yildiran - shafts and shaft components

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Shafts and

Shaft

ComponentsLecture Notes

Prepared by H. Orhan

YILDIRAN

MECE 304 Mechanical Machine Elements-Shafts

LECTURE NOTES- MECE 304 Mechanical Machine Elements

Chapter 4- Shafts and Shaft Components

(Notes from: Chapter 7, Budynas R.G., Nisbett J.K., Shigley’s Mechanical Engineering Design,

Mc Graw Hill, 8th Edition and special notes)

Spring Semester 2007/2008

Halil Orhan YILDIRAN, MS2

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7-1 Introduction 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.

Approach to shaft sizing:*Determine stress

*Establish tentative size (consider also axial location of elements)

*Make deflection and slope analysis

7-1 Shaft materials

Common: SAE 1020-1050 steels w/o heat treatmentHeat treatment: SAE 1040, 4140, 5140, 8640 steels

For surface hardening: SAE 1020, 8620 steels


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7-3 Shaft layout*Use shoulders for axial locating

*The length of the cantilever should be kept short to minimize the

deflection which is made for ease of mounting

*Use 2 support bearings

*Keep the length as short as possible

* Use shoulders, retaining rings, and pins, to transmit the axial load into the shaft

*One bearing carry axial load

*Common torque-transfer elements are:• Keys

• Splines

• Setscrews

• Press fitting*Consideration should be given to the method of assembling the

components onto the shaft, and the shaft assembly into the frame.

*For press fitting design a short slide length


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Figure 7-3

Tapered roller

bearings usedin a mowing

machine

spindle.

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Figure 7-4 A

bevel-gear drive

in whichboth pinion and

gear are

straddle-

mounted.

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Figure 7-5 Arrangement

showing bearinginner rings press-fitted to

shaft while outer rings float

in the housing

Figure 7-6 Arrangement as

of Fig. 7--5 except that the

outer bearing rings are preloaded.

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Figure 7-7 In this

arrangement the inner ring of

the left-hand bearing is locked to the shaft between

a nut and a shaft shoulder

Figure 7-8 Arrangement is

similar to Fig. 7-7 in that the

bearing positions the entire

shaft assembly.

Shield

Groove for grinding

Retaining ring

Lock nut

Floating RS bearing

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7-4 Shaft design for stress

The fluctuating stresses due to bending and torsion are:


For solid shafts with round sections

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The von Mises stresses for rotating round, solid shafts, neglecting

axial loads, are given by.

Expressions for any of the common failure criteria obtained by

substituting 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. 298.

The resulting equations for several of the commonly used failure curves are summarized below. The names given indicates:

First is the significant failure theory, followed by a fatigue failure locus

name. Example; DE-Gerber indicates the stresses are combined using the distortion

energy (DE) theory, and the Gerber locus is used for the fatigue failure.

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12


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The Gerber and modified Goodman criteria do not guard against yielding, requiring a separate check for yielding. Therefore von Mises maximum

stress is calculated as:

Checking for yielding

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SAMPLE PROBLEM 7-1 At a machined shaft shoulder the small diameter d is 28 mm, the large diameter D is 42 mm, and the fillet radius is 2.8 mm.

The bending moment is 142.4 N.m and the steady torsional moment is 124.3 N.m. The heat-treated steel shaft has an ultimate strength of Sut =

735 MPa and a yield strength of Sy = 574 MPa. The reliability goal is 0.99.

(a) Determine the fatigue factor of safety of the design using each of the fatigue failure criteria described in this section.(b) Determine the yielding factor of safety.

Estimating Stress Concentration

*Finding stress concentration factor is an iterative process

*Shoulders for bearing and gear support should match the catalog

recommendation for the specific bearing or gear*For the standard shoulder fillet, for estimating Kt, the worst end of the

spectrum, with r/d = 0.02 and D/d = 1.5, Kt values from the stress

concentration charts for shoulders indicate 2.7 for bending, 2.2 for torsion,

and 3.0 for axial.

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Figure 7-9 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

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*For keyways estimate the stress concentration for keyways regardless of the actual shaft dimensions by assuming a typical ratio of r/d = 0.02.

This gives

Kt = 2.2 for bending and Kts = 3.0 for torsion, assuming the key is in place.

*Retaining rings. If the groove width is slightly greater than the groove

depth, and the radius at the bottom of the groove is around 1/10 of the groove width then from Figs. A–13–16 and A–13–17, stress concentration

factors for typical retaining ring dimensions are around;

Kt=5 for bending and axial, and

Kts=3 for torsion.

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Table 7-1 Summary of first Iteration Estimates for Stress Concentration Factors Kt.

Do not use if actual dimensions are known!

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7-5 Deflection ConsiderationsDeflection of the shaft, both linear and angular, should be checked at

gears and bearing locations against allowable values.

Table 7-2 Typical Maximum Ranges for Slopes and Transverse Deflections

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If any value is larger of the found deflection and slope than the allowable deflection or slope at that point, a new diameter can be found

from:

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For a stepped shaft with individual cylinder length li and torque Ti , theangular deflection can be estimated from

for a constant torque throughout homogeneous material, from

Torsional stiffness of the shaft k in terms of segment stiffnesses is

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7-6 Critical Speeds of ShaftsCritical speeds: at certain speeds the shaft is unstable, with deflections

increasing without upper bound. Designers seek first critical speeds at

least twice the operating speed.

The shaft, because of its own mass, has a critical speed.

When geometry is simple, as in a shaft of uniform diameter, simplysupported, the task is easy. It can be expressed as:

where m is the mass per unit length,

A is the cross-sectional area, and

γ the specific weight.

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For an ensemble of attachments, Rayleigh’s method for lumped masses gives

where wi is the weight of the ith location and

yi is the deflection at the ith body location.

Figure 7-12 (a) A uniform-

diameter shaft for Eq. (7–22).

(b) A segmented

uniform-diameter shaft for

Eq. (7–23)

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Use of influence coeeficients to find crital speeds:An influence coefficient is the transverse deflection at location i on a

shaft due to a unit load at location j on the shaft. From Table A–9–6 we

obtain, for a simply supported beam with a single unit load as shown in

Fig. 7–13.

Figure 7-13 The

influence coefficient δi j

is the deflection at i due to a unit load at j.

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For three loads the influence coefficients may be displayed as

From the influence coefficients above, one can find the deflections y1,

y2, and y3 of Eq. (7–23) as follows (δij = δji ):

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Taking Fi=mi ω2 yi and for non trivial solution

Expanding the determinant

The three roots of Eq. (7–27) can be expressed as 1/ω12, 1/ω22, and

1/ω32

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Comparing 7-27 and 7-28

If we order the critical speeds such that ω1 < ω2 < ω3, then 1/ω12» 1/ω2

2 , and 1/ω32. So the first, or fundamental, critical speed ω1 can be

approximated by

Define 1/ω112 = m1 δ11 and so forth

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For n body shaft

Eq. 7-32 is called Dunkerleys equation (speed is below actual)

For the load at station 1, placed at the center of span, denoted with the

subscript c, the equivalent load is found from:

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SAMPLE PROBLEM 7-5 Consider a simply supported steel shaft as

depicted in Fig. 7–14, with 25 mm diameter and a 775 mm span between

bearings, carrying two gears weighing 175 and 275 N.(a) Find the influence coefficients.

(b) Find ∑ wy and ∑ wy2 and the first critical speed using Rayleigh’s equation, Eq. (7–23).

(c) From the influence coefficients, find ω11 and ω22.(d) Using Dunkerley’s equation, Eq. (7–32), estimate the first critical speed.

(e) Use superposition to estimate the first critical speed.

( f ) Estimate the shaft’s intrinsic critical speed. Suggest a modification to

Dunkerley’s equation to include the effect of the shaft’s mass on the first critical speed of the attachments.

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Figure 7-14 (a) A 25 mm uniform-diameter shaft for Ex. 7-5

(b) Superposing of equivalent loads at the center of the shaft for the purpose

of finding the first critical speed

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7-7 Miscellaneous Shaft ComponentsSetscrews

Figure 7-15 Socket

setscrews: (a) flat point;

(b) cup point; (c) oval point; (d) cone point; (e)

half-dog point

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Table 7-4 Typical holding force for cup point socket setscrews

Size, mm Seating Torque, N.m Holding force, N

#0 0.11 222

#1 0.2 289

#2 0.2 378

#3 0.5 534

#4 0.5 712

#5 1.1 890

#6 1.1 1112

#8 2.2 1713

#10 4.0 2403

6 9.8 4450

8 18.6 6675

10 32.8 8900

11 48.6 11125

12 70.0 13350

14 70.0 15575

16 149.7 17800

20 271.2 22250

22 587.6 26700

25 813.6 31150

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Keys and pins

Figure 7-16 (a) Square

key; (b) round key;

(c and d) round pins; (e)

taper pin; (f) split tubular spring pin.

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Table 7-6 Millimeter dimensions for some standard square and rectangular

key applications

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Figure 7-17(a) Gib head key, (b) woodruff key

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Table 7-7 Dimensions of woodruff key,

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Table7-8 Sizes of Woodruff key suitable for various shaft diameter

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For fillets cut by standard milling-machine cutters, with a ratio of r/d = 0.02, Peterson’s charts give Kt = 2.14 for bending and Kts = 2.62 for

torsion without the key in place, or Kts = 3.0 for torsion with the key in

place

Retaining rings

Is frequently used instead of a shaft shoulder or a sleeve to axially position a component on a shaft or in a housing bore.

Appendix Tables A–13–16 and A–13–17 give values for stress

concentration factors for flat-bottomed grooves in shafts, suitable for

retaining rings.

Figure 7-17 Retaining rings: (a,b)

external, (c,d) internal

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SAMPLE PROBLEM 7-6 A UNS G10350 steel shaft, heat-treated to a

minimum yield strength of 525 MPa, has a diameter of 36 mm. The

shaft rotates at 600 rev/min and transmits 36 kW through a gear. Select an appropriate key for the gear.

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7-8 Limits and FitsThe definitions illustrated in Fig. 7–20 are explained as follows:

• Basic size is the size to which limits or deviations are assigned and is

the same for both members of the fit.

• Deviation is the algebraic difference between a size and the

corresponding basic size.

• Upper deviation is the algebraic difference between the maximum limit and the corresponding basic size.

• Lower deviation is the algebraic difference between the minimum limit

and the corresponding basic size.

• Fundamental deviation is either the upper or the lower deviation, depending on which is closer to the basic size.

• Tolerance is the difference between the maximum and minimum size

limits of a part.

• International tolerance grade numbers (IT) designate groups of

tolerances such that the tolerances for a particular IT number have the

same relative level of accuracy but vary depending on the basic size.

• Hole basis represents a system of fits corresponding to a basic hole size. The fundamental deviation is H.

• Shaft basis represents a system of fits corresponding to a basic shaft size. The fundamental deviation is h.

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Figure 7-20 Definitions applied to a cylindrical fit

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Table 7-9

Preferred

Fits

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Stress and Torque capacity in Interference FitsThe pressure p generated at the interface of the interference fit, is given

by

If both member are of the same material

where d is the nominal shaft diameter, di is the inside diameter (if any)

of the shaft, do is the outside diameter of the hub, E is Young’s modulus, and v is Poisson’s ratio, with subscripts o and i for the outer

member (hub) and inner member (shaft), respectively.

δ is the diametral interference between the shaft and hub.

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Since there is tolerance on both diameters, the min. and max. interferences will be

Tangential stress at the interference of the shaft and hub

The radial stress at the interference of the shaft and hub

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A stress element on the surface of a rotating shaft will experience a completely reversed bending stress in the longitudinal direction, as well

as the steady compressive stresses in the tangential and radial

directions. This is a three-dimensional stress element. Shear stress

due to torsion in shaft may also be present. Since the stresses due to

the press fit are compressive, the fatigue situation is usually actually

improved. For this reason, it may be acceptable to simplify the shaft analysis by ignoring the steady compressive stresses due to the press

fit. There is, however, a stress concentration effect in the shaft bending

stress near the ends of the hub, due to the sudden change from

compressed to uncompressed material. For first estimates, values are typically not greater than 2.

The friction force Ff and the torque capacity of the joint is

Where l is the length of the hub and f is the coefficient of friction.

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MECE 304 Mechanical Machine Elements-Shafts-Supplements

4-2 Tension, compression and Torsionδ = Fl/AE 4-3

θ = Tl/JG 4-5

4-3 Deflection due to bending

q/EI=d^4y/dx^4 4-10V/EI=d^3y/dx^3 4-11

M/EI=d^2y/dx^2 4-12

θ =dy/dx 4-13

y=f(x) 4-14

4-4 Beam deflection methods

Equations 4-10 to 4-14 are the basis for relating the intensity of loading

q, vertical shear V, bending moment M, slope of neutral surface θ and

transverse deflection y.

The methods used to solve the integration problem for beam deflection

are.*Superposition

*The moment area method*Singularity function*Integration (numerical, graphical)

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4-6 Beam deflection by superpositionSuperposition resolves the effect of combined loading on a structure by

determining the effect of each load separately and adding the results

algebraically. Conditions for superposition

*each effect is linearly related to the load that produces it

*the load does not create a condition that effects the results of another

load*the deformation resulting from any specific load are not large enough to

appeciably alter the geometric relations of the parts of the structural

system

(See Sample Problem S_2)

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MECE 304 Mechanical Machine Elements-Shaft-Supplements

ADDITIONAL INFORMATION

Figure- Dimensions for some standard square and rectangular key

applications

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Table- Designation and Tolerances for shafts and hubs for keys (see Fig S-47)

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SPLINES (there are standards for splines as DIN, SAE, TS)

Fig. Parallel

side splines

Fig. Involute

splines

Fig. Involute

serrations

Fig. Parallel side splined

shaft with six splines

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*Parallel side splines may be major or minor dia fit

*The torque which an integral multispline shaft can transmit

Mt =(1/2) phLi(D- h)

Where

Mt= Torque capacity N.mm

P= bearing pressure, Mpah=Spline heigth, mm

L= Length of spline contact, mm

i= number of splines

D=Diameter, mm

Failure of involute splines may be with shear or bearing pressures. So

must be checked for shear and for bearing pressures.

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Table Proportions of SAE standard parallel side splines

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The theoretical torque capacity of straight-sided spline with sliding according to SAE

Mt = 6.895* 10^6*i*((D+d)/4)* L*h

Where

Mt= torque in N.m

i =number of splinesd =inside diameter of spline, m

D =pitch diameter of spline, m

L =length of spline contact, m

h =minimum height of contact in one tooth of spline, m

yildiran - shafts and shaft components

Documents

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