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ME 452: Machine Design II Spring Semester 2017
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 4th, 8:00 am – 10:00 am
MTHW 210 AND LILY 3118
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. In general, do not write your work on the exam pages unless instructed to do so.
Draw any free body diagrams, and other diagrams, clearly and to a reasonable size.
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 2017 Name of Student: ____________________________ Circle one: Division 1 Division 2 Problem 1 (25 points). For the spur gearset shown in Figure 1, the gear is rotating at 500 rpm and transmits 6 horsepower to the pinion. The pitch diameter of the pinion is 5 inches, the diametral pitch of the gear is 1,5 inches and the gear ratio is 2. The commercial gearset has full-depth teeth with a pressure
angle of 20o and the face widths of the gear and pinion are 0.8 inches. The gear is subjected to a light shock load and the pinion a moderate shock load where the loads are applied at the tips of the teeth. The gear and pinion operate at room temperature. The gear is nitrided through-hardened Grade 2 steel with a Brinell hardness BH 300. The known AGMA factors for the gear are: dynamic factor vK 1.25, size
factor sK 1.0, and load distribution factor mK 1.0. The backup ratio for the gear is Bm 1.5.
(i) Determine the load transmitted by the gearset. (ii) Determine the AGMA bending 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 2017 Name of Student: ____________________________ Circle one: Division 1 Division 2 Problem 2 (25 points). The mean coil diameter of a linear helical compression spring is D = 2.25 inch and the spring index is C = 9. The spring has squared and ground ends and is placed between two flat parallel plates as shown in Figure 2. The spring material is music wire A228 and the spring is peened. When the spring is assembled between the plates, the spring is subjected to a preload of 45 lbs and the spring length is 5 inches (shown on the left). The spring is then subjected to a fluctuating load. At the maximum working load of 95 lbs, the spring length is 3.75 inches (shown on the right). Determine: (i) The number of active coils (round up to the nearest quarter of a coil). (ii) The free length and the shut height of the spring. (iii) The alternating component and the mean component of the shear stress. (iv) The fatigue factor of safety using the Gerber-Zimmerli failure criterion.
Figure 2. The loads acting on the compression spring (Not drawn to scale).
F = 45 lbs
F = 45 lbs
5 in
3.75 in
F = 95 lbs
F = 95 lbs
4
ME 452: Machine Design II FINAL EXAM SPRING 2017 Name of Student: ____________________________ Circle one: Division 1 Division 2 Problem 3 (25 points). A UNF 3/ 4 inch - 16 SAE grade 8 steel bolt with rolled threads is used to clamp together two identical plates. Each plate is 1.5 inches thick as shown in Figure 3. The length of the bolt is 3.75 inches and the diameter of the washer face under the bolt head is wd 1.5 d (where d is the diameter of the bolt shank). The half-apex angle of the pressure cone distribution in the plates is
0.30 The bolt is preloaded to 330 x 10 lbs and then the bolted connection is subjected to an external
load which fluctuates between 36 x10 lbs and 39 x10 lbs. The modulus of elasticity of the bolt is
6 2b
E 30 x10 lbs/in and the modulus of elasticity of each plate is 6 2p
E 36 x10 lbs/in . Determine:
(i) The stiffness of the bolt and the total stiffness of the plates (use the pressure-cone method). (ii) The factor of safety guarding against joint separation. (iii) The load factor, that is, the factor of safety guarding against overloading. (iv) The factor of safety guarding against fatigue using the modified Goodman criterion.
Figure 3. Two identical plates clamped together by a bolt.
1.5 in
1.5 in
5
ME 452: Machine Design II FINAL EXAM SPRING 2017 Name of Student: ____________________________ Circle one: Division 1 Division 2 Problem 4 (25 points). An actuating force is applied at point A on the arm of an external rim-type brake as shown in Figure 4. The radius of the drum is 5 inches, the distance from the hinge pin O to the center of the drum C is 7 inches, and the distance from point C to point A is 9 inches. The drum is rotating counterclockwise, the shoe lining material is cermet, and the face width of the shoe is 3 inches. (i) Determine the magnitude of the limiting actuating force F acting at point A. (ii) Determine the braking capacity of the brake. (iii) Determine the magnitude and direction of the reaction force at pin O. Show this force on Figure 4.
Figure 4. The external contracting rim-type brake.
O
F
Drum Rotation
30o
120o
30o
C
r = 5 in
A
60o
OC = 7 inches CA = 9 inches
6
Solution to Problem 1. (i) 5 Points. The transmitted (or tangential) load can be written from Eq. (13-35), see page 707, as
33000tt
HW
V (1)
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
(2)
The gear ratio (or speed ratio), see page 694 and Eq. (14-22), see page 754, can be written as
G GG
P P
N dm
N d (3)
Substituting the given gear ratio 2Gm and pitch diameter of the pinion 5 inPd into Eq. (3), and
rearranging, the pitch diameter of the gear is
2 x 5 10 inchesG G Pd m d (4)
Substituting Eq. (4) and the given speed of the gear 500 rpmGn into Eq. (2), the pitch line velocity is
10 5001309 ft/ min
12tV
(5)
Substituting Eq. (5) and the given horsepower 6 hpH into Eq. (1), the transmitted load is
33000 6151.26 lb
1309tW
(6)
Check: The horsepower input can be written from Eq. (13-33), see page 706, as
63000GinH
nT (7)
Rearranging Eq. (7), the input torque can be written as
63000
Gin
HT
n (8)
Substituting the given horsepower input 6 hpH and speed of the gear 500 rpmGn into Eq. (8), the
input torque is 63000 x 6
756 lbs-in500inT (9)
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 (10a)
7
Substituting Eq. (9) and the radius of the pitch circle of the gear 5 in,Gr see Eq. (4), and into Eq.
(10a), and rearranging, the transmitted load acting on the pinion is
756151.2 lbs
5tW (10b)
Note that this answer is in good agreement with Eq. (6). (ii) 10 Points. The AGMA bending stress in the gear can be written in US customary units from Eq. (14-15), see page 746, as
( )to v s m B
W PK K K K K
F J (11)
The diametral pitch of the gear and the pinion can be written from Eq. (13-1), see page 676, as
G P
G P
N NP
d d (12)
Substituting Eq. (4) and the given diametral pitch 15 inchesP into Eq. (12), and rearranging, the number of teeth on the gear is
5 10 50G GN P d (13a)
and the number of teeth on the pinion is
5 5 25p pN P d (13b)
Since the loads are applied at the tips of the teeth and the number of teeth on the gear is 50,GN see
Eq. (13a), then the bending strength geometry factor for the gear teeth from Figure 14-6, see page 753, is
0.28J (14)
The problem states that the input gear is subjected to a light shock load and the output pinion is subjected 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
1.5oK (15a)
The dynamic factor for the gear (see Section 14.7, page 756) is given in the problem statement as
1.25vK (15b)
The size factor for the gear (see Eq. (a), page 759) is given in the problem statement as
1.0sK (15c)
The load distribution factor for the gear (see Eq. (14-30), page 759) is given in the problem statement as
mK = 1.0 (15d)
The backup ratio for the gear is given as Bm 1.5, therefore, the rim-thickness factor for the gear
from Eq. (14-40), see page 764, is
8
BK = 1.0 (15e)
Substituting the given face width of the gear F 0.8 inches, diametral pitch 15 inchesP and Eqs. (6), (14), and (15) into Eq. (11), the AGMA bending stress in the gear can be written as
151.26 5(1.5 1.25 1.0 1.0 1.0) kpsi
0.8 0.28
(16a)
Therefore, the AGMA bending stress in the gear is
26.33 kpsi = 6331 lbs/in (16b)
(iii) 10 points. The AGMA safety factor guarding against bending fatigue failure for the gear can be written from Eq. (14-41), see page 765, as
/ ( )t N T RF
S Y K KS
(17)
First, determine the bending strength geometry factor for the gear. For nitrided through-hardened Grade 2 steel the repeatedly applied bending strength (for 810 load cycles and 95 percent reliability) from Figure 14-3, see page 747, is
2B108.6H 15890 lbs/instS (18a)
Substituting the specified Brinell hardness BH 300 into Eq. (18a), the repeatedly applied bending
strength for the gear is 248470 lbs/ins 48.47 kpsitS (18b)
Second, determine the modification factors for the gear. 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, is
0.0178nY 1.3558 N (19a)
Substituting the number of load cycles specified in the problem statement, that is, 810 cycles,N into Eq. (19a), the bending strength stress-cycle factor for the gear is
8 0.0178nY 1.3558 (1x10 ) 0.9768 (19b)
Note the procedure of obtaining the bending strength stress-cycle factor from Figure 14.14, see page 763, is not acceptable since a scale is not provided.
Since the gear is operating at room temperature then the temperature factor is
KT = 1 (20)
The reliability factor for the specified 95 percent reliability can be written from Eq. (14-38), see page 763, is
0.658 0.0759ln (1 )RK R (21a)
9
Substituting the reliability specified in the problem statement, that is 0.95,R into Eq. (21a), the reliability factor is
0.658 0.0759ln (1 0.95) 0.8854RK (21b)
Finally, determine the AGMA safety factor guarding against bending fatigue failure for the gear. Substituting Eqs. (16b), (18b), (19b), (20), and (21b) into Eq. (17), the AGMA safety factor guarding against bending fatigue failure for the gear can be written as
1 48.47 0.9768
6.33 1.0 0.8854FS
(22)
Therefore, the AGMA safety factor guarding against bending fatigue failure for the gear is
8.5FS (23)
10
Solution to Problem 2. (i) 5 Points. The spring stiffness can be written from Eq. (10-9), see page 520, as
4
38 a
d Gk
D N (1a)
Rearranging this equation, the number of active coils can be written as
4
38a
d GN
D k (1b)
The stiffness of the spring can also be written as
max min
working
F FFk
x y
(2)
The maximum load and the minimum load (that is, the preload) are specified, respectively, as
max 95 lbsF and min 45 lbsF (3)
The working deflection of the spring can be written as
working a my l l (4a)
Substituting the assembled length 5.0 inchesal and minimum working length 3.75 inches,ml see
Figure 2, into Eq. (4a), the working deflection is
5.0 3.75 1.25 inchesworkingy (4b)
Substituting Eqs. (3) and (4b) into Eq. (2), the stiffness of the spring is
95 4540 lbs/in
1.25k
(5)
The spring index can be written from Eq. (10.1), see page 519, as
DC
d (6a)
Substituting 9C and 2.25 inchesD into Eq. (6a), and rearranging, the wire diameter is
2.250.25 in
9d (6b)
The modulus of elasticity and the modulus of rigidity of music wire A228 with a wire diameter 0.25 ind from Table 10-5, see page 526, respectively, are
28.0 MpsiE and G = 11.6 Mpsi (7)
Substituting Eqs. (5), (6b), and (7) into Eq. (1b), the number of active coils is
11
4 6
3
0.25 11.6 1012.43
8 2.25 40aN
(8a)
Therefore, the number of active coils (rounded up to the nearest quarter of a coil) is
12.5aN (8b)
(ii) 5 points. The free length of the spring (see Example 10.1, page 527) can be written as
0 a initialL l y (9)
The initial deflection of the spring can be written from Eq. (2) as
mininitial
Fy
k (10a)
Substituting the preload min 45 lbsiF F and Eq. (5) into Eq. (10a), the initial deflection is
45in 1.125 in
40initialy (10b)
Substituting the assembled length 5.0 inchesal and Eq. (10b) into Eq. (9), the free length is
0 5.0 1.125 6.125 inL (11)
From Table 10-1, see page 521, the shut height of a spring with squared and grounded ends is
s tL d N (12)
where the total number of coils from Table 10-1, see page 521, is
2t aN N (13a)
Substituting Eq. (8b) into Eq. (13a), the total number of coils (rounded up to the nearest quarter) is
12.5 2 14.5 coilstN (13b)
Substituting Eqs. (6b) and (13b) into Eq. (12), the shut height of the spring is
0.25 14.5 3.625 insL (14)
(iii) 5 Points. The alternating and the mean components of the shear stress can be written from Eqs. (10.32) and (10.33), see page 537, as
3
8 aa B
F DK
d
and 3
8 mm B
F DK
d
(15)
The Bergstrasser factor can be written from Eq. (10-5), see page 519, as
12
4 2
4 3B
CK
C
(16a)
Substituting the spring index C = 9 into Eq. (16a), the Bergstrasser factor is
4 9 21.1515
4 9 3BK
(16b)
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 min
2m
F FF
(17)
Substituting Eq. (3) into Eqs. (17), the alternating and mean components of the load, respectively, are
95 4525 lb
2aF
and 95 45
70 lb2mF
(18)
Substituting Eqs. (6b), (16b) and (18) into Eqs. (15), the alternating and the mean components of the shear stress are
3
8 25 2.251.1515 10.556 kpsi
0.25a
(19a)
and
3
8 70 2.251.1515 29.557 kpsi
0.25m
(19b)
(v) 5 Points. The fatigue factor of safety using the Gerber-Zimmerli failure criterion, see Example 10-4, page 538, can be written as
saf
a
Sn
(20)
The alternating component of the shear strength can be written from Table 6.7, see page 307, as
22 2 21 1
2su se
sase su
r S SS
S r S
(21)
The torsional modulus of rupture of the music wire can be written from Eq. (10.30), see page 537, as
0.67su utS S (22)
The ultimate tensile strength of the music wire can be written from Eq. (10-14), see page 523, as
ut m
AS
d (23a)
For A228 music wire with a wire diameter 0.004 in 0.256 ind from Table 10-4, see page 525,
m201 kpsi.inA and 0.145m (23b)
13
Substituting Eqs. (23b) into Eq. (23a), the ultimate tensile strength of the music wire is
0.145
201245.75 kpsi
(0.25)utS (24)
Substituting Eq. (24) into Eq. (22), the torsional modulus of rupture of the music wire is
0.67 245.75 164.65 kpsisuS (25)
The torsional endurance strength (that is, the Gerber ordinate intercept for the Zimmerli data for a peened spring) can be written from Eq. (6.42), see page 306, or from the equation on page 536 as
21 /
sase
sm su
SS
S S
(26)
The alternating and mean components of the shear strength for a peened spring from Eq. (10-29), see page 536, are
57.5 kpsisaS and 77.5 kpsismS (27)
Substituting Eqs. (25) and (27) into Eq. (26), the torsional endurance strength of the music wire is
2
57.573.87 kpsi
1 77.5 /164.65seS
(28)
The slope of the load line, see page 537 and Example 10-4, page 538, is defined as
a
m
r
(29a)
Substituting Eqs. (19) into Eq. (30a), the slope of the load line is
10.5560.3571
29.557r (29b)
Substituting Eqs. (25), (28), and (29b) into Eq. (21), the alternating component of the shear strength is
22 2(0.3571) (164.65) 2(73.87)1 1 39.88 kpsi
2(73.87) (0.3571)(164.65)saS
(30)
Substituting Eqs. (20a) and (30) into Eq. (20), the fatigue factor of safety is
39.883.8
10.556fn (31)
14
Solution to Problem 3. (i) 10 Points. The stiffness of the bolt can be written from Eq. (8-17), see page 427, is
d t bb
d t t d
A A Ek
A l A l
(1)
The nominal major diameter (or bolt shank diameter) of the UNF 3/4 in – 16 bolt, see Table 8.2, page 413, is
0.7500 ind (2a)
Also, the tensile stress area, see Table 8.2, page 413, is
20.373 intA (2b)
Since the length of the bolt is specified as 3.75 inL then the length of the threaded portion of the bolt can be written from Table 8-7, see page 426, as
12 in
4TL d (3a)
Substituting Eq. (2a) into Eq. (3a), the length of the threaded portion of the bolt is
12(0.7500) 1.75 in
4TL (3b)
The length of the unthreaded portion of the bolt (that is, the bolt shank) can be written from Table 8-7, see page 426, as
d Tl L L (4a)
Substituting the length of the bolt 3.75 inchL and Eq. (3b) into Eq. (4a), the length of the unthreaded portion of the bolt is
3.75 1.75 2.00 indl (4b)
The grip is the total thickness of the two identical plates, that is
2(1.5) in 3.00 inl (5)
The length of the threaded portion of the bolt within the grip can be written from Table 8-7, see page 426, as
t dl l l (6a)
Substituting Eqs. (4b) and (5) into Eq. (6a), the length of the threaded portion of the bolt within the grip is
3.00 2.00 1.00 intl (6b)
The cross sectional area of the unthreaded portion of the bolt (that is, the bolt shank) is
2 22(0.7500)
0.4418 in4 4d
dA
(7)
15
Substituting Eqs. (2b), (4b), (6b), (7) and 6 230 10 lbs / inbE into Eq. (1), the stiffness of the bolt is
66(0.4418)(0.373) 30 10
4.162 10 lbs / in(0.4418 1.00) (0.373 2.00)bk
(8)
The stiffness of the two identical plates (using the diameter of the washer face under the bold head
wd 1.5 d and the half-apex angle 030 ) can be written from Eq. (8-22), see page 429, as
p0.5774
0.5774 0.52ln 5
0.5774 2.5
m
E dk
l dl d
(9a)
Substituting the modulus of elasticity of the plates 6 2pE 36 10 lbs / in , the bolt shank diameter 0.7500 in,d and Eq. (5), into Eq. (9a), the total stiffness of the two plates is
6
60.5774 36 10 0.750022.847 10 lbs / in
0.5774 3.00 0.5 0.75002 ln 5
0.5774 3.00 2.5 0.7500
mk
(9b)
(ii) 5 Points. The factor of safety guarding against joint separation can be written from Eq. (8-30), see page 441, as
0max (1 )
iFn
P C
(10)
The stiffness constant of the joint can be written from Eq. (f), see page 436, as
b
b m
kC
k k
(11a)
Substituting Eqs. (8) and (9b) into Eq. (11a), the stiffness constant of the joint is
6
6 6
4.162 100.154
4.162 10 22.857 10C
(11b)
which implies that the percentage of the external load taken by the bolt is 15.4%. Note that this value is acceptable since the joint constant should, in general, be less than 0.20, see Table 8-12, page 437.
Substituting the preload 30 kips,iF the maximum value of the external load max 9 kips,P P and
Eq. (11b) into Eq. (10), the factor of safety against joint separation is
3
0 3
30 103.94
9 10 (1 0.154)n
(12)
(iii) 5 Points. The load factor, that is, the factor of safety guarding against overloading, can be written from Eq. (8-29), see page 440, as
max max
p t i p iL
S A F F Fn
CP CP
(13)
16
The minimum proof strength of an SAE grade 8 steel bolt from Table 8-9, see page 433, is
120 kpsipS (14)
The proof load can be written from Eq. (8.32), see page 442, as
P p tF S A (15a)
Substituting Eqs. (2b) and (14) into Eq. (15a), the proof load is
120 x 0.373 44.76 kipsPF (15b)
Note that the ratio of the preload 30 kipsiF to the proof load in Eq. (15b) is
300.67
44.76i
P
F
F (16)
Note that this ratio is less than the recommended ratios in Chapter 8, see Eq. (8.31), page 442. Substituting Eqs. (2b), (11b), (14), the preload 30 kips,iF and the maximum value of the external
load max 9 kipsP P into Eq. (13), the load factor is
3 3
3
120 10 0.373 30 1010.64
0.1541 9 10Ln
(17)
Check: The factor of safety guarding against the static stress exceeding the proof strength (commonly referred to as the yielding factor of safety) can be written from Eq. (8-28), see page 440, as
max
p tp
i
S An
CP F
(18a)
Substituting Eqs. (2b), (11b), and (14) into Eq. (18a), the yielding factor of safety is
120 0.3731.4
0.154 9 30pn
(18b)
Comparing the answers for the three factors of safety, that is, Eqs. (12), (17), and (18b), note that
L O pn n n (19)
(iv) 5 Points. The fatigue factor of safety using the modified Goodman line can be written from Eq. (8.38), see page 446, as
( )
( )e ut i
fut a e m i
S Sn
S S
(20)
The ultimate tensile strength of the SAE grade 8 steel bolt from Table 8-9, see page 433, is
150 kpsiutS (21)
The fully corrected endurance strength of the bolt with 0.75 ind from Table 8-17, see page 445, is
17
23.2 kpsieS (22)
The preload stress can be written from Example 8.3, see page 439, as
ii
t
F
A (23a)
Substituting the preload 30 kipsiF and Eq. (2b) into Eq. (23a), the preload stress is
330 1080.429 kpsi
0.373i
(23b)
The mean component of the normal stress can be written from Eq. (8.36), see page 445, as
max min( )
2i
mt t
C P P F
A A
(24)
Substituting the maximum external load max 9 kips,P the minimum external load min 6 kips,P the
preload 30 kips,iF and Eqs. (2b) and (11b) into Eq. (24), the mean component of the normal stress
can be written as 3 3 30.1541(9x10 6x10 ) 30 x10
psi psi2 x 0.373 0.373m
(25a)
Therefore, the mean component of the normal stress is
3098.5 psi 80429 psi 83.53 kpsim (25b)
The alternating component of the normal stress can be written from Eq. (8-35), see page 445, as
max min( )
2at
C P P
A
(26a)
Substituting the maximum external load max 9 kips,P the minimum external load min 6 kips,P and
Eqs. (2b) and (11b) into Eq. (26a), the alternating component of the stress is
0.1541(9000 6000)619.71psi
2 x 0.373a
(26b)
Comparing the alternating component of the normal stress with the mean stress and the preload stress, that is, Eqs. (23b), (25b) and (26b), note that
a m i (27a)
that is 0.619 kpsi k83.53 80.429psi kpsi (27b)
This indicates that the slope of the load line is not 1, see Figure 8.20, page 446. In fact, the slope of the load line for this problem is
18
0.6190.1996
83.53 80.429a
m i
r
(28)
Substituting Eqs. (21), (22), (24b), (25b), and (26b) into Eq. (20), the fatigue factor of safety using the modified Goodman line can be written as
23.2(150 80.429)
150 x 0.619 23.2(83.53 80.429)fn
(29a)
Therefore, the fatigue factor of safety using the modified Goodman line is
16149.8
92.85 71.94fn
(29b)
19
Solution to Problem 4. (i) 10 Points. The free body diagram of the shoe is shown in Figure 1. The x axis points from the center of the drum, point C, to the hinge pin O and the y axis is directed from point C toward the shoe.
Figure 1. The free body diagram of the shoe with the x and y axes.
From static equilibrium, the sum of the moments about the hinge pin O can be written as
OM = 0 (1a)
From the free body diagram, see Figure 1, the sum of the moments about the hinge pin O is
F 0f Nc M M (1b)
Rearranging Eq. (1b), the actuating force on the shoe, see Eq. (16-4), see page 835, can be written as
N fM MF
c
(2)
Therefore, the shoe is self-energizing for the given counterclockwise rotation of the drum. The moment arm of the actuating force F about the hinge pin O, see Figure 1, is
16sin 60 13.8564 insoc (3)
O
MN
C A
12
y
yR
60ofFNF
OA = 16 inches
c
xRx
Mf
60o
D
F
a = 7 inches
20
The moment due to the normal force about the hinge pin O can be written from Eq. (16-3), see page 835, as
2
1
2sinsin
aN
a
p b r aM d
(4a)
where the integral, see Eq. (16-8), see page 836, can be written as
2
1
2 2 12 1)
1 1sin sin )(2 sin(2
2 4 2 4B d
(4b)
The angles to the heel of the shoe and the toe of the shoe, respectively, are
1 30 and 2 1120 150 (5)
Substituting Eqs. (5) into Eq. (4b) gives
(150 1 ( 1sin (300 sin (
2 4 2 4
) 30 )) 60 )
o oooB
(6a)
that is (1.5255 0.453) 1.4802B (6b)
Since 2 90o then the location of the maximum pressure acting on the shoe, see page 834, is
max 90a (7)
Since the shoe is self-energizing then the limiting pressure is equal to the maximum pressure of the frictional material, that is
maxap p (8a)
For cermet material, the maximum pressure from Table 16.3, see page 862, is
2max 150 lbs/insp (8b)
The face width of the shoe is specified as 3 insb (9)
The radius of the drum and the distance from the center of the drum to the hinge pin are specified as
5 insr and 7 insa OC (10)
Substituting Eqs. (6b), (7), (8), (9), and (10) into Eq. (4a), the moment due to the normal force about the hinge pin O is
3150 3 5 71.4802 23.313 10 lbs.ins
1NM
(11)
The moment due to the friction force about the hinge pin O can be written from Eq. (16-2), see page 835, as
21
2
1
sin ( cos )sin
af
a
f p b rM r a d
(12)
The coefficient of friction for cermet, see Table 16.3, page 862, is
0.32f (13) The integral in Eq. (12) can be written as
2
1
150
30
sin ( cos ) sin (5 7cos ) 8.6603
o
o
D r a d d
(14)
Substituting Eqs. (5b), (7), (8), (12b), and (13) into Eq. (12), the moment of the friction force about the hinge pin O is
30.32 150 3 58.6603 6.235 10 lbs.ins
1fM
(15)
Check: Equation Eq. (12) can also be written as
( )sin
af
a
f p b rM r C a A
(16)
where the integrals can be written as
22
11
21sin cos sin
2A d
(17a)
and
2
1
2
1
1 2sin cos cos cosC d
(17b)
Substituting Eqs. (5) into Eqs. (17) gives
2 21sin 150 sin 30 0
2o oA and cos 30 cos150 1.732o oC (18)
Then substituting Eqs. (18) into Eq. (16), the moment of the friction force about the hinge pin O is
30.32 150 3 5(5 x1.732) 6.235 10 lbs.ins
1fM
(19)
Note that Eq. (19) is in complete agreement with Eq. (15). Substituting Eqs. (2), (11), and (14) into Eq. (3), the limiting actuating force for the shoe is
3 323.313 10 6.235 10
1232.5 lbs13.8564
F
(20)
Note that the x and y components of the limiting actuating force on the shoe are
01232.5cos 60 616.25 lbsxF and 1232.5sin 60 1067.4 lbsoyF (21)
22
(ii) 5 Points. The braking torque (or the braking capacity) generated by the shoe can be written from Eq. (16-6), see page 835, as
21 2(cos cos )
sina
a
f p b rT
(22)
Substituting Eqs. (5b), (6), (7), (8) and (13) into Eq. (22), the braking torque generated by the shoe can be written as
230.32 150 3 5 (cos30 cos150 )
6.235 10 lbs.ins1
o o
T
(23)
(iii) 10 points. For an external contracting self-energizing shoe, the x and y components of the reaction force at the hinge pin O (that is, xR and yR ) can be written from Eqs. (16-14), see page 842, as
( )sin
ax x
a
p b rR F A f B
(24a)
and
( )sin
ay y
a
p b rR F f A B
(24b)
Note that the x and y components of the reaction force are assumed to be positive as shown in the free body diagram, see Figure 1. Substituting Eqs. (6b), (18), and (21) into Eqs. (24), the x and y components of the reaction force are
616.25 (0 1.4802 )sin
ax
a
p b rR f
(25a)
and
1067.4 (0 1.4802)sin
ay
a
p brR
(25b)
Substituting Eqs. (7), (8b), (9), and (13) into Eq. (25a), and rearranging, the x-component of the reaction force is
150 3 5616.25 (0 0.32 1.4802) 1065.75 lbs
1xR
(26a)
The negative sign on the right hand side of this equation indicates that the x-component of the resultant of the friction force and normal force on the shoe is in the positive x–direction, that is
616.25 1065.75 0xR (26b)
Rearranging Equation (26b), the x-component of the reaction force is
616.25 1065.75 1682 lbsxR (26c)
The negative sign indicates that the x-component of the reaction force at the hinge pin O is acting to the right (that is, in the opposite direction to the known x-component of the actuating force xF ).
Substituting Eqs. (7), (8b), and (9) into Eq. (25b) gives
23
150 3 51067.4 (0 1.4802) 3330.45 lbs
1yR
(27a)
The negative sign on the right hand side of this equation indicates that the y-component of the resultant of the friction force and normal force on the shoe is in the positive y–direction, that is
1067.4 3330.45 0yR (27b)
Rearranging Eq. (27b), the y-component of the reaction force is
3330.45 1067.4 2263 lbsyR (27c)
The negative sign indicates that the y-component of the reaction force is acting upward (that is, in the same direction as the known y-component of the actuating force yF ).
The magnitude of the reaction force at the hinge pin O, from Eqs. (26b) and (27b), is
2 2 2 2( 1682) ( 2263) 2819.6 lbsx yR R R (28)
The direction of the reaction force at the hinge pin O is
tan y
x
R
R (29a)
Substituting Eqs. (26) into Eq. (29a), the direction of the reaction force at the hinge pin O is
1 12263tan ( ) tan ( 1.3454) 53.4
1682o
(29b)
The magnitude and direction of the reaction force at the hinge pin O are shown in Figure 2.
Figure 2. The magnitude and direction of the reaction force at the hinge pin O.
F = 1232.5 lbs
C A
1
2
y
60o
fF
NF
R = 2819.6 lbs
x
O
Ry = 22635 lbs
Rx = 1682 lbs
24
Check. From static equilibrium, the sum of the forces in the x-direction can be written as
xF = 0 (30a)
From the free body diagram, see Figure 1, Eq. (28a) can be written as
x x x xN fR F (F ) (F ) = 0 (30b)
Substituting Eq. (21) into Eq. (30b), the x-component of the hinge pin reaction force can be written as
x x xN fR 616.25 (F ) (F ) = 0 (31)
The x-component of the normal force can be written as
xN(F ) ( )cossin
a
a
p b rA
(32a)
Note from the symmetry of the shoe that the angle 090 , therefore
xN150 3 5
(F ) (0)(0) 01
(32b)
The x-component of the friction force can be written as
xf(F ) ( )sinsin
a
a
p b rf B
(33a)
or as
xf150 3 5
(F ) ( 0.32 1.4802)1 1065.74 lbs1
(33b)
The negative sign indicates that the x-component of the friction force is acting to the right (which is intuitively correct). Substituting Eq. (32b) and (33b) into Eq. (31), the x-component of the hinge pin reaction force is
xR 616.25lbs 0 1065.74 lbs 0 (34a)
Therefore, the x-component of the hinge pin reaction force is
xR 1065.74 lbs 616.25lbs 449.49 lbs (34b)
The positive sign indicates that the x-component of the reaction force at the hinge pin O is acting to the left.
From static equilibrium, the sum of the forces in the y-direction can be written as
yF = 0 (35a)
From the free body diagram, see Figure 1, Eq. (35a) can be written as
y y y yN fR F (F ) (F ) = 0 (35b)
25
Substituting Eq. (21) into Eq. (35b), the y-component of the hinge pin reaction force can be written as
y y yN fR 1067.4 (F ) (F ) 0 (36)
The y-component of the normal force can be written as
yN(F ) ( ) sinsin
a
a
p b rB
(37a)
or as
yN150 3 5
(F ) (1.4802)(1) 3330.45 lbs1
(37b)
The positive sign indicates that the y-component of the normal force is acting downward (which is correct).
The y-component of the friction force can be written as
yf(F ) ( ) cossin
a
a
p b rf A
(38a)
or as
yf150 3 5
(F ) (0)(0) 01
(38b)
Substituting Eqs. (37b) and (38b) into Eq. (36), the y-component of the hinge pin reaction force is
yR 1607.4 0 3330.45 0 (39a)
Therefore, the y-component of the hinge pin reaction force is
yR 1607.4 3330.45 1723.05 lbs (39b)
Note that Eqs. (34b) and (39b) are in complete agreement with Eqs. (26b) and (27b). Check: The graphical approach. The arm and shoe is a three-force member. The three forces are the two known forces: (i) the actuating force F acting at point A; and (ii) the contact force RF acting on the
shoe (this is the resultant of NF and fF ). The third force is the unknown reaction force R acting at the
hinge pin O. The force polygon is shown in Figure 3.
26
Figure 3. The force polygon.
F = 1232.5 lbs
R
FR
