term-by-term linearization
of the EOM’s about this equilibrium state.
!
Terms with no linear components
The terms N jkp q˙ j q˙ k are quadratic in the q̇ ’s, and therefore will be nonlinear regardless
of the number of terms retained in the Taylor series expansion for Njkp . As a result,
these terms will be completely dropped in the linearization process.
!
Mass matrix
Writing the m pk coefficients in terms of their Taylor series expansions:
N
!
m pk = m pk
q
0
+#
"m pk
j=1
"q j
zj +K
q
0
Therefore, the linearization of the first term in equation (5a) gives:
!
$
&
" m pk ˙z˙k = "& m pk
k=1
k=1&
%
N
N
N
q
0
+"
j=1
#m pk
#q j
'
)
z j + K)˙z˙k
)
q
(
0
N
= " m pk
k=1
N
q
0
˙z˙k
= " M pk ˙z˙k
k=1
where
!
I.2-8
M pk = m pk
q
0
= M kp = SYMMETRIC mass matrix
!
Gyroscopic matrix
Similarly, writing the g pk coefficients in terms of their Taylor series expansions:
N
"g pk
g pk = g pk
+#
! q 0 j=1 "q j
zj +K
q
0
Therefore, the linearization of the second term in equation (5a) gives:
!
$
&
" g pk z˙k = "& g pk
k=1
k=1&
%
N
N
N
q
0
#g pk
+"
j=1
#q j
'
)
z j + K) z˙k
)
q
(
0
N
= " g pk
k=1
N
q
0
z˙k
= " G pk z˙k
k=1
where
!
G pk = g pk
q
0
= "gkp
q
0
= "Gkp = SKEW " SYMMETRIC "gyroscopic" matrix
Conservative stiffness matrix
First we write the dynamic potential function U in its Taylor series expansion:
!
( )
N
"U
"q
k=1 k
N
U (q) = U q 0 + #
zk +
q
0
N
1
" 2U
##
2 j=1 k=1"qk"q j
Therefore, differentiating produces:
!
I.2-9
zk z j + K
q
0
N
"U
"U
=0+ #
"q p
"q
k=1 k
=
=
"U
"q p
"U
"q p
N
q
N
"z k 1
" 2U
+ ##
"q p 2 j=1 k=1"qk"q j
0
N
+
q
0
1
" 2U
#
2 j=1 "q p"q j
N
+
q
0
q
0
$ "z
"z '
&& k z j + zk j )) + K
"q p (
% "q p
N
zj +
q
0
1
" 2U
#
2 k=1 "qk"q p
zk + K
q
0
N
1
1
K pj z j + # K kp zk + K
#
2 j=1
2 k=1
where
!
K pj =
!
" 2U
"q p"q j
= K jp = SYMMETRIC "stiffness matrix"
q
0
Then, truncating the above after the linear terms and using the symmetry of the stiffness
coefficients give:
"U "U
=
"q p "q p
"U
$
"q p
N
q
0
N
1
1
+ # K pj z j + # K pk zk
2 j=1
2 k=1
N
+ # K pj z j
q
0
j=1
Note that we needed to retain up to, and including, the quadratic terms in U for the
linearization of the EOM’s.
!
I.2-10
Damping matrix
In a similar fashion, we can write the truncated Taylor series expansion for the third term
on the left hand side of equation (5a) as:
N
"R
= # C pj z˙ j
"q˙ p j=1
!
where
C pj =
" 2R
"q˙ p"q˙ j
= C jp = SYMMETRIC "damping matrix"
q
0
!
Nonconservative stiffness matrix
Writing Pp in their Taylor series expansions:
N #
"Pp &
(( z j
Pp = Pp q 0 + )%%
"
q
j
'q
j=1$
( )
0
Then, truncating Pp after the linear terms gives:
!
( )
N
Pp = Pp q 0 " # D pj z j
j=1
where
!
$ #P '
p
))
D pj = "&&
% #q j ( q
0
!
I.2-11
Final form of linearized EOM’s
Substituting our linearized terms into the explicit EOM’s (5a) gives:
N
N
N
N
" M pj ˙z˙ j + " G pj z˙ j + " C pj z˙ j + " K pj z j +
j=1
!
j=1
j=1
j=1
#U
#q p
( )
N
= f p ( t ) + Pp q 0 $ " D pj z j
q
0
j=1
Since the linearization was performed about an equilibrium state q 0 , we know from
equation (6) that:
"U
"q p
( )
!
= Pp q 0
q
0
Therefore, the linearized EOM’s take on the form of:
!
N
N
N
j=1
j=1
j=1
" M pj ˙z˙ j + " (G pj + C pj )z˙ j + " (K pj + D pj )z j = f p (t )
In matrix-vector form, these equations are:
!
[ M ]˙z˙ + [[G] + [C]] z˙ + [[K ] + [ D]] z = f ( t )
T
T
T
T
where [ M ] = [ M ] , [G] = "[G] , [C ] = [C ] and [K ] = [K ] . In general, the symmetry
! properties of [D] are not known.
For the systems of the type abiding by the assumptions stated early in this section, the
! solution of!the above linearized
!
! represents small-amplitude motion about a
equations
given equilibrium state.
I.2-12
Remarks
(a)
The “mass matrix” [M] is always symmetric ( [M] = [M]T ) regardless of the
system.
(b)
The “gyroscopic matrix” [G] will not appear in the EOM’s for scleronomic
systems. To see this, note that the [G] coefficients arise from the hj terms in the
T1 portion of the kinetic energy expression. Recall that the appearance of T1
requires the position vectors to be explicit functions of time (that is, for
rheonomic systems). However, the [G] matrix may be absent from many
rheonomic systems. When it does appear, [G] will be skew-symmetric ( [G] = [G]T ) . Since [G] is skew-symmetric, there is no power in the term [G] ż ; that
is, no energy is lost or gained in this term.
(c)
The “damping matrix” [C] originates from the non-conservative generalized
forces. When the damping forces [C] ż are derivable from a Rayleigh
dissipation function (as above), the damping matrix will be symmetric and
positive semi-definite. With [C] being positive semi-definite, the damping terms
[C] ż are generally dissipative; that is, energy is lost from the system through
these terms.
(d)
The matrix [K] is the contribution to the “stiffness matrix” from the dynamic
potential function, U = V - T0 and will always be symmetric.
(e)
The matrix [D] is the contribution to the stiffness matrix from the nonconservative generalized forces. The symmetry properties of [D] are not known
in general. We will therefore assume that [D] will be non-symmetric unless
known to be otherwise.
(f)
Recall that the z coordinates represent small motion of the system about an
equilibrium state. The z coordinates will necessarily be small whereas the
corresponding generalized coordinates in our original problem q may NOT
necessarily be small. To illustrate, consider a representative time history for q
and z below where z is offset from q by the equilibrium value q0 :
q(t)
z(t)
q0
t
t
I.2-13
(g)
A set of EOM’s derived for a system via Newton-Euler will be completely
equivalent to the above Lagrangian EOM’s, although they may appear to be
different in form. In particular, the symmetry/skew-symmetry properties
observed above may not be present in the Newton-Euler equations.
I.2-14
Method for producing the linearized form of Lagrange’s Equations
The above matrix equation not only shows us some general properties of the coefficients
that appear in the Lagrangian formulation, but also provides us with “formulas” that can
be used to find these coefficients. That is, we can construct these EOM’s of motion
without actually writing down the kinetic energy and virtual work expressions and
without using the original set of Lagrange’s equations. The following provides a guide
that can be followed in such a formulation of the linearized EOM’s for a system of
particles (an equivalent approach for handing systems of rigid bodies can also be
established):
(i)
Write down the position vectors for all n particles in terms of N independent
generalized coordinates, qj :
ri = ri (q1,q2 ,K,qN ,t )
; i = 1, 2, …, n
For later use, note any explicit time dependence in these vectors.
∂r
∂r
! the partial derivatives i and i . From these, form the coefficients h
(ii) Form
j
∂t
∂qj
and mjk , the non-conservative generalized forces Qj , the potential energy V, the
Rayleigh dissipation function, the function T0 and the dynamic potential U using:
n
hk = # mi
i=1
" ri " ri
•
"t "qk
n
m jk = # mi
i=1
!
" ri " ri
•
"q j "qk
N
"W =
$ Q p "q p
#
Q p = Pp + f p ( t )
p=1
!
n
!
1
"r "r
T0 = # mi i • i
2 i=1
"t " t
U = V - T0
(iii) Form
! the equilibrium equations:
"U
"q p
( )
= Pp q 0
q
0
(iv) Solve the equilibrium equations for q0 . Although these equations are simply
algebraic equations, they are generally very nonlinear, and you will probably need
!
I.2-15
to solve them numerically. Furthermore, it is likely that more than one
equilibrium state may exist and therefore multiple solutions of the equilibrium
equations may exist.
(v)
Form the mass, gyroscopic, damping and the stiffness matrices:
M pk =
"m pk
"q j
G pk = g pk
!
!
!
q
0
0
= "Gkp
# " 2R &
(( = C pk
C pk = %%
$ "q˙ p "q˙ k ' q
0
K pk =
!
q
= M kp
" 2U
"q p"qk
= K kp
q
0
$ #Pp ' ??
D pk = "&
) = Dkp
% #qk ( q
0
!
I.2-16
Procedural notes
The partial differentiation and algebra in forming the T0 , h j and m jk terms in step (ii)
above can become a bit tedious even for relatively simple systems. My
recommendation is to replace the formality of step (ii) by the following:
•
Write down the kinetic energy, T, for
!the!system!directly from velocities and
angular velocities.
•
Separate the terms of T into those which are independent of the q˙ j ’s (call this
T0 ), linear in the q˙ j ’s (call this T1 ) and quadratic in the q˙ j ’s (call this T2 ):
T = T0 + T1 + T2
!
•
!
! by comparing the above
Identify!h j and m jk terms
! with:
!
N
T1 = " h j q˙ j
!
j= 1
!
! 1 N
T2 = " m jkq˙ j q˙ k
2 j= 1
!
This is the procedure that has been used in the lecture examples for this course. Note that
this recommended procedure works for systems containing both particles and rigid
bodies.
!
I.2-17
Example I.6
A two-particle system is acted upon by a constant-magnitude force F that always acts
along the line of the massless link. All springs are unstretched when x = θ = 0. Using
the explicit Lagrangian formulation, find the EOM’s in terms of the generalized
coordinates x and θ. Motion for the system lies in a horizontal plane.
F
x
θ
m
L
k
M
c
K
I.2-18
Example I.7
A mass-spring-dashpot system moves in a vertical plane. The coordinate x is used to
represent the position of the particle when the spring is unstretched (i.e., x = 0 when the
spring is unstretched). The restoring force in the spring, Fs, is given by the nonlinear
function:
Fs = kx + "x 2
Determine the linearized differential equation that describes the small motion of the
system about the equilibrium state of the system.
!
Fs
c
g
x
kx
m
x
I.2-19
Example I.8
Particle P, having a mass of m, is attached to a turntable through a set of fur springs. The
turntable rotates about a vertical axis at a constant rate of Ω. Derive the EOM’s for this
two-DOF system using the coordinates x and y, where x and y are coordinates of P as
measured by an observer on the turntable.
Y
y
x
ky/2
m
kx/2
P
Ωt
kx/2
X
ky/2
I.2-20
Example I.9
The four-particle system moves in a horizontal plane. Assume all masses to be m, all
stiffnesses to be k and the damping constant for the damper to be c. Let x1, x2 and x3 to be
absolute coordinates, and x4 represents the motion of particle 4 relative to particle 1.
Use the explicit Lagrangian formulation to directly determine the mass, damping and
stiffness matrices for the system.
x1
x4
1
4
3
2
x3
x2
I.2-21