1

Copy of Final Exam Cover Sheet. ME 452 Machine Design II Spring Semester, 2021


Name of Student (Please Print): ____________________________________________________

=============================================================================

Item 1: Honor System for the ME 452 Final Exam.

With my signature below, I acknowledge the following policies for the ME 452 Final Exam.

(i) I referred only to my video lecture notes, crib sheets and examples, and the course textbook.
(ii) I did not exceed the stipulated continuous time of TWO hours to work on the exam.
(iii) I have neither given, nor received, any unauthorized assistance during the exam.

I have read the policies (i), (ii), and (iii) above and agree to honor these conditions.
Note that you do not need to print this page, the page can be signed digitally in the Gradescope
link of the exam. You can use any device with a camera to scan and upload your work. Please
ensure that your scan of the solution is readable and contains all of the necessary information.


Signature of Student (to guarantee Honor System):_____________________________________

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Item 2: Instructions for the ME 452 Final Exam.

(i) The exam is only available on Gradescope. The exam will NOT be posted on the course website.

(ii) The link to Gradescope will open at 9:00 am EST, Monday morning, May 3rd, and close at
9:00 am EST, Tuesday morning, May 4th. (EST is the West Lafayette Campus time).

(iii) Include your name clearly at the top of each page of your exam. There will be FOUR problems
and 25 points will be assigned to each problem. Each problem with the accompanying solution
must be uploaded to Gradescope as a separate PDF file.

(iv) Draw any diagrams or free body diagrams clearly and to a reasonable scale. If you believe that
assumptions are required to solve a problem then you must clearly state your assumptions. Credit
will be given if your assumptions are essential to the solution.

(v) For full credit you must show all of your work and calculations. Please do not write on the
exam pages. Present your solutions clearly and logically on blank pages, you can use engineering
paper, plain white paper, or copy paper. Write on one side of the paper only. Any work that cannot
be followed by the graders is assumed to be in error.

(vi) You are allowed TWO hours to attempt the exam. The time begins when you open the link to
access the exam on Gradescope. The link will expire after 2 hours and 30 minutes. The additional
30 minutes is to allow you to sign the cover sheet and upload your four PDF files to Gradescope.
You do not need any electronic devices other than the ME approved calculator, a computer to view
and submit your exam, and a camera to scan your exam.
2

ME 452 Machine Design II MOCK EXAM

Name of Student: __________________________________________

Problem 1 (25 points). The rotating shaft shown in Figure 1 is simply supported by bearings at
points O and C. The radius of the groove at the center of the shaft is r = 15 mm. The constant force
at point A is AF = 450 N, the constant force at point B is BF = 450 N, and the constant shaft
torque is T = 215 N.m. The shaft has a tensile yield strength ytS = 175 MPa, a compressive yield
strength ycS = 220 MPa, and an ultimate tensile strength utS = 315 MPa. The fully corrected
endurance strength of the shaft is eS = 115 MPa. The fatigue stress concentration factors for
normal stress are mf fK = K = 2.25 and for shear stress are fs fsmK = K = 1.85. The infinite life
fatigue factor of safety for the critical element of the shaft is fn = 4.
(i) Show the location of the critical element of the shaft on Figure 1 below. Briefly explain your
reasons why the element you have shown is the critical element.
(ii) Determine the diameter d at the groove using the Goodman failure criterion.
(iii) Using the Langer line, determine the factor of safety guarding against first-cycle yielding for
the critical element.





















Figure 1. A simply supported rotating shaft. (Not drawn to scale).
O
d
30 mm 90 mm 30 mm
D
A B C
FA = 450 N FB = 450 N
Y
Z
r = 15 mm
T = 215 Nm
90 mm
3

ME 452 Machine Design II MOCK EXAM

Name of Student: __________________________________________

Problem 2 (25 points). For the commercial quality spur gearset shown in Figure 2, the diametral
pitch of the gear is 10 teeth per inch, the diameter of the pitch circle of the gear is 8 inches, and
the gear ratio is 5. The teeth of the gearset are full-depth with a pressure angle of 20o and the face
widths of the gear and the pinion are 1.25 inches. The speed of the input gear is 1250 rpm and
transmits 9 horsepower to the output pinion. The input gear has a light shock load and the output
pinion has moderate shock load and the loads are applied at the tips of the teeth. The gear and the
pinion are solid gears, made of carburized and hardened Grade 1 steel, and operate at room
temperature. The known AGMA factors for the gear and the pinion are: the dynamic factor
vK =1.25, the size factor sK =1.0, and the load distribution factor mK =1.0. Determine:
(i) The tangential load and the radial load acting on the meshing gear teeth.
(ii) The AGMA bending stress acting on the pinion teeth.
(iii) The AGMA bending fatigue factor of safety for the pinion for 1010 cycles at 0.999 reliability.































Figure 2. A spur gearset. (Not drawn to scale).
Input
Output
Gear
Pinion
4

ME 452 Machine Design II MOCK EXAM

Name of Student: __________________________________________

Problem 3 (25 points). The torsion spring shown in Figure 3 has straight tangent-ends and is
designed to hold a cabinet door closed. The wire diameter d 0.05inches,= the spring index C 10,=
and the diameter of the solid steel circular pin through the center of the spring is 0.30 inches. The
lengths of the straight ends of the spring are both 0.75 inches. The spring is ASTM A232 wire with
a modulus of elasticity E 30 Mpsi= and a yield strength y utS 0 .85 S= (where utS is the ultimate
tensile strength) and the spring is shot peened.
When the door is closed, the spring has an angular deflection of o120 due to a bending moment
of 1.5 lbs.ins acting on the spring . When the door is open, the spring has an angular deflection of
o240 due to the maximum bending moment of 2.0 lbs.ins acting on the spring.
Determine the following (do not use an iteration procedure):
(i) The number of active turns and the number of body turns in the spring.
(ii) The minimum inside diameter of the coil.
(iii) The diametral clearance between the spring and the pin when the cabinet door is open.
(iv) The static factor of safety guarding against yielding when the cabinet door is open.






Figure 3. A torsion spring with straight ends. (Not drawn to scale).


L = 0.75 inches
L = 0.75 inches
Pin Diameter = 0.30 inches
5

ME 452 Machine Design II MOCK EXAM

Name of Student: __________________________________________

Problem 4 (25 points). An external contracting drum brake is shown in Figure 4. The drum is
rotating counterclockwise and the radius of the drum is r = 115 mm. The face width of the friction
pad on the arm is 30 mm. The mean coefficient of friction of the brake pad is 0.35 and the
maximum pressure of the frictional material of the shoe is 750 kPa.
(i) Is the brake self-energizing or self-deenergizing? Briefly explain your answer.
(ii) Determine the magnitude of the horizontal limiting actuating force, ,aF that is acting on the
arm. The arm is vertical when the brake pad is in contact with the drum as shown in Figure 4.
(iii) Determine the magnitude and the direction of the braking torque.






















Figure 4. An external contracting drum brake. (Not drawn to scale).
r = 115 mm
360 mm
180 mm
120°
Fୟ
Drum rotation about the fixed
shaft G is counterclockwise.
30°
O
G
30°
6

Solution to Problem 1 (25 Points). Mock Exam, April 27th, 2021
(i) The location of the critical element of the shaft (denoted as E) is shown on Figure 1 below. The
reasons why E is the critical element include: (i) The stress raiser, that is, the groove; (ii) the
bending moment and the torque are large at the mid-section. The shear force at the mid-section is
not large and is not significant; and (iii) the element is in tension, and the yield strength in tension
is less than the yield strength in compression.




















Figure 1. The location of the critical element E.

(ii) The diameter of the shaft can be written from the distortion energy-Goodman criterion of
failure, see Eqs. (7-6) and (7-8), pages 381 and 382, as

( ) ( ) ( ) ( )
1/3
1/2 1/22 2 2 216 1 14 3 4 3f f a fs a f m fs m
e ut
n
d K M K T K M K T
S Sπ
     
= + + +            
(1)

Since the bending moment is fully reversed and the torque is constant then the mean component
of the bending moment acting on the critical element E, see Figure 6-17c, page 302, and the
alternating component of the torque are

0mM = and 0aT = (2a)

Substituting Eq. (2a) into Eq. (1), the diameter of the shaft can be written as

( ) ( )
1/3
16 1 12 3f f a fs m
e ut
n
d K M K T
S Sπ
  
= +     
(2b)
O
d
30 mm 180 mm 30 mm
D
A B C
FA = 450 N FB = 450 N
Y
Z
r
T = 215 Nm
The critical element E.
7

The alternating component of the bending moment acting on element E can be written as

O AaM EO x R EA x F= − (3a)

From the sum of the forces in the y-direction, the reaction forces at the bearings O and C are

450 x 210 450 x 30R 450 N
240O
+
= = and R 900 450 450 NC = − = (3b)

Substituting Eqs. (3b) into Eq. (3a), the alternating component of the fully reversed bending
moment acting on the critical element E is

aM 120 x 450 90 x 450 13500 N.mm= − = (4)

Since the torque is constant then the mean and alternating components of the torque are

m 215000 N.mmT = and a 0T = (5)

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

fK = 2.25 and fsK = 1.85 (6)

Substituting the infinite life fatigue factor of safety f 4n = , the endurance limit, the ultimate
strength, and Eqs. (4), (5), and (6) into Eq. (2b), the shaft diameter at the groove can be written as

( ) ( )
1/3
16 x 4 1 12 2.25x13500 3 1.85x 215000
115 315
d
π
  
= +     (7a)
or as
{ }
1/316 x 4 528.26 2187.06d
π
 
= +   (7b)

Therefore, the diameter of the shaft at the groove is

38.1 mmd = (7c)

(iii) The static factor of safety guarding against yielding for the critical element using the Langer
line, see Eq. (7-16), page 382, can be written as

max
yt
y
S
n
σ
=
′
or as
max
yc
y
S
n
σ
=
′
(8)

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. (8).
The maximum von Mises stress acting on the critical element E is given by Eq. (7-15), see
page 382, can be written as

1/2 1/22 2 2 2
amax max max m am3 ) 3( )σ σ′    = + τ = + + τ + τ(σ σ    (9a)
8

or as
1/222
3 3max
16 (T )32 (M M )
3 sf
m afm a K TK
d d
σ
π π
 + +  ′ = +          
(9b)

The mean component of the normal stress due to bending and the alternating component of the
shear stress due to the constant torque acting on the critical element E, respectively, are

0mσ = and 0aτ = (10)

The alternating component of the normal stress acting on the critical element E can be written as

3
af32 K
a
M
d
σ
π
= (11a)

Substituting Eqs. (4), (6), and (7) into Eq. (11a), the alternating component of the normal stress is

3
32(2.25)(13500) 5.594 MPa
(38.1)a
σ
π
= = (11b)

The mean component of the shear stress acting on the critical element E can be written as

3
mfs
m
16 K T
d
τ
π
= (12a)

Substituting Eqs. (4) and (5) into Eq. (12a), the mean component of the shear stress is

3m
16(1.85)(215000) 36.627 MPa
(38.1)
τ
π
= = (12b)

Substituting Eqs. (10), (11b) and (12b) into Eq. (9a), the von Mises maximum stress for the critical
element is
1/22 2
max 5.594 3(36.627) 63.687 MPaσ ′  = + =  (13)

Substituting the tensile yield strength 175 MPaytS = and Eq. (13) into the first equation of Eq.
(8), the static factor of safety for the critical element E guarding against yielding is

175 2.75
63.687y
n = = (14)

As an aside for purpose of comparison. For a conservative approach, the static factor of safety for
the critical element E guarding against yielding can be written for as
m
yt
y
a
S
n
σ σ
=
′ ′+
(15)

The von Mises mean stress can be written as

9

1/222
mm m3σ ′  = + τσ  (16a)

Substituting Eqs. (10) and (12b) into Eq. (16a), the von Mises mean stress is

1/22
m 0 3x 36.627 63.44 MPaσ ′  = + =  (16b)

The von Mises alternating stress can be written as

1/222 3aa aσ ′  = + τσ  (17a)

Substituting Eqs. (10) and (11b) into Eq. (17a), the von Mises alternating stress is

1/225.594 0 5.594 MPaaσ ′  = + =  (17b)

Substituting the tensile yield strength and Eqs. (16b) and (17b) into Eq. (15), the static factor of
safety for the critical element guarding against yielding is

175 175 2.53
63.44 5.594 69.03y
n = = =
+
(18)

Note that the answer from Eq. (18) is indeed more conservative than the answer given by Eq. (14).
As an aside. An alternative approach.
The diameter of the shaft at the groove can be written from the distortion energy-Goodman
criterion, see Eq. (2b) and Eq. (7.8), page 368, as

1/3
f16 n A B( )
e ut
d
S Sπ
 
= +  
(19)

The coefficients in Eq. (19) are given by Eq. (7-6), see page 381, namely

( ) ( )22f fA 4 3 N.ma asK M K T= + (20a)
and
( ) ( )22f fB 4 3 N.mm msK M K T= + (20b)

Substituting the given values and Eqs. (4) and (5) into Eqs. (20), the coefficients are

( ) ( )2 2A 4 2.25 x13500 3 1 60.75 N.m.85 x 0= + = (21a)
and
( ) ( )2 2B 4 2.25 x 3 1.85 x 20 68815 .9 N m000 23 .= =+ (21b)

Substituting the given factor of safety, endurance limit, ultimate tensile strength and Eqs. (21) into
Eq. (19), the diameter of the shaft at the groove can be written as
10

1/3
6 6
16 x 4 60.75 688.923( )
115x10 315x10
md
π
 
= +  
(22a)

which can be written as

1/3 1/3
6 6 m
16 x 4 16 x 4 x 2.715(0.528 2.187)
x10 x10
d
π π
   
= + =      
(22b)

Therefore, the diameter of the shaft at the groove is

38.1 mmd = (23)

Note that this answer agrees with the answer in Eq. (7c).



11

Solution to Problem 2.
(i) The transmitted load can be written from Eq. (13-35), see page 712, as

33000t
t
HW
V
= (1)

The pitch line velocity can be written from Eq. (13-34), see page 711, as

ft/min
12
G G
t
d nV π= (2a)

Substituting the diameter of the pitch circle of the gear 8 inGd = and the speed of the gear
1250 rpmGn = into Eq. (2a), the pitch line velocity is

8 1250 2618 ft/ min
12t
V π × ×= = (2b)

Substituting the given horsepower 9hpH = and Eq. (2b) into Eq. (1), the transmitted load is

33000 9 113.45lb
2618t
W ×= = (3)

The radial component of the load can be written from Figure 13-29, see page 711, as

tanr tW W φ= (4a)

Substituting Eq. (3) and 020φ = into Eq. (4a), the radial component of the load is

0113.45 tan 20 41.29 lbrW = × = (4b)

Check: The horsepower input can be written from Eq. (13-33), see page 711, as

63000
GinH
nT
= (5a)

Rearranging Eq. (5a), the input torque can be written as

63000
in
G
HT
n
= (5b)

Substituting the horsepower 9hpH = and speed 1250 rpmGn = into Eq. (5b), the input torque is

63000 x 9 453.6 lbs-in
1250in
T = = (5c)

The torque can also be written from Eq. (b) and Figure 13-29, see page 711, as

tin GT r W= (6a)
12

Substituting the pitch radius of the gear 4inGr = and Eq. (5c) into Eq. (6a), and rearranging the
equation, the tangential load (or the transmitted load) acting on the pinion from the gear is

453.6 113.45 lbs
4t
W = = (6b)

Note this answer is in complete agreement with Eq. (3).
(ii) The AGMA bending stress in the pinion can be written in US customary units from Eq. (14-
15), see page 751, as
( )t o v s m B
W P
K K K K K
J F
σ = (7)

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

G
G
N
P
d
= (8)

Substituting the diametral pitch of the gear, 1,10 inchesP −= and the diameter of the pitch circle of
the gear, 8 in,Gd = into Eq. (8), and rearranging the equation, the number of teeth on the gear is

10 8 80GN = × = (9)

The gear ratio can be written from Eq. (14-22), see page 759, as

G
G
p
d
m
d
= (10a)

Substituting the given gear ratio 5Gm = and the diameter of the pitch circle of the gear 8 in,Gd =
into Eq. (10a) and rearranging the equation, the diameter of the pitch circle of the gear is

8 1.6inches
5p
d = = (10b)

Therefore, the number of teeth on the pinion from Eq. (13-1), see page 684, is

10 1.6 16p pN P d= = × = (11)

Since the number of teeth on the pinion is 16pN = and the loads are applied at the tips of the teeth
then the bending strength geometry factor for the pinion teeth from Figure 14-6, see page 759, is

0.226J = (12)

To determine the modification factors for the pinion. Since the input gear is subjected to a light
shock load and the output pinion is subjected to a moderate shock lock then the overload factor
from the table on Figure 14-17, see page 772, is

13

1.50oK = (13a)

The dynamic factor for the pinion is given in the problem statement as

1.25vK = (13b)

The size factor for the pinion, see Section 14-10, page 765, is given in the problem statement as

1.0sK = (13c)

The load distribution factor for the pinion, see Section 14-11, page 765, is given in the problem
statement as
1.0mK = (13d)

Since the pinion is solid then the rim thickness factor for the pinion, see Section 14-16, page
770, is
1.0BK = (13e)

Substituting the given face width and Eqs. (3), (12), and (13) into Eq. (7), the AGMA bending
stress acting on the pinion can be written as

113.45 10 (1.5 1.25 1.0 1.0 1.0) kpsi
0.226 1.25
σ
×
= × × × × ×
×
(14a)

Therefore, the AGMA bending stress acting on the pinion is

7.53 kpsiσ = (14b)

(iii) The AGMA bending fatigue factor of safety for the pinion can be written from Eq. (14-41),
see page 771, as
n / ( )t T R
F
S Y K K
S
σ
= (15)

For carburized and hardened Grade 1 steel gears, the repeatedly applied bending strength (at 710
cycles and 0.99 reliability) from Table 14-3, see page 754, is

255000 lbs/ins 55 kpsitS = = (16)

For carburized and hardened Grade 1 steel commercial gears, the repeatedly applied bending
strength stress-cycle factor (at 1010 load cycles) from Figure 14-14, see page 769, is

nY 0.9= (17)

Since the gears are operating at room temperature then the temperature factor is

1TK = (18)

The reliability factor for a 99.9% reliability from Table 14-10, see page 770, is
14

1.25RK = (19)

Substituting Eqs. (14), (16), (17), (18), and (19) into Eq. (15), the AGMA bending fatigue
factor of safety for the pinion is

55 0.9 / (1 1.25) 5.3
7.53F
S × ×= = (20)


15

Solution to Problem 3.
(i) The equivalent number of active turns can be written from Eq. (10-51), see page 561, as

4
10.8
=
′
a
d EN
D k
(1)

The spring index can be written from Eq. (10-1), see page 526, as
DC
d
= (2a)

Substituting 10C = and 0.05 inchesd = into Eq. (2a), the mean coil diameter (at the no load
condition) is
10 0.05 = 0.5 inchesD C d= × = × (2b)

The spring stiffness can be written from Eq. (10-45), see page 560, as
2 1
2 1θ θ
−
′ =
′ ′−
M Mk (3)
The angular deflection of the spring is

2 1 240 120 120θ θ− = ° − ° = ° (4a)

Therefore, the number of turns is
2 1
120 1 turns turns
360 3
θ θ °′ ′− = =
°
(4b)
Substituting 1 1.5lbs-ins,M = 2 2.0 lbs-ins,M = and Eq. (4b) into Eq. (3), the spring stiffness is

2.0 1.5 1.5 lbs-ins / turn
1/ 3
k −′ = = (5)

Substituting Eqs. (2b), (5), and the modulus of elasticity E = 30 Mpsi into Eq. (1), the equivalent
number of active turns is
4 60.05 (30)10 23.15 turns
10.8 0.5 1.5a
N ×= =
× ×
(6)

The number of body turns can be written from Eq. (10-48), see page 560, as

1 2
3π
+
= −b a
l lN N
D
(7)

The lengths of the straight ends from the given figure are

1 2 0.75 inl l= = (8)

16

Substituting Eqs. (2b), (6), and (8) into Eq. (7), the number of body turns is

0.75 0.7523.15 22.83 turns
3 x 0.5b
N
π
+
= − = (9)

(ii) The minimum inside coil diameter, see Figure 10-1, page 526, is

′ ′= −iD D d (10)

where the helix diameter of the coil, see Eq. (10-53), page 561, is

θ
′ =
′+
b
b C
N DD
N
(11)

The angular deflection of the body of the coil, see Eq. (10-54), page 561, is

4
10.8θ ′ = bC
M D N
d E
(12a)

The maximum angular deflection will occur when 2 240θ = ° ; i.e., when the bending moment
is 2 2.0 lbs-ins,M = therefore, the angular deflection of the body of the coil is
4 6
10.8 2.0 0.5 22.83 1.315 turns
0.05 (30)10C
θ × × ×′ = =
×
(12b)

Substituting Eqs. (5b), (9), and (12b) into Eq. (11), the helix diameter of the coil is

22.83 0.5 0.473 inches
22.83 1.315
D ×′ = =
+
(13)

Substituting Eq. (13) and 0.05 inchesd = into Eq. (10), the minimum inside coil diameter is

0.473 0.05 0.423 inchesiD′ = − = (14)

(iii) The diametral clearance between the spring and the pin, see Eq. (10-55), page 561, can be
written as
′Δ = −i pD D (15a)
Substituting Eq. (14) and 0.3 inches.pD = into Eq. (15a), the diametral clearance is
0.423 0.3 0.123 inchesΔ = − = (15b)
(iv) The static factor of safety guarding against yielding is defined as

17

y
s
S
N
σ
= (16)

The ultimate tensile strength, see Eq. (10-14), page 532, can be written as
mut
AS
d
= (17a)

For ASTM A232, see Table 10-4, page 532, the coefficient A and the exponent m are

A = 169 kpsi-inm and m = 0.168 (17b)

Substituting Eq. (17b) and 0.05 inchd = into Eq. (17a), the ultimate tensile strength is

0.168
169 279.55 kpsi
0.05ut
S = = (18a)
Therefore, the yield strength is

0.85 0.85 279.55 237.62 kpsiutyS S= = × = (18b)

The bending stress, see Eq. (10-44), page 560, can be written as
3
32
σ
π
= i
MK
d
(19)
The bending stress concentration factor (i.e., the Wahl factor), see Eq. (10-43), page 559, is
24 1
4 ( 1)
− −
=
−
i
C CK
C C
(20a)

Substituting the spring index C =10 into Eq. (20a), the bending stress concentration factor is

24 10 10 1 1.081
4 10(10 1)i
K × − −= =
× −
(20b)
Substituting Eq. (20b) and 0.05 inchesd = M = 2.0 lbs.ins into Eq. (19), the maximum bending
stress is
3
32 21.081 x 176.18 kpsi
0.05
σ
π
×
= =
×
(21)
Substituting Eq. (18b) and (21) into Eq. (16), the static factor guarding against yielding is

237.62 1.34
176.18s
N = = (22)

18

Solution to Problem 4.
(i) The brake is self-energizing for the given counterclockwise rotation of the drum. The reason is
that the friction force is helping to reduce the actuating force. Page 831 states the reason as follows:
“A brake shoe is self-energizing its moment sense helps set the brake, self-deenergizing if the
moment resists setting the brake.”
(ii) The free body diagram of the shoe is shown in Figure 4.1. Note that the x axis points from the
center of the drum, point G, to the hinge pin O and the y axis is directed from point G toward the
shoe. Also, note that the x and y components of the reaction force are chosen to be positive.






















Figure 4.1. The free body diagram of the shoe and the arm.

From static equilibrium, the sum of the moments about the hinge pin O can be written as

OM = 0 (1a)
R୷
Fୟ
F୒
F୤
R୶
O
G
Y
X
r = 115 mm
30°
120°
Drum rotation is
counterclockwise
c = 0.360 m
A
a = 0.180 m
19

From the free body diagram, see Figure 4.1, the sum of the moments about the hinge pin O is

F 0N fac M M+ − + = (1b)

The moment arm of the actuating force about the hinge pin O is given in Figure 4, that is

= 360 mm = 0.360 m (2)

Rearranging Eq. (1b), the actuating force, see Eq. (16-13), see page 834, can be written as

F N fa
M M
c
−
= (3)

Therefore, the shoe is self-energizing for the given counterclockwise rotation of the drum.
The moment due to the normal force about the hinge pin O can be written from Eq. (16-3), see
page 827, as
2
1
2sin
sin
a
N
a
p b r aM d
θ
θ
θ θ
θ
=  (4)

where the integral, see Eq. (16-8), see page 828, can be written as

2
1
2 2 1
2 1
1 1sin sin 2 sin 2
2 4 2 4
B d
θ
θ
θ θθ θ θ θ   = = − − −       (5a)

From Figure 4.1, the angles to the heel of the shoe and the toe of the shoe, respectively, are

1 30θ =  and 2 1 120 150θ θ= + =  (5b)

Substituting Eq. (5b) into Eq. (5a) gives

150 1 1sin 300 sin
2
0 0
2
6
4
3
4
o
o
o
oB
   
= − − −      
(6a)
that is
1.5255 0.0453 1.4802B = − = (6b)

Since 2 90
oθ > then the location of the maximum pressure acting on the shoe, see page 826, is

90aθ =  (7)

Since the shoe is self-energizing then the limiting pressure is the maximum pressure of the
frictional material, that is
3750 kPa = 750 x 10 Paap = (8)

The face width of the friction pad is given in the problem statement as

= 30 mm = 0.030 m (9)

The radius of the drum and the distance from the center of the drum, G, to the hinge pin, O, are
20


= 115 mm = 0.115 m and = GO = 180 mm = 0.180 m (10)
Substituting Eqs. (6b), (7), (8b), (9), and (10) into Eq. (4), the moment due to the normal force
about the hinge pin O is

ே = ቀ଻ହ଴×ଵ଴య୶଴.ଷ଴×ଵ଴షయ×଴.ଵଵହ×଴.ଵ଼଴ଵ ቁ1.4802 = 689.4 N ∙ m (11)

The moment due to the friction force about the hinge pin O can be written from Eq. (16-2),
see page 827, as
2
1
sin ( cos )
sin
a
f
a
f p b rM r a d
θ
θ
θ θ θ
θ
= − (12)

The mean coefficient of friction of the brake pad is specified as

= 0.35 (13)

The integral in Eq. (12) can be written as
= ׬ ( − )ఏమఏభ = ׬ (0.115 − 0.180 )ଵହ଴೚ଷ଴೚ = 0.2078 (14)

Substituting Eqs. (5b), (7), (8), (9), (10), and (13) into Eq. (12), the moment of the friction force
about the hinge pin O is

௙ = ቀ଴.ଷହ×ଽ଴଴×ଵ଴య୶଴.ଷ଴×଴.ଵଶଵ ቁ × 0.2078 = 180.39 N ∙ m (15)

Check: Equation (12) can also be written as

( )
sin
a
f
a
f p b rM r C a A
θ
= − (16)

where the integral, see Eq. (16-8), see page 828, can be written as

22
11
21sin cos sin
2
A d
θθ
θθ
θ θ θ θ = =    (17a)
and
( )
2
1
2
1
1 2sin cos cos cosC d
θ
θ
θ
θ
θ θ θ θ θ= = = − (17b)

Substituting Eq. (5b) into Eqs. (17) gives
= ଵଶ (sinଶ 150° − sinଶ 30°) = 0 and = 30° − cos 150° = 1.732 (18)

Substituting Eqs. (18) into Eq. (16), the moment of the friction force about the hinge pin O is

௙ = ቀ଴.ଷହ×ଽ଴଴×ଵ଴య×଴.ଷ଴×଴.ଵଶଵ ቁ × 0.12 × 1.732 = 180.39 N ∙ m (19)

21

Note that Eq. (19) is in complete agreement with Eq. (15).
Substituting Eqs. (2), (11), and (15) into Eq. (2), the limiting actuating force for the shoe is
௔ = ଺଼ଽ.ସ଴ିଵ଼଴.ଷଽ଴.ଷ଺଴ = 1413.9 N (20)

(iii) The braking torque (or the braking capacity) generated by the shoe can be written from Eq.
(16-6), see page 827, as

= ௙ ௣ೌ ௕ ௥మ(௖௢௦ ఏభି௖௢௦ ఏమ)௦௜௡ఏೌ (21)

Substituting Eqs. (5), (8b), (10) and (13) into Eq. (21), the braking torque generated by the shoe
can be written as

2 0 0(0.35x 750 x 0.03x 0.115 )(cos30 cos150 )T = Nm
1
− (22a)

That is, the braking torque generated by the shoe is

T = 180.39 Nm (22b)

The direction the braking torque on the drum is clockwise.
Note that the braking torque given by Eq. (22b) is equal to the moment of the friction force
about the hinge pin O, see Eq. (19). The reason is due to the symmetry of the design of this single
brake shoe.



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