1

ME 452: Machine Design II Spring Semester 2016 Name of Student: ____________________________________________________ Circle your Lecture Division Number: Lecture 1 Lecture 2

FINAL EXAM

Thursday, May 5th, 2016

OPEN BOOK AND CLOSED NOTES For full credit you must show all your work and calculations clearly and logically on the sheets of paper attached to the end of each problem.

Please use only the blank pages for all your work and write on one side of the paper only.

Please draw your plots and figures (such as free body diagrams) neatly and clearly labelled. At the end of the exam, please staple each problem separately. Staple this cover sheet and your crib sheet in front of Problem 1.

Problem 1

Problem 2

Problem 3

Problem 4

Total

2

ME 452: Machine Design II FINAL EXAM SPRING 2016 Name of Student: _____________________________ Circle one: Lecture 1 Lecture 2 Problem 1 (25 points). The spur pinion shown in Figure 1 has a diametral pitch of 8 teeth/in, a face width of 1.25 in, 32 teeth, and a 20 pressure angle. The pinion rotates at 2000 rpm and transmits a load of 400 lbs to the meshing spur gear. The pinion and gear are commercial quality and the load is applied at the highest point of single-tooth contact. Poisson’s ratio and modulus of elasticity for the pinion and the gear are P 0.292, PE 30 Mpsi, G 0.211, and GE 14.5 Mpsi, respectively.

For the pinion, the life is 810 cycles, the AGMA bending strength and wear strength are 30 kpsi and

95 kpsi, respectively, the repeatedly applied bending strength stress cycle factor nY 0.9768 and the

pitting resistance stress cycle factor NZ 0.9484. Also, for the pinion, the AGMA geometry factor

J 0.38 and the AGMA modification factors are:

OK 1, VK 1.414, SK 1, mK 1.414, BK 1, TK 1, and RK 1

(i) Determine the horsepower transmitted by the gearset.

(ii) Determine the AGMA bending stress and the AGMA contact stress acting on the pinion.

(iii) Determine the AGMA factor of safety guarding against bending fatigue failure of the pinion.

(iv) Determine the AGMA factor of safety guarding against wear of the pinion.

Figure 1. The spur gear and pinion.

3

ME 452: Machine Design II FINAL EXAM SPRING 2016

Name of Student: ___________________________ Circle one: Lecture 1 Lecture 2 Problem 2 (25 points). The spring index of a linear helical compression spring is 8 and the mean coil diameter of the spring is 4 inches. The spring material is ASTM No. A228 music wire. The spring has squared and ground ends and is not set and is not peened. The spring is subjected to an initial load of 202.5 lbs and the corresponding spring length is 6 inches, see Figure 2. The spring is then subjected to a maximum working load of 405 lbs and the corresponding spring length is 5 inches.

Determine the following (do not use an iteration procedure):

(i) The total number of coils in the spring.

(ii) The load required to shut the spring (that is, the load at the shut height of the spring).

(iii) Plot the load acting on the spring against the length of the spring. On your plot, clearly show the

values of: (a) the free length, (b) the solid length, and (c) the load required to shut the spring.

(iv) The static factor of safety at the shut height of the spring. Is failure predicted to have occured?

Figure 2. Loads on the compression spring (Not drawn to scale).

F = 202.5 lbs

F = 202.5 lbs

6 in

5 in

F = 405 lbs

F = 405 lbs

4

ME 452: Machine Design II FINAL EXAM SPRING 2016 Name of Student: _______________________________ Circle one: Lecture 1 Lecture 2 Problem 3 (25 points). Plate A is bolted to flange B by a 3/4 in - 10 UNC-2A, SAE grade 5, coarse

series, steel bolt, as shown in Figure 3. The stiffness of the plate is 645 x 10 lbs / in, the stiffness of the

flange is 625 x 10 lbs / in, and the stiffness of the confined gasket is 65 x 10 lbs / in. The bolt is

preloaded to 320 x 10 lbs and then subjected to a repeated tensile load which has a maximum value of 34.8 x 10 lbs. The fully corrected endurance limit of the bolt is eS 17.5 kpsi. The factor of safety

guarding against joint separation is specified as 5. Determine the following (do not use an iteration procedure):

(i) The maximum value of the external load that is taken by the bolt.

(ii) The total stiffness of the members (plate A and flange B). Neglect the stiffness of the two washers.

(iii) The factor of safety guarding against overloading the bolt.

(iv) The infinite life fatigue factor of safety for the bolt using the modified Goodman criterion. Assume

that the slope of the load line is 1.

Figure 3. A plate bolted to a flange.

Plate A

Flange B

A confined gasket

5

ME 452: Machine Design II FINAL EXAM SPRING 2016 Name of Student: ____________________________ Circle one: Lecture 1 Lecture 2 Problem 4 (25 points). An external contracting rim-type brake has two symmetric shoes acting on the drum as shown in Figure 4. The brake lining material for both the left shoe and the right shoe is woven cotton. The drum is rotating counterclockwise and the radius of the drum is 6 inches. The face width of each shoe is 3 inches. The figure shows that the limiting value of the actuating force F is horizontal and the magnitude is the same on both the left shoe and the right shoe. (i) Determine the moments of the normal force and the friction force on the self-energizing shoe.

(ii) Determine the moments of the normal force and the friction force on the self-deenergizing shoe.

(iii) Determine the braking capacity of the left shoe and the right shoe.

4 in 4 in 8 in

12 in

12 in

120◦

FF

DrumRotation

Left Shoe Right Shoe

OQ

Drum

30◦

Figure 4. The geometry of the rim-type drum brake. (Not drawn to scale).

2 in. 2 in.

6 in.

6 in.

Left Shoe Drum

Right Shoe

Drum Rotation

F F

G H

30°

120°

3 in 3 in 5 in

8 in

8 in

Left Shoe Drum

Right Shoe

Drum Rotation

F F

A B

r = 6 in 30°

120°

5 in

8 in

8 in

6

Solution to Problem 1 (25 points). (i) 6 Points. The horsepower transmitted by the gearset, see Example 14-1, page 740, is

tW Vhp

33,000 (1)

The pitch line velocity, see Eq. (13.34), page 707, and Example 14-1, see page 740, can be written as

V12

P Pd n (2)

The pitch circle diameter of the pinion, see Eq. (13-1), page 676, and Example 13-1, page 683, is

324 in

8P

P

Nd

P (3)

Substituting Eq. (3) and the pinion speed 2000 rpmPn into Eq. (2), the pitch line velocity is

4 2000V 2094.4 ft/min

12

(4)

Substituting Eqs. (4) and the transmitted load tW 400 lbs into Eq. (1), the horsepower is

400 2094.4hp 25.39 hp

33,000

(5)

Alternative Method: The horsepower transmitted by the gearset can also be written from Eq. (13-33), see page 706, as

P Php = T ω (6)

The torque transmitted by the pinion can be written from Eq. (b), see page 706, as

tP PT = (d / 2) W (7a)

Substituting Eqs. (3) and tW 400 lbs into Eq. (7a), the torque transmitted by the pinion is

PT (4 / 2) 400 800 in-lbs (7b)

The angular velocity of the pinion is 2

209.44 rad/s60

PP

n (8)

Substituting Eqs. (7b) and (8b) into Eq. (6), the horsepower transmitted by the spur gearset is

800 209.44hp 25.39 hp

6600

(9)

7

(ii) 10 Points. The AGMA bending stress for the pinion from Eq. (14-15), see page 746, is

t m Bo v s

P K Kσ = W K K K

F J (10a)

Substituting tW 400 lbs , OK 1, VK 1.414, SK 1, 1P 8 in , F 1.25 in, mK 1.414, BK 1,

and J 0.38 into Eq. (10a), the AGMA bending stress for the pinion is

(8) ( (1)σ = (400) (1) ( 13.47 kpsi

(1.25) (0.38

1.414)1.414)(1)

) (10b)

The AGMA contact stress for the pinion can be written from Eq. (14-16), see page 746, as

cσ = C ft mp o v s

P

CKK K K

d FW

I (11)

The AGMA elastic coefficient can be written from Eq. (14-13), see page 744, as

22 11

1p

GP

p G

C

E E

(12a)

Substituting 0.292,P 30 Mpsi,PE 0.211,G and 14.5 MpsiGE into Eq. (12a), the elastic

coefficient is

2 2

6

1

6

/2

1 0.292 1 0.21130 10 14.5 10

11817.26 psipC

(12b)

The surface condition factor, from Fig. 14-18, see page 767, and Example 14-4, see page 770, is

1fC (13)

The geometry factor for external gears can be written from Eq. (14-23), see page 755, as

cos sin

2 1G

N G

mI

m m

(14)

Note that Eq. (14) includes the load sharing ratio, however, Eqs. (b) and (c) on page 755 do not. The current practice for spur gears is to assume that the load sharing ratio, see page 753, is

1Nm (15)

The gear ratio can be written from Eq. (14-22), see page 754, as

G GG

P P

d Nm

d N (16)

8

The radius of the pitch circle of the gear, see Example 13-1, page 683, is

5.5 2 3.5 inG Pr C r (17a)

Therefore, the pitch circle diameter of the gear is

2 2 x 3.5 7 inG Gd r (17b)

Substituting Eqs. (3) and (17b) into Eq. (16), the gear ratio is

7 3.51.75

4 2 Gm (18)

Substituting Eqs. (15), (18), and the pressure angle 20o into Eq. (14), the geometry factor is

o ocos 20 sin 20 (1.75)0.102

2 x 1 1.75 1I

(19)

Substituting Eqs. (12b), (18), and (19) into Eq. (11), the AGMA contact stress for the pinion is

c

(1. (1)σ = 1

414)(400)(1)(1.414)(1) psi

(817.26

(0.1024)( 2 ) )1. 5 (20a)

that is

c 1568.627 psiσ 1817.26 = 71.97 kpsi (20b)

(iii) 4 Points. The factor of safety guarding against bending fatigue failure of the pinion can be written from Eq. (14-41), page 765, as

( / )t N T RF

S Y K KS

(21a)

Substituting tS 30 kpsi, NY 0.9768, 1,TK 1,RK and Eq. (10b) into Eq. (21a), the factor of

safety guarding against bending fatigue failure of the pinion is

30x 0.97682.18

13.47FS (21b)

(iv) 5 Points. The factor of safety guarding against pitting failure of the pinion, see Eq. (14-42), page 765, can be written as

2

NZ / ( )cH

c

T RS K KS

(22a)

Substituting cS 95 kpsi, NZ 0.9484, 1,TK 1,RK and Eq. (20b) into Eq. (22a), the factor of

safety guarding against pitting failure of the pinion is

9

2295 x 0.9484

1.252 1.5771.97HS

(22b)

10

Solution to Problem 2 (25 points). (i) 6 points. The total number of coils for a spring with squared and ground ends, see Table 10-1, see page 521, is

atN N 2 (1)

The number of active coils for the spring can be written from Eq. (10-9), see page 520, is

4a 3

N8

d G

D k (2)

The spring stiffness can be written as Fy

k

(3a)

Substituting the given data into Eq. (3a), the spring stiffness is

405 202.5 202.5 lbs/in6 5

k

(3b)

The spring index can be written from Eq. (10-1), see page 519, is

DCd

(4a)

Rearranging this equation, the wire diameter can be written as

4 0.5 in8

DdC

(4b)

The modulus of elasticity and the modulus of rigidity (for ASTM No. A228 music wire and a wire diameter 0.125 inchesd ) from Table 10-5, see page 526, are

28.0 MpsiE and 11.6 MpsiG (5)

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

4 6a 3

0.5 x 11.6 x 10N 6.99 78 x 4 x 202.5

(6)

Note that the recommended range, see page 533, is a3 N 15. Substituting Eq. (6) into Eq. (1), the total number of coils in the spring is

tN 6.99 2 8.99 (7a)

In general, the total number of coils is specified to the nearest 1/4 of a turn, see page 533, i.e.,

tN 9 (7b)

(ii) 6 points. The load required to shut the spring can be written as

( )sshut fF k L L (8)

11

The free length of the spring can be written as

af initialL L y (9)

where the assembled length of the spring is specified as

6 inaL (10a) and the initial deflection is

min 202.5 1 in202.5 initial

Fy

k (10b)

Substituting Eqs. (10a) and (10b) into Eq. (9), the free length of the spring is

6 in +1in = 7 infL (11)

The shut height (or the solid length) of the spring with squared and ground ends, see Table 10-1, page 521, can be written as

L d Ns t (12a)

Substituting Eqs. (4b) and (7a) into Eq. (12a), the shut height of the spring is

0.5 x 9 4.5 insLs (12b)

Substituting Eqs. (3b), (11), and (12b) into Eq. (8), the load required to shut the spring is

202.5 (7 4.5) 506.25 lbsshutF (13)

(iii) 6 Points. A plot of the load F against the length L of the spring is shown in the figure below.

The plot shows: (i) the free length, (ii) the assembled length, (iii) the minimum working length, (iv) the solid length, (v) the initial deflection, (vi) the working deflection, and (vii) the clash deflection. The plot also shows the load required to shut the spring.

yinitial = 1 ins

ywork

yclash

0

Load, F

La = 6 ins Length, L

Fshut = 506.25 lbs

Lf = 7 ins

Fmin = 202.5 lbs

Ls = 4.5 ins Lm = 5 ins

Fwork = 405 lbs

12

(iv) 7 points. The static factor of safety at the shut height of the spring can be written from Eq. (a), see page 530, as

sys

SN (14)

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

mutS Ad (15)

The coefficient and the exponent from Table 10-4, see page 524, respectively, are

A = 201 kpsi.inm and m = 0.145 (16)

Substituting Eqs. (4b) and (16) into Eq. (15), the ultimate tensile strength of the spring material is

0.145201 0.5 222.25 kpsiutS (17)

Note that the diameter of the wire must be expressed in inches in this equation. The torsional yield strength for a not set spring can be written from Table 10-6, see page 526, as

0.45sy utS S (18a)

Substituting Eq. (17) into Eq. (18a), the torsional yield strength for the not set spring is

0.45 222.25 100.01 kpsisyS (18b)

Since the spring is not set then the shear stress at the shut height can be written from Eq. (a), see page 530, as

3

8shut

B

F DK

d

(19)

where the Bergstrasser factor can be written from Eq. (10-5), see page 519, is

4 2 4 8 2 1.17244 3 4 8 3B

CKC

(20)

Substituting Eqs. (4b), (13), and (20) into Eq. (19), the shear stress at the shut height is

8 506.25 41.1724 48.365 kpsi

30.5

(21)

Substituting Eqs. (18b) and (21) into Eq. (14), the factor of safety at the shut height is

100.01 2.148.365sN (22)

Since the factor of safety at the shut height is greater than 1 then failure is predicted not to have occurred.

13

Solution to Problem 3 (25 Points). (i) 7 Points. The maximum value of the external load that is taken by the bolt can be written from Eq. (d), see page 436, as

maxbP C P C P (1a)

and the maximum value of the external load that is taken by the members is

m maxP (1 ) (1 )P C P C (1b)

The maximum external tensile load acting on the bolted connection is specified as

maxP = P 4800 lbs = 4.8 kip (1c)

The factor of safety guarding against joint separation can be written from Eq. (8-30), see page 441, as

0 (1 )iF

nP C

(2a)

Rearranging this equation, the joint constant can be written as

0

1 iFC

P n (2b)

The preload on the bolt is specified as 20,000 lbsiF (3a)

and the factor of safety guarding against joint separation is specified as

0 5n (3b)

Substituting Eqs. (1b), (3a), and (3b), into Eq. (2b), the joint constant is

200001 1 0.833 0.167

4800 5C

(4)

Substituting Eqs. (1c) and (4) into Eq. (1a), the maximum value of the external load that is taken by the bolt is

bP 0.167 4800 800 lbs (5a)

Substituting Eqs. (1c) and (4) into Eq. (1b), the maximum value of the external load that is taken by the members is

mP 0.833 4800 4000 lbs (5b)

Check: The resultant load acting on the members can be written from Eq. (8-25), see page 436, as

0m m iF P F

(ii) 4 Points. The stiffness of the members can be written from Eq. (8-18), see page 427, as

14

1 2

1 1 1

m m mk k k (6a)

The stiffness of plate A and the stiffness of flange B are specified, respectively, as

1645 x 10 lbs / inmk and 2

625 x 10 lbs / inmk (6b)

Substituting Eqs. (6b) into Eq. (6a), the stiffness of the members is

161 1

16.07 x10 lbs / in45 25mk

(7)

Wrong Solution. It is wrong to include the stiffness of the confined gasket in the stiffness of the members. If the confined gasket is included in Eq. (6a) then the stiffness of the members is written as

1 2

1 1 1 1

m gk k k k (8a)

In this case, the stiffness of the members would be

161 1 1

3.81 x 10 lbs / in45 25 5mk

(8b)

Equation (7) gives the correct value for the stiffness of the members. Equation (8b) does not give the correct value for the stiffness of the members (i.e., it is close to the stiffness of the gasket). Aside: The stiffness constant of the joint can be written from Eq. (f), see page 436, as

b

mb

kC

k k

(9a)

Substituting Eqs. (3) and (6) into Eq. (9a) gives

60.167

16.07 x10b

b

k

k

(9b)

Rearranging this equation, the stiffness of the bolt is

63.21 x 10 lbs / inbk (10a)

Comparing Eqs. (7) and (10a), the ratio of the stiffness of the members to the stiffness of the bolt is

6

616.07 x 10

= 53.21 x 10

m

b

k

k (10b)

Notes: (i) The load to cause joint separation can be written from Eq. (e), see page 441, as

o oP n P (11)

15

Substituting Eqs. (1c) and (3b) into Eq. (11), the load to cause joint separation is

5 4800 24,000 lbsoP (12)

(ii) The resultant load acting on the members can be written from Eq. (8-25), see page 436, as

m m iF P F (13)

The portion of the external load taken by the members, as determined in Eq. (5b), is

(1 0.167) 4800 4000 lbsmP (14)

Substituting Eqs. (3a) and (14) into Eq. (13), the resultant load acting on the members is

4000 20000 16000 lbsmF (15)

Note that the members are in compression. (The factor of safety is greater than 1). Aside: A plot of the resultant load acting on the members mF against the external load P is shown in the

figure below. When the external load 0P then the load in the members is 20,000 lbsm iF F and

when the maximum external tensile load max 4,800 lbsP P then the load in the members is

16,000 lbs.mF Also, from Eq. (12) the load required to cause joint separation is 0 24,000 lbs.P

Figure. Plot of the resultant load acting on the members mF against the external load P.

(iii) 7 Points. The factor of safety guarding against overloading (or the load factor), see Eq. (8-29), page 440, can be written as

p t iL

S A Fn

CP

(16)

The proof strength for a 3/4 in SAE Grade No. 5 steel bolt, see Table 8-9, page 433, is

85 kpsipS (17)

Fm (kip)

Joint Separation

0

Fi = - 20 kip

Fmax = - 16 kip

Pmax = 4.8 kip Po = 24 kip

1-C = 0.833

P (kip)

16

The tensile stress area of 3 / 4 in-10 UNC bolt, see Table 8-2, page 413, is

20.334 intA (18)

The preload force is specified as

F 20,000 lbs = 20 kipi (19)

Substituting Eqs. (1c), (4), (17), (18), and (19), into Eq. (16), the factor of safety guarding against overloading is

(85 0.334) 2010.47

0.167 4.8Ln

(20)

(iv) 7 Points. The infinite life fatigue factor of safety of the bolt using the modified Goodman criterion can be written from Eq. (8-48), see page 447, as

t2 ( A )

( )e ut i

feut

S S Fn

C P S S

(21)

Note that this equation is valid for a slope of the load line 1r (as specified in the problem statement). The tensile strength for a 3 / 4 in SAE Grade No. 5 steel bolt from Table 8-9, see page 433, is

120 kpsiutS (22)

The fully corrected endurance strength for a SAE Grade 5 steel bolt with rolled threads from Table 8-17, see page 445, is

18.6 kpsieS (23)

Substituting Eqs. (4), (17), (18), (19), (22), and (23) into Eq. (21), the infinite life fatigue factor of safety of the bolt using the modified Goodman criterion can be written as

2 x 18.6 (120 x 0.334 20

4.8

)

0.167 (120 18.6)fn

(24a)

Therefore, the infinite life fatigue factor of safety of the bolt using the modified Goodman criterion is

746.986.72

110.10fn (24b)

Alternative Approach: The infinite life fatigue factor of safety of the bolt using the modified Goodman criterion can also be written from Eq. (8-37), see page 446, as

af

a

Sn

(25)

The alternating component of the strength using the modified Goodman criterion can be written from Eq. (c), see page 445, as

17

( )

( )e a ut i

aa e mut i

S SS

S S

(26)

For the given slope of the load line is 1, see Figure 8.20, page 446, we can write

m ai (27)

Substituting Eq. (27) into Eq. (26), and rearranging, the alternating component of the strength is

( )e ut ia

eut

S SS

S S

(28)

The minimum stress in the bolt (i.e., the preload stress), see Example 8-5, page 448, can be written as

ii

t

F

A (29a)

Substituting Eqs. (18) and (19) into Eq. (29a), the minimum stress in the bolt is

2059.88 kpsi

0.334i (29b)

Substituting Eqs. (22), (23), and (29b) into Eq. (28), the alternating component of the strength is

18.6(120 59.88)8.068 kpsi

120 18.6aS

(30)

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

2at

C P

A (31a)

Substituting Eqs. (1c), (4), and (18), into Eq. (31a), the alternating component of the bolt stress is

0.167 4.81.2 kpsi

2 0.334a

(31b)

Substituting Eqs. (30) and (31b) into Eq. (25), the infinite life fatigue factor of safety is

8.0686.72

1.2fn (32)

Note that this answer is in good agreement with Eq. (24b).

18

Solution to Problem 4. (i) 9 Points. The x and y axes for the left shoe and the right shoe are as shown in Figure 2.

4 in 4 in

12 in

12 in

120◦

FF

DrumRotation

OQ

1

a8 in

22

1

xx

yy

c = 24 in

Figure 2. The x and y axes for each shoe.

The angle to the heel of each shoe is

11

o330 tan 9.44

8

(1a)

The angle to the toe of each shoe is

2 1o o o120 120 9.44 129.44 (1b)

The location of the maximum pressure acting on each shoe, see page 834, is

maxo90a (2)

The face width of each shoe is specified as

3 inb (3)

The distance from the center of the drum to the hinge pins is

3 in 3 in

5 in

8 in

8 in

A B

c = 16 in

F F

Drum Rotation

19

2 23 8 8.54 insa (4)

The moment arm for the actuating force F is

16 insc (5)

Note that for the counterclockwise rotation of the drum the left shoe is self-energizing and the right shoe is self-deenergizing. Based on the fact that the limiting actuating force is acting on the left shoe and this shoe is self-energizing then the limiting pressure is equal to the maximum pressure of the frictional material, that is

maxap p (6a)

The maximum pressure for woven cotton material from Table 16.3, see page 862, is

2max 100 lbs/inp (6b)

and this is the maximum pressure acting on the left shoe. Since the right shoe is self-deenergizing then the limiting pressure on the right shoe is less than the

maximum pressure of the frictional material (cannot use the value in Table 16.3 to obtain the maximum pressure). The limiting pressure of the right shoe is unknown at this point in the analysis. See Eq. (19b).

For the left shoe (the self-energizing shoe), the moment due to the normal force about the hinge pin A can be written from Eq. (16-3), see page 841, is

2

1

2sinsina

Na

p b r aM d

(7a)

where the integral can be written from Eq. (16-8), see page 836, as

0

2

1 0

129.442 2 1

2 19.44

( 1sin sin(2 2

)) 1.3734

2 4B d

(7b)

Substituting Eqs. (2), (3), (4), (6b), and (7b), into Eq. (7a), the moment due to the normal force about the hinge pin A can be written as

3100 3 6 8.541.3734 21.122 10 lbs.ins

1NM

(8)

For the left shoe (the self-energizing shoe), the moment due to the frictional force about the hinge pin A can be written from Eq. (16-2), see page 841, as

2

1

sin ( cos )sin

af

a

f p b rM r a d

(9)

The first integral in Eq. (9) can be written from Eq. (16-8), see page 836, as

20

00

0 0

129.449.44 2 22 1

9.4

1

4 9 4

2

. 4

sinsin

inc

so

2sA d

(10a)

The second integral in Eq. (9) can be written as

0

0 0

0 0

0

9.44129.44 129.44

1 2 2 19.44 9.449. 4

12

4

cos cos cos ssin coC d (10b)

Substituting Eqs. (10) into Eq. (9), the moment due to the frictional force about the hinge pin A can be written as

( )sin

af

a

f p b rM rC aA

(11a)

where the integrals in Eqs. (10) are

0.2982 0.0345 0.2848A (12a) and

0.9865 0.6358 1.6217C (12b)

and the coefficient of friction for woven cotton, see Table 16.3, page 862, is

0.47f (12c)

Substituting Eqs. (2), (3), (4), (6b), and (12), into Eq. (11a), the moment of the frictional force about the hinge pin A can be written as

0.47 100 3 6(6 x1.6217 8.54 x 0.2848) lbs.ins

1fM

(13a)

that is 3846 7.298 6.174 10 lbs.insfM (13b)

(ii) 12 Points. To determine the moments of the normal force and the frictional force about the hinge pin B. First, the maximum pressure must be determined for the right shoe. Note that the limiting pressure on this shoe is not the maximum pressure of woven cotton given in Table 16.3.

Recall that the limiting actuating force for the left shoe (that is, the self-energizing shoe) can be written from Eq. (16-4), see page 835, as

N fM MF

c

(14a)

Substituting Eqs. (8) and (13b) into (14a), the limiting actuating force for the left shoe is

3 321.122 10 6.174 10934.25 lbs

16F

(14b)

Note that the actuating force on the right shoe is equal to the limiting actuating force for the left shoe, that is

934.25 lbsF (15a)

21

However, the right shoe is self-deenergizing, therefore, the limiting actuating force on this shoe, see Eq. (16-7), page 836, can be written as

N fM MF

c

(15b)

Rearranging this equation, the moment about the hinge pin B due to the actuating force is

N fF c M M (16)

Substituting Eqs. (7b) and (9) into Eq. (16), the moment about the hinge pin B due to the actuating force on the right shoe can be written as

2 2

1 1

2sin sin ( cos )sin sina a

a a

p b r a f p b rF c d r a d

(17a)

or as

( )sin sina a

a a

p b r a f p b rF c B rC aA

(17b)

Rearranging this equation in terms of the limiting pressure gives

( )sin sina

a a

b r a f b rF c p B rC aA

(18)

Substituting Eqs. (2), (3), (4), (5), (7b), (11a), (11b), and (12) into Eq. (18), the moment about the hinge pin B due to the actuating force on the right shoe (that is, the self-deenergizing shoe) can be written as

3 6 8.54 0.47 3 6934.25 16 (1.3734) (6 x1.6217 8.54 x 0.2848)

1 1ap

(19a)

or as 14948 (211.12 61.758)ap (19b)

Then rearranging this equation, the maximum pressure on the right shoe is

254.77 lbs/inap (20)

Note that the maximum pressure on the right shoe is, indeed, less than the maximum pressure on the left shoe, see Eqs. (6b) and (20).

Substituting Eqs. (1), (2), (3), (4), (7b), and (20), into Eq. (7a), the moment due to the normal force about the hinge pin B is

354.77 3 6 8.541.3734 11.563 10 lbs.ins

1NM

(21)

Substituting Eqs. (1), (2), (3b), (6b), (11), (12), and (20) into Eq. (9), the moment of the frictional force about the hinge pin B can be written as

22

30.47 54.77 3 67.2973 3.381 10 lbs.ins

1fM

(22)

Check: The actuating force on the left shoe and the right shoe is the same and can be written as

N f N f

left right

M M M MF

c c

(23a)

or ( ) ( )N f N fleft rightM M M M (23b)

Note that ( ) ( )N Nleft rightM M and ( ) ( ) .f fleft rightM M

Substituting Eqs. (8), (13), (21), and (22) into Eq. (23b), gives

3 3 3 3(21.122 x10 6.174 x10 ) (11.563x10 3.381x10 )left right (24a)

that is 3 3(14.948x10 ) (14.944 x10 )left right (24b)

The difference in the left side and the right side of the equation is due to round off error. (iii) 4 Points. The total braking torque (generated by both shoes) can be written as

total rlT T T (25a)

The braking torque generated by each shoe can be written from Eq. (16-6), see page 835, as

21 2(cos cos )

sina

a

f p b rT

(25b)

Substituting the maximum pressure 2max 100 lbs/insap p , see Eq. (6b), into Eq. (25b), the braking

torque generated by the left shoe (that is, the self-energizing shoe) can be written as

20.47 100 3 6 (cos9.44 cos129.44 )lbs.ins

1

o o

lT

(26a)

Therefore, the braking torque generated by the left shoe (self-energizing shoe) is

38.232 10 lbs.inslT (26b)

Substituting Eq. (20) into Eq. (25b), the braking torque generated by the right shoe (that is, the self-deenergizing shoe) can be written as

20.47 54.77 3 6 (cos9.44 cos129.44 )lbs.ins

1

o o

rT

(27a)

Therefore, the braking torque generated by the right shoe (that is, the self-deenergizing shoe) is

34.508 10 lbs.insrT (27b)

23

Check: Note that the braking torque generated by the left shoe is, indeed, greater than the braking torque generated by the right shoe, see Eqs. (26b) and (27b). The maximum braking torque must occur on the self-energizing shoe (the left-hand shoe).

Substituting Eqs. (26b) and (27b) into Eq. (25a), the total braking torque generated by both shoes is

3 3(8.232 4.508) 10 12.74 10 lbs.instotalT (28)

Check: Comparing Eq. (26a) with Eq. (27a) we note that the ratio of the braking torque on each shoe is equal to the ratio of the maximum pressure on each shoe, that is

)

)

(

(righta

left

r

l a

pT

T p (29a)

Rearranging this equation, the braking torque on the right-hand shoe can be written as

)

)

(

(right

l

a

efr l

a t

pT T

p (29b)

Substituting Eqs. (6b), (20), and (26b) into Eq. (29b), the braking torque on the right-hand shoe is

3 38.232 1054.77

4.509 10 lbs.ins100rT (30)

Note that Eqs. (27b) and (30) are in complete agreement.