mechanical design an integrated approach Part 7 pps

Chapter 9 Shafts and Associated Parts Bearings and Lubrication Helical, Bevel, and Worm Gears Belts, Chains, Clutches, and Be kes Springs Power Screws; Fasteners, and Connections -A

Given: S, = 150 ksi, S, = 72 ksi

Design Decision: Employ the Goodman criterion, based on the maximum normal stress

———läin —

Figure P8.21 8.22 Accantilever spring is subjected to a concentrated load P varying continuously from 0 to Po (Figure P8.22) What is the greatest allowable load P, for n = 4?

Given: S, = 850 MPa, 5S, == 175 MPa, b=5mm, h = 1Ô mm, K; =2

Assumption: Failure occurs due to bending stress at the fillet

Design Decision: Use the Soderberg criterion

Figure P8.22 8.23 Resolve Problem 8.22 for the load varying from P,,/2 upward to P, downward, n = 2

8.24 A24-mm wide, 4-mm thick, and 3-mm long leaf spring, made of AISI L050CD steel, is straight

and unstressed when the cam and shaft are removed (Figure P8.24) Use the Goodman theory

to calculate the factor of safety n for the spring

Given: S, = 250 MPa, E =: 200 GPa, p= 0.3

Assumption: The cam rotates continuously Leaf spring is considered as a wide cantilever beam

8.25 Repeat Problem 8.24 for the case in which the cantilevered spring is made of normalized AIST

1095 steel and employing the Soderberg criterion

8.26 Figure P8.26 shows a circular aluminum bar having two shoulder fillets supporting a concentrated load P at its midspan Determine the allowable value for diameter D if stress con- ditions at the fillets are to be satisfactory for conditions of operation Dimensions shown are in

8.27

CHAPTERS @ FATIGUE

Assumptions: The load P varies from 2 KN to 6 KN The Soderberg relation is employed

Figure P8.27

341

Chapter 9 Shafts and Associated Parts

Bearings and Lubrication

Helical, Bevel, and Worm Gears

Belts, Chains, Clutches, and Be kes

Springs

Power Screws; Fasteners, and Connections

-Axisymmetric Problemis in Design

Finite Element’ Analysis in Design

SHAFTS AND ASSOCIATED PARTS:

Critical Speed of Shafts Mounting Parts’

Stresses in Keys Splines

Couplings Universal Jotats

where the rotational axes are at an angle with respect to one another The customary shaft

types are straight shafts of constant or stepped cross section and crankshafts (Figure 9.1) The former two carry rotating members such as gears, pulleys, grooved pulleys (sheaves), or other wheels The latter are used to convert reciprocating motion into rotary motion or vice versa

Most shafts are under fluctuating loads of combined bending and torsion with various degrees of stress concentration Many shafts are not subjected to shock or impact loading;

however, some applications arise where such load takes place (Section 9.5) Thus, the as- sociated considerations of static strength, fatigue strength, and reliability play a significant role in shaft design A shaft designed from the preceding viewpoint satisfies strength re- quirements Of equal importance in design is the consideration of shaft deflection and

rigidity requirements Excessive lateral shaft deflection can cause bearing wear or failure

and objectionable noise The operating speed of a shaft should not be close to a critical speed (Section 9.7), or large vibrations are likely to develop

In addition to the shaft itself, the design usually must include calculation of the nec- essary keys and couplings Keys, pins, snap rings, and clamp collars are used on shafts to secure rotating elements The use of shaft shoulder is an excellent means of axially posi- tioning the shaft elements Figure 9.2 shows a stepped shaft supporting a gear, a crowned pulley, and a sheave, The mounting parts, discussed in Section 9.8, as well as shaft shoul- ders, are a source of stress raisers, and they must be properly selected and located to min- imize the resulting stress concentrations Press and shrink fits (Section 9.6) are also used for mounting Shafts are carried in bearings, in a simply supported form, cantilevered or overhang, depending on the machine configuration Couplings connect a shaft to a shaft

Figure 9.1 Common shaft types: (a) constant

diameter; (b) stepped; () crank shaft

CHAPTER? © SHAFTS AND ASSOCIATED ParTs

Figure 9.2 A stepped shaft with various elements attached

of power source or load Parameters that must be considered in the selection of a coupling

to connect two shafts include the angle between the shafts, transmitted power, vibrations,

and shock loads The websites www.pddnet.com, www.powertransmission.com, and

www.grainger.com present general information on shaft couplings

345

9.2 MATERIALS USED FOR SHAFTING

To minimize deflections, shaft materials are generally cold-drawn or machined from hot-

rolled, plain-carbon steel The shaft ends should be made with chamfers to facilitate forc-

ing on the mounted parts and to avoid denting the surfaces Cold drawing improves the physical properties It raises considerably the values of ultimate tensile and yield strengths

of steel Where toughness, shock resistance, and greater strength are needed, alloy steels are used The foregoing materials can be heat treated to produce the desired properties If the service requirements demand resistance to wear rather than extreme strength, it is cus- tomary to harden only the surface of the shaft, and a carburizing grade steel can be used

Note that the hardening treatment is applied to those surfaces requiring it; the remainder of the shaft is left in its original condition

Thick-walled seamless tubing is available for simpler, smaller shafts Large-diameter members (> about 75-mm diameter), such as railroad axles, press cranks, and so on, are usually forged and machined to the required size In addition to steels, high strength nodu- lar cast iron is used to make shaped shafts; for example, automotive engine crankshafts

Bronze or stainless steel is sometimes used for marine or other corrosive environments

Because keys and pins are loaded in shear, they are made of ductile materials Soft, low-

carbon steel is in widespread usage Most keys and pins are usually made from cold-rolled bar stock, cut to length and tapered if needed

9.3 DESIGN OF SHAFTS IN STEADY TORSION

In the design of circular slender shafts that transmit power at a specified speed, the mate-

rial and the dimensions of the cross section are selected to not exceed the allowable shear- ing stress or a limiting angle of twist when rotating Therefore, a designer needs to know

the torque acting on the power-transmitting shaft (see Section 1.12) Equations (1.15)

through (1.17) may be used to convert the power supplied to the shaft into a constant torque

1 Assume that, as is often the case, shear stress is closely associated with failure Note, however, that in some materials, the maximum tensile and compression stresses oc- curring on planes at 45° (see Figure 3.27) to the shaft axis may cause the failure

An important value of the shear stress is defined by tnax = Tc/J

3 The maximum usable value of ta,, without failure is the yield shear strength S,, or ultimate shear strength Si

4 A factor of safety is applied to tna, to determine the allowable stress tay = Sys /7 or

Ta = Sus /n The required parameter J/c of the shaft based on the strength specifica- tion of is

A solid shaft is to transmit 500 kW at n = 1200 rpm without exceeding the yield strength in shear of

Sys or 8 twisting throtigh more than 4° in a length of 2 m Calculate the required diameter of the shaft

Design Decisions: The shaft is made of steel having S,, = 300 MPa and G = 80 GPa A safety factor of 1.5 is used

Solution: : The torque, applying Eq (1.15),

The foregoing gives c = 23.3 mm

‘Distortion Specification The size of the shaft is now obtained from Eq (4.9):

ại T-

CHAPTER? @ SHAFTS AND ASSOCIATED Parts

Substituting the: given numerical values,

9.4 COMBINED STATIC LOADINGS ON SHAFTS

The shaft design process is far simpler when only static loads are present than when the loading fluctuates However, even with the fatigue loading, a preliminary estimate of shaft diameter may be needed many times, as is observed in the next section Hence, the results

of the rational design procedure of Section 3.2, presented here, is useful in getting the first

estimate of shaft diameter for any type of combined static loading conditions

BENDING, TORSION, AND AXIAL Loaps

Consider a solid circular shaft of diameter D, acted on by bending moment M, torque T, and

axial load P To begin with, we determine the maximum normal and shear stresses occur- ring in the outer fibers at a critical section:

= 32M + 4P T”xÐD3 ° xD?

ty = p3

in which the axial component of o, may be either additive or subtractive The foregoing equations are used with a selected design criterion Note that, for a hollow shaft, the pre- ceding expressions become

32M

Ty * Dal ĐI x29 ng tr ee 8-9 :

16T

Try = x D3{1 — (d/D)*] (9.5)

The quantities D and d represent the outer and inner diameters of the shaft, respectively

Substituting Eqs (9.3) into Eq (7.11), a shaft design formula based on the maximum shear theory of failure is

S 4

TẺ “zpl8M + PD)?+ sry? (9.6)

BENDING AND TORSION

Under many conditions, the axial force P in the preceding expressions is either 0 or so small that it can be neglected Substituting P = 0 into Eqs (9.6) and (9.7), we have the fol- lowing shaft design equations based on the maximum shear stress theory of failure:

9.5 DESIGN OF SHAFTS FOR FLUCTUATING

AND SHOCK LOADS

Shafts are used in a wide variety of machine applications The design process for circular torsion members is described in Section 9.3; we are now concerned with the members car- rying fluctuating and shock loads of combined bending and torsion, which is the case for most transmission shafts [1-10] Referring to Section 8.8, the definitions of the mean and alternating moments and torques, are

: aK + Man) and Ma peas 5 (Monax = Moin)

For a solid round shaft of diameter D subjected to bending moment M and torsion T,

we have on an outermost element

ea Md te Ep

We can replace Oy, Txay Trym» and Tyyq by these formulas (using the appropriate subscripts ono, t, M, and T) to express the equations developed in Section 8.12 in terms of the bend-

CHAPTER 9 9 SHAFTS AND ASSOCIATED Parts

The maximum shear stress theory combined with Goodman fatigue criterion, applying

Eq (8.28), is thus obtained as

S32 Ñ ° „11⁄2

—“=—_l| n al mF Š 1) + G + 3 1) | M„+— , uw (9.11)

Ina similar manner, the maximum energy of distortion theory incorporated with Goodman fatigue criterion, from Eq (8.32):

BB = aD Cc Sus) tưng s} Ì ` TY 3G 8 VIC (9.12)

The quantities Š„ and S, represent the ultimate strength and endurance limit, respectively

Note that an alternate form of the maximum energy of distortion theory associated with Goodman fatigue relation, through the use of Eq (8.35), may be expressed in the form

equipment, where operation is jerky, this condition requires special consideration To ac- count of shock condition, multiplying coefficients (that is, correction factors Ky» in bend- ing and K, in torsion) may be used in the foregoing equations Thus, the maximum shear stress theory associated with Goodman fatigue relation, Eq (9.11) becomes

Operation of shafts under steady loads involves a completely reversed alternating bend-

ing stress (o,) and an approximate torsional mean stress (t,,) This is the case of a rotating

shaft with constant moment M = max = ~Myin and torque T = Trax = Thin Therefore,

349

350 PART IL) @ APPLICATIONS

Table 9.1 Shock factors in bending and torsion

Gradually applied or steady 1.0

The preceding expressions could also be written on the basis of the Soderberg

criterion, replacing the quantity S, by the yield strength Sy, as needed In so doing, for instance, Eq (9.16a) becomes

The foregoing discussion shows that the design of shafts subjected to fluctuating and

_ shock loads cannot be carried out in a routine manner, as in the case of static loads Usually, the diameter of a shaft must be assumed and a complete analysis performed at a critical

section where the maximum stress occurs A design of this type may require several revi- sions The FEA is in widespread use for such cases for final design, Alternatively, experi- mental methods are used, since formulas of solid mechanics may not be sufficiently accurate

DISPLACEMENTS

Shaft deflections frequently can be a critical factor, since excessive displacements cause

rapid wear of shaft bearings, misalignments of gears driven from the shaft, and shaft vibrations (see Section 9.6) Deflection calculations require that the entire shaft geometry

be defined Hence, a shaft typically is first designed for strength, then the displacements are calculated once the geometry is completely prescribed Both transverse and twisting dis- placements must be analyzed Approaches used in obtaining the deflections of a shaft

CHAPTER? © SHAFTS AND ASSOCIATED ParTs 351

Shaft Design for Repeated Torsion and Bending

Power is transmitted from a motor through a gear at £ to pulleys at D and C of a revolving solid shaft

AB with ground surface Figure 9.3a shows the corresponding load diagram of the shaft The shaft is mounted on bearings at the ends A and B Determine the required diameter of the shaft by employing

the maximum energy of distortion theory of failure incorporating the Soderberg fatigue relation

Given: The shaft is made of steel with an ultimate strength of 810 MPa and a yield strength of

605 MPa Torque fluctuates 10% each way from the mean value, The fatigue stress-concentration fac- tor for bending and torsion is equal to 1.4 The operating temperature is 500°C maximum

Design Assumptions: Bearings act as simple supports, A factor of safety of 2 = 2 fs used, The

survival rate is taken to be 50%

Solution: The reactions at A and B, as obtained from the equations of statics, are noted in Fig-

ure 9,3a, The determination of the resultant bending moment of (M? + M?)'/? is facilitated by using

the moment diagrams (Figures 9.3b and 9.3c) At point C, we have

Me ={(.1)? + (1.5971! == 1.503 kN-m

Similarly, at D and E,

EXAMPLE 9.2

‘The maximum bending monient is at D Note from Figure 9.3d that the torque is also maximum at D,

‘Ty EKN-m: The exact: location along the shaft where the maximum stress occurs, the critical section, is therefore:at D Hence, at point D,

My = 0 Mg = 2:121kN-m

Te = KN: mM, Ty = 0.101) = 0.1 KN-m Using Eq: (8:1), the enthirance limit of the material is

Because the loading is smooth, Ky, = Ky = 1 ftom Table 9.1

Substituting the S, = 605 MPa for S, and the numerical values obtained into Eq, (9.12), we have

Comments: Since this is larger than 51, mm, our assun ption is correct A diameter of 70 mm is

therefore quite satisfactory

EXAMPLE 9.3

Detetmining Factor of Safety for a Stepped Shaft under Torsional Shock Loading

A‘Stepped shaft of diameters D and d with a shoulder fillet radius r has been machined from AISI

1095 ahnealed steel and fixed at end A (Figure 9.4) Determine the factor of safety nở, using the max-

` imum shear stress theory incorporated with the Goodman fatigue relation

Given: Free’end C of the shaft is made to rotate back and forth between 1.0° and 1.5° under tor- sional minor shock loading The shaft is at room temperature

Ky = 15° (by Table 9.1), G = 79 GPa (from Table B.1)

Sq, = 658 and Hy = 192 (by Table B.4)

Design Assumption: - A reliability of 95% is used, Solution:: From the geometry of the shaft, D = 2d and Lap = Lac = L The polar moment of inertia of the shaft segments are

path (at ~ G@ \W6See * Jac

Cj = 1) (for normal temperature)

Si = 0.295, = 190.8 MPa (applying Eq (8.4))

and

K, = 1.6 © (from Figure C.8, for D/d = 2 and r/d = 0.067)

3 =0.92 (from Figure 8.9, for r == 2mm and Hp = 192 annealed steel)

Kp =l +0.920.6~—1)= 155 (using Eq (8.13b)) Therefore

The preferred limits and fits for cylindrical parts are given by the American National Standards Institute (ANSI) Standard B4.1-1967 This is widely used for establishing toler- ances for various fits The American Gear Manufacturers Association (AGMA) Standard 9003-A9L, Flexible Couplings-Keyless Fits, contains formulas for the calculation of inter-

CHAPTER 9 ® SHAFTS AND ASSOCIATED PARTS

The interference fits are usually characterized by maintenance of constant pressures between two mating parts through the range of sizes The amount of interference needed to create a tight joint varies directly with the diameter of the shaft A simple rule of thumb is

to use 0.001 in of interference for diameters up to 1 in and 0.002 in for diameters between

1 through 4 in A detailed discussion of the state of stresses in shrink fits is found in Chap-

ter 16, where we consider applications to various members.*

355

9.7 CRITICAL SPEED OF SHAFTS

A rotating shaft becomes dynamically unstable at certain speeds, and large amplitudes of

lateral vibration develop stresses to such a value that rupture may occur The speed at

which this phenomenon occurs is called a critical speed Texts on vibration theory show

that the frequency for free vibration when the shaft is nor rotating is the same as its critical speed That is, the critical speed of rotation numerically corresponds to the lateral natural frequency of vibration, which is induced when rotation is stopped and the shaft center is displaced laterally, then suddenly released Hence, a natural frequency is also called a crit- ical frequency or critical speed (11]

Equating the kinetic energy due to the rotation of the mounted shaft masses to the po- tential energy of the deflected shaft results in an expression, called the Rayleigh equation

This expression defines the critical speed of the shaft A shaft has as many critical speeds

as there are rotating masses Unless otherwise specified, the term critical speed is used to

refer to the lowest or fundamental critical speed The critical speed m., (in cycles per sec- ond, cps) for a shaft on two supports and carrying multiple masses is defined as follows:

nok xin +W?ôa+ + a SE [ews 9.48)

SS Dae Wilt+.W2ð2 +: Why 52 2z.\.3) W8? ‘

The quantity W,, represents the concentrated weight (including load) of a rotating mass and ổ„ is the respective static deflection of the weight, as shown in Figure 9.5 The accel- eration of gravity is represented by g In SI units, g = 9.81 m/s’

Note that the Rayleigh equation only estimates the critical speed It ignores the effects

of the weight of the shaft, self-damping of the material, the flexibility of the bearings or supports, and assumes that all weights are concentrated Tests have shown that the forego- ing factors tend to lower the calculated critical speed, Thus, the approximate values of ry calculated from Eq (9.18) are always higher than the true fundamental frequency A good rule of thumb in practice is to keep the actual operating speed about 25% lower than the calculated critical speed More accurate approaches for determining the critical frequency, such as a modified Rayleigh’s method (Rayleigh-Ritz) and Holzer’s method, exist but are somewhat more complicated to implement [12, 13]

Only rotations sufficiently below or above the critical speed result in dynamic stabil- ity of the shafts In unusual situations, in very high-speed turbines, sometimes satisfactory operation is provided by quickly going through the critical speed, then running well above the critical speed This practice is to be avoided if possible, as vibration may develop from

1 *Some readers may prefer to study Sections 16.3 and 16.4 as a potential assembly method

Figure 9.5 Simply supported shaft with concen-

trated loads (deflection greatly exaggerated)

other causes in the operation above critical speeds, even though the operation is stable

Interestingly, the critical speed of a shaft on three supports is also equal to the natural fre- quency of the shaft in lateral vibration [12]

Shaft critical speeds may readily be estimated by calculating static deflections at one

or several points The maximum allowable deflection of a shaft is usually determined by the critical speed and gear or bearing requirements Critical speed requirements vary greatly with the specific application

EXAMPLE 9.4 Determining Critical Speed’ of a Hollow Shaft

A shaft: with inner arid outer diameters of d and D, respectively, is mounted between bearings and

supporting two wheels as shown in Figure 9.6 Calculate the critical speed in rpm

Assumptions: : The’ shaft is made of steel having & = 210 GPa The weight of the shaft is ignored

Bearings act as simple supports

“ Solution: | The nioment of inertia of the cross section is J = %(254 ~ 15%) = 267 x 10° mm* The

concéntrated forces ate We = 20 x 9.81 = 196.2 N and Wp = 30 x 9.81 = 294.3 N Static deflec-

tions at C and D‘can be obtained by the equations for Case 6 of Table A.9

Deflection at C Diie to the load at C;

The total deflection is then ðc,= 0.194 + 0.215 = 0.409 mm Deflection at D Sitnilarly, we obtain foes 196.2(0.4) (0.5) (1.5% — 0.42 ~ 0.52) 7:

1 [9.81 (196.2 x 0.409 x 10-F + 294.3 x 0.369 x 1073) 1"

Leg So

Bee Oa | 196.2(0.409 & 10-3)? -F 294.3(0.369 x 10-5)

= 25.37 cps = 1522 rpm

Determination of Critical Speed of a Stepped Shaft EXAMPLE 9.5

Figure 9.7a shows a stepped round shaft supported by two bearings and carrying the flywheel weight

W Calculate the critical speed in rpm

Given: The moment of inertia / of the shaft in its central region is twice that the moment of inertia

in the end parts:

Assumptions: The shaft is made of steel with E = 200 GPa The shaft weight is ignored Bear- ings áct 4s simple supports

Solution: The application of the moment-aréa method to obtain the static deflection at the midspan

Cis illustrated in Figure 9.7 The bending moment diagram is given in Figure 9.7b and the M/ET di- agram in Figure 9.7e Note that, in the latter figure, C, and Cy denote the centroids of the triangular and trapezoidal areas, réspectively :

The first moment of the various parts of the ă/ET điagram are used tọ ñnd the defection, From the symmetry of the beam, the tangent to the deflection curve at C is horizontal Hence, according to the sécond-monient area theorem defined by Eq (4.23), the deflection d¢is obtained by taking the

moment of the M/EI area diagram between A and C about point A That is,

Sc = (first moment of triangle) + (first moment of trapezoid)

CHAPTER9 ® SHAFTS AND ASSOCIATED Parts

Tables of dimensions for the mounting parts may be found in engineering handbooks and

manufacturer’s catalogs

Keys

A key enables the transmission of torque from the shaft to the hub, Numerous kinds of keys are used to meet various design requirements They are standardized as to size and shape in several styles Figure 9.8 illustrates a variety of keys The grooves in the shaft and hub into which the key fits form the keyways or key seats, The square, flat type of keys are most

The gib-head key is tapered so that, when firmly driven, it prevents relative axial

motion Another advantage is that the hub position can be adjusted for the best location

A tapered key may have no head or a gib head (as in Figure 9.8d) to facilitate removal

(e)

Figure 9.8 | Common types of shaft keys: (a) square key (w » D/4);

(b) flat key (w * D/4, h ~ 3w/4); (c) round key (often tapered); (d) gib-head key; (e) Woodruff key,

359

the, automotive and machine tool industries Woodruff keys yield better concentricity after

lính A pin is employed for axial positioning and the transfer of relatively light torque or axial medium loads by nuts, pins, and clamp joints; heavy loads by press or shrink fits Interfer-

RINGS AND COLLARS Tek (9.20}

Retaining rings, commonly referred to as snap rings, are available in numerous varieties where r is the shaft radius

and require that a small groove of specific dimensions be machined in the shaft Keys, pins, Shear and compressive or bearing stresses are calculated for the keys from force F, and snap rings can be avoided by the use of clamp collars that squeeze the outside diame- using a sufficiently large factor of safety For steady loads, a factor of safety of 2 is com-

ter of the shaft with high pressure to clamp something to it The hub bore and clamp collar monly applied On the other hand, for minor to high shock loads, a factor of safety of 2.5

have a matching slight taper The clamp collar with axial slits is forced into the space be- to 4.5 should be used

tween hub and shaft by tightening the bolts For keyways, the concentration of stress depends on the values of the fillet radius at the

METHODS OF AXIALLY POSITIONING OF Huss bending or torsion loading, the theoretical stress concentration factors range from 2 tO paiegay Forces

Figure 9.10 shows common methods of axially positioning and retaining hubs into about 4, contingent on the ratio r of r/D [15] The quantity r represents the fillet radius (see ¿n2 key tightly fitted

shafts [14] Axial loads acting on shafts or members mounted on the shaft are transmitted Figure 9.8d) and D is the shaft diameter The approximate values of the fatigue stress con- at top and bottom

as follows: light loads by clamp joints, setscrews, snap rings, and tapered keys (Figure 9.8); centration factor range between 1.3 and 2 [4]

A shaft of diameter D rotates at 600 ‘y¥pm and transmits 100 hp through a gear A square key of width /was.to be uséd (Figure 9.8a) asa mounting part Determine the required length of the key

Design Decisions: The shaft and key will be made of AISI 1035 cold-drawn steel having the

same: tensile: and compressive yield strength and that yield strength in shear is Sy, = S,/2 The

transmitted power produces intermittent minor shocks and a factor of safety of n = 2.5 is used

Solution: - From Table B.3, for AIST 1035 CD steel, we find S, = 460 MPa Through the use of

Inasmuch as Sy = Sy, this also results in L = 43 mm

9.10 SPLINES

When axial movement between the shaft and hub is required, relative rotation is prevented

by means of splines machined on the shaft and into the hub For example, splines are used

to connect the transmission output shaft to the drive shaft in automobiles, where the sus-

pension movement causes axial motion between the components Splines are essentially

“built-in keys.” They can transform more torque than can be handled by keys There are two forms of splines (Figure 9.12): straight or square tooth splines and involute tooth splines The former is relatively siraple and employed in some machine tools, automatic equipment, and so on The latter has an involute curve in its outline, which is in widespread use on gears The involute tooth has less stress concentration than the square tooth and,

Figure 9.12 Some common types of splines:

{a) straight-sided; (b) involute

hence, is stronger Also easier to cut and fit, the involute splines are becoming the promi- nent spline form

Formulas for the dimensions of splines are based on the nominal shaft diameter

(6, 16] Figure 9.12a shows the standard SAE 6 and 10 straight spline fittings Note that the values of root diameter d, width w, and depth # of the internal spline are based on the nom- inal shaft diameter D or about the roct diameter of the external spline According to the

SAE, the torque capacity (in lb -in.) of straight-sided splines with sliding is

where

T = theoretical torque capacity

n = number of splines

lm = (D+ d)/4, mean or pitch radius (see Figure 9.12)

Ah = depth of the spline

L, = length of the spline contact

p = spline pressure The SAE states that, in actual practice, owing to the inaccuracies in spacing and tooth form, the contact length L, is about 25% of the spline length

Involute splines (Figure 9.12b) have a general form of internal and external involute gear teeth, discussed in detail in Chapter 11, with modified dimensions The length L, of spline contact required to transmit a torque, as suggested by the SAE, is

Here, T represents the torque on the shaft and L, is given by Eq (9.23) or (9.24) If bend-

ing is present, the flexure stress in the spline must also be calculated

9.11 COUPLINGS

Couplings are used to connect two shafts A wide variety of commercial shaft couplings are

available They may be grouped into two broad classes: rigid and flexible A rigid coupling

locks the two shafts together, allowing no relative motion between them, although some axial adjustment is possible at assembly No provision is made for misalignment between the two shafts connected, nor does it reduce shock or vibration across it from one shaft to the other However, shafts are often subject to some radial, angular, and axial misalign- ment In these situations, flexible couplings must be used Severe misalignment must be corrected; slight misalignment can be absorbed by flexible couplings This prevents fatigue failure or destruction of bearings

CLAMPED RIGID COUPLINGS

Collinear shafts can be connected by clamp couplings that are made in several designs The most-common two-piece split coupling clamps around both shafts by means of bolts and

transmits torque It is necessary to key the shafts to the coupling The torque is transmitted

mainly by friction due to the clamping action and partially by the key Clamp couplings are widely used in heavy-duty service

FLANGED RIGID COUPLINGS

Collinear shafts can also be connected by flanged couplings, similar to those shown in Fig- ure 9.13 The flanged portion at the outside diameter serves a safety function by shielding the bolt heads and nuts The load is taken to be divided equally among the bolts Rigid cou- plings are simple in design They are generally restricted to relatively low-speed applica- tions where good shaft alignment or shaft flexibility can be expected

Keyed couplings are the most widely used rigid couplings (Figure 9.13a) They can transmit substantial torques The coupling halves are attached to the shaft ends by keys As can be seen in the figure, flange alignment is obtained by fitting a shallow machined pro- jection on one flange face to a female recess cut in the face of the other flange Another common way to obtain flange alignment is to permit one shaft to act as a pilot and enter the mating flange Keyed couplings employ standard keys as discussed in Section 9.8

Compression couplings have a split double cone that does not move axially but is

squeezed against the shaft by the wedging of the flanges, as shown in Figure 9.13b This kind of coupling transmits torque only by the frictional force between the shaft and the split

double cone, eliminating the need for a key and keyway in the coupling

In specifying a rigid coupling using ground and fitted flange bolts, the designer should

Figure 9.13 Flanged rigid couplings: (a) keyed type; (b) compression type

of the projected area of the bolt in contact with the side of the hole, shear of the flange at the hub, and shear or crushing of the key Note that, in contrast to fitted bolts, a flange cou- pling designed on the basis of friction-torque capacity requires a somewhat different analy- sis than that just described [17]

For flanged rigid couplings, it is usually assumed that shear stress in any one bolt is uniform and governed by the distance from its center to the center of the coupling Friction between the flanges is disregarded Then, if the shear stress in a bolt is multiplied by its cross-sectional area, the force in the bolt is ascertained The moment of the forces devel- oped by the bolts around the axis of a shaft estimates the torque capacity of a coupling

365

Determining the ‘Torque Capacity of a Rigid Coupling

A flanged keyed coupling is keyed to a: shaft (Figure: 9.13a) Calculate the torque that can be transmitted

Given: There are 6 bolts of 25-mm diameter The bolt circle diameter is Dy = 150 mm

Assumptions: | ‘The torque capacity is controlled by an allowable shear strength of 210 MPa in the bolts

Solution: Area in shear for one bolt is

The insert cushions the effect of shock and impact loads that could be transferred between

shafts A shear type of rubber-inserted coupling can be used for higher speeds and horse-

powers A chain coupling type consists of two identical sprockets coupled by a roller chain

Figure 9.14 shows the two identical hubs of a square-jawed coupling with an elas- tomer (i.e., rubber) insert In operation, the halves slide along the shafts on which they are mounted until they engage with the elastomer The clearances permit some axial, angular, and parallel misalignment Clearly, the jaws are subjected to bearing and shear stresses

The force acting on the jaw producing these stresses depends on the horsepower and speed that the coupling is to transmit

Many other types of flexible couplings are available Details, dimensions, and Joad rat- ings may be found in the catalogs of various manufacturers or mechanical engineering handbooks

9.12 UNIVERSAL JOINTS

‘A tiniversal joint is a kinematics linkage used to connect two shafts that have permanent in- tersecting axes Universal joints permit substantial misalignment of shafts They come in two common types: the Hooke or Cardan coupling, which does not have constant velocity across a single joint, and the Rzeppa coupling, which does Both types can deal with very large angular misalignment Shaft angles up to 30° may be used [18]

Hooke’s coupling is the simplest kind of universal joint It consists of a yoke on each shaft connected by a central cross-link Figure 9.15 depicts a double-Hooke joint, where plain bearings are used at the yoke-to-cross connections These joints are employed mostly with equal yoke alignment angles (@) in the ‘two joints, as shown in the figure The use of equal angles provides uniform angular velocity in the driven shaft A pair of Hooke’s

couplings is often used in a rear-drive automobile drive shaft, Note that a familiar applica-

tion of the Rzeppa coupling is in front-wheel-drive automobiles, where the drive shaft is short and shaft angles can be large For further information, see texts on mechanics of

Figure 9.15 Two arrangements of a pair of Hooke’s couplings for achieving constant velocity ratio

Case Study 9-1 | SHAFTING DESIGN OF A WINCH CRANE GEAR Box

Figure 9.16 shows the input shaft of the crane gear box, supported in the gear box by bearings A and B and driven

by electric motor (see Case Study 1-1) Determine (a) The factor of safety for the shaft using the

maximum energy of distortion theory incorporated with the Goodman criterion

(b) The rotational displacements or slopes at the

bearings

(c) The stresses in the shaft key

Given: The geometry and dimensions of the hollow shaft and square shaft key are known

Data: Refer to Figure 9.16a and Case Study 11-1

4 The factor of safety ism = 3 against shear of shaft key

A survival rate of 99.9% is used

Solution: See Figure 9.16 and Table A.9

(a) The reactions at A and B, as determined by the con- ditions of equilibrium, are indicated in Figure 9.16b

The moment and torque diagrams are obtained in the

usual manner and drawn in Figure 9.16c Observe that the critical section is at point C We have

Mc = (2.77) + (7.627 }'? = 8.11 N-m

Te = 2.06 N.m

(continued)

The mean and alternating moments and torques, by where

Eq (9.10), are then

My, =O My = 8.11 Nem C, 0.75 (by Table 8.3)

Ty = 2.06 N-m Tạ =0 ŒC, 0.85 (using Eq, (8.9)) The modified endurance limit, through the use €Œ =1

of Eqs (8.1) and (8.6) and referring to Section 8.7, is Ky =2

Se = CCC, (z) s S! = 0.5(520) = 260 MPa (from Eq (8.1)

Se = (0.86) (0.75) (0.85)(1)( = |(260) = 71.27 (0.86)(0.75)(0.85)(} (5) ) MPa Comments: Inasmuch as the bearing and gear :

stiffnesses are ignored, the negligibly small values of

O, and 8, estimated by the preceding equations rep- resent higher angles than the true slopes Therefore, self-aligning bearings are not necessary

Since the loading is steady, the shock factors Ky, =

Ky, = 1 by Table 9.1

Substituting the numerical values into Eq (9.12)

and replacing D? with D°[1 ~ (d/D)*), we obtain

520(10% 32 (c) The compressive forces acting on the sides of the

= shaft key equal F, = 206 N (Figure 9.16) The shear

given by Case 6 of Table A.9 Note that The allowable shear stress in the shaft key is

Lb? =(L+b)(L—b) =(L+b)a and simi-

larily L2 ~ a? = (L+a)b Introducing the given

data, the results are

f,abtL + b) Since ty >> t, shear should not occur at shaft key

Ôa =~— 6EIL We obtain the same result on the basis of compres-

sion or bearing on key (see Section 9.9)

206(66)(84)(150 + 84) Boe

maining three shafts and the associated keys in the

Design of Transmission Shafting, ANS/ASME B106.1M-1985,

Loewenthal, S H “Proposed Design Procedure for Transmission Shafting Under Fatigue Loading.” Technical Note TM-78927, NASA, 1978