Solution to Problem 1.

Review. The free body diagram, shear force diagram, bending moment diagram, and torque
diagram of the shaft are shown in Figures 1, 2, 3, and 4, respectively.

FA = 225 N FB = 275 N
T = 150 N.m

O C
A E B

RO = 235.715 N RC = 264.285 N
Figure 1. The free body diagram of the shaft.
Y 264.285 N

A
O Z
- 10.715 N B C
- 235.715 N
V (N)
Figure 2. The shear force diagram of the shaft.

A E B C
O Z

-14.143 N.m
M (N.m) -15 N.m -15.857 N.m
Figure 3. The bending moment diagram of the shaft.
T (N.m)

150 N.m

Z
Figure 4. The torque diagram of the shaft.

From the sum of the moments about the bearing O and the sum of the forces in the Y-direction,
the reaction forces at bearings C and O are
225 x 60 + 275 x 220
RC = = 264.285 N and R O = 500 − 264.285 = 235.715 N (1)
280

1
(i) The shaft diameter at the groove can be written from the Goodman criterion, see Eq. (7.8), page
382, as
1/3
 16 n f A B 
d = ( + ) (2)
 π S e S ut 
The coefficients in Eq. (2) are given by Eq. (7.6), see page 381, namely

( ) ( )
2 2
A = 4 K f Ma + 3 K fs Ta (3a)
and
B = 4 ( K f M m ) + 3 ( K fs Tm )
2 2
(3b)

The rotating shaft is subjected to fully reversed bending, therefore, the mean component of the
bending moment at the mid-point of the shaft, point E, is
Mm = 0 (4)

The alternating component of the bending moment acting at the mid-point of the shaft, point E,
can be written as
M a = − OE x R O + AE x FA (5a)

Substituting Eq. (1) into Eq. (5a), the alternating component of the bending moment is

M a = −140 x 235.715 + 80 x 225 = −15000 N.mm = 15 N.m (5b)

The mean and alternating components of the constant torque, see Figure 4, are
Tm = 150 N.m and Ta = 0 (6)

The fatigue stress concentration factors for the normal stress and the shear stress are

K f = 2.25 and K fs = 1.95 (7)

Substituting Eqs. (4), (5), (6), and (7) into Eqs. (3), the coefficients are

A = 4 ( 2.25 x15 ) + 3 (1.95 x 0 ) = 67.5 N.m

2 2
(8a)
and
B = 4 ( 2.25 x 0 ) + 3 (1.95 x 150 ) = 506.625 N.m
2 2
(8b)

Substituting n f = 3 , Se = 125 MPa, Sut = 335 MPa, and Eqs. (8) into Eq. (2), the diameter of the
shaft at the groove can be written as
1/3
 16 x 3 67.5 506.625 
d = ( + )
6 
m (9a)
 π 125 x10 335 x10 
6

2
that is, the diameter of the shaft at the groove can be written as
1/3 1/3
 16 x 3   16 x 3 x 2.052 
d = (0.54 + 1.512)  =  m (9b)
 π x10  π x10
6 6
 

Therefore, the diameter of the shaft at the groove, see Figure 1, is

d = 31.53 mm (10)

Check: The diameter of the shaft at the groove can also be written from Eq. (2) as
1/3
 16 n f  1  2 1/2 1   
2 1/2
4 ( K f M a ) + 3 ( K fsTa )  + 4 ( K M ) + 3 ( K T )
2 2
d =   (11)
 π      
 Se   f m fs m
 Sut

Substituting the infinite life fatigue factor of safety, the endurance limit, the ultimate tensile
strength, and Eqs. (4), (5), and (6) into Eq. (11) the diameter of the shaft at the groove is
d = 31.53 mm (12)

Note that this answer is in complete agreement with Eq. (10).

(ii) The von Mises mean stress for critical element at E can be written from Eq. (7.4) see page 381,
as
1/ 2
σ m′ =  σ m 2 + 3τm 2  (13a)

and the von Mises alternating stress can be written from Eq. (7.5) see page 381, as
1/ 2
σ a′ =  σ a 2 + 3τa 2  (13b)

The mean component of the normal stress and the alternating component of the shear stress,
respectively, are
σm = 0 and τa = 0 (14a)

The alternating component of the normal stress is

32 K f M a 32(2.25)(15000)
σa = = = 10.967 MPa (14b)
πd3 π (31.53)3
The mean component of the shear stress is
16 K fsTm 16(1.95)(150000)
τm = = = 47.525 MPa (14c)
πd3 π (31.53)3
Substituting Eqs. (14a) and (14c) into Eq. (13a), the von Mises mean stress for the critical element
at E is
1/ 2
σ m′ =  0 + 3 x 47.5252  = 82.316 MPa (15a)

3
Substituting Eqs. (14a) and (14b) into Eq. (13b), the von Mises alternating stress at E is
1/ 2
σ a′ = 10.967 2 + 0  = 10.967 MPa (15b)

The maximum von Mises stress at E can be written from Eq. (7.15), see page 382, as
1/ 2 1/2
′ = σ max
σ max 2
+ 3 τ2max  = (σ m + σ a ) 2 + 3( τm + τa ) 2  (16a)
or as
1/ 2
 32 K (M m + M a )  2  16 K fs (Tm + Ta )  
2
f
′
σ max =   + 3   (16b)
 πd3 


 π d 3  
 

Substituting Eqs. (4) and (5) into Eq. (16b), the maximum von Mises stress at E is
1/ 2
′ = 10.967 2 + 3(47.525) 2 
σ max = 83.043 MPa (17)

(iii) The factor of safety guarding against first-cycle yielding for the critical element at E using the
Langer line, see Eq. (7.16), page 382, can be written as
S yt S yc
ny = or as ny = (18)
′
σ max ′
σ max

Since the tensile yield strength is less than the compressive yield strength then the static factor of
safety for the critical element guarding against yielding is given by the first equation of Eq. (18).
Substituting the tensile yield strength S yt = 190 MPa and Eq. (17) into Eq. (18), the factor of
safety guarding against first-cycle yielding for the critical element at E is
190
ny = = 2.28 (19)
83.043
Using the conservative approach, see page 383. The factor of safety guarding against first-cycle
yielding for the critical element at E can be written as
S yt
ny = (20)
σ m′ + σ a′

Substituting the tensile yield strength and Eqs. (15) into Eq. (20), the factor of safety guarding
against first-cycle yielding for the critical element at E is
190 190
ny = = = 2.03 (21)
82.316 + 10.967 93.283
Note that this answer given is, indeed, more conservative than the answer given by Eq. (19).

4
Solution to Problem 2.
(i) For the given position, the cam and the follower can be modeled as two cylinders. Therefore,
the half-width of the contact patch can be written from Eq. (3-73), see page 148, as

 (1 − ν12 ) (1 − ν 2 2 ) 
+
 2 F   E1 E2 

b=   (1)
 π l  1 1 
 + 
 d1 d 2 

The subscript 1 is used here to denote the cam and the subscript 2 is used to denote the follower.
Since the follower is flat-faced then the diameter of the follower is infinite, that is

d2 = ∞ (2)

The elastic material properties of the carbon steel cam, see Table A-5, page 1023, are

E1 = 207 GPa and ν1 = 0.292 (3a)

The elastic material properties of the titanium alloy follower, see Table A-5, page 1023, are

E2 = 114 GPa and ν 2 = 0.340 (3b)

Substituting and the given geometry, the force F = 8000 N, and Eqs. (2) and (3) into Eq. (1), the
half-width of the contact patch can be written as
1/2
  2 × 8000   (1 − 0.2922 ) / (207 ×103 ) + (1 − 0.3402 ) / (114 ×103 )  
b =     mm (4a)
  π × 40  1/100 + 1/ ∞ 

Therefore, the half-width of the contact patch is

b = 0.394 mm (4b)

(ii) The maximum pressure on the contact patch can be written from Eq. (3-74), see page 148, as
 2  F 
pmax =     (5a)
 π  bl 

Substituting the given geometry, the force F = 8000 N, and Eq. (4b) into Eq. (5a), the maximum
pressure acting on the contact patch is

2 × 8000
pmax = = 323.157 MPa (5b)
π × 0.394 × 40

5
(iii) The x, y, and z components of the normal stress acting on the element of the contact patch on
the Z-axis can be written from Eqs. (3-75), (3-76), and (3-77), see page 148, as

 z
2
z 
σ x = −2ν pmax  1 +   −  (6a)
 b b


 2  
  1 + 2  z   
 b −2 z 
σ y = − pmax    (6b)
2  b
 z 
 1 +  b   
    
and
pmax
σZ = − (6c)
2
z
1+  
b

(a) Substituting Z = 0 , b = 0.394 mm, Poisson’s ratio for the cam, that is, and ν = ν 1 = 0.292, and
Eq. (5b), into Eqs. (6a), the x component of the normal stress on element O of the cam on is

σ x = − 2 × 0.292 × 323.157(1 − 0) = −188.724 MPa (7)

Substituting Z = 0 , b = 0.394 mm, and Eq. (5b), into Eqs. (6b) and (6c), the y and z components
of the normal stress acting on element O of the cam are

 2  
 1 + 2 ( 0)  
σ y = − 323.157  − 2(0)  = −323.157 MPa (8a)
  1 + ( 0 ) 
2

and
323.157
σz = − = −323.157 MPa (8b)
1 + (0)
2

The principal normal stresses acting on element O of the cam, written in ordered form, are

σ1 ≥ σ 2 ≥ σ 3 (9)

Therefore, from Eqs. (7) and (9), the maximum principal normal stress acting on element O of the
cam is
σ 1 = σ x = −188.724 MPa (10a)

and the minimum principal normal stresses acting on element O of the cam is

6
 σ 2 = σ 3 = σ y = σ z = − 323.157 MPa (10b)

Check. From Eqs. (5b), (10a), and (10b), the ratios of the principal normal stresses and the
maximum pressure are
σ1 σ2 σ3
= − 0.584 and = = −1 (11)
pmax pmax pmax

The maximum shear stress acting on the element O of the cam on the Z-axis, see Eq. (3.72),
page 147, can be written as
σ −σ
τ max = 1 3 (12a)
2
Substituting Eqs. (10a) and (10b) into Eq. (12a), the maximum shear stress acting on element O of
the cam on the Z-axis is
− 188.724 − (− 323.157)
τ max = = 67.217 MPa (12b)
2

(b) Substituting Z = 0 , b = 0.394 mm, ν = ν 2 = 0.340, and Eq. (5b) into Eq. (6a), the x component
of the normal stress acting on element O of the follower is

σ x = − 2 × 0.340 × 323.157 (1 − 0) = − 219.747 MPa (13a)

Substituting Z = 0 , b = 0.394 mm, and Eq. (5b) into Eqs. (6b) and (6c), the y and z components
of the normal stress acting on element O of the follower are the same as Eqs. (8a) and (8b), that is

σ y = −323.157 MPa and σ z = −323.157 MPa (13b)

Therefore, the maximum principal normal stresses acting on element O of the follower is

σ 1 = σ x = − 219.747 MPa (14a)

and the minimum principal normal stresses acting on element O of the follower on the Z-axis are

σ 2 = σ 3 = σ y = σ z = − 323.157 MPa (14b)

Check: Substituting Eqs. (14a) and (14b) into Eq. (12a), the maximum shear stress acting on
element O of the follower is
− 219.747 − ( − 323.157)
τ max = = 51.705 MPa (15)
2
Equations (12b) and (15) show that the maximum shear stress at the surface of the cam (at point
O) is greater than the maximum shear stress at the surface of the follower (at point O). The reason
is that Poisson’s ratio for the cam material is less than Poisson’s ratio for the follower material.

7
(iv) Since the depth OB = 0.25 mm does not correspond to the location of the maximum shear
stress, that is, Z = 0.786 b = 0.310 mm, see page 148, then Figure 3-40, see page 149, cannot be
used in the solution to this part of the problem.
Substituting Z = OB = 0.25 mm , b = 0.394 mm, and ν = ν 1 = 0.292 into Eq. (7a), the X-
component of the normal stress on element B is

  0.25 
2
0.25 
σ x = −2 × 0.292 × 323.157  1 +   −  = − 103.761MPa (16a)
  0.394  0.394 

Substituting Z = 0.25 mm into Eq. (7b), the Y-component of the normal stress on element B is

 2  
  1 + 2  0.25   
  0.394   − 2 0.25  = − 82.482 MPa
σ y = − 323.157    (16b)
2  0.394 
  0.25  
  1 +  0.394   
    

Substituting Z = 0.25 mm into Eq. (7c), the Z-component of the normal stress on element B is

323.157
σz = − = − 272.863MPa (16c)
2
 0.25 
1+  
 0.394 

Again, the principal normal stresses on element B are written in ordered form, see Eq. (9), that is,
σ1 ≥ σ 2 ≥ σ 3 . Therefore, the minimum principal normal stress on element B is

σ 3 = σ z = − 272.863MPa (17a)

the maximum principal normal stress on element B is

σ 1 = σ y = − 82.482 MPa (17b)

and the third principal normal stress acting on element B on the Z-axis is

σ 2 = σ x = −103.760 MPa (17c)

Substituting Eqs. (17a) and (17b) into Eq. (12a), the maximum shear stress acting on element B is

− 82.482 − ( − 272.863)
τ max = = 95.191MPa (18)
2
Comparing Eq. (18) with Eq. (12b), the observation is that the maximum shear stress acting on
element B of the cam is greater than the maximum shear stress acting on element O of the cam.
This agrees with the Hertzian theory of contact stress.

8
Solution to Problem 3.
(i) The Sommerfeld number from Eq. (12-10), see page 632, can be written as
2
 r  μN
S =  (1)
c P

The absolute viscosity of the SAE 50 lubricant at an operating temperature of 100°F, see Figure
12-2, page 628, is
μ = 40 μreyn (2)

The nominal bearing pressure (that is, the load acting on the projected area of the bearing), see Eq.
(12.7), page 630, can be written as
W
P= (3a)
2rl

Substituting the load W = 400 lbs , the radius of the journal r = 2 in, and the length of the bearing
l = 5 in into Eq. (3a), the nominal bearing pressure is
400 lb
P= = 20 psi (3b)
2 × 2 in × 5 in

The speed of the journal specified in the problem statement is 1200 rpm, that is
1200
N= = 20 rev / s (4)
60
Substituting r = 2 in, the radial clearance c = 0.02 in, and Eqs. (2), (3), and (4), into Eq. (1),
the Sommerfeld number can be written as
2
 2in   (40×10 reyn) × (20 rev/s) 
-6
S=   (5a)
 0.02 in   20 psi 
Therefore, the Sommerfeld number is
S = 0.400 (5b)
(ii) The friction torque acting on the journal can be written from Eq. (12.6), see page 629, as
4π 2 r3 l μ N
T= (6a)
c
Substituting the given geometry and Eqs. (2) and (4) into Eq. (6a), the friction torque acting on the
journal is
4 π 2 x 23 x 5 x 40 x10−6 x 20
T= = 63.165 lbs.in (6b)
0.02

Check: The friction torque acting on the journal can be written from Eq. (12.8), see page 630, as

9
 T = 2 r2 f l P = f W r (7)

The coefficient of friction variable specified in the problem statement (based on Figure 12.17, see
page 642) is
r
f = 7.7 (8a)
c
Substituting r = 2 in and c = 0.02 in into Eq. (8a), and rearranging, the coefficient of friction is
0.02
f = 7.7 × = 0.077 (8b)
2
Check: Substituting r = 2 in, W = 400 lbs, and Eq. (8b) into Eq. (7), the friction torque acting on
the journal is
T = 0.077 × 400 × 2 = 61.60 lbs.ins (9a)

This answer is in good agreement with Eq. (6b). Also, the coefficient of friction can be written
from Eq. (12.9), see page 631, as
μN r
f = 2π 2 ( )( ) (9b)
P c
Substituting r = 2 in, c = 0.02 in, and Eqs. (2), (3b), and (4) into Eq. (9b), the coefficient of
friction is
40 × 10−6 × 20 2
f = 2π 2 ( )( ) = 0.079 (9c)
20 0.02
Note this answer is in good agreement with Eq. (8b). However, this answer is based on Petroff’s
equation and does not agree with the given problem statement.
(iii) The thermal energy loss at steady state (that is, the heat rate) can be written from page 648 as
2π N T
H Loss = (10a)
J
where for common petroleum lubricants, the Joulean heat equivalent, see page 648, is
J = 778 ft.lb / Btu = 9336 in.lb / Btu (10b)

Substituting Eqs. (4), (6b), and (10b) into Eq. (10a), the thermal energy loss at steady state is
2 π × 20 × 63.165
H Loss = = 0.850 Btu (11a)
9336
Substituting Eqs. (4), (9a), and (10b) into Eq. (10a), the thermal energy loss at steady state is
2 π × 20 × 61.60
H Loss = = 0.829 Btu (11b)
9336
Check: The thermal energy loss at steady state can also be written from Eq. (b) on page 648 as

10
  4 π P r l N c  f r 
H Loss =    (12a)
 J  c 
Substituting Eqs. (3b), (4), (9c), and (10b) into Eq. (12a), the thermal energy loss at steady state
is
4 π × 20 × 2 × 5 × 20 × 0.02 0.0079 × 2
H Loss = ( ) = 0.851 Btu (12b)
9336 0 .0 2
Substituting Eqs. (3b), (4), and (10b) and the flow variable specified in the problem statement into
Eq. (12a), the thermal energy loss at steady state is
4 π × 20 × 2 × 5 × 20 × 0.02
H Loss = (7.7) = 0.8295 Btu (12c)
9336
Note that Eq. (12b) is in good agreement with Eq. (11a) and Eq. (12c) is in good agreement with
Eq. (11b).
(iv) The thermal energy loss at steady state can also be written in terms of the temperature rise and
the flow parameters from Eq. (a), see page 647, as

 Q 
H loss = ρ C p Q ΔT 1 − 0.5 s  (13a)
 Q
Rearranging Eq. (13a), the flow ratio can be written as

Qs  H loss 
= 2 1 −  (13b)
Q  ρ C p Q ΔT
 
where the density and the specific heat capacity of common petroleum lubricants are specified on
page 648, are

ρ = 0.0311 lbm / in 3 and C p = 0.42 Btu / lbm o F (14)

The flow variable, see Figure 12.18, page 643, is specified in the problem statement as
Q
= 3.6 (15a)
rcNl

Substituting r = 2 in , c = 0.02 in, l = 5 in, and Eq. (4) into Eq. (15a), and rearranging, the flow
rate (that is, the volumetric oil flow) is

Q = 3.6 (2 × 0.02 × 20 × 5) =14.4in 3 /s (15b)

Substituting the specified temperature rise ΔT = 5°F and Eqs. (11), (12b), and (15b) into Eq.
(13b), the flow ratio can be written as

Qs  0.829 Btu 
= 2 1 −  (16a)
Q 
 (
0.0311 lbm/in 3
( )
0.42 Btu / lbm ° F ) (
14.4 in 3
/ s ()5 ° F ) 


11
Therefore, the flow ratio is
Qs
= 0.24 (16b)
Q

Check: The flow ratio can also be obtained from Eq. (12-19), see page 648, that is
9.70 ΔTF f ( r / c)
= (17a)
P (1 − 0.5 Qs / Q )( Q / r c N l )
Rearranging this equation, the flow ratio can be written as

Qs  f ( r / c) P 
= 2 1 − (17b)
Q  9.70 ΔT ( Q / r c N l ) 
 F 
Substituting the temperature rise ΔT = 5°F, and Eqs. (1b), (7a), and (14), into Eq. (17b), the flow
ratio can be written as
Qs  (7.7)(20) 
= 2 1 − (18a)
Q  9.70 (5) ( 3.6 ) 
 

Therefore, the flow ratio is

Qs
= 0.24 (18b)
Q

Note that Eq. (18b) is in complete agreement with Eq. (16b).

12
Solution to Problem 4.
(i) The transmitted load can be written from Eq. (13-35), see page 712, as
H
Wt = 33000 (1)
Vt

The pitch line velocity can be written from Eq. (13-34), see page 711, as
π dG nG
Vt = ft/min (2a)
12
Substituting the pitch circle diameter of the gear dG = 5 in and the speed of the gear nG = 500 rpm
into Eq. (2a), the pitch line velocity is
π × 5 × 500
Vt = = 654.5 ft/ min (2b)
12
Substituting the horsepower H = 5 hp and Eq. (2b) into Eq. (1), the transmitted load is

33000 × 5
Wt = = 252.1 lb (3)
654.5
The radial component of the load can be written from Figure 13-30, see page 713, as
Wr = Wt tan φ (4a)

Substituting Eq. (3) and pressure angle φ = 200 into Eq. (4a), the radial component of the load is

Wr = 252.1× tan 200 = 91.76 lb (4b)

(ii) The AGMA normal stress due to bending of the pinion teeth can be written in US customary
units from Eq. (14-15), see page 751, as
Wt P
σ= (Ko Kv K s Km K B ) (5)
FJ

To determine the bending strength geometry factor J for the pinion. The diametral pitch can be
written from Eq. (13-1), see page 684, as
N
P= (6)
d

Substituting the diametral pitch P = 8 inches−1 and the pitch circle diameter of the gear dG = 5 in
into Eq. (6), and rearranging, the number of teeth on the gear is
N G = P dG = 8 × 5 = 40 (7a)

Therefore, the number of teeth on the pinion is

13
 N p = P d p = 8 × 3 = 24 (7b)

Since N p = 24 and the loads are applied at the tip of the teeth then the bending strength geometry
factor for the pinion teeth from Figure 14-6, see page 759, is
J = 0.25 (8)
To determine the overload factor. For light shock input and moderate shock output the overload
factor from the table on page 772, see Figure 14-17, is
K o = 1.5 (9a)

The dynamic factor for the pinion is given in the problem statement as
K v = 1.25 (9b)

The size factor for the pinion, see Eq. (a), page 765, is given in the problem statement as
K s = 1.0 (9c)

The load distribution factor for the pinion, see Eq. (14-30), page 765, is given in the problem
statement as
K m = 1.0 (9d)

Since the pinion is solid then the rim thickness factor for the pinion, see Section 14-16, page 770,
is
K B = 1.0 (9e)

Substituting the face width F = 1.25 inches, the diametral pitch P = 8 inches−1 , and Eqs. (3), (8),
and (9) into Eq. (5), the AGMA normal stress due to bending of the pinion teeth can be written as
252.1× 8
σ= × (1.5 × 1.25 × 1.0 × 1.0 ×1.0) kpsi (10a)
1.25 × 0.25

Therefore, the AGMA normal stress due to bending of the pinion teeth is
σ = 12.1 kpsi (10b)

(iii) The AGMA bending factor of safety for the pinion can be written from Eq. (14-41), see page
771, as
S fb
SF = (11)
σ
where the fully corrected bending fatigue strength can be written as

 Y 
S fb =  N  St (12)
 KT K R 

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For commercial quality carburized and hardened Grade 1 steel gears, the repeatedly applied
bending strength stress-cycle factor, for number of load cycles greater than 107 , from Figure 14-
14, see page 769, can be written as
YN = 1.3558 N −0.0178 (13a)

Substituting the number of load cycles N = 109 cycles into Eq. (13a), the repeatedly applied
bending strength stress-cycle factor is

YN = 1.3558(109 ) −0.0178 = 0.937 (13b)

Note that the larger the value of YN then the larger the bending factor of safety for the pinion. This
is consistent with using commercial quality gears. If precision gears are used then a lower factor
of safety would be acceptable.
The gearset is operating at room temperature, therefore, the temperature factor from page 770,
is
KT = 1 (14)
The reliability factor, for R = 0.995 , can be written from the least-squares regression fit, see
Eq. (14-38), page 769, as
K R = 0.50 − 0.109 ln (1 − R) (15a)

Substituting the specified reliability R = 0.995 into Eq. (15a), the reliability factor is

K R = 0.50 − 0.109 ln (1 − 0.995) = 1.077 (15b)

Substituting the repeatedly applied bending strength St = 55 kpsi and Eqs. (13b), (14), and (15b)
into Eq. (12), the fully corrected bending strength is

 0.937 
S fb =   x 55 = 47.85 kpsi (16)
 1.0 × 1.077 
Substituting Eqs. (10b) and (16) into Eq. (12), the AGMA bending factor of safety for the pinion
is
47.85
SF = = 3.95 (17)
12.1

15