Force, Pressure and Friction



~~~~MECHANICAL BASIC CONCEPTS~~~~ | |

|~~~A Brief Review of Fundamental Concepts related to Mechanical Stuff~~~ |

INDEX

Force, Pressure and Friction

Velocity and Acceleration

Work and Energy

Stress and Strain

Metal Fatigue

Fretting Corrosion

Predicting the Life of Rolling Element Bearings

Properties of Belt Drives

Basics of Vibration

Torsional Vibration Overview

Transmissibility

- Force, Pressure and Friction -

FORCE

A FORCE can be defined as "A push or a pull on an object". The FORCE (push or pull) may result from a contact between two objects, or from an influence in which no contact takes place, such as magnetism or gravitation. A FORCE can cause a change in motion of the object. If the object is not acted upon by other pushes and / or pulls which combine to form an equal and opposite counteracting action, then the FORCE will change the motion of the object to which it is applied.

Force is a vector quantity, meaning that it has both magnitude and direction. Forces are sometimes described in terms of magnitude only, and in many of those cases, the direction is self-evident.

Sir Isaac Newton, the 17th century English mathematician, formulated a series of observations about the basic behavior of forces on objects. Those observations have become known as "Newton's Laws of Motion", and are fundamental to the study of forces acting on objects. They are:

1. Every object continues in a state of rest or of uniform motion until it is compelled by a force to change its state of rest or motion.

2. The change in motion of an object is proportional to the net magnitude of the combination of the applied forces, and takes place along the straight line in which the ombination of the applied forces acts (sometimes stated as: F = MA, or  force = mass x acceleration).

3. For every action, there is an equal and opposite reaction. In other words, when two objects exert forces on each other, the forces are equal in magnitude, opposite in direction, and collinear.

The equation "F = MA" is a simplification of Newton's second law, but it has extreme significance. It means that the force required to accelerate an object is equal to the mass of the object multiplied by the desired acceleration. This simple equation forms the basis for determining the loads applied to objects as the result of motion ("dynamics").

Another common example of Newton's second law is the calculation of the force required to lift an object (its weight) The weight of an object is the acceleration of gravity (32.2 ft-per-second-per-second average on earth; quite different on other planets) times the mass of the object.

PRESSURE

A PRESSURE is the result of a FORCE being applied to a specific cross-sectional area, and is defined as FORCE per unit AREA, as in POUNDS per SQUARE INCH. For example, if a downward FORCE of 1000 pounds is applied evenly to a square plate of steel which measures 2" by 2" (4 square inches of area), then the PRESSURE applied to that block (Force per unit AREA) is determined by dividing the FORCE (1000 pounds) by the AREA (4 square inches), which is 250 pounds per square inch ("psi").

If the same 1000 pound FORCE was applied to a plate which measured 2" x 4" (8 square inches), then the PRESSURE would be reduced to 125 psi because the area of the plate doubled. The same force is being applied over a greater area, resulting in a LOWER force per unit area.

Taking it a step further, suppose you have a hydraulic cylinder with a 1/2" diameter piston. The area of that piston = diameter x diameter x 0.785, or in this case, 0.5 x 0.5 x 0.785 = 0.196 square inches. Now, if you apply 1000 pounds to the rod of that cylinder, the 1000 pound FORCE is applied by the rod to the piston, which acts against the oil in the cylinder to produce a pressure in the oil of 5102 (1000 / 0.196 = 5102) psi. If that oil is routed through some tubing to another hydraulic cylinder which has a 2.5 inch diameter piston, then the 5102 psi will be applied to the 4.91 square inch piston (2.5 x.2.5 x .785 = 4.91) and results in a 25, 050 pound force being available at the end of the rod on that cylinder.

FRICTION

FRICTION is an especially interesting example of a force. It is the resistance to motion which takes place when one body is moved upon another. Friction is generally defined as "that force which acts between two bodies at their surface of contact, so as to resist their sliding on each other".

Suppose that a block of metal, weighing 40 pounds, is resting on a flat, horizontal table top. If, using an accurate tension scale, you exert a small horizontal force on the block, the block will not move. Now suppose you increase the horizontal force until the block moves, and you notice that the value of the force is 8 pounds.

You now have enough data to calculate an important friction parameter known as the coefficient of friction (μ), which defines the nature of the resistance to motion these two bodies exert on each other. The value of the coefficient of friction (μ) is the horizontal force needed to move the block ( 8 lbs.) divided by the vertical force pressing the block and the table together (40 lbs.) μ = 8 / 40 = 0.20 )

There are several interesting properties of friction between dry, unlubricated surfaces, summarized as follows:

1. At low velocities, the friction is independent of the velocity of rubbing. As the velocity increases, the friction decreases. In other words, the force required to overcome friction and start a body into motion is greater than the force required to sustain the resulting motion. That fact is reflected in the existence of two different coefficients of friction for each material pair: the static coefficient and the dynamic coefficient.

2. For low contact pressures (normal {perpendicular} force per unit area), friction is directly proportional to the normal force between the two surfaces. As the contact pressure increases, the friction does not rise proportionately, and when the pressure becomes very high, friction increases rapidly until seizing takes place.

3. For a constant normal force, the friction, in both its total amount and its coefficient, is independent of the surface area in contact (as long as the pressure is not high enough to enter the seizing region).

Now suppose you apply a thin film of oil on the table under the block. The oil reduces the coefficient of friction to somewhere in the neighborhood of 0.025, so the block can now be moved with a horizontal force of about 1 pound (0.025 * 40 = 1).

The properties of friction between well-lubricated surfaces are considerably different from those above for dry surfaces.

1. The frictional resistance is almost independent of the contact pressure if the surfaces are flooded with oil.

2. For low contact pressures, the friction varies directly with velocity. For high contact pressures, the friction is very high at low velocities, dropping to a minimum at about 2 feet-per-second, then increasing as the square root of velocity.

3. For well-lubricated surfaces, the friction decreases dramatically with increasing temperature, from the influence of (a) rapidly-decreasing oil viscosity and (b) for a journal bearing, increasing diametral clearance.

4. If the bearing surfaces are flooded with oil, the friction is almost independent of the nature of the materials of the contact surfaces. As the lubrication diminishes, the coefficient of friction becomes more dependent on the materials.

60 Velocity and Acceleration –

VELOCITY

Linear Velocity is the measure of the linear (straight-line) distance something moves in a specified amount of time. It is typically calculated as distance divided by time, and has units such as feet-per-second, miles-per-hour, and furlongs-per-fortnight.

Velocity, in a strict definition, is a vector quantity, meaning that it has both magnitude and direction. Technically, "speed" is the magnitude portion of a velocity. However, velocity is sometimes given in terms of magnitude only, and in many of those cases, the direction is irrelevant. In others, the direction is self-evident.

Angular velocity is the measure of rotational distance something moves in a specified amount of time. It is typically expressed in units such as revolutions-per-minute (RPM) and degrees-per-second.

For many engineering calculations, it is necessary to express angular velocity in units of radians-per-second, rather than degrees-per-second or RPM. (The explanation of "why" requires more math than is appropriate here, but suffice it to say it is necessary in order to make the numbers work out right).

A radian is an angular measurement equal to approximately 57.3 degrees. It is defined as the angle formed by an arc on the circumference of a circle, the length of which is equal to the radius of that circle. Since the circumference of a circle is the radius times 2π, then obviously the value of a radian is the angle 360° divided by 2π, or 57.29578 degrees.

ACCELERATION

Acceleration is the measurement of how quickly the velocity of an object is changing, usually with respect to time. If you measure the velocity of an object at a particular time (Time1), then again at a subsequent time (Time2), then the average acceleration which the object has experienced will be:

Acceleration = (Velocity2 - Velocity1) / (Time2-Time1)

Clearly, the longer the period of time over which the measurements are taken, the more that value becomes an average, and the less will be known about the instantaneous acceleration of the object.

Acceleration is a critically important value for dynamic systems, because it is the instantaneous acceleration imposed on moving (dynamic) components, along with the mass of the components, which determines the actual forces raquired or applied in order to get components within the a system to change velocity from one value to another (Newton's second law), covered previously in Force, Pressure and Friction).

Linear acceleration is typically expressed in inches-per-second-per-second and feet-per-second-per-second(velocity per unit time). Common units of angular acceleration are degrees-per-second-per-second, radians-per-second-per-second and RPM-per-second.

However, acceleration (and velocity as well) need not be expressed with respect to time. For example, the acceleration value typically used in camshaft lobe design is inches-per-degree-per-degree or inches-per-degree² . This value is the acceleration which a cam lobe applies to the cam follower it is driving. In order to calculate the forces a cam applies to its mating components, the cam lobe angular velocity with respect to time must be known. Using that value, the lobe acceleration value can then be converted into inches-per-second-per-second, from which the forces are then calculated. In 2004, for some undiscernible, but most certainly politically-correct, reason, the cam design community apparently switched to metric units { velocity in mm/deg and lobe acceleration in mm/deg² }.

- Work and Energy -

WORK

Suppose the engine of your car stalled while you were in line to exit from a flat, level parking lot. You try several times to restart it, but it just won't start.

Since you are a considerate person, you decide to push your car out of the way of the people behind you. You get out and go round back and begin to push on the car. Suppose also that you are a fairly strong person, so you exert a horizontal force of 100 pounds on the rear of the car. The car doesn’t move. But you are also a persistent person, so you continue to push on the car for two whole minutes, exerting the same 100 pounds of force. The car still won’t move. Although you will probably be quite tired, you will have done NO WORK.

WHY? Because WORK is defined as a FORCE operating through a DISTANCE. The car didn’t move, so although there was FORCE, there was no MOTION.

Now you get smart and release the parking brake, and again push with the same constant 100 pound force. The car moves, and it travels 165 feet in two minutes. In that case, you will have produced 16,500 foot-pounds ofWORK (100 pounds of force x 165 feet of distance).

ENERGY

Later, you have recovered from the parking lot ordeal, and are working in your shop. You need to install a 3-inch long spring into a 2-inch space. The nature of this particular spring is that it takes 600 pounds of force to compress it one inch.

Using a lever-operated spring compressor, you pull on the lever with a force of 100 pounds and you move the lever 6 inches, causing the compressor to squeeze the spring and shorten it by 1 inch. The spring is now pushing on the compressor with a force of 600 pounds. You have stored the WORK you did on the compressor lever (100 pounds x 1 inch = 600 inch-pounds) in the spring, in the form of ENERGY (600 pounds and 1 inch).

ENERGY is defined as the CAPACITY of a body to do WORK, by virtue of the position or condition of the body.

Now suppose there is a 150-pound plate of steel on your bench, resting on four blocks which are 2 inches tall (so the space between the bottom of the plate and the bench is 2 inches). You install the compressed spring into that space and locate it at exactly the CG of the plate, and release the spring compressor.

The spring will lift the steel plate 3/4 of an inch, so the spring has done WORK on the plate, thereby releasing some of the ENERGY stored in the spring.

There are many different forms of energy. There are a few which are of particular interest with respect to powerplants: kinetic energy (the energy contained in a body by virtue of its velocity), potential energy (the energy contained in a body by virtue of its position), chemical energy (energy which can be released by a chemical reaction, such as combustion), and heat energy (energy which can be used to make machines operate).

- Stress and Strain -

n order to better understand the various discussions about loads, stresses and the expected life of components, it helps to have an understanding of the fundamental terms used in those discussions. The following paragraphs give a very brief background on those terms. Please note that this is a very basic explanation of subjects which some engineers spend their entire careers studying.

TENSILE and COMPRESSIVE STRESS

Stress is a value which describes the amount of load carried by each unit of cross sectional area of a component. For example, suppose the block shown in Figure 1 weighs 10,500 pounds and it is suspended from the shaft (with the arrowhead). The shaft diameter is 0.504", so it has a cross-sectional area of 0.200 square inches (in²).

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The stress that the 10,500 pound load applies to the shaft is defined asthe load divided by the cross-sectional area of the shaft, which is 10,500 pounds ÷ 0.200 square inches = 52,500 pounds per square inch (psi). Note that this "load per unit area" is the PRESSURE concept, presented inForce, Pressure and Friction.

If the diameter of the shaft is increased to 3/4", then the area increases to 0.4418 square inches (in²) and the stress decreases to 10,500 ÷ 0.4418 = 23,766 psi.

Since the load shown in the picture is trying to lengthen the shaft, it is called tensile stress. If we flipped the picture over so that the shaft was supporting the block from underneath, the block would be trying to shorten the shaft, and the stress would be called compressive stress.

The loads which produce tensile and compressive stresses are acting perpendicular to the areas on which they act. Tensile and compressive stresses are often referred to as normal stresses. Here, "normal" is not a behavioral term, but a geometric one, which, in this context means "acting perpendicular to" a particular area or plane.

SHEAR STRESS

Forces which act parallel to the areas resisting them are known as shear forces, and produce shear stress in the elements which carry those loads.

For example, suppose the two clevises (clevi?) in this picture have equal tensile forces acting on their ends ("normal" to the end faces), as depicted by the arrows. Those forces are trying to pull the clevises apart, and in so doing, they apply a shear force (parallel to the cross-sectional area of the bolt) on the bolt holding the parts together. The shear force is trying to cut the bolt in half across its diameter in two places, where the two outer faces of the lower clevis meet the two inner faces of the upper clevis. This instance is known as "double shear" because there are two separate areas of the bolt exposed to the shear force.

Suppose the bolt in this example is 3/4" diameter, and the tensile loads are both 20,000 pounds. The cross-sectional area of the bolt is 0.4418 square inches (in²). Then shear stress applied to the bolt would be 20,000 pounds divided by twice the bolt area (because the load is shared by two different cross-sectional areas of the bolt), or 22,634 psi.

It is important to note that the shear stress capacity of most metallic materials is considerably less than the tensile or compressive capacities.

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BENDING STRESS

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Bending Stress occurs when a component is loaded by forces which, instead of trying to stretch or shrink the component, are trying tobend it. Those bending forces generate a combination of tensile and compressive stress in the load-carrying components, known asbending stress.

An example of that type of loading is shown in tFigure 3, where the tubular shaft is resting on two triangular supports. The two vertical arrows represent downward forces applied to the tube. The tube deflects, or bends, downward under the influence of those forces, as illustrated.

If you cut a section through the tube at the right hand support and were able to visualize the internal stresses between metal molecules at the cut face, they would appear as tensile and compressive stresses, illustrated by the green arrows in Figure 4. The stress magnitude is largest at the outer extremities and decrease to zero at the geometric center

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The bending stress is calculated from theproperties of the cross-section and the magnitude of the bending moment. The bending moment is the magnitude of the applied force times the distance from the point where the force is applied to the cross section being examined. That horizontal distance from the point of application of the force (the purple arrow) to the cross-section is called the moment arm. (This is a massive simplification, but it illustrates the concept.)

STRAIN

Strain is the measure of how much a material deforms when a load is applied to it, expressed in inches of deformation per inch of material length. For example, if the 1/2" diameter shaft supporting the 10,000 pound load in Figure 1 is 12" long, it will stretch about 0.020" (20 thousandths) from its unloaded length, which you can measure. The strain is the measured deflection (0.020) divided by the length of the shaft, or 0.020 ÷ 12 = 0.00167 inches per inch.

Suppose you measure a specimen which has no load applied to it. Then you apply a load to the specimen, then release the load and measure the specimen again. If you find that it has returned to its original length, then the specimen experienced elastic deformation when it stretched under the load.

Most materials are elastic. That is, if you apply a load to the material, it will deform in some way, by an amount which is proportional to the load. When you remove the load, the material will return to its original shape, as long as the load wasn't too large. The deformation might be too small to measure, but it still occurs.

If, after the load is released, you can measure some permanent deformation in the specimen, (the specimen does not return to its original length), then the material has been stressed beyond its elastic limit and has experienced plastic deformation.

YIELD STRESS and ULTIMATE STRESS

There are two basic values which characterize the strength of a metal. Each of those values is a stress level at which a particular event occurs.

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The dark line in Figure 5 shows the stretch of a specimen in response to an applied tensile stress. In this example, from 0 to to 70,000 psi the stretch (strain) is proportional to the stress, and if the load is removed, the specimen returns to its original length. This is known as elasticdeformation.

However, if the load (in this example) exceeds 70,000 psi, notice that the strain becomes greater per unit increase in stress, and is no longer proportional to the stress. Once this specimen has been loaded beyond 70,000 psi, the specimen does not return to its original length (red line) when the load is removed. The specimen has been permanently deformed (plastic deformation).

The stress level at which a material no longer behaves elastically, but instead experiences a small permanent (plastic) deformation is known as the Yield Stress (YS), (also known as theproportional limit). That is the stress level at which the elastic limit of the material has been exceeded.

The second interesting value is called the Ultimate Tensile Stress (UTS). It is the stress value at which the material will break under the influence of pure tensile stress.

A specialized test machine is used to measure those values. A small sample of the subject metal (usually 0.357" or 0.504" diameter) is installed in the machine, as shown in Figures

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The machine applies a tension load on the specimen, and increases the load until the specimen breaks. The computer which controls the machine plots the load vs. deflection in real time, and detects the yield point. At some point after the material begins to yield, the operator pauses the test and removes the deflection gauge (to prevent it from being destroyed when the specimen fractures). The operator then resumes the test and the machine continues to increase the load until the specimen breaks. The load at which the specimen broke is divided by the original cross-sectional area (0.100 in² for 0.357" diameter or 0.200 in² for 0.504" diameter) to find the UTS value. As the specimen approaches the failure point, a portion of the diameter (where the failure will occur) begins to plastically-deform and reduce in diameter ("necking-down"). The percentage that the cross sectional area of the material reduces is a measurement of the ductility (lack of brittleness) of the material, and is stated as %-reduction-in-area (ROA).

STRESS CONCENTRATION

It has been known for a long time that the presence of irregularities or discontinuities in a part (holes, rapid changes in diameter, shoulders, grooves, notches, etc.) significantly increases the value of the actual stress which occurs in a part when compared to the stress value calculated based on the cross section of the part. These increases in stress, called stress concentrations, occur in the immediate vicinity of the discontinuity.

The ratio of the actual stress to the calculated stress is known as a stress concentration factor. The magnitude of stress concentration factors can be 3, 4, or more depending on the severity of the particular discontinuity.

This phenomenon can be demonstrated in tests. There is a large body of accumulated data relating the physical characteristics of various types of discontinuities to the increase in observed stress they cause. There are several books which present the methods to calculate these factors. One of the most accepted works on this subject is Peterson's Stress Concentration Factors (ref-2:4). Some FEA systems have incorporated the effect of discontinuities into their calculations.

There are also several accepted methods to diminish the stress-increasing effect of discontinuities, including tapers, large fillet radii, radiused undercuts, and gentle discontinuities surrounding a necessarily abrupt one.

As an example of theses techniques, consider the picture below, showing a typical input shaft for an automotive transmission, with a shoulder for the supporting bearing and a snap-ring retaining the bearing on the shoulder.

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The following two pictures show a cross section through the shaft pictured above, without the bearing or snap ring installed. Figure 8 shows the typical shaft implementation, with a sharp-cornered snap ring groove (necessary to clear the edges of the snap ring), and a sharp-cornered shoulder, required to clear the small radius on the edge of the bearing, typically less than 0.040". The sharp corners in the groove and the shoulder provide severe concentrations (as much as 10 to 1) for the applied stresses, making the material behave in fatigue (next topic)as if the applied stress was significantly greater. Compare theshaft in Figure 8 to the better configuration shown in Figure 9, which has teardrop, large-radius grooves on both sides of the sharp-cornered snap ring groove, and which has a large radius undercut into the bearing shoulder. These provisions dramatically reduce the stress concentrations (down to as low as 1.5 to 1) as compared to following figure.

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Metal Fatigue –

Long ago, engineers discovered that if you repeatedly applied and then removed a nominal load to and from a metal part (known as a "cyclic load"), the part would break after a certain number of load-unload cycles, even when the maximum cyclic stress level applied was much lower than the UTS, and in fact, much lower than the Yield Stress (UTS and YS are explained in Stress and Strain). These relationships were first published by A. Z. Wöhler in 1858.

They discovered that as they reduced the magnitude of the cyclic stress, the part would survive more cycles before breaking. This behavior became known as "FATIGUE" because it was originally thought that the metal got "tired". When you bend a paper clip back and forth until it breaks, you are demonstrating fatigue behavior.

The following information on this page attempts to explain  metal fatigue by answering several common questions:

1. What is fatigue loading?

2. How do you determine the fatigue strength of a material?

3. Does the strength of a material affect its fatigue properties?

4. Why is the surface of a part so important?

5. Is fatigue life an exact number?

6. Do real-world parts behave the same as laboratory tests?

7. Are fatigue cycles cumulative?

WHAT IS FATIGUE LOADING?

There are different types of fatigue loading. One type is zero-to-max-to zero, where a part which is carrying no load is then subjected to a load, and later, the load is removed, so the part goes back to the no-load condition. An example of this type of loading is a chain used to haul logs behind a tractor.

Another type of fatigue loading is a varying load superimposed on a constant load. The suspension wires in a railroad bridge are an example of this type. The wires have a constant static tensile load from the weight of the bridge, and an additional tensile load when a train is on the bridge.

The worst case of fatigue loading is the case known as fully-reversing load. One cycle of this type of fatigue loading occurs when a tensile stress of some value is applied to an unloaded part and then released, then a compressive stress of the same value is applied and released.

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A rotating shaft with a bending load applied to it is a good example of fully reversing load. In order to visualize the fully-reversing nature of the load, picture the shaft in a fixed position (not rotating) but subjected to an applied bending load (as shown here). The outermost fibers on the shaft surface on the convex side of the deflection (upper surface in the picture) will be loaded in tension (upper green arrows), and the fibers on the opposite side will be loaded in compression (lower green arrows). Now, rotate the shaft 180° in its bearings, with the loads remaining the same. The shaft stress level is the same, but now the fibers which were loaded in compression before you rotated it are now loaded in tension, and vice-versa.

In fact, the laboratory mechanism used to test the fatigue life of materials is a rotating shaft with an applied bending load.

To illustrate how damaging fully-reversing load is, take a paper clip, bend it out straight, then pick a spot in the middle, and bend the clip 90° back and forth at that spot (from straight to "L" shaped and back). Because you are plastically-deforming the metal, you are, by definition, exceeding its yield stress. When you bend it in one direction, you are applying a high tensile stress to the fibers on one side of the OD, and a high compressive stress on the fibers on the opposite side. When you bend it the other way, you reverse the stresses (fully reversing fatigue). It will break in about 25 cycles.

The number of cycles that a metal can endure before it breaks is a complex function of the static and cyclic stress values, the alloy, heat-treatment and surface condition of the material, the hardness profile of the material, impurities in the material, the type of load applied, the operating temperature, and several other factors.

HOW IS THE FATIGUE STRENGTH OF A METAL DETERMINED?

The fatigue behavior of a specific material, heat-treated to a specific strength level, is determined by a series of laboratory tests on a large number of APPARENTLY IDENTICAL SAMPLES of that specific material.

This picture shows a laboratory fatigue specimen. These laboratory samples are optimized for fatigue life. They are machined with shape characteristics which maximize the fatigue life of a metal, and are highly polished to provide the surface characteristics which enable the best fatigue life.

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A single test consists of applying a known, constant tbending stress to a round sample of the material, and rotating the sample around the bending stress axis until it fails. As the sample rotates, the stress applied to any fiber on the outside surface of the sample varies from maximum-tensile to zero to maximum-compressive and back. The test mechanism counts the number of rotations (cycles) until the specimen fails. A large number of tests is run at each stress level of interest, and the results are statistically massaged to determine the expected number of cycles to failure at that stress level.

The cyclic stress level of the first set of tests is some large percentage of the Ultimate Tensile Stress (UTS), which produces failure in a relatively small number of cycles. Subsequent tests are run at lower cyclic stress values until a level is found at which the samples will survive 10 million cycles without failure. The cyclic stress level that the material can sustain for 10 million cycles is called the Endurance Limit (EL).

In general, steel alloys which are subjected to a cyclic stress level below the EL (properly adjusted for the specifics of the application) will not fail in fatigue. That property is commonly known as "infinite life". Most steel alloys exhibit the infinite life property, but it is interesting to note that most aluminum alloys as well as steels which have been case-hardened by carburizing, do not exhibit an infinite-life cyclic stress level (Endurance Limit).

IS THERE ANY RELATIONSHIP BETWEEN UTS AND FATIGUE STRENGTH?

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The endurance limit of steel displays some interesting properties. These are shown, in a general way, in this graph, (Figure 3) and briefly discussed below.

It is a simplistic rule of thumb that, for steels having a UTS less than 160,000 psi, the endurance limit for the material will beapproximately 45 to 50% of the UTS if the surface of the test specimen is smooth andpolished.

That relationship is shown by the line titled "50%". A very small number of special case materials can maintain that approximate 50% relationship above the 160,000 psi level.

However, the EL of most steels begins to fall away from the 50% line above a UTS of about 160,000 psi, as shown by the line titled "Polished".

For example, a specimen of SAE-4340 alloy steel, hardened to 32 Rockwell-C (HRc), will exhibit a UTS around 150,000 psi and an EL of about 75,000 psi, or 50% of the UTS. If you change the heat treatment process to achieve a hardness of about 50 HRc, the UTS will be about 260,000 psi, and the EL will be about 85,000 psi, which is only about 32% of the UTS.

Several other alloys known as "ultra-high-strength steels" (D-6AC, HP-9-4-30, AF-1410, and some maraging steels) have been demonstrated to have an EL as high as 45% of UTS at strengths as high as 300,000 psi. Also note that these values are EL numbers for fully-reversing bending fatigue. EL values for hertzian (contact) stress can be substantially higher (over 300 ksi).

The line titled "Notched" shows the dramatic reduction in fatigue strength as a result of the concentration of stress which occurs at sudden changes in cross-sectional area (sharp corners in grooves, fillets, etc.). The highest EL on that curve is about 25% of the UTS (at around 160,000 psi).

The surface finish of a material has a dramatic effect on the fatigue life. That fact is clearly illustrated by the curve titled "Corroded". It mirrors the shape of the "notched" curve, but is much lower. That curve shows that, for a badly corroded surface (fretting, oxidation, galvanic, etc.) the endurance limit of the material starts at around 20 ksi for materials of 40 ksi UTS (50%), increases to about 25 ksi for materials between 140 and 200 ksi UTS, then decreases back toward 20 ksi as the material UTS increases above 200 ksi.

WHY IS THE SURFACE SO IMPORTANT?

Fatigue failures almost always begin at the surface of a material. The reasons are that (a) the most highly-stresses fibers are located at the surface (bending fatigue) and (b) the intergranular flaws which precipitate tension failure are more frequently found at the surface.

Suppose that a particular specimen is being fatigue tested (as described above). Now suppose the fatigue test is halted after 20 to 25% of the expected life of the specimen and a small thickness of material is machined off the outer surface of the specimen, and the surface condition is restored to its original state. Now the fatigue test is resumed at the same stress level as before. The life of the part will be considerably longer than expected. If that process is repeated several times, the life of the part may be extended by several hundred percent, limited only by the available cross section of the specimen. (ref-3:8:6) That proves fatigue failures originate at the surface of a component.

IS THE ENDURANCE LIMIT AN EXACT NUMBER?

It is important to remember that the Endurance Limit of a material is not an absolute nor fully repeatable number. In fact, several apparently identical samples, cut from adjacent sections in one bar of steel, will produce differentEL values (as well as different UTS and YS) when tested, as illustrated by the S-N diagram below. Each of those three properties (UTS, YS, EL) is determined statistically, calculated from the (varying) results of a large number of apparently identical tests done on a population of apparently identical samples.

The plot below shows the results of a battery of fatigue tests on a specific material. The tests at each stress level form statistical clusters, as shown. a curve is fitted through the clusters of points, as shown below. The curve which is fitted through these clusters, known as an "S-N Diagram" (Stress vs. Number), represents the statistical behavior of the fatigue properties of that specific material at that specific strength level. The red points in the chart represent the cyclic stress for each test and the number of cycles at which the specimen broke. The blue points represent the stress levels and number of cycles applied to specimens which did not fail. This diagram clearly demonstrates the statistical nature of metal fatigue failure.

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DO REAL-WORLD COMPONENTS EXHIBIT THE "LABORATORY" EL?

Unfortunate experience has taught engineers that the value of the Endurance Limit found in laboratory tests of polished, optimized samples does not really apply to real-world components.

Because the EL values are statistical in nature, and determined on optimized, laboratory samples, good design practice requires the determination of the actual EL will be for each specific application, known as theAPPLICATION-SPECIFIC ENDURANCE LIMIT (ASEL).

In order to design for satisfactory fatigue life (prior to testing actual components), good practice requires that the "laboratory" Endurance Limit value be reduced by several adjustment factors. These reductions are necessary to account for:

        (a) the differences between the application and the testing environments, and

        (b) the known statistical variations of the material.

This procedure is to insure that both the known and the unpredictable factors in the application (including surface condition, actual load, actual temperature, tolerances, impurities, alloy variations, heat-treatment variations, stress concentrations, etc. etc. etc.) will not reduce the life of a part below the required value. Please read that paragraph again, and understand it well.

An accepted contemporary practice (ref-2:3:328) to estimate the maximum fatigue loading which a specific design can survive is the Marin method, in which the laboratory test-determined EL of the particular material (tested on optimized samples) is adjusted to estimate the maximum cyclic stress a particular part can survive (the ASEL).

This adjustment of the EL is the result of six fractional factors. Each of these six factors is calculated from known data which describe the influence of a specific condition on fatigue life.

Those factors are:

1. Surface Condition (ka): such as: polished, ground, machined, as-forged, corroded, etc. Surface is perhaps the most important influence on fatigue life;

2. Size (kb): This factor accounts for changes which occur when the actual size of the part or the cross-section differs from that of the test specimens;

3. Load (kc): This factor accounts for differences in loading (bending, axial, torsional) between the actual part and the test specimens;

4. Temperature (kd): This factor accounts for reductions in fatigue life which occur when the operating temperature of the part differs from room temperature (the testing temperature);

5. Reliability (ke): This factor accounts for the scatter of test data. For example, an 8% standard deviation in the test data requires a ke value of 0.868 for 95% reliability, and 0.753 for 99.9% reliability.

6. Miscellaneous (kf): This factor accounts for reductions from all other effects, including residual stresses, corrosion, plating, metal spraying, fretting, and others.

These six fractional factors are applied to the laboratory value of the material endurance limit to determine the allowable cyclic stress for an actual part:

Real-World Allowable Cyclic Stress = ka * kb * kc * kd * ke * kf * EL

IS FATIGUE LOADING CUMULATIVE?

It is important to realize that fatigue cycles are accumulative. Suppose a part which has been in service is removed and tested for cracks by a certified aircraft inspection station, a place where it is more likely that the subtleties of Magnaflux inspection are well-understood. Suppose the part passes the inspection, (i.e., no cracks are found) and the owner of the shaft puts it on the "good used parts" shelf.

Later, someone comes along looking for a bargain on such a part, and purchases this "inspected" part. The fact that the part has passed the inspection only proves that there are no detectable cracks RIGHT NOW. It gives no indication at all as to how many cycles remain until a crack forms. A part which has just passed a Magnaflux inspection could crack in the next 100 cycles of operation and fail in the next 10000 cycles (which at 2000 RPM, isn't very long!).

- Fretting Corrosion -

Fretting corrosion has been the cause of countless failures at the contact points of machinery components. (Clickhere to see several clear photos of fretting which resulted in fatigue fracture). The ASM Handbook on Fatigue and Fracture defines fretting as:

"Fretting is a special wear process that occurs at the contact area between two materials under load and subject to minute relative motion by vibration or some other force." (ref-3:9:321).

When two pieces of material, pressed together by an external static load, (for example, bolted flanges, riveted lap-joints, press-fits such as a gear or bearing on a shaft) are subjected to a transverse cyclic loading, so that one contacting face is relatively displaced cyclically parallel to the other face, in the presence of high contact stress, wear on the mating surfaces occurs. If the magnitude of the displacement is less than about 0.003 inches, the wear is termed "fretting".

Fretting occurs by contacting asperities on the mating surfaces continually welding together then breaking. That leads to surface pitting and the transfer of metal particles from one surface to another. In addition, the small fragments of metal which are broken off oxidize, forming oxide particles which, for most engineering metals, are harder than the mated parts. These particles become trapped between the mating surfaces and cause abrasive damage and scoring.

Briefly, the characteristics of fretting are:

1. It is most serious when oxygen is present, although it can also occur in an inert gas;

2. It is worst under perfectly dry conditions;

3. It increases with contact load, slip amplitude, and number of oscillations;

4. Soft materials generally exhibit more susceptibility to fretting than do hard materials of a similar type;

5. Lubricants, particularly when used with surface treatments such as phosphating, reduce fretting damage.

It is interesting to note that there is disagreement in the reference literature on the effectiveness of lubrication. Reference 3:9:324 says: "The introduction of a lubricant into the interface can make matters worse by increasing the relative slip".

Fretting appears to be particularly aggressive in cases of disks (gears, pulleys, wheels, flywheels, bearings, hubs, etc.) which are press-fit (shrink-fit) onto shafts which are subjected to reversing bending stress, and worse yet under the added influence of vibration. The stress concentration which occurs where the shaft just meets the disk compounds the problem.

Under fretting conditions, fracture cracks can initiate at very low stresses, well below the fatigue limit of non-fretted specimens. Fretting corrosion can reduce the endurance limit of steels to as little as 18% of their original values. The greatest reduction in fatigue strength occurs when the fretting process AND cyclic stressing are applied simultaneously. The fact that fatigue cracks can form under low nominal cyclic stresses in areas where fretting is occurring is dramatically illustrated by the well-known low fatigue limit of a shaft having a pressed-on bearing.(ref-3:8:365)

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This plot illustrates the dramatic effect that fretting has on the fatigue life of steels. The line titled "Corroded" mirrors the shape of the "notched" curve, but is much lower. The corroded curve shows that, for a badly corroded surface (fretting, oxidation, galvanic, etc.) the endurance limit of the material starts at around 20 ksi for materials of 40 ksi UTS (50%), increases to about 25 ksi for materials between 140 and 200 ksiUTS, then decreases back toward 20 ksi as the material UTS increases above 200 ksi.

Prevention of fretting fatigue in the design process is essential. Although there is ample descriptive material on the mechanism and examples of fretting, there is limited availability of generalized techniques or modeling methodology for the prediction of crack initiation due to fretting. Testing is usually required to find and validate a solution to a fretting problem.

- Predicting the Life of Rolling Element Bearings-

The prediction of the life of a rolling-element bearing (ball, roller, needle) is a statistical calculation of the fatigueproperties of the bearing components, in which life is stated as the number of hours that a specified percentageof a large population of apparently-identical bearings will survive under a specified load with a specified set ofoperating conditions.

From that definition, it should be clear that the prediction of the expected life of a rolling element bearing in a specific application involves a bit more analysis than simply plucking a load rating from a bearing catalog. Thedynamic load rating listed in bearing catalogs is the load at which 90% of a large population of apparently-identical bearings will survive one million cycles. A shaft turning at 6000 RPM will produce one million cycles in 2 hours and 46 minutes. That's probably OK for a race car, but not exactly an aircraft-quality life expectancy.

The life of a given bearing is a nonlinear function of the applied load. For a ball bearing, it is related to load to the3.00 power (i.e. load x load x load); For roller and needle bearings, it is the 3.33 power. That means that a relatively small increase in bearing load can cause a dramatic reduction in bearing life.

The usual life rating for industrial applications is called "L-10" life. The L-10 life is the number of hours in service that 90% of a large population of apparently-identical bearings will survive when subjected to the boundary conditions (load, speed, lubrication, material and cleanliness) that are specific to the application. Stated another way, 10% of that population will have failed in the L-10 number of service hours.

We think that the L-5 life (5% failure) criterion is more appropriate than L-10 for aircraft design. To achieve the 5% failure rate with a given set of boundary conditions requires 1.64 times greater bearing capacity (dynamic load rating); to achieve a 2% failure rate requires 3.0 times greater dynamic load rating. Put another way, for a given L-10 life, the L-5 life of the same bearing is 61% of the L-10 value, and the L-2 life is 33% of the L-10 value.

Most rolling element bearing manufacturers publish detailed life-load analysis procedures. The bearing life calculations which the manufacturers publish take into account factors including:

1. the combined effect of applied radial and thrust loads,

2. RPM,

3. pitchline velocity,

4. lubricant viscosity,

5. contamination,

6. bearing load ratings, and

7. desired probability of survival (failure rate).

All of them are important, but the effects of lubricant viscosity and cleanliness are huge.

These calculations have been implemented in a computer program which EPI wrote several years ago for doing bearing life analysis. This program uses a user-defined load model of the expected service to evaluate bearing life in an actual application. The load model is a set of different operating loads and speeds which the actual machine is designed to handle, and an estimate of the % of service that each load and speed represents.

For example, the load model of an an aircraft propeller gearbox on a normally-aspirated engine might be described as 5% of the time at max power (takeoff), 10% at climb power, 70% at cruise power, and 15% in aerobatic maneuvers. Each one of those conditions impose significantly different loads and speeds on the gearbox bearings. The combination of those different conditions in a realistic load model allows for a more reasonable selection of bearings for an application.

Certain bearing manufacturers (SKF, for example, in late 2003) have made available, on their websites interactive programs for predicting "L-10" bearing life.

The SKF website provides a life-calculation program which produces THREE different values for an L-10 life rating:

1. the original (old) Arvid Palmgren method, which predicts a relatively short life,

2. the seasoned, proven "A23" method (the one implemented in the EPI bearing program), and

3. the new, SKF-"marketing-friendly" rating.

The one you pick depends on how realistic you want to be.

Our bearing life calculation program matches the SKF-website values calculated for an L-10 life under the "A23" method for the same input values. However, our program also includes the ability to calculate life ratings for complete load models (described above) as well as for survival probabilities more appropriate to aircraft applications (95, 96, 97, 98 and 99%).

- Properties of Belt Drives -

This section describes the general properties of V-belt and Toothbelt transmission systems, and provides a detailed explanation of PRELOAD, a necessary component of all types of belt drives.

BELT DRIVE BASICS

In order for a belt drive to operate properly, the residual tension in the "loose" span (the non-driving span) of the belt can never be allowed to get near zero (unlike a chain drive, where the "loose" span can actually be loose.). That requirement is accomplished by establishing a static "preload" on the belt. The term "preload" means the establishment of a static tension value in all the spans of the belt.

For V-belt drives, preload maintains the contact force between the belt and the surface of the pulley grooves so that friction can transmit the power. For toothbelt drives, preload maintains the correct contact pattern between the belt teeth and sprocket grooves. The preload for V-belt drives is usually greater than that required for toothbelt drives.

Any belt will stretch when a load is applied to it, although the amount of stretch is usually very small. The amount of force to produce a specified stretch is known as the belt modulus.

The static preload in the system stretches each span (spring) of the belt equally, as illustrated in the picture below.

[pic]

The amount of preload required in a belt drive depends on a number of factors, including:

1. the maximum amount of torque which the drive must transmit;

2. the diameter of the driving sprocket (pulley);

3. the minimum arc of contact between the belt and sprockets ("wrap");

4. the properties of the belt.

Preload can be established by means of adjusting the centerline distance between shafts (shimming) or by using an idler which imposes a side load on the "loose" span.

Everyone is probably familiar with establishing preload on an automotive fan-belt system by wedging on the pivoting alternator and locking it in place with the slotted locator arm. You are also probably familiar with the result of not having enough preload in the drive: the belt will slip and squeal under high load. That happens frequently with power steering belts when you turn the steering wheel up against one of the stops.

When torque is applied to the driving sprocket (pulley) of a belt drive, one span of the belt gets tighter and stretches slightly in response to the additional load applied by the driving torque. At the same time, the load in the loose span reduces by the same amount, and the loose span shortens by the same amount the tight side stretched.

Clearly then, the tension in the tight (driving) span becomes greater than the preload, and conversely, the tension in the loose span becomes less than the preload. This behavior is illustrated in the picture below.

[pic]

When a belt drive is operating and transmitting power, the amount of force being transmitted from the driving sprocket (pulley) to the driven sprocket (pulley) is the difference between the tension in the " tight" (driving) span of the belt and the tension in the "loose" (following) span of the belt.

NOTE that the sum of the tight and loose strand tensions is always the same, which means that the bearing loads and shaft bending load caused by a belt drive will be essentially constant, regardless of the torque being transmitted. (Only the angle of application changes with applied torque).

The value of the tension in the tight (driving) span is the sum of two values: (a) the "driving tension" plus (b) the existing "loose-side" tension, as shown above. The "driving tension" is the tension in the tight span produced by the torque applied to the driving sprocket (pulley), and has the value:

Driving Tension = applied drive torque / drive sprocket pitch radius

For example, a torque of 2760 lb-in (230 lb-ft) applied to a sprocket of 2.155 pitch radius (4.311 pitch diameter) produces a driving tension of 1281 pounds (2760 / 2.155). If a smaller diameter driving sprocket is used, say 1.705 pitch radius (3.409 pitch diameter) then the driving tension increases to 1620 pounds (2760 / 1.705).

The amount of preload which a belt drive requires is specified by a value known as "tension ratio" (TR), which is expressed as:

TR = Tight-Side Tension / Loose-Side Tension

Knowing that Tight-Side Tension is the sum of the Driving Tension plus the Loose-Side Tension, that equation can be rewritten as:

TR = (drive tension + loose-side tension) / loose-side tension

or (by means of some high-school algebra)

TR = 1 + (drive tension / loose-side tension)

Rearranging that equation produces the more useful form:

Loose-side tension = Drive-tension / (TR - 1)

The value of the preload can be measured by applying a side force to one span and measuring the force required to deflect the belt a specified amount. Note that the side-load force you apply at midspan is NOT the preload. The force required to produce the specified deflection just a rough measure of the belt preload.

The force and deflection values used to measure preload are a function of the "springiness" of the belt and the length of the span, and can be calculated from data provided by belt manufacturers.

V-BELT DRIVES

The transmission of force from a pulley to a V-belt (and the inverse) depends on friction between the belt and the pulley. The friction force between the belt and the pulley depends on three factors:

1. the normal force between the belt and the pulley (that is, the force which is perpendicular to the side surface of the groove),

2. the coefficient of friction between the belt and the pulley, and 

3. the arc of contact between the belt and the pulley.

You already know (explained in Force and Friction) that the frictional force between two objects is the product of the coefficient of friction and the normal force. The coefficient of friction is independent of the shape of the belt, but the normal force between the belt and sides of the pulley sheave depends on:

1. the included angle between the sheave sides, and

2. the tension on the belt.

Because of the wedging action between the belt and the pulley, the normal force is a significant multiple of the belt tension.

In order for the system to transmit the required torque, the belt(s) must be wedged into the groove(s) tightly enough to transmit the applied force.

In order to achieve the necessary wedging, it is necessary to apply a static preload to the drive system. The required preload is calculated from three factors:

1. the maximum torque to be transmitted by the drive,

2. the minimum arc of contact, and

3. the pitchline velocity of the belt(s) (to account for the effect of centrifugal force, which reduces the wedging effect).

Belt manufacturers recommend that the tension ratio for a system with 180° of contact on each sheave should be no tighter than 5:1 (when a new belt has been installed) and no looser than 8:1 (after the belt has run-in).

If the arc of contact on any load-transmitting pulley in the system is less than 180°, then the required tension ratio decreases, which means more preload.

TOOTH-BELT DRIVES

Preload in a toothed-belt system is required in order to keep the teeth from attempting to climb up the sides of the grooves in the sprockets, and, in extreme cases, from jumping ("ratcheting") under the most severe loading conditions.

In a static (non-moving) toothed-belt drive, the tension force in each span of the belt is equal, and is determined by the required preload. The required preload is determined by four factors:

the number of teeth engaged on the driving sprocket,

1. the pitch diameter of the driving sprocket,

2. the maximum torque to be transmitted by the drive, and

3. the recommended tension ratio.

If the preload of a toothbelt is slightly less than required, the teeth will try to climb the sides of the grooves at high torque loadings, which leads to (a) rapid wear of the belt, and (b) very high shaft bending loads (higher than if the preload were correct).

Belt manufacturers recommend that the tension ratio for a system with more than a defined minimum number of teeth in contact on the smallest sprocket should be in the range of 8:1 to 10:1 (after the belt has run-in).

- Basics of Vibration -

This section crams a lot of information into a short space, so a little extra time spent reading it can pay dividends later on. (To engineering readers, please forgive the liberties taken in the interest of brevity.)

Components in a vibrating system have three properties of interest. They are: MASS (weight), ELASTICITY(springiness) and DAMPING (dissipation). Most physical objects have all three properties, but in many cases one or two of those properties are relatively insignificant and can be ignored (for example, the damping of a block of steel, or in some cases, the mass of a spring).

The property of mass (weight) causes an object to resist acceleration. It also enables an object to store energy, in the form of velocity (kinetic) or height (potential). You must expend energy to accelerate a mass or to lift a mass. If you decelerate a moving mass, or drop a lifted mass, the energy it took to accelerate it or to lift it (as applicable) can be recovered.

The property of elasticity enables an object to store energy in the form of deflection. A common example is a spring, but any piece of metal has the property of elasticity. That is, if you apply two equal and opposite forces to opposite sides of it, it will deflect. Sometimes that deflection can be seen; sometimes it is so small that it can't be measured with a micrometer. The size of the deflection depends on the size of the applied force and the dimensions and properties of the piece of metal. The amount of deflection caused by a specific force determines the "spring rate" of the metal piece. Note that all metals (in the solid state) have some amount of elasticity.

You must expend energy to deflect a spring. The spring stores most of the energy required to deflect it. When you release a deflected spring, the stored energy can be recovered.

The term damping is frequently misunderstood. The property of damping enables an object to DISSIPATE energy, usually by conversion of kinetic (motion) energy into heat energy. The misnamed automotive device known as a "shock absorber" is a common example of a damper. If you push on the ends of a fully extended "shock absorber" (so as to collapse it) the rod moves into the body at a velocity related to how hard you are pushing. Double the force and the velocity doubles. When the "shock" is fully collapsed, and you release your hand pressure, nothing happens (except maybe you drop it). The rod does not spring back out. The energy (defined as a force applied over a distance) which you expended to collapse the damper has been converted into heat which is dissipated through the walls of the shock absorber.

The RESONANT FREQUENCY of an object (or system) is the frequency at which the system will vibrate if it is excited by a single pulse. As an example, consider a diving board. When a diver bounces on the end of the board and commences a dive, the board will continue to vibrate up and down after the diver has left it. The frequency at which the board vibrates is it’s resonant frequency, also known as it’s natural frequency. Another example is a tuning fork. When struck, a tuning fork "rings" at it’s resonant frequency. The legs of the fork have been carefully manufactured so as to locate their resonant frequency at exactly the acoustic frequency at which the fork should ring.

The resonant frequency of a system (symbolized "ω n", pronounced "omega-sub-n") is determined by both the mass properties and the elasticity properties of the system.

If you increase the mass property or decrease the elasticity property of a system, the resonant frequency will decrease according to the relationship:

ωn = √(k/m)

where "k" is the appropriate elasticity value (spring rate) and "m" is the appropriate mass value.

Note that damping has no effect on the system resonant frequency. Damping simply dissipates energy from a vibrating system, and thereby limits the amount of energy which can feed back into the system, and so limits the amount of "out-of-control" vibration which can occur near resonance.

A WAVEFORM is a pictorial representation of a vibration. For illustration purposes, Figure 1 shows waveform representing the instantaneous torque of a typical 8-cylinder engine.

The term FREQUENCY occurs often in the discussion of vibration. The frequency of a waveform (torque variation, in this case) is the number of times per second that the waveform repeats itself, while the order of a waveform is the number of times that the waveform repeats itself during a particular event (such as one revolution of a crankshaft).

[pic]

The waveform shown in Figure 1 is a 4th order torsional vibration (four spikes per rotation). The frequency of the vibration changes with engine RPM. For example, at 4600 RPM the frequency of the 4th order vibration is 307 cycles per second, or "Hertz" (HZ), the units of cycles per second.

(The frequency value 307 is calculated as follows: 4600 revolutions-per-minute divided by 60 seconds-per-minute = 76.7 revolutions-per-second, multiplied by 4 pulses-per-revolution = 307 HZ. Using the same arithmetic, the excitation frequency of a 4th-order vibration at 800 RPM is 53 HZ.)

NOTE: Torsional Excitation by Piston Engines is covered in Engine Technology.

- Torsional Vibration Overview -

Consider the system represented schematically in Figure 1. It consists of a heavy flywheel, rigidly mounted on a round shaft. The shaft is supported in a frictionless bearing at the flywheel end, and rigidly attached to a big strong wall at the opposite end.

[pic]

Single Degree of Freedom Torsional System

Suppose you grabbed the rim of the flywheel with both hands and twisted it like a steering wheel. The wheel turns a bit, but the rotation of the wheel is resisted by the twisting action of the shaft. When shaft has twisted enough to exert the same resisting torque on the center of the wheel as you are applying to the rim, the wheel turns no further. As you slowly relax your twisting force, the wheel returns to its neutral (untwisted) position.

The shaft (if it has not been twisted excessively) has acted like a particular kind of spring known as a torsion spring or torsion bar. The torsional spring rate of the shaft can be defined as the amount of torque it takes to twist it one degree (symbolized by "Kt"). The torsional rate can be calculated if the dimensions and material of the shaft are known.

Now suppose you twist the wheel again, but instead of releasing it slowly, you suddenly let go. What happens? The energy stored in the shaft (torsion spring) exerts a torque on the flywheel, which accelerates the flywheel toward its neutral position. When the flywheel reaches the neutral position, the flywheel has reached its maximum velocity. The shaft has released all its stored energy, and the flywheel has recovered that energy (which had been stored in the spring) and has stored it as kinetic (motion) energy.

Because the flywheel is in motion, it continues to rotate past neutral. After the flywheel passes the neutral position, the shaft begins to deflect again, opposing the motion of the flywheel, and begins to decelerate it. When the resistance of the shaft stops the flywheel, it has rotated (theoretically) to the same angular distance past neutral as when you released it, but in the opposite direction. The shaft has recovered the wheel’s kinetic energy and converted it (again) to deflection energy. The shaft now begins to accelerate the wheel back in the other direction.

This back-and-forth motion of the flywheel and shaft is an example of torsional vibration. In an idealized world, this vibration goes on forever. In the real world, there is always some energy lost, so the vibration eventually stops (there are no practical frictionless bearings, and springs have some amount of hysteresis loss, and everyone knows about aerodynamic drag).

In this example, we are interested in how many times per second (how frequently) the wheel oscillates (cycles) back and forth. That frequency is the resonant frequency of the torsional system. The value of the resonant frequency can be calculated if you know two values: (1) the torsional rate of the shaft (Kt) and (2) the mass moment of inertia (MMOI) of the flywheel (symbolized "Jm").

The MMOI of a flywheel is a calculated value, related to it's weight, which takes into account how far the weight is concentrated from the center of rotation. The MMOI is a measure of how difficult it is to accelerate or decelerate the flywheel, and can be thought of as "FLYWHEEL EFFECT".

Now suppose the Jm and the Kt of the system pictured in Figure 2 are such that the resonant frequency is one cycle per second (1.0 HZ). Also suppose that the we get a Very Coordinated Person to apply a 10 LB. yank on the flywheel rim, in the direction the wheel is starting to move, every time it changes direction (every half-second). The input from the VCP has changed the system into what is known as a Forced Vibration System. The 10-LB. yank every half-second is called a driving force, or an excitation force.

Now what will happen to the system? Because it is being driven at its resonant frequency and is essentially without damping, the number of degrees the flywheel moves will increase with every oscillation until the weakest part of the system (perhaps the shaft, perhaps the attachment of the shaft to the wall, perhaps the attachment of the flywheel to the shaft) fails. This situation is an example of undamped, forced vibration at resonance. It is a system which is out of control.

The study of forced vibration systems shows that when the excitation force is sinusoidal (in the shape of a sine wave) and is periodic (repeats itself over and over), then the system will vibrate at the frequency of the excitation.

- Transmissibility -

Figure 1 shows the torsional excitation generated by an even-fire 8-cylinder engine (explained in ENGINE TORSIONAL EXCITATION). This is a "fourth order" excitation, which at an 800 RPM idle, produces 53 pulses per second (Hz), and at 5000 RPM, produces 333 pulses per second (Hz). Note that the four torque peaks are nearly twice as high as the mean torque produced by the engine (the torque measured by the dynamometer).

[pic]

One might think it would be sufficient to design a PSRU to sustain the loads produced by the input torque peaks. However, depending on the resonant frequencies of the engine-PSRU-propeller system, the PSRU and the propeller can be subjected to vibratory forces many times HIGHER than the peaks produced by the engine. The following discussion explains how that can happen. (The fundamentals of vibration are presented in Vibration Basics and Torsional Vibration Overview.)

The ratio between the amplitude of the excitation torque (the size of the pulses) and the amplitude of the output torque is expressed by a value called TRANSMISSIBILITY. For the Engine-PSRU-Prop system, the engine torque variation (as shown in Figure 1) multiplied by the transmissibility equals the torque variation applied to the PSRU, which is then multiplied by the gear ratio and applied to the prop.

Figure 2 shows how transmissibility changes with excitation frequency and with damping. The horizontal axis is frequency ratio (r), which is the excitation frequency divided by the resonant frequency. The symbol "β" shows the amount of damping in the system. β=0 is none, β=2 is a lot.

[pic]

The different curves in Figure 2 show how transmissibility will change as the excitation frequency changes from zero to seven times the resonant frequency, and with different amounts of damping. (You might notice the apparent discontinuity at r=3.0. The curves are continuous, but the x-axis granularity halves at r=3.0 to enhance the appearance of the plot.) Note that at r=1 (resonance) the system with no damping (β=0) allows the transmissibility to become huge (theoretically infinite).

Notice that below crossover, the more damping the system has, the less the input waveform will be amplified, butamplified it will still be. Understand this: No matter how much damping is added, if the system is operating below crossover, the PSRU (thus the gears, shafts and propeller) feels torque pulses greater than what the engine produces. If operation below crossover is unavoidable, then adding damping significantly reduces the amplification around resonance, but also diminishes the pulse suppression available above crossover. Further, it can be problem to implement damping in a reliable manner, and to cool it. Remember: dampers convert motion energy to heat energy.

The following example shows the practical implications of transmissibility in a PSRU system. Figure 3 shows the transmissibility curves of two different PSRU systems for the same V-8 engine and propeller. The first-mode resonant frequency for one system is at 450 RPM; the other is at 3700 RPM.

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In order to examine the effect those different resonance points have on gear and propeller loads, let's determine the torque variation seen by the PSRU in each system while being driven at 4000 RPM by an engine such as the one shown in Figure 1 (625 lb.-ft. mean torque, with peaks of 1235 lb.-ft. and valleys of 68 lb.-ft.).

First, consider the system with the resonant frequency at 3700 RPM. At 4000 RPM, the transmissibility is 5.93, which means the engine torque pulses (from 68 to 1235 lb-ft) are amplified nearly 6 times before reaching the PSRU, and range from valleys of -2678 lb-ft. to peaks of +4243 lb-ft. It is easy to see why this gearbox and prop did not last very long.

Now consider the system with the resonant frequency at 450 RPM. At 4000 RPM, the transmissibility is .013. That means the engine torque pulses (which range from 68 to 1235 lb-ft) are nearly erased. The torque which is applied to the PRSU varies from a low of 618 to a peak of  633 lb-ft. This is "turbine-smooth".

In order to design and evaluate Engine-PSRU-Propeller systems, EPI has implemented several proprietary computer programs which assist us in the design of our proprietary coupling system which has transmissibility numbers below 0.10 in the normal operating ranges (nearer to 0.010 at cruise and takeoff power settings). This coupling system takes into account the properties of the whole powerplant (engine, gearbox, reduction ratio, propeller characteristics) in order to locate the system first mode resonant frequency well below engine idle speed.

One of the most useful of that suite is our program which calculates, with great accuracy, the effective engine MMOI values from data which include the crankshaft stroke and weight, number of journals, bobweight value, and the dimensions of the flywheel, torsional absorber, and accessories. That program uses known measured values from the industry as a validity cross-check.

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LAST BUT NOT LEAST I REQUEST ALL TELUGU PEOPLE TO LEARN & SHARE KNOWLEDGE ABOUT TELUGU NUMBERS

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