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EG-260 A.K. Dynamics (1)
EG-260 DYNAMICS I Lagrange's Equation 1. 2. 3. 4. 5. 6. 7. 8. Introduction ......................................................2 An overview of the procedure. ........................3 Useful energy expressions................................4 A mass-spring example ....................................4 A trolley mounted pendulum example ...........7 Two pendulum example.................................13 Damping ..........................................................16 Advantages of Lagrange's method ...............19
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1. Introduction Lagrangian methods use an energy approach to obtain the equations of motion in terms of generalised coordinates. The equation of motion in terms of the kinetic energy (T), potential energy (U) and generalised coordinated qi for an n degree of freedom system is given by
d T T U - + = Qi( n ) i = 1,2,....... n q q q dt &i i i
q
(1)
& where qi = t is the generalised velocity and Qi(n) is the non-conservative force corresponding to the generalised co-ordinates qi .The forces represented by Qi(n) may be either damping, or dissipative forces or other external forces.
If Fxk , Fyk and Fzk represent the external forces acting on the kth mass if the system in the x, y and z directions, respectively, the generalised force is given by:
x y z Qi(n ) = Fxk k + Fyk k + Fzk k qi qi qi (2)
where xk , yk , zk are the displacements of the kth. mass in the x , y and z directions, respectively. For a torsional system the force Fxk is replaced by the moment Mxk about the x axis and the displacement xk is replaced by the angular displacement qxk about the x axis
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If the Lagrangian, L, is defined as L = T- U, then equation (1) may be expressed as:
d L dt q i &
L - q = Qi i
(3)
For a conservative system Qi(n) = 0, so that equation (1) becomes:
d T dt qi &
T U - n q + q = 0 i = 1,2 ,....... (4) i i
2. An overview of the procedure. An overview of the procedure is given below.
Determine a set of generalized co-ordinates, denoted by q, which are an independent set of variables to record position. Express the kinetic and potential energy in terms of the generalised co-ordinates q. Express the work done by forces and moments. Derive the equation of motion using equations of the type expressed by equations (1) or (3).
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3. Useful energy expressions
KE of a mass
T=
1 2 & mx 2
(5)
KE of body rotating about centre of gyration
1 &2 T = I G 2
PE of a spring
(6)
U=
PE due to gravity
1 2 k (extension ) 2
(7) (8)
U = mgh
4. A mass-spring example
k1
m1
k2
m2
k3
Figure 1 MDOF mass-spring system
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Figure 1 shows a 2 degree of freedom mass-spring system, the kinetic energy is given by:
T= 1 1 & &2 m1q12 + m2q2 2 2
(9)
The potential energy is given by:
U= 1 2 1 1 2 2 k1q1 + k2 (q2 - q1 ) + k3 q2 2 2 2
(10)
The work done by external forces is given by:
f1 (q1 ) + f 2 (q2 )
(11)
From equation (9) the partial derivatives of the kinetic energy are given by:
T & = m1q1 & q1 T T = =0 q1 q2 T & = m2 q2 & q2
(12)
and from equation(10) the derivatives of potential energy are:
U = k1q1 - k 2( q2 - q1 ) q1 U = k 2( q 2 - q1 ) + k3q2 q2
(13)
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Applying Lagrange's equation (1) to each mass in turn gives:
d T T U q - q + q = Q1 dt &1 1 1
(14)
Substituting from equations (12)and (13) gives:
d & (m1q1 ) - 0 + (k1 + k 2 )q1 - k2 q2 = Q1 dt
(15)
which gives the equation of motion for mass m1 as:
&& m1q1 + (k1 + k2 )q1 - k 2q2 = Q1
(16)
and for mass m2 Lagrange's equation is :
d T T U q - q + q = Q2 dt & 2 2 2
(17)
Substituting from equations (12)and (13) gives:
d & (m2 q2 ) - k2q1 + (k 2 + k3 )q2 = Q2 dt
(18)
which gives the equation of motion for mass m2 as:
&& m2q2 - k2q1 + (k2 + k3 )q2 = Q2
(19)
Thus from equations (16): && m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = f1 and
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(20)
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EG-260 Dynamics (1)
&& m2 q2 - k2 q1 + (k2 + k3 )q 2 = f 2
which for q1 = x1 and q2 = x2 are the same equations of motion as given previously.
(21)
5. A trolley mounted pendulum example Figure 2 shows a trolley mounted pendulum. The surface over which the trolley moves is horizontal and the surface between the rollers and the surface is assumed to be friction free. Find the natural frequency of the system.
x
M
q
l
l cos(q) m l sin(q) Figure 2 Trolley mounted pendulum
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Solution Kinetic energy of trolley is:
Tt = 1 1 & & Mx 2 = Mq12 2 2
(22)
To calculate the kinetic energy of pendulum, first calculate the velocity in terms of the generalised co-ordinates x and q:
vx =
d & (x + l sin( )) = x + l& cos dt
vy = d (- l cos( )) = l& sin dt
(23)
(24)
Thus the kinetic energy of pendulum is:
Tp = 1 2 m v2 + vy x 2 1 & & & = m l 2 2 sin 2 + x + l cos 2
(
)
(
(
))
2
(25)
Simplifying gives:
Tp =
1 & & m l 2& 2 + x 2 + 2 xl& cos 2
(
)
(26)
The total kinetic energy is T = Tp + Tt and from equations (22) and (25) is:
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T =
1 1 & & & & M x 2 + m l 2& 2 + x 2 + 2 xl cos 2 2
(
)
(27)
T=
1 & & (M + m )x 2 + 1 ml 2& 2 + mlx& cos 2 2
(28)
Next consider the potential energy of the system. The pendulum has moved a vertical distance of l l cos (q) and the trolley has no movement in the vertical direction thus the potential energy of the system is:
U = mg (l - l cos ) = mgl (1 - cos ( ))
(29)
The external forces are zero i.e. Qi(n) = 0 so for Lagrange's equation (1) with q1 = x
d T T U + =0 - & dt x x x
From equation (28)
T & & = (M + m )x + ml cos & x
(30)
(31)
and
T =0 x
(32)
From equation (29)
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U =0 x
(33)
Substituting equations (31), (32) and (33) in equation (30) gives: d & (M + m )x + ml& cos + 0 + 0 = 0 dt which gives: & (M + m)&& + ml& cos - ml& 2 sin = 0 x For Lagrange's equation (1) with = q2 q.
d T T U + =0 &- dt
(
)
(34)
(35)
(36)
From equation (28)
T & & = ml 2 + mlx cos &
(37)
and
T && = -mlx sin
(38)
From equation (29)
U = mgl sin ( )
(39)
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Substituting equations (37), (38) and (39) in equation (36) gives:
d & & & ml 2 + mlx cos - - mlx& sin + mgl sin = 0 (40) dt
Hence: & & && ml 2& + ml&& cos - mlx sin + mlx sin + mgl sin = 0 (41) x Hence:
(
) (
)
& ml 2& + ml&&cos + mgl sin = 0 x
For small angular displacements: sin(q) = q and cos(q) = 1 Substituting the small angle approximation in equation (35) gives:
(42)
& (M + m)&& + ml& - ml& 2 = 0 x
(43)
& Since the last term is quadratic in , linearising equation (43) gives
& (M + m)&& + ml& = 0 x
(44)
Similarly, substituting the small angle approximation in equation (42) gives:
& && mlx + ml 2& + mgl = 0
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(45)
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Equations (44) and (45) may be written in matrix notation:
&& M + m ml x 0 0 x 0 ml 2 & & + 0 mgl = 0 ml 4 244 4 4 (46) 1 4 3{ 1 2 3 {
Mass M && x Stiffneass K x
To find the natural frequency use:
det - 2 M + K = 0
So
(47)
- 2 (M + m ) - ml
2
- 2 ml - ml + mgl
2 2
=0
(48)
Hence
- 2 (M + m) - 2 ml 2 + mgl - 2 ml
Simplifying gives:
(
) (
)
2
=0
(49)
2 - (M + m ) - 2 ml 2 + mgl - 2 m 2 l 2 = 0
So either w2 = 0, i.e. rigid body motion or:
{
(
)
}
(50)
- (M + m ) - 2 ml 2 + mgl - 2 m 2 l 2 = 0
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(
)
(51)
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Simplifying:
(M + m )(- 2l + g ) + 2 ml = 0
- Ml + Mg + mg = 0
2
- 2 Ml + Mg - 2 ml + mg + 2 ml = 0
So
(52)
2 =
(m + M )g
Ml
(53)
g M 2 = , which is the result for a simple As l pendulum.
6. Two pendulum example Figure 3 shows two pendulums of length l, separated by a spring of stiffness k N/m2
q1
a
q2
l
m1
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m1 Figure 3 Double pendulum
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The generalised co-ordinates are q1 = q1 and q2 =q2 . The kinetic energy of the system is:
T=
2 1 1 & 2 m1 l&1 + m 2 l 2 2 2 1 &2 1 &2 = m1 l 21 + m 2 l 2 2 2 2
( )
( )
(54)
So that:
T & = m1l 21 & 1 T & = m1l 2 2 & 2
(55)
and (56)
and
T T = =0 1 2
U = m1 gl(1 - cos 1 ) + m2 gl( 1 - cos 2 ) + 1 2 2 k (sin 1 - sin 2 ) 2
(57)
The potential energy is:
(58)
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U = m1 gl sin 1 + k 2 cos 1 (sin 1 - sin 2 ) 1 U = m2 gl sin 2 - k 2 cos 2 (sin 1 - sin 2 ) 2
(59)
(60)
Substituting equations (55) and (59) in Lagrange's equation:
d T & dt i
gives for q1:
T U - + = 0 i i
(61)
d m1 l 2&1 + m1 gl sin1 + k 2 cos 1 (sin1 - sin 2 ) = 0 (62) dt
(
)
hence:
& m1l 2& + m1 gl sin1 + k 2 cos1 (sin1 - sin 2 ) = 0 (63) 1
and for q2
d & m2 l 2 2 + m2 gl sin 2 - k 2 cos 2 (sin 2 - sin 2 ) = 0 (64) dt
(
)
hence:
& m2 l 2&2 + m2 gl sin 2 - k 2 cos 2 (sin1 - sin 2 ) = 0 (65)
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Using the small angle assumption, equations (63) and (65) may be expressed as & m1l 2& + m1 gl1 + k 2 (1 - 2 ) = 0 (66) 1 & m2 l 2&2 + m2 gl 2 - k 2 (1 - 2 ) = 0 (67) Collecting terms in q1 and q2 gives:
& m1l 2& + m1 gl + k 2 1 - k 2 2 = 0 1 & m2 l 2&2 - k 21 + m2 gl + k 2 2 = 0
(
(
)
)
(68) (69)
Equations (68)and (69) may be expressed in matrix form as:
m1l 2 0 & 0 &1 m1 gl + k 2 + 2 && m2 l 2 - k 2 1 0 = 2 2 0 m 2 gl + k - k 2
(70) From which the frequencies and mode shapes may be calculated in the normal manner. 7. Damping Lagrange's equations are based on the principle of conservation of energy and therefore are not applicable to non-conservative systems such as those including damping. However, if damping is expressed in terms of Rayleigh & dissipation function, i.e. damping is a function of q , such that the force due to damping is given by:
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1 2 Fd = c(velociy) 2 then Lagrange's equation may used.
(71)
For Lagrange's equation as given by equation (1) the generalized force is given by F Qi = - d (72) & qi
k1 c1
k2
k3
c2
c3
Figure 4 MDOF mass-spring-damper system
For the spring mass-damper system shown in Figure 4, the damping force is given by:
1 1 & & & 2 &2 Fd = 1 c1 q12 + c2 (q2 - q1 ) + c3 q 2 2 2 2
(73)
The generalised force for Lagrange's equation is then given by:
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Q1 = -
Fd & & & = -{c1 q1 - c2 (q 2 - q1 )} &1 q & & = -(c1 + c2 )q1 + c2 q 2 Fd & & & = -{c 2 (q 2 - q1 ) + c3 q2 } &2 q & & = c2 q1 - (c2 + c3 )q 2
(74)
and
Q2 = -
(75)
The generalised forces are given by equations (74) and (75). Equations (20)and (21) give the equations of motion for the undamped system as : && m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = f1 (76) and && m2 q2 - k2 q1 + (k2 + k3 )q 2 = f 2 (77) Substituting for the friction force as given equations (74) by equations (76) gives: && & & m1 q1 + (k1 + k 2 )q1 - k 2 q 2 = - (c1 + c 2 )q1 + c 2 q 2 hence the equation of motion for mass m1 is: && & & m1 q1 + (c1 + c2 )q1 - c2 q 2 + (k1 + k 2 )q1 - k 2 q 2 = 0 (79) Substituting for the friction force as given equations (75) by equations (77) gives: (78)
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&& & & m2 q2 - k 2 q1 + (k 2 + k 3 )q 2 = c2 q1 - (c2 + c3 )q2 Hence the equation of motion for mass m2 is: && & & m2 q2 - c2 q1 + (c2 + c3 )q2 - k 2 q1 + (k 2 + k 3 )q2 = 0
(80)
(81)
In matrix form the equations of motion with q1 = x1 and q2 = x2 are:
m1 0 0 &&1 (c1 + c2 ) x + m 2 &&2 - c2 x & - c 2 x1 (k1 + k 2 ) + & (c2 + c3 ) x2 - k 2 - k2 =0 (k 2 + k 3 )
(82) 8. Advantages of Lagrange's method Equations contain only scalar quantities One equation for each degree of freedom. The resulting equations are independent of the choice of coordinate system since kinetic and potential energy does not depend on coordinates
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