ch_11

1-1

Chapter 11 • Shafting

11.1 A shaft assembly shown in sketch a is driven by a flat belt at location A and drives a flatbelt at location B. The drive belt pulley diameter is 300mm; the driven belt pulley is 500mm. The belts are long relative to the sheave diameter. The distance between sheaves is800 mm, and the distance from each sheave to the nearest bearing is 200 mm. The beltsare horizontal and load the shaft in opposite directions. Determine the size of the shaftand the types of steel that should be used. Assume a safety factor of 5.

Notes: This is an open-ended problem in that the shaft diameter will depend on the materialchosen. For this solution, AISI 1080 steel is used. The axes referenced in the solution have beenadded in green on the figure, as have the labels "1" and "2" for the bearings. These labels are allarbitrary. Also, the reactions at the bearings have been added in red.

Approach: This problem will require determination of the stresses in the critical section of theshaft. The first step, as always, is to find the bearing reactions. One must then calculate thenormal stress due to bending, the shear stress due to torsion and obtain the principal stresses, allas a function of shaft diameter. From this, one uses a static failure criterion (this solution uses thedistortion energy criterion) to find the minimum shaft diameter.

I. Determination of ReactionsFrom equilibrium of forces and moments in the xy plane, it can be easily shown that R1y=R2y=0.Taking equilibrium in the z-direction:

ΣFz=0=R1z-700N-200N+300N+600N+R2z

ΣM1=0=-(200N+700N)(0.2m)+(300N+600N)(1.m)+R2z(1.2m); R2z=-600NThe negative sign indicates it acts in the negative z-direction. Substituting this value into the forceequilibrium equation gives R1z=600N.

II. Identification of Critical SectionThe pulley diameters are given as 300 and 500 mm, so the radii are 150 and 250 mm. The

torque applied at point A is (700N-200N)(0.15m) = 75Nm. The torque at B is (300-600)(0.25m)=-75Nm. Therefore, the shaft sees a torque between the pulleys, but there is no torque in the shaftelsewhere. The shear force, bending moment and torque diagrams are shown. From the diagrams,it can be seen that the critical location is just to the inside of either pulley, so we take M=120Nmand T=75Nm. Note that we are neglecting shear; the shear stresses from vertical shear are a

2

maximum at the center of the shaft and zero where thebending stresses are maximum. Shafts are considered tobe slender and more likely to fail from areas ofmaximum normal stress due to bending than from shear.This can be verified for this case by the student byrepeating the following analysis at the center of the shaftcross section.

III. Determination of Stress Components andPrincipal Stresses

Remember that since the shaft cross section iscircular, from Table 4.1 on page 148,

I =πd4

64, J =

πd4

32, c =

d

2

Therefore, from Equations (4.48) and (4.34), respectively:

σx =Mc

I=

120Nm( ) d / 2( )πd4 / 6 4( ) =

3840

πd3 Nm

τ =cT

J=

d / 2( ) 75Nm( )πd4 / 3 2( ) =

1200

πd3 Nm

Note that σy and σz are zero, so that from Equation (2.16), the principal normal stresses can befound as:

σ1,σ2 =σx + σ y

2± τxy

2 +σ x − σy( )2

4=

1

πd 3 Nm3840

2+ 12002 +

38402

4

so that

σ1 =4184Nm

πd3 , σ2 = −344

πd3 Nm

IV. Application of Failure CriterionAs discussed in the note above, this is an open ended problem, and this solution will

consider the special case of AISI 1080 steel as the shaft material. From the inside front cover, theyield strength of AISI 1080 is 380MPa. From Equations (6.10) and (6.11) on page 238, failureoccurs when

σe = σ12 + σ2

2 − σ1σ2( )1 / 2>

Sy

ns

1

πd3 Nm 41842 + −344( )2 − 4184( ) −344( )( )1 / 2>

380 × 106 N / m2

5This can be solved for d as d<0.02635m or d<26.35mm. Since shaft diameters are generallyexpressed in integers of millimeters for small shafts, it is good design practice to specify a shaftdiameter of 30 mm.

11.2 The gears in the shaft assembly shown in Fig. 11.1(a) transmit 100 kW of power androtate at 3600 rpm. Gear wheel 1 is loaded against another gear such that the force P1 acts

3

in a 45° upward direction at a radius of 80 mm from the shaft center. The force P2 actsvertically downward at a radius of 110mm from the shaft center. The distance frombearing A to gear 1 is 100 mm, that from gear 1 to gear 2 is 85mm, and that from gear 2to bearing B is 50 mm. Do the following:

a) Draw a free-body diagram with forces acting on the shaft when bearings A and Btransmit only radial forces.

b) Give values of force components as well as the resultant force at locations A and B.c) Give the transmitted torque.d) Draw a bending moment diagram in the x-y and x-z planes along with a torque

diagram. Also, indicate what the maximum bending moment and the maximum torqueshould be.

e) Find the safety factor according to the distortion-energy-theory (DET) and themaximum shear stress theory (MSST) if the shaft has a diameter of 35 mm and ismade of high-carbon steel (AISI 1080).

Notes: As usual, we will assume that the power loss in the bearings is negligible.

Solution: The solution follows the order of the assigned questions.

I. Free Body Diagram

The sketch for the problem, with the coordinate system used in the solution as well as the forcereactions are shown above left. Above right are the xy and xz plane views of the free bodydiagram when the forces P1 and P2 act radially, that is, without the associated torques.

II. Determination of ReactionsBefore we can determine the reactions, we have to obtain P1and P2. Since the speed is

3600rpm, ω=(3600)(2π)/60=377rad/s. Therefore, from Equation (4.41),

T =hp

ω=

100,000 Nm / s

377/ s= 265Nm

We will assume that there are no torque losses in the bearings or in the gear teeth. Therefore, thetransmitted torque is applied equally and oppositely to both gears. The forces P1 and P2 can becalculated as

P1 =T

r1=

265Nm

0.08m= 3315N

P2 =T

r2=

265Nm

0.110m= 2410N

4

We can then derive the reactions from force and moment equilibrium as follows. First, in the xyplane, using counterclockwise moments as positive, the moment and force equations yield:

MA = 0 = P1 cos45°( ) 0.1m( )∑ − P2 0.185m( ) + By 0.235m( ) ;By = 900N

Fy = 0 = Ay + P1 c o s 4 5° −P2 + By = Ay + 2344N −∑ 2410N + 900N ; Ay = −834NThen in the xz plane,

MA = 0 = P1 sin45°( ) 0.1m( ) − Bz 0.235m( );Bz = 998N∑Fz = 0 = Az − P1 s i n 4 5°+ Bz = Az − 2344N + 998N ;Az =1350N∑

The shear, bending and torque diagrams are as follows:

The resultant forces at A and B are the vector sums of their components, or

A = Az2 + Ay

2 = 1350N( )2 + −834N( )2 = 1590N

B = Bz2 + By

2 = 998N( )2 + 900N( )2 =1340N

The maximum moment occurs at either gear 1 or gear 2; in reviewing the moment diagrams, thelarger components are at gear 1, so we calculate the maximum moment as:

Mmax = 83.4Nm( )2 + 135Nm( )2 = 159Nm

III. Safety Factor of ShaftFor AISI 1080 steel, the yield strength is found on the inside front cover as 380MPa. For

the distortion-energy theory, equation (11.14) gives the safety factor for a shaft diameter ofd=35mm=0.035m as:

ns =πd3Sy

32 M2 + 3

4T 2

=π 0.035m( )3 380MPa( )

32 159Nm( )2 + 3

4265Nm( )2

= 5.73

For the maximum shear stress theory, equation (11.18) gives:

ns =πd3Sy

32 M2 + T 2=

π 0.035m( )3 380MPa( )32 159Nm( )2 + 265Nm( )2

= 5.18

5

11.3 The shaft assembly given in Problem 11.2 has an extra loading from thermal expansionof the shaft. The bearings are assumed to be rigid in the axial direction, so that when theshaft heats up it cannot elongate but instead compressive stress builds up. Determine thethermal stresses and find the safety factor by the DET if the shaft heats up

a) 5°Cb) 15°C

Notes: There are a number of approaches for solving this problem. One could calculate thecomponent of thermal stress, bending stress and torsion shear stress and use a Mohr's circleapproach to determine the principal stresses for direct substitution in the distortion-energy theory.A simpler approach is to take advantage of the derivation in the book and use Equation (11.22).

Solution: The linear thermal expansion coefficient for steel is found from Table 3.5 on page 118as 11x10-6/°C. The stiffness of steel is found on the inside front cover as 207GPa. Taking thetemperature rise of 5°, the axial thermal stress is given by

σ = εE = αE∆T = 11 × 10−6 / °C( ) 207GPa( ) 5°C( ) = 11.4MPa

The axial load is

P = σA =σπd 2

4=

11.4MPa( )π 0.035m( )2

4= 11.0kN

Equation (11.22) can be solved for the safety factor as:

ns =πd3Sy

4 8M + Pd( )2 + 48T2=

π 0.035m( )3 380MPa( )4 8 1 5 9Nm( ) + 11.0kN( ) 0.035m( )( )2 + 48 265Nm( )2

= 5.17

If the temperature rise is 15°C, then the same approach is followed, so that P=32.9kN andns=4.24.

11.4 Given the shaft assembly in Problem 11.2, but calculating as if the AISI steel were brittle[using the maximum-normal-stress theory (MNST)], find the safety factor ns for fatigueby using the information in problem 11.2.

Notes: The solution shown here incorporates correction factors for surface finish and size onfatigue strength. Most shafts are machined or drawn; it is rare that a shaft is polished or ground,so this is the basis for the surface finish correction factor.

Solution: First, the fatigue strength of AISI 1080 steel is determined. Then, Equation (11.40)gives the solution.

I. Fatigue Properties of ShaftFrom Equation (7.7), we know that the endurance limit for steel (unmodified) is 0.5Sut,

where Sut for 1080 steel is 615MPa as found on the inside front cover. The only correction factorsthat we know must be included are the surface finish factor and the size factor. If the shaft ismachined or cold drawn then the surface finish factor is given by Equation (7.21) usingcoefficients from Table 7.3 on page 274:

k f = eSutf = 4.51( ) 615MPa( )−0.265 = 0.82

The size factor is given by Equation (7.22) on page 275 as

ks = 1.189d−0.112 =1.189 35( )−0.112 = 0.80The modified endurance limit is given by Equation (7.16) as:

6

Se = k f ks ′ S e = k f ks 0.5Sut( ) = 0.82( ) 0.80( ) 0.5( ) 615MPa( ) = 202MPa

II. Fatigue Analysis of Shaft.The safety factor is given by Equation (11.40), where the following values are taken fromProblem 11.2:

Mm=0, Ma=159Nm, Ta=0, Tm=265Nm, d=0.035mAlso, we will assume there are no stress concentration associated with the shaft. Therefore thesafety factor is

ns = πd3Su / 1 6

Kc Mm + Su

SeMa

+ Kc

2 Mm + Su

SeMa

2

+ Kcs2 Tm + Su

SeTa

2

= π 0.035m( )3 615MPa( ) /16

615

202159Nm + 615

202159Nm

2+ 265Nm( )2

= 5.0

11.5 A bridge used for sightseeing is built in the form of a T, where the bottom of the T isfastened into the rock face. The top of the T is situated 7m horizontally from thefastening point, and the length of the cross bar is 8m, half on each side of the fasteningpoint. The bridge was built to carry 40 persons with an average weight of 75kg perperson. Assume that the weight is evenly distributed along the cross bar with a safetyfactor of 8. How large will the safety factor be if all the people gather on one end of thecross bar? The load-carrying structure in the bridge is a circular tube with outer diameterdo and inner diameter di. Use the DET and neglect the weight of the bridge.

Notes: This problem is challenging because no more data has been given than is necessary tosolve the problem. By keeping the diameters of the tube in the derivation, they eventually cancel,allowing determination of the safety factor. The instructor may wish to alter this problem byspecifying a material and a wall thickness and asking the students to determine the member'souter diameter.

Solution: The approach is to calculate the stress state for the first case and use the known safetyfactor to define a relationship between the diameters which remain. The second case is thenanalyzed, compared to the first to eliminate the known function, and the safety factor isdetermined.

I. Uniformly Distributed LoadIn this case, there is no torque, only bending, where the moment is given by:

M = 40( ) 75kg( ) 9.81m / s2( ) 7m( ) = 206,000 Nm

For the tube

c =d0

2; I =

π64

do4 − di

4( )so that the bending stress is

7

σx =Mc

I=

206,000 Nm( ) do

2π64

do4 − di

4( )= 2.1MNm

d0

do4 − d i

4

In this case, this is the only stress and we know we achieve a safety factor of 8. Therefore weknow:

ns =Sy

σ x= 8 =

Sy

2.1MNmd0

do4 − di

4

;Sy do

4 − di4( )

do= 16.8MNm

II. Concentrated LoadIf the people are concentrated at one end of the bridge, there is, in addition to the moment, atorque given by

T=(40)(75kg)(9.81kgm/s2)(8m)/2=118kNmThe shear stress which results at the outer diameter is

τxy =Tc

J=

118kNm( ) do / 2( )π32

do4 − d i

4( )

= 0.60MNmdo

do4 − di

4

Therefore the principal stresses are, from Equation (2.16),

σ1,σ2 =σx + σ y

2± τxy

2 +σx − σy

2

2

=do 106 Nm( )

do4 − di

42.1+ 0

2± 0.602 +

2.1− 0

2

2

So that

σ1 =do 106 Nm( )

do4 − di

4 2.26( ), σ2 = −do 106 Nm( )

do4 − di

4 0.159( )

Using Equation (11.11) on page 429 gives

σ12 + σ 2

2 − σ1σ2( )1 / 2=

Sy

ns;ns =

Sy

σ12 + σ2

2 − σ1σ2( )1 / 2

=Sy 10−6 Nm( ) do

4 − di4( )

do

1

2.262 + 0.1592 + 2.26( ) 0.159( )( )1 / 2=16.8

2.38= 7.17

Alternate Solution:Use a 1 subscript for the uniformly distributed load case, and a 2 subscript for the concentratedload. As determined above, M1=M2=206kN-m, T1=0, and T2=118kN-m. Equation (11.13) gives:

d =32ns

πSyM2 +

3

4T2

1 / 3

Comparing the two loading cases, and realizing the diameter is the same for the two cases,

32ns1

πSyM1

2 +3

4T1

2

1 / 3

=32ns2

πSyM2

2 +3

4T2

2

1 / 3

Solving for ns2 and substituting for the moments and torques gives:

8

ns2 = ns1

M12 +

3

4T1

2

M22 + 3

4T2

2

= 8( )206kNm( )2 +

3

40( )2

206kNm( )2 + 3

4118kNm( )2

= 7.17

11.6 Find the change in safety factor if the load-carrying structure in the bridge in Problem11.5 is a beam with a solid rectangular section of height h and width b instead of thehollow cylinder. Assume that h/b=6 and that the shear stress is

τmax =T b / 2( )

J* =T b / 2( )

0.298hb3 =T

0.596hb2

Notes: This is similar to problem 11.5, and follows a similar path of solution.

Solution:

I. Uniform Loading The loading hasn't changed from Problem 11.5, so the bending moment for this case is stillM=206kNm. The stress calculation is modified by using the proper formulae for a rectangularcross section:

c =h

2, I =

bh3

12so that the bending stress is

σx =206kNm h / 2( )

bh3 /12=

34kNm

b3

where we have used the relationship h/b=6 or h=6b to eliminate h.

II. Concentrated LoadingThe torque is, as in Problem 11.5, T=118kNm. The shear stress must be calculated for arectangular cross section, so using the equation for shear stress given in the problem statement,

τmax =T

0.596hb2 =118kNm

0.596 6b( )b2 =33.0kNm

b3

The stresses σx and τmax are in different locations, the tensile stress at the top of the beam and theshear stress at the mid-height of the beam. The safety factor for the tensile stress is given byσ1=σx and σ2=0. Thus the DET gives ns=Sy/σx=8. At the mid-height of the beam, the shear stressis

σ1,σ2 = ± τxy2

Thus using the distortion energy theory gives

σ12 + σ 2

2 − σ1σ2( )1 / 2= τxy

2 + τxy2 + τxy

2( )1 / 2= 3τxy

Therefore, the safety factor is given by

ns =Sy

3τxy= 4.82

9

11.7 A ladder to a pleasure boat has two vertical aluminum sides to carry the load from thesteps bolted to them. By mistake the bolts on one side of the step were changed to blacksteel and therefore corroded away. Calculate how much the safety factor decreases ifpeople walk at the middle of the steps and all steps except the bad one are fixed to bothvertical sides. The vertical sides are much less elastic than the steps. Use MSST. Thelength of a step (width of the ladder) is 0.600m, and its cross section is 15mm thick and150mm wide.

Notes: This problem is open-ended and has many subtle aspects which would allow differentapproaches and assumptions in its solution. For example, ladders used on pleasure craft are oftenvery close to walls, so that a person will naturally place their weight on the step away from thewall end (applying a torque to the step). Also, the weight of a person has not been defined, so thedimensions of a 95th or 99th percentile person could be obtained from a human factors text. Allof these considerations are fine, but this solution merely demonstrates the general approach to thisproblem.

Solution:Moment equilibrium gives

M0-M+Tx=0; M=M0+Tx=M0-Px/2Integrating Equation (5.3) gives

dy

dx= −

M0x

EI+

Px2

2EI 2( )+ C1

The boundary conditions are, first, that at x=0, dy/dx=0, so that C1=0. The second boundarycondition is that at x=l/2, dy/dx=0, so that

0 = −M0

l

2

EI+

Pl2

4EI 4( ); M0 =

Pl

8Therefore, the moment at x=l/2 is

Mx= l / 2 =Pl

8−

Pl

4= −

Pl

8Therefore,

τ1=2τ2=P/A

σ1,σ2 =σx

2± τxy

2 +σx

2

2

If correctly mounted, this becomes:

σ1,σ2 =Plc

2 8( )I±

P

2 A

2

+Plc

16I

2

; σ1 − σ 2 = 2P

2A

2

+Plc

16I

2

If corroded, this is

σ1,σ2 =Plc

2 2( )I±

P

A

2

+Plc

16I

2

; σ1 − σ 2 = 2P

A

2

+Plc

4I

2

The safety factor decreases for the corroded bolts to

σ1 − σ2( )corroded

σ1 − σ 2( )correct

=

1

A

2+ lc

4I

2

1

2 A

2+ lc

16I

2=

1

0.15 0.15( )

2

+0.6( ) 0.0075( )

4 0.15( ) 0.015( ) 3 / 1 2( )

2

1

0.03 0.15( )

2

+ 0.6( ) 0.0075( )16 0.15( ) 0.015( ) 3 / 1 2( )

2 = 3.998

10

11.8 Gears 3 and 4 act on the shaft shown in sketch b. The resultant gear force, PA=600lbf,acts at an angle of 20° from the y axis. The yield stress for the shaft, which is made ofcold-drawn steel, is 71,000 psi and the ultimate stress is 85,000 psi. The shaft is solid andof constant diameter. The safety factor is 2.6. Assume the DET throughout. Also, forfatigue loading conditions assume completely reversed bending with a bending momentamplitude equal to that used for static conditions. The alternating torque is zero and theGoodman relationship is assumed. Determine the safe diameter due to static and fatigueloading. Show shear and moment diagrams in the various places.

Notes: The reaction forces and their labels have been added in red to the problem sketch. It isassumed that there is no torque loss in the bearings.

Solution: The first step is to determine the force on Gear C. This is done by equating torquesapplied at the gears:

ΣT=0=PAcos20°(12in) – PCcos20°(5in); PC=1440 lbsThe torque between the gears is T=PAcos20°(12in)=6766inlb = 6.76 kipin.

I. Determination of Reactions and Shear and Moment Diagrams.Now obtain the reactions be looking at the x-y and x-z planes separately. In the xy plane:

ΣMO=0=(PAcos20°)(20in)+By(36in)-PCsin(20°)(46in); By=316 lbs

ΣFy=0=Oy+PAcos20°+By-PCsin20°; Oy=-387 lbs.

And in the x-z plane:ΣMO=0=PAsin20°(20in)-Bz(36in)-PCcos20°(46in); Bz=-1615lbs

ΣFz=0=Oz+Bz-PAsin20°+PCcos20°; Oz=467 lbs

The shear and bending moment diagrams can then be constructed for the xy and xz planes as:

-11

The moments at A and B are calculated from the Pythagorean theorem and are 12.1 kip-in at Aand 14.4 kipin at B. This means that the section of the shaft just inside the gear at B is critical, sowe take T=6.76 kipin, and M=14.4 kipin. For the shaft material, Sy=71,000 psi, Su=85,000psi.

II. Static AnalysisUsing the distortion-energy theory, Equation (11.13) gives:

d =32ns

πSyM2 +

3

4T 2

1 / 3

=32 2.6( )

π 71000psi( ) 14400inlb( )2 +3

46760 inlb( )2

= 1.8in

III. Fatigue AnalysisFor fatigue analysis, using MM=0 and MA=14.4 kip-in, TA=0 and TM=6.76 kipin, we need theendurance limit:

′ S e = 0 . 5Su = 0.5(85,000psi) =42,500 psiKf=0.58; ks=0.869(2)-0.112=0.8041, so Se=19.821 ksi

From Equation (11.38):

d = 32ns

πSyMm +

Sy

SeK f Ma

2

+ 3

4Tm +

Sy

SeKfsTa

2

1 / 3

= 32ns

πSy

Sy

SeK f Ma

2

+ 34

Tm2

= 2.18in

Rounding this to the nearest quarter inch shaft suggests using a 2.25 inch diameter.

11.9 Derive Equation (11.39) of the text. Start by showing the stresses acting on an obliqueplane at angle φ. The MNST should be used, implying that the critical stress line extendsfrom Se/ns to Su/ns.

Notes: The approach is very similar to the derivation of Equation 11.31. The derivation starts onpage 432.

Solution:For a brittle material, Figure 11.4(b) becomes the modified sketch shown to the left below.

12

Where Kc is the stress concentration factor for normal stresses and Kcs is the stress concentrationfactor for shear. Summing forces in the normal direction (that is, in the direction of σφ),

σφA-Kcs(τm+τa)sinφAcosφ-Kc(σm+σa)sinφAsinφ-Kcs(τm+τa)cosφAsinφ=0A can be eliminated; then using the double angle formulas from geometry and simplifying, thisequation becomes

σφ = Kc

2σm +σa( )+ Kcs τm + τa( )s i n 2φ − Kc

2σm + σa( )c o s 2φ

If this is re-organized by mean stress and alternating stress components, this can be written as

σφ =Kcσ m

2+ Kcsτm sin2φ −

Kcσ m

2cos2φ

+

Kcσa

2+ Kcsτa sin2φ −

Kcσ a

2cos2φ

= σ φm + σφa

The Soderberg line for this case has been drawn to the right above. Note the reference pointslabeled in the sketch. From similar triangles,

HF

OF=

OJ

OD=

HG

ODor

Su / ns −σφm

Su / ns=

σφa

Se / ns;

1ns

=σφm

Su+

σφa

SeSubstituting for σφm and σφa,

1ns

= Kc

2σm

Su+ σa

Se

+ Kcs

τ m

Su+ τa

Se

s i n 2φ − Kc

2σm

Su+ σa

Se

c o s 2φ

To find the maximum safety factor, take the derivative of this expression with respect to φ and setequal to zero:

ddφ

1ns

= 2 Kcs

τm

Su+ τa

Se

c o s 2φ + Kc

σm

Su+ σa

Se

sin2 φ = 0

Therefore,

sin2 φcos2φ

= t a n 2φ =−2Kcs

τm

Su+ τa

Se

KcσmSu

+ σaSe

From simple trigonometry, the sine and cosine can now be evaluated as:

13

sin2 φ =2Kcs

τm

Su+ τa

Se

Kc2 σ m

Su+ σa

Se

2

+ 4Kcs2 τm

Su+ τa

Se

2

cos2φ =−Kc

σm

Su+ σa

Se

Kc2 σ m

Su+ σa

Se

2

+ 4Kcs2 τm

Su+ τa

Se

2

Substituting into the relationship for ns above gives

1ns

= Kc

2σm

Su+ σa

Se

+ Kcs

τ m

Su+ τa

Se

2Kcsτm

Su+ τa

Se

Kc2 σm

Su+ σa

Se

2

+ 4Kcs2 τm

Su+ τa

Se

2− Kc

2σm

Su+ σa

Se

−Kcσm

Su+ σa

Se

Kc2 σ m

Su+ σa

Se

2

+ 4Kcs2 τm

Su+ τa

Se

2

1

ns=

Kc

2

σm

Su+

σa

Se

+

Kc2 σ m

Su+

σa

Se

2

+ 4Kcsτ m

Su+

τa

Se

2

2 Kc2 σ m

Su+

σa

Se

2

+ 4Kcs2 τm

Su+

τa

Se

2

or, rearranging

2Su

ns= Kc σm +

Suσ a

Se

+ Kc

2 σ m +Su

Seσa

2

+ 4Kcs2 τm +

Su

Seτa

2

which is the desired relationship.

11.10 The shaft shown in sketch c rotates at 1000 rpm and transfers 6kW of power from inputgear A to output gears B and C. The spur gears A and C have pressure angles of 20°. Thehelical gear has a pressure angle of 20° and a helix angle of 30° and transfers 70% of theinput power. All important surfaces are ground. All dimensions are in millimeters. Theshaft is made of annealed carbon steel with Sut=636 MPa and Sy=365 MPa.a) Draw a free body diagram as well as the shear and moment diagrams of the shaft.b) Which bearing should support the thrust load?c) Determine the minimum shaft diameter for a safety factor of 2 and 99% reliability.

Notes: The drawing only shows the shaft; the bearings would be mounted on the shouldersmachined into the shaft. We will merely refer to the bearings as D and E according to the sketchabove. Also, the sketch has the reactions added in red and the applied force components as

14

indicated, which will then serve as a free body diagram. Note that gear C's matching gear has notbeen drawn; this solution places it as shown. Also, the problem statement doesn't indicate adirection of rotation for the shaft; the direction of the forces acting tangent to the shaft can beopposite of what is in this solution, but this does not affect the end result nor approach to theproblem. Finally, we are not told how the shaft loads vary or even if they do vary. This can leadto some differences in approaches for solving the problem; in this solution we will note that byspecifying surface finish, fatigue must be a consideration, so we treat the loads as mean loads,take the load variation to be zero and use the fatigue equations for shaft design.

Solution:I. Determination of Reactions, Shear and Moment Diagrams

First of all, let us note that the angular velocity of the shaft is 1000rpm=104.7rad/s. Sincethere is 6 kilowatts transmitted from the drive gear A, then the torque applied at this location isobtained from:

hp=Tω;6000Nm/s=TA(104.7rad/s);TA=57.3NmSince gear B transmits 70% of the power, and the shaft speed is constant, then the torquetransmitted at B is (0.7)(57.3)=40.1Nm, and the torque at gear C is 17.2Nm. The forces can thenbe calculated. For gear A, since 57.3Nm is transmitted and the diameter of the gear is 700mm, thetangential force is

Az=TA/rA=(57.3Nm)/(0.35m)=163.7NSince the pressure angle is 20°, then

tan20°=Ay/Az;Ay=Aztan20°=(163.7N)tan20°=59.6NSimilarly,

Bz=TB/rB=(40.1Nm)/(0.2m)=200N, By=Bztan20°=73N, Bx=Bztan30°=116N,Cz=TC/rC=(17.2Nm)/(0.09m)=191N, Cy=Cztan20°=69.56N.

Dx will be equal and opposite to Bx (see the discussion below regarding why the thrust force istaken at D). Dy and Ey are obtained by moment and force equilibrium in the xy plane:

ΣMD=0=-Cy(90mm)+By(270mm)+Ay(380mm)-Ey(460mm); Ey=78.5N

ΣFy=0=Dy+Cy-By-Ay+Ey=Dy+69.56N-73N-59.6N+78.5N; Dy=-15.5NFrom equilibrium in the xz plane:

ΣMD=0=Cz(90mm)-Bz(270mm)+Az(380mm)+Ez(460mm);Ez=-55N

ΣFz=0=Dz+Cz-Bz+Az+Ez=Dz+191N-200N+163.7N-55N;Dz=-100NThe desired shear, axial force, moment and torque diagrams are as follows.

15

II. Thrust LoadThe axial force arises because of the use of a helical gear at B. Thrust loads are not

unusual and will be discussed in much greater detail in Chapter 13. In this case, notice that theaxial force acts to the left according to the drawing of the helical gear. If the thrust load is takenby the bearing on the right, this may excessively load the bearing mounting. If the thrust load istaken by the bearing on the right, then the load is transferred through the shaft shoulder that hasbeen machined to facilitate this transfer. For this reason, the left bearing should take the thrustload.

III. Selection of Shaft Diameter.From the shear and bending moment diagrams, it is clear that the critical location for the

shaft is at point B. We will ignore the contribution of the axial force P, since the stress itcontributes is going to be very small (on the order of one megapascal for a diamter of 10millimeters - clearly negligible). The loading at this location is Tm=57.3Nm, Ta=Mm=0 and thebending moment amplitude is

MB = 7.59Nm( )2 + 8.36Nm( )2 = 11.3NmWe will use Equation (11.35) on page 435 to get the shaft diameter, but we first need themodified endurance limit for the shaft. Equation (11.35) gives the diameter but note that we needto determine the endurance limit for the shaft first. Using the approach described in detail inChapter 7, the endurance limit is (0.5)Sut=0.5(636MPa)=318MPa. kf is given by Eq. (7.21) as

k f = eSutf = 1.58( ) 636( )−0.085 = 0.91

ks is obtained from Equation (7.22) as ks=1.189d-0.112

, and kr is obtained from Table 7.4 as 0.82.Therefore the modified endurance limit is

Se = k f kskr ′ S e = 0.91( ) 1.189d −0.112( ) 0.82( ) 318MPa( ) = 279MPa( )d −0.112

Note that this is true for d in millimeters only (which was the requirement for application ofEquation (7.22)). Therefore, so that d is in meters we can write:

Se = kbkskr ′ S e = 0.91( ) 1.189d −0.112( ) 0.82( ) 318MPa( ) = 279MPa( ) 1000d( )−0.112

From Equation (11.35),

16

d = 32ns

πSyMm +

Sy

SeK f Ma

2

+ Tm +Sy

SeK fsTa

2

1 / 3

= 32ns

πSy

Sy

SeMa

2

+ Tm2

1 / 3

= 32 2( )π 365MPa( )

365

279 1000d( )−0.112 11.3Nm( )

2

+ 57.3Nm( )2

1 / 3

Evaluating this numerically gives d=0.0150m. A value of 15 mm could be specified, which mayeven allow the use of standard bar stock for manufacturing the shaft, but it is common practice toround up to the next highest five millimeter diameter, so a better design practice would be tospecify a 20mm shaft diameter.

11.11 In Problem 11.10 if the shaft diameter is 30 mm and the shaft is made of AISI 1080 cold-drawn steel, what is the safety factor while assuming 90% reliability?

Notes: This problem is straightforward after Problem 11.10 has been completed.

Solution:The forces, shear and moment diagrams and loading are all the same. However, AISI 1080 has anultimate strength of 615MPa and a yield strength of 380MPa. Calculating a modified endurancelimit as in Problem 11.10, the unmodified endurance limit is (0.5)Sut=0.5(615MPa)=307.5MPa. kf

is given by Eq. (7.21) as

k f = eSutf = 1.58( ) 615( )−0.085 = 0.91

ks is obtained from Equation (7.22) as ks=1.189d-0.112

,=1.189(30)-0.112

=0.81 and kr is obtained fromTable 7.4 as 0.90. Therefore the modified endurance limit is

Se = k f kskr ′ S e = 0.91( ) 0.81( ) 0.90( ) 307.5MPa( ) = 204MPaTherefore, Equation (11.34) gives:

ns =πd3Sy

32 Mm +Sy

SeK f Ma

2

+ Tm +Sy

SeK fsTa

2=

πd3Sy

32Sy

SeMa

2

+ Tm( )2

= π 0.030m( )3 380 MPa( )

32380 × 11.3Nm

204

2+ 57.3Nm( )2

= 16.5

11.12 The rotor shown in sketch d has a stiff bearing on the left. Find the critical speed whenthe shaft is made of steel with E=210GPa.

17

Note: This is a single mass system; the critical speed is given by Equation 11.50 as soon as theend deflection is determined. This is the most work in the problem; the approach is the same as inChapter 5, or else one can use the solution, readily available in any solid mechanics textbook, of

ymax = W6 EI

x2( ) 3l − x( )

Solution: This is a single mass system, and the critical speed is given by Equation (11.50) onpage 442 as

ω = gδ

δ is the maximum deflection of the beam, and for the system shown in the sketch, it is thedeflection caused by the 100kg mass (981N force). We need to derive the deflection of thecantilevered beam.

I. Calculation of Mass DeflectionThe moment as a function of x can be found from statics (and is left for the student to derive) as

M=W(l-x)Where W is the 981N load at the free end and l is the beam length (500mm or 0.5m). FromEquation (5.3) on page 183,

d2y

dx 2 = − MEI

= − W l − x( )EI

Integrating two successive times yields:dydx

= − WEI

lx − 12

x2

+ A

y = − WEI

12

lx2 − 16

x3

+ Ax + B

The boundary conditions for a clamped end are that the slope and the displacement are zero;therefore, A and B are zero. At x=l, which is the position of the mass and also of maximumdeflection,

y = − WEI

12

l3 − 16

l3

= − Wl3

3EIThe negative sign indicates the deflection is downwards; for determination of the critical

frequency the absolute deflection is used, so we use the value of δ=Wl3/3EI.

II. Determination of Critical FrequencyFrom Equation (11.50) and noting that W=mg and I=πd4/64,

ω = gδ

= 3gEI

Wl3= 3Eπd4

64ml3= 3 210GPa( )π 0.08m( )4

64 100kg( ) 0.5m( )3= 318rad / s

The shaft speed is then easily calculated asNa = 318rad / s( ) 60s / m i n( ) 1rev / 2πrad( ) = 3040rpm

11.13 Determine the critical speed in bending for the shaft assembly shown in sketch e. Themodulus of elasticity of the shaft is E=207GPa, its length l=350mm, its diameter d=8mm,and the rotor mass ma=2.3kg.

18

Notes: This is a single mass system; the critical speed is given by Equation 11.50 as soon as themass deflection is determined. This is the most work in the problem. This solution will usesingularity functions to get the deflection at the location of the mass, but other approaches can bepursued.

Solution: The left bearing is referred to as A, the right as B as shown in the figure. The reactionscan be found through statics as Ay=2W/3, By=W/3.

I. Determination of Mass DeflectionThe load intensity function can be written as (neglecting the force at B, since our interest onlyextends to the location of the mass):

q(x) = − 2W3

x −1 + W x − l3

−1

From Equation (2.4) on page 41,

V(x) = − q(x)dx−∞

x

∫ = 2W3

x 0 − W x − l3

0

+ C1

M(x) = V(x)dx−∞

x

∫ = 2W3

x 1 − W x − l3

1

+ C1x + C2

where C1 and C2 are constants of integration, They can be evaluated by noting that because of thesimple supports represented by the bearings, the moment is zero at x=0 (therefore C2=0) and thatthe moment is zero at x=l (therefore C1=0). Then applying Equation (5.3),

d2y

dx 2 = − MEI

= − 2W3EI

x 1 + WEI

x − l3

1

Integrating twice yields

y = − W9EI

x 3 + W6EI

x − l3

3

+ C3x + C4

Where C3 and C4 are constants. The boundary conditions are that y=0 at x=0 (yielding C4=0) andy=0 at x=l, yielding

0 = − W9EI

l3 + W6EI

23

l3

+ C3l ;C3 = 5Wl2

81EISubstituting this into the expression for deflection and evaluating this function at x=l/3 gives:

y x = l3

= δ = − W

9EIl3

3

+ 5Wl2

81EIl3

= − Wl3

243EI+ 5Wl3

243EI= 4Wl3

243EI

Now, since W=(2.3kg)(9.81m/s2)=22.6N, l=0.35m, E=207GPa amd I=πd

4/64 with d=0.008m, δ is

calculated as 0.000383m

19

II. Calculation of Critical FrequencyThe critical frequency is given by Equation (11.50):

ω = gδ

= 9.81m / s2

0.000383m= 160rad / s

The critical shaft speed is then 1530rpm.

11.14 Calculate the diameter of the shaft in the assembly shown in sketch f so that the firstcritical speed is 9000 rpm. The shaft is made of steel with E=207GPa. The distancea=300mm and the mass ma=100kg. The mass of the shaft is neglected. Use both theRayleigh and Dunkerley methods.

Notes: This is a multiple mass system, and the solution can be found from Equations (11.57) and(11.58). To use either of these equations, we must calculate deflections of the shaft under avariety of conditions. This is the lengthiest part of the problem.

Solution:I. Calculation of DeflectionsFirst of all, note that the loads are PA=200kg(9.81m/s2)=1960N, PB=981N, and I=πd

4/64. The

desired critical frequency is 9000rpm=942rad/s. This solution takes advantage of the deflectionequation given on page 193 combined with the principle of superposition. To differentiate effects,we will refer to deflections as "yij" where the deflection is at location i due to mass j. If we breakthe beam into two problems, where only one mass exists, we get the following:Deflections at A: Deflections at B:a) Use a=0.3m, b=0.6m, l=0.9m, P=1960N.Using x=l/3=0.3m gives the deflection under

gear A as yAA=2.31x10-9m

5/d

4.

b) Use a=0.6m, b=0.3m, l=0.9m, P=981N,

x=0.3m gives yAB=0.1013x10-9m

5/d

4. (Note that

the <x-a> term is zero)

a) use a=0.3, b=0.6, l=0.9, P=1960, x=0.6, so

that yBA=2.03x10-9m

5/d

4

b) Use a=0.6, b=0.3, l=0.9, P=981, x=0.6 so

that yBB=1.16x10-9m

5/d

4

Therefore, yA=yAA+yAB=3.33x10-9m

5/d

4 and yB=yBA+yBB=3.19x10

-9m

5/d

4.

II. Rayleigh MethodEquation (11.57) on page 442 gives the critical frequency as a function of load and deflections.For this case, Equation (11.57) becomes:

20

ωcr = 942rad / s =g WAδA + WBδB( )

WAδ A2 + WBδB

2

=9.81m / s2( ) 1960N( ) 3.33 ×10−9m5 / d4( )+ 981N( ) 3.19 ×10−9m 5 / d4( )[ ]

1960N( ) 3.33×10−9 m5 / d 4( )2+ 981N( ) 3.19 ×10−9m5 / d 4( )2

This is solved numerically as d=0.131m=131mm.

III. Dunkerly MethodEquation (11.58) and (11.50) combine to give the estimate of critical speed as:

1

ωcr2 =

1

ω12 +

1

ω 22 =

1

g /δ1( )2 +1

g /δ2( )2 =δ1 + δ 2

g;ωcr =

g

δ1 + δ2

Substituting for δ1 and δ2 gives d=0.133m=133mm.

11.15 The simple support shaft shown in sketch g has two weights on it. Neglect the shaft

weight. Using the Rayleigh method for the stainless steel (E=26 x 106 psi) shaft,

determine the safe diameter to ensure that the first critical speed is no less than 3600rpm.

Note: The approach is the same as for Problem 11.14.

Solution:I. Calculation of DeflectionsFirst of all, note that the loads are PA=300lb, PB=500lb, and I=πd

4/64. The desired critical

frequency is 3600rpm=377rad/s. This solution takes advantage of the deflection equation givenon page 193 combined with the principle of superposition. To differentiate effects, we will referto deflections as "yij" where the deflection is at location i due to mass j. If we break the beam intotwo problems, where only one mass exists, we get the following:Deflections at A: Deflections at B:a) Use a=8in, b=12in, l=20in, P=300lb. Usingx=8in gives the deflection under gear A as

yAA=0.036in5/d

4.

b) Use a=15in, b=5in, l=20in, P=500lb, x=8in

gives yAB=0.041in5/d

4.

a) use a=8, b=12, l=20, P=300, x=15, so that

yBA=0.024in5/d

4

b) Use a=15, b=5, l=20, P=500, x=15 so that

yBB=0.0367in5/d

4

Therefore, yA=yAA+yAB=0.077in5/d

4 and yB=yBA+yBB=0.0607in

5/d

4.

II. Rayleigh Method

21

Equation (11.57) on page 442 gives the critical frequency as a function of load and deflections.For this case, Equation (11.57) becomes:

ωcr = 377rad / s =g WAδA + WBδ B( )

WAδ A2 + WBδB

2

=386.4in / s2( ) 300 lb( ) 0.077in5 / d4( ) + 500lb( ) 0.0607in5 / d 4( )[ ]

300 lb( ) 0.077in5 / d4( )2+ 500 lb( ) 0.0607in5 / d4( )2

This is solved numerically as d=2.23in


1

ωcr2 =

1

ω12 +

1

ω 22 =

1

g /δ1( )2 +1

g /δ2( )2 =δ1 + δ 2

g;ωcr =

g

δ1 + δ2

Substituting for δ1 and δ2 gives d=2.26in.

11.16 Determine the first critical speed by the Dunkerly and Rayleigh methods for the steelshaft shown in sketch h. Neglect the shaft mass. The moment of inertia is I=πr4/4, wherer is the shaft radius. The method of superposition may be used with the following given:

δ = P6 EI

bx3

l− x − a 3 −

xb l2 − b2( )l

Notes: The approach is the same as for Problem 11.14

Solution:I. Calculation of DeflectionsFirst of all, note that the loads are PA=120lb, PB=80lb, and I=πd

4/64=πr4/4in

4. To differentiate

effects of the two loads, we will refer to deflections as "yij" where the deflection is at location idue to mass j. If we break the beam into two problems, where only one mass exists, we get thefollowing:Deflections at A: Deflections at B:a) Use a=20in, b=70in, l=90in, P=120lb. Usingx=20in gives the deflection under gear A asyAA=0.0367in

b) Use a=60in, b=30in, l=90in, P=80lb, x=20ingives yAB=0.0256in

a) use a=20, b=70, l=90, P=120, x=60, so thatyBA=0.0385in

b) Use a=60, b=30, l=90, P=80, x=60 so thatyBB=0.0407in

Therefore, yA=yAA+yAB=0.0626in and yB=yBA+yBB=0.0792in

22


ωcr =g WAδA + WBδ B( )

WAδ A2 + WBδ B

2 =386.4in / s2( ) 120 lb( ) 0.0626in( )+ 80lb( ) 0.0792in( )[ ]

120lb( ) 0.0626in( )2 + 500lb( ) 0.0792in( )2

This is solved as ωcr=74.2rad/s=708rpm.


1

ωcr2 =

1

ω12 +

1

ω 22 =

1

g /δ1( )2 +1

g /δ2( )2 =δ1 + δ 2

g;ωcr =

g

δ1 + δ2

Substituting for δ1 and δ2 gives ωcr=70.52rad/s=673rpm.

11.17 Calculate the diameter d of the shaft in sketch f so that the lowest critical speed becomes9000rpm. The modulus of elasticity E=107GPa, the distance a=300mm, and the massma=100kg. Neglect the mass of the shaft. Use the Rayleigh and Dunkerly methods.

Notes: The approach is the same as for Problem 11.14

Solution:I. Calculation of DeflectionsThe only change from Problem 11.14 is that the stiffness has changed from 207GPa to 107 GPa.Since the deflections are linearly related to the stiffness, we can directly write the deflectionsfrom the data in problem 11.14 as:

yA=(3.33x10-9m

5/d

4)(207/107)=6.44x10

-9m

5/d

4

yB=(3.19x10-9m

5/d

4)(207/107)=6.17x10

-9m

5/d

4

(The alternative is to repeat the approach in problem 11.14)


ωcr = 942rad / s =g WAδA + WBδ B( )

WAδ A2 + WBδB

2

=9.81m / s2( ) 1960N( ) 6.44 ×10−9m5 / d4( ) + 981N( ) 6.17 ×10−9m5 / d4( )[ ]

1960N( ) 6.44 ×10−9 m5 / d 4( )2+ 981N( ) 6.17 × 10−9 m5 / d4( )2

This is solved numerically as d=0.155m=155mm.


1

ωcr2 =

1

ω12 +

1

ω 22 =

1

g /δ1( )2 +1

g /δ2( )2 =δ1 + δ 2

g;ωcr =

g

δ1 + δ2

Substituting for δ1 and δ2 gives d=0.156m=156mm.

23

11.18 The rotor in sketch i has a shaft diameter of 32 mm, and the disk mass is 170kg. Calculatethe critical speed if the spring rates are 1668 and 3335 N/m, and are the same in alldirections. The elastic modulus E=206GPa. For a shaft without springs, the influence

coefficient α11=l3/6EI.

Solution:The centrifugal force is

P=mayωcr2

The deformation is y=αP, so ωcr2=1/(maα)

The deformation comes from the springs and shaft in bending

α = 14C1

+ 14C2

+ l3 64( )6 Eπd4

Therefore,

ωcr2 =

1

170kg( ) 0.16m( )3 64( )6 206GPa( )π 0.032m( )4 +

1

4 1.668MN / m( )+

1

4 3.335MN / m( )

= 2 0 , 3 3 8 /s2

or ωcr=143rad/s=1360rpm.

11.19 Calculate the critical speed for a rotor shown in sketch j that has two moment-freebearings. The shaft has a diameter of 20mm and is made of steel with E=206GPa.

Notes: The difficult part of this problem is obtaining the deflection at the mass. This can be doneusing the approach in Chapter 4, or else a solution can be obtained from a solid mechanics text orstress analysis handbook. The deflection at the mass is given by

δ = 2Pl3

3EIand this will be used in the solution below.

Solution: From Equation (11.50), the critical speed is given by

ω = gδ

= g 3EI( )2 Pl3

= 3EI

2mal3=

3 206GPa( ) π64

0.020m( )4

2 7kg( ) 0.2m( )3= 208rad / s

24

where the equation for deflection given above has been used, as has been the substitution P=mag.The critical speed of 208 rad/s can also be expressed as 1990rpm.

11.20 A flywheel has a hub made of aluminum alloy 2014. The hub is connected to the AISI1040 steel shaft with a flat, 15-mm long key made of AISI 1020 steel. Dimension the keyso that it can transmit 20Nm of torque with a safety factor of 3.

Notes: Often, the key is sized arbitrarily given the torque that must be transmitted. This problemintroduces an additional constraint, namely that a soft hub is used, so the key must not causefailure of the hub. One can foresee this occurring if a very short key is used - the bearing stresseswould cause the hub to plastically deform. We will also assume that the same safety factor is usedfor the hub and the key, which is not normally the case. Usually, the hub uses a much highersafety factor to ensure that no damage will occur in the case of excessive torque. The point of akey is to protect the expensive components, which will not occur if it fails at the same time asthese components. However, the solution scheme for a different safety factor is straightforwardafter doing this problem.

Solution: First we obtain the material properties from the inside front cover as Sy,steel=295MPa,Sy,al=97MPa.

I. Analysis of keyThe key shear criterion is given by Equations 11.60 and 11.61, which when combined give:

τ = 2Tdwl

=Ssy

ns=

0.40Sy

ns;dw = 2Tns

0.40dlSy= 2 20Nm( ) 3( )

0.40 0.015m( ) 295MPa( ) = 6.78 ×10−5m2

II. Analysis of hubThe bearing stress on the softer hub must be considered. Equations 11.63 and 11.64 give

σ = 4Tdlh

=0.90Sy

ns;dh = 4Tns

0.90lSy= 4 20Nm( ) 3( )

0.90 0.015m( ) 97MPa( ) =1.83 ×10−4m2

This gives the dimensions of the key as a function of shaft diameter. If the diameter of the shaft is20 mm, then the key dimensions are w=3.39mm and h=9.16mm

11.21 A jaw crusher is used to crush iron ore down to the particle size needed for iron ore pelletproduction. The pellets are later used in a blast furnace to make steel. If the iron orepieces coming into the crusher are too large, the crusher is protected by a torque-limitingkey made of copper. The torque on the 400-mm diameter shaft connecting the flywheelwith the jaw mechanism should never go over 700kN-m. Dimension the copper key sothat the shaft is not damaged when an oversized ore particle comes into the crusher.

Notes: This is an open ended problem, since the shear pin will ultimately require an area basedon the material’s ultimate strength. Good design practice would be to use the same length of pinas the hub on which it is attached - this minimizes the bearing stresses on the hub for a given pinheight.

Solution:Note that for shear pin failure, the ultimate strength should be used, which is 220 MPa for copper(see inside cover of text). The shear force is given by τwl=0.5Suwl (using the maximum shearstress theory). The torque developed is T=0.5Suwlr=700kN-m. The area of the key is then

25

wl =700kN − m

0.5 220MPa( ) 0.2m( ) = 0.032m2

Any combination which gives this area will work; arbitrarily setting w=d/4=100mm gives l=320mm. Note: if the distortion energy theory is used, l=275mm.

11.22 The output torque of a flywheel for each revolution of a shaft is 10 Nm from 0 to π and120Nm from π to 2π. The coefficient of fluctuation is 0.04 at an average speed of 2000rpm. Assume that the flywheel disk is an AISI 1040 steel plate of 30-mm constantthickness. Determine the following:

a) Average load torque.b) Locations θωmax and θωmin

c) Energy requiredd) Outside diameter of flywheel

Notes: This problem is straightforward and is very similar to Example 11.7 on page 451.The plot of torque as a function of phase from 0 to 2π is:

Since the intervals are equal, Tave is easily calculated as 65Nm. If this torque is applied, the shaftwill accelerate over the 0 to π interval, and decelerate between π and 2π. Therefore, the maximumangular velocity will be at π, the minimum at 2π. The energy needed is

Ke = T − Tave( )dφπ

2π

∫ = 55πNm

The mass moment of inertia required is given by Equation (11.72):

Im =Ke

C f ωave2 =

55πNm

0.04( ) 2000 ×2π60

2

sec−2= 0.098Nmsec2 = 0.098kgm2

If the flywheel is a disk of thickness 0.030m, and recognizing for steel ρ=7850kg/m3,

Im = 0.098kgm2 =πd2tm ρ

4

d2

8

;d = 0.255m = 255mm

11.23 The output torque of a flywheel for each revolution of a shaft is 200 in-lbf from 0 to π/3,1600 in-lbf from π/3 to 2π/3 , 400in-lbf from 2π/3 to π, 900 in-lbf from π to 5π/3, and200in-lbf from 5π/3 to 2π. Input torque is assumed to be constant. The average speed is860rpm. The coefficient of fluctuation is 0.10. Find the diameter of the flywheel if it iscut from a 1-in thick steel plate.

26

Notes: This problem is relatively straightforward. The steps on pages 450-451 can serve as auseful guide to solving this problem. A useful material constant which is needed is the density of

steel, which from the inside front cover is ρ=7840kg/m3=0.28lb/in

3.

Solution:1. Plot of load torque Tl versus θ for one cycle:The plot is given below. To aide in the determination of the average torque, the areas of eachsection have been marked in the curve.

2. Determination of average torqueThe average torque is the total area under the curve divided by the base length (2π). Therefore theaverage torque is 700inlb.3. From the plot of torque versus angular position, the critical area seems to be the sectionof high torque between π/3 and 5π/3. This could be confirmed by considering all possiblesections and taking the one which maximizes the kinetic energy change, but intuition in this caseis correct. Therefore, we will evaluate the kinetic energy change between π/3 and 5π/3.4. The kinetic energy change over this interval is

Ke = Tl − Tave( )dθ∫ = 1600 − 700( )inlbπ3

+ 400 − 700( )inlbπ3

+ 900 − 700( )inlb2π3

= 1000π3

inlb

5. The average speed is given as ωave=860rpm=90.1rad/s.6. The coefficient of fluctuation is given as 0.10. The required mass moment of inertia is,from Equation (11.72)

Im =Ke

C f ωave2 =

1000πinlb

3 0.10( ) 90.1rad / s( )2 = 0.411πinlbs2

And, from Table 4.2 on page 150 for a disk with a thickness of 1 in,

Im = 0.411πinlbs2 =mad2

8=

ρ π4

d2 t

d 2

8=

ρπd4 t

32Solving for d,

d4 =32 0.410πinlbs2( )

ρπt=

32 0.410πinlbs2( ) 386lbm

lbs2 / in

0.28lbm / in3( ) 1in( )=18128in4

or d=11.6 in. A value of 12 inches would be a good design designation.

11.24 A 20-mm thick flywheel is made of aluminum alloy 2014 and runs at 9000 rpm in aracing car motor. What is the safety factor if the aluminum is stressed to a quarter of its

27

yield limit at 9000rpm? To decrease the flywheel outer diameter, highly dense exoticmaterials are investigated. Find the best material from Table A.2 that can be substitutedfor aluminum alloy 2014 but with the same safety factor. The flywheel is machined froma solid piece of aluminum alloy 2014 with no central hole, and the thickness cannot belarger than 20mm.

Notes: To minimize the outer diameter while maintaining the same thickness, one has to obtain adenser material. However, given a material, there is an additional requirement that maintain asafety factor of 4 at this speed. Therefore, we will calculate the stress as a function of radius forthe materials in Table A.2 in order of decreasing density until a candidate is found which gives animproved and safe design.

Solution:First of all, note that ω=9000rpm=942rad/s.I. Aluminum flywheelWe note from the text on page 452 that σθ is maximum at r=ri. With ri=r=0, Equations (11.75)and (11.76) become:

σθ = 3+ ν( )8

ρω2ro2

σr = 3+ ν( )8

ρω2ro2

Since these are principal stresses, Equation (6.10) on page 238 gives

σe = σθ2 +σ r

2 − σθσ r( )1 / 2=

3 + ν( )8

ρω2ro2

We can get the material properties for aluminum 2014 from the inside front cover as ν=0.33 and

ρ=2710kg/m3 and Sy=97MPa. Since σe=Sy/4, we can obtain ro as

ro2 =

2Sy

3 +ν( )ρω 2 = 2 97MPa( )3.33( ) 2710kg / m3( ) 942rad / s( )2 ;ro = 0.155m =155mm

or a flywheel diameter of 310mm. The mass moment of inertia is

Im = mad2

8= πd4 tmρ

32=

π 0.31m( )4 0.020m( ) 2710kg /m3( )32

= 0.049kgm2

II. Silver flywheel.Silver is a possible material in Table A.2 because of its high density. The important properties are

ρ=10490kg/m3, ν=0.37, Sy=55MPa. To get a mass moment of inertia of 0.049kgm

2, the diameter

must be

0.049kgm2 = πd 4tmρ32

;d4 =32 0.049kgm2( )

π 0.02m( ) 10490kg / m3( ) ;d = 0.221m

The stress in the silver is given by

σe = 3 + ν( )8

ρω2ro2 = 3+ 0.37( )

810490kg( ) 942rad / s( )2 0.1105m( )2 = 47.9MPa

Since this is more than one-fourth the yield strength, silver cannot be substituted for thealuminum.

III. Molybdenum flywheel

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Molybdenum is the next best material, with ρ=10220kg/m3, ν=0.3, Sy=565MPa. The diameter is

calculated as before as 0.223m, and the resulting stress is 41.7MPa. Since this is less than one-fourth the yield strength, molydbenum will work.

11.25 The aluminum flywheel considered in Problem 11.24 has a mass moment of inertia of

0.0025 kgm2. By mistake the motor is accelerated to 7000 rpm in neutral gear, and the

throttle sticks in the fully open position. The only way to stop the motor is to disconnectthe electric lead from the spark plug. It takes 6s before the motor is motionless.Neglecting the inertia of all movable parts in the motor except the flywheel, calculate theinternal friction moment in the motor and the (mechanical part of the) friction losses inthe motor at 5000 rpm. The friction moment is assumed to be constant at all speeds.

Notes: This problem only needs Equation (11.66) for a solution. This problem is actually anadvanced dynamics problem, where a constant torque causes a flywheel to stop from 7000rpm insix seconds.

Solution:Note that 7000rpm=733rad/s and 5000rpm=524rad/s. Equation (11.66) gives:

Tl − Tm = Imdωdt

;Tl − Tm

Imdt = dω

Here we set Tl to the frictional torque, Tm is the mean torque applied by the motor, which is zeroonce the spark plug is disengaged. Tl and Im are constants, so this equation can be integrateddirectly:

Tl

Imdt

0

6s

∫ = dω733rad /s

0

∫ ;Tl

0.0025kgm2( ) 6s( ) = −733rad / s;Tl = −0.305Nm

The negative sign indicates that it causes decelleration. The friction losses are the product oftorque and angular velocity; since the friction torque is taken as constant, this is:

hp = Tlω = 0.305Nm( ) 524rad / s( ) =160W

11.26 A one-cylinder ignition bulb motor to an old fishing boat has a flywheel that gives themotor a coefficient of fluctuation of 25% when it idles at 180 rpm. The mass moment ofinertia for the flywheel is 1.9kg-m

2. Determine the coefficient of fluctuation at 500 rpm if

the compression stroke consumes equally large energy at all speeds. Also, calculate themass moment of inertia needed to get 20% coefficient of fluctuation at 500 rpm.

Solution:First of all, at idle speed, ω=(180rpm)(2π/60)=18.8 rad/sec. If the energy is constant, fromEquation (11.71),

Ke = ImC f ωave2 = Const.; ImC f1ωave1

2 = ImC f 2ωave22

C f 2 = C f1ωave1

2

ωave22 = 0.25( ) 1802

5002

= 0.032

The kinetic energy is (1.9kgm2)(0.25)(18.8sec-1)2=169Nm. For Cf=0.2 at 500 rpm,

169Nm = Im 0.2( ) 500× 2π60

2

; Im = 0.31kgm2

29

11.27 A flywheel for a city bus drive should store as much energy as possible for a givenflywheel weight. The diameter must be less than 1.5 m, and the mass must be smallerthan 250kg. Find which material from Tables A.1 and A.2 gives the highest possiblestored energy for a safety factor of 4 if the flywheel has constant thickness.

Notes: Equation (11.82) gives the requirement for the best performing flywheels. This equationcan be derived using an approach as in problem 11.24. However, this problem merely requiresevaluation of the performance index for the materials in Tables A.1 and A.2.

Solution:The following values are obtained for the materials in Tables A.1 and A.2:Material σmax=Sy/4, MPa ρ, kg/m

3 Mf=σmax/ρIronGray cast ironNodular cast ironMalleable cast ironAISI 1020 steelAISI 1040 steelAISI 1080 steelType 446 stainlessType 316 stainlessType 410 stainless

32.531.2568.75

5573.7587.595

86.2551.7568.75

7870715071207200786078507840750080007800

0.004130.004370.009660.007640.009380.01110.01210.0115

0.006470.00881

Aluminum (>99.5%)2014 AlCopper (99.95%)Brass (70Cu-30Zn)Bronze (92Cu-8Sn)MagnesiumMolybdenumNickelSilverTitanium

4.2524.2517.2518.75

3810.25

141.2534.513.75

60

27102800894085308800174010208900104904510

0.001570.008660.001930.002200.004320.005890.138

0.003880.001310.0133

The maximum value occurs with Molybdenum.

11.28 A flywheel on an AISI 1080 steel shaft is oscillating due to disturbances from acombustion engine. The engine is a four-stroke engine with six cylinders, so that thetorque disturbances come three times per revolution. The flywheel has a shaft diameter of

20mm and is 1 m long, and the flywheel moment of inertia is 0.5 kg-m2. Find the engine

speed at which the large torsional vibrations will appear.

Notes: This problem requires some knowledge of differential equations. The approach is to applyEquations (11.66) and (11.35) to get a differential equation in terms of angular deflection. Thesolution of this equation is sinusoidal with a characteristic frequency which yields the solution.

Solution:From Equation (11.66) and noting that dθ/dt=ω,

Tl − Tm = Imdωdt

= Imd2θdt2

30

Where θ is the angular deflection. We take θ as zero at Tm, so that Tm is our reference and is setequal to zero. Our concern is finding what frequencies yield the extreme value of θ. Notehowever, that T is related to θ by equation (4.35) on page 155 as

T = JGθl

so combining these two equations gives a differential equation for θ as:

T = JGθl

= Imd2θdt 2 ;

d2θdt2 − GJ

Im lθ = 0

this has a solutionθ=Asinωet+Bcosωet

where

ωe =GJ

Iml=

79.62GPa( ) π32

0.02m( )4

0.5kgm2( ) 1m( )= 50.0rad / s = 478rpm

but if we take θ=0 at t=0 then B must be zero. If A is zero, there are no vibrations, which weknow is not true. Therefore, A must be non-zero and the shaft vibrates at the critical frequency of478 rpm. Note though that the engine applies torque three times per engine revolution; thereforethe critical engine rpm is one-third this answer or 160 rpm.

ch_11

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