MODELING OF SDOF SYSTEMS
2.2 SPRINGS
2.2.1 INTRODUCTION
A spring is a flexible mechanical link between two particles in a mechanical system. In real- ity a spring itself is a continuous system. However, the inertia of the spring is usually small compared to other elements in the mechanical system and is neglected. Under this assump- tion the force applied to each end of the spring is the same.
The length of a spring when it is not subject to external forces is called its unstretched length. Since the spring is made of a flexible material, the force F that must be applied to the spring to change its length by x is some continuous function of x,
(2.1) The appropriate form of f(x) is determined by using the constitutive equation for the spring’s material. Since f(x) is infinitely differentiable at x 0, it can be expanded by a Taylor series about x 0 (a MacLaurin expansion):
(2.2) Since x is the spring’s change in length from its unstretched length, when x0, F0.
Thus When x is positive, the spring is in tension. When x is negative, the spring is in compression. Many materials have the same properties in tension and compression.
That is, if a tensile force F is required to lengthen the spring by , then a compressive force of the same magnitude F is required to shorten the spring by . For these materials, or fis an odd function of x. The Taylor series expansion of an odd func- tion cannot contain even powers. Thus, Equation (2.2) becomes
(2.3) F = k1x + k3x3 + k5x5 + Á
f (-x) = -f (x),
d d k0 = 0.
F = k0 + k1x + k2x2 + k3x3 + Á F = f (x)
All springs are inherently nonlinear. However in many situations xis small enough that the nonlinear terms of Equation (2.3) are small compared with k1x. A linear spring obeys a force-displacement law of
(2.4) where kis called the spring stiffness or spring constant and has dimensions of force per length.
Thus, for a linear spring, , which is illustrated in Figure 2.1.
The work done by a force is calculated according to Equation (1.40). For a linear system where the spring force is applied to a particle whose displacement is x, in the hori- zontal direction the force is represented by –kxi, and the differential displacement vector is dxi. The work done by the spring force as its point of application moves from a position described by to a position described by x2is
(2.5) Since the work depends upon the initial and final position of the point of application of the spring force and not the path of the system, the spring force is conservative. A poten- tial energy functioncan be defined for a spring as
(2.6) where xis the change in the length of the spring from its unstretched length.
A torsional spring is a link in a mechanical system where application of a torque leads to an angular displacement between the ends of the torsional spring. A linear torsional spring has a relationship between an applied moment M and the angular displacement of
(2.7) where the torsional stiffness kthas dimensions of force times length. The potential energy function for a torsional spring is
(2.8)
2.2.2 HELICAL COIL SPRINGS
The helical coil spring is used in applications such as industrial machines and vehicle sus- pension systems. Consider a spring manufactured from a rod of circular cross section of diameter D. The shear modulus of the rod is G. The rod is formed into a coil of N turns of radius r. It is assumed that the coil radius is much larger than the radius of the rod and that the normal to the plane of one coil nearly coincides with the axis of the spring.
Consider a helical coil spring when subject to an axial load F. Imagine cutting the rod with a knife at an arbitrary location in a coil, slicing the spring in two sections. The cut exposes an internal shear force F and an internal resisting torque Fr, as illustrated in
V = 1 2ktu2 M = ktu
u V(x) =
1 2 kx2 U1:2 =
L
x2 x1
(-kx)dx = k x21 2 - k x22
2 x1
k =
df dx|
x=0
F = kx
f
x k = (0) = slope of
tangent df
dx tangent at x = 0
Actual force deflection curve
FIGURE 2.1
The spring stiffness is the derivative of the force displacement relation atx=0.
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Figure 2.2. Assuming elastic behavior, the shear stress due to the resisting torque varies linearly with distance from the center of the rod to a maximum of
(2.9) where is the polar moment of inertia of the rod. The shear stress due to the shear force varies nonlinearly with distance from the neutral axis. For the maxi- mum shear stress due to the internal shear force is much less than the maximum shear stress due to the resisting torque, and its effect is neglected.
Principles of mechanics of materials can be used to show that the total change in length of the spring due to an applied force F is
(2.10) Comparing Equation (2.10) with Equation (2.4) leads to the conclusion that under the assumptions stated a helical coil spring can be modeled as a linear spring of stiffness
(2.11) k = GD4
64Nr3 x = 64Fr3N
GD4
r/DW1 J = (pD4)/32
tmax= FrD
2J = 16Fr
pD3
F
F
T = Fr FIGURE 2.2
A spring is subject to a forceFalong its axis. A section cut of the spring reveals its cross section has a shear force Fand a torqueFrwhere ris the coil radius.
E X A M P L E 2 . 1
A tightly wound spring is made from a 20-mm-diameter bar of 0.2% C-hardened steel (G80109N/m2). The coil diameter is 20 cm. The spring has 30 coils. What is the largest force that can be applied such that the elastic strength in shear of 220 106N/m2 is not exceeded? What is the change in length of the spring when this force is applied?
S O L U T I O N
Assuming the shear stress due to the shear force is negligible, the maximum shear stress in the spring when a force F is applied is
Thus the maximum allowable force is
The stiffness of this spring is calculated by using Equation (2.11):
The total changes in length of the spring due to application of the maximum allowable force is
¢ = F
k = 0.518 m k =
(80 * 109N/m2)(0.02m)4
(64)(30)(0.1m3) = 6.67 * 103N m Fmax = 220 * 106 N/m2
6.37 * 104
= 3.45 * 103 N t = FrD
2J = F (0.1 m)(0.02m) 2p
32(0.02m)4
= 6.37 * 104F
2.2.3 ELASTIC ELEMENTS AS SPRINGS
Application of a force F to the block of mass m of Figure 2.3 results in a displacement x. The block is attached to a uniform thin rod of elastic modulus E, unstretched length L, and cross- sectional area A. Application of the force results in a uniform normal strain in the rod of
(2.12) The strain energy per volume is the area under the stress–strain curve, which for an elastic bar:
(2.13) The total strain energy is
(2.14) If the force is suddenly removed, the block will oscillate about its equilibrium position. The initial strain energy is converted to kinetic energy and vice versa, a process which contin- ues indefinitely. If the mass of the rod is small compared to the mass of the block, then inertia of the rod is negligible and the rod behaves as a discrete spring. From strength of materials, the force F required to change the length of the rod by xis
(2.15) A comparison of Equation (2.15) with Equation (2.4) implies that the stiffness of the rod is
(2.16) The motion of a particle attached to an elastic element can be modeled as a particle attached to a linear spring, provided the mass of the element is small compared to the mass of the particle and a linear relationship between force and displacement exists for the ele- ment. In Figure 2.4, a particle of mass m is attached to the midspan of a simply supported beam of length L, elastic modulus E, and cross-sectional moment of inertia I. The trans- verse displacement of the midspan of the beam due to an applied static load F is
(2.17) Thus a linear relationship exists between transverse displacement and static load. Hence if the mass of the beam is small, the vibrations of the particle can be modeled as the vertical motion of a particle attached to a spring of stiffness
(2.18) k = 48EI
L3 x = L3
48EI F k = AE
L F = AE
L x S = sV =
1
2EE2AL = 1
2(EA/L)x2 s = 1
2sE = 1 2EE2 E = F
AE = x L
x
L
A, E FIGURE 2.3
Longitudinal vibrations of a mass attached to the end of a uniform thin rod can be modeled as a linear mass-spring system withk = AE/L.
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In general the transverse vibrations of a particle attached to a beam can be modeled as those of a particle attached to a linear spring. Let w(z) represent the displacement function of the beam due to a concentrated unit load applied at z a. Then the displacement at za due to a load Fapplied at zais
(2.19) Then the spring stiffness for a particle placed at z ais
(2.20) k = 1
v(a) x = v(a)F
x
x
L/2 L/2
m
m 48EI (a)
(b) k =
L3
FIGURE 2.4
The transverse vibrations of a machine attached to the midspan of a simply supported beam (a) mod- eled by a mass-spring system and the stiffness of the spring is 48EI/L3. (b) provided the mass of the beam is small in com- parison to the mass of the machine.
E X A M P L E 2 . 2
A 200-kg machine is attached to the end of a cantilever beam of length L 2.5 m, elastic modulus E 200 109N/m2, and cross-sectional moment of inertia 1.8 10–6 m4. Assuming the mass of the beam is small compared to the mass of the machine, what is the stiffness of the beam?
S O L U T I O N
From Table D.2 the deflection equation for a cantilever beam with a concentrated unit load at z Lis
(a) The deflection at the end of the beam is
(b) The stiffness of the cantilever beam at its end is
k = 3 EI (c) L3 =
3 (200 * 109 N/m2) (1.8 * 10-6 m4)
(2.5 m)3 = 6.91 * 104 N/m v(L) = 1
EIa-L3 6 + L
2L2b = L3 3EI v(z) =
1 EIa-
1 6z3 +
L 2z2b
Equation (2.18) is used for the stiffness of a pinned-pinned beam at its midspan. The equation for the stiffness of a cantilever beam at its end is
(2.21) The equivalent stiffness of a fixed-fixed beam at its midspan is
(2.22)
2.2.4 STATIC DEFLECTION
When a spring is not in its unstreched length when a system is in equilibrium, the spring has a static deflection. When the system of Figure 2.5(b) is in equilibrium a static force in the spring is necessary to balance the gravity force. From the FBD of Figure 2.5(b) the force in the spring is Fsmg. Since the force is the stiffness times the change in length from its unstretched length, the static deflection is calculated as
(2.23)
¢s = mg
k k =
192EI L3 k = 3EI
L3
k
m
mg k∆s
(a) (b)
FIGURE 2.5
(a) The spring has a static spring force when the system is in static equilibrium.
(b) FBD of the mass when the system is in equilibrium.
r2 r1
m1 m2
m2g T2 T1
k∆s
(a)
R
m1g
(b)
FIGURE 2.6
(a) System of Example 2.3. (b) FBDs of system when it is in equilibrium.
E X A M P L E 2 . 3
Determine the static deflection of the spring in the system of Figure 2.6(a).
S O L U T I O N
The FBDs of the system in its equilibrium position are shown in Figure 2.6(b). Summing forces to zero on the FBD of the left hand block leads to
(a) Summing moments about the center of the disk leads to as
(b) from which the static deflection is determined as
(c)
¢s =
m1gr1 - m2gr2 kr1 m2gr2 - (m1g - k¢
s)r1 = 0
πMO = 0, T1 = m1g - k¢
s
πF = 0
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Torsional oscillations occur in the system of Figure 2.7. A thin disk of mass moment of inertia I is attached to a circular shaft of length L, shear modulus G, and polar moment of inertia J. When the disk is rotated through an angle from its equilibrium position, a moment
(2.24) develops between the disk and the shaft. Thus, if the polar mass moment of inertia of the shaft is small compared with I, then the shaft acts as a torsional spring of stiffness
(2.25)