Springs are one of the most important components in
vibration analysis because they are mechanical link between two particles in a
mechanical system. In general a spring is a continuous system. As usual the
spring has inertia but when we compare a vale of spring inertia to the other
elements in the mechanical system we can see that this value is small so we can
neglect it.
The length of a spring when it is not subject to
external forces is called the un-stretched length. Since the spring is made of
a flexible material, the applied force F that must be applied to the spring to
change its length by x is some continuous function of x.
The appropriate form of the f(x) is determined by
using the constitutive equation for the spring’s material. Since f(x)
id infinitely differentiable at x =
0, it can be expanded by a Taylor series about x = 0. This procedure is called MacLaurin expansion:
Since x is
the spring’s change in length from its unscratched length, when x=0, F=0.
Thus k0=0. When x is positive, the
spring is in tension. When x is negative, the spring 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, f(x) must be an odd function and cannot
contain even powers.
In reality all the springs are nonlinear. However in
many situations x is small enough that the nonlinear terms of previous equation
are small compared to the first variable that is k1*x. A linear spring obeys a
force displacement law of
where k is
called the spring stiffness or spring constant and has dimensions of force pre
length. The force in a spring whose force displacement law is given by (1.3) is
conservative that is the work done by the spring force between two arbitrary
displacement is independent of the path taken between the two displacement.
Thus a potential energy function exists and for a linear spring is determined
as:
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 fi of
where a torsional stiffness kt has dimensions of force
times length. The potential energy function for a torsional spring is
Example 1.1
The numerical values for the parameters in the system
are:
Determine the potential energy in the spring when the
system is in the equilibrium. Develop an expression for the potential energy in
the spring when the pulley is rotated clockwise from the equilibrium position
Figure 1.1 – Free body diagram of
static equilibrium position
Let's start solving the example by
defining Δ as statitc deflection of the spring, its change in length from its
unstreched length when the system is in equilibrium. The static deflection is
determined by applying hte equations of equilibrium to the free body diagram.
The free body diagram show that it
was assumed that the spring is compressed when the system is in equilibrium. The
negative sign indiates that the spring is actually stretched. Its potential
energy when the system is in equilibirim is:
When the pulley is rotated through a
clockwise angle θ from the system's equilibrium position, the spring is
shortened in lenght 2rθ from its length when the system is in equilibrium. Thus
the total change in lenght of the spring is
and the potential energy in the
spring is:
1.1.2.Helical coil springs
The helical coil spring is used in applications such
as industrial machines and vehicle suspension systems. Consider a spring
manufacture from a rod of circular cross section of diameter D. The shear
modulus of the rod is G. The rod is formed into 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.
Figure 1.2. Free body diagram of cut coil spring exposes
resultant shear force and resultant torque.
Consider a helical coil spring when subject to an axis
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 previous figure.
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
Where J = Pi*D^4/32 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 r/D>>1 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:
Example 1.2
A tightly wound spring is made from a 20 mm diameter
bar of 0.2% C hardened steel (G=80e9 N/m^2). 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 e6 N/m^2 is not exceeded? What is the
change in length of the spring when this force is applied?
Assuming 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:
The total change in length of the spring due to
application of the maximum allowable force is:
1.1.3.Elastic elements as spring
Application of a force F to the block of mass m of
figure 1.3, results in a displacement x. The block is attached to a uniform
thin rod of elastic modulus E, un-stretched length L, and cross sectional area
A. Application of the force results in uniform normal strain in the rod of:
The total strain energy developed due to the work of the force in stretching the rod is:
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 continues 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 x is:
A comparison of Eq. (1.13) and Eq. (1.4) implies that the stiffness of the rod is:
Figure1.3 – Longitudinal vibrations of mass attached
to end of uniform thin rod can be modeled as a linear mass-spring system with
k=AE/L
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 element. In figure
1.4 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
transverse displacement of the midspan of the beam due to applied static load F
is
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.
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 z=a due to a load F applied at z =a is:
Then the spring stiffness for a particle placed at z =
a is
Figure 1.4 – a) The transverse vibrations of a block
attached to a simply supported beam, b) are modeled by the mas spring system,
provided the mas of the block beam is small compared to the mass of the block
Figure 1.5 – Torsional stiffness of the shaft is JG/L
Torsional oscillations occur in the system of Fig 1.5.
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
develops between the dis 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:
Example 1.3
A 200 kg machine is attached to the end of a
cantilever beam of length L =2.5 m. Elastic modulus E=200e9 N/m^2, and cross
sectional moment of inertia 1.8e-6m^4. Assuming the mass of the beam is small
compared to the mass of the machine, what is the stiffness of the beam?
1.1.4. Spring in combination
Most of vibrational problems include more than one
spring. That means they are in some combination. In this section will talk
about what combinations of springs exist and how to solve them. The solution is
very simple. You need to replace this combination of springs with one which has
equivalent stiffness. The equivalent stiffness is determined such that the
system with a combination of springs has the same displacement x as the
equivalent system when both systems are subject to the same force F. A model of
one degree of freedom system consisting of a block attached to a spring of an
equivalent stiffness is illustrated in next figure.
The resultant force acting on the block is :
The springs in the system are parallel. The displacement
of each spring in the system is the same, but the resultant force acting on the
block is the sum of the forces developed in the parallel springs. If x is the
displacement of the block, then the force developed in the i-th spring is ki*x
and the resultant is
Equating the forces form leads to
The springs shown in next figure are in series. The
force developed in each spring is the same and equal to the force acting on the
block. The displacement of the block is the sum of the changes in length of the
springs in the series combination. If xi is the change in length of the i-th spring then
Since the force is the same in each spring, xi=F/ki
and eq becomes
Since the series combinations is to be replaced by a
spring of an equivalent stiffness is used leading to
We can see similarity with electrical components
because electrical circuit component can also be placed in series and parallel
and effect of the combination replaced by a single component with and
equivalent value. The equivalent capacitance of capacitors in parallel or
series is calculated like that of springs in parallel or series. The equivalent
resistance of resistors in series is the sum of the resistances, whereas the
equivalent resistance of resistors in parallel is calculated by using an
equation similar to Eq.(1.26).
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