RMBC
MACHINE DESIGN: SHAFT DESIGN
MODULE 2
SHAFT DESIGN
DEFINITIONS AND USEFUL INFORMATION

Shaft - is a rotating or stationary member, usually of circular cross section much smaller in diameter than its length, used
to transmit motion or power; having mounted on it such power-transmitting elements as gears, pulleys, belts, chains,
cam, flywheels, cranks, sprockets, and rolling-element bearings.

Axle - is a non-rotating member that carries no torque and is used to support rotating wheels, pulleys, and the like.

Spindle - is a short shaft.

Line shaft, or main shaft, - is one driven by a prime mover, power is taken from it by belts or chain, usually at several
points along the shaft.

Counter shaft, or Jackshaft, or Head shaft - is a shaft intermediate between a line shaft and a driven machine.

Transmission shaft – is a shaft that is used to transmit power between the source and the machine absorbing the power.

Machine Shaft – a shaft forming on integral part of the machine itself.

The maximum-shear-stress hypothesis states that yielding begins whenever the maximum shear stress in any element
becomes equal to the maximum shear stress in a tension-test specimen of the same material when that specimen begins
to yield (Shigley & Mischke, 2001).

The maximum-shear-strain-energy hypothesis predicts that failure by yielding occurs when the total strain energy in a
volume reaches or exceeds the strain energy in the same volume corresponding to the yield strength in tension or
compression (Shigley & Mischke, 2001).

The distortion-energy theory predicts that yielding will occur whenever the distortion energy in a unit volume equals the
distortion energy in the same volume when uniaxially stressed to the yield strength.

Distortion-energy theory (DET), also known as the vom Mises criterion, postulates that failure is caused by the elastic
energy associated with shear deformation. This theory is valid for ductile materials and predicts yielding when combined
loading with greater accuracy than any other recognized theory (Hamrock, Jacobson, & Schmid, 1999).

Maximum-Normal-Stress Theory (MNST) states that a part subjected to any combination of loads will fail whenever the
greatest positive principal stress exceeds the tensile yield strength or whenever the greatest negative principal stress
exceeds the compressive yield strength. This theory works best for fibrous brittle materials and some glasses, but works
reasonably well for brittle materials in general and is therefore popular (Hamrock, Jacobson, & Schmid, 1999).

Coulomb-Mohr Theory is a theory of failure prediction identical to internal friction theory.

Internal Friction Theory is a failure prediction theory accounting for difference between compressive and tensile
strengths of brittle materials.

Maximum-Shear-Stress Theory (MSST) is a failure prediction theory stating that yielding occurs when largest shear
stress exceeds critical value.
DESIGN EQUATIONS

1.
The following design equations for shaft design are used for static loadings, based on strength, and are used mostly
of all standard books in machine design.
Shaft Under Pure Torsion Load Only
a)
For Solid Shaft,
b) For Hollow Shaft,
ss 
T c 16 T

J
 D3
ss 
16 T D o
Tc

,
J
 D o4  D i4


ss 
16T
4
 D 1     


3
o
Di
= ratio of the inside diameter to the outside diameter
Do
ss = maximum allowable torsional or shear stress, in psi or Mpa
D = diameter of a solid shaft, in inches, cm or mm.
Where,  
12
RMBC
MACHINE DESIGN: SHAFT DESIGN
Do = outside diameter of a hollow shaft, in inches, cm or mm.
Di = inside diameter of a hollow shaft, in inches, cm or mm.
c = distance of the outermost fiber from the neutral axis, in inches, cm or mm
J = polar moment of inertia, in inches4, cm4 or mm4
J
 D4
(For Solid Shaft)
32
J


16 D o4  D i4
(For Hollow Shaft)
32
T = torque or torsional moment, in in-lb or kN-m
T
63000 Hp
, in-lb  English System
n
T
Hp = transmitted Hp
n = shaft speed, in rpm
30 P
, kN-m  SI Units
n
P = transmitted power, kW
Counter
Shaft
Figure 1. Machine Shafting
Prime
Mover
Machin
e
Shaft
Main Shaft
Or Line
Shaft
Machi
ne
Shaft
Driven
Machin
e
2.
Driven
Machin
e
Shaft Under Pure Bending Load Only
a)
For Solid Shaft,
b) For Hollow Shaft,
s
M c 32 M

I
 D3
s
32 M D o
Mc
,

I
 D o4  D i4


ss 
16T
4
 D3o 1     


Di
= ratio of the inside diameter to the outside diameter
Do
s = maximum allowable bending stress or design bending stress, in psi or Mpa
M = bending moment, in in-lb or kN-m
I = moment of inertia, in inches4, cm4 or mm4
Where,  
I
3.
 D4
(For Solid Shaft)
64
I

Shaft Under Combined Torsion and Bending Loads Only
a)
Based on the Maximum Shear Stress Theory (MSST)
a.1. For Solid Shaft,
a.2. For Hollow Shaft,
 16 
16TE
ss(max)  
M2  T2 
3 
 D3
 D 
16Do
16Do TE
ss(max) 
T2  M2 
4
4
  Do  Di 
  Do4  D14 
ss(max) 
Where,  
16TE
4
 D3o 1    


Di
= ratio of the inside diameter to the outside diameter
Do
TE = Equivalent twisting moment, TE 
M2  T2
ss(max) = maximum shear stress, in psi or Mpaa
13

 4
D o  D i4 (For Hollow Shaft)
64
RMBC
MACHINE DESIGN: SHAFT DESIGN
s s (max) 
s s (max)
sy

s ys
(Based on Yield Strength)
2N N
s
s
 n  ns (Based on Endurance Strength)
2N N
N = Factor of Safety
b) Based on Octahedral Shear Stress Theory or Maximum Distortion-Energy Theory
e 
Where,  e 
e 
c)
sy
N
32
 D3
M2 
3 2
T
4
 equivalent octahedral shear stress, based on yield strength
sn
 equivalent octahedral shear stress, based on endurance strength
N
Based on Maximum Normal Stress Theory (MNST)
 16 
max  
M
3 
 D 
c.2. For hollow shaft,
smax 
ss(max)
Where,  


c.1. For solid shaft,
16
M2  T2 
M 
D  

 D3o  D3o  i  

 Do  

32ME

4
 D3o 1    


4
32M E
 D3
M2  T 2

16
  D 4 
 D3o 1   i  
  Do  


M 
M2  T 2

Di
= ratio of the inside diameter to the outside diameter
Do
ME = Equivalent bending moment, ME 
M  TE
2
smax  maximum normal stress, caused by induced stresses, in psi or Mpa
smax 
4.
sy
or max 
N
sn
N
Shaft Under Combined Bending and Axial Loads
a.
Maximum Normal Stress (Solid Shaft),
s max 
If  = 1,
b.
Maximum Normal Stress (Hollow Shaft),
s max 
If  = 1, s max 
32

 D 1 
3
o
4
o
32

 D3o 1  o4
s max 
DP 

M  8 


32
 D3
32 
DP 
M
3 
8 
D 
Let,  
Di
Do



Do P 1  2
M  
8




Do P 1  2
M 
8



 


 


Where,  = ratio of the maximum intensity of stress resulting from the axial load to the average axial stress
 = 1, if the shaft is subject to tensile load, say, a propeller shaft of an airplane engine
P = axial load, tensile or compressive
Straight-Line Formula,
L
 115 ,
k

14
1
L
1  0.0044  
k
RMBC
MACHINE DESIGN: SHAFT DESIGN
Euler’s Equation,
L
 115 ,
k
L
2
n  E  k 
sy

L
 slenderness ratio of the shaft
k
I
k = radius of gyration, in. =
A
A = cross sectional area of the shaft, in.2
E = modulus of elasticity of the shaft material, psi
n = 1, for hinged ends
n = 1.6, for ends partly restrained, as in bearings.
5.
I = rectangular moment of inertia of the shaft, in.4
sy = yield stress of the shaft in compression, psi
n = constant for the type pf column end support
n = 2.25, for fixed ends
ss(max) 
Maximum Shear Stress Equation for Solid Shaft,
If  = 1, ss(max) 
If  = 1, ss max  
2
DP 

2
M  8  T


16
 D3
2
DP 

2
M  8  T


16
 D3
ss max  
b) Maximum Shear Stress for Hollow Shaft,
16

 D3o 1  4
16


 D3o 1  4
2
Do P

2 
2
M   8 1     T




2
Do P

2 
2
M  8 1     T





Shaft Under Torsion and Axial Loads (Based on MSST)
a)
ss(max) 
Maximum Shear Stress for Solid Shaft,
If  = 1, ss(max) 
ss(max) 
If  = 1, ss(max) 
16

 D3o 1  4
 DP 
T2   

 8 
16
 D3
 DP 
T2  

 8 
16
 D3
b) Maximum Shear Stress for hollow Shaft,
7.
L = length of the shaft, inches
Shaft Under Combined Torsion, Bending, and Axial Loads [Based on the Maximum Shear Stress Theory (MSST)]
a)
6.
2
2
2
16

 D3o 1  4

2
 Do P
2 
2
 8 1     T




2
 Do P
2 
2
 8 1    T





Shaft Under Torsion and Bending Loads, Considering Shock and Fatigue Factors (ASME Code)

The following equations are the same equations introduced above but with the consideration of the combined shock
and endurance (or fatigue) factors for both torsion and bending. These equations are in accordance with the
American Society of Mechanical Engineers (ASME) Standards on shafting design.
a)
Shaft Under Torsion Only
a.1. For Solid Shaft,
ss 
T c 16 k t T

J
 D3
a.2. For Hollow shaft,
ss 
16 k t T Do
16k t T
Tc


4
4
J
 D o  Di
 D3o 1  4



b) Shaft Under Bending Only
b.1. For Solid Shaft,
s
M c 32 k m M

I
 D3
15

RMBC
MACHINE DESIGN: SHAFT DESIGN
s
b.2. For Hollow Shaft,
c)
16k m M
M c 32 k m M D o


4
I
 Do4  Di4
 D3o 1     




Shaft Under Combined Bending and Torsion
c.1. For Solid Shaft
s s (max) 
s(max) 
k m M2  k t T2
16
 D3
k M 
 m
16
 D3
kmM  ktT
2
2


c.2. For Hollow Shaft
ss(max) 
smax) 
16
 D 1  4 
 kmM
3
o
16

D 1 
3
o
4

k M 
 m
 ktT
2
k mM
2
2
2
  k t T  

d) Shaft Under Combined Bending and Axial Loads
d.1. For Solid Shaft, s max 
32
 D3
DP 

 kmM   8 


If  = 1, s max 
DP 

M  8 


32
 D3
d.2. For Hollow Shaft
s max 
e)
32

 D3o 1  o4



Do P 1  2
k m M  
8


 
If  = 1, s max 



 D3o 1  o4



Do P 1  2
k m M 
8


Shaft Under Torsion and Axial Loads
e.1. For Solid Shaft, ss(max)
16

 D3
e.2. For Hollow Shaft, ss(max) 
ktT
2
16

 D3o 1  4
 DP 
 

 8 

2
If  = 1, ss(max)
16

 D3
ktT
2
 DP 


 8 
2
2
2
 Do P
2 
 8 1      k t T 



If  = 1, ss(max) 
f)
32

16

 D3o 1  4

2
2
 Do P
2 
 8 1    ktT




Shaft Under Bending, Torsion, and Axial Loads
f.1. For Solid Shaft, ss(max)
16

 D3
2
DP 
2

 kmM   8   ktT


If  = 1, ss(max)
f.2. For Hollow Shaft,
ss max  
If  = 1, ss max  
16

 D3o 1  4
16

 D3o 1  4
16

 D3
2
DP 
2

 kmM  8   ktT


2

Do P
2

2 
k m M   8 1      k t T 



Do P
2

2 
k m M  8 1      k t T 






2
Where, km = combined shock and endurance factor for bending
kt = combined shock and endurance factor for torsion
ss(max) = maximum shear stress caused by induced stresses, psi or Mpa
smax = maximum tensile or compressive stress caused by induced stresses, in psi or Mpa
16
 


RMBC
MACHINE DESIGN: SHAFT DESIGN
Table 1. Shock and Fatigue Factors (Black & Adams, p271)
Nature of Loading
Stationary Shafts
Load Gradually Applied
Load Suddenly Applied
Rotating Shafts
Load Gradually Applied
Load Suddenly Applied, minor shock
Load Suddenly Applied, Heavy Shock
8.
km
kt
1.0
1.5 to 2.0
1.0
1.5 to 2.0
1.5
1.5 to 2.0
2.0 to 3.0
1.0
1.0 to 1.5
1.5 to 3.0
1993 PSME Code Equations (p18) for Shaft with Assumed Allowable Stresses
a)
Main Transmitting shaft with Assumed Stress of 4000 psi
D3 N
P
80
1
and
 80 P  3
D

 N 
Where, P = transmitted power, in Hp
D = shaft diameter, inches
N = shaft rpm
b) Line shaft Carrying Pulleys with Assumed Stress of 6000 psi
D3 N
P
53.5
1
and
 53.5 P  3
D

 N 
Where, P = transmitted power, in Hp
D = shaft diameter, inches
c)
N = shaft rpm
Small Shaft, Short Shaft, and Counter Shaft with Assumed Stress of 8500 psi
D3 N
P
38
1
and
 38 P  3
D

 N 
Where, P = transmitted power, in Hp
D = shaft diameter, inches
N = shaft rpm
COMMON SHAFT SIZES

The common available sizes for steel circular shaft are listed in Table 1.2 and Table 1.3 below for English unit and
metric unit, respectively.
Table 2. Common Shaft Sizes, in inches, for Circular Shaft in English Unit
(Source: Machine Design by Robert H. Creamer, p159)
½
1-1/16
1-5/8
2-3/16
2-15/16
3-7/8
5-1/4
9/16
1-1/8
1-11/16
2-1/4
3
3-15/16
5-7/16
5/8
1-3/16
1-3/4
2-5/16
3-1/8
4
5-1/2
11/16
1-1/4
1-13/16
2-3/8
3-1/4
4-1/4
5-3/4
¾
1-5/16
1-7/8
2-7/16
3-3/8
4-7/16
5-15/16
13/16
1-3/8
1-15/16
2-1/2
3-7/16
4-1/2
6
7/8
1-7/16
2
2-5/8
3-1/2
4-3/4
15/16
1-1/2
2-1/16
2-3/4
3-5/8
4-15/16
1
1-9/16
1-1/8
2-7/8
3-3/4
5
Table 1.3. Typical Metric Shaft Diameters, mm, (Source: Machine Design by Robert H. Creamer, p160)
4
12
40
75
110
180
5
15
45
80
120
190
6
17
50
85
130
200
7
20
55
90
140
220
8
25
60
95
150
240
9
30
65
100
160
260
10
35
70
105
170
280
DESIGN EQUATIONS BASED ON DEFLECTION AND RIGIDITY


Another ways of determining the sizes of shafts are the torsional and lateral rigidity or deflections.
Torsional rigidity or deflection of transmission shafts, as a rule of thumb, should be limited to 1 o in 20 diameters.
17
RMBC
MACHINE DESIGN: SHAFT DESIGN

The lateral deflection caused by bending should not to exceed 0.01 inch per foot of length.

Permissible angle of twist: For drive shaft of machine tools,

 0.75 o to 1.0 o per ft.
L
1.

 0.08 o per ft ; and for line shafts,
L
Torsional Deflection or Angle of Twist
o
The total torsional deflection or angle of twist, in radians, for both solid and hollow circular shaft could be
determined by the used of the following equations.
1
a.
For Solid Shaft,
 32T L  4
T L 32T L


and D  

4
JG D G
 G  
b.
For Hollow shaft,

TL
32 T L

J G  D o4  D i4 G


Where,  = angle of twist or angular or torsional deflection, radian
D = shaft diameter, inches
Do = outside diameter of hollow shaft, inches
Di = inside diameter of hallow shaft, inches
G = shear modulus of elasticity or modulus of rigidity in shear, psi or Mpa
G = 12 x 106 psi (for steel)
L = length of shaft, m or ft.
2.
Empirical Equations from Machinery’s Handbook, 26 th Edition, page 279
a)
For allowable twist not exceeding 0.08 degree per foot length.
1
D  0.29  T 
1
4
or
 Hp  4
D  4.6 

 N 
Where, D = shaft diameter, inches
Hp = transmitted Horsepower
T = torque, in-lb
N = shaft speed, rpm
S.I Units (allowable twist of 0.26 degree per meter length)
1
D  2.26  T  4
P
or D  125.7  
N
0.25
Where, D = shaft diameter, mm
P = transmitted power, kW
T = torque, N-mm
N = shaft speed, rpm
b) For Allowable twist not exceeding 1 degree per 20D length
1
D  0.1 T 
1
3
 Hp  3
or D  4.0 

 N
Where, D = shaft diameter, inches
Hp = transmitted Hp
c)
T = torque, in-lb
N = shaft speed, rpm
For short, solid shaft, subjected only to heavy transverse shear
D
1.7 V
ss
Where, D = shaft diameter, inches
V = maximum transverse shearing loads, lb
MATERIALS USED FOR SHAFTING
1.
For the service requirements are too severe
 Hot-rolled, plain-carbon steel, the least expensive material
 For maximum machineability, a normalizing or annealing treatment may be necessary to improve the grain structure
and to secure uniformity.
18
RMBC




MACHINE DESIGN: SHAFT DESIGN
Cold-drawn bars have a smooth, bright finish and have diameters held to tolerances of a few thousandths of an inch;
it is sometimes erroneously called cold-rolled shafting; available in both plain-carbon and alloy compositions;
widely used in the field of general power transmission.
Cold drawing improves the physical properties; it raises the values for tensile strength and the yield point.
For greater strength needed, use a steel of somewhat high-carbon content.
For forged shaft, such as used in internal combustion engine and railroad cars, the carbon content is usually 0.45 or
0.50 %. The widely used steel for such service is plain carbon steel 1045.
2.
For the service requirements are more severe, or certain desirable physical properties are to be obtained
 Alloy steel can be used.
 General-purpose alloy steels: chromium-molybdenum steel 4140, and chromium-nickel-molybdenum 4340.
 When heat-treated to high strength and hardness, alloy steels are tougher, more ductile, and better adopted to shock
and impact loads.
3.
For the service requirements demand resistance to wear rather than extreme strength.
 It is customary to harden only the surface of a shaft.
 The case hardening or carburizing process is in wide use.
4.
ASME Code Recommendation for Commercial Shafting (without any definite specifications for physical and chemical
properties of material)
 Maximum working shear stress, 8000 psi
 Range of Ultimate strength: 4500 to 70000 psi
 Elastic limits: 22500 to 55000 psi
 When there is keyway at the section for which the stress calculations are made, the working stress is to be reduced to
75 % of the value for a solid circular shaft.
Important Notes for Commercial Shafting

Transmission shafts are commonly made of cold-rolled stock.

Hot-rolled stock is used for diameters over 3 inches.

Forged stock is generally used for diameters over 6 inches.

Cold-rolled shafting is stronger than hot-rolled of the same analysis, but with the same resistance to deformation.

Sizing for cold-rolled shafting:
 Increment of 1/16” from ½” to 2-1/2”
 Increment of 1/8” from 2-1/2” to 4”
 Increment of ¼” from 4” to 6”


Standard sizing for transmission shaft
 Increment of ¼” from 5/16” to 2-7/16”
 Increment of ½” from 2-7/16” to 5-15/16”
Standard length for transmission shaft: 16, 20, and 24 ft.
PRACTICE PROBLEMS (Module 2)
1.
A shaft carries a steady torque of 30 000 in-lb at a shearing stress of 8 000 psi. What is the diameter of the shaft? [Ans.
2.67 inches]
2.
Compute the diameter of a solid shaft that rotates 100 rpm and transmits 1.2 Hp. The design stress for shear is to be 6
000 psi and the shaft is subjected to torsion only. [Ans. 0.863 inch or 7/8 inch]
3.
A solid steel torsion shaft needed with a capacity of 2.5 kW when rotating at 1200 rpm. Find the minimum diameter if
the allowable shear stress is 40 MPa. [Ans. 13.6 mm]
4.
A 2-inch line shaft transmits 20 Hp and rotates at 200 rpm. Two pulleys spaced 30 inches apart cause a torsional
deflection. If the shear modulus of elasticity is 12 000 000 psi, find the angle of twist in degrees. [ Ans. 0.573 o]
5.
Find the diameter of a shaft for which torsional deflection will not exceed one degree in a length of 20 diameters. The
shaft is to transmit 40 Hp at 200 rpm. Assume that the shear modulus of elasticity is 12000000 psi. [Ans. 2.31 inches]
6.
The torsional deformation of a steel shaft is to be 1 o in a length of 600 mm when the shearing stress is equal to 69 MPa.
Find the diameter of the shaft. Consider a modulus of elasticity of 79 300 MPa. [Ans. D = 59.84 mm]
7.
A 1-inch sold shaft is to be replaced with a hollow shaft of equal torsional strength having an outside diameter of 2
inches. Find the inside diameter and the percentage of weight saved. [Ans. 1.93 inches, 72.5 %]
19
RMBC
MACHINE DESIGN: SHAFT DESIGN
8.
A solid transmission shaft is 88.9 mm in diameter. It is desired to replace it with a hollow shaft of the same material and
same torsional strength but its weight should only be half as much as the solid shaft. Find the outside diameter and inside
diameter of the hollow shaft. [Ans.: 107.31 mm, 86.97 mm]
9.
A shaft transmits 10 Hp at 500 rpm. Assume that the design stresses are 6000 psi (shear) and 8000 psi (tension) and the
shaft is subject to a bending moment of 3000 in-lb. Compute the diameter of the shaft. [Ans. Based on shear, D = 1.40
inches; Based on flexure or bending, D = 1.59 inches]
10. A centrifugal pump is directly coupled to a motor. The pump rating is 3600 lpm against a total head of 8 m of water.
Pump efficiency is 65 % at shaft speed of 550 rpm. Calculate the torsional stress induced on the 40-mm diameter motor
shaft. [Ans: 10 003 kPaa]
11. A line shaft is to transmit 200 Hp at 900 rpm. a) Find the diameter of the shaft. b) If the line shaft is connected to a 2”
counter shaft with speed of 1600 rpm, find the Hp transmitted. [Ans: a) 2.282”, say 2 7/16”; b) 337 Hp]
12. A steel shaft is subjected to a constant torque of 2260 kN-m. The ultimate strength and yield strength of the shafting
material are 668 MPa and 400 MPa, respectively. Assume a factor of safety of 2 based on yield and endurance strength
in shear. Determine the diameter of the shaft. [Ans. 1.915”]
13. A 2-inch diameter solid shaft has a maximum bending moment of 6000 in-lb and an applied torque of 3000 in-lb.
Compute the equivalent bending moment. [Ans. 6354.1 in-lb]
14. A 48 inches diameter saw blade is mounted on a pulley driven steel shaft, requiring a blade peripheral linear speed of
150 fps. Motor drive is 25 Hp and 1200 rpm, with 6 inches diameter pulley. Determine a) the shaft rpm to attain blade
peripheral speed required; b) the shaft diameter; c) the shaft pulley diameter. [Ans: a) 716.2 rpm, b) D = 2.106”, and c)
10.053”]
15. A hollow shaft has an inner diameter of 0.035 m and an outer diameter of 0.06 m. Compute for the torque if the shear
stress is not to exceed 120 Gpa, in kN-m. [Ans. 4500.09 kN-m]
16. Design the size of solid steel shaft to be used for a 500 Hp, 250 rpm application if the allowable torsional deflection is 1o
per foot and the allowable stress is 10 000 psi and modulus of rigidity is 13 x 10 6 psi. [Ans. 4.004 inches]
17. How fast is the gear running in a shaft 57.15 mm in diameter receiving from an engine of 400 kW? [Ans. 3 765.9 rpm]
18. A transmission shaft sustains 7 000 psi of stress. If the diameter of shaft is 1.43375 inches. What torque can safely be
transmitted? [Ans. 4 051 in-lb]
19. What factor of safety is needed for a 1.998-inch diameter shaft with an ultimate strength of 50 000 psi to transmit 40000
in-lb of torque. [Ans. 1.96]
20. A 7.5 MW gas turbine running at 4500 rpm delivers power to a municipality. What is the diameter of the transmission
shaft? [Ans. 143 mm]
21. What is the maximum torque that can be applied to a hollow circular steel shaft of 120 mm outside diameter and 80 mm
inside diameter without exceeding a shearing stress of 80 Mpa? [Ans. 21.78 kN-m]
22. What is the polar section modulus of a 4-inch solid shaft?
a) 25.13 in3
b) 12.57 in4
c) 12.57 in3
d) 25.13 in4
23. A hollow shaft carries a torque 3.4 kN-m at a shearing stress of 55 MPa. The outside diameter is 1.25 times that of the
inside diameter. Find the inside diameter, in mm.
a) 64.87
b) 46.87
c) 84.67
d) 74.64
24. A round steel shaft transmits 0.75 Hp at 1750 rpm. The shaft is subjected to torsion only and the design stress is 7000
psi. Determine the diameter.
a) 0.27 inch
b) 0.37 inch
c) 0.47 inch
d) 0.57 inch
25. A round steel shaft rotates at 200 rpm and is subjected to a torque of 275 N-m and a bending moment of 415 N-m.
Determine the equivalent twisting moment and the equivalent bending moment.
a) 597.84 N-m; 546.42 N-m
b) 456.42 N-m; 497.85 N-m
c) 546.42 N-m; 597.84 N-m
d) 497.85 N-m; 456.42 N-m
26. A shaft supported on bearings 200 mm apart transmits 187 kW at 200 rpm. The maximum bending moment is 2712 Nm. The allowable shearing stress is 53.3 MPa and the allowable bending stress is 46.7 MPa because of the unusual
loading. Find the shaft diameter.
a) 96.25 mm
b) 109.5 mm
c) 102.9 mm
d) 100.2 mm
27. A ship’s propeller shaft is 5 inches in diameter. The thrust load is 12000 lb and the torque is 150000 in-lb. Find the
resultant maximum shearing and compressive stresses.
a) 6434.76 psi and 6119.18 psi
b) 6244.76 psi and 6191.18 psi
c) 6119.18 psi and 6424.76 psi
d) 6191.18 psi and 6244.76 psi
20