Machine Design 10

0.11 SPLINES When more torque must be transmitted than can be handled by keys, splines can be used instead. Splines are essentially "built-in keys" formed by contouring the outside of the shaft and inside of the hub with toothlike forms. Early splines had teeth of square cross section but these have been supplanted by involute spline teeth, as shown in Figure 10-17. The involute tooth form is universally used on gears, and the same cutting techniques are used to manufacture splines. In addition to its manufacturing advantage, the involute tooth has less stress concentration than a square tooth and is stronger. The SAE defines standards for both square and involute spline tooth forms and ANSI publishes involute spline standards.† The standard involute spline has a pressure angle of 30° and half the depth of a standard gear tooth. The tooth size is defined by a fraction whose numerator is the diametral pitch (which defines tooth width—see Chapter 12 for more information on these terms) and whose denominator controls tooth depth (and is always double the numerator). Standard diametral pitches are 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 24, 32, 40, and 48. Standard splines can have from 6 to 50 teeth. Splines may have either a flat root or a filleted root, both shapes being shown in Figure 10-17. See reference 8 for complete dimensional information on standard splines. Some advantages of splines are maximum strength at the root of the tooth, accuracy of tooth form due to the use of standard cutters, and good machined surface finish from the standard gear-cutting (hobbing) process, which eliminates the need for grinding. A major advantage of splines over keys is their ability (with proper clearances) to accommodate large axial movements between shaft and hub while simultaneously transmitting torque. They are used to connect the transmission output shaft to the driveshaft

0.14 CRITICAL SPEEDS OF SHAFTS All systems containing energy-storage elements will possess a set of natural frequencies at which the system will vibrate with potentially large amplitudes. Any moving mass stores kinetic energy and any spring stores potential energy. All machine elements are made of elastic materials and thus can act as springs. All elements have mass, and if they also have a velocity, they will store kinetic energy. When a dynamic system vibrates, a transfer of energy from potential to kinetic to potential, etc., repeatedly takes place within the system. Shafts meet these criteria, rotating with some velocity and deflecting both in torsion and in bending. If a shaft, or any element for that matter, is subjected to a time-varying load, it will vibrate. Even if it receives only a transient load, such as a hammer blow, it will vibrate at its natural frequencies, just as a bell rings when struck. This is called a free vibration. Such a transient or free vibration will eventually die out due to the damping present in the system. If the time-varying loading is sustained, as, for example, in a sinusoidal manner, the shaft or other element will continue to vibrate at the driving function's forcing frequency. If the forcing frequency happens to coincide with one of the element's natural frequencies, then the amplitude of the vibratory response will be much larger than the amplitude of the driving function. The element is then said to be in resonance. Figure 10-26a shows the amplitude response of a forced vibration, and Figure 10- 26b a self-excited vibration, as a function of the ratio of the forcing frequency to the system's natural frequency ωf / ωn. When this ratio is 1, the system is in resonance and the amplitude of the response approaches infinity in the absence of damping. The amplitude responses in Figure 10-26 are shown as a dimensionless ratio of the output to input amplitudes. Any damping, shown as a damping ratio ζ, reduces the amplitude ratio at resonance. A natural frequency is also called a critical frequency or critical speed. Exciting a system at or near its critical (resonant) frequencies must be avoided, since the resulting deflections will often cause stresses large enough to quickly fail the part. A system consisting of discrete lumps of mass connected with discrete spring elements can be considered to have a finite number of natural frequencies equivalent to its number of kinematic degrees of freedom. But a continuous system such as a beam

1 To minimize both deflections and stresses, the shaft length should be kept as short as possible and overhangs minimized. 2 A cantilever beam will have a larger deflection than a simply supported (straddle mounted) one for the same length, load, and cross section, so straddle mounting should be used unless a cantilevered shaft is dictated by design constraints. (Figure 10-2 shows a situation in which an overhung or cantilevered section of shaft is required for serviceability. The sheave on the right-hand end of the shaft carries an endless V-belt. If the sheave were mounted between the bearings, then the shaft assembly would have to be disassembled to change a belt, which is undesirable. In such cases, the cantilevered shaft can be the lesser of the evils.) 3 A hollow shaft has a better stiffness/mass ratio (specific stiffness) and higher natural frequencies than a comparably stiff or strong solid shaft, but will be more expensive and larger in diameter. 4 Try to locate stress-raisers away from regions of large bending moment if possible and minimize their effects with generous radii and reliefs. 5 If minimizing deflection is the primary concern, then low-carbon steel may be the preferred material, since its stiffness is as high as that of more expensive steels and a shaft designed for low deflection will tend to have low stresses. 6 Deflections at gears carried on the shaft should not exceed about 0.005 in and the relative slope between the gear axes should be less than about 0.03°. [1] 7 If plain (sleeve) bearings are used, the shaft deflection across the bearing length should be less than the oil-film thickness in the bearing.[1] 8 If non-self-aligning rolling element bearings are used, the shaft's slope at the bearings should be kept to less than about 0.04°. [1] 9 If axial thrust loads are present, they should be taken to ground through a single thrust bearing per load direction. Do not split axial loads between thrust bearings, as thermal expansion of the shaft can overload the bearings. 10 The first natural frequency of the shaft should be at least three times the highest forcing frequency expected in service, and preferably much more. (A factor of 10X or more is preferred, but this is often difficult to achieve in mechanical systems). Design for Fully Reversed Bending and Steady Torsion This loading case is a subset of the general case of fluctuating bending and fluctuating torsion, and, because of the absence of an alternating component of torsional stress, is considered to be a simple multiaxial fatigue case. (The presence of local stress concentrations can cause complex multiaxial stresses, however.) This simple loading case has been experimentally investigated, and data exist for failure of parts so loaded as shown in Figure 10-3. The ASME has defined an approach for the design of shafts loaded in this manner. THE ASME METHOD An ANSI/ASME Standard for the Design of Transmission Shafting is published as B106.1M-1985. This standard presents a simplified approach to the design of shafts. The ASME approach assumes that the loading is fully reversed bending (zero mean bending component) and steady torque (zero alternating torque component) at a level that creates stresses below the torsional yield strength of the material. The standard makes the case that many machine shafts are in this category. I

10.0 INTRODUCTION Transmission shafts, or just shafts, are used in virtually every piece of rotating machinery to transmit rotary motion and torque from one location to another. Thus, the machine designer is often faced with the task of designing shafts. This chapter will explore some of the common problems encountered in that endeavor. Table 10-0 shows the variables used in this chapter and notes the equations or sections in which they occur. At a minimum, a shaft typically transmits torque from the driving device (motor, or engine) through the machine. Sometimes shafts will carry gears, sheaves (pulleys), or sprockets, which transmit the rotary motion via mating gears, belts, or chains from shaft to shaft. The shaft may be an integral part of the driver, such as a motor shaft or engine crankshaft, or it may be a freestanding shaft connected to its neighbor by a coupling of some design. Automated production machinery often has line shafts that extend the length of the machine (10 m or more) and carry the power to all the workstations. Shafts are carried in bearings, in a simply supported (straddle-mounted) configuration, cantilevered, or overhung, depending on the machine configuration. The pros and cons of these mounting and coupling arrangements will also be discussed. 10.1 SHAFT LOADS The loading on rotating transmission shafts is principally one of two types: torsion due to the transmitted torque or bending from transverse loads at gears, sheaves, and sprockets. These loads often occur in combination, since, for example, the transmitted torque may be associated with forces at the teeth of gears or sprockets attached to the shafts.

10.10 KEYS AND KEYWAYS The ASME defines a key as "a demountable machinery part which, when assembled into keyseats, provides a positive means for transmitting torque between the shaft and hub." Keys are standardized as to size and shape in several styles.* A parallel key is square or rectangular in cross section and of constant height and width over its length. (See Figure 10-14a.) A tapered key is of constant width but its height varies with a linear taper of 1/8 in per foot, and it is driven into a tapered slot in the hub until it locks. It may either have no head or have a gib head to facilitate removal. (See Figure 10- 14b.) A Woodruff key is semicircular in plan and of constant width. It fits in a semicircular keyseat milled in the shaft with a standard circular cutter. (See Figure 10-14c.) The tapered key serves to lock the hub axially on the shaft, but the parallel or Woodruff keys require some other means for axial fixation. Retaining rings or clamp collars are sometimes used for this purpose. Parallel Keys Parallel keys are the most commonly used. The ANSI and ISO standards each define particular key cross-sectional sizes and keyseat depths as a function of shaft diameter at the keyseat. A partial reproduction of this information is provided in Table 10-2 for the lower range of shaft diameters. Consult the respective standards for larger shaft sizes. Square keys are recommended for shafts up to 6.5-in dia (US), or 25 mm-dia (ISO), and rectangular keys for larger diameters. The parallel key is placed with half of its height in the shaft and half in the hub, as shown in Figure 10-14a. Parallel keys are typically made from standard cold-rolled bar stock, which is conventionally "negatively toleranced," meaning it will never be larger than its nominal dimension, only smaller. For example, a nominal 1/4-in square bar will have a tolerance on width and height of +0.000, -0.002 in. Thus, the keyseat can be cut with a standard 1/4-in milling cutter and the bar stock key will fit with slight clearance. A special keystock is also available, which is positively toleranced (e.g., 0.250 +0.002, -0.000). FIGURE 10-13 Deflection Functions for Example 10-3 x Deflection 02468 0 1 2 X 10-3 in shaft length (in) z shaft length (in) y Deflection 02468 0 1 2 X 10-3 in z shaft length (in) Deflection Magnitude 02468 X 10-3 in 0 1 2 z * ANSI standard B17.1-1967, Keys and Keyseats, and B17.2- 1967, Woodruff Keys and Keyseats, available from the American Society of Mechanical Engineers, 345 East 47th St., New York, N.Y. 10017. Ch 10 4ed Final 570 10/18/09, 6:19 PM Chapter 10 SHAFTS, KEYS, AND COUPLINGS 571 10 It is used when a closer fit between key and keyseat is desired and may require machining of the keystock to final dimension. The key fit can be of concern when the torque loading is alternating from positive to negative each cycle. When the torque changes sign, any clearance between key and keyway will be suddenly taken up with resulting impact and high stresses. This is termed backlash. A setscrew in the hub, placed at 90° from the key, can both hold the hub axially and stabilize the key against backlash. The ANSI Standard also defines the size of setscrew to be used with each size key as shown in Table 10-2. Key length should be less than about 1.5 times the shaft diameter to avoid excessive twisting with shaft deflection. If more strength is needed, two keys can be used, oriented at 90° and 180° for example. Tapered Keys The width of a tapered key for a given shaft diameter is the same as for a parallel key as shown in Table 10-2. The taper and gib-head size are defined in the standard. The

10.12 INTERFERENCE FITS Another common means of coupling a hub to a shaft is to use a press or shrink fit, also called an interference fit. A press fit is obtained by machining the hole in the hub to a slightly smaller diameter than that of the shaft as shown in Figure 10-18. The two parts are then forced together slowly in a press, preferably with oil lubricant applied to the joint. The elastic deflection of both shaft and hub act to create large normal and frictional forces between the parts. The friction force transmits the shaft torque to the hub and resists axial motion as well. The American Gear Manufacturers Association (AGMA) publishes a standard AGMA 9003-A91, Flexible Couplings—Keyless Fits which defines formulas for the calculation of interference fits. Only relatively small parts can be press-fitted without exceeding the force capacity of a typical shop press. For larger parts, a shrink fit can be made by heating the hub to expand its inside diameter and/or an expansion fit can be made by cooling the shaft to reduce its diameter. The hot and cold parts can be slipped together with little axial force, and when they equilibrate to room temperature, their dimensional change creates the desired interference for frictional contact. Another method is to hydraulically expand the hub with pressurized oil delivered through passageways in the shaft or hub. This technique can be used to remove a hub as well. The amount of interference needed to create a tight joint varies with the diameter of the shaft. Approximately 0.001 to 0.002 units of diametral interference per unit of shaft diameter is typical (the rule of thousandths), the smaller amount being used with larger shaft diameters. For example, the interference for a diameter of 2 in would be about 0.004 in, but a diameter of 8 in would receive only about 0.009 to 0.010 in of interference. Another (and simpler) machinist's rule of thumb is to use 0.001 in of interference for diameters up to 1 in, and 0.002 in for diameters from 1 to 4 in.

10.13 FLYWHEEL DESIGN* A flywheel is used to smooth out variations in the speed of a shaft caused by torque fluctuations. Many machines have load patterns that cause the torque-time function to vary over the cycle. Piston compressors, punch presses, rock crushers, etc., all have timevarying loads. The prime mover can also introduce torque oscillations to the transmission shaft. Internal-combustion engines with one or two cylinders are an example. Other systems may have both smooth torque sources and smooth loads, such as an electrical generator driven by a steam turbine. These smooth-acting devices have no need for a flywheel. If the source of the driving torque or the load torque has a fluctuating nature, then a flywheel is usually called for. A flywheel is an energy-storage device. It absorbs and stores kinetic energy when speeded up and returns energy to the system when needed by slowing its rotational speed. The kinetic energy Ek in a rotating system is EI a k = m 1 2 10 17 2 ω (. ) where Im is the mass moment of inertia of all rotating mass on the shaft about the axis of rotation and ω is the rotational velocity. This includes the Im of the motor rotor and anything else rotating with the shaft plus that of the flywheel. Flywheels may be as simple as a cylindrical disk of solid material, or may be of spoked construction with a hub and rim. The latter arrangement is more efficient of material, especially for large flywheels, as it concentrates the bulk of its mass in the rim, which is at the largest radius. Since the mass moment of inertia Im of a flywheel is proportional to mr2, mass at larger radius contributes much more. If we assume a soliddisk geometry with inside radius ri and outside radius ro, the mass moment of inertia is I m m oi = + ( ) rr b 2 10 17 2 2 (. ) The mass of a solid circular disk of constant thickness t and having a central hole is m W g g r rt

10.15 COUPLINGS A wide variety of commercial shaft couplings are available, ranging from simple keyed, rigid couplings to elaborate designs that utilize gears, elastomers, or fluids to transmit the torque from one shaft to another or to other devices in the presence of various types of misalignment. Couplings can be roughly divided into two categories, rigid and compliant. Compliant in this context means that the coupling can absorb some misalignment between the two shafts and rigid implies that no misalignment is allowed between the connected shafts. igid Couplings Rigid couplings lock the two shafts together allowing no relative motion between them, though some axial adjustment is possible at assembly. These are used when accuracy and fidelity of torque transmission are of paramount importance, as, for example, when the phase relationship between driver and driven device must be accurately maintained. Automated production machinery driven by long line-shafts often uses rigid couplings between shaft sections for this reason. Servomechanisms also need zero backlash connections in the drive train. The trade-off is that the alignment of the coupled shaft axes must be adjusted with precision to avoid introducing large side forces and moments when the coupling is clamped in place. Figure 10-33 shows some examples of commercial rigid couplings. There are three general types: setscrew couplings, keyed couplings, and clamp couplings. SETSCREW COUPLINGS use a hard setscrew that digs into the shaft to transmit both torque and axial loads. These are not recommended for any but light-load applications and can loosen with vibration. KEYED COUPLINGS use standard keys as discussed in an earlier section and can transmit substantial torque. Setscrews are often used in combination with a key, the screw being located 90° from the key. For proper holding against vibration, a cup-point setscrew is used to dig into the shaft. For better security, the shaft should be dimpled with a shallow drilled hole under the setscrew to provide a mechanical interference against axial slip rather than relying on friction. CLAMP COUPLINGS are made in several designs, the most common being one- or two-piece split couplings that clamp around both shafts and transmit torque through friction as shown in Figure 10-33. A taper-lock coupling uses a split-tapered collet which is squeezed between shaft and the tapered coupling housing to clamp the shaft as shown in Figure 10-34.

10.5 SHAFT LOADS The most general shaft-loading case is that of a fluctuating torque and a fluctuating moment in combination. There can be axial loads as well, if the shaft axis is vertical or is fitted with helical or worm gears having an axial force component. (A shaft should be designed to minimize the portion of its length subjected to axial loads by taking them to ground through suitable thrust bearings as close to the source of the load as possible.) Both torque and moment can be time varying, as shown in Figure 10-1, and can have both mean and alternating components. The combination of a bending moment and a torque on a rotating shaft creates multiaxial stresses. The issues discussed in Section 6.12 (p. 376) on multiaxial stresses in fatigue are then germane. If the loadings are asynchronous, random, or misphased, then it will be a complex multiaxial stress case. But, even if the moment and torque are in phase (or 180° out of phase), it may still be a complex multiaxial stress case. The critical factor in determining whether it has simple or complex multiaxial stresses is the direction of the principal alternating stress on a given shaft element. If its direction is constant with time, then it is considered a simple multiaxial stress case. If it varies with time, then it is a complex multiaxial stress case. Most rotating shafts loaded in both bending and torsion will be in the complex category. While the direction of the alternating bending stress component will tend to be constant, the torsional component's direction varies as the element rotates around the shaft. Combining them on the Mohr's circle will show that the result is an alternating principal stress of varying direction. One exception to this is the case of a constant torque superposed on a time-varying moment. Since the constant torque has no alternating component to change the direction of the principal alternating stress, this becomes a simple multiaxial stress case. However even this exception cannot be taken if there are stress concentrations present, such as holes or keyways in the shaft, since they will introduce local biaxial stresses and require a complex multiaxial fatigue analysis. Assume that the bending moment function over the length of the shaft is known or is calculable from the given data and that it has both a mean component Mm and an alternating component Ma. Likewise, assume the torque on the shaft is known or calculable from given data and has both mean and alternating components, Tm and Ta. Then the general approach follows that outlined in the list labeled Design Steps for Fluctuating Stresses in Section 6.11 (p. 360) in combination with the multiaxial stress issues addressed in Section 6.12 (p. 376). Any locations along the length of the shaft that appear to have large moments and/or torques (especially if in combination with stress concentrations) need to be examined for possible stress failure and the cross-sectional dimensions or material properties adjusted accordingly. 10.6 SHAFT STRESSES With the understanding that the following equations will have to be calculated for a multiplicity of points on the shaft and their combined multiaxial effects also considered, we must first find the applied stresses at all points of interest. The largest alternating and mean bending stresses are at the outside surface and are found from

10.8 SHAFT DESIGN Both stresses and deflections need to be considered in shaft design. Often, deflection can be the critical factor, since excessive deflections will cause rapid wear of shaft bearings. Gears, belts, or chains driven from the shaft can also suffer from misalignment introduced by shaft deflections. Note that the stresses in a shaft can be calculated locally for various points along the shaft based on known loads and assumed cross sections. But, the deflection calculations require that the entire shaft geometry be defined. So, a shaft is typically first designed using stress considerations and then the deflections calculated once the geometry is completely defined. The relationship between the shaft's natural frequencies (in both torsion and bending) and the frequency content of the forceand torque-time functions can also be critical. If the forcing functions are close in frequency to the shaft's natural frequencies, resonance can create vibrations, high stresses, and large deflections. General Considerations Some general rules of thumb for shaft design can be stated as follows:

A shaft is a beam that deflects transversely and is also a torsion bar that deflects torsionally. Both modes of deflection need to be analyzed. The principles of deflection analysis were reviewed in Chapter 4 and will not be detailed again here. Section 4.10 (p. 162) developed an approach for calculating beam deflections using singularity functions and Section 4.12 (p. 177) investigated torsional deflection.

A taper pin creates a truly tight torque coupling and locates axially as well as radially with phasing but weakens the shaft. It can be disassembled with slightly more difficulty than a key. A clamp collar is easy to install but has no repeatable phasing. This is only a disadvantage if timing of the shaft rotation to other shafts in the system is required. It allows easy (though inaccurate) adjustment of phasing if desired. Press fits are semipermanent connections that require special equipment to disassemble. They do not provide repeatable phasing. 10.3 SHAFT MATERIALS In order to minimize deflections, steel is the logical choice for a shaft material because of its high modulus of elasticity, though cast or nodular iron is sometimes also used, especially if gears or other attachments are integrally cast with the shaft. Bronze or stainless steel is sometimes used for marine or other corrosive environments. Where the shaft also serves as the journal, running against a sleeve bearing, hardness can become an issue. Through- or case-hardened steel may be the material of choice for the shaft in these cases. See Chapter 11 for a discussion of desired relative hardness and material combinations for shafts and bearings. Rolling-element bearings do not need hardened shafts. Most machine shafts are made from low- to medium-carbon steel, either cold rolled or hot rolled, though alloy steels are also used where their higher strengths are needed. Cold-rolled steel is more often used for smaller-diameter shafts (< about 3-in dia) and hot-rolled used for larger sizes. The same alloy when cold rolled has higher mechanical properties than if hot rolled due to the cold working, but this comes at the cost of residual tensile stresses in the surface. Machining for keyways, grooves, or steps relieves these residual stresses locally and can cause warping. Hot-rolled bars must be machined all over to remove the carburized outer layer, whereas portions of a cold-rolled surface can be left as-rolled except where machining to size is needed for bearings, etc. Prehardened (30HRC) and ground precision (straight) steel shafting can be purchased in small sizes and can be machined with carbide tools. Full-hard, ground, precision shafting (60HRC) is also available but cannot be machined. 10.4 SHAFT POWER The power transmitted through a shaft can be found from first principles. In any rotating system, instantaneous power is the product of torque and angular velocity, PT a = ω (.) 10 1 where ω must be expressed in radians per unit time. Whatever the base units used for calculations, power is usually converted to units of horsepower (hp) in any English system or to kilowatts (kW) in any metric system. (See Table 1-5 on p. 24 for conversion factors.) Both torque and angular velocity can be time varying, though much of rotating machinery is designed to operate at constant or near-constant speeds for large blocks of time. In such cases, the torque will often vary with time. The average power is found from

Compliant Couplings A shaft as a rigid body has six potential degrees of freedom (DOF) with respect to a second shaft. However, due to symmetry only four of these DOF are of concern. They are axial, angular, parallel, and torsional misalignment, as shown in Figure 10-35. These can occur singly or in combination and may be present at assembly due to manufacturing tolerances or may occur during operation due to the relative motions of the two shafts. The final driveline of an automobile has relative motion between the ends of the driveshaft. The drive end is affixed to the frame and the driven end is on the road. The frame and road are separated by the car's suspension, so the driveshaft couplings must absorb both angular and axial misalignment as the car traverses bumps. Unless care is taken to align two adjacent shafts, there can be axial, angular, and parallel misalignment in any machinery. Torsional misalignment occurs dynamically when a driven load attempts to lead or lag the driver. If the coupling allows any torsional clearance, there will be backlash when the torque reverses sign. This is undesirable if accurate phasing is needed, as in servomechanisms. Torsional compliance in a coupling may be desirable if large shock loads or torsional vibrations must be isolated from the driver. Numerous designs of compliant couplings are manufactured and each offers a different combination of features. The designer can usually find a suitable coupling available commercially for any application. Compliant couplings can be roughly divided into several subcategories, which are listed in Table 10-7 along with some of their characteristics. The torque ratings are not shown, as these vary widely with size and materials. Various size couplings can handle power levels from subfractional horsepower to thousands of horsepower. JAW COUPLINGS have two (often identical) hubs with protruding jaws, as shown in Figure 10-36. These jaws overlap axially and interlock torsionally through a compliant insert of rubber or soft-metal material. The clearances allow some axial, angular, and parallel misalignment, but can also allow some undesirable backlash. FLEXIBLE-DISK COUPLINGS are similar to jaw couplings in that their two hubs are connected by a compliant member (disk) of elastomeric or metallic-spring material, as shown in Figure 10-37. These allow axial, angular, and parallel misalignment, with some torsional compliance but little or no backlash.

For the three-mass case, the squares of the natural frequencies are the two roots of II I k II II k I I II kk I I I 123 n n 2 2 2 12 13 1 23 13 2 ( ) ω ω − + [ ] ( ) + + ( ) + ++ 12 1 2 3 ( ) = 0 10 29 (.) Higher-order polynomials can be derived for additional masses and an iterative rootfinding method can be used to solve them. Approximate methods are also available to solve for the torsional natural frequencies with any number of masses. These allow the shaft mass to be easily accounted for if desired by breaking it into discrete masses. Holzer's method is commonly used for torsional as well as for lateral vibration of shafts. See reference 10 or any vibrations text for a derivation and discussion of these methods. Space does not allow a complete treatment of them here. Controlling Torsional Vibrations When shafts are long and/or have a number of masses distributed along their length, torsional vibrations can be a serious design problem. Internal-combustion engine crankshafts are an example. The crank-throw geometry severely reduces the torsional stiffness, which lowers their natural frequency. This, in combination with the presence of strong higher harmonics from the cylinder explosions in the torque, can lead to early failure from torsional fatigue. The straight-eight engine popular in the 1930s through 1940s was less successful than its straight-six cousin, due in part to the problems of torsional vibrations in the long, eight-throw crankshaft. The V-8 engine with its shorter, stiffer, four-throw crankshaft has completely supplanted the straight-eight. Even in these shorter engines, torsional crankshaft vibrations can be a problem. Several methods can be used to counter the effects of an unwanted correspondence between the forcing frequencies and the system natural frequencies. The first line of defense is to redesign the mass and stiffness properties of the system to get the critical frequencies as far above the highest forcing frequency as possible. This usually involves increasing stiffness while removing mass, something not always easy to do. Effective use of geometry to obtain the maximum stiffness with a minimum of material is required. The term specific stiffness refers to the stiffness-to-mass ratio of an object. We want to maximize the specific stiffness to increase the natural frequencies. Finite element analysis can be very useful in refining a design's geometry to alter its natural frequencies because of the detailed information obtainable from that analysis. Another approach is the addition of a tuned absorber to the system. A tuned absorber is a mass-spring combination added to the system whose presence alters the set of natural frequencies away from any dominant forcing frequencies. The system is effectively tuned away from the undesired frequencies. This approach can be quite effective in some cases and is used in linear-motion as well as in torsional systems. A torsional damper is usually added to the end of an engine crankshaft to reduce its oscillations. This device, also called a Lanchester damper after its inventor, is a disk coupled to the shaft through an energy-absorbent medium such as rubber or oil. The oil-coupling provides viscous damping and the rubber has significant internal hysteresis damping. Its effect is to reduce the peak amplitude at resonance, as can be seen in Figure 10-26 (p. 594) for larger values of ζ. The reader is referred to reference 11 for more information on all of these methods.

GEAR AND SPLINE COUPLINGS use straight or curved external gear teeth in mesh with internal teeth, as shown in Figure 10-38. These can allow substantial axial movement between shafts and, depending on the tooth shapes and clearances, can absorb small angular and parallel misalignment as well. They have high torque capacity due to the number of teeth in mesh. HELICAL AND BELLOWS COUPLINGS are one-piece designs that use their elastic deflections to allow axial, angular, and parallel misalignment with little or no backlash. Helical couplings (Figure 10-39 and Chapter 10 title page photograph) are made from a solid metal cylinder cut with a helical slit to increase its compliance. Metal-bellows couplings (Figure 10-40) are made of thin sheet metal by welding a series of cupped washers together, by hydraulically forming a tube into the shape, or by electroplating a thick coating on a mandrel. These couplings have limited torque capability compared to other designs but offer zero backlash and high torsional stiffness in combination with axial, angular, and parallel misalignment. LINKAGE COUPLINGS or Schmidt couplings (Figure 10-41) connect two shafts through a network of links that allow significant parallel misalignment with no side loads or torque losses and no backlash. Some designs allow small amounts of angular and axial misalignment as well. These couplings are often used where large parallel adjustments or dynamic motions are needed between shafts. UNIVERSAL JOINTS are of two common types, the Hooke coupling (Figure 10-42), which does not have constant velocity (CV), and the Rzeppa coupling, which does. Hooke couplings are generally used in pairs to cancel their velocity error. Both types can handle very large angular misalignment, and in pairs provide large parallel offsets

It is common to size the key so that it will fail before the keyseat or other location in the shaft fails in the event of an overload. The key then acts like a shear pin in an outboard motor to protect the more expensive elements from damage. A key is inexpensive and relatively easy to replace if the keyseat is undamaged. This is one reason to use only soft, ductile materials for the key, having lower strength than that of the shaft so that a bearing failure will selectively affect the key rather than the keyway if the system sees an overload beyond its design range. Stress Concentrations in Keyways Since keys have relatively sharp corners (< 0.02-in radius), keyseats must also. This causes significant stress concentrations. The keyway is broached in the hub and runs through its length, but the keyway must be milled into the shaft and has one or two ends. If an end-mill is used, the keyway will look like Figure 10-15a and will have sharp corners in the side view at one or both ends as well as along each side. If, instead, a sledrunner keyway is cut as shown in Figure 10-15c, the sharp corner at the end is eliminated and the stress concentration reduced. A Woodruff keyseat in the shaft also has a large radius in the side view but it (and every keyseat) suffers from sharp corners on the sides. Peterson[7] shows experimentally derived stress-concentration curves for end-milled keyseats in shafts under either bending or torsional loading. These are reproduced in Figure 10-16. These factors range from about 2 to about 4 depending on the ratio of the corner radius to the shaft diameter. Curve-fits to Figure 10-16 have been done and functions created for these curves so that the stress-concentration factor can be determined "on the fly" during a shaft-design computation. See the file SHFTDES for example. These factors should be applied to the bending and shear stresses in the shaft at the keyway location as was done in Examples 10-1 and 10-2

Lateral Vibration of Shafts and Beams—Rayleigh's Method A complete analysis of the natural frequencies of a shaft or beam is a complicated problem, especially if the geometry is complex, and is best solved with the aid of Finite Element Analysis software. A so-called modal analysis can be done on a finite element model of even complex geometries and will yield a large number of natural frequencies (in three dimensions) from the fundamental up. This is the preferred and frequently used approach when analyzing a completed or mature design in detail. However, in the early stages of design, when the part geometries are still not fully defined, a quick and easily applied method for finding at least an approximate fundamental frequency for a proposed design is very useful. Rayleigh's method serves that purpose. It is an energy method that gives results within a few percent of the true ωn. It can be applied to a continuous system or to a lumped-parameter model of the system. The latter approach is usually preferred for simplicity. RAYLEIGH'S METHOD equates the potential and kinetic energies in the system. The potential energy is in the form of strain energy in the deflected shaft and is maximum at the largest deflection. The kinetic energy is a maximum when the vibrating shaft passes through the undeflected position with maximum velocity. This method assumes that the lateral vibrating motion of the shaft is sinusoidal and that some external excitation is present to force the lateral vibration (Figure 10-26a). To illustrate the application of this method, consider a simply-supported shaft with three disks (gears, sheaves, etc.) on it as shown in Figure 10-27. We will model this as three discrete lumps of known mass on a massless shaft. The shaft's geometry will define the bending spring constant, thus lumping all the "spring" into the shaft. The total potential energy stored at maximum deflection is the sum of the potential energies of each lumped mass: E g p = ++ ( ) mm m a 2 11 2 2 33 δδδ (. ) 10 25 where the deflections are all taken as positive regardless of the local shape of the deflection curve because the strain energy is not affected by the external coordinate system. The energy of the deflected shaft is ignored as small compared to the disk energy. The total kinetic energy is the sum of the individual kinetic energies: E mm m b k n = ++ ( ) ω δδδ 2 1 1 2 2 2 2 3 3 2 2 (. ) 10 25 where the velocities are taken as positive. Equating these gives ω δ δ δ δ δ δ n i i i n i i i n i i i n i i i n i i i n i i i n g m m g W g W g g W W = = c ( ) ( ) = = = = = = = ∑ ∑ ∑ ∑ ∑ ∑ 1 2 1 1 2 1 1 2 1 (. ) 10 25 the second version resulting from substituting m = W / g, where the Wi are the weight forces of the discrete lumps into which we divided the system and the δi are the dynamic deflections at the locations of the weights due to their vibration. The weight forces and their deflections are all taken as positive to represent the maximum stored energies

SHEAR FAILURE The average stress due to direct shear was defined in equation 4.9, repeated here: τ xy shear F A = ( .10) 10 where F is the applied force and Ashear is the shear area being cut. In this case Ashear is the product of the key's width and length. The force on the key can be found from the quotient of the shaft torque and the shaft radius. If the shaft torque is constant with time, the force will be also and the safety factor can be found by comparing the shear stress to the shear yield strength of the material. If the shaft torque is time varying, then a fatigue failure of the key in shear is possible. The approach then is to compute the mean and alternating shear-stress components and use them to compute the mean and alternating von Mises stresses. These can then be used in a modified-Goodman diagram to find the safety factor as described in Section 6.13 (p. 381). BEARING FAILURE The average bearing stress is defined as σx bearing F A = ( .11) 10 where F is the applied force and the bearing area is the area of contact between the key side and the shaft or the hub. For a square key this will be its half-height times its length. A Woodruff key has a different bearing area in the hub than in the shaft. The hub's Woodruff bearing area is much smaller and will fail first. The bearing stress should be calculated using the maximum applied force, whether constant or time varying. Since compressive stresses do not cause fatigue failures, bearing stresses can be considered static. The safety factor is found by comparing the maximum bearing stress to the material yield strength in compression. Key Materials Because keys are loaded in shear, ductile materials are used. Soft, low-carbon steel is the most common choice unless a corrosive environment requires a brass or stainless steel key. Square or rectangular keys are often made from cold-rolled bar stock and merely cut to length. The special keystock mentioned above is used when a closer fit is required between key and keyway. Tapered and Woodruff keys are also usually made from soft, cold-rolled steel. Key Design Only a few design variables are available when sizing a key. The shaft diameter at the keyseat determines the key width. The key height (or its penetration into the hub) is also determined by the key width. This leaves only the length of the key and the number of keys used per hub as design variables. A straight or tapered key can be as long as the hub allows. A Woodruff key can be had in a range of diameters for a given width, which effectively determines its length of engagement in the hub. Of course, as the Woodruff key diameter is increased, it further weakens the shaft with its deeper keyseat. If a single key cannot handle the torque at reasonable stresses, an additional key can be added, rotated 90° from the first.

Shafts as Beams The methods of Section 4.10 are directly applicable. The only complicating factor is the usual presence of steps in a shaft that change the cross-sectional properties along its length. The integration of the M / EI function becomes much more complicated due to the fact that both I and M are now functions of the dimension along the shaft-beam. Rather than do an analytical integration as was done in Section 4.10 for the case of constant I, we will use a numerical integration technique such as Simpson's rule or the trapezoidal rule to form the slope and deflection functions from the M / EI function. This will be demonstrated in an example. If the transverse loads and moment are time varying, then the absolute maximum magnitudes should be used to calculate the deflections. The deflection function will depend on the loading and the beam boundary conditions, i.e., whether simply supported, cantilevered, or overhung. Shafts as Torsion Bars The methods of Section 4.12 are directly applicable, particularly equation 4.24 (p. 178), since the only practical shaft cross section is circular. The angular deflection θ (in radians) for a shaft of length l, shear modulus G, polar moment of inertia J, with torque T is θ = Tl GJ (.) 10 9a from which we can form the expression for the torsional spring constant: k T GJ l t = = b θ (.) 10 9 If the shaft is stepped, the changing cross sections complicate the torsional deflection and spring constant calculation due to the changing polar moment of inertia J. Any collection of adjacent, different-diameter sections of shaft can be considered as a set of springs in series since their deflections add and the torque passes through unchanged. An effective spring constant or an effective J can be computed for any segment of shaft in order to find the relative deflection between its ends. For a segment of a shaft containing three sections of differing cross sections J1, J2, and J3 with corresponding lengths l1, l2, l3, the total deflection is merely the sum of the deflections of each section subjected to the same torque. We assume that the material is consistent throughout.

Stresses in Interference Fits An interference fit creates the same stress state in the shaft as would a uniform external pressure on its surface. The hub experiences the same stresses as a thick-walled cylinder subjected to internal pressure. The equations for stresses in a thick-walled cylinder were presented in Section 4.17 (p. 203) and depend on the applied pressures and the radii of the elements. The pressure p created by the press fit can be found from the deformation of the materials caused by the interference. p r E r r r r r E r r r r a o o o o i i i i = + − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + + − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 0 5 10 14 2 2 2 2 2 2 2 2 . δ ν ν (. ) where δ = 2Δr is the total diametral interference between the two parts, r is the nominal radius of the interface between the parts, ri is the inside radius (if any) of a hollow shaft, and ro is the outside radius of the hub as shown in Figure 10-18. E and ν are the Young's moduli and Poisson's ratios of the two parts, respectively. The torque that can be transmitted by the interference fit can be defined in terms of the pressure p at the interface, which creates a friction force at the shaft radius.

Substituting in equation 10.17b gives an expression for Im in terms of the disk geometry: I g m oi = − ( ) r rt d π γ 2 10 17 4 4 (. ) where γ is the material's weight density and g is the gravitational constant. There are two stages to the design of a flywheel. First, the amount of energy required for the desired degree of smoothing must be found and the moment of inertia needed to absorb that energy determined. Then a flywheel geometry must be defined that both supplies that mass moment of inertia in a reasonably sized package and is safe against failure at design speeds. Energy Variation in a Rotating System Figure 10-21 shows a flywheel, designed as a flat circular disk, attached to a motor shaft. The motor supplies a torque magnitude Tm, which we would like to be as constant as possible, i.e., to be equal to the average torque Tavg. Assume that the load on the other side of the flywheel demands a torque Tl, which is time varying, as shown in Figure 10-22. This torque variation can cause the shaft speed to vary depending on the torquespeed characteristic of the driving motor. We need to determine how much Im to add in the form of a flywheel to reduce the speed variation of the shaft to an acceptable level. Write Newton's law for the free-body diagram in Figure 10-21. The left-hand side of this expression represents the change in kinetic energy Ek between the maximum and minimum shaft ω's and is equal to the area under the torquetime diagram of Figure 10-22 between those extreme values of ω. The right-hand side of equation 10.18c is the change in kinetic energy stored in the flywheel. The only way

The ASME has abandoned this shaft design standard, so it has only historical value now. If used at all, equation 10.6 should be applied only to situations where the loads are as it assumes them to be, namely constant torque and fully reversed moment. The ASME standard gives nonconservative results if either of the loading components that it assumes to be zero is in fact nonzero in a given case. We recommend using the more general approach of equation 10.8 (see below) for shaft design as it covers all loading situations. Figure 10-4 shows the Gough elliptical failure line of Figure 10-3 superposed on the Gerber, Soderberg, and modified-Goodman lines. Note that the elliptical line closely matches the Gerber line at the left-hand end but diverges to intersect the yield strength on the mean stress axis. The elliptical line has the advantage of accounting for possible yielding without needing to introduce an additional constraint involving the yield line. However, the Gough elliptical line, while a good fit to the failure data, is less conservative than the combination of Goodman line and yield line used as a failure envelope. Design for Fluctuating Bending and Fluctuating Torsion When the torque is not constant, its alternating component will create a complex multiaxial stress state in the shaft. Then the approach described in Section 6.12 (p. 376), which computes the von Mises components of the alternating and mean stresses using equations 6.22 (p. 379), can be used. A rotating shaft in combined bending and torsion has a biaxial stress state, which allows the two-dimensional version of equation 6.22b to be used. σ στ σ σσ τ ' ' a aa m mm m a axial = + = + ( ) + 2 2 2 2 3 10 3 ( .7 ) These von Mises stresses can now be entered into a modified-Goodman diagram (MGD) for a chosen material to find a safety factor, or equations 6.18 (pp. 367-368) can be applied without drawing the MGD. For design purposes, where the diameter of the shaft is the desired quantity to be found, equations 10.2, 10.3, and 10.6 as presented require iteration to find a value for d, given some known loads and assumed material properties. This is not a great difficulty if an equation solver with iterative ability is used. However, a hand-calculator solution is cumbersome with the equations in this form. If a particular failure case is assumed for the MGD, the equations can be manipulated to provide a design equation (similar to equation 10.6) for shaft diameter d at the section of interest. If the failure model used is Case 3 from Section 6.11 (p. 360), which assumes that the mean and alternating loads maintain a constant ratio,* failure will occur at point R in Figure 6-46c (p. 368). The safety factor as defined in equation 6.18e (p. 368) is then 1 10 NSS b f a f m ut = + σ σ ' ' ( .7 ) where Nf is the desired safety factor, Sf is the corrected fatigue strength at the selected cycle life (from equation 6.10 on p. 338), and Sut is the ultimate tensile strength of the material. * Note that this assumption is also implicit in equation 10.6 for t

The character of both the torque and bending loads may be either steady (constant) or may vary with time. Steady and time-varying torque and bending loads can also occur in any combination on the same shaft. Ch 10 4ed Final 550 10/18/09, 6:18 PM Chapter 10 SHAFTS, KEYS, AND COUPLINGS 551 10 If the shaft is stationary (nonrotating) and the sheaves or gears rotate with respect to it (on bearings), then it becomes a statically loaded member as long as the applied loads are steady with time. However, such a nonrotating shaft is not a transmission shaft, since it is not transmitting any torque. It is merely an axle, or round beam, and can be designed as such. This chapter is concerned with rotating, transmission shafts and their design for fatigue loading. Note that a rotating shaft subjected to a steady, transverse-bending load will experience a fully reversed stress state as shown in Figure 10-1a. Any one stress element on the shaft surface goes from tension to compression each cycle as the shaft turns. Thus, even for steady bending loads, a rotating shaft must be designed against fatigue failure. If either or both the torque and transverse loads vary with time, the fatigue loading becomes more complex, but the fatigue-design principles remain the same, as outlined in Chapter 6. The torque, for example, could be repeated or fluctuating as shown in Figures 10-1b and c, as could the bending loads. We will deal primarily with the general case, which allows for the possibility of both steady and time-varying components in both bending and torsion loads. If either load lacks a steady or time-varying component in a given case, it will merely force a term in the general equations to zero and simplify the calculation. 10.2 ATTACHMENTS AND STRESS CONCENTRATIONS While it is sometimes possible to design useful transmission shafts that have no changes in section diameter over their length, it is most common for shafts to have a number of steps or shoulders where the diameter changes to accommodate attached elements such as bearings, sprockets, gears, etc., as shown in Figure 10-2, which also shows a collection of features commonly used to attach or locate elements on a shaft. Steps or shoulders are necessary to provide accurate and consistent axial location of the attached elements as well as to create the proper diameter to fit standard parts such as bearings. Keys, snap rings, or cross-pins are often used to secure attached elements to the shaft in order to transmit the required torque or to capture the part axially. Keys require a groove in both shaft and part and may need a setscrew to prevent axial motion. Snap rings groove the shaft, and cross-pins create a hole through the shaft. Each of these changes in contour will contribute some stress concentration and this must be accounted for in the fatigue-stress calculations for the shaft. Use generous radii where possible, and techniques such as those shown in Figures 4-37 (p. 191), 4-38 (p. 192), and 10-2 (at the sheave and snap ring) to reduce the effects of these stress concentrations. Keys and pins can be avoided by using friction to attach elements (gears, sprockets) to a shaft. Many designs of clamp collars (keyless fits*) are available, which squeeze the outside diameter (OD) of the shaft with high compressive force to clamp something to it, as shown on the sprocket hub in Figure 10-2 and in Figure 10-34 (p. 605). The hub has a gently tapered bore, and a matching taper on this type of clamp collar is forced into the space between hub and shaft by tightening the bolts. Axial slits in the tapered portion of the collar allow it to change diameter and squeeze the shaft, creating sufficient friction to transmit the torque. Another type of clamp collar, called a split collar, uses a screw to close a radial slit and clamp the collar to the shaft. Press and shrink fits are also used for this purpose and will be discussed in a later section of this

The most efficient flywheel design in terms of maximizing Im for minimum material used is one in which the mass is concentrated in its rim and its hub is supported on spokes, like a carriage wheel. This puts the majority of the mass at the largest radius possible and minimizes the weight for a given Im. Even if a flat, solid circular disk flywheel design is chosen, either for simplicity of manufacture or to obtain a flat surface for other functions (such as an automobile clutch), the design should be done with an eye to reducing weight and thus cost. Since, in general, Im = mr2, a thin disk of large diameter will need fewer pounds of material to obtain a given Im than will a thicker disk of smaller diameter. Dense materials such as cast iron and steel are the obvious choices for a flywheel. Aluminum is seldom used. Though many metals (lead, gold, silver, platinum) are more dense than iron and steel, one can seldom get the accounting department's approval to use them in a flywheel. Figure 10-23 shows the change in the torque of Figure 10-22 after the addition of a flywheel sized to provide a coefficient of fluctuation of 0.05. The oscillation in torque about the unchanged average value is now 5%, much less than what it was without the flywheel. Note that the peak value is now 87 instead of 372 lb-in. A much smallerhorsepower motor can now be used because the flywheel is available to absorb the energy returned from the load during the cycle. Stresses in Flywheels As a flywheel spins, the centrifugal force acts upon its distributed mass and attempts to pull it apart. These centrifugal forces are similar to those caused by an internal pressure in a cylinder. Thus, the stress state in a spinning flywheel is analogous to a thickwalled cylinder under internal pressure (see Section 4.17 on p. 203). The tangential stress of a solid-disk flywheel as a function of its radius r is σ γ ω ν ν ν t io i o g r r r r r = r a ⎛ + ⎝ ⎞ ⎠ ++ − + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 22 2 2 2 3 2 8 1 3 3 (. ) 10 23 and the radial stress is σ γ ω ν r io i o g r r r r r = r b ⎛ + ⎝ ⎞ ⎠ +− − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 22 2 2 2 3 2 8 (. ) 10 2

The problem is that we typically do not know the dynamic deflections of the system a priori. Rayleigh showed that virtually any estimate of the deflection curve, provided that it reasonably represents the maximum deflection and the boundary conditions of the actual dynamic curve, will suffice. The static deflection curve due to the weights of the assumed lumps (including the weight of the shaft or not as desired) is a very suitable estimate. Note that any applied external loads are not included in this deflection calculation, only the gravitational forces. The resulting approximate ωn will always be higher than the true fundamental frequency by a few percent regardless of the assumed deflection-curve shape. If more than one estimated deflection curve is tried, the one yielding the lowest value of ωn should be used, as it will be the closest approximation. Equation 10.25c can be applied to a system of any complexity by breaking it into a large number of lumps. If gears, pulleys, etc., are on the shaft, they make logical lumped masses. If the shaft mass is significant or dominant, it can be broken into discrete elements along its length with each piece providing a term in the summation. Rayleigh's method can theoretically be used to find higher frequencies than the fundamental but to do so is difficult without a good estimate of the shape of the higher-order deflection curve. More accurate methods for approximating both the fundamental frequency and the higher frequencies exist but are somewhat more complicated to implement. Ritz modified Rayleigh's method (Rayleigh-Ritz) to allow iteration to the higher frequencies. Holzer's method is more accurate and can find multiple frequencies. See reference 10 for more information. Shaft Whirl Shaft whirl is a self-excited vibratory phenomenon to which all shafts are potentially subject. While it is common and recommended practice to dynamically balance all rotating elements in machinery (especially if operated at high speeds), it is not possible to achieve exact dynamic balance except by chance. (See reference 12 for a discussion of dynamic balancing.) Any residual unbalance of a rotating element causes its true mass center to be eccentric from the axis of shaft rotation. This eccentricity creates a centrifugal force that tends to deflect the shaft in the direction of the eccentricity, increasing it and thus further increasing the centrifugal force. The only resistance to this force

The torsional spring constant kt for a solid circular shaft is k GJ l t = lb - in/rad or N - m/rad ( . ) 10 27b where G is the material's modulus of rigidity, and l is the shaft length. The polar second moment of area J of a solid circular shaft is J d = c π 4 32 in or m ( . ) 10 27 4 4 If the shaft is stepped, then an equivalent polar second moment of area Jeff is found from J l l J d eff i i i n = ∑ =1 (. ) 10 27 where l is total shaft length, and Ji, and li are the polar moments and lengths of the subsections of shaft of differing diameters, respectively. The mass moment of inertia of a solid circular disk about its axis of rotation is: I mr m = e 2 2 in - lb - sec or kg - m ( . ) 10 27 2 2 where r is the disk radius and m is its mass. These equations are sufficient to find the critical frequency of a single disk mounted on a fixed axle, as shown in Figure 10-30. Two Disks on a Common Shaft A more interesting problem is that of two (or more) disks displaced on a common shaft as shown in Figure 10-31. The two disks shown will oscillate torsionally at the same natural frequency, 180° out of phase. There will be a point, called a node, somewhere on the shaft, at which there will be no angular deflection. On either side of the node,

as well. These are used in automobile driveshafts, a pair of Hooke couplings in a reardrive driveshaft and Rzeppas (called CV joints) in a front-drive automobile. The variety of couplings available makes it necessary for the designer to seek more detailed information on their capabilities from their manufacturers, who are always willing to help in the selection of the proper type of coupling for any application. Manufacturers can often supply test data on the load and alignment capabilities of their particular couplings. 10.16 CASE STUDY We will now address the design of shafts in one of the Case Study assemblies that were defined in Chapter 9. Designing Driveshafts for a Portable Air Compressor The preliminary design of this device is shown in Figure 9-1 (repeated opposite). There are two shafts, input and output. The output shaft will have the larger torque in this case since it is at the slower speed of 1 500 rpm. The torque on that shaft was defined in the earlier Case Study 8A in Chapter 9 and is shown in Figure 9-3 (repeated opposite). Since this is a time-varying torque, the shaft must be designed for fatigue loading. In addition to the torque on the shafts, there will be side loads from the gears that apply bending moments, making a combined-loading situation. Note in Figure 9-1 that the shafts are shown as short, being only long enough to accommodate the gear and bearings. This is done to minimize the bending moments from the gear forces. Since the gears are yet to be designed, we will have to make some assumptions about their diameters and thicknesses to do a preliminary design of the shafts. Later selection of bearings may also dictate some changes to our shaft design. This is typical of design problems, since all of their elements interact. Iteration is necessary to refine all the elements' designs.

chapter. However, as we will see, these friction couplings also create stress concentrations in the shaft and can cause fretting corrosion as described in Section 7-6 (p. 434). A standard taper pin is sometimes used to couple elements to shafts as seen in the sheave of Figure 10-2. The hole is reamed to match the standardized pin-taper and the purchased pin is driven into place. The shallow taper locks it by friction. It must be driven out for disassembly. This technique should be used with caution in locations of large bending moment, as it weakens the shaft as well as creating stress concentration. Rolling-element bearings as shown in Figure 10-2 are intended to have their inner and outer races be press-fitted to shaft and housing, respectively. This requires closetolerance machining of the shaft diameter and requires a step shoulder to provide a stop for the press fit and for axial location. Thus, one must start with a larger stock shaft diameter than the bearing inside diameter (ID) and machine the shaft to fit the selected bearing whose sizes are standardized (and are metric). A snap ring is sometimes used to guarantee no axial movement of the shaft versus the bearing as shown at the sheave end of the shaft in Figure 10-2. Snap rings are commercially available in a variety of styles and require that a small, close-tolerance groove of specific dimension be machined in the shaft. Note in Figure 10-2 how the axial location of the shaft is achieved by capturing only one of the bearings (the right one) axially. The other bearing at the left-hand end has axial clearance between it and the step. This is to prevent axial stresses being generated by thermal expansion of the shaft between the two bearings. So, it appears that we cannot escape the problems of stress concentration in practical machinery. In the case of shafts, we need to provide steps, snap rings, or other means to accurately locate components axially on the shaft, and we have to key, pin, or squeeze the shaft to transmit the torque. Each of these methods of attachment has its own advantages and disadvantages. A key is simple to install and sizes are standardized to the shaft diameter. It provides accurate phasing* and is easily disassembled and repaired. It may have no resistance to axial movement and it does not always provide a truly tight torque coupling due to the slight clearance between key and keyway. Torque reversals can cause slight backlash.

comes from the elastic stiffness of the shaft as shown in Figure 10-28. The initial shaft eccentricity is labeled e and the dynamic deflection is δ. A free-body diagram shows the forces acting to be km e a δ δω = + ( ) 2 (. ) 10 26 δ ω ω = ( ) − e k m b 2 2 (. ) 10 26 The dynamic deflection of the shaft from this centrifugal force causes it to whirl about its axis of rotation with points at the center of the deflected shaft describing circles about the axis. Note in equation 10.26b that the deflection becomes infinite when ω2 = k / m. As the rotational speed of the shaft approaches the fundamental natural (or critical) speed of lateral vibration, a similar resonance phenomenon to that of lateral vibration occurs. Note in equation 10.26b that when ω2 = k / m, δ = ∞. Equation 10.26b can be normalized to a nondimensional form, which clearly shows the relationship: δ ω ω e ω ω c n n = ( ) − ( ) 2 2 1 (. ) 10 26 Equation 10.26c and Figure 10-29 show the amplitude of shaft deflection normalized to the original eccentricity (δ / e) as a function of the ratio between the rotational frequency and critical frequency ω / ωn. Note that at ω / ωn = 0, there is no response, unlike the forced vibration of the previous section. This is because no centrifugal force exists unless the shaft is rotating. As the shaft speed increases, the deflection rapidly increases. If no damping is present (ζ = 0), at ω / ωn = 0.707, the deflection of the shaft is equal to the eccentricity and it becomes theoretically infinite at resonance (ω / ωn = 1). Of course, there will always be some damping present, but if ζ is small, the deflections will be very large at resonance and can cause stresses large enough to fail the shaft. Note what happens when the shaft speed passes through ωn. The phase shifts 180°, which means that the deflection switches sides abruptly at resonance. At higher ratios of ω / ωn, the deflection approaches -e, which means that the system is then rotating about the mass center of the eccentric mass, and the shaft centerline is eccentric. Conservation of energy makes the system want to rotate about its true mass center. In any

e must now determine how large a flywheel is needed to absorb this energy with an acceptable change in speed. The change in shaft speed during a cycle is called its fluctuation Fl and is equal to Fl = − ω ω max min ( .19a) 10 From ωmax @ C ωmin @ B C to D D to A B to C A to B ΔArea = ΔE Accumulated Sum = E Min & Max Total ΔEnergy = E @ ωmax - E @ ωmin = (-6 032) - (+20 073) = -26 105 in-lb +9 356 +154 -6 032 +20 073 +15 388 -9 202 -26 105 +20 073 Table 10-5 Accumulation of Energy Pulses Under a Torque-Time Curve Ch 10 4ed Final 588 10/18/09, 6:20 PM Chapter 10 SHAFTS, KEYS, AND COUPLINGS 589 10 We can normalize this to a dimensionless ratio by dividing it by the average shaft speed. This ratio is called the coefficient of fluctuation Cf. Cf max min avg = ( ) ω ω− ω ( .19b) 10 This coefficient of fluctuation is a design parameter to be chosen by the designer. It is typically set to a value between 0.01 and 0.05 for precision machinery, and as high as 0.20 for crushing or hammering machinery, which correspond to a 1 to 5% fluctuation in shaft speed. The smaller this chosen value, the larger the flywheel will have to be. This presents a design trade-off. A larger flywheel will add more cost and weight to the system, which are factors that have to be considered against the smoothness of operation desired. We found the required change in kinetic energy Ek by integrating the torque curve, TT d E a l avg k min max ( ) − = ∫ θ θ ω θ ω @ @ (. ) 10 20 and can now set it equal to the right-hand side of equation 10.18c: EI b k = − m max min ( ) 1 2 10 20 2 2 ω ω (. ) Factoring this expression: EI c k = + m max min max min ( )( ) − 1 2 ω ωω ω (. ) 10 20 If the torque-time function were a pure harmonic, then its average value could be expressed exactly as ω ω ω avg max min = ( ) + 2 ( .21) 10 Our torque functions will seldom be pure harmonics, but the error introduced by using this expression as an approximation of the average is acceptable. We can now substitute equations 10.19b and 10.21 into equation 10.20c to get an expression for the mass moment of inertia Is needed in the entire rotating system in order to obtain the selected coefficient of fluctuation. EI C I E C k s avg f avg s k f avg = ( )( ) = 1 2 2 10 2 ω ω ω ( .22) Equation 10.22 can be used to design the physical flywheel by choosing a desired coefficient of fluctuation Cf, and using the value of Ek from the numerical integration of the torque curve (see Table 10-5 for an example) and the average shaft ω to compute the needed system Is. The physical flywheel's mass moment of inertia Im is then set equal to the required system Is. But, if the moments of inertia of the other rotating elements on the same shaft (such as the motor) are known, the physical flywheel's required Im can be reduced by those amounts.

factors are applied in the same manner as before. For static loading they need to be used to determine whether local yielding will compromise the interference fit. For dynamic loading, they are modified by the material's notch sensitivity to get a fatigue stress-concentration factor to use in equation 10.8 (p. 560) for shaft design. Fretting Corrosion This problem was discussed in Chapter 7. Interference fits are the primary victims of fretting problems. Though the fretting mechanism is not yet fully understood, certain precautions are known to help reduce its severity. See Section 7.6 (p. 434) for more details.

in automobiles and trucks where the suspension movement causes axial motion between the members. They are also used within nonautomatic, nonsynchromesh truck transmissions to couple the axially shiftable gears to their shafts. In addition, engine torque is usually passed into the transmission through a spline that connects the engine clutch to the transmission input shaft and allows the axial motion necessary to disengage the clutch from the flywheel. The loading on a spline is typically pure torsion, either steady or fluctuating. While it is possible to have bending loads superposed, good design practice will minimize the bending moments by proper placement of bearings and by keeping a cantilevered spline as short as possible. As with keys, two failure modes are possible, bearing or shear. Shear failure is usually the limiting mode. Unlike keys, many teeth are available to share the load to some degree. Ideally, the spline length l needs to be only as long as is required to develop a combined tooth shear strength equal to the torsional shear strength of the shaft itself. If the spline were made perfectly with no variation in either tooth thickness or spacing, all teeth would share the load equally. However, the realities of manufacturing tolerances prohibit this ideal condition. The SAE states that "actual practice has shown that due to inaccuracies in spacing and tooth form, the equivalent of about 25% of teeth are in contact, so that a good approximate formula for a splined shaft (length) is" l d dd d r ir p ≅ ( ) − 3 44 2 1 ( .12) 10 where dr is the root diameter of the external spline, di is the internal diameter (if any) of a hollow shaft, and dp is the pitch diameter of the spline, which is approximately at mid-tooth. The variable l represents the actual engaged length of the spline teeth with one another and should be considered as the minimum value needed to develop the strength in the teeth of an equivalent-diameter shaft. The shear stress is calculated at the pitch diameter of the splines, where the shear area is A d l a shear p = π 2 (. ) 10 13 The shear stress can be calculated using the SAE assumption that only 25% of the teeth are actually sharing the load at any one time by considering only 1/4 of the shear area to be stressed: τ π ≅= = = 4 4 8 16 10 13 2 F A T r A T d A T d l b shear p shear p shear p (. ) where T is the torque on the shaft. Any bending stresses on the spline must also be calculated and properly combined with this shear stress. If the loading is pure torsion and static, then the shear stress of equation 10.13b is compared to the shear yield strength of the material to obtain a safety factor. If the loads are fluctuating or if bending is present, then the applied stresses should be converted to equivalent von Mises tensile stresses and compared to the appropriate strength criteria using the modified-Goodman diagram.

or shaft has an infinite number of particles, each of which is capable of elastic motion versus its neighboring particles. Thus a continuous system has an infinity of natural frequencies. In either case, the lowest, or fundamental, natural frequency is usually of most interest. The natural frequencies of vibration of a system can be expressed either as a circular frequency ωn with units of rad/sec or rpm, or as a linear frequency fn with units of hertz (Hz). They are the same frequencies expressed in different units. The general expression for the fundamental natural frequency is ωn k m = rad/sec ( . ) 10 24a f k m n = b 1 2 10 24 π Hz ( . ) where k is the spring constant of the system and m is its mass. The natural frequencies are a physical property of the system; once built, it retains them essentially unchanged unless it loses or gains mass or stiffness during its useful life. Equations 10.24 define the undamped natural frequency. Damping reduces the natural frequency slightly. Shafts, beams, and most machine parts tend to be lightly damped, so the undamped value can be used with little error. The usual design strategy is to keep all forcing, or self-exciting, frequencies below the first critical frequency by some comfort margin. The larger this margin, the better, but a factor of at least 3 or 4 is desirable. This keeps the amplitude-response ratio close to one or zero, as shown in Figures 10-26a and 10-26b. In some cases, the fundamental frequency of a shaft system cannot be made higher than the required rotating frequency. If the system can be accelerated rapidly enough through resonance, before the vibrations have a chance to build up amplitude, then the system can be run at a speed higher than resonance. Stationary power plants are in this category. The massiveness of the turbines and generators gives a low fundamental frequency (see equation 10.24), but they must be run at a high speed to generate the proper line frequency of AC current. Thus they operate to the right of the peak in Figure 10-14b where the amplitude ratio approaches one at high ratios of ωf / ωn. Their start-ups and shutdowns may be infrequent but must always be accomplished rapidly to get through the resonance peak before any damage is caused by excessive deflections. Also, sufficient driving power must be available to provide the energy absorbed at resonance by the oscillations in addition to accelerating the rotating mass. If the driver lacks sufficient power, then the system may stall in resonance, unable to increase its speed in the face of the potentially destructive vibrations. This is called the Sommerfeld effect.[9] There are three types of shaft vibration of concern: 1 Lateral vibration 2 Shaft whirl 3 Torsional vibration The first two involve bending deflections and the last torsional deflection of the shaft.

points on the shaft rotate in opposite angular directions when vibrating. The system can be modeled as two separate, single-mass systems coupled at this stationary node. One has mass moment and spring constant I1, k1 and the other I2, k2. Their common natural frequency is then ωn k I k I = a 1 1 2 2 = (. ) 10 28 The spring constants of the shaft segments are each found from kt = JG / l assuming that the J is constant across the node. JG l I JG l I lI l I I l l l I l I I b 11 2 2 11 2 2 2 1 1 2 1 2 10 28 = and = so ( . ) = − ( ) = + Equation 10.28b allows the location of the node to be found. Substituting this expression into equation 10.28a gives ω ω ω n n t nt k I JG l I JG l I I I I k I I I I k I I I I c = = + = + = + 1 1 11 1 2 1 2 1 2 1 2 2 1 2 1 2 10 28 = or ( . ) which defines the critical speed for torsional vibration in terms of the known mass properties of the two disks and the overall spring constant of the shaft. The critical frequency needs to be avoided in any forcing functions applied to the shaft in order to avoid torsional resonance that will overload it. Devices attached to the shaft, such as piston engines or piston pumps, will have frequencies in their torque-time functions that correspond to the pulses of their operation multiplied by their rotational frequency. For example, a four-cylinder engine will have a strong forcing frequency component at four times its rpm. If this fourth harmonic coincides with the shaft's critical frequency, there could be a problem. When designing a shaft, the frequency characteristics of the driving and driven rotating devices attached to it must be taken into account along with their primary rotational frequency. Multiple Disks on a Common Shaft Two disks on a common shaft have one node and one torsional natural frequency. Three disks will have two nodes and two natural frequencies. This pattern will hold for any number of disks, assuming in all cases that the masses of the disks dominate the shaft mass, allowing it to be ignored. N disks will have N - 1 nodes and natural frequencies. The degree of the equation for the natural frequencies will also be N - 1 if we consider the variable to be ωn 2 rather than ωn. Note that equation 10.28c for two masses is linear with this assumption. The equation for three masses is quadratic in ωn 2 and for four masses is a cubic in ωn 2.

system where the rotating elements are eccentric and large compared to the shaft this will occur. Perhaps you have observed a pivoted ceiling fan rotating with its motor center orbiting about the axis of rotation. The fan blades are usually not in perfect balance and the assembly rotates about the mass center of the blade assembly rather than about the motor/shaft centerline. It should be clear that rotation of a system at or near its critical frequency is to be strictly avoided. The critical frequency for shaft whirl is the same as for lateral vibration and can be found using Rayleigh's method or any other suitable technique. Because this shaft-whirl vibration amplitude ratio starts at zero rather than at one (as with forced vibrations) the forcing frequency can be closer to the critical frequency than for the lateral vibration case. Keeping the operating speed below about half the shaft-whirl critical frequency will usually provide good results unless the initial eccentricity is excessive (which should not be allowed anyway). Note the difference between shaft lateral vibration and shaft whirl. Lateral vibration is a forced vibration, requiring some outside source of energy such as vibrations from other parts of the machine to precipitate it, and the shaft then vibrates in one or more lateral planes whether or not it is rotating. Shaft whirl is a self-excited vibration caused by the shaft rotation acting on an eccentric mass. It will always occur when both rotation and eccentricity are present. The shaft assumes a deflected shape, which then rotates or whirls about the axis much like a jump-rope being swung by children. Torsional Vibration Just as a shaft can vibrate laterally, it can also vibrate torsionally and will have one or more torsional natural frequencies. The same equations that describe lateral vibrations can be used for torsional ones. The systems are analogous. Force becomes torque, mass becomes mass moment of inertia, and linear spring constant becomes torsional spring constant. Equation 10.24 (p. 595) for the circular natural frequency becomes, for a single-degree-of-freedom rotating system: ωn t m k I = rad/sec (

taper is a locking one, which means that the friction force between the surfaces holds the key in place axially. The gib head is optional and provides a surface for prying the key out when the small end is not accessible. Tapered keys tend to create eccentricity between hub and shaft, as they drive all the radial clearance to one side. Woodruff Keys Woodruff keys are used on smaller shafts. They are self-aligning, so are preferred for tapered shafts. The penetration of a Woodruff key into the hub is the same as that of a square key, i.e., half the key width. The semicircular shape creates a deeper keyseat in the shaft, which resists key-rolling, but weakens the shaft compared to a square or tapered keyseat. Woodruff key widths as a function of shaft diameter are essentially the same as those for square keys shown in Table 10-2. The other dimensions of the Woodruff key are defined in the ANSI Standard, and keyseat cutters are readily available to match these dimensions. Table 10-3 reproduces a sample of the key-size specifications from the standard. Each key size is given a key number, which encodes its dimensions. The ANSI Standard states: "The last two digits give the nominal key diameter in eighths of an inch and the digits preceding the last two give the nominal width in thirty-seconds of an inch." For example, the key number 808 defines a key size of 8/32 X 8/8 or 1/4 wide X 1-in dia. See reference 6 for complete dimensional information on keys. Stresses in Keys There are two modes of failure in keys: shear and bearing. A shear failure occurs when the key is sheared across its width at the interface between the shaft and hub. Bearing failure occurs by crushing either side in compression

to extract kinetic energy from the flywheel is to slow it down, as shown in equation 10.17a. Adding kinetic energy speeds it up. It is impossible to obtain exactly constant shaft velocity in the face of changing energy demands by the load. The best we can do is to minimize the speed variation (ωmax - ωmin) by providing a flywheel with sufficiently large Im.

uses the elliptical curve of Figure 10-3 fitted through the bending endurance strength on the σa axis and the tensile yield strength on the σm axis as the failure envelope. The tensile yield strength is substituted for the torsional yield strength by using the von Mises relationship of equation 5.9 (p. 251). The derivation of the ASME shaft equation is as follows. Starting with the relationship for the failure envelope shown in Figure 10-3a: σ τ a e m S Sys a ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 2 1 10 ( .5 ) introduce a safety factor Nf N S N S b f a e f m ys ⎛ σ τ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 2 1 10 ( .5 ) Recall the von Mises relationship for Sys from equation 5.9 (p. 251): SS c ys y = 3 10 ( .5 ) and substitute it in equation 10.5b. N S N S d f a e f m y ⎛ σ τ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2 2 3 1 10 ( .5 ) Substitute the expressions for σa and τm from equations 10.2c and 10.3c, respectively: k M d N S k T d N S e f a f e fsm m f y 32 16 3 1 10 3 2 3 2 π π ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ( .5 ) which can be rearranged to solve for the shaft diameter d as d N k M S k T S a f f a f fsm m y = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⎫ ⎬ ⎪ ⎪ ⎭ ⎪ ⎪ 32 3 4 10 6 2 2 1 2 1 3 π (.) The notation used in equation 10.6 is slightly different than that of the ANSI/ASME standard in order to remain consistent with the notation used in this text. The standard uses the approach of reducing the fatigue strength Sf by the fatigue-stress-concentration factor kf rather than using kf as a stress increaser as is done consistently in this text. In most cases (including this one) the result is the same. Also, the ASME standard assumes the stress concentration for mean stress kfsm to be 1 in all cases, which gives d N k M S T S b f f a f m y = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⎫ ⎬

where kf and kfm are the bending-fatigue stress-concentration factors for the alternating and mean components, respectively (see equations 6.11 and 6.17 on pp. 339 and 360). Since the typical shaft is a solid-round cross section,* we can substitute for c and I: c r d I d == = b 2 64 10 2 4 π (.) giving σ π σ π a f a m fm m k M d k M d = = c 32 32 10 2 3 3 (.) where d is the local shaft diameter at the section of interest. The alternating and mean torsional shear stresses are found from τ τ a fs a m fsm m k T r J k T r J = = (.) 10 3a where kfs and kfsm are the torsional fatigue stress-concentration factors for the alternating and mean components, respectively (see equation 6.11 on p. 343 for kfs and use the applied shear stresses and shear yield strength in equation 6.17 on p. 364 to get kfsm). For a solid-round cross section,* we can substitute for r and J: r d J d = = b 2 32 10 3 4 π (.) giving τ π τ π a fs a m fsm m k T d k T d = = c 16 16 10 3 3 3 (.) A tensile axial load Fz, if any is present, will typically have only a mean component (such as the weight of the components) and can be found from σ π m fm z fm z axial k F A k F d = = 4 10 4 2 ( .) 10.7 SHAFT FAILURE IN COMBINED LOADING Extensive studies of fatigue failure of both ductile steels and brittle cast irons in combined bending and torsion were done originally in England in the 1930s by Davies[3] and Gough and Pollard.[5] These early results are shown in Figure 10-3, which is taken from the ANSI/ASME Standard B106.1M-1985 on the Design of Transmission Shafting. Data from later research is also included on these plots.[2, 4] The combination of torsion and bending on ductile materials in fatigue was found to generally follow the elliptical relationship as defined by the equations in the figure. Cast brittle materials (not shown) were found to fail based on the maximum principal stress. These findings are similar to those for combined torsional and bending stresses in fully reversed loading shown in Figure 6-15 (p. 321)

where l is the length of the hub engagement, r is the shaft radius, and μ is the coefficient of friction between shaft and hub. The AGMA standard suggests a value of 0.12 ≤μ≤ 0.15 for hydraulically expanded hubs and 0.15 ≤μ≤ 0.20 for shrink or pressfit hubs. AGMA assumes (and recommends) a surface finish of 32 μin rms (1.6 μm Ra), which requires a ground finish on both diameters. Equations 10.14a and 10.14b can be combined to give an expression that defines the torque obtainable from a particular deformation, coefficient of friction, and geometry. T lr E r r r r E r r r r c o o o o i i i i = + − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + + − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ π μδ ν ν 1 1 10 14 2 2 2 2 2 2 2 2 (. ) The pressure p is used in equations 4.47 (p. 204) to find the radial and tangential stresses in each part. For the shaft: σt i i shaft p r r r r = − a + − 2 2 2 2 ( .15 ) 10 σrshaft = − p b ( .15 ) 10 where ri is the inside radius of a hollow shaft. If the shaft is solid, ri will be zero. For the hub: σt o o hub p r r r r = a + − 2 2 2 2 (. ) 10 16 σrhub = − p b (. ) 10 16 These stresses need to be kept below the yield strengths of the materials to maintain the fit. If the materials yield, the hub will become loose on the shaft. Stress Concentration in Interference Fits Even though there may be no disruption of the smooth surface of the press-fit shaft by shoulders or keyways, an interference fit nevertheless creates stress concentrations in the shaft and hub at the ends of the hub due to the abrupt transition from uncompressed to compressed material. Figure 10-19a shows a photoelastic study of a press-fit hub on a shaft. The fringes show stress concentration at the corners. Figure 10-19b shows how the stress concentration can be reduced by providing circumferential relief grooves in the faces of the hub close to the shaft diameter. These grooves make the material at the edge of the hub more compliant enabling it to deflect away from the shaft and reduce the stress locally. This approach is similar to the techniques for stress-concentration reduction shown in Figure 4-38 (p. 192). Figure 10-20 shows curves of stress-concentration factors for interference fits between hubs and shafts developed from the photoelastic study of Figure 10-19a. The values on the abscissa are ratios of hub length to shaft diameter. These geometric stress-concentration

where γ = material weight density, ω = angular velocity in rad/sec, ν = Poisson's ratio, r is the radius to a point of interest, and ri, ro are inside and outside radii of the soliddisk flywheel. Figure 10-24 shows how these stresses vary over the radius of the flywheel. The tangential stress is a maximum at the inner radius. The radial stress is zero at inside and outside radii and peaks at an interior point but is everywhere less than the tangential stress. The point of most interest is then at the inside radius. The tangential tensile stress at that point is what fails a flywheel, and when it fractures at that point, it typically fragments and explodes with extremely dangerous results. Since the forces causing the stress are a function of rotational speed, there will always be some speed that will fail the flywheel. A maximum safe operating speed should be calculated for a flywheel and some means taken to preclude its operation at higher speeds, such as a speed control or governor. A safety factor against overspeeding can be determined as the quotient of the speed that will cause yielding over the operating speed, Nos = ω / ωyield. Failure Criteria If the flywheel spends most of its life operating at essentially constant speed, then it can be considered to be statically loaded and the yield strength used as a failure criterion. The number of start-stop cycles in its operating regime will determine whether a fatigueloading situation needs to be considered. Each runup to operational speed and rundown to zero constitutes a fluctuating stress cycle. If the number of these start-stop cycles is large enough over the projected life of the system, then fatigue-failure criteria should be applied. A low-cycle fatigue regime may require a strain-based fatigue failure analysis rather than a stress-based one, particularly if there exists the possibility of any transient overloads that may cause the local stresses to exceed the yield stress at stress concentrations

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