FREE VIBRATIONS OF SDOF SYSTEMS
3.7 COULOMB DAMPING
The right-hand side of Equation (3.56) is a nonlinear function of the generalized coordi- nate. Thus the free vibrations of a one-degree-of-freedom system with Coulomb damping are governed by a nonlinear differential equation. However, an analytical solution exists and is obtained by solving Equation (3.55).
Without loss of generality, assume that free vibrations of the system of Figure 3.15 are ini- tiated by displacing the mass a distance to the right, from equilibrium, and releasing it from rest. The spring force draws the mass toward equilibrium; thus the velocity is initially negative.
Equation (3.55) applies over the first half-cycle of motion, until the velocity again becomes zero.
The solution of Equation (3.55) subject to and with mgon the right-hand side is
(3.57) Equation (3.57) describes the motion until the velocity changes sign at t⫽ nwhen
(3.58) Equation (3.55) with – mg on the right-hand side governs the motion until the velocity next changes sign. The solution of Equation (3.55) using Equation (3.58) and
as initial conditions is
(3.59) x(t) = ad -
3mmg
k bcosvnt - mmg
k p
vn … t … 2p vn x#
A
vpnB
= 0m xap
vnb = -d + 2mmg
k
>v p x(t) = ad -
mmg
k bcosvnt + mmg
k
x# m (0) = 0 x(0) = d
d
k
x m
µ
External forces (a)
(b)
(c)
Effective forces kx
mg
x˙ > 0 F = µmg
N
=
kx mg
F = µmg
N
= x˙ < 0
FIGURE 3.15
(a) A mass slides on a surface with a coefficient of friction . (b) FBDs at an arbitrary instant for > 0. (c) FBDs at an arbitrary instant for < 0.mx# x#
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The velocity again changes sign at t⫽2 nwhen
(3.60) The motion during the first complete cycle is described by Equations (3.57) and (3.59). The amplitude change between the beginning and the end of the cycle is
(3.61) The motion is cyclic. The analysis of the subsequent and each successive cycle continues in the same fashion. The initial conditions used to solve for the displacement during a half- cycle are that the velocity is zero and the displacement is the displacement calculated at the end of the previous half-cycle.
The period of each cycle is
(3.62) Thus Coulomb damping has no effect on the natural frequency.
Mathematical induction is used to develop the following expressions for the displace- ment of the mass during each half-cycle:
(3.63)
(3.64)
(3.65) Equation (3.65) shows that the displacement at the end of each cycle is 4 mg/kless than the displacement at the end of the previous cycle. Thus the amplitude of free vibra- tion decays linearly as shown, when Equations (3.63) and (3.64) are plotted in Figure 3.16.
The amplitudes on successive cycles form an arithmetic sequence. If xnis the ampli- tude at the end of the nth cycle then
(3.66) with x0⫽ . The solution of this difference equation is Equation (3.65).
The motion continues with this constant decrease in amplitude as long as the restor- ing force is sufficient to overcome the resisting friction force. However, since the friction
d xn - xn-1 =
4mmg k
m xa2np
vnb = d - a4mmg k bn 2an - 1
2b p
vn … t … 2n p vn x(t) = cd - (4n - 1)mmg
k dcosvnt - mmg
k 2(n - 1)p
vn … t … 2an - 1 2bp
vn x(t) = cd - (4n - 3)mmg
k dcosvnt + mmg
k T =
2p vn x(0) - xa2p
vnb = 4mmg
k xa2p
vnb = d - 4mmg
k
>v p
FIGURE 3.16
Free response of a system with Coulomb damping. The motion is cyclic with a linear decay of amplitude. The period is the same as the natural period with motion ceasing with a permanent displacement.
–6 –3 0 3 6
8.0 6.0
Time (10–1 s)
µ = 0.1 m = 100 kg ωn = 100 rad/s x0 = 0.005 m
Displacement (10–3 m)
4.0 2.0
causes a decrease in amplitude, the restoring force eventually becomes less than the friction force. This occurs when
(3.67) Motion ceases during the nth cycle, where nis the smallest integer such that
(3.68) When motion ceases a constant displacement from equilibrium of mg/kis maintained.
The effect of Coulomb damping differs from the effect of viscous damping in these respects:
1. Viscous damping causes a linear term proportional to the velocity in the governing dif- ferential equation, while Coulomb damping gives rise to a nonlinear term.
2. The natural frequency of an undamped system is unchanged when Coulomb damp- ing is added, but is decreased when viscous damping is added.
3. Motion is not cyclic if the viscous damping coefficient is large enough, whereas the motion is always cyclic when Coulomb damping is the only source of damping.
4. The amplitude decreases linearly because of Coulomb damping and exponentially because of viscous damping.
m n 7 kd
4mmg - 1 4 k`xa2nvp
n
b` … mmg
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5. Coulomb damping leads to a cessation of motion with a resulting permanent displace- ment from equilibrium, while motion of a system with only viscous damping contin- ues indefinitely with a decaying amplitude.
Since the motion of all physical systems ceases in the absence of continuing external excitation, Coulomb damping is always present. Coulomb damping appears in many forms, such as axle friction in journal bearings and friction due to belts in contact with pul- leys or flywheels. The response of systems to these and other forms of Coulomb damping can be obtained in the same manner as the response for dry sliding friction.
The general form of the differential equation governing the free vibrations of a linear system where Coulomb damping is the only source of damping is
(3.69)
where Ffis the magnitude of the Coulomb damping force. The decrease in amplitude per cycle of motion is
(3.70)
⌬A = 4Ff meqv2n x$
+ v2nx = e
Ff meq x#
6 0 -
Ff meq x#
7 0
E X A M P L E 3 . 1 0
An experiment is run to determine the kinetic coefficient of friction between a block and a surface. The block is attached to a spring and displaced 150 mm from equilibrium. It is observed that the period of motion is 0.5 s and that the amplitude decreases by 10 mm on successive cycles. Determine the coefficient of friction and how many cycles of motion the block executes before motion ceases.
S O L U T I O N
The natural frequency is calculated as
(a) The decrease in amplitude is expressed as
(b) which is rearranged to yield
(c) From Equation (3.68) the motion ceases during the 15th cycle. The mass has a permanent displacement of 2.5 mm from its original equilibrium position.
m =
¢A 4gv2n =
(0.01 m)(12.57rad/s)2 4(9.81m/s2) = 0.04
¢A = 4mmg
k =
4mg v2n vn =
2p T =
2p
0.5 s = 12.57 rad/s
FIGURE 3.17
(a) Tree swing of Example 3.11. (b) The tension developed in opposite sides of a rope are unequal due to friction. (c) FBDs of swing at an arbitrary instant.
T1 T2 M
T1
mg
External forces Effective forces (c)
(b) (a)
= mlθ¨
mlθ˙2 T2 M d = 8.2 cm µ = 0.1
2T1
2T2
3.5 m
A father builds a swing for his children. The swing consists of a board attached to two ropes, as shown in Figure 3.17. The swing is mounted on a tree branch, with the board 3.5 m below the branch. The diameter of the branch is 8.2 cm and the kinetic coefficient of friction between the ropes and the branch is 0.1. After the swing is installed and his child is seated, the father pulls the swing back 10 and releases. What is the decrease in angle of each swing and how many swings will the child receive before Dad needs to give another push?
S O L U T I O N
Because of the friction between the tree branch and the ropes, the tension on opposite sides of a rope will be different. These tensions can be related using the principles of belt fric- tion. When the swing is swinging clockwise,
T2 = T1emb (a)
°
E X A M P L E 3 . 1 1
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where is the angle of contact between the tree branch and the rope. As the child swings the angle of contact may vary. However, this complication is too much to handle with a simplified analysis. A good approximation is to assume is constant and ⫽ rad.
When the swing is swinging counterclockwise
(b) Let be the clockwise angular displacement of the swing from equilibrium. Summing forces in the direction of the tensions gives ∑Fext⫽∑Feff
(c) The swing is pulled back only 10 . Thus the usual small-angle approximation is valid, with cos 1 and the nonlinear inertia term ignored in comparison to the tensions and grav- ity. The belt friction relations and the normal force equation are solved simultaneously to yield
(d)
(e)
Summing moments about the center of the tree branch, using the free-body diagrams of Figure 3.17(c) and the small-angle assumption yields
(f)
Substituting for the tensions into the preceding equation and rearranging leads to
(g)
The frequency of the swinging is
(h) which is the same as it would be in the absence of friction.
vn = A
g
l = 1.67 rad/s
u
# 7 0 u
# 6 0 u
$ +
g l u = d
gd 2l2
1 - emp 1 + emp -
gd 2l2
1 - emp 1 + emp (2T1 - 2T2) d
2 - mglu = ml2 u
$ b
eff
MO
= aa MOb
ext
aa
T2 = mg 2(1 + emp) T1 =
mgemp 2(1 + emp) u
# 7 0,
T2 =
mgemp 2(1 + emp) T1 =
mg 2(1 + emp) u
# 7 0, u L
° 2T1 + 2T2 - mg cosu = mlu
#2
u
T1 = T2emb
p b b
b
The governing differential equation is of the form of Equation (3.69). Thus, from Equation (3.70), the decrease in amplitude per swing is
Motion ceases when, at the end of a cycle, the moment of the gravity force about the center of the branch is insufficient to overcome the frictional moment. This occurs when
or
Thus, if Dad does not give the swing another push after 23 swings, the swing will come to rest with an angle of response of 0.1 .°
u 6 d 2l
emp - 1
emp + 1 = 0.10°
mglu 6 |T2 - T1|d 2d
l
emp - 1
emp + 1 = 2a0.082 m
3.5 m be0.1p - 1
e0.1p + 1 = 0.0073 rad = 0.42°
–σy
⑀ σy
σ
FIGURE 3.18
Stress strain diagram for a linearly elastic isotropic material with the same behavior in compression and tension. Material behavior is linear for |s|6sy.
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