Machine element design is the important common course fir Mechanical Engineering streams.

1. MACHINE ELEMENT TWO
SHAFT DESIGN
Lec.1
By: Design Team
Faculty of Manufacturing
Mechanical Engineering
CHAPTER ONE

2. 1. INTODUCTION
- Shafts are used in all kinds of machinery and
mechanical equipment to transmit power.
- Most shafts are subjected to fluctuating loads of
combined bending and torsion with various degree
of stress concentration.
- Beside the strengthening of the shaft, the operating
speed should not be close to a critical speed.
- For more safety and smoother operating conditions
shafts should be checked against deflection.
2

3. 2. TYPES OF SHAFTS
(A) Transmission shafts:
This type transmit power between the source
(e.g. electric motor) and the machine absorbing
power (e.g. machine tool, pump,...).
3

4. (B) Special machine shafts
have short lengths.
- Shaft with a special construction such as the
crankshafts.
- From the other hand, there are circular machine
elements, which remain stationary (i.e. without
transmitting power) and called axles. Here, the
axles is normally suffering from bending moment
and normal forces without shear stress exceeded
from the twisting moment.
N.B.: Spindles are shafts which transmit power but
4

5. 3. STANDRAD SIZES OF SHAFTS
Shafts can be found as row materials in the following
sizes:
- 25 mm to 60 mm with 5 mm step.
- 60 mm to 110 mm with 10 mm step.
- 110 mm to 140 mm with 15 mm step.
- 140 mm to 500 mm with 20 mm step.
The standard lengths of the row materials are:
5 m , 6 m , 7 m.
5

6. 4. STRESS IN SHAFTS
According to the type of loads acting on the shafts, the
shafts can rotate under the action of one of the
following state of stresses:
1 Shear stress due to the transmitted power (i.e. due
to the twisting torque.).
2 Normal stress due to bending moment which can be
tensile or compressive. This type generated from the
forces acting upon machine elements such as gears
and pulley.
6

7. 3- Normal stress due to normal force which can be
tensile or compressive. This type of stress can be
found from an axial component of an inclined
load. The normal stresses from bending moment
and normal force must be added algebraically.
4- Combined normal and shear stresses.
7

8. 5. DESIGN OF SHAFTS
Design of shafts can be carried out on the basis of:
(1) Strength:
In the design proceeding according to the strength,
the following conditions can be considered:
- Shafts under the action of twisting moment or
torque only.
- Shafts under the action of bending moment only
or/and axial loads.
- Shafts under the action of combined torsion and
bending.
8

9. (2) Rigidity and Stiffness:
Here, the stability of the shafts from the point of view
of deflection (in the case of bending moment) or/and
buckling (in the case of compressive axial force).
F
Deflected Shaft
9

10. 5.1 DESIFN OF SHAFTS UNDER TWISTING
MOMENT
For circular shafts under the action of twisting
moment or torque, the general torsion equation can
T



G
J r L
 D
L

T
Y
Y
Y
Cross-section
R=D/2
X X
be formulated as follows:
Y
Stress Distribution

11

11. T



G
J r L
where:
T :The twisting moment or torque acting on the
shaft in N.mm
J : The polar moment of inertia of the shaft in mm .
 : The shear stress generated in the shaft in N/mm.
r : The radius of the shaft (also the position at which
the shear stress generated) in mm.
i.e. r = d / 2
 : The angle of rotation of the section at which the
shear stress is generated with respect to the
stationary (fixed) section in radian.
…………………………..(1)
12

12. CIRULAR (ROUND) SOLID SHAFTS
J 
d4
32
From general torque equation
Y
Cross-section
R=D/2
X X
The polar moment of inertia can be expressed as
follows:
Y
………..(2)
d
2
d4
32 T


...(3)
13

13. i.e.,:
T 
d3
16

or :
d  3
16T
………..(4)
From equation 4, the shaft diameter (d) can be
determined for solid shafts subjected to a twisting
moment (T).
14

14. For hollow shafts the polar moment of inertia “Ip” can
be determined from the following equation:
CIRULAR (ROUND) HOLLOW SHAFTS
J 

32
0 i
4
( d  d4
) ………..(5)
L
z
z
do
di
x x
y
y
where,
d0 : Outer diameter of the shaft.
di : Inner diameter of the shaft.
15

15. Then, by substituting in equation “1”, the twisting
moment can be expressed as follows:
Assuming that:
k 
d i
d 0
T 

16d 0
( d4
 d4
)
0 i
T

16d0
0
d4
( 1  k4
)
or,
16T
(1 k4
)
do  3 ……..(6)
16

16. The size of a hollow shaft ( i.e., determination of do &
di) subjected to twisting moment can be determined
from the last equation after assuming the diameters
ratio (k).
It must be noticed that:
1. Hollow shafts are usually used in marine work (high
transmitted power).
2. Hollow shafts are more stable against deflection
than solid shafts.
17

17. The twisting moment (T) can be determined from the
relation which connect the power with both of the
twisting moment and the angular velocity as follows:
P = T .  (7)
i.e.,
4500
P 
2NT
T
4500 P
2N
where;
T : Twisting moment in kp.m
N : Speed in r.p.m.
P :Transmitting power in Horse power (hp)
N.B. : 1 kW = 1.34 hp
18

18. 2 500
T 
45006.7
 9.6 kp.m  96 Nm
EXAMPLE:
Shaft transmits:
- A power of 5 kW
- At 500 rpm
P = 5 kw = 5 x 1.34 = 6.7 HP
T
4500 P
2N
19
T= P/2  N =5000/2*3.14*(500/60)
T = 95.65 Nm

19. SUMMARY (SHAFT UNDER SHEAR STRESS)
Solid shaft
16T

d  3
Y
R=D/2
X X
Y
Stress Distribution

Hollow shaft
Y
Cross-section
R=Do /2
X X
Cross-section
Y
Stress Distribution

16T
(1 k4
)
do  3
4500
P 
2NT T
4500 P
2N
22

20. 5.2 SHAFTS UNDER BENDING MOMENT
A
B
C
Gear 1
Gear 2
Gear 3
The shafts; which rotate under the action of pure
bending moment; are subjected to a uni-axial state of
tensile and compressive stress.
This also the case of the axles, which transmit bending
moment only The design of both of them should be
based on the following bending equation:
23

21. where;
M :Maximum bending moment acting on the critical
section in N.mm
I : Second moment of inertia of the cross section
about the axis of rotation in mm4 .
b : Normal stress due to bending moment in kp/cm
y : The distance from the neutral axis to the outer
most fiber on which the stress is maximal in mm.
 b
I

My
Y
Cross-section
R=D/2
X X
Y
Stress Distribution

A
25

22. For solid circular shaft:
64
I 

d4
y 
d
2
&
b
d4
64

Mb (d 2)

32Mb
b
d  3
Therefore, the diameter of solid shafts can be
determined from the following equation:
26

23. For hollow circular shaft:
4
4
o i
(d  d )

64
I  2
y 
do
&
Therefore, the diameter of solid shafts can be determined
from the following equation:
b
o
32Mb
 (1 k4
)
d  3 k 
di
i.e.,k  1
do
64
4 4
o i
Mb(d 2)
b
(d d )

 
6 4
M b (do 2)
b
o
(1  k 4
)
 d 4
 
or,
27

24. 5.3 SHAFTS SUBJECTED TO COMBINED STATE OF STRESS
In most practical cases, the shafts are subjected in
general into a combination of bending and twisting
moments as well as normal force.
The design of these shafts should be based on the
simultaneously effect of all the loading elements.
The theories of elastic failure must be carefully utilized.
According to practical experience, the following two
theories are important in the design of these shafts:
1 Maximum shear stress theory of elastic failure
(Guest`s theory)
2 Maximum normal stress theory
(Rankine`s theory).
28

25. Maximum Shear Theory
According to the maximum shear stress theory of
elastic failure, the maximum shear stress in the shaft
can be expressed as follows:
1 2
 42
b s
s (max) 
2
Where,
s : Shear stress due to twisting moment.
b : Normal stress due to bending moment.
29

26. Replacing the normal and shear stresses with their
equations, which are functions of the shaft diameter
and the loads (i.e. bending and twisting moments), the
following relation can be obtained:
Te is known as the equivalent twisting moment.
Te  M2
 T2

 4 d3 
1 32M
2
16T 
2
d3
(max) 
2
(M2
)  (T2
)
d3

16
30

27. Therefore, the following equation can be used to
evaluate the diameter of the shaft according to the
maximum shear stress theory of elastic failure.
3
16 M2
 T2
d 
max
d  3
16Te
max
31

28. o
16Te
max
 (1 k4
)
d  3
In The case of hollow shafts, the external diameter can
be determined according to the following equation:
32

29. Maximum normal stress theory
According to the maximum normal stress theory of
elastic failure, the maximum normal stress in the shaft
can be expressed as follows:
Where,
s : Shear stress due to twisting moment, N/mm2.
b : Normal stress due to bending moment, N/mm2.
2
s
2
b
yt b
2 2
1 1
 (max)    (  )  
yt  1
Where,
yt : Yield stress of shaft material, N/mm2
1 : Maximum principale stress, N/mm2
33

30. Using the same treatment that has been used in the
maximum shear stress theory, the following equation can
be obtained:
3
yt  (M M2
 T2
)
 d

32
M2
2
 T2

The term 1
M is defined as the equivalent
bending moment, i.e.:
2
2
e
1
2 M
 M T 
M 
34

31. yt
d  3
32Me
In The case of hollow shafts, the external diameter can be
determined according to the following equation:
yt
o
32Me
 (1 k4
)
d  3
35

32. The following table summarize the equations used in design
of shaft using the allowable stress as material property.
max
d  3
16T
32Mb
 b
d  3
Method Solid Shaft Hollow Shaft
Pure Shear Stress
t
(M )
16T
do  3
(1 k4
)
Pure Normal Stress
b
(M )
32Mb
 (1 k4
)
b
do  3
Max. Shear Theory
(Mb &Mt)
3
16 M2
 T2
max
d  3
o
max (1  k4
)
16 M2
 T2
d 
Principle Normal
Stress Theory
(Mb &Mt) yt
32Me

d  3
4
yt
e
32M
 (1 k )
do  3
36

33. SOLVED PROPLEM (1)
SOLUTION
Given
N = 200 r.p.m.
P = 25 hp.
(all ) = 420 k.P/cm2
37

34. I. Determination of thetransmitted torque:
Let T = Torque transmitted by the shaft
T 
4500P
i.e. T=8950 kp.cm
2N
T 
25x4500
 89.5
2x200
T  8950kp.cm
41

35. I. II. Diameter of the shaft
Let d = shaft diameter
i.e. d =47.7 mm
16T
all
d  3
x420
d  3
16x8950
 4.77cm
Therefore,
d = 50 mm
39

36. SOLVED PROPLEM (2)
Find the diameter of a solid steel shaft transmits
25 HP at 200 r.p.m. The ultimate shear stress is
3600 kp/cm2 and a factor of safety as 8 .
If a hollow shaft is assumed to replace it, find the
inside and outside diameters when a ratio of
inside diameter to outside diameter of 0.5 is
assumed.
SOLUTION
Given
P =25 h.p.
N =200 r.p.m.
u =3600 kp/cm2
Horse power transmitted,
Speed of shaft,
Ultimate shear stress,
Factor of safety n = 8
Allowable shear stress, all = 3600/8 =450 kp/cm2
40

37. I. Determination of thetransmitted torque:
Let T = Torque transmitted by the shaft
T 
4500P
i.e. T=8950 kp.cm
2N
T 
25x4500
 89.5
2x200
T  8950kp.cm
41

38. II. Diameter of the shaft
Let d = shaft diameter
d =47.7 mm
d  3
16T
all
x420
d  3
16x8950
 4.77cm
i.e.
Therefore,
d = 50 mm
42

39. III. Diameters of hollow shaft:
Let di = Inside diameter
do = Outside diameter
Therefore,
do = 50 mm
di = 25 mm
3 4
o
e
16T
16
do  3
(1 k4
)
d (1  k )

T 
16x8950
0
d 
d0  4.76  5cm
x4501 0.54

but, k = 0.5
Then,
di = 0.5 x 50 = 25 mm
43

40. 6. SHAFTS SUBJECTED TO FLUCTUATING LOADS
In practices, shafts are subjected to fluctuating
torque and bending moment. In this case, the design
should be based on the consideration of the
combined shock and fatigue factors.
Thus for a shaft Subjected to combined bending and
torsion, the equivalent twisting moment can be
expressed as follows:
Te  (KmM)2
 (Kt T)2
Also, the equivalent bending moment is as follows:
m t
e m
2
M 
1
[K M  (K M)2
 (K T)2
44

41. where;
Km = Combined shock and fatigue factor for
bending moment.
Kt = Combined shock and fatigue factor for
twisting moment.
The recommended values for these factors are
given in the following table:
Nature of Load Km Kt
1. Stationary shafts:
(a)Gradually applied load
(b)Suddenly applied load
2. Rotating shafts:
(a)Gradually applied load
(b)Suddenly applied load with minor shock.
(c)Suddenly applied load with major shock.
1.0
1.5 : 2.0
1.5
1.5 : 2.0
2.0 : 3.0
1.0
1.5 : 2.0
1.5
1.5 : 2.0
2.0 : 3.0
45

42. Solved Examples
A solid circular shaft is subjected to a bending moment of 3000 N-m
and a torque of 10,000 N-m. The shaft is made of 45 C 8 steel having
ultimate tensile stress of 700 MPa and a ultimate shear stress of 500
MPa. Assuming a factor of safety as 6, determine the diameter of the
shaft.
Solutions
Given : M = 3000 N-m = 3 × 106 N-mm ; T = 10,000 N-m = 10 × 106
N-mm ;
σtu = 700 MPa = 700 N/mm2 ; τu = 500 MPa = 500 N/mm2
We know that the allowable tensile stress,

43. We know that the allowable tensile stress,
and allowable shear stress,
Let d = Diameter of the shaft in mm.
According to maximum shear stress theory, equivalent twisting
moment,
We also know that equivalent twisting moment (Te),

45. Example.
A mild steel shaft transmits 20 kW at 200 rpm. It carries a
central load of 900N and is simply supported between the
bearings 2.5 meters apart. Determine the size of the shaft, if
the allowable shear stress is 42 MPa and the maximum
tensile or compressive stress is not to exceed 56 MPa.
What size of the shaft will be required, if it is subjected to
gradually applied loads?
Solution.
Given: P = 20 kW; N = 200rpm.; W = 900 N; L = 2.5 m;
τ = 42 MPa = 42 N/mm2; σb = 56 MPa = 56 N/mm
Size of the shaft
Let d = Diameter of the shaft, in mm.
We know that torque transmitted by the shaft,

49. EN
D 50