MACHINE DESIGN AND SHOP PRACTICE PROBLEMS MACHINE DESIGN AND SHOP PRACTICE PROBLEMS ELEMENTS (with answers and solutions)
ELEMENTS (with answers and solutions) 1.
1. What is the polar section modulus of a solid shaft with a diameter of 101.6 mm?What is the polar section modulus of a solid shaft with a diameter of 101.6 mm? a) 209.5 cm a) 209.5 cm33 b) b) 209.5 209.5 cmcm44 c) 205.9 c) 205.9 cmcm33 d) d) 205.9 205.9 cmcm44 Answer: c) 205.9 cm Answer: c) 205.9 cm33 Solution: Solution: j j 33
1010..1616
33 205205..99cmcm33 16 16 D D 16 16 D D JJ 2 2 c c JJ Z Z
2.2. With the water interruptions prevailing in your town, you have been asked to design an upright cylindrical waterWith the water interruptions prevailing in your town, you have been asked to design an upright cylindrical water tank 6 meters in diameter and 6 meters high, vented, and to be filled completely with water. Determine the tank 6 meters in diameter and 6 meters high, vented, and to be filled completely with water. Determine the minimum thickness of the tank plate if the stress is limited to 40 Mpa.
minimum thickness of the tank plate if the stress is limited to 40 Mpa. a)
a) 3.3 3.3 mm mm b) 4.4 b) 4.4 mm mm c) c) 5.5 5.5 mm mm d) d) 8.8 8.8 mmmm Answer: b) Thickness, t = 4.4 mm
Answer: b) Thickness, t = 4.4 mm Solution:
Solution: p p
ghgh
10001000
99..80668066
66
5858839839..66 PaPa
5858..83968396 kPakPa
4040000000
00..0044100441mm 44..4141mmmm 2 2 6 6 8396 8396 .. 58 58 ss 2 2 pD pD tt tt
3.3. A steel shaft 1.75 inches in diameter transmits 40 Hp at 1800 rpm. Assuming a modulus of rigidity of 12 x 10A steel shaft 1.75 inches in diameter transmits 40 Hp at 1800 rpm. Assuming a modulus of rigidity of 12 x 1066 psi, find the torsional deflection of the shaft i
psi, find the torsional deflection of the shaft i n degrees per foot length.n degrees per foot length. a) a) 0.0073 0.0073 b) b) 0.0037 0.0037 c) c) 0.0063 0.0063 d)d) 0.00013 0.00013 Solution: Solution:
11..7575
1212 xx1010
00..0001300013 radianradian 00..00730073 1800 1800 40 40 63000 63000 32 32 G G D D n n Hp Hp 000 000 63 63 32 32 G G D D T T 32 32 JG JG T T L L 44 44
44 66
4.4. A helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the curveA helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the curve correction factor of the spring.
correction factor of the spring. a)
a) 1.144 1.144 b) b) 1.44 1.44 c) c) 1.1144 1.1144 d) d) 1.14141.1414 Answer: Curve Correction Factor, K
Answer: Curve Correction Factor, K cc = 1.1144 = 1.1144
Solution: Solution: 11..11441144 0625 0625 .. 1 1 184 184 .. 1 1 C C 2 2 1 1 C C 2 2 C C 615 615 .. 0 0 4 4 C C 4 4 1 1 C C 4 4 K K K K K K ss w w c c
5.5. A vise is equipped with a 1-inch single square thread, with 4 threads per inch. The frictional radius of the collarA vise is equipped with a 1-inch single square thread, with 4 threads per inch. The frictional radius of the collar is 0.5 inch. The coefficient of friction for both the collar and threads is 0.20. How much external torque must be is 0.5 inch. The coefficient of friction for both the collar and threads is 0.20. How much external torque must be applied to produce a force of 200 lb against the jaws of the vise?
applied to produce a force of 200 lb against the jaws of the vise? a)
a) 39.73 39.73 in-lb in-lb b) 33.97 b) 33.97 in-lb in-lb c) c) 37.93 37.93 in-lb in-lb d) d) 39.37 39.37 in-lbin-lb Answer: c) 37.93 in-lb Answer: c) 37.93 in-lb Vise Jaw Vise Jaw Collar Collar Screw Screw Nut Nut Handle Handle F = 200 lb F = 200 lb
Solution: Solution:
For the thread pitch,
For the thread pitch,
p
p
.
.
inc
inch
h
Nu
Numb
mber of
er of Thr
Thread
eads per Inc
s per Inch
h
1 1
11 0 0 2525 44
For the minor diameter, D
For the minor diameter, Dii = D = Doo – – p = 1 p = 1 – – 0.25 = 0.75 inch 0.25 = 0.75 inch
For the mean or pitch diameter of the screw,
For the mean or pitch diameter of the screw,
D
D
mm
D
D
o o
D
D
ii
1 0 1
0 7..
755
0 8750 875.
.
iin
nc
ch
h
2
2 22
For the lead angle, For the lead angle,
o o m m mm
L
L
p
p
..
tta
an
n
tta
an
n
tta
an
n
tta
an
n .
.
..
D
D
D
D
..
1 1 11 11 0 0 2525 11 0 0 090909095 5 5 5 191977 0 875 0 875Solving for the torque required to overcome the collar friction, Solving for the torque required to overcome the collar friction,
c c c c.
.
..
fFD
fFD
T
T
0 0 2 2 22000 0 0 0 55
2020iin
n
llb
b
2 2 22Solving for the torque required to overcome the thread friction, Solving for the torque required to overcome the thread friction,
m m ss..
FD
FD
tta
an
n
f
f
.
.
..
T
T
.
.
iin
n
llb
b
f
f
tta
an
n
.
.
..
200 200 0 80 87575 0 0 090909095 5 0 0 22 25 93 25 93 2 2 1 1 2 2 1 1 0 0 2 2 0 0 99009955 Solving for the total torque,Solving for the total torque, T = T
T = Tss + T + Tcc= = 25.93 25.93 + + 20 20 = = 45.93 45.93 inches inches ans.ans.
6.
6. A helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the WahlA helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the Wahl factor of the spring.
factor of the spring. a)
a) 1.148 1.148 b) b) 1.184 1.184 c) c) 1.418 1.418 d) 1.814d) 1.814 Answer: b) Wahl Factor, K
Answer: b) Wahl Factor, K ww = 1.184 = 1.184
Solution: Solution: 88 8 8 1 1 1 1 d d D D C C
mm
88 11..184184 615 615 .. 0 0 8 8 8 8 4 4 1 1 8 8 4 4 C C 615 615 .. 0 0 4 4 C C 4 4 1 1 C C 4 4 K K ww
7.7. A helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the value ofA helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the value of Bergstrassar factor of the spring.
Bergstrassar factor of the spring. a)
a) 1.172 1.172 b) b) 1.712 1.712 c) c) 1.217 1.217 d) 1.271d) 1.271 Answer: a) Bergstrassar Factor, K
Answer: a) Bergstrassar Factor, K BB = 1.1724 = 1.1724
Solution: Solution: 88 8 8 1 1 1 1 d d D D C C
mm
88 33 11..17241724 4 4 2 2 8 8 4 4 3 3 C C 4 4 2 2 C C 4 4 K K BB
8.8. Compute the tooth thickness of a 14.5Compute the tooth thickness of a 14.5oo spur gear with diametral pitch of 5. spur gear with diametral pitch of 5. a)
Answer: b) Tooth thickness, t = 0.31416 inch Answer: b) Tooth thickness, t = 0.31416 inch
Solution:
Solution: 00..3141631416 inchinch 5 5 5708 5708 .. 1 1 p p 5708 5708 .. 1 1 tt d d
9.9. Compute the speed of the gear mounted on a 52.5 mm diameter shaft receiving power from a driving motorCompute the speed of the gear mounted on a 52.5 mm diameter shaft receiving power from a driving motor with 250 Hp.
with 250 Hp. a)
a) 2 2 182 182 rpm rpm b) b) 2 2 071 071 rpm rpm c) c) 2 2 265 265 rpm rpm d) d) 2 2 341 341 rpmrpm Answer: c) The speed, N = 2265 rpm
Answer: c) The speed, N = 2265 rpm
Solution: From PSME Code:
Solution: From PSME Code:
22264264..9292 rprpmm 4 4 .. 25 25 5 5 .. 52 52 250 250 80 80 D D P P 80 80 N N 3 3 3 3
10.10. Compute the working strength of 1 inch bolt which is screwed up tightly in packed joint when the allowableCompute the working strength of 1 inch bolt which is screwed up tightly in packed joint when the allowable working stress is 13000 psi.
working stress is 13000 psi. a)
a) 3 3 600 600 lb lb b) b) 3 3 950 950 lb lb c) c) 3 3 900 900 lb lb d) d) 3 3 800 800 lblb Answer c) The working strength, W = 3 900 lb
Answer c) The working strength, W = 3 900 lb
Solution: From
Solution: From machinery’s machinery’s Handbook, Handbook, page page 1149:1149:
00..5555dd 00..2525dd
1313000000
00..5555
11
00..2525
11 33900900 lblb ssW
W
tt 22
22
11.
11. Compute the nominal shear stress at the surface, in MPa, for a 50 mm diameter shaft that is subjected to aCompute the nominal shear stress at the surface, in MPa, for a 50 mm diameter shaft that is subjected to a torque of 0.48 kN-m.
torque of 0.48 kN-m. a)
a) 16.95 16.95 b) b) 21.65 21.65 c) c) 19.56 19.56 d) 25.12d) 25.12 Answer: Shear Stress, s
Answer: Shear Stress, sss = 19.56 MPa = 19.56 MPa
Solution:
Solution:
0
0
..
050
050
19
19
556
556
..
96
96
kPa
kP
a
19
19
..
56
56
MPa
MPa
48
48
..
0
0
16
16
D
D
T
T
16
16
ss
ss 33 33
12.12. A hollow iron pipe to be designed as a column has an outside diameter of 240 mm and is subjected to a force ofA hollow iron pipe to be designed as a column has an outside diameter of 240 mm and is subjected to a force of 80 KN. Find the pipe thickness if the compressive stress is limited to 16 MPa.
80 KN. Find the pipe thickness if the compressive stress is limited to 16 MPa. a.
a. 5.85 5.85 mm mm b. b. 6.85 6.85 mm mm c. c. 7.85 7.85 mm mm d. d. 8.85 8.85 mmmm Answer: b) Pipe thickness, t = 6.85 mm
Answer: b) Pipe thickness, t = 6.85 mm
Solution: Inside Diameter,
Solution: Inside Diameter,
16
16
000
000
0
0
..
2263
2263
m
m
226
226
..
3
3
mm
mm
80
80
4
4
24
24
..
0
0
ss
F
F
4
4
D
D
D
D
ii 22oo 22
Thickness of the Pipe,
Thickness of the Pipe,
6
6
..
85
85
mm
mm
2
2
3
3
..
226
226
240
240
2
2
D
D
D
D
tt
oo
ii
13.13. A uniform beam 12 meters long is fixed at one end. It has a uniform weight of 50 kg/m along its length. A loadA uniform beam 12 meters long is fixed at one end. It has a uniform weight of 50 kg/m along its length. A load of 20 kgs. is suspended on the beam 4 m from the free end. The moment at the fixed end is
of 20 kgs. is suspended on the beam 4 m from the free end. The moment at the fixed end is a.
a. 3760 3760 kg-m kg-m b. b. 0.0 0.0 kg-m kg-m c. c. 60 60 kg-m kg-m d. d. 4800 4800 kg-mkg-m Answer: a) 3760 kg-m
Answer: a) 3760 kg-m Solution: Bending Moment, Solution: Bending Moment,
600
600
8
8
20
20
3
3
760
760
kg
kg
m
m
6
6
M
M
W = 50 (12) = 600 kg W = 50 (12) = 600 kg F = 20 kgs F = 20 kgs 6 m 6 m 8 m 8 m14.
14. A A 2020oo straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/inch drives a 42-tooth gear. straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/inch drives a 42-tooth gear. The two shafts are at right angles and in the same plane. Find the pitch angle of the pinion.
The two shafts are at right angles and in the same plane. Find the pitch angle of the pinion. a) 18.4
a) 18.4oo b) b) 2020oo c) 14.5c) 14.5oo d) d) 20.520.5oo Answer: a) Pitch angle of the pinion = 18.4
Answer: a) Pitch angle of the pinion = 18.4oo
Solution: Solution: 11 oo g g p p 1 1 1818..44 42 42 14 14 tan tan T T T T tan tan
15.15. A A 2020oo straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/inch drives a 32-tooth gear. straight-tooth bevel pinion having 14 teeth and a diametral pitch of 6 teeth/inch drives a 32-tooth gear. The two shafts are at right angles and in the same plane. The pinion is to transmit 1800 rpm and transmitting 50 The two shafts are at right angles and in the same plane. The pinion is to transmit 1800 rpm and transmitting 50 hp.
hp. Determine the Determine the pitch diameters of pitch diameters of the gears.the gears. a)
a) 2.33 2.33 inches inches and and 5.36 5.36 inches inches b) b) 3.23 3.23 inches inches and and 3.56 3.56 inchesinches c)
c) 5.36 5.36 inches inches and and 6.36 6.36 inches inches d) d) 2.33 2.33 inches inches and and 2.33 2.33 inchesinches Answer: a) 2.33 inches and 5.36 inches
Answer: a) 2.33 inches and 5.36 inches
Solution: Solution: 11 1 1 d d T T 1414 D D 22..333 3 iinncchheess P P 66
22
2 2 11 1 1 T T 3232 D D D D 22..333 3 55..336 6 iinncchheess T T 1414
16.16. A 1-inch diameter shaft has a single disc weighing 75 lb mounted midwA 1-inch diameter shaft has a single disc weighing 75 lb mounted midw ay between two bearings 20” apart. Finday between two bearings 20” apart. Find the lowest critical speed in rpm. Neglect the weight of the shaft. Assume that the modulus of elasticity is 30 x the lowest critical speed in rpm. Neglect the weight of the shaft. Assume that the modulus of elasticity is 30 x 10 1066 psi. psi. a) a) 2038 2038 rpm rpm b) b) 2308 2308 rpm rpm c) c) 2380 2380 rpm rpm d) d) 2803 2803 rpmrpm Solution: Solution: D
D = = 1 1 inch inch E E = = 30 30 x x 101066psi psi L L = = 20 20 inchesinches
Moment of Inertia, Moment of Inertia,
4 4 4 4 4 4 1 1 D D I I 00..00449911 iinn.. 6 64 4 6644
Solving for the lowest critical speed,
Solving for the lowest critical speed,
6 6 cr cr 3 3 33 57 576 6 330 0 x x 10 10 0.0.040491 91 3232.2.2 576EIg 576EIg 21 213.33.39 rad/9 rad/ ss W W LL 775 5 2200
cr cr cr cr 30 213.39 30 213.39 30 30 N N
22037037.68 rp.68 rpmm
17.17. A flywheel weighing 457 kg has a radius of 375 mm. How much energy, in N-m, does the flywheel loss from 3A flywheel weighing 457 kg has a radius of 375 mm. How much energy, in N-m, does the flywheel loss from 3 rps to 2.8 rps?
rps to 2.8 rps? a)
a) 368 368 b) b) 150 150 c) c) 1474 1474 d) d) 3838
Answer: c) Flywheel Energy =1474 Answer: c) Flywheel Energy =1474 Solution: Solution: VV11
22
00..375375
33
77..069069mm//ssV
V
22
2
2
0
0
..
375
375
2
2
..
8
8
6
6
..
597
597
m
m
//
ss
m m N N 91 91 .. 473 473 1 1 2 2 597 597 .. 6 6 069 069 .. 7 7 457 457 2 2 V V V V m m KE KE 2 2 2 2 2 2 2 2 2 2 1 1
18.18. A triple threaded worm has a pitch diameter of 3 inches. The wheel has 25 teeth and a pitch diameter of 5A triple threaded worm has a pitch diameter of 3 inches. The wheel has 25 teeth and a pitch diameter of 5 inches. Material for both the wheel and the wheel is phosphor bronze. Determine the helix angle of the gear. inches. Material for both the wheel and the wheel is phosphor bronze. Determine the helix angle of the gear. a) 11.31 a) 11.31oo b) b) 13.1113.11oo c) c) 11.1311.13oo d)d) 10.13 10.13oo 10” 10” 10”10” W = 75 lb W = 75 lb
Answer: Gear Helix angle = 11.31 Answer: Gear Helix angle = 11.31oo
Solution:
Solution: Circular pitch Circular pitch of the wof the worm gear,orm gear, cc gg
g g D D 55 P P 00..6622883 3 iinncchh T T 2525
Where, PWhere, Pcc = P = Paa = pitch of the worm = pitch of the worm
Solving for the lead, L = N
Solving for the lead, L = Ntt P Paa = 3(0.6283) = 1.8849 inches = 3(0.6283) = 1.8849 inches
Solving for the lead angle of the worm,
Solving for the lead angle of the worm, 1 1 1 1 oo
w w L L 11..88884499 ttaan n ttaan n 1111..3311 D D 33
For the helix angle, and considering that the shafts angle is 90 For the helix angle, and considering that the shafts angle is 90oo,,
gg = = = 11.31 = 11.31 o o ans. ans. 19.
19. A single square thread power screw is to raise a load of 70 kN. The screw has a major diameter of 36 mm and aA single square thread power screw is to raise a load of 70 kN. The screw has a major diameter of 36 mm and a pitch of
pitch of 6 mm. 6 mm. The coefficient The coefficient of thread of thread friction friction and collar and collar friction are friction are 0.13 0.13 and 0.10 and 0.10 respectively. If respectively. If the collarthe collar mean diameter is 90 mm and the screw turns at 60 rpm, find the axial linear speed of the screw.
mean diameter is 90 mm and the screw turns at 60 rpm, find the axial linear speed of the screw. a)
a) 5 5 mm/s mm/s b) b) 6 6 mm/s mm/s c) c) 7 7 mm/s mm/s d) d) 5.5 5.5 mm/smm/s Answer: b) 6 mm/s
Answer: b) 6 mm/s
Solution: For the linear speed of the screw, V
Solution: For the linear speed of the screw, Vnn = n (L) = (60 rpm)(6 mm/rev) = 360 mm/min = 6 mm/s = n (L) = (60 rpm)(6 mm/rev) = 360 mm/min = 6 mm/s
20.
20. A double thread ACME screw driven by a motor at 400 rpm raises the attached load of 900 kg at a speed of 10A double thread ACME screw driven by a motor at 400 rpm raises the attached load of 900 kg at a speed of 10 m/min. The screw has a pitch diameter of 36 mm; the coefficient of friction on threads is 0.15. The friction m/min. The screw has a pitch diameter of 36 mm; the coefficient of friction on threads is 0.15. The friction torque on the thrust bearing of the motor is taken as 20 % of the total input. Determine the lead angle.
torque on the thrust bearing of the motor is taken as 20 % of the total input. Determine the lead angle. a) 12.465
a) 12.465oo b) b) 14.26514.265oo c) 15.462c) 15.462oo d) d) 16.45216.452oo Answer: a) 12.465
Answer: a) 12.465oo
Solution: For the lead,
Solution: For the lead, 00..025025mm 2525 mmmm 400 400 10 10 n n V V L L
For the lead angle,
For the lead angle, 11 oo
m m 1 1 465 465 .. 12 12 36 36 25 25 tan tan D D L L tan tan
21.21. What Hp can a 1-inch diameter short shaft transmit at 380 rpm?What Hp can a 1-inch diameter short shaft transmit at 380 rpm? a) a) 3 3 Hp Hp b) b) 18 18 Hp Hp c) c) 10 10 Hp Hp d) d) 7.1 7.1 HpHp Answer: c) 10 Hp Answer: c) 10 Hp Solution: Solution:
10
10
Hp
Hp
38
38
380
380
1
1
38
38
N
N
D
D
P
P
3 3 3 3
22.22. A spur pinion rotates at 600 rpm and transmits 25 kW to a mating gear. The pitch diameter of the pinion is 100A spur pinion rotates at 600 rpm and transmits 25 kW to a mating gear. The pitch diameter of the pinion is 100 mm, and the pressure angle is 20
mm, and the pressure angle is 20oo. Determine the tangential load, in N.. Determine the tangential load, in N. a) a) 7660 7660 b) b) 6790 6790 c) c) 3900 3900 d) d) 30983098 Solution: Solution:
600
600
7
7
957
957
..
75
75
N
N
000
000
25
25
30
30
10
10
..
0
0
2
2
n
n
P
P
30
30
D
D
2
2
F
F
tt
23.
23. A multiple-disk clutch, composed of three plates with a small diameter of 150 mm and large diameter of 200A multiple-disk clutch, composed of three plates with a small diameter of 150 mm and large diameter of 200 mm, is designed to transmit 100 kW at 3000 rpm at a coefficient of friction of 0.5. Determine the spring force mm, is designed to transmit 100 kW at 3000 rpm at a coefficient of friction of 0.5. Determine the spring force needed to engage the clutch.
needed to engage the clutch. a) a) 2820 2820 N N b) b) 2428 2428 N N c) c) 5460 5460 N N d) d) 3638 3638 NN Answer: 3638 N Answer: 3638 N Solution: Solution:
30003000
00..31833183kNkN mm 318318..33 N N mm 100 100 30 30 n n P P 30 30 T T
fs fs fs fs f f NN 4 4 d d D D P P f f N N r r P P f f T T
200200 150150
22 33637637..7171NN 5 5 .. 0 0 300 300 318 318 4 4 N N d d D D f f T T 4 4 P P fs fs
24.24. A wire rope lifts a load of 10 kips at a maximum speed of 1000 feet per minute, attained in 5 seconds startingA wire rope lifts a load of 10 kips at a maximum speed of 1000 feet per minute, attained in 5 seconds starting from rest. The rope has a metallic cross sectional area of 0.4 in
from rest. The rope has a metallic cross sectional area of 0.4 in22. Determine the maximum tensile stress on the. Determine the maximum tensile stress on the rope in ksi.
rope in ksi. a)
a) 26.2 26.2 b) b) 25.0 25.0 c) c) 27.6 27.6 d) d) 32.432.4 Answer: c) maximum tensile stress = 27.6 MPa
Answer: c) maximum tensile stress = 27.6 MPa
Solution: Solution: 11 oo
33..3333 ffpsps22 5 5 0 0 60 60 // 1000 1000 tt V V V V a a
1111034034..1616 lblb 2 2 .. 32 32 33 33 .. 3 3 1 1 000 000 10 10 g g a a 1 1 W W F F LL
ksi ksi 6 6 .. 27 27 psi psi 4 4 .. 585 585 27 27 4 4 .. 0 0 16 16 .. 034 034 ,, 11 11 S Stt
25.25. What is the bursting steam pressure of a hemispherical steel shell with a diameter of 100 inches and made ofWhat is the bursting steam pressure of a hemispherical steel shell with a diameter of 100 inches and made of 0.0635-m thick steel plate, if the joint efficiency is 70 % and the tensile strength is 60 000 psi?
0.0635-m thick steel plate, if the joint efficiency is 70 % and the tensile strength is 60 000 psi? a)
a) 4 4 020 020 psi psi b) b) 4 4 200 200 psi psi c) c) 2 2 400 400 psi psi d) d) 2 2 040 040 psipsi Answer: b) 4200 psi Answer: b) 4200 psi Solution: Solution:
psi psi 200 200 4 4 .. in in 100 100 70 70 .. 0 0 in in // lb lb 000 000 60 60 in in 4 4 .. 25 25 5 5 .. 63 63 4 4 D D E E ss tt 4 4 p p 2 2 j j ll
Where,Where, p p = = bursting bursting pressure, pressure, psi psi ssll = shell stress, psi = shell stress, psi
t
t = = shell shell thickness, thickness, inches inches D D = = shell shell diameter, diameter, inches inches EE j j = = jointjoint
efficiency efficiency
Note: For the longitudinal stress of the thin
Note: For the longitudinal stress of the thin -walled cylinder, and the stress for spherical tank:-walled cylinder, and the stress for spherical tank:
j j L L E E tt 4 4 D D p p ss
26.26. A cylinder having an internal diameter if 508 mm and external diameter if 914.4 mm is subjected to an internalA cylinder having an internal diameter if 508 mm and external diameter if 914.4 mm is subjected to an internal pressure of
pressure of 69 69 MPa MPa and an and an external prexternal pressure of essure of 14 14 MPa. MPa. Determine Determine the the hoop stress hoop stress at tat t he inner he inner surface of surface of thethe cylinder.
cylinder. a)
a) 90.11 90.11 MPa MPa b) b) 91.10 91.10 MPa MPa c) c) 911.0 911.0 MPa MPa d) d) 19.10 19.10 MpaMpa Answer: a) 90.11 MPaa
Solution: Solution:
2 2 2 2 2 2 2 2 2 2 2 2 ii 2 2 o o 2 2 o o o o 2 2 ii 2 2 o o ii ti ti mm mm 2 2 508 508 mm mm 2 2 4 4 .. 914 914 mm mm 2 2 4 4 .. 914 914 MPa MPa 14 14 2 2 mm mm 2 2 508 508 mm mm 2 2 4 4 .. 914 914 MPa MPa 69 69 r r r r r r p p 2 2 r r r r p p ss
sstiti = 90.11 Mpa = 90.11 Mpa Where, sWhere, stiti= = maximum maximum tangential tangential or or hoop hoop stress stress at at the the inside inside r r ii = inside radius, mm = inside radius, mm
p
pii= = internal internal pressure, pressure, Mpa Mpa r r oo = outside radius, mm = outside radius, mm
p
poo = external pressure, Mpa = external pressure, Mpa
Note: For the maximum tangential or hoop stress at the outside,
Note: For the maximum tangential or hoop stress at the outside,
2 2 ii 2 2 o o 2 2 ii 2 2 o o o o 2 2 ii ii to to r r r r r r r r p p r r p p 2 2 ss
27.27. What length of a square key is required for a 4-in diameter shaft transmitting 1000 hp at 1000 rpm? TheWhat length of a square key is required for a 4-in diameter shaft transmitting 1000 hp at 1000 rpm? The allowable shear and compressive stresses in the key are 15 ksi and 30 ksi, respectively.
allowable shear and compressive stresses in the key are 15 ksi and 30 ksi, respectively. a)
a) 2.1 2.1 inches inches b) b) 2.8 2.8 inches inches c) c) 3.2 3.2 inches inches d) d) 4.2 4.2 inchesinches Answer: a) 2.1 inches
Answer: a) 2.1 inches
Solution: Transmitted torque,
Solution: Transmitted torque,
63
63
000
000
in
in
lb
lb
1000
1000
1000
1000
63000
63000
n
n
Hp
Hp
63000
63000
T
T
Key width,Key width,
,,
for
for
good
good
propo
proport
rtion
ion
4
4
D
D
b
b
Key length based on shear,
Key length based on shear,
1
1
4
4
2
2
..
1
1
inches
inches
000
000
15
15
000
000
63
63
2
2
b
bD
D
ss
T
T
2
2
L
L
ss
Key length based on compression,Key length based on compression,
1
1
4
4
2
2
..
1
1
000
000
30
30
000
000
63
63
4
4
D
D
tt
ss
T
T
4
4
L
L
c c
Therefore, use L = 2.1 inches Therefore, use L = 2.1 inches 28.
28. The root diameter of a double square thread is 0.55 inch. The screw has a pitch of 0.2 inch. Find the number ofThe root diameter of a double square thread is 0.55 inch. The screw has a pitch of 0.2 inch. Find the number of thread per inch.
thread per inch. a)
a) 0.2 0.2 threads/inch threads/inch b) b) 10 10 threads/inch threads/inch c) c) 5 5 threads/inch threads/inch d) 2.5 d) 2.5 threads/inchthreads/inch Answer: TPI = 5 threads per inch
Answer: TPI = 5 threads per inch Solution:
Solution:
The number of threads per inch,
The number of threads per inch, 55 threadsthreads //inchinch 2 2 .. 0 0 1 1 p p 1 1 TP TPII
29.29. A single square thread power screw is to raise a load of 70 kN. The screw has a major diameter of 36 mm and aA single square thread power screw is to raise a load of 70 kN. The screw has a major diameter of 36 mm and a pitch of
pitch of 6 mm. 6 mm. The coefficient The coefficient of thread of thread friction friction and collar and collar friction are friction are 0.13 and 0.13 and 0.10 respectively. 0.10 respectively. If the If the collarcollar mean diameter is 90 mm and the screw turns at 60 rpm, find the combined efficiency of the screw and collar. mean diameter is 90 mm and the screw turns at 60 rpm, find the combined efficiency of the screw and collar. a) a) 15.32 15.32 % % b) b) 12.53 12.53 % % c) c) 13.52 13.52 % % d) d) 15.97 15.97 %% Answer: b) 12.53 % Answer: b) 12.53 % p p W = 70 N W = 70 N
Solution: For the depth of the thread,
Solution: For the depth of the thread,
66 33mmmm 2 2 1 1 p p 2 2 1 1 h h
For the mean diameter of the screw,
For the mean diameter of the screw, DDmm
DDoo
hh
3636
33
3333mmmm For the lead angle,For the lead angle, 11 oo
m m 1 1 m m 1 1 3123 3123 .. 3 3 33 33 6 6 tan tan D D p p tan tan D D L L tan tan
Solving for the Efficiency,
Solving for the Efficiency,
tan tan f f 1 1 D D D D f f f f tan tan % % 100 100 tan tan f f 1 1 tan tan e e m m c c c c
1212..5353 %% 3125 3125 .. 3 3 tan tan 13 13 .. 0 0 1 1 33 33 90 90 10 10 .. 0 0 10 10 .. 0 0 3125 3125 .. 3 3 tan tan % % 100 100 3125 3125 .. 3 3 tan tan 13 13 .. 0 0 1 1 3125 3125 .. 3 3 tan tan e e o o o o o o o o
30.30. Determine the power capacity of a cone clutch under uniform pressure and assuming the following conditions:Determine the power capacity of a cone clutch under uniform pressure and assuming the following conditions: major diameter = 250 mm; minor diameter = 200 mm; length of conical elements in contact = 125 mm; major diameter = 250 mm; minor diameter = 200 mm; length of conical elements in contact = 125 mm; rotational speed = 870 rpm; coefficient of friction = 0.30; and allowable pressure = 70 kPa.
rotational speed = 870 rpm; coefficient of friction = 0.30; and allowable pressure = 70 kPa. a)
a) 25.74 25.74 Hp Hp b) b) 24.75 24.75 Hp Hp c) c) 27.45 27.45 Hp Hp d) d) 24.57 24.57 HpHp Answer: a) 25.74 Hp
Answer: a) 25.74 Hp
Solution:
Solution: Friction Friction radius,radius,
125125 100100 112112..9696 mmmm 100 100 125 125 3 3 2 2 r r r r r r r r 3 3 2 2 r r 2 2 2 2 3 3 3 3 2 2 ii 2 2 o o 3 3 ii 3 3 o o f f
Surface Area of contact,
Surface Area of contact, AAf f
22
r r f f b b
22
00..1129611296
00..125125
00..08870887 mm22 Force normal to the surface of contact,Force normal to the surface of contact, FFnn
pA pAf f
7070
00..08870887
66..209209 kNkN Power Capacity, Power Capacity,
00..3030
66..209209
00..1129611296
1919..22 kWkW 2525..7474 HpHp 30 30 870 870 r r F F f f 30 30 n n 30 30 T T n n P P
f f
nn f f
31.31. Three extension springs are hooked in series that support a single weight of 100 kg. The first spring is rated at 4Three extension springs are hooked in series that support a single weight of 100 kg. The first spring is rated at 4 kN/m and the other two springs are rated 6 kN/m each. Determine the equivalent stiffness of the three springs. kN/m and the other two springs are rated 6 kN/m each. Determine the equivalent stiffness of the three springs. a) a) 1.71 1.71 kN/m kN/m b) b) 5 5 kN/m kN/m c) c) 2.71 2.71 kN/m kN/m d) d) 3.71 3.71 kN/mkN/m Answer: a) 1.71 kN/m Answer: a) 1.71 kN/m Solution: Solution: 12 12 7 7 12 12 4 4 3 3 3 3 1 1 4 4 1 1 6 6 1 1 6 6 1 1 4 4 1 1 k k 1 1 k k 1 1 k k 1 1 k k 1 1 3 3 2 2 1 1 e e
11..7171kNkN//mm 7 7 12 12 k k ee
32.32. A flat belt is 6 inches wide and 1/3 inch thick and transmits 20 Hp. The center distance is 8 ft. The drivingA flat belt is 6 inches wide and 1/3 inch thick and transmits 20 Hp. The center distance is 8 ft. The driving pulley is
pulley is 6 inches i6 inches in diameter n diameter and rotates at and rotates at 2 000 2 000 rpm such rpm such that the that the loose side loose side of the of the belt is belt is on top. on top. The drivenThe driven pulley
pulley is is 18 18 inches inches in in diameter. diameter. The The belt belt material material is is 0.035 0.035 lb/inlb/in33 and the coefficient of friction is 0.30. and the coefficient of friction is 0.30. Determine the belt net tension.
Determine the belt net tension. a)
Answer: b) 210 lb Answer: b) 210 lb Solution: Solution:
210210lblb 000 000 2 2 20 20 000 000 63 63 6 6 2 2 n n Hp Hp 000 000 63 63 D D 2 2 D D T T 2 2 F F F F F F 11 22
Other Solution: Other Solution:
22000000
31413141..5959 ffpmpm 12 12 6 6 n n D D V Vmm
lb lb 210 210 59 59 .. 3141 3141 20 20 000 000 33 33 V V Hp Hp 000 000 33 33 F F F F F F m m 2 2 1 1
33.33. A pulley 600 mm in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley isA pulley 600 mm in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is 155
155oo, the coefficient of friction between belt and pulley is 0.35 and the safe working stress of the belt is 2.1, the coefficient of friction between belt and pulley is 0.35 and the safe working stress of the belt is 2.1 MPa. Determine the belt tensions ratio, neglecting the effect of centrifugal force.
MPa. Determine the belt tensions ratio, neglecting the effect of centrifugal force. a) a) 2.578 2.578 b) b) 2.857 2.857 c) c) 5.287 5.287 d) 5.782d) 5.782 Solution: Solution: ee
ee 22..578578 F F F F 180 180 155 155 35 35 .. 0 0 f f 2 2 1 1
Answer: a) 2.578 Answer: a) 2.578 34.34. Select a deep-groove ball bearing to carry a radial load FSelect a deep-groove ball bearing to carry a radial load Fxx = 800 lb and a thrust load F = 800 lb and a thrust load Fzz = 700 lb at 1750 rpm. = 700 lb at 1750 rpm.
The service is 8 hr/day, but it is not continuous; design for 20 000 hr. The operation is smooth with little The service is 8 hr/day, but it is not continuous; design for 20 000 hr. The operation is smooth with little vibration; the outer ring rotates. Determine the design life in mr with no more than 10 % failure.
vibration; the outer ring rotates. Determine the design life in mr with no more than 10 % failure. a) a) 20100 20100 mr mr b) b) 2100 2100 mr mr c) c) 2001 2001 mr mr d) d) 1200 1200 mrmr Answer: b) 2100 mr Answer: b) 2100 mr Solution: Solution:
21002100 mr mr 10 10 750 750 1 1 60 60 000 000 20 20 rp rpmm hr hr // ss min min 60 60 Hr Hrss B B 6 6 10 10
35.35. Determine the Hp lost when a collar is loaded with 2000 lb, rotates at 50 rpm, and has a coefficient of friction ofDetermine the Hp lost when a collar is loaded with 2000 lb, rotates at 50 rpm, and has a coefficient of friction of 0.15. The outside diameter of the collar is 4 inches and the inside diameter is 2 inches.
0.15. The outside diameter of the collar is 4 inches and the inside diameter is 2 inches. a) a) 0.7314 0.7314 Hp Hp b) b) 0.3714 0.3714 Hp Hp c) c) 0.4713 0.4713 Hp Hp d) d) 0.4371 0.4371 HpHp Answer: b) 0.3714 Hp Answer: b) 0.3714 Hp Solution: Solution:
00..37143714 63000 63000 rp rpmm 50 50 inches inches 56 56 .. 1 1 lb lb 2000 2000 15 15 .. 0 0 63000 63000 n n r r W W f f 63000 63000 n n T T ffHpHp
f f
f f
Where, Where,
33
11 11..5656 inchesinches 1 1 2 2 3 3 2 2 r r r r r r r r 3 3 2 2 r r 2 2 2 2 3 3 3 3 2 2 ii 2 2 o o 3 3 ii 3 3 o o f f
36.36. What load in N must be applied to a 25 mm round steel bar 2.5 m long (E = 207 Gpa) to stretch the bar 1.3 mm?What load in N must be applied to a 25 mm round steel bar 2.5 m long (E = 207 Gpa) to stretch the bar 1.3 mm? a)
a) 42 42 000 000 N N b) b) 52 52 840 840 N N c) c) 53 53 000 000 N N d) d) 60 60 000 000 NN Answer: b) The load, 52 840 N
Answer: b) The load, 52 840 N Solution: The load,
Solution: The load,
10001000
44
00..025025
207207000000000000000000
5252837837..6666 NN 5 5 .. 2 2 3 3 .. 1 1 E E D D 4 4 L L L L E E A A F F 22
22
37.37. A 50-mm diameter shaft is to transmit 12 kW power at a speed of 500 rpm, determine the mean diameter of theA 50-mm diameter shaft is to transmit 12 kW power at a speed of 500 rpm, determine the mean diameter of the pin, under double shear, for a material having a safe unit stress of 40 N/
a)
a) 11.08 11.08 mm mm b) b) 12.08 12.08 mm mm c) c) 13.08 13.08 mm mm d) d) 10.08 10.08 mmmm Answer: b) Diameter of the pin = 12.08 mm
Answer: b) Diameter of the pin = 12.08 mm
Solution:
Solution: Transmitted Transmitted Torque,Torque,
500500 229229..183183 N N mm 229229183183 N N mmmm 000 000 12 12 30 30 n n P P 30 30 T T
Pin Shearing Force,
Pin Shearing Force,
99167167..3232 N N 50 50 183 183 229 229 2 2 D D T T 2 2 F Fss
Pin Mean Diameter,
Pin Mean Diameter, ss22 22ss
ss ss ss d d F F 2 2 d d 2 2 F F 4 4 A A 2 2 F F ss
4040 1212..0808 mmmm 32 32 .. 167 167 9 9 2 2 ss F F 2 2 d d ss ss
38.38. A 1200 mm cast iron pulley is fastened to 112.5 mm shaft by means of a 28.13 mm square key 175 mm long.A 1200 mm cast iron pulley is fastened to 112.5 mm shaft by means of a 28.13 mm square key 175 mm long. The key and shaft have a shearing stress of 14 000 psi. Determine the force acting at the pulley that will shear The key and shaft have a shearing stress of 14 000 psi. Determine the force acting at the pulley that will shear the key.
the key. a)
a) 10 10 015 015 lb lb b) b) 11 11 005 005 lb lb c) c) 11 11 050 050 lb lb d) d) 10 10 501 501 lblb Answer: a) The force acting on the pulley = 10 015 lb
Answer: a) The force acting on the pulley = 10 015 lb
Solution: Solution:
1010014014..7474 lblb 200 200 1 1 5 5 .. 112 112 4 4 .. 25 25 175 175 4 4 .. 25 25 13 13 .. 28 28 000 000 14 14 D D D D bL bL ss D D 2 2 bLD bLD ss 2 2 D D T T 2 2 F F p p ss p p ss p p p p
Where,Where, TorqueTorque ba basedsed onon shear shear ,, inin lblb 2 2 bLD bLD ss T T
ss
39.39. A 75-mm diameter shaft is transmitting 350 kW at 650 rpm. A flange coupling is used and has 6 bolts, each 18A 75-mm diameter shaft is transmitting 350 kW at 650 rpm. A flange coupling is used and has 6 bolts, each 18 mm in diameter. Find the required diameter of the bolts circle based on an average shearing stress of 27.5 MPa. mm in diameter. Find the required diameter of the bolts circle based on an average shearing stress of 27.5 MPa. a)
a) 245 245 mm mm b) b) 254 254 mm mm c) c) 452 452 mm mm d) d) 425 425 mmmm Answer: a) Bolt Circle diameter = 245 mm
Answer: a) Bolt Circle diameter = 245 mm
Solution: Solution:
00..018018
2727500500
650650
66 00..245245mm 245245mmmm 350 350 30 30 8 8 n n n n ss d d P P 30 30 8 8 n n d d ss T T 8 8 D D 2 2 2 2 B B ss 2 2 2 2 B B 2 2 ss B B
Note:Note: Torque Based on Shear,Torque Based on Shear,
n n P P 30 30 8 8 n n n n D D d d ss T T BB BB 2 2 ss
40.40. If a sleeve bearing has an outside diameter of 38.1 mm and a length of 50.8 mm, the wall thickness is 3/16 inch.If a sleeve bearing has an outside diameter of 38.1 mm and a length of 50.8 mm, the wall thickness is 3/16 inch. The bearing is subjected to a radial load of 500 kg. What is the bearing pressure, in psi?
The bearing is subjected to a radial load of 500 kg. What is the bearing pressure, in psi? a)
a) 904 904 psi psi b) b) 409 409 psi psi c) c) 490 490 psi psi d) d) 940 940 psipsi Answer: c) Bearing pressure = 490 psi
Answer: c) Bearing pressure = 490 psi
Solution:
Solution: For bearing For bearing or or projected projected area,area, BB inin.. 22
331616 inin 22..2525 inin22 4 4 .. 25 25 1 1 .. 38 38 .. in in 4 4 .. 25 25 8 8 .. 50 50 D D L L A A
Bearing pressure,Bearing pressure,
490490 psi psi in in 25 25 .. 2 2 kg kg // lb lb 205 205 .. 2 2 kg kg 500 500 A A W W p p 2 2 B B
T T Pin Pin Shaft Shaft41.
41. Determine the minimum whole depth of spur gear of 14.5Determine the minimum whole depth of spur gear of 14.5oo involute type with diametral pitch of 24 and circular involute type with diametral pitch of 24 and circular pitch of 0.1309 inch.
pitch of 0.1309 inch. a)
a) 0.09000 0.09000 inch inch b) b) 0.08900 0.08900 inch inch c) c) 0.0899 0.0899 inch inch d) d) 0.089758 0.089758 inchinch Answer: c) Whole depth = 0.0899 inch
Answer: c) Whole depth = 0.0899 inch
Solution: From Vallance, page 262:
Solution: From Vallance, page 262: 00..089875089875 inchinch 24 24 157 157 .. 2 2 p p 157 157 .. 2 2 h h d d
42.42. A parallel helical gear set was a 17-tooth pinion driving a 34-tooth gear. The pinion has a right-hand helix angleA parallel helical gear set was a 17-tooth pinion driving a 34-tooth gear. The pinion has a right-hand helix angle of 30
of 30oo, a normal pressure angle of 20, a normal pressure angle of 20oo, and a normal diametral pitch of 5 teeth/in. Find the axial circular pitches., and a normal diametral pitch of 5 teeth/in. Find the axial circular pitches. a)
a) 1.2566 1.2566 inches/tooth inches/tooth b) b) 1.6625 1.6625 inches/tooth inches/tooth c) c) 1.6526 1.6526 inches/tooth inches/tooth d) d) 1.6256 1.6256 inches/toothinches/tooth Answer: a) Axial circular pitch = 1.2566 inches/tooth
Answer: a) Axial circular pitch = 1.2566 inches/tooth
Solution: Solution: cncn c c oo P P 0.628320.62832 P
P 00..772255552 2 iinncchh // ttooootthh c coos s ccoos 3s 300
c c a a oo P P 0.72552 0.72552 PP 11..2255666 6 iinncchheess / t/ tooootthh ttaan n ttaan 3n 300
43.43. Determine the Poisson’s ratio of a material whose modulus of elasticity is 200 GPa and whose modulus ofDetermine the Poisson’s ratio of a material whose modulus of elasticity is 200 GPa and whose modulus of rigidity is 80 GPa.
rigidity is 80 GPa. a)
a) 0.33 0.33 b) b) 0.25 0.25 c) c) 0.38 0.38 d) d) 0.220.22 Answer: b) Poisson’s ratio = 0.25
Answer: b) Poisson’s ratio = 0.25
Solution: Solution: )) 1 1 (( 2 2 E E G G
E = E = 200GPa, 200GPa, G G = = 80GPa:80GPa:
0
0
..
25
25
44.
44. A steel has a BHN = 300. What is its approximate ultimate strength in ksi?A steel has a BHN = 300. What is its approximate ultimate strength in ksi? a)
a) 300 300 ksi ksi b) b) 150 150 ksi ksi c) c) 75 75 ksi ksi d) d) 200 200 ksiksi Answer: b) Ultimate Strength = 150 ksi
Answer: b) Ultimate Strength = 150 ksi Solution: S
Solution: Suu 0.5(BHN), ksi 0.5(BHN), ksi
45.
45. If the angular deformation of a solid shaft should not to exceed 1If the angular deformation of a solid shaft should not to exceed 1oo in a length of 1.8 m and the allowable in a length of 1.8 m and the allowable shearing stress is 83 MMa, what is the diameter of the shaft? Assume that the shaft material has G = 77 x 10 shearing stress is 83 MMa, what is the diameter of the shaft? Assume that the shaft material has G = 77 x 1066 kPa.
kPa. a)
a) 222.34 222.34 mm mm b) b) 234.22 234.22 mm mm c) c) 23.42 23.42 cm cm d) d) 24.22 24.22 cmcm Answer: a) Shaft diameter = 222.34 mm
Answer: a) Shaft diameter = 222.34 mm
Solution: Solution: G G D D L L ss 2 2 G G D D L L 16 16 ss D D 32 32 G G D D L L T T 32 32 JG JG T TLL ss 4 4 ss 3 3 4 4
7777xx1010
222222..3434mmmm 180 180 1 1 8 8 .. 1 1 000 000 83 83 2 2 G G L L ss 2 2 D D 6 6 o o o o ss
46.46. Determine the Hp lost when a collar is loaded with 2000 lb, rotates at 50 rpm, and has a coefficient of friction ofDetermine the Hp lost when a collar is loaded with 2000 lb, rotates at 50 rpm, and has a coefficient of friction of 0.15. The outside diameter of the collar is 4 inches and the inside diameter is 2 inches.
0.15. The outside diameter of the collar is 4 inches and the inside diameter is 2 inches. a)
Answer: b) Frictional Horsepower = 0.374 Hp Answer: b) Frictional Horsepower = 0.374 Hp
Solution: Solution:
00..37143714 63000 63000 rpm rpm 50 50 inches inches 56 56 .. 1 1 lb lb 2000 2000 15 15 .. 0 0 63000 63000 n n r r W W f f 63000 63000 n n T T ffHpHp
f f
f f
Where, Where,
33
11 11..5656inchesinches 1 1 2 2 3 3 2 2 r r r r r r r r 3 3 2 2 r r 2 2 2 2 3 3 3 3 2 2 ii 2 2 o o 3 3 ii 3 3 o o f f
47.47. Two shafts 3.6 m between centers carry pulleys 1.2 m in diameter and 0.91 m in diameter respectively,Two shafts 3.6 m between centers carry pulleys 1.2 m in diameter and 0.91 m in diameter respectively, connected by a crossed belt. It is desired to put the belt on as an open belt. How long a piece must be cut of it? connected by a crossed belt. It is desired to put the belt on as an open belt. How long a piece must be cut of it? a)
a) 303.3 303.3 mm mm b) b) 330 330 mm mm c) c) 333.0 333.0 mm mm d) d) 330.3 330.3 mmmm Answer: Length to be cut off = 303.3 mm
Answer: Length to be cut off = 303.3 mm
Solution: For the length of an open belt connection, Solution: For the length of an open belt connection,
3600
3600
10
10
520
520
..
22
22
mm
mm
4
4
910
910
1200
1200
3600
3600
2
2
910
910
1200
1200
2
2
C
C
4
4
D
D
D
D
C
C
2
2
D
D
D
D
2
2
L
L
2 2 2 2 1 1 2 2 2 2 1 1 o o
For the length of belt for crossed belt connection, For the length of belt for crossed belt connection,
3600
3600
10
10
823
823
..
55
55
mm
mm
4
4
910
910
1200
1200
3600
3600
2
2
910
910
1200
1200
2
2
C
C
4
4
D
D
D
D
C
C
2
2
D
D
D
D
2
2
L
L
2 2 2 2 1 1 2 2 2 2 1 1 c c
Solving for the difference of lengths, Solving for the difference of lengths,
mm mm 33 33 .. 303 303 22 22 .. 520 520 10 10 55 55 .. 823 823 10 10 L L L L L L
cc
oo
48.48. A 20-tooth motor sprocket, running at 1200 rpm, drives a blower at a speed ratio of 4:1. Using the largestA 20-tooth motor sprocket, running at 1200 rpm, drives a blower at a speed ratio of 4:1. Using the largest permissible chain
permissible chain size and size and the largest the largest permissible center permissible center distance of distance of 80 pitches, 80 pitches, what length what length of chain of chain in pitchesin pitches is required to connect the sprockets?
is required to connect the sprockets? a)
a) 200 200 pitches pitches b) b) 212 212 pitches pitches c) c) 216 216 pitches pitches d) d) 220 220 pitchespitches Answer: b) Chain length = 21 pitches
Answer: b) Chain length = 21 pitches Solution:
Solution:
212212 pitch pitch es.. es C C 40 40 N N N N C C 2 2 2 2 N N N N L L p p 2 2 1 1 tt 2 2 tt p p 2 2 tt 1 1 tt c c
49.49. A 20-kW motor, running at 1200 rpm, drives a 400 mm diameter pulley at a belt tension ratio of 2.4. If theA 20-kW motor, running at 1200 rpm, drives a 400 mm diameter pulley at a belt tension ratio of 2.4. If the belt’s
belt’s tight side tension is only 1200 N, determine the transmission efficiency.tight side tension is only 1200 N, determine the transmission efficiency. a)
a) 87.97 87.97 % % b) b) 84.58 84.58 % % c) c) 85.66 85.66 % % d) d) 86.55 86.55 %% Answer: a) Transmission Efficiency = 87.97 %
Answer: a) Transmission Efficiency = 87.97 %
Solution: Solution:
8797 8797 .. 0 0 000 000 20 20 4 4 .. 2 2 1200 1200 1200 1200 2 2 40 40 .. 0 0 30 30 1200 1200 P P F F F F 2 2 D D 30 30 n n P P F F 2 2 D D 30 30 n n P P 30 30 n n T T P P P P input input 2 2 1 1 input input net net input input out out input input output output
50.50. A right-handed single-thread hardened-steel worm has a catalog rating of 2.25 kW at 650 rpm when meshedA right-handed single-thread hardened-steel worm has a catalog rating of 2.25 kW at 650 rpm when meshed with a 48-tooth cast-steel gear. The axial pitch of the worm is 25 mm, normal pressure angle is 14.5
with a 48-tooth cast-steel gear. The axial pitch of the worm is 25 mm, normal pressure angle is 14.5oo, and the, and the pitch diameter of the worm is 100 mm. The coefficient of
pitch diameter of the worm is 100 mm. The coefficient of friction is 0.085. Determine the friction is 0.085. Determine the shafts center distance.shafts center distance. a)