1

ME 452: Machine Design II Spring Semester 2018

Name of Student: ____________________________________________________

Circle your lecture division number: Division 1 Division 2

9:30 am – 10:20 am 1:30 pm – 2:20 pm

FINAL EXAM

Thursday, May 3rd, 3:30 pm – 5:30 pm

CL50

OPEN BOOK AND CLOSED NOTES

For full credit you must show your solutions clearly and logically on the sheets of paper attached to the

end of each problem.

Use only the blank pages for your solutions and write on one side of the paper only. Please do not write

your work on the exam pages.

Draw any free body diagrams, and other diagrams, clearly and to a reasonable size in your solution.

Staple each problem separately at the end of the exam with this cover sheet in front of Problem 1.

Problem 1 (25 points)

Problem 2 (25 points)

Problem 3 (25 points)

Problem 4 (25 points)

Total (100 points)

2

ME 452: Machine Design II FINAL EXAM SPRING 2018

Name of Student: ______________________________ Circle one: Division 1 Division 2

Problem 1 (25 points). For a commercial spur gearset, see Figure 1, the pitch diameter of the pinion is 6

inches, the diametral pitch of the pinion is 1,4 inches− and the gear ratio is 3. The gear and the pinion

have full-depth teeth with a pressure angle of 20o

and the face widths are 0.75 inches. The gear speed is

750 rpm and transmits 50 horsepower to the pinion. The gear is subjected to a light shock load and the

pinion a moderate shock load. The gear and pinion are nitrided through-hardened Grade 2 steel with a

modulus of elasticity of 30Mpsi, a Poisson’s ratio of 0.28, and a Brinell hardness of 300. The gearset

operates at room temperature, the gear backup ratio is 1.5, and the known AGMA factors are:

Geometry Dynamic Size Reliability Load Distribution Surface Condition

J = 0.43 KV = 1.25 KS = 1.0 KR = 0.88 Km = 1.0 Cf = 1

(i) Determine the AGMA bending stress on the teeth of the gear.

(ii) Determine the AGMA wear stress on the teeth of the gear.

(iii) Determine the AGMA safety factor guarding against bending fatigue failure for the gear for 810

load cycles and 95 percent reliability.

Figure 1. A spur gearset (Not drawn to scale).

Input

Output

Gear

Pinion

3

ME 452: Machine Design II FINAL EXAM SPRING 2018

Name of Student: ______________________________ Circle one: Division 1 Division 2

Problem 2 (25 points). A linear helical compression spring with plain and ground ends has a mean coil

diameter D = 30 mm and a spring index C = 12. The spring material is A228 music wire and the spring

is peened but is not set. Figure 2 shows that when the spring is assembled in place it is subjected to a

preload of 45 N and the length is 65 mm and when the spring is subjected to a maximum working load

of 100 N then the length is 40 mm. Determine the following:

(i) The shut height of the spring.

(ii) The static factor of safety at the shut height of the spring using the Bergstrasser factor.

(iii) The fatigue factor of safety using the Bergstrasser factor and the Gerber failure criterion with the

Zimmerli data.

Figure 2. The loads acting on the compression spring (Not drawn to scale).

F = 45 N

F = 45 N

65 mm

40 mm

F = 100 N

F = 100 N

4

ME 452: Machine Design II FINAL EXAM SPRING 2018

Name of Student: ______________________________ Circle one: Division 1 Division 2

Problem 3 (25 points). A cylindrical pressure vessel where the head is connected to the body by six

identical bolts evenly spaced as shown in Figure 3. Each bolt is 5 /8′′ – 12 UNC – grade 5 with rolled threads, the lengths of the bolts are 3 inches, and the modulus of elasticity of the bolt material is

30 Mpsi. The diameters are A = 8 in, B = 11 in, and C = 14 in; and the thicknesses are D = 1.05 in, E =

1.30 in, and G = 0.05 in. The preload on each bolt is 12000 lbs and the pressure in the cylinder fluctuates

between 1250 psi p 1500 psi.≤ ≤ The stiffness of the cylinder head is 120 Mlb/in; the stiffness of the cylinder body is 180 Mlb/in; and the stiffness of the gasket is 20 Mlb/in. Determine the following:

(i) The stiffness constant of the joint.

(ii) The factors of safety guarding against: (a) joint separation; and (b) overloading a bolt.

(iii) The fatigue factor of safety for a bolt using the modified Goodman criterion.

Figure 3. A cylindrical pressure vessel. (Not drawn to scale).

Cylinder head

p

A stiff gasket

cylinder body

A

B

D

G

E

C 6 bolts evenly spaced

around the cylinder

5

ME 452: Machine Design II FINAL EXAM SPRING 2018

Name of Student: ______________________________ Circle one: Division 1 Division 2

Problem 4 (25 points). A vertical actuating force Fa

is acting at the end of each arm of the two

symmetric shoes of the external contracting drum brake shown in Figure 4. The dimensions are

7 inches,a = 4 inches,b = 1.75 inches,e = and the radius of the drum is 3 inches.r = The face widths

of the two shoes are 1.5 inches and the subtended angle of each shoe is 30 .oθ = The coefficient of friction for the brake lining material is 0.35 and the maximum pressure of the lining material is

2225 lbs/in . Determine the following:

(i) The actuating force Fa

acting on each arm.

(ii) The torque capacity of the brake.

(iii) The vertical distance c that would make the self-energizing shoe become a self-locking shoe.

Figure 4. An external contracting drum brake. (Not drawn to scale).

a

b

Fa

Fa

c

e

e

c

r

rotation of drum

is clockwise

shoe

shoe

O2

O1

b

a

θ

θ

6

Solution to Problem 1.

(i) 10 Points. The AGMA bending stress on the teeth of the gear can be written in US customary units

from Eq. (14-15), see page 746, and the roadmap on page 766, as

( )t o v s m BW P

K K K K KF J

σ = (1)

The diametral pitch 1,4 inchesP −= the face width 0.75 in,F = the geometry factor 0.43,J = the dynamic factor 1.25,

vK = the size factor 1.0,

sK = and the load distribution factor

mK = 1.0. The gear

(input) is subjected to a light shock load and the pinion (output) to a moderate shock load. Therefore, the

overload factor for the gearset from the table at the bottom of page 766, see Figure 14-17, is

1.5o

K = (2a)

For the gear backup ratio Bm 1.5,= the gear rim-thickness factor from Eq. (14-40), see page 764, is

BK = 1.0 (2b)

The transmitted (or tangential) load can be written from Eq. (13-35), see page 707, as

33000tt

HW

V= (3a)

The pitch line velocity can be written from Eq. (13-34), see page 707, as

ft/min12 12

G G p pt

d nd nV

ππ= = (3b)

The gear ratio (or speed ratio), see page 694 and also see Eq. (14-22), see page 754, can be written as

G GG

P P

N dm

N d= = (4a)

Substituting the gear ratio 3G

m = and the diameter of the pitch circle (pitch diameter) of the pinion 6 in

Pd = into Eq. (4a), and rearranging, the diameter of the pitch circle of the gear is

3x 6 18 inchesG G P

d m d= = = (4b)

Substituting Eq. (4b) and the speed of the gear 750 rpmG

n = into Eq. (3b), the pitch line velocity is

18 7503534.29 ft/ min

12tV

π × ×= = (5)

Substituting Eq. (5) and the horsepower 50 hpH = into Eq. (3a), the transmitted load is

33000 50466.9 lb

3534.29tW

×= = (6)

Check: The horsepower transmitted through the gear can be written from Eq. (13-33), see page 706, as

7

63000

GinHnT

= (7a)

Rearranging Eq. (7a), the input torque can be written as

63000

G

in

HT

n= (7b)

Substituting the input horsepower 50 hpH = and the speed of the gear 750 rpmGn = into Eq. (7b), the input torque is

63000 x 504200 lbs-in

750inT = = (7c)

Check: The input torque can be written from Eq. (b), see page 706, and Figure 13-13, see page 707, as

Gin tT r W= (8a)

Substituting Eq. (7c) and the radius of the pitch circle of the gear 9 in,Gr = see Eq. (4b), into Eq. (8a), and rearranging, the transmitted load acting on the pinion is

4200466.7 lbs

9tW = = (8b)

Note that this answer is in good agreement with Eq. (6). The difference is due to round-off error.

The diametral pitch of the spur gearset can be written from Eq. (13-1), see page 676, as

G P

G P

N NP

d d= = (9)

Substituting Eq. (4b) and the diametral pitch 14 inchesP −= into Eq. (9), and rearranging, the number of teeth on the gear is

4 18 72G G

N P d= = × = (10a)

and the number of teeth on the pinion is

4 6 24p p

N P d= = × = (10b)

Substituting the given data and Eqs. (2) and (6) into Eq. (1), the AGMA bending stress on the teeth

of the gear can be written as

466.9 4(1.5 1.25 1.0 1.0 1.0) kpsi

0.75 0.43σ ×= × × × × ×

× (11a)

Therefore, the AGMA bending stress on the teeth of the gear is

210.858 kpsi = 10858 lbs/inσ = (11b)

(ii) 7 Points. The AGMA pitting resistance formula for the gear teeth can be written from Eq. (14-16),

see page 746, and the roadmap on page 767, as

8

c p o v s m f= C ( K K K K C )F I p

t

d

Wσ (12)

The modification factors o v sK , K , K , mK , and fC for the gear are known, therefore, there are only

two new AGMA factors here; i.e., (i) the elastic coefficient pC ; and (ii) the surface geometry factor I.

The AGMA elasticity coefficient for the gear teeth can be written from Eq. (14-13), see page 744, as

1/2

22

1

11p

gP

P g

C

E Eπ

= − ν− ν +

(13a)

Substituting Poisson’s ratio 0.28P g

ν ν= = and the modulus of elasticity 630 1 psi0P gE E= = × into Eq. (13a), the elasticity coefficient for the gear teeth is

1/2

2

6

12276 psi

1 0.282

30 10

pC

π

= = − ×

(13b)

(ii) The surface geometry factor for the external contact of the gear and pinion teeth (see Figure 1) can

be written from Eq. (14-23), see page 755, as

G

n G

mcos sinI ( )

2 m m 1

φ φ=+

(14a)

where nm 1.0= for a spur gearset, see pages 753 and 755. Substituting the known values into Eq. (14a), the surface geometry factor is

cos sin 3I ( ) 0.1205

2 x1.0 3 1

ο ο20 20= =+

(14b)

Substituting the given values and Eqs. (2), (6), (13b), and (14b) into Eq. (12), the AGMA pitting

resistance formula for the gear teeth can be written as

c466.9

= 2276 (1.00 ×1.5×1.00 x 1.25×1.0)0.75× 0.1205× 6

σ (15a)

Simplifying this equation, the AGMA pitting resistance for the gear teeth is

2c = 91450.3 lbs/in 91.450 kpsiσ = (15b)

(iii) 8 points. The AGMA factor of safety guarding against bending fatigue failure for the gear can be

written from Eq. (14-41), see page 765, as

9

/ ( )t N T R

F

S Y K KS

σ= (16)

For nitrided through-hardened Grade 2 steel, the repeatedly applied bending strength (for 810 load

cycles and 95 percent reliability) is written from Figure 14-3, see page 747, as

2108.6H 15890 lbs/ins

BSt = + (17a)

Substituting the given Brinell hardness BH 300= into Eq. (17a), the repeatedly applied bending strength for the gear is

248470 lbs/ins 48.47 kpsiSt = = (17b)

Check: From Figure 14-3, see page 747, the repeatedly applied bending strength for the gear is given as

50 kpsiSt = (17c)

Note that Table 14.3, see page 748, is not valid for the given information.

For commercial gears (nitrided through-hardened Grade 2 steel), the repeatedly applied bending

strength stress-cycle factor (for load cycles equal to, or greater than, 810 ) from Figure 14-14, see page

763, the curve fit equation is given as 0.0178

Y 1.3558 Nn−= (18a)

Substituting the number of load cycles 810 cycles,N = into Eq. (18a), the repeatedly applied bending strength stress-cycle factor for the gear is

0.01788Y 1.3558(1x10 ) 0.9768n

−= = (18b)

Incorrect approach: Note that the bending strength stress-cycle factor obtained from reading the plot on

Figure 14.14, see page 763, would not give an accurate answer (a scale is not provided).

Since the gear is operating at room temperature then the temperature factor is

KT = 1 (19)

and the reliability factor is specified as 0.88R

K = . Substituting Eqs. (15b), (17b), (18b), and (19) into Eq. (16), the AGMA factor of safety guarding against bending fatigue failure for the gear is

1 48.47 0.97684.96

10.858 1.0 0.88F

S ×= = ×

(20)

10

Solution to Problem 2.

(i) 7 points. The shut height of a compression spring with plain and ground ends can be written from

Table 10-1, see page 521, is

L d Ns t= (1a)

where the total number of coils from Table 10-1, see page 521, is

1t a

N N= + (1b)

The number of active coils can be written, by rearranging Eq. (10-9), see page 520, as

4

38a

d GN

D k= (2)

The wire diameter from Eq. (10-1), page 519, is

302.5 mm

12D

dC

== = (3)

The modulus of rigidity of A228 music wire with 2.5 mmd = (that is, 0.0984 in),d = from Table 10.5, see page 526, is

81 GPa=G (4)

The spring stiffness can be obtained from the given forces and deflections, see Figure 2, that is

F 100 45 2.2 N/mm

y 65 40 k

∆ −= = =∆ − (5)

Substituting Eqs. (3), (4), and (5) into Eq. (2), the number of active coils is

4

3

2.5 816.66

8 30 2.2a

N×= =

× × (6)

Substituting Eq. (6) into Eq. (1b), the total number of coils is

6.66 1 7.66 coilst

N = + = (7a)

Substituting Eqs. (3) and (7a) into Eq. (1a), the shut height of the spring is

2.5 7.66 19.15 mms

L = × = (7b)

(ii) 8 points. The static factor of safety, see Eq. (a), page 530, can be written as

S

Ssyns τ= (8)

The ultimate tensile strength can be written from Eq. (10-14), see page 523, as

ut m

AS

d= (9)

11

The material specific constants for A228 music wire with wire diameter 2.5 mmd = that is, 0.0984 in,d = see Eq. (3), that is, 0.004in 0.256 ind< < from Table 10-4, see page 525, are

m2211 MPa.mmA = and 0.145m = (10a)

Since the spring is not set then the torsional yield strength of the wire from Table 10-6, see page 526,

can be written as

0.45sy utS S= × (10b)

Substituting Eqs. (3) and (10a) into Eq. (9), the ultimate tensile strength is

22111935.9 MPa

0.1452.5utS = = (10c)

Substituting Eq. (10c) into Eq. (10b), the torsional yield strength of the wire is

0.45 1935.9 871.16 MPasyS = × = (10d)

The shear stress in the spring (not set) at the shut height, see Table 10-6, page 526, and Eq. (10-2),

page 519, can be written as

8

3S

F DshutKB

dτ

π

=

(11a)

The Bergstrasser factor, see Eq. (10-5), page 519, is

4 2 4 12 21.1111

4 3 4 12 3

+ × += = =− × −C

KB C (11b)

The load at the shut height of the spring, see Example 10-1, page 527, can be written as

( )F k L Lo sshut= − (12)

The free length of the spring (see the plot of the loads against the lengths) can be written as

L L yo a initial= + (13a)

where the assembled length of the spring is specified (see Figure 2) as

65 mmLa = (13b) The initial deflection can be written as

minF

yinitial k

= (14a)

Then substituting the given minimum force and Eq. (5b) into Eq. (14a), the initial deflection is

45 20.45 mm

2.2 yinitial

= = (14b)

Substituting Eqs. (13b) and (14b) into Eq. (13a), the free length of the spring is

12

65 mm + 20.45 mm = 85.45 mmLo = (15)

Substituting Eqs. (5), (7b), and (15) into Eq. (12), the load at the shut height of the spring is

2.2 (85.45 19.15) 145.86 NshutF = × − = (16)

A plot of the load F against the length L of the spring is shown in the figure below.

The plot shows: (i) the solid length; (ii) the minimum working length; (iii) the assembled length; and

(iv) the free length. The plot also shows: (i) the initial deflection; (ii) the working deflection; and (iii) the

clash deflection.

Substituting Eqs. (3), (11b), and (16) into Eq. (11a), the shear stress at the shut height is

8 145.86 0.031.1111 792.375 MPa

30.0025S

τπ

× ×= = ×

(17)

Substituting Eqs. (10d) and (17) into Eq. (8), the static factor of safety of the spring at the shut height is

871.161.1

792.375ns = = (18)

This result indicates that the spring is marginally safe at the shut height.

(iii) 10 Points. The fatigue factor of safety using the Gerber-Zimmerli fatigue failure criterion can be

written from Example 10-4, see page 538, as

sa

f

a

Sn

τ= (19)

The alternating and mean components of the load can be written from Eqs. (10-31), see page 537,

respectively, as

max min

2a

F FF

−= and max min2

m

F FF

+= (20a)

Lm = 40 mm

yinitial = 20.45 mm

ywork

yclash

0

Load

La = 65 mm Length

Fshut = 145.86 N

LO = 85.45 mm

Fmin = 45 N

Ls = 19.15 mm

Fmax = 100 N

13

Substituting the maximum and minimum loads into Eqs. (20a), the alternating and mean components of

the load, respectively, are

100 4527.5 N

2aF

−= = and 100 45 72.5 N2

mF+= = (20b)

Substituting Eqs. (3), (11b), and (20b) into Eq. (11a), the alternating and mean components of the shear

stress are

3

8 27.5 301.1111 149.39 MPa

(2.5)a

τπ

× ×= =

(21a)

and

3

8 72.5 301.1111 393.85 MPa

(2.5)m

τπ

× ×= =

(21b)

The ultimate shear strength of the spring material can be written from Eq. (10.30), see page 537, as

0.67su ut

S S= (22a)

Substituting Eq. (10c) into Eq. (22a), the ultimate shear strength of the A228 music wire is

0.67 1935.9 1297 MPasu

S = × = (22b)

The torsional endurance strength (that is, the Gerber ordinate intercept for the Zimmerli data for a

peened spring) can be written from the equation on page 536 as

( )21 /sa

se

sm su

SS

S S=

− (23)

The alternating and mean components of the endurance strength (for peened spring) from Eq. (10-29),

see page 536, are

398 MPasa

S = and 534 MPasm

S = (24)

Substituting Eqs. (22b) and (24) into Eq. (23), the torsional endurance strength is

( )2398

479.2 MPa1 534 /1297

seS = =

− (25)

The alternating component of the endurance strength can be written from page Table 6.7, see page 307,

or from Example 10-4, see page 538, as

22 2 2

1 12

su se

sa

se su

r S SS

S r S

= − + + (26)

where the slope of the load line can be written as

a

m i

Fr

F F=

− (27a)

14

Substituting the preload and Eqs. (20b) into Eq. (27a), the slope of the load line is

27.51

72.5 45r = =

− (27b)

Substituting Eqs. (22b), (25), and (27b) into Eq. (26), the alternating component of the endurance

strength is

22 2(1) (1297) 2(479.2)1 1 553.36 MPa

2(479.2) (1)(1297)sa

S

= − + + = (28)

Substituting Eqs. (21a) and (28) into Eq. (19), the fatigue factor of safety using the Gerber-Zimmerli

fatigue failure criterion is

553.363.7

149.39f

n = = (29)

Check: The slope of the load line can be written as

a

m i

rτ

τ τ=

− (30)

The preload shear stress can be written from Eq. (10-2), see page 519, as

8

3i

F DiKBd

τπ

=

(31a)

Substituting the known values into this equation, the preload shear stress is

8 x 45 x 301.1111 244.46 MPa

3(2.5)i

τπ

= =

(31b)

Substituting Eqs. (21b) and (31b) into Eq. (30), the slope of the load line is

149.391

393.85 244.46

a

m i

rτ

τ τ= = =

− − (32)

Substituting Eqs. (22b), (25), and (32) into Eq. (26), the alternating component of the endurance

strength is

22 2(1.00) (1297) 2(479.2)1 1 418.29 MPa

2(479.2) (1.00)(1297)sa

S

= − + + = (33)

Substituting Eqs. (21a) and (33) into Eq. (19), the fatigue factor of safety using the Gerber-Zimmerli

fatigue-failure criterion is

418.292.8

149.39f

n = = (34)

15

Alternative method. If the preload is ignored then the slope of the load line can be written from Eqs.

(21), see Example 10-4, page 538, as

149.390.3793

393.85

a

m

rττ

= = = (35a)

Check: The slope of the load line can be written as

27.50.3793

72.5

a

m

Fr

F= = = (35b)

Substituting Eqs. (22b), (25), and (35b) into Eq. (26), the alternating component of the endurance

strength is

22 2(0.3793) (1297) 2(479.2)1 1 299.98 MPa

2(479.2) (0.3793)(1297)sa

S

= − + + = (36)

Substituting Eqs. (3b) and (21a) into Eq. (19), the fatigue factor of safety using the Gerber-Zimmerli

fatigue failure criterion is

299.982.0

149.39f

n = = (37)

16

Solution to Problem 3.

(i) 10 points. The stiffness constant of the joint, see Eq. (f), page 436, can be written as

b

b m

kC

k k=

+ (1)

The stiffness of each bolt can be written from Eq. (8-17), see page 427, as

d tb

d t t d

A A Ek

A l A l=

+ (2)

The modulus of elasticity is 30 Mpsi.E = The tensile stress area for a 5 / 8′′ – 12 UNC bolt from Table 8.2, see page 413, is

20.226 intA = (3a)

and the cross-sectional area of the bolt shank is

2 220.625 0.3068 in

4 4d

dA

π π× ×= = = (3b)

The length of the threaded portion of the bolt within the grip can be written from Table 8.7, page 426, as

t dl l l= − (4a)

where the length of the grip is

1.05 1.30 0.05 2.40 inchesl M N G= + + = + + = (4b)

The length of the bolt shank can be written from Table 8-7, page 426, as

d Tl L L= − (5a)

where the length of each bolt is specified as

L = 3 inches (5b)

and the length of the threaded portion of the bolt (for 6 inches),L ≤ see Table 8-7, page 426, is

2 0.25 2 0.625 0.25 1.50 inchesT

L d= + = × + = (5c)

Substituting Eqs. (5b) and (5c) into Eq. (5a), the length of the bolt shank is

3 1.50 1.5 inchd

l = − = (5d)

Substituting Eqs. (4b) and (5d) into Eq. (4a), the length of the threaded portion of the bolt within the grip

is

2.40 1.5 0.9 inchest

l = − = (6)

Substituting Eqs. (3), (4), (5), and (6) into Eq. (2), the stiffness of each bolt is

17

660.3068 0.226 30 10 3.381 10 lb/in

0.3068 0.9 0.226 1.5b

k× × ×= = ×× + ×

(7)

The total stiffness of the three members can be written from Eq. (8.18), see page 427, as

m m m G1 2

1 1 1 1

k k k k= + + (8a)

Substituting 1

120 Mlb/in,mk = 2

180 Mlb/in,mk = and 20 Mlb/inGk = into Eq. (8a), the total stiffness of

the three members is

6 66

61 10 23x 10 in/lbs

120 180 20

0

60

1 10

3m

k= + + = (8b)

Therefore, the total stiffness of the three members is

615.652 10 lbs/inmk = × (8c)

Note that the total stiffness is close to the stiffness of the gasket alone. Substituting Eqs. (7) and (8c) into

Eq. (1), the stiffness constant of the joint is

6

6 63.381 10

C 0.1783.381 10 15.562 10

×= =× + ×

(9a)

Recall that the recommended range of the stiffness constant of a bolted connection, see the Table 8-12,

page 437, is

0.10 0.20C≤ ≤ (9b)

(ii) 6 Points. The factor of safety guarding against joint separation can be written from Eq. (8-30), see

page 441, as

0

max(1 )

iFnC P

=−

(10)

The maximum force acting on each bolt can be written as

max

max6

cylp A

P = (11a)

The inner diameter of the cylinder body is J 8 inches,= therefore, the cross-sectional area is

2 22J 8 50.266 in

4 4cyl

Aπ π× ×= = = (11b)

Substituting the maximum pressure and Eq. (11b) into Eq. (11a), the maximum force acting on a bolt is

max

0.261500 x 512566.5 lbs

6

6P == (12)

18

Substituting the preload 12000 lbi

F = and Eqs. (9a) and (12) into Eq. (10), the factor of safety guarding

against joint separation is

0 1.17(1 0.178)12566.5

12000n = =

− (13)

Therefore, the bolts are safe against joint separation.

The factor of safety guarding against overloading (the overload factor of safety) can be written from

Eq. (8-29), see page 440, as

max

p t i

L

S A Fn

CP

−= (14a)

The proof strength of each bolt from Table 8-9, see page 433, is

85pS = kpsi (14b)

Substituting Eqs. (3a), (9a), (12), and (14b) into Eq. (14a), the overload factor of safety is

3 0.226 120003.22

0.17

85 x 10 x

x 12566.58L

n−= = (14c)

Therefore, the bolts are safe against overloading.

(iii) 9 Points. The fatigue factor of safety using the modified Goodman criterion can be written from Eq.

(8-38), see page 446, as

( )

( )

e ut i

f

ut a e m i

S Sn

S S

σσ σ σ

−=+ −

(15)

The fully corrected endurance strength for a bolt with rolled threads from Table 8-17, see page 445, is

18.6e

S = kpsi (16a)

The ultimate tensile strength of a bolt from Table 8-9, see page 433, is

120ut

S = kpsi (16b)

The preload stress can be written from Example 8-3, see page 439, as

i

i

t

F

Aσ = (17a)

Substituting the preload 12000 lbi

F = and Eq. (3a) into Eq. (17a), the preload stress is

212000 53097 lbs/in0.226

iσ = = (17b)

The alternating component of the stress can be written from Eq. (8-35), see page 445, as

max min( )

2a

t

C P P

Aσ −= (18)

19

The minimum force acting on each bolt can be written as

min

min6

cylp A

P = (19a)

Substituting the minimum pressure and Eq. (12b) into Eq. (19a), the minimum force acting on each bolt

is

min

0.21250 x 510472 lbs

6

66P == (19b)

Substituting Eqs. (3a), (9a), (12b), and (19b) into Eq. (18), the alternating component of the stress is

0.178(12566.5 104720.826 kpsi

2 x 0.22

)

6a

σ −= = (20)

The mean component of the stress can be written as

max min )(

2

i

m

t t

C P P F

A Aσ ++= (21a)

Substituting Eqs. (3a), (9a), (12), and (19b) into Eq. (21a), the mean component of the stress is

20.178(12566.5 10472 53097 62169 lbs/in2 x 0.226

)m

σ += + = (21b)

Substituting Eqs. (16a), (16b), (17b), (20), and (21b) into Eq. (15), the fatigue factor of safety is

18.6(120 53.097)4.64

120x0.826 18.6(62.169 53.097)f

n−= =

+ − (22)

Therefore, the bolts are safe against fatigue failure.

Incorrect Solution. For a repeated load situation (which is not the case here), the fatigue factor of safety

using the modified Goodman criterion can be written from Eq. (8-45), see page 447, as

(

)(

)e ut if

a ut e

S Sn

S S

σσ

−=+

(23)

Then substituting Eqs. (15b), (16b), (18), (19b), (20b) into Eq. (23), the fatigue factor of safety is

18.6(120 53.097) 1244.410.86

0.826(120 18.6) 114.48f

n−= = =

+ (24)

Note that the answers from Eqs. (22) and (24) are not in agreement. The reason is that the given load P

in this problem is not a repeated load, therefore, Eq. (23) is not valid in this case.

20

Solution to Problem 4. (i) 5 Points. The free-body diagram of the lower shoe is shown in Figure 1 and

the free-body diagram of the upper shoe is shown Figure 2.

Figure 1. The free body diagram of the lower shoe.

Figure 2. The free body diagram of the upper shoe.

Note from the free body diagram of the lower shoe, see Figure 1, the moment about the hinge pin O2

produced by the normal force is counterclockwise and the moment produced by the friction force is

clockwise. Since the two moments are acting in opposite directions then the lower shoe is “self-

energizing” and under certain conditions can become self-locking.

The normal force acting on the lower shoe can be written as

( ) max maxArea w ( )n lowerF p p rθ= = (1a)

where the face width of the shoe is given as w 1.5 inches.= Substituting the maximum pressure of the lining material

max225 psip = (which acts on the self-energizing shoe) and the given data into Eq. (1a),

the normal force acting on the lower shoe is

(Fn)lower = 225 x 1.5 x 3 x (30o x π/180o) = 530.13 lb (1b)

The frictional force acting on the lower shoe can be written as

(Ff)lower = µ(Fn)lower (2a)

a

b

Fa

c

O1 F

n

Ff

Ry

Rx

Fa

b

c

a

Ry

Rx

Fn

Ff

O2

21

Substituting Eq. (1b) and the coefficient of friction for the brake lining material µ = 0.35 into Eq. (2a),

the frictional force acting on the lower shoe is

(Ff)lower = 0.35 x 530.13 = 185.55 lb (2b)

The frictional torque on the brake drum due to the lower shoe can be written as

Tlower = r (Ff)lower (3a)

Substituting Eq. (2b) into Eq. (3a), the frictional torque due to the lower shoe is

Tlower = 3 x 185.55 = 556.65 in-lb (3b)

For static equilibrium, the sum of the moments about the hinge pin O2, see Figure 1, is

O2

0M =∑ (4a)

which can be written as

aFa – b(Fn)lower + c(Ff)lower = 0 (4b)

Substituting Eq. (2a) into Eq. (4b), the sum of the moments about the hinge pin O2 can be written as

aFa – b(Fn)lower + c(µFn)lower = 0 (4c)

Rearranging Eq. (4c), the actuating force acting on the lower shoe can be written as

( )a n lowerb c

F Fa

µ− =

(5a)

where the vertical distance c is

3 1.75 1.25 inchesc r e= − = − = (5b)

Substituting Eqs. (1b) and (5b) into Eq. (5a), the actuating force acting on the lower shoe is

4 (0.35)(1.25)x 530.13 269.8 lb

7a

F− = =

(6)

Note that this is the same force that acts on the upper shoe (which is a self-deenergizing shoe), therefore,

the pressure on the upper shoe cannot be the maximum pressure of the friction lining material (it must be

less than max

225 psip = ).

(ii) 5 Points. The free body diagram of the upper shoe is shown in Figure 2. The moments about the

hinge pin O1 produced by the friction force and the normal force are in the same direction (that is,

clockwise). Therefore, the shoe is “self-deenergizing” and cannot become self-locking. From the free

body diagram, the sum of the moments about the hinge pin O1 can be written as

– aFa + b(Fn)upper + c(Ff)upper = 0 (7a)

Substituting Eq. (2a) into Eq. (7a), the sum of the moments about the hinge pin O1 can be written as

aFa – b(Fn)upper – c(µFn)upper = 0 (7b)

Rearranging Eq. (7b), the actuating force acting on the upper shoe can be written as

22

( )a n upperb c

F Fa

µ+ =

(8a)

Then rearranging Eq. (8a), the normal force acting on the upper shoe can be written as

( )n auppera

F Fb cµ = +

(8b)

Substituting Eq. (6) and the known data into Eq. (8b), the normal force acting on the upper shoe is

( ) 7 269.8 425.6 lb4 (0.35)(1.25)

n upperF

= = +

(9)

Therefore, the frictional force acting on the upper shoe is

( ) ( ) (0.35)(425.6) 148.96 lbf n upperupperF Fµ= = = (10)

The frictional torque on the brake drum due to the upper shoe is

( ) ( )upper f n upperupperT r F r Fµ= = (11a)

Substituting Eq. (10) into Eq. (11a), the frictional torque due to the upper shoe is

(3)(148.96) 446.88 in-lbupper

T = = (11b)

Aside: The pressure on the upper shoe is not the same as the pressure on the lower shoe. Recall from

Homework 12 that the ratio of the braking torque on each shoe is equal to the ratio of the pressure that is

on each shoe, that is

(

(

)

)

upper a uppe

l

r

loweo e rw r a

T p

T p= (12a)

Rearranging this equation, the pressure on the upper shoe can be written as

( )()upper

a a

lower

upper lower

Tp p

T= × (12b)

Substituting Eqs. (3b) and (11b) into Eq. (12b), the pressure on the upper shoe is

2446.88)556.

( 225 180.63 lb/i65

nuppera

p = × = (13)

Note that the pressure on the upper shoe is, as expected, less than the pressure on the lower shoe.

The torque capacity of the brake can be written as

lower upperT T T= + (14a)

Substituting Eq. (3b) and (11b) into Eq. (14a), the torque capacity of the brake is

556.65 446.88 1003.53 in-lbT = + = (14b)

23

(iii) 5 Points. The condition for the lower shoe (the self-energizing shoe) to be self-locking is

0a

F = (15a)

Substituting Eq. (15a) into Eq. (5a), the condition for self-locking is

0b c

a

µ− = or 0b cµ− = (15b)

Rearranging Eq. (15b), the vertical distance from the hinge pin O2 that will make the lower shoe be self-

locking is

411.43 inches

0.35

bc

µ= = = (16)

This answer indicates that the location of the hinge pin O2 must be moved from the original position

shown in Figure 4. The original vertically downward distance from the hinge pin O2 to the drum surface

(that is, 1.25 inchesc = ) must now be increased vertically upward to 11.43 inchesc = (which places the hinge pin O2 above the hinge pin O1). In this case, the vertically downward distance from the hinge pin

O2 to the center of the drum, see Eq. (5b), must be rewritten as

e c r= − that is 11.43 3 8.43 inchese = − = (17)

Aside: Substituting the actuating force on the lower shoe 0a

F = into Eq. (4b), and rearranging, the frictional force on the lower shoe can be written as

( ) ( )f n lowerlowerb

F Fc

=

(18a)

Then substituting Eqs. (1b) and (16) into Eq. (18a), the frictional force on the lower shoe is

( ) 4 530.13 185.55 lb11.43

f lowerF

= =

(18b)

Note that the frictional force given by Eq. (18b) is the same as the frictional force given by Eq. (2b).

Therefore, the frictional torque due to the lower shoe is as given by Eq. (3b), that is

Tlower = 556.65 in-lb (19)

Also, when the actuating force is 0a

F = then the normal force acting on the upper shoe, see Eq. (8b), is

( ) 0n auppera

F Fb cµ = = +

(20)

Therefore, the frictional torque due to the upper shoe, see Eq. (11a), is also zero, that is

Tupper = r µ (Fn)upper = 0 (21)

So, if the lower shoe is made self-locking then this drum brake will act like a single-short shoe drum

brake with only the lower shoe providing the braking torque. In this case, the total braking torque of the

drum brake is given by Eq. (19), that is

Ttotal = Tlower = 556.65 in-lb (22)