Vibration Two

Kingdom of Saudi Arabia

Ministry of Higher Education

Umm Al-Qura University College of Engineering & Islamic Architecture

Mechanical Engineering Department

Mechanical Vibrations

804 420 3

Lecture No. 2

Dr. Mohammad S. Alsoufi BSc, MSc PhD

Room No.: 1080

Tel.: 00966 (012) 5270000 Ext.: 1163

E-mail: [email protected]





2. Spring Elements

A spring is a type of mechanical link, which in most applications is assumed to have negligible mass and damping.

The most common type of spring is the helical-coil spring used in retractable pens and pencils, staplers, and suspensions

of freight trucks and other vehicles (see Figure (a)).

A spring is said to be linear if the elongation or reduction in length x is related to the applied force F as

helical-coil spring

F = kx

Where k is a constant, known as the spring constant or spring stiffness or spring rate. The spring constant k is always positive and denotes the force (positive or negative) required

causing a unit deflection (elongation or reduction in length) in the spring.

When the spring is stretched (or compressed) under a tensile (or compressive) force F, according to Newtons 3rd Law of motion, a restoring force or reaction of magnitude is developed opposite to the applied force.

This restoring force tries to bring the stretched (or compressed) spring back to its original unstretched or free length as shown in Figure below (b) or (c).

If we plot a graph between F and x, the result is a straight line according to F = kx.

The work done (U) in deforming a spring is stored as strain or potential energy in the spring, and it is given by

Deformation of a spring

Hooks Law: is that a principle of physics that state the force F needed to extend or compress a spring by some

distance x is proportional to that distance.





2.1 Nonlinear Springs

Most springs used in practical systems exhibit a nonlinear force-deflection relation, particularly when the deflections are large.

In vibration analysis, nonlinear springs whose force-deflection relations are given by

a denotes the constant associated with the linear part

b indicates the constant associated with the (cubic) nonlinearity.

Nonlinear and linear springs

The spring is said to be Hard b > 0 Linear b = 0 Soft b < 0

The force-deflection relations for various values of b are shown in Figure

Some systems, involving two or more springs, may exhibit a nonlinear force-

displacement relationship although the

individual springs are linear (see Figure

below).

In Figure (a), the weight (or force) W travels freely through the clearances c1

and c2 present in the system.

Once the weight comes into contact with a particular spring, after passing through

the corresponding clearance, the spring

force increases in proportion to the

spring constant of the particular spring as

shown in Figure (b).

It can be seen that the resulting force

displacement relation, although

piecewise linear, denotes a nonlinear

relationship.





Nonlinear spring force-displacement relation

In Figure (a), the two springs, with stiffnesses k1 and k2 have different lengths.

Note that the spring with stiffness k1 is shown, for simplicity, in the form of two parallel springs, each with a stiffness of k1/2.

Spring arrangement models of this type can be used in the vibration analysis of packages and suspensions used in aircraft landing gears.

When the spring k1 deflects by an amount x = c, the second spring starts providing an additional stiffness k2 to the system.

The resulting nonlinear force displacement relationship is shown in Figure (b).

2.2 Linearization of a Nonlinear Spring

Actual springs are nonlinear and follow F = kx only up to a certain deformation.

Beyond a certain value of deformation (after point A in Figure below), the stress exceeds the yield point of the material and the force-deformation relation becomes nonlinear.

Nonlinearity beyond proportionality limit





In many practical applications, we assume

that the deflections are

small and make use of

the linear relation in

F = kx

Even, if the force-deflection relation of a

spring is nonlinear, as

shown in Figure below,

we often approximate it

as a linear one by using a

linearization process.

Linearization process

To illustrate the linearization process:

Let the static equilibrium load F acting on the spring causes a deflection of x*.

If an incremental force F is added to F, the spring deflects by an additional quantity x.

The new spring force F + F can be expressed using Taylors series expansion about the static equilibrium position x* as

( )

( )

| ( )

|

( )

For small values of x, the higher-order derivative terms can be neglected to obtain

( )

| ( )

Since F = F (x*), we can express F as

Where k is the linearized spring constant at x* given by

|

We may use for simplicity, but sometimes the error involved in the approximation may be very large.





Example No.1: Equivalent Linearized Spring Constant A precision milling machine, weighing 4500 N, is supported on a rubber mount. The force

deflection relationship of the rubber mount is given by

Where the force F and the deflection x are measured in pounds and inches, respectively.

Determine the equivalent linearized spring constant of the rubber mount at its static equilibrium position.

Solution:

The static equilibrium position of the rubber mount x* under the weight of the milling machine, can be determined

( )

( )

The roots of the cubic equation above, can be found (for example, using the function roots in MatLAB

) as

The static equilibrium position of the rubber mount is given by the real root of ( ) , The equivalent linear spring constant of the rubber mount at its static equilibrium position can be determined using

| ( ) ( )

The equivalent linear spring constant , predicts the static deflection of the milling machine as

The result is slightly different from the true value of 1.2547 cm. The error is due to the truncation of the higher-order derivative terms in

( )

( )

| ( )

| ( )





2.3 Spring Constants of Elastic Elements

As stated earlier, any elastic or deformable member (or element) can be considered as a spring.

The equivalent spring constants of simple elastic members such as rods, beams, and hollow shafts are given later.

The procedure of finding the equivalent spring constant of elastic members is illustrated through the following examples.

Example No.2: Spring Constant of a Rod Find the equivalent spring constant of a uniform rod of length l, cross-sectional area A, and Youngs modulus E subjected to an axial tensile (or compressive) force F as shown in Figure below.

Spring constant of a rod

Solution:

The elongation (or shortening) of the rod under the axial tensile (or compressive) force F can be expressed as

is the strain

induced in the rod.

is the stress induced in the

rod.

Using the definition of the spring constant k, we obtain above.

The significance of the equivalent spring constant of the rod is shown in Figure (b).





Example No.3: Spring Constant of a Cantilever Beam

Find the equivalent spring constant of a cantilever beam subjected to a concentrated load F at its

end as shown in Figure (a) below.

Spring constant of a cantilever beam

Solution:

We assume, for simplicity, that the self-weight (or mass) of the beam is negligible and the concentrated load F is due to the weight of a point mass (W = mg).

From strength of materials, we know that the end deflection of the beam due to a concentrated load F = W is given by

Where E is the Youngs modulus and I is the moment of inertia of the cross section of the beam about the bending or z-axis (i.e., axis perpendicular to the page).

Hence the spring constant of the beam is Figure (b) above:

It is possible for a cantilever beam to be subjected to concentrated loads in two directions at its end one in the y direction Fy and the other in the z

direction Fz as shown in Figure (a) below.

When the load is applied along the y direction, the beam bends about z-axis Figure (b) and hence the equivalent spring constant will be equal to





When the load is applied along the z direction, the beam bends about the y-axis Figure (c) below and hence the equivalent spring constant will be equal to

Spring constants of a beam in two directions

The spring constants of beams with different end conditions can be found in a similar manner using results from strength of materials.

For example, to find the spring constant of a fixed-fixed beam subjected to a concentrated force P at (x = a), first we express the deflection of the beam at the

load point (x = a) using b = l a, as

Fixed-fixed Beam

( )

[ ( )]

( ) ( )

Then, find the spring constant k as

( ) ( )

Where I = Izz

The effect of the self-weight (or mass) of the beam can also be included in finding the spring constant of the beam (will discuss later).





2.4 Combination of springs

Case No.1 Case No.2

Springs in Parallel

Springs in Series

Case No.1: Springs in Parallel

To derive an expression for the equivalent spring constant of springs

connected in parallel, consider the two

springs shown in Figure (a) below.

When a load W is applied, the system undergoes a static deflection as shown in

Figure (b).

Then, the free-body diagram, shown in Figure (c), gives the equilibrium

equation

Springs in parallel

If keq denotes the equivalent spring constant of the combination of the two springs, then for the same static deflection st, we have

Then,

In general, if we have n springs with spring constants k1, k2, k3, , kn, in parallel, then the equivalent spring constant keq can be obtained





Case No.2: Springs in Series

To derive an expression for the equivalent spring constant of springs connected in

series by considering the two springs

shown in Figure (a).

Under the action of a load W, springs 1 and 2 undergo elongations 1 and 2 respectively, as shown in Figure (b).

The total elongation (or static deflection) of the system, st, is given by

Springs in series

Since both springs are subjected to the same force W, we have the equilibrium shown in Figure (c)

If keq denotes the equivalent spring constant, then for the same static deflection

Equations above give

Or

Substituting these values of 1 and 2 into , we obtain

That is

The equation above can be generalized to the case of n springs in series





In certain applications, springs are connected to rigid components such as pulleys, levers, and gears.

In such cases, an equivalent spring constant can be found using energy equivalence, as illustrated later in examples.

Example No.4: Equivalent k of a Suspension System Figure below shows the suspension system of a freight truck with a parallel spring arrangement.

Find the equivalent spring constant of the suspension if each of the three helical springs is made of steel with a shear modulus G = 80 10

9 N/m

2 and has five effective turns, mean

coil diameter D = 20 cm, and wire diameter d = 2 cm

Parallel arrangement of springs in a freight truck

(Courtesy of Buckeye Steel Castings Company)

Solution:

The stiffness of each helical spring is given by

( ) ( )

( ) ( )

(See formula sheet)

Since the three springs are identical and parallel, the equivalent spring constant of the suspension system is given by

( )





Example No.5: Torsional Spring Constant of a Propeller Shaft

Determine the torsional spring constant of the steel propeller shaft shown in Figure below

Propeller shaft

Solution:

We need to consider the segments 12 and 23 of the shaft as springs in combination.

The torque induced at any cross section of the shaft (such as AA or BB) can be seen to be equal to the torque applied at the propeller, T.

The elasticities (springs) corresponding to the two segments 12 and 23 are to be considered as series springs.

The spring constants of segments 12 and 23 of the shaft ( kt12 and kt23) are given by

(

)

( ) ( )

( )

(

)

( ) ( )

( )

Since the springs are in series

( )( )

( ) ( )





Example No.6: Equivalent k of Hoisting Drum

A hoisting drum, carrying a steel wire rope, is mounted at the end of a cantilever beam as shown in

Figure (a) below.

Determine the equivalent spring constant of the system when the suspended length of the wire rope is l.

Assume that the net cross-sectional diameter of the wire rope is d and the Youngs modulus of the beam and the wire rope is E.

Solution:

The spring constant of the cantilever beam is given by

(

)

Where,

The stiffness of the wire rope subjected to axial loading is

Where,

Since both the wire rope and the cantilever beam experience the same load W, as shown in Figure (b), they can be modeled as springs in series, as shown in Figure (c).

The equivalent spring constant is given by

(

)





Problem No.1: Determine the equivalent spring constant of the system shown in Figure below?

Springs in series-parallel

Solution No.1:

Problem No.2: find the equivalent spring constant of the system in the direction of





Solution No.2:

Problem No.3: Find the equivalent torsional spring constant of the system shown in Figure below.

Assume that k1, k2, k3, and k4 are torsional and k5 and k6 are linear spring constants

Solution No.3:





Problem No.4: A machine of mass m = 500 kg is mounted on a simply supported steel beam of

length l = 2 m having a rectangular cross section (depth = 0.1 m, width = 1.2 m) and Youngs modulus E = 2.06 10

11 N/m

2. To reduce the vertical deflection of the beam, a spring of stiffness k

is attached at mid-span, as shown in Figure below.

Determine the value of k needed to reduce the deflection of the beam by

A. 25 percent of its original value. B. 50 percent of its original value. C. 75 percent of its original value.

Note: Assume that the mass of the beam is negligible.

Solution No.4:





Problem No.5: Find the length of the equivalent uniform hollow shaft of inner diameter d and

thickness t that has the same axial spring constant as that of the solid conical shaft shown in Figure

below?

Solution No.5:





Problem No.6: Four identical rigid bars each of length a are connected to a spring of stiffness k to

form a structure for carrying a vertical load P, as shown in Figures (a) and (b).

Find the equivalent spring constant of the system for each case, disregarding the masses of the bars and the friction in the joints.

Solution No.6:





Problem No.7: The tripod shown in Figure below is used for mounting an electronic instrument

that finds the distance between two points in space. The legs of the tripod are located symmetrically

about the mid-vertical axis, each leg making an angle with the vertical. If each leg has a length l and axial stiffness k, find the equivalent spring stiffness of the tripod in the vertical direction.

Solution No.7:

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Vibration Two

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