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7/29/2019 Dynamics Generalized coordinates
1/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Generalized Single Degree of FreedomSystems
Giacomo Boffi
Dipartimento di Ingegneria Strutturale, Politecnico di Milano
April 13, 2011
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Outline
Introductory Remarks
Assemblage of Rigid Bodies
Continuous Systems
Vibration Analysis by Rayleighs Method
Selection of Mode Shapes
Refinement of Rayleighs Estimates
7/29/2019 Dynamics Generalized coordinates
2/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Introductory Remarks
Until now our SDOFs were described as composed by asingle mass connected to a fixed reference by means of aspring and a damper.
While the mass-spring is a useful representation, manydifferent, more complex systems can be studied as SDOFsystems, either exactly or under some simplifyingassumption.
1. SDOF rigid body assemblages, where flexibility isconcentrated in a number of springs and dampers, canbe studied, e.g., using the Principle of Virtual
Displacements and the DAlembert Principle.2. simple structural systems can be studied, in an
approximate manner, assuming a fixed pattern ofdisplacements, whose amplitude (the single degree offreedom) varies with time.
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Further Remarks on Rigid Assemblages
Today we restrict our consideration to plane, 2-D systems.In rigid body assemblages the limitation to a single shape ofdisplacement is a consequence of the configuration of the
system, i.e., the disposition of supports and internal hinges.When the equation of motion is written in terms of a singleparameter and its time derivatives, the terms that figure ascoefficients in the equation of motion can be regarded as thegeneralised properties of the assemblage: generalised mass,damping and stiffness on left hand, generalised loading onright hand.
mx + cx + kx = p(t)
7/29/2019 Dynamics Generalized coordinates
3/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Further Remarks on Continuous Systems
Continuous systems have an infinite variety of deformation
patterns.By restricting the deformation to a single shape of varyingamplitude, we introduce an infinity of internal contstraintsthat limit the infinite variety of deformation patterns, butunder this assumption the system configuration ismathematically described by a single parameter, so that
our model can be analysed in exactly the same way as astrict SDOF system,
we can compute the generalised mass, damping,stiffness properties of the SDOF system.
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Final Remarks on Generalised SDOF Systems
From the previous comments, it should be apparent that
everything we have seen regarding the behaviour and theintegration of the equation of motion of proper SDOFsystems applies to rigid body assemblages and to SDOFmodels of flexible systems, provided that we have the meansfor determining the generalised properties of the dynamicalsystems under investigation.
7/29/2019 Dynamics Generalized coordinates
4/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Assemblages of Rigid Bodies
planar, or bidimensional, rigid bodies, constrained tomove in a plane,
the flexibility is concentrated in discrete elements,
springs and dampers, rigid bodies are connected to a fixed reference and to
each other by means of springs, dampers and smooth,bilateral constraints (read hinges, double pendulumsand rollers),
inertial forces are distributed forces, acting on eachmaterial point of each rigid body, their resultant can bedescribed by
a force applied to the centre of mass of the body,proportional to acceleration vector and total massM =
dm
a couple, proportional to angular acceleration and themoment of inertia J of the rigid body,J =
(x2 +y2)dm.
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Rigid Bar
x
G
L
Unit mass m= constant,
Length L,
Centre of Mass xG = L/2,
Total Mass m= mL,
Moment of Inertia J = mL2
12
= mL3
12
7/29/2019 Dynamics Generalized coordinates
5/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Rigid Rectangle
G
y
a
b
Unit mass = constant,
Sides a, b
Centre of Mass xG = a/2, yG = b/2
Total Mass m= ab,
Moment of Inertia J = ma2 + b2
12=
a3b + ab3
12
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Rigid Triangle
For a right triangle.
y
G
a
b
Unit mass = constant,
Sides a, b
Centre of Mass xG = a/3, yG = b/3
Total Mass m= ab/2,
Moment of Inertia J = ma2
+ b2
18= a
3
b + ab3
36
7/29/2019 Dynamics Generalized coordinates
6/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Rigid Oval
When a = b = D = 2R the oval is a circle.
x
y
a
b
Unit mass = constant,
Axes a, b
Centre of Mass xG = yG = 0
Total Mass m= ab
4,
Moment of Inertia J = mL2
12= m
L3
12
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
trabacolo1
c k c k 2 211 N
m , J2 2
p(x,t) = P x/a f(t)
a 2 a a a a a
The mass of the left bar is m1 = m4a and its moment of
inertia is J1 = m1(4a)2
12= 4a2m1/3.
The maximum value of the external load isPmax = P 4a/a = 4P and the resultant of triangular load is
R = 4P 4a/2 = 8Pa
7/29/2019 Dynamics Generalized coordinates
7/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Forces and Virtual Displacements
c1Z4
m1Z2
3k1Z4
c2Z2m2Z
3kZ
3
NZ(t)
J2Z3a
8Pa f(t)J1Z4a
Z4
Z2
3Z4
Z 2Z3
Z3
u
2 = Z/(3a)1 = Z/(4a)
u= 7a4a cos 13a cos 2, u= 4a sin 11+3a sin 22
1 = Z/(4a), 2 = Z/(3a)
sin 1 Z/(4a), sin 2 Z/(3a)
u=
14a +
13a
Z Z = 7
12aZ Z
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Principle of Virtual Displacements
c1Z4
m1Z2
3k1Z4
c2Z2m2Z
3kZ
3
NZ(t)
J2Z3a
8Pa f(t)J1Z4a
Z4
Z2
3Z4
Z 2Z3
Z3
u
2 = Z/(3a)1 = Z/(4a)
The virtual work of the InertialDampingElasticExternal
forces:
WI = m1Z
2
Z
2 J1
Z
4a
Z
4a m2
2Z
3
2Z
3 J2
Z
3a
Z
3a
=
m1
4+ 4
m2
9+
J1
16a2+
J2
9a2
Z Z
WD = c1Z
4
Z
4c2Z Z = (c2 + c1/16) Z Z
WS = k13Z
4
3Z
4 k2
Z
3
Z
3=
9k116
+k2
9
Z Z
WExt = 8Pa f(t)2Z
3+ N
7
12aZ Z
7/29/2019 Dynamics Generalized coordinates
8/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Principle of Virtual Displacements
For a rigid body in condition of equilibrium the total virtualwork must be equal to zero
WI + WD + WS + WExt = 0
Substituting our expressions of the virtual work contributionsand simplifying Z, the equation of equilibrium is
m1
4+ 4
m2
9+
J1
16a2+
J2
9a2
Z+
+ (c2 + c1/16) Z +
9k116
+ k29
Z =
8Pa f(t)2
3+ N
7
12aZ
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Principle of Virtual Displacements
Collecting Z and its time derivatives give us
mZ + cZ + kZ = pf(t)
introducing the so called generalised properties, in ourexample it is
m = 14
m1 + 49
9m2 + 116a2
J1 + 19a2
J2,
c =1
16c1 + c2,
k =9
16k1 +
1
9k2
7
12aN,
p =16
3Pa.
It is worth writing downthe expression of k: k =
9k116
+k2
9
7
12aN
Geometrical stiffness
7/29/2019 Dynamics Generalized coordinates
9/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Lets start with an example...
Consider a cantilever, with varying properties m and EJ,subjected to a load that is function of both time t andposition x,
p = p(x, t).
The transverse displacements v will be function of time andposition,
v = v(x, t)
H
x m= m(x)
N
EJ = EJ(x)v(x, t)
p(x, t)
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
... and an hypothesis
To study the previous problem, we introduce an approximatemodel by the following hypothesis,
v(x, t) = (x)Z(t),
that is, the hypothesis of separation of variablesNote that (x), the shape function, is adimensional, whileZ(t) is dimensionally a generalised displacement, usuallychosen to characterise the structural behaviour.In our example we can use the displacement of the tip of thechimney, thus implying that (H) = 1 because
Z(t) = v(H, t) and
v(H, t) = (H)Z(t)
7/29/2019 Dynamics Generalized coordinates
10/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Principle of Virtual Displacements
For a flexible system, the PoVD states that, at equilibrium,
WE = WI.
The virtual work of external forces can be easily computed,the virtual work of internal forces is usually approximated bythe virtual work done by bending moments, that is
WI
M
where is the curvature and the virtual increment ofcurvature.
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
WE
The external forces are p(x, t), N and the forces of inertiafI; we have, by separation of variables, that v = (x)Zand we can write
Wp =H
0 p(x, t)v dx =H
0 p(x, t)(x) dx
Z = p
(t) Z
WInertia =
H0
m(x)vv dx =
H0
m(x)(x)Z(x) dx Z
=
H0
m(x)2(x) dx
Z(t) Z = mZ Z.
The virtual work done by the axial force deserves a separatetreatment...
7/29/2019 Dynamics Generalized coordinates
11/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
WN
The virtual work of N is WN = Nu where u is thevariation of the vertical displacement of the top of thechimney.We start computing the vertical displacement of the top ofthe chimney in terms of the rotation of the axis line, (x)Z(t),
u(t) = H
H0
cos dx =
H0
(1 cos ) dx,
substituting the well known approximation cos 1 2
2in
the above equation we have
u(t) =
H
0
2
2dx =
H
0
2(x)Z2(t)
2dx
hence
u=
H0
2(x)Z(t)Z dx =
H0
2(x) dx ZZ
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
WInt
Approximating the internal work with the work done bybending moments, for an infinitesimal slice of beam we write
dWInt =1
2Mv(x, t) dx =
1
2M(x)Z(t) dx
with M = EJ(x)v(x)
(dWInt) = EJ(x)2(x)Z(t)Z dx
integrating
WInt = H0
EJ(x)2(x) dx ZZ = k Z Z
7/29/2019 Dynamics Generalized coordinates
12/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Remarks
the shape function must respect the geometricalboundary conditions of the problem, i.e., both
1 = x2 and 2 = 1 cos x2H
are accettable shape functions for our example, as1(0) = 2(0) = 0 and
1(0) =
2(0) = 0
better results are obtained when the second derivativeof the shape function at least resembles the typicaldistribution of bending moments in our problem, so that
between
1 = constant and 2 =2
4H2cos
x
2H
the second choice is preferable.
7/29/2019 Dynamics Generalized coordinates
13/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Example
Using (x) = 1 cos x2H , with m= constant and
EJ = constant, with a load characteristic of seismic
excitation, p(t) = mvg(t),
m = m
H0
(1 cosx
2H)2 dx = m(
3
2
4
)H
k = EJ4
16H4
H0
cos2x
2Hdx =
4
32
EJ
H3
k
G
= N2
4H2H
0
sin2x
2Hdx =
2
8HN
pg = mvg(t)
H0
1 cosx
2Hdx =
1
2
mH vg(t)
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Vibration Analysis
The process of estimating the vibration characteristicsof a complex system is known as vibration analysis.
We can use our previous results for flexible systems,
based on the SDOF model, to give an estimate of thenatural frequency 2 = k/m
A different approach, proposed by Lord Rayleigh, startsfrom different premises to give the same results but theRayleighs Quotient method is important because itoffers a better understanding of the vibrationalbehaviour, eventually leading to successive refinements
of the first estimate of 2
.
7/29/2019 Dynamics Generalized coordinates
14/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Rayleighs Quotient Method
Our focus will be on the free vibration of a flexible,undamped system.
inspired by the free vibrations of a proper SDOF we
write
Z(t) = Z0 sin t and v(x, t) = Z0(x) sin t,
the displacement and the velocity are in quadrature:when v is at its maximum v = 0 (hence V= Vmax,T= 0) and when v = 0 v is at its maximum (henceV= 0, T= Tmax,
disregarding damping, the energy of the system isconstant during free vibrations,
Vmax + 0 = 0 + Tmax
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Rayleigh s Quotient Method
Now we write the expressions for Vmax and Tmax,
Vmax =1
2Z20
S
EJ(x)2(x) dx,
Tmax = 12
2Z20S
m(x)2(x) dx,
equating the two expressions and solving for 2 we have
2 =
S EJ(x)
2(x) dxS m(x)
2(x) dx.
Recognizing the expressions we found for k
and m
wecould question the utility of Rayleighs Quotient...
7/29/2019 Dynamics Generalized coordinates
15/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Rayleighs Quotient Method
in Rayleighs method we know the specific timedependency of the inertial forces
fI = m(x)v = m(x)2Z0(x) sin t
fI has the same shape we use for displacements. if were the real shape assumed by the structure in
free vibrations, the displacements v due to a loadingfI =
2m(x)(x)Z0 should be proportional to (x)through a constant factor, with equilibrium respected inevery point of the structure during free vibrations.
starting from a shape function 0(x), a new shape
function 1 can be determined normalizing thedisplacements due to the inertial forces associated with
0(x), fI = m(x)0(x), we are going to demonstrate that the new shape
function is a better approximation of the true modeshape
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Selection of mode shapes
Given different shape functions i and considering the trueshape of free vibration , in the former cases equilibrium isnot respected by the structure itself.To keep inertia induced deformation proportional to i we
must consider the presence of additional elastic constraints.This leads to the following considerations
the frequency of vibration of a structure with additionalconstraints is higher than the true natural frequency,
the criterium to discriminate between different shapefunctions is: better shape functions give lower estimatesof the natural frequency, the true natural frequency
being a lower bound of all estimates.
7/29/2019 Dynamics Generalized coordinates
16/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Selection of mode shapes 2
In general the selection of trial shapes goes through two
steps,
1. the analyst considers the flexibilities of different parts ofthe structure and the presence of symmetries to devisean approximate shape,
2. the structure is loaded with constant loads directed asthe assumed displacements, the displacements arecomputed and used as the shape function,
of course a little practice helps a lot in the the choice of aproper pattern of loading...
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Selection of mode shapes 3
p = m(x)
P = M
p = m(x)
p=
m(x)
p = m(x)
(a)
(b) (c)
(d)
7/29/2019 Dynamics Generalized coordinates
17/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Refinement R00
Choose a trial function (0)(x) and write
v(0) = (0)(x)Z(0) sin t
Vmax =1
2Z(0)2
EJ(0)2 dx
Tmax =1
22Z(0)2
m(0)2 dx
our first estimate R00 of 2 is
2 =
EJ(0)2 dxm(0)2 dx
.
GeneralizedSDOFs
Giacomo Boffi
IntroductoryRemarks
Assemblage of
Rigid Bodies
Continuous
Systems
Vibration Analysis
by RayleighsMethod
Selection of Mode
Shapes
Refinement ofRayleighs
Estimates
Refinement R01
We try to give a better estimate of Vmax computing theexternal work done by the inertial forces,
p(0) = 2m(x)v(0) = Z(0)2(0)(x)
the deflections due to p(0) are
v(1) = 2v(1)
2= 2(1)Z
(1)
2= 2(1)Z(1),
where we write Z(1) because we need to keep the unknown2 in evidence. The maximum strain energy is
Vmax =1
2
p(0)v(1) dx =
1
24Z(0)Z(1)
m(x)(0)(1) dx
Equating to our previus estimate of Tmax we find the R01estimate
2 =Z(0)
Z(1)
m(x)(0)(0) dxm(x)(0)(1) dx
7/29/2019 Dynamics Generalized coordinates
18/18
Generalized
SDOFs
Giacomo Boffi
Introductory
Remarks
Assemblage of
Rigid Bodies
ContinuousSystems
Vibration Analysis
by Rayleighs
Method
Selection of Mode
Shapes
Refinement of
Rayleighs
Estimates
Refinement R11
With little additional effort it is possible to compute Tmaxfrom v(1):
Tmax =
1
2 2
m(x)v(1)2
dx =
1
2 6
Z(1)2
m(x)(1)2
dx
equating to our last approximation for Vmax we have the R11approximation to the frequency of vibration,
2 =Z(0)
Z(1)
m(x)(0)(1) dxm(x)(1)(1) dx
.
Of course the procedure can be extended to compute betterand better estimates of 2 but usually the refinements arenot extended beyond R11, because it would be contradictorywith the quick estimate nature of the Rayleighs Quotientmethod and also because R11 estimates are usually verygood ones.
Refinement Example
m
1.5m
2m
k
2k
3k
(0)
1
11/15
6/15
1
1
1
1
1.5
2
(1)p(0)
2m
T=1
22 4.5 mZ0
V=
1
2 1 3kZ0
2 =3
9/2
k
m=
2
3
k
m
v(1) =15
4
m
k2(1)
Z(1) =15
4
m
k
V(1) =1
2m
15
4
m
k4 (1 + 33/30 + 4/5)
=1
2m
15
4
m
k4
87
30
2 =92m
m 878mk
=12
29
k
m= 0.4138
k
m
Recommended