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Sample Problems - Machine Design

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MACHINE DESIGN AND SHOP PRACTICE PROBLEMS
ELEMENTS (with answers and solutions)
1.
What is the polar section modulus of a solid shaft with a diameter of 101.6 mm?
a) 209.5 cm3
b) 209.5 cm4
c) 205.9 cm3
d) 205.9 cm4
Answer: c) 205.9 cm3
Solution:
2.
Zj 
J 2J  3 


D  10.163  205.9 cm 3
c D 16
16
With 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
minimum thickness of the tank plate if the stress is limited to 40 Mpa.
a) 3.3 mm
b) 4.4 mm
c) 5.5 mm
d) 8.8 mm
Answer: b) Thickness, t = 4.4 mm
Solution: p  gh  10009.80666  58 839.6 Pa  58.8396 kPa
pD 58.83966

 0.00441 m  4.41 mm
2s t
2 40 000
A steel shaft 1.75 inches in diameter transmits 40 Hp at 1800 rpm. Assuming a modulus of rigidity of 12 x 10 6
psi, find the torsional deflection of the shaft in degrees per foot length.
a) 0.0073
b) 0.0037
c) 0.0063
d)
0.00013
t
3.
 63 000 Hp 
 6300040 
32

32

n

T
32T


 1800 
Solution:




 0.00013 radian  0.0073
L JG D 4 G
 D4 G
 1.754 12 x 10 6
A 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.
a) 1.144
b) 1.44
c) 1.1144
d) 1.1414

4.

Answer: Curve Correction Factor, Kc = 1.1144
5.
4C  1 0.615

Kw
4
C  1.184  1.1144
 C4
Solution: K c 
2C  1
Ks
1.0625
2C
A 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
applied to produce a force of 200 lb against the jaws of the vise?
a) 39.73 in-lb
b) 33.97 in-lb
c) 37.93 in-lb
d) 39.37 in-lb
Answer: c) 37.93 in-lb
Vise Jaw
Screw
Collar
F = 200 lb
Nut
Handle
Solution:
For the thread pitch, p 
1
1
  0.25 inch
Number of Threads per Inch 4
For the minor diameter, Di = Do – p = 1 – 0.25 = 0.75 inch
For the mean or pitch diameter of the screw, Dm 
Do  Di 1 0.75

 0.875 inch
2
2
For the lead angle,
 L 
p 
1 
1  0.25 
1
o
  tan 1 
  tan 
  tan 
  tan 0.09095  5.197

D

D
0
.
875



 m
 m
Solving for the torque required to overcome the collar friction,
Tc 
fFDc  0.2  200  0.5 

 20 in  lb
2
2
Solving for the torque required to overcome the thread friction,
Ts 
FDm  tan   f  200  0.875   0.09095  0.2 

  25.93 in  lb


2  1 f tan  
2
 1 0.2  0.9095  
Solving for the total torque,
T = Ts + Tc = 25.93 + 20 = 45.93 inches
6.
ans.
A 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.
a) 1.148
b) 1.184
c) 1.418
d) 1.814
Answer: b) Wahl Factor, Kw = 1.184
Solution: C 
7.
Dm
1

8
1
d
8
Kw 
4C  1 0.615 48  1 0.615



 1.184
4C  4
C
48  8
8
A 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.
a) 1.172
b) 1.712
c) 1.217
d) 1.271
Answer: a) Bergstrassar Factor, KB = 1.1724
Dm
4C  2 48  2
1
KB 

 1.1724

8
1
4C  3 48  3
d
8
Compute the tooth thickness of a 14.5o spur gear with diametral pitch of 5.
a) 0.23140 inch
b) 0.31416 inch
c) 0.43140 inch
Solution: C 
8.
d) 0.03140 inch
Answer: b) Tooth thickness, t = 0.31416 inch
1.5708 1.5708

 0.31416 inch
pd
5
Compute the speed of the gear mounted on a 52.5 mm diameter shaft receiving power from a driving motor
with 250 Hp.
a) 2 182 rpm
b) 2 071 rpm
c) 2 265 rpm
d) 2 341 rpm
Solution: t 
9.
Answer: c) The speed, N = 2265 rpm
Solution: From PSME Code: N 
80 P
D
3
80 250

 52.5 


 25.4 
3
 2 264.92 rpm
10. Compute the working strength of 1 inch bolt which is screwed up tightly in packed joint when the allowable
working stress is 13000 psi.
a) 3 600 lb
b) 3 950 lb
c) 3 900 lb
d) 3 800 lb
Answer c) The working strength, W = 3 900 lb
Solution:

From

machinery’s


Handbook,
page
1149:
W  s t 0.55d  0.25d  13 000 0.551  0.251  3 900 lb
11. Compute 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.
a) 16.95
b) 21.65
c) 19.56
d) 25.12
2
2
Answer: Shear Stress, ss = 19.56 MPa
Solution: ss 
16 T
16 0.48

 19 556.96 kPa  19.56 MPa
3
3
D
 0.050
12. A 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.
a. 5.85 mm
b. 6.85 mm
c. 7.85 mm
d. 8.85 mm
Answer: b) Pipe thickness, t = 6.85 mm
Solution: Inside Diameter, D i 
D o2 
Thickness of the Pipe, t 
4F

s
0.242 
4 80
 0.2263m  226.3 mm
 16 000
D o  D i 240  226.3

 6.85 mm
2
2
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 load
of 20 kgs. is suspended on the beam 4 m from the free end. The moment at the fixed end is
a. 3760 kg-m
b. 0.0 kg-m
c. 60 kg-m
d. 4800 kg-m
Answer: a) 3760 kg-m
W = 50 (12) = 600 kg
F = 20 kgs
Solution: Bending Moment,
M  6 600  8 20  3 760 kg  m
6m
8m
14. A 20o 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.
a) 18.4o
b) 20o
c) 14.5o
d) 20.5o
Answer: a) Pitch angle of the pinion = 18.4o
 Tp
Solution:   tan 1 
 Tg


  tan 1  14   18.4 o

 42 

15. A 20o 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
hp. Determine the pitch diameters of the gears.
a) 2.33 inches and 5.36 inches
b) 3.23 inches and 3.56 inches
c) 5.36 inches and 6.36 inches
d) 2.33 inches and 2.33 inches
Answer: a) 2.33 inches and 5.36 inches
T 
 32 
D2  D1  2    2.33    5.36 inches
 14 
 T1 
16. A 1-inch diameter shaft has a single disc weighing 75 lb mounted midway 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
106 psi.
a) 2038 rpm
b) 2308 rpm
c) 2380 rpm
d) 2803 rpm
Solution: D1 
T1 14

 2.33 inches
Pd
6
W = 75 lb
Solution:
D = 1 inch
E = 30 x 106 psi
L = 20 inches
 D4  1

 0.0491 in.4
Moment of Inertia, I 
64
64
4
576 E I g

Solving for the lowest critical speed, cr 
W L3
10”
10”
 576  30 x 106   0.0491 32.2
3
 75 20
 213.39 rad / s
30 cr 30  213.39 

 2037.68 rpm


17. A 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?
a) 368
b) 150
c) 1474
d) 38
Ncr 
Answer: c) Flywheel Energy =1474
Solution: V1  2  0.3753  7.069 m / s



V2  2  0.3752.8  6.597 m / s

m V12  V22
457 7.0692  6.5972

 1 473.91 N  m
2
2
18. A 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.
a) 11.31o
b) 13.11o
c) 11.13o
d)
o
10.13
KE 
Answer: Gear Helix angle = 11.31o
Solution: Circular pitch of the worm gear, Pc 
Dg
Tg

  5
25
 0.6283 inch
Where, Pc = Pa = pitch of the worm
Solving for the lead, L = Nt Pa = 3(0.6283) = 1.8849 inches
 L 
1  1.8849 
o
Solving for the lead angle of the worm,   tan 1 
  tan 
  11.31
 3 
 Dw 
For the helix angle, and considering that the shafts angle is 90o,
g =  = 11.31o
ans.
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 a
pitch of 6 mm. The coefficient of thread friction and collar friction are 0.13 and 0.10 respectively. If the collar
mean diameter is 90 mm and the screw turns at 60 rpm, find the axial linear speed of the screw.
a) 5 mm/s
b) 6 mm/s
c) 7 mm/s
d) 5.5 mm/s
Answer: b) 6 mm/s
Solution: For the linear speed of the screw, Vn = n (L) = (60 rpm)(6 mm/rev) = 360 mm/min = 6 mm/s
20. A 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
torque on the thrust bearing of the motor is taken as 20 % of the total input. Determine the lead angle.
a) 12.465o
b) 14.265o
c) 15.462o
d) 16.452o
Answer: a) 12.465o
Solution: For the lead, L 
V 10

 0.025 m  25 mm
n 400
 L
For the lead angle,   tan 1 
 D m

 25 
o
  tan 1 
  12.465
36




21. What Hp can a 1-inch diameter short shaft transmit at 380 rpm?
a) 3 Hp
b) 18 Hp
c) 10 Hp
d) 7.1 Hp
Answer: c) 10 Hp
Solution: P 
D3 N 1 380

 10 Hp
38
38
3
22. A 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 20o. Determine the tangential load, in N.
a) 7660
b) 6790
c) 3900
d) 3098
 2  30 P   2   30 25000
  


  7 957.75 N
 D   n   0.10    600 
Solution: Ft  
23. A 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
needed to engage the clutch.
a) 2820 N
b) 2428 N
c) 5460 N
d) 3638 N
Answer: 3638 N
Solution: T 
30 P 30 100

 0.3183 kN  m  318.3 N  m
 n  3000
 D  d  
T  f  P  rf  N fs  f  P  
  N fs
 4 
4T
4 318 300
P

 3 637.71 N
f  D  d   N fs 0.5 200  1502
24. A 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 in2. Determine the maximum tensile stress on the
rope in ksi.
a) 26.2
b) 25.0
c) 27.6
d) 32.4
Answer: c) maximum tensile stress = 27.6 MPa
Solution: a 
V1  Vo 1000 / 60  0

 3.33 fps 2
t
5
St 

a
3.33 

F  WL 1    10 000 1 
  11 034.16 lb
g
32.2 


11,034.16
 27585.4 psi  27.6 ksi
0.4
25. What 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?
a) 4 020 psi
b) 4 200 psi
c) 2 400 psi
d) 2 040 psi
Answer: b) 4200 psi
Solution:
p
4 t sl E j
D


 63.5 
4
in  60 000 lb / in 2 0.70
 25.4 

 4 200 psi
100 in.
Where, p = bursting pressure, psi
t = shell thickness, inches
sl = shell stress, psi
D = shell diameter, inches
Ej
=
joint
efficiency
Note: For the longitudinal stress of the thin-walled cylinder, and the stress for spherical tank: s L 
pD
4tEj
26. A cylinder having an internal diameter if 508 mm and external diameter if 914.4 mm is subjected to an internal
pressure of 69 MPa and an external pressure of 14 MPa. Determine the hoop stress at the inner surface of the
cylinder.
a) 90.11 MPa
b) 91.10 MPa
c) 911.0 MPa
d) 19.10 Mpa
Answer: a) 90.11 MPaa
Solution:
s ti 


p i ro2  ri2  2p o ro2
ro2
 ri2

69 MPa 914.4


2
2
2
2
  508
 
 914.4

mm   
mm    214 MPa 
mm 
  2
 
 2

2
 914.4
  508

mm   
mm 

 2
  2

2
sti = 90.11 Mpa
Where, sti = maximum tangential or hoop stress at the inside
pi = internal pressure, Mpa
po = external pressure, Mpa
Note: For the maximum tangential or hoop stress at the outside, s to 
ri = inside radius, mm
ro = outside radius, mm

2p i ri2  p o ro2  ri2
ro2

 ri2
27. What 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.
a) 2.1 inches
b) 2.8 inches
c) 3.2 inches
d) 4.2 inches
Answer: a) 2.1 inches
Solution: Transmitted torque, T 
63000Hp 630001000

 63000 in  lb
n
1000
D
, for good proportion
4
2T
2 63000

 2.1 inches
Key length based on shear, L 
ss bD 1500014
Key width, b 
Key length based on compression, L 
4T
4 63000

 2.1
sc t D 30 00014
Therefore, use L = 2.1 inches
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 of
thread per inch.
a) 0.2 threads/inch
b) 10 threads/inch
c) 5 threads/inch
d) 2.5 threads/inch
Answer: TPI = 5 threads per inch
Solution:
The number of threads per inch, TPI 
1
1

 5 threads / inch
p 0.2
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 a
pitch of 6 mm. The coefficient of thread friction and collar friction are 0.13 and 0.10 respectively. If the collar
mean diameter is 90 mm and the screw turns at 60 rpm, find the combined efficiency of the screw and collar.
a) 15.32 %
b) 12.53 %
c) 13.52 %
d) 15.97 %
Answer: b) 12.53 %
W = 70 N

p
1
1
Solution: For the depth of the thread, h    p    6  3 mm
2
2
For the mean diameter of the screw, D m  D o  h  36  3  33 mm
 L
For the lead angle,   tan 1 
 D m
Solving for the Efficiency, e 
e

 p
  tan 1 

 D m

 6 
  tan 1 
  3.3123 o
33




tan 1  f tan  100 %
f D
tan   f   c c
 Dm

 1  f tan  

tan 3.3125 1  0.13 tan 3.3125 100 %  12.53 %
 0.10 90 
1  0.13 tan 3.3125 
tan 3.3125  0.10  
33 
o
o
o
o


30. 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;
rotational speed = 870 rpm; coefficient of friction = 0.30; and allowable pressure = 70 kPa.
a) 25.74 Hp
b) 24.75 Hp
c) 27.45 Hp
d) 24.57 Hp
Answer: a) 25.74 Hp
Solution:
Friction radius, rf 
3
3
2  ro  ri
3  ro2  ri2
  2  1253  1003 
   
 112.96 mm
  3  1252  1002 



Surface Area of contact, A f  2rf b  20.112960.125  0.0887 m 2
Force normal to the surface of contact, Fn  pA f  700.0887  6.209 kN
Power Capacity,
 n Tf
n
f Fn rf   870 0.306.2090.11296  19.2 kW  25.74 Hp
P

30
30
30
31. Three 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.
a) 1.71 kN/m
b) 5 kN/m
c) 2.71 kN/m
d) 3.71 kN/m
Answer: a) 1.71 kN/m
Solution:
1
1
1
1
1 1 1 1 1 3 4 7



     

k e k1 k 2 k 3 4 6 6 4 3
12
12
ke 
12
 1.71 kN / m
7
32. A 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 6 inches in diameter and rotates at 2 000 rpm such that the loose side of the belt is on top. The driven
pulley is 18 inches in diameter. The belt material is 0.035 lb/in3 and the coefficient of friction is 0.30.
Determine the belt net tension.
a) 201 lb
b) 210 lb
c) 102 lb
d) 120 lb
Answer: b) 210 lb
Solution:
F  F1  F2 
2 T 2  63 000 Hp   2  63 00020 
 
   
  210 lb
D
D
n
  6  2 000 
6
Vm   D n   2 000  3141.59 fpm
 12 
33 000 Hp 33 00020
F  F1  F2 

 210 lb
Vm
3141.59
Other Solution:
33. A pulley 600 mm in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is
155o, 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.
a) 2.578
b) 2.857
c) 5.287
d) 5.782
Solution:
  
F1
 e f  e0.35155 180   2.578
F2
Answer: a) 2.578
34. Select a deep-groove ball bearing to carry a radial load Fx = 800 lb and a thrust load Fz = 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
vibration; the outer ring rotates. Determine the design life in mr with no more than 10 % failure.
a) 20100 mr
b) 2100 mr
c) 2001 mr
d) 1200 mr
Answer: b) 2100 mr
Solution: B10  Hrs 60 min s / hr rpm  
20 000601 750  2100 mr
10 6
35. Determine 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.
a) 0.7314 Hp
b) 0.3714 Hp
c) 0.4713 Hp
d) 0.4371 Hp
Answer: b) 0.3714 Hp
Solution: fHp 
Tf n
f W rf n 0.152000 lb1.56 inches 50 rpm 


 0.3714
63000
63000
63000
3
3
2  ro  ri  2  23  13 
 
  1.56 inches
3  ro2  ri2  3  32  12 
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?
a) 42 000 N
b) 52 840 N
c) 53 000 N
d) 60 000 N
Where, rf 
Answer: b) The load, 52 840 N
Solution: The load,
 1.3    
AE    2
2
F
  D E  
   0.025 207 000 000 000  52 837.66 N


L
L4
2
.
5
1000
4

 
37. A 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/mm 2.
a) 11.08 mm
b) 12.08 mm
c) 13.08 mm
d) 10.08 mm
Shaft
Answer: b) Diameter of the pin = 12.08 mm
Solution:
Transmitted Torque, T 
30 P 30 12 000

 229.183 N  m  229183 N  mm
n
 500
2 T 2 229183

 9167.32 N
D
50
F
4 Fs
2 Fs

Pin Mean Diameter, ss  s 
2
2 As 2  d
 d2
Pin Shearing Force, Fs 
d
2 9 167.32
 12.08 mm
 40
2 Fs

 ss
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.
The key and shaft have a shearing stress of 14 000 psi. Determine the force acting at the pulley that will shear
the key.
a) 10 015 lb
b) 11 005 lb
c) 11 050 lb
d) 10 501 lb
Answer: a) The force acting on the pulley = 10 015 lb
Solution:
 s bLD 
2 s

 D
2T
 2 
Fp 

 s s bL
 Dp
Dp
Dp

Where, T 
s s bLD
2

  14 000 28.13  175  112.5   10 014.74 lb



 25.4  25.4  1 200 

 Torque based on shear , in  lb
39. A 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.
a) 245 mm
b) 254 mm
c) 452 mm
d) 425 mm
Answer: a) Bolt Circle diameter = 245 mm
Solution: D B 
8T
 ssd n B
2

830P
 d ss n n B
2
2

830350
 0.0182 27 5006506
2
 0.245 m  245 mm
 s s d 2 D B n n B 30 P

8
n
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.
The bearing is subjected to a radial load of 500 kg. What is the bearing pressure, in psi?
a) 904 psi
b) 409 psi
c) 490 psi
d) 940 psi
Note:
Torque Based on Shear, T 
Answer: c) Bearing pressure = 490 psi
Solution:
 
 50.8
 38.1

For bearing or projected area, A B  L D  
in.
in.  2 3
in   2.25 in 2
16
 25.4
 25.4

Bearing pressure, p 
W 500 kg 2.205 lb / kg 

 490 psi
AB
2.25 in 2
Pin
T
41. Determine the minimum whole depth of spur gear of 14.5 o involute type with diametral pitch of 24 and circular
pitch of 0.1309 inch.
a) 0.09000 inch
b) 0.08900 inch
c) 0.0899 inch
d) 0.089758 inch
Answer: c) Whole depth = 0.0899 inch
Solution: From Vallance, page 262: h 
2.157 2.157

 0.089875 inch
pd
24
42. A parallel helical gear set was a 17-tooth pinion driving a 34-tooth gear. The pinion has a right-hand helix angle
of 30o, a normal pressure angle of 20o, and a normal diametral pitch of 5 teeth/in. Find the axial circular pitches.
a) 1.2566 inches/tooth b) 1.6625 inches/tooth
c) 1.6526 inches/tooth
d) 1.6256 inches/tooth
Answer: a) Axial circular pitch = 1.2566 inches/tooth
Solution: Pc 
Pcn
0.62832

 0.72552 inch / tooth
cos  cos30o
Pc
0.72552

 1.2566 inches / tooth
tan  tan 30o
43. Determine the Poisson’s ratio of a material whose modulus of elasticity is 200 GPa and whose modulus of
rigidity is 80 GPa.
a) 0.33
b) 0.25
c) 0.38
d) 0.22
Pa 
Answer: b) Poisson’s ratio = 0.25
Solution: G 
E
2(1  )
E = 200GPa,
G = 80GPa:
  0.25
44. A steel has a BHN = 300. What is its approximate ultimate strength in ksi?
a) 300 ksi
b) 150 ksi
c) 75 ksi
d) 200 ksi
Answer: b) Ultimate Strength = 150 ksi
Solution: Su  0.5(BHN), ksi
45. If the angular deformation of a solid shaft should not to exceed 1 o 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 6
kPa.
a) 222.34 mm
b) 234.22 mm
c) 23.42 cm
d) 24.22 cm
Answer: a) Shaft diameter = 222.34 mm
 D3ss 
L
32
 16 
TL 32T L

  2 ss L


Solution:  
JG D4G
DG
 D 4G
D
2 ss L

G
2830001.8
 222.34 mm
 
6
1 
77
x
10


 180o 
o


46. Determine 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.
a) 0.7314 Hp
b) 0.3714 Hp
c) 0.4713 Hp
d) 0.4371 Hp
Answer: b) Frictional Horsepower = 0.374 Hp
Solution:
fHp 
Tf n
f W rf n 0.152000 lb1.56 inches50 rpm


 0.3714
63000 63000
63000
3
3
2  r  r  2  23  13 

 1.56 inches
 3  32  12 



Where, rf   o2 i2
3 ro  ri

47. 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?
a) 303.3 mm
b) 330 mm
c) 333.0 mm
d) 330.3 mm
Answer: Length to be cut off = 303.3 mm
Solution: For the length of an open belt connection,
D  D1    
1200  9102  10 520.22 mm

L o  D1  D 2   2C  2
  1200 910  23600 
2
4C
43600
2
2
For the length of belt for crossed belt connection,
D  D1     1200 910  23600  1200 910  10823.55 mm

Lc  D1  D2   2C  2
 
2
4C
43600
2
2
2
Solving for the difference of lengths,
L  L c  L o  10 823.55  10 520.22  303.33 mm
48. A 20-tooth motor sprocket, running at 1200 rpm, drives a blower at a speed ratio of 4:1. Using the largest
permissible chain size and the largest permissible center distance of 80 pitches, what length of chain in pitches
is required to connect the sprockets?
a) 200 pitches
b) 212 pitches
c) 216 pitches
d) 220 pitches
Answer: b) Chain length = 21 pitches
Solution: L  N t1  N t 2  2C  N t 2  N t1   212.pitches
c
p
2
2
40C p
49. A 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 tight side tension is only 1200 N, determine the transmission efficiency.
a) 87.97 %
b) 84.58 %
c) 85.66 %
d) 86.55 %
Answer: a) Transmission Efficiency = 87.97 %
Solution:  
Poutput
Pinput
1200
  n  D 
 1200   0.40  
Tout  n    n   D  F

   net 
   F1  F2  

 1200 

30
2
30
2
30
2
2.4 
 
 


30




 0.8797
Pinput
Pinput
Pinput
20 000
50. A 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 o, and the
pitch diameter of the worm is 100 mm. The coefficient of friction is 0.085. Determine the shafts center distance.
a) 241 mm
b) 142 mm
c) 412 mm
d) 124 mm
Answer: a) Center distance = 241 mm
Solution:
Speed Ratio, SR 
Tan  
Tg
D g cos 
Dg
w n w




g
ng
Tw D w sin  D w tan 
p
L
25


 0.07958
D w D w 100
T 
   4.55o
T 


T 
g
D w tan    g D w  p    g  p    48  25   381.97 mm
Pitch diameter of the gear, D g  
 D   T  

T 
w   w    1   

 Tw 
 w
Center Distance, C 
D w  Dg
2

100  381.97
 241 mm
2
51. A 20o 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.
a) 18.4o
b) 20o
c) 14.5o
d) 20.5o
Answer: a) Pitch angle = 18.4o
 Tp 
  tan 1  14   18.4 o
 Tg 
 42 


Solution:   tan 1 
52. A triple-thread worm has a lead angle of 17 o and a pitch diameter of 2.2802 inches. Find the center distance
when the worm is mated with a wheel of 48 teeth.
a) 6.72 inches
b) 7.26 inches
c) 6.27 inches
d) 7.62 inches
Answer: a) Center Distance = 6.72 inches
Solution:
 Tg
Dg  
T
 w

D w tan    48 2.2802 tan17o  11.154 inches

 3 

D w  D g 2.2802 11.154
C

 6.72 inches
2
2
53. A double-thread worm has a pitch diameter of 3 inches. The wheel has 20 teeth and a pitch diameter of 5
inches. Find the gear helix angle.
a) 4.69o
b) 9.46o
c) 6.49o
d) 6.94o
Answer: b) Gear helix angle = 9.46o
Solution:
 T
  tan 1  w
 Tg
 D g


 D w



  tan 1  2  5   9.46o

20
3
  

54. A 36-tooth pinion turning at 300 rpm drives 120-tooth gear of 14.5o involute full depth pressure angle.
Determine the rpm of the driven gear.
a) 60 rpm
b) 45 rpm
c) 75 rpm
d) 90 rpm
Answer: d) 90 rpm
 Tp 
  300 36   90 rpm
 Tg 
 120 


Solution: n g  n p 
55. If two parallel shafts are connected by cylinders in pure rolling contact and turning in the same direction, and
having a speed ratio of 2.75, what is the Center distance of the two shafts assuming that the diameter of the
smaller cylinder is 22 cm?
a) 18.25 cm
b) 19.25 cm
c) 20.25 cm
d) 17.25 cm
Answer: b) Center Distance = 19.25 cm
Solution:
Diameter of the bigger cylinder, D 2  SRD1   2.7522  60.5 cm
Center distance, C 
D 2  D1 60.5  22

 19.25 cm
2
2
56. Two extension coil 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 spring is rated at 6 kN/m. Determine the total deflection of the springs.
a) 408.6 mm
b) 486.0 mm
c) 480.6 mm
d) 460.8 mm
Answer: a) Spring Deflection = 408.6 mm
Solution: t  1  2 
 k  k2 
4  6
F F
  100 kg9.8066 N / kg

 F 1
  408.6 mm
k1 k 2
 46 
 k1k 2 
57. If the ultimate shear strength of a steel plates is 42 000 psi, what force is necessary to punch a 0.75 inch
diameter hole in a 0.625 inch thick plate?
a) 61 850 lb
b) 65 810 lb
c) 61 580 lb
d) 60 185 lb
Answer: a) Punching force = 61 850 lb


Solution: F  su As  su d t   42000lb / in 2 0.75 in 0.625 in   61850.1 lb
58. A stepped torsion shaft has diameters of 16 mm and 12 mm and a fillet radius of 2 mm. The shaft is subjected to
a torque of 12.5 N-m. Find the maximum induced stress caused by the fillet. Consider a stress concentration
factor of 1.25.
a) 36.84 MPaa
b) 46.05 MPa
c) 38.64 MPa
d) 45.06 MPa
Answer: b) 46.05 MPa
Solution: Solving for the average induced shear stress in the shaft
ss 
16T 16 12.5

 36841422.01 Pa  36.84 MPa
 d3   0.012 3
Solving for the maximum induced shear stress,
r = 2 mm
D = 16 mm
d = 12 mm
ss max   k ts ss  1.25 36.84   46.05 MPa
59. A steam engine that has a stroke of 12 inches has an overhung crank of 11 inches. The maximum tangential
force, P, on the crank may be assumed as 75000 lb. Assuming an allowable stress in shear as 4400 psi,
determine the crankshaft diameter.
a) 4.77 inches
b) 3.77 inches
c) 2.77 inches
d) 1.77 inches
Answer: a) Crankshaft Diameter = 4.77 inches
Solution:
Solving for the torque, T  P  R  75006  45 000 in  lb
B
A
Crankpin
6”
11”
Crankshaft
Solving for the bending moment, M  750011  82 500 in  lb
Solving for the shaft diameter,
 16 

D  
  s s 
1
 3  16
M2  T2   

  4400
1
3
45 000  82 500   4.77"

2
2
60. A parallel helical gear-set consists of a 19-tooth pinion driving a 57-teeth gear. The pinion has a left-hand helix
angle of 20o, a normal pressure angle of 14½o, and a normal diametal pitch of 10 teeth/inch. If the pinion is to
transmit 50 Hp at a speed of 1750 rpm. Determine the center distance of the two gears.
a) 2.02 inches
b) 6.06 inches
c) 4.04 inches
d) 2.06 inches
Answer: c) 4.04 inches
Solution:
Pdn 
Tp
Tg
Pd


cos  D p cos  D g cos 
Pitch diameter of the pinion, D p 
Tp
Pdn cos 

19
10 cos 20 o
 2.02 inches
 Tg 
  2.02 57   6.06 inches
Pitch Diameter of the gear, D g  D p 
 Tp 
 19 
 
D p  Dg 2.02  6.06

 4.04 inches
Center-to-center distance, C 
2
2
61. A 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 6 mm. The coefficient of thread friction and collar friction are 0.13 and 0.10 respectively. If the collar
mean diameter is 90 mm and the screw turns at 60 rpm, find the axial linear speed of the screw.
a) 5 mm/s
b) 6 mm/s
c) 7 mm/s
d) 5.5 mm/s
Answer: b) 6 mm/s
Solution: For the linear speed of the screw,
Vn = n (L) = (60 rpm)(6 mm/rev) = 360 mm/min = 6 mm/s
62. Select a deep-groove ball bearing to carry a radial load Fx = 800 lb and a thrust load Fz = 700 lb at 1800 rpm.
The service is 8 hr/day, but it is not continuous; design for 18 250 hr. The operation is smooth with little
vibration; the outer ring rotates. Determine the design life in million revolution (mr) with no more than 10 %
failure.
a) 1791 mr
b) 1971 mr
c) 1197 mr
d) 1917 mr
63. If the ultimate shear strength of a steel plates is 42 000 psi, what force is necessary to punch a 0.75 inch
diameter hole in a 0.625 inch thick plate?
a) 61 850 lb
b) 65 810 lb
c) 61 580 lb
d) 60 185 lb
64. Find the pressure required to punch a 1-inch square hole in ¼ -inch thick steel.
a) 20.67 tons
b) 26.76 tons
c) 26.67 tons
d) 26.67 tons
65. What 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.
a) 2.1 inches
b) 2.8 inches
c) 3.2 inches
d) 4.2 inches
66. A flange coupling is to transmit 15,000 in-lb between two 2.5” diameter shafts. How many ½” diameter bolts in
a 6” diameter bolt circle are required if the shear stress in each bolt is limited to 3000 psi?
a) 9 bolts
b) 7 bolts
c) 6 bolts
d)
8
bolts
67. Determine the radius of gyration of a wheel that has an outside and inside diameters of 1 meter and 0.75 meter,
respectively? Neglect its hub and arms.
a) 0.5125 m
b) 0.6125 m
c) 0.3125 m
d) 0.4125 m
68. What is the polar section modulus of a solid shaft with a diameter of 101.6 mm?
a) 209.5 cm3
b) 209.5 cm4
c) 205.9 cm3
d) 205.9 cm4
69. If the angular deformation of a solid shaft should not to exceed 1 o 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 6
kPa.
a) 222.34 mm
b) 234.22 mm
c) 23.42 cm
d) 24.22 cm
70. What modulus of elasticity in tension is required to obtain a unit deformation of 0.00105 from a load producing
a unit tensile stress of 3163.27 kg/cm2?
a) 40 x 106 psi
b) 43 x 106 psi
c) 45 x 106 psi
d) 46 x 106 psi
71. If the ultimate shear strength of a steel plates is 42 000 psi, what force is necessary to punch a 0.75 inch
diameter hole in a 0.625 inch thick plate?
a) 61 850 lb
b) 65 810 lb
c) 61 580 lb
d) 60 185 lb
72. Determine the tensile stress area of an American Standard Screw Threads 6-32 UNC with basic major diameter
of 0.1380 inch.
a) 9.085 x 10-3 in2
b) 9.085 x 10-2 in2
c) 9.085 x 10-4 in2
d) 9.085 x 10-5
2
in
73. What is the number of threads per mm and the tensile stress area of a standard Metric screw Thread designated
by M10 x 1.5?
a) 0.555 and 57.99 mm2
b) 0.667 and 57.99 mm2 c) 1.5 and 57.99mm2
d) 1.75 and 57.99
2
mm
74. A line shaft is to transmit 200 Hp at 900 rpm. Find the diameter of the shaft.
a) 2.18 inches
b) 2.28 inches
c) 3.18 inches
d) 3.28 inches
75. A round steel shaft rotates at 200 rpm and is subjected to a torque of 275 N-m and a bending moment of 415 Nm. Determine the equivalent twisting moment.
a) 597.84 N-m
b) 456.42 N-m
c) 546.43 N-m
d) 497.85 N-m
76. A 75-mm diameter shaft is transmitting 300 kW at 600 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.
a) 227.4 mm
b) 477.2 mm
c) 274.7 mm
d) 247.7 mm
77. A 1.75-inch-diameter shaft is supported by two sleeve bearings. The total load on the two bearings is 2800 lb.
Find the friction power loss, in Hp, if the coefficient of friction between shaft and bearing is 0.10 and the shaft
rotates 200 rpm.
a) 0.88 Hp
b) 0.78 Hp
c) 0.98 Hp
d) 0.68 Hp
78. Determine 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.
a) 0.7314 Hp
b) 0.3714 Hp
c) 0.4713 Hp
d) 0.4371 Hp
79. A vertical steel cylinder water tank is 30 m in diameter and 45 m high. The allowable stress of the steel plate is
1224 kg/cm2. Without reinforcing angle bars and rods, what is the thickness of the steel plate?
a) 55.15 mm
b) 51.55 mm
c) 65.15 mm
d) 61.55 mm
80. What 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?
a) 4 020 psi
b) 4 200 psi
c) 2 400 psi
d) 2 040 psi
81. The root diameter of a double square thread is 0.55 inch. The screw has a pitch of 0.2 inch. Find the outside
diameter and the number of thread per inch.
a) 0.2 threads/inch
b) 10 threads/inch
c) 5 threads/inch
d)
2.5
threads/inch
82. 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?
a) 303.3 mm
b) 330 mm
c) 333.0 mm
d) 330.3 mm
83. A flat belt is 6 inches wide and 1/3 inch thick and transmits 15 Hp. The center distance is 8 ft. The driving
pulley is 6 inches in diameter and rotates at 2 000 rpm such that the loose side of the belt is on top. The driven
pulley is 18 inches in diameter. The belt material is 0.035 lb/in3 and the coefficient of friction is 0.30.
Determine the belt net tension.
a) 175.5 lb
b) 157.5 lb
c) 155.7 lb
d) 165.7 lb
84. A pulley 600 mm in diameter transmits 40 kW at 500 rpm. The arc of contact between the belt and pulley is
144o, 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.
a) 2.41
b) 2.14
c) 1.24
d) 4.12
85. A roller chain and sprocket is to drive vertical centrifugal discharge bucket elevator. The pitch of chain
connecting sprockets is 1.75”. The driving sprocket is rotating at 120 rpm and has 11 teeth while the driven
sprocket is rotating at 38 rpm. Determine the number of teeth of driven sprocket.
a) 33 teeth
b) 35 teeth
c) 30 teeth
d) 34 teeth
86. A helical steel spring has a maximum load of 800 lb and a corresponding deflection of 2 inches. If it has 8
active coils and an index of 6, what minimum shear strength of the spring material is required?
a) 57 ksi
b) 47 ksi
c) 67 ksi
d) 37 ksi
87. A helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the value of
Bergstrasssar factor of the spring.
a) 1.172
b) 1.712
c) 1.217
d) 1.271
88. A precision cut gear transmits 25 Hp at a pitch line velocity of 6000 fpm. If the service is intermittent, find the
dynamic load.
a) 247.05 lb
b) 274.05 lb
c) 275.04 lb
d) 247.05 lb
89. What is the difference of the values of the Wahl factor and the Bergstrasser factor, in percentage?
a) 1 %
b) Less than 1 %
c) Greater than 1 %
d) 0.5 %
90. A double-thread worm has a pitch diameter of 3 inches. The wheel has 20 teeth and a pitch diameter of 5
inches. Find the gear helix angle.
a) 4.69o
b) 9.46o
c) 6.49o
d) 6.94o
91. A helical-coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8 inch. Determine the shearstress augmentation factor of the spring.
a) 1.625
b) 1.0625
c) 1.0256
d)
1.0526
92. A disc clutch has 6 pairs of contacting friction surfaces with an outside diameter of 200 mm and an inside
diameter of 100 mm. The coefficient of friction of the clutch materials is 0.4 and the axial force is 1500 N. The
shaft speed is 1200 rpm. Determine the Hp that can be transmitted by the clutch assuming uniform pressure.
a) 35.2 Hp
b) 23.5 Hp
c) 47.2 Hp
d) 27.4 Hp
93. A flywheel has a mean diameter of 4 ft and is required to handle 2250 ft-lb of kinetic energy. It has a width of 8
inches, mean operating speed is 300 rpm and the coefficient of fluctuation is to be 0.05. Find the weight of rim,
assuming that the arms and hub are equivalent to 10 % of the total rim weight. The flywheel is made up of cast
iron with specific weight of 0.26 lb per cubic inch.
a) 333.7 lb
b) 373.3 lb
c) 337.3 lb
d) 733.3 lb
94. A 20o involute spur gear has a tooth whole depth of 16.95 mm, a tooth thickness of 13.2 mm, and a pitch of 3.
Determine the circular pitch of the gear.
a) 26.6 mm
b) 16.6 mm
c) 25.6 mm
d) 24.6 mm
95. A parallel helical gear-set consists of a 19-tooth pinion driving a 57-teeth gear. The pinion has a left-hand helix
angle of 20o, a normal pressure angle of 14½o, and a normal diametal pitch of 10 teeth/inch. If the pinion is to
transmit 50 Hp at a speed of 1750 rpm. Determine the center distance of the two gears.
a) 2.02 inches
b) 6.06 inches
c) 4.04 inches
d) 2.06 inches
96. A 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 o, and the
pitch diameter of the worm is 100 mm. The coefficient of friction is 0.085. Determine the shafts center distance.
a) 241 mm
b) 142 mm
c) 412 mm
d) 124 mm
97. A 20o 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.
a) 18.4o
b) 20o
c) 14.5o
d) 20.5o
98. A triple-thread worm has a lead angle of 17 o and a pitch diameter of 2.2802 inches. Find the center distance
when the worm is mated with a wheel of 48 teeth.
a) 6.72 inches
b) 7.26 inches
c) 6.27 inches
d) 7.62 inches
99. A double-thread worm has a pitch diameter of 3 inches. The wheel has 20 teeth and a pitch diameter of 5
inches. Find the gear helix angle.
a) 4.69o
b) 9.46o
c) 6.49o
d) 6.94o
100. A 36-tooth pinion turning at 300 rpm drives 120-tooth gear of 14.5o involute full depth pressure angle.
Determine the rpm of the driven gear.
a) 60 rpm
b) 45 rpm
c) 75 rpm
d)
90
rpm
101. Three 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.
a) 1.71 kN/m
b) 5 kN/m
c) 2.71 kN/m
d) 3.71 kN/m
102. Three extension springs are hooked in parallel that support a single weight of 100 kg. The springs are rated 4
kN/m, 5 kN/m, and 6 kN/m, respectively. Determine the equivalent spring constant of the three springs.
a) 10 kN/m
b) 15 kN/m
c) 9 kN/m
d) 11 kN/m
103. A 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 6 mm. The coefficient of thread friction and collar friction are 0.13 and 0.10 respectively. If the collar
mean diameter is 90 mm and the screw turns at 60 rpm, find the axial linear speed of the screw.
a) 5 mm/s
b) 6 mm/s
c) 7 mm/s
d) 5.5 mm/s
104. A 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
torque on the thrust bearing of the motor is taken as 20 % of the total input. Determine the lead angle.
a) 12.465o
b) 14.265o
c) 15.462o
d) 16.452o
105. What is the minimum clearance allowed for meshing spur gears with diametral pitch of 20? The spur gear has
25 teeth.
a) 0.00785 inch
b) 0.00758 inch
c) 0.00857 inch
d) 0.00758 inch
106. A 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 2. Determine the maximum tensile stress on the
rope in ksi.
a) 26.2
b) 25.0
c) 27.6
d) 32.4
107. A casting weighing 300 pounds is to be lifted by means of an overhead crane. The casting is lifted 10 feet in 12
seconds. What is the horsepower developed?
a) 0.54
b) 0.84
c) 0.95
d) 0.45
108. What wall thickness is required for a 500mm diameter cylinder under an internal pressure of 20 MPa? Use a
design stress of 80 MPa.
a) 73 mm
b) 53 mm
c) 63 mm
d)
83
mm
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