FREE VIBRATIONS OF SDOF SYSTEMS
3.8 HYSTERETIC DAMPING
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.
-
be irreversible. A more realistic stress-strain curve for the loading-unloading process is shown in Figure 3.19 when | | y.
The curve in Figure 3.19 is a hysteresis loop. The area enclosed by the hysteresis loop from a force–displacement curve is the total strain energy dissipated during a loading–unloading cycle. In general, the area under a hysteresis curve is independent of the rate of the loading- unloading cycle.
In a vibrating mechanical system an elastic member undergoes a cyclic load-displacement relationship as shown in Figure 3.19. The loading is repeated over each cycle. The existence of the hysteresis loop leads to energy dissipation from the system during each cycle, which causes natural damping, called hysteretic damping. It has been shown experimentally that the energy dissipated per cycle of motion is independent of the frequency and proportional to the square of the amplitude. An empirical relationship is
(3.71) where Xis the amplitude of motion during the cycle and his a constant, called the hysteretic damping coefficient.
The hysteretic damping coefficient cannot be simply specified for a given material. It is dependent upon other considerations such as how the material is prepared and the geom- etry of the structure under consideration. Existing data cannot be extended to apply to every situation. Thus it is usually necessary to empirically determine the hysteretic damp- ing coefficient.
Mathematical modeling of hysteretic damping is developed from a work-energy analy- sis. Consider a simple mass-spring system with hysteretic damping. Let X1be the ampli- tude at a time when the velocity is zero and all energy is potential energy stored in the spring. Hysteretic damping dissipates some of that energy over the next cycle of motion.
Let X2be the displacement of the mass at the next time when the velocity is zero, after the
¢E = pkhX2
s s 6
FIGURE 3.19
Behavior of a real engineering material as a system executes one cycle of motion. The area enclosed by the curve is the dissipated strain energy per unit volume. This dissipated energy is the basis for hys- teretic damping.
x F
system executes one half-cycle of motion. Let X3 be the displacement at the subsequent time when the velocity is zero, one full cycle later. Application of the work-energy princi- ple over the first half-cycle of motion gives
(3.72) The energy dissipated by hysteretic damping is approximated by Equation (3.71) with Xas the amplitude at the beginning of the half-cycle.
(3.73) This yields
(3.74) A work-energy analysis over the second half-cycle leads to
(3.75) Thus the rate of decrease of amplitude on successive cycles is constant, as it is for vis- cous damping. By analogy a logarithmic decrement is defined for hysteretic damping as
(3.76) which for small his approximated as
(3.77) By analogy with viscous damping an equivalent damping ratio for hysteretic damping is defined as
(3.78) and an equivalent viscous damping coefficient is defined as
(3.79) The free vibrations response of a system subject to hysteric damping is the same as the response of the system when subject to viscous damping with an equivalent viscous damp- ing coefficient given by Equation (3.79). This is true only for small hysteretic damping, as subsequent plastic behavior leads to a highly nonlinear system. The analogy between vis- cous damping and hysteretic damping is also only true for linearly elastic materials and for materials where the energy dissipated per unit cycle is proportional to the square of the amplitude. In addition, the hysteretic damping coefficient is a function of geometry as well as the material.
The response of a system subject to hysteretic or viscous damping continues indefi- nitely with exponentially decaying amplitude. However, hysteretic damping is significantly different from viscous damping in that the energy dissipated per cycle for hysteretic damp- ing is independent of frequency, whereas the energy dissipated per cycle increases with fre- quency for viscous damping. Thus while the mathematical treatments of viscous damping and hysteretic damping are the same they have significant physical differences.
ceq = 2z2mk = hk vn z =
d 2p =
h 2 d = ph d = lnX1
X3 = -ln(1 - ph) X3 = 21 - phX2 = (1 - ph)X1 X2 = 21 - phX1
1
2kX21 = 1
2kX22 + 1 2pkhX21 T1 + V1 = T2 + V2 +
¢E 2
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E X A M P L E 3 . 1 2
The force-displacement curve for a structure of Figure 3.20(a) modeled by the system of Figure 3.20(b) is shown in Figure 3.20(c). The structure is modeled as a one-degree-of-free- dom system with an equivalent mass 500 kg located at the position where the measure- ments are made. Describe the response of this structure when a shock imparts a velocity of 20 m/s to this point on the structure.
S O L U T I O N
The area under the hysteresis curve is approximated by counting the squares inside the hys- teresis loop. Each square represents (1 104 N)(0.002 m) 20 N m of dissipated energy. There are approximately 38.5 squares inside the hysteresis loop resulting in 770 N m dissipated over one cycle of motion with an amplitude of 20 mm.
#
=
#
*
FIGURE 3.20
(a) One-story frame structure modeled as a SDOF system. (b) Hysteretic damping leads to an equivalent viscous-damping coefficient of 6100 N s/m. (c) Force-displacement curve over one cycle for the system of Example 3.12.
#
meq = 500 kg
ceq = 6100 N · s/m keq = 5 × 106 N/m
x x
(b) (a)
5 × 104
–5 × 104 –1 × 105 1 × 105
1.5 × 105
(c) 10
–20 –10 20
Displacement (mm) Force (N)
The equivalent stiffness is the slope of the force deflection curve and is determined as 5 106N/m. Application of Equation (3.71) leads to
(a) The logarithmic decrement, damping ratio, and natural frequency are calculated by using Equations (3.77) and (3.78)
(b) (c)
(d) The response of this structure with hysteretic damping is approximately the same as the response of a simple mass-spring-dashpot system with a damping ratio of 0.0615 and a nat- ural frequency of 100 rad/s. Then from Equation (3.28) with and x0⫽0, the response is
x(t) = 0.20e-6.13tsin(99.81t)m (e)
x#
0 = 20 m/s vn =
A k m =
A
5 * 106 N/m
500 kg = 100 rad/s z =
h
2 = 0.0613 d = ph = 0.385 h =
¢E pkX2 =
770 N
#
mp(5 * 106 N/m)(0.02 m)2 = 0.123
*