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Lecture notes with problems in Mechanical Engineering focused on the Shafts and Shaft Components | Shigley’s Mechanical Engineering Design Ch.7. - 9th ed.
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Chapter Outline 7–1 Introduction 360 7–2 Shaft Materials 360 7–3 Shaft Layout 361
7–4 Shaft Design for Stress 366
7–5 Deflection Considerations 379 7–6 Critical Speeds for Shafts 383 7–7 Miscellaneous Shaft Components 388 7–8 Limits and Fits 395
359
360 Mechanical Engineering Design
A shaft is a rotating member, usually of circular cross section, used to transmit power or motion. It provides the axis of rotation, or oscillation, of elements such as gears, pulleys, flywheels, cranks, sprockets, and the like and controls the geometry of their motion. An axle is a nonrotating member that carries no torque and is used to support rotating wheels, pulleys, and the like. The automotive axle is not a true axle; the term is a carryover from the horse-and-buggy era, when the wheels rotated on nonrotating members. A nonrotating axle can readily be designed and analyzed as a static beam, and will not warrant the special attention given in this chapter to the rotating shafts which are subject to fatigue loading. There is really nothing unique about a shaft that requires any special treatment beyond the basic methods already developed in previous chapters. However, because of the ubiquity of the shaft in so many machine design applications, there is some advantage in giving the shaft and its design a closer inspection. A complete shaft design has much interdependence on the design of the components. The design of the machine itself will dictate that certain gears, pulleys, bearings, and other elements will have at least been partially analyzed and their size and spacing tentatively determined. Chapter 18 provides a complete case study of a power transmission, focusing on the overall design process. In this chapter, details of the shaft itself will be examined, including the following:
In deciding on an approach to shaft sizing, it is necessary to realize that a stress analy- sis at a specific point on a shaft can be made using only the shaft geometry in the vicinity of that point. Thus the geometry of the entire shaft is not needed. In design it is usually possible to locate the critical areas, size these to meet the strength requirements, and then size the rest of the shaft to meet the requirements of the shaft-supported elements. The deflection and slope analyses cannot be made until the geometry of the entire shaft has been defined. Thus deflection is a function of the geometry everywhere , whereas the stress at a section of interest is a function of local geometry. For this rea- son, shaft design allows a consideration of stress first. Then, after tentative values for the shaft dimensions have been established, the determination of the deflections and slopes can be made.
Deflection is not affected by strength, but rather by stiffness as represented by the mod- ulus of elasticity, which is essentially constant for all steels. For that reason, rigidity cannot be controlled by material decisions, but only by geometric decisions.
362 Mechanical Engineering Design
The geometric configuration of a shaft to be designed is often simply a revision of existing models in which a limited number of changes must be made. If there is no existing design to use as a starter, then the determination of the shaft layout may have many solutions. This problem is illustrated by the two examples of Fig. 7–2. In Fig. 7–2 a a geared countershaft is to be supported by two bearings. In Fig. 7–2 c a fanshaft is to be configured. The solutions shown in Fig. 7–2 b and 7–2 d are not neces- sarily the best ones, but they do illustrate how the shaft-mounted devices are fixed and located in the axial direction, and how provision is made for torque transfer from one element to another. There are no absolute rules for specifying the general layout, but the following guidelines may be helpful.
Figure 7– A vertical worm-gear speed reducer. (Courtesy of the Cleveland Gear Company.)
Figure 7– ( a ) Choose a shaft configuration to support and locate the two gears and two bearings. ( b ) Solution uses an integral pinion, three shaft shoulders, key and keyway, and sleeve. The housing locates the bearings on their outer rings and receives the thrust loads. ( c ) Choose fan-shaft configuration. ( d ) Solution uses sleeve bearings, a straight- through shaft, locating collars, and setscrews for collars, fan pulley, and fan itself. The fan housing supports the sleeve bearings.
( a ) ( b )
( c )
Fan
( d )
Shafts and Shaft Components 363
Axial Layout of Components
The axial positioning of components is often dictated by the layout of the housing and other meshing components. In general, it is best to support load-carrying components between bearings, such as in Fig. 7–2 a, rather than cantilevered outboard of the bear- ings, such as in Fig. 7–2 c. Pulleys and sprockets often need to be mounted outboard for ease of installation of the belt or chain. The length of the cantilever should be kept short to minimize the deflection. Only two bearings should be used in most cases. For extremely long shafts carrying several load-bearing components, it may be necessary to provide more than two bearing supports. In this case, particular care must be given to the alignment of the bearings. Shafts should be kept short to minimize bending moments and deflections. Some axial space between components is desirable to allow for lubricant flow and to provide access space for disassembly of components with a puller. Load bearing components should be placed near the bearings, again to minimize the bending moment at the loca- tions that will likely have stress concentrations, and to minimize the deflection at the load-carrying components. The components must be accurately located on the shaft to line up with other mating components, and provision must be made to securely hold the components in position. The primary means of locating the components is to position them against a shoulder of the shaft. A shoulder also provides a solid support to minimize deflection and vibration of the component. Sometimes when the magnitudes of the forces are reasonably low, shoulders can be constructed with retaining rings in grooves, sleeves between components, or clamp-on collars. In cases where axial loads are very small, it may be feasible to do without the shoulders entirely, and rely on press fits, pins, or col- lars with setscrews to maintain an axial location. See Fig. 7–2 b and 7–2 d for examples of some of these means of axial location.
Supporting Axial Loads
In cases where axial loads are not trivial, it is necessary to provide a means to transfer the axial loads into the shaft, then through a bearing to the ground. This will be partic- ularly necessary with helical or bevel gears, or tapered roller bearings, as each of these produces axial force components. Often, the same means of providing axial location, e.g., shoulders, retaining rings, and pins, will be used to also transmit the axial load into the shaft. It is generally best to have only one bearing carry the axial load, to allow greater tolerances on shaft length dimensions, and to prevent binding if the shaft expands due to temperature changes. This is particularly important for long shafts. Figures 7– and 7–4 show examples of shafts with only one bearing carrying the axial load against a shoulder, while the other bearing is simply press-fit onto the shaft with no shoulder.
Providing for Torque Transmission
Most shafts serve to transmit torque from an input gear or pulley, through the shaft, to an output gear or pulley. Of course, the shaft itself must be sized to support the torsional stress and torsional deflection. It is also necessary to provide a means of transmitting the torque between the shaft and the gears. Common torque-transfer elements are:
Shafts and Shaft Components 365
Splines are essentially stubby gear teeth formed on the outside of the shaft and on the inside of the hub of the load-transmitting component. Splines are generally much more expensive to manufacture than keys, and are usually not necessary for simple torque transmission. They are typically used to transfer high torques. One feature of a spline is that it can be made with a reasonably loose slip fit to allow for large axial motion between the shaft and component while still transmitting torque. This is useful for connecting two shafts where relative motion between them is common, such as in connecting a power takeoff (PTO) shaft of a tractor to an implement. SAE and ANSI publish standards for splines. Stress-concentration factors are greatest where the spline ends and blends into the shaft, but are generally quite moderate. For cases of low torque transmission, various means of transmitting torque are available. These include pins, setscrews in hubs, tapered fits, and press fits. Press and shrink fits for securing hubs to shafts are used both for torque transfer and for preserving axial location. The resulting stress-concentration factor is usually quite small. See Sec. 7–8 for guidelines regarding appropriate sizing and tolerancing to transmit torque with press and shrink fits. A similar method is to use a split hub with screws to clamp the hub to the shaft. This method allows for disassembly and lateral adjustments. Another similar method uses a two-part hub consisting of a split inner member that fits into a tapered hole. The assembly is then tightened to the shaft with screws, which forces the inner part into the wheel and clamps the whole assembly against the shaft. Tapered fits between the shaft and the shaft-mounted device, such as a wheel, are often used on the overhanging end of a shaft. Screw threads at the shaft end then permit the use of a nut to lock the wheel tightly to the shaft. This approach is useful because it can be disassembled, but it does not provide good axial location of the wheel on the shaft. At the early stages of the shaft layout, the important thing is to select an appro- priate means of transmitting torque, and to determine how it affects the overall shaft layout. It is necessary to know where the shaft discontinuities, such as keyways, holes, and splines, will be in order to determine critical locations for analysis.
Assembly and Disassembly Consideration should be given to the method of assembling the components onto the shaft, and the shaft assembly into the frame. This generally requires the largest diame- ter in the center of the shaft, with progressively smaller diameters towards the ends to allow components to be slid on from the ends. If a shoulder is needed on both sides of a component, one of them must be created by such means as a retaining ring or by a sleeve between two components. The gearbox itself will need means to physically posi- tion the shaft into its bearings, and the bearings into the frame. This is typically accom- plished by providing access through the housing to the bearing at one end of the shaft. See Figs. 7–5 through 7–8 for examples.
Figure 7–
Arrangement showing bearing inner rings press-fitted to shaft while outer rings float in the housing. The axial clearance should be sufficient only to allow for machinery vibrations. Note the labyrinth seal on the right.
366 Mechanical Engineering Design
Figure 7– Similar to the arrangement of Fig. 7–5 except that the outer bearing rings are preloaded.
Figure 7– In this arrangement the inner ring of the left-hand bearing is locked to the shaft between a nut and a shaft shoulder. The locknut and washer are AFBMA standard. The snap ring in the outer race is used to positively locate the shaft assembly in the axial direction. Note the floating right-hand bearing and the grinding runout grooves in the shaft.
When components are to be press-fit to the shaft, the shaft should be designed so that it is not necessary to press the component down a long length of shaft. This may require an extra change in diameter, but it will reduce manufacturing and assembly cost by only requiring the close tolerance for a short length. Consideration should also be given to the necessity of disassembling the compo- nents from the shaft. This requires consideration of issues such as accessibility of retaining rings, space for pullers to access bearings, openings in the housing to allow pressing the shaft or bearings out, etc.
Critical Locations It is not necessary to evaluate the stresses in a shaft at every point; a few potentially critical locations will suffice. Critical locations will usually be on the outer surface, at axial locations where the bending moment is large, where the torque is present, and where stress concentrations exist. By direct comparison of various points along the shaft, a few critical locations can be identified upon which to base the design. An assessment of typical stress situations will help.
Figure 7– This arrangement is similar to Fig. 7–7 in that the left-hand bearing positions the entire shaft assembly. In this case the inner ring is secured to the shaft using a snap ring. Note the use of a shield to prevent dirt generated from within the machine from entering the bearing.
368 Mechanical Engineering Design
Combining these stresses in accordance with the distortion energy failure theory, the von Mises stresses for rotating round, solid shafts, neglecting axial loads, are given by
σ (^) a ′ = (σ (^) a^2 + 3 τ (^) a^2 )^1 /^2 =
32 K (^) f M (^) a π d^3
16 K (^) f s Ta π d^3
(7–5)
σ (^) m ′ = (σ (^) m^2 + 3 τ (^) m^2 )^1 /^2 =
32 K (^) f M (^) m π d^3
16 K (^) f s Tm π d^3
(7–6)
Note that the stress-concentration factors are sometimes considered optional for the midrange components with ductile materials, because of the capacity of the ductile material to yield locally at the discontinuity. These equivalent alternating and midrange stresses can be evaluated using an appropriate failure curve on the modified Goodman diagram (See Sec. 6–12, p. 303, and Fig. 6–27). For example, the fatigue failure criteria for the modified Goodman line as expressed previously in Eq. (6–46) is 1 n
σ (^) a ′ Se
σ (^) m ′ Sut Substitution of σ (^) a ′ and σ (^) m ′ from Eqs. (7–5) and (7–6) results in
1 n
π d^3
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
Sut
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
For design purposes, it is also desirable to solve the equation for the diameter. This results in
d =
16 n π
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
Sut
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
Similar expressions can be obtained for any of the common failure criteria by sub- stituting the von Mises stresses from Eqs. (7–5) and (7–6) into any of the failure criteria expressed by Eqs. (6–45) through (6–48), p. 306. The resulting equations for several of the commonly used failure curves are summarized below. The names given to each set of equations identifies the significant failure theory, followed by a fatigue failure locus name. For example, DE-Gerber indicates the stresses are com- bined using the distortion energy (DE) theory, and the Gerber criteria is used for the fatigue failure. DE-Goodman
1 n
π d^3
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
Sut
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
(7–7)
d =
16 n π
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
Sut
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
Shafts and Shaft Components 369
DE-Gerber
1 n
π d^3 Se
2 B Se ASut
(7–9)
d =
⎝ 8 n A π Se
2 B Se ASut
1 / 3 (7–10)
where
A =
4( K (^) f M (^) a ) 2 + 3( K (^) f s Ta ) 2
4( K (^) f M (^) m ) 2 + 3( K (^) f s Tm ) 2
DE-ASME Elliptic
1 n
π d^3
K (^) f M (^) a Se
K (^) f s Ta Se
K (^) f M (^) m S (^) y
K (^) f s Tm S (^) y
(7–11)
d =
16 n π
K (^) f M (^) a Se
K (^) f s Ta Se
K (^) f M (^) m S (^) y
K (^) f s Tm S (^) y
1 / 3
(7–12)
DE-Soderberg
1 n
π d^3
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
S (^) yt
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
(7–13)
d =
16 n π
Se
4 ( K (^) f M (^) a )^2 + 3 ( K (^) f s Ta )^2
S (^) yt
4 ( K (^) f M (^) m )^2 + 3 ( K (^) f s Tm )^2
For a rotating shaft with constant bending and torsion, the bending stress is com- pletely reversed and the torsion is steady. Equations (7–7) through (7–14) can be sim- plified by setting M (^) m and Ta equal to 0, which simply drops out some of the terms. Note that in an analysis situation in which the diameter is known and the factor of safety is desired, as an alternative to using the specialized equations above, it is always still valid to calculate the alternating and mid-range stresses using Eqs. (7–5) and (7–6), and substitute them into one of the equations for the failure criteria, Eqs. (6–45) through (6–48), and solve directly for n. In a design situation, however, having the equations pre-solved for diameter is quite helpful. It is always necessary to consider the possibility of static failure in the first load cycle. The Soderberg criteria inherently guards against yielding, as can be seen by noting that its failure curve is conservatively within the yield (Langer) line on Fig. 6–27, p. 305. The ASME Elliptic also takes yielding into account, but is not entirely conservative
Shafts and Shaft Components 371
Table 6–6: k (^) e = 0. 814 Se = 0 .787(0.870)0.814(52.5) = 29 .3 kpsi
For a rotating shaft, the constant bending moment will create a completely reversed bending stress.
M (^) a = 1260 lbf · in Tm = 1100 lbf · in M (^) m = Ta = 0
Applying Eq. (7–7) for the DE-Goodman criteria gives
1 n
π( 1. 1 )^3
Answer n = 1. 63 DE-Goodman
Similarly, applying Eqs. (7–9), (7–11), and (7–13) for the other failure criteria,
Answer n = 1. 87 DE-Gerber
Answer n = 1. 88 DE-ASME Elliptic
Answer n = 1. 56 DE-Soderberg
For comparison, consider an equivalent approach of calculating the stresses and apply- ing the fatigue failure criteria directly. From Eqs. (7–5) and (7–6),
σ (^) a ′ =
π (1.1) 3
= 15 235 psi
σ (^) m ′ =
π ( 1. 1 )^3
= 9988 psi
Taking, for example, the Goodman failure critera, application of Eq. (6–46) gives 1 n
σ (^) a ′ Se
σ (^) m ′ Sut
n = 1. 63
which is identical with the previous result. The same process could be used for the other failure criteria. ( b ) For the yielding factor of safety, determine an equivalent von Mises maximum stress using Eq. (7–15).
σ (^) max′ =
π ( 1. 1 )^3
π ( 1. 1 )^3
= 18 220 psi
Answer n (^) y =
S (^) y σ (^) max′
372 Mechanical Engineering Design
For comparison, a quick and very conservative check on yielding can be obtained by replacing σ (^) max′ with σ (^) a ′ + σ (^) m ′. This just saves the extra time of calculating σ (^) max′ if σ (^) a ′ and σ (^) m ′ have already been determined. For this example,
n (^) y =
S (^) y σ (^) a ′ + σ (^) m ′
which is quite conservative compared with n (^) y 4.50.
Estimating Stress Concentrations The stress analysis process for fatigue is highly dependent on stress concentrations. Stress concentrations for shoulders and keyways are dependent on size specifications that are not known the first time through the process. Fortunately, since these elements are usually of standard proportions, it is possible to estimate the stress-concentration factors for initial design of the shaft. These stress concentrations will be fine-tuned in successive iterations, once the details are known. Shoulders for bearing and gear support should match the catalog recommendation for the specific bearing or gear. A look through bearing catalogs shows that a typical bearing calls for the ratio of D / d^ to be between 1.2 and 1.5. For a first approximation, the worst case of 1.5 can be assumed. Similarly, the fillet radius at the shoulder needs to be sized to avoid interference with the fillet radius of the mating component. There is a significant variation in typical bearings in the ratio of fillet radius versus bore diameter, with r / d^ typically ranging from around 0.02 to 0.06. A quick look at the stress con- centration charts (Figures A–15–8 and A–15–9) shows that the stress concentrations for bending and torsion increase significantly in this range. For example, with D / d^ =^1.^5 for bending, K^ t =^2 .7 at r / d^ =^0 .02, and reduces to K^ t =^2 .1 at r / d^ =^0 .05, and further down to K^ t =^1 .7 at r / d^ =^0 .1. This indicates that this is an area where some attention to detail could make a significant difference. Fortunately, in most cases the shear and bending moment diagrams show that bending moments are quite low near the bearings, since the bending moments from the ground reaction forces are small. In cases where the shoulder at the bearing is found to be critical, the designer should plan to select a bearing with generous fillet radius, or consider providing for a larger fillet radius on the shaft by relieving it into the base of the shoulder as shown in Fig. 7–9 a. This effectively creates a dead zone in the shoulder area that does not
Sharp radius
Bearing
Shaft
Large radius undercut Stress flow
( a )
Shoulder relief groove
( b )
Large radius relief groove
( c ) Figure 7– Techniques for reducing stress concentration at a shoulder supporting a bearing with a sharp radius. ( a ) Large radius undercut into the shoulder. ( b ) Large radius relief groove into the back of the shoulder. ( c ) Large radius relief groove into the small diameter.
374 Mechanical Engineering Design
EXAMPLE 7–
This example problem is part of a larger case study. See Chap. 18 for the full context. A double reduction gearbox design has developed to the point that the general layout and axial dimensions of the countershaft carrying two spur gears has been proposed, as shown in Fig. 7–10. The gears and bearings are located and supported by shoulders, and held in place by retaining rings. The gears transmit torque through keys. Gears have been specified as shown, allowing the tangential and radial forces transmitted through the gears to the shaft to be determined as follows.
W (^) 23 t = 540 lbf W (^) 54 t = 2431 lbf W (^) 23 r = 197 lbf W (^) 54 r = 885 lbf
where the superscriptst andr represent tangential and radial directions, respectively; and, the subscripts 23 and 54 represent the forces exerted by gears 2 and 5 (not shown)on gears 3 and 4, respectively. Proceed with the next phase of the design, in which a suitable material is selected, and appropriate diameters for each section of the shaft are estimated, based on providing sufficient fatigue and static stress capacity for infinite life of the shaft, with minimum safety factors of 1.5.
Datum^ 0.250.751.251.752.0 2.753.50 7.50 8.50 9.509.7510.2510.7511.2511. C A D E F
D 1 D 2
D 3 D
D (^5) D 6 D (^7)
Gear 3 d 3 12
Bearing A Bearing B Gear 4 d 4 2.
G H I J K L M B N
Figure 7– Shaft layout for Ex. 7–2. Dimensions in inches.
Shafts and Shaft Components 375
Start with Point I, where the bending moment is high, there is a stress con- centration at the shoulder, and the torque is present. At I , M (^) a = 3651 lbf in , Tm = 3240 lbf in , M (^) m = Ta = 0
A B G I J K
RBy
RAz RBz
RAy
W 5^ t 4
W 5^ r 4
655
(^33413996)
230 2220
x-z Plane^ ^1776
V 115
M
160
1632
1472
713 907
725 x-y Plane
V^357
M
4316
3651
749
2398
M TOT
3240
T
y
z
x
W 2^ t 3
Perform free body diagram analysis to get reaction forces at the bearings.
R (^) Az = 115 .0 lbf
R (^) Ay = 356 .7 lbf
R (^) Bz = 1776 .0 lbf
R (^) By = 725 .3 lbf
From M (^) x , find the torque in the shaft between the gears, T = W (^) 23 t ( d 3 /2) = 540 (12/2) = 3240 lbf · in.
Generate shear-moment diagrams for two planes.
Combine orthogonal planes as vectors to get total moments, e.g., at J ,
4316 lbf · in.
Shafts and Shaft Components 377
Note that we could have used Eq. (7–7) directly. Check yielding.
n (^) y =
S (^) y σ (^) max′
S (^) y σ (^) a ′ + σ (^) m ′
Also check this diameter at the end of the keyway, just to the right of point I, and at the groove at point K. From moment diagram, estimate M at end of keyway to be M 3750 lbf-in. Assume the radius at the bottom of the keyway will be the standard r d 0.02, r 0.02 d 0.02 (1.625) 0.0325 in.
K (^) t = 2 .14 (Table 7–1), q 0.65 (Fig. 6–20) K (^) f = 1 + 0 .65(2. 14 − 1) = 1. 74 K (^) ts = 3 .0 (Table 7–1), q (^) s = 0 .71 (Fig. 6–21) K (^) f s = 1 + 0. 71 ( 3 − 1 ) = 2. 42
σ (^) a ′ =
32 K (^) f M (^) a π d^3
π(1.625) 3
= 15 490 psi
σ (^) m ′ =
K (^) f s Tm π d^3
π( 1. 625 )^3
= 16 120 psi
1 n (^) f
σ (^) a ′ Se
σ (^) m ′ Sut
n (^) f = 1. 17
The keyway turns out to be more critical than the shoulder. We can either increase the diameter or use a higher strength material. Unless the deflection analysis shows a need for larger diameters, let us choose to increase the strength. We started with a very low strength and can afford to increase it some to avoid larger sizes. Try 1050 CD with Sut = 100 kpsi. Recalculate factors affected by Sut , i.e., k (^) a → Se ; q → K (^) f → σ (^) a ′ k (^) a = 2 .7(100)−^0.^265 = 0 .797, Se = 0 .797(0.835)(0.5)(100) = 33 .3 kpsi q = 0 .72, K (^) f = 1 + 0 .72(2. 14 − 1) = 1. 82
σ (^) a ′ =
π(1.625) 3
= 16 200 psi
1 n (^) f
n (^) f = 1. 54
Since the Goodman criterion is conservative, we will accept this as close enough to the requested 1.5. Check at the groove at K , since K (^) t for flat-bottomed grooves are often very high. From the torque diagram, note that no torque is present at the groove. From the moment diagram, M (^) a = 2398 lbf in, M (^) m = Ta = Tm = 0. To quickly check if this location is potentially critical, just use K (^) f = K (^) t = 5. 0 as an estimate, from Table 7–1.
σ a =
32 K (^) f M (^) a π d^3
π(1.625) 3
= 28 460 psi
n (^) f =
Se σ a
378 Mechanical Engineering Design
This is low. We will look up data for a specific retaining ring to obtain K (^) f more accurately. With a quick online search of a retaining ring specification using the website www.globalspec.com, appropriate groove specifications for a retaining ring for a shaft diameter of 1.625 in are obtained as follows: width, a = 0 .068 in; depth, t = 0 .048 in; and corner radius at bottom of groove, r = 0. 01 in. From Fig. A–15–16, with r / t = 0. 01 / 0. 048 = 0. 208 , and a / t = 0. 068 / 0. 048 = 1. 42 K (^) t = 4. 3 , q = 0 .65 (Fig. 6–20) K (^) f = 1 + 0 .65(4. 3 − 1) = 3. 15
σ a =
32 K (^) f M (^) a π d^3
π(1.625) 3
= 17 930 psi
n (^) f =
Se σ a
Quickly check if point M might be critical. Only bending is present, and the moment is small, but the diameter is small and the stress concentration is high for a sharp fillet required for a bearing. From the moment diagram, M (^) a = 959 lbf · in, and M (^) m = Tm = Ta = 0. Estimate K (^) t = 2.7 from Table 7–1, d = 1 .0 in, and fillet radius r to fit a typical bearing. r / d = 0 .02, r = 0 .02(1) = 0. 02 q = 0 .7 (Fig. 6–20) K (^) f = 1 + (0.7)(2. 7 − 1) = 2. 19
σ a =
32 K (^) f M (^) a π d^3
π(1) 3
= 21 390 psi
n (^) f =
Se σ a
Should be OK. Close enough to recheck after bearing is selected. With the diameters specified for the critical locations, fill in trial values for the rest of the diameters, taking into account typical shoulder heights for bearing and gear support. D 1 = D 7 = 1 .0 in D 2 = D 6 = 1 .4 in D 3 = D 5 = 1 .625 in D 4 = 2 .0 in
The bending moments are much less on the left end of shaft, so D 1 , D 2 , and D 3 could be smaller. However, unless weight is an issue, there is little advantage to requiring more material removal. Also, the extra rigidity may be needed to keep deflections small.
