 A shaft is a rotating member,
usually of circular cross
section, used to transmit
power or motion. It provides
the axis of rotation, or
oscillation, of elements such
as gears, pulleys, flywheels,
cranks, sprockets, and the
like and controls the
geometry of their motion.
 An axle is a non-rotating
member that carries no
torque and is used to support
rotating wheels, pulleys, and
the like.
 A spindle is a short rotating
shaft.
 It is not necessary to evaluate the stresses in a shaft at
every point; a few potentially critical locations will
suffice.
 Critical locations will usually be on the outer surface,
at axial locations where the bending moment is large,
where the torque is present, and where stress
concentrations exist.
 A free body diagram of the shaft will allow the torque
at any section to be determined.
 The bending moments on a shaft can be determined by
shear and bending moment diagrams.
 Axial stresses on shafts due to the axial components
will almost always be negligibly.
 Shaft Design for Stress due to Static loads
A. Solid Shaft with single load
• Where the shaft is subjected to axial force
 axial
4F

d 2
• Where the shaft is subjected to pure bending moment
 bending 
32 M
d 3
• Where the shaft is subjected to pure torsional moment
16 T
 
d 3
B. Solid Shaft with combined loads
 When the shaft is subjected to combination of axial,
bending and torsional loads, the principles stresses
and principle shear stresses are obtained by
constructing Mohr’s circle as shown in below figure.

 x   axial   bending
x
x 
 1       zx2
2
 2 
2
x 
 max      zx2
 2 
2
1
 max

 zx
x
 When the shaft is subjected to combination of bending
and torsional moments (σaxial = 0) which is the common
case, the principle stresses will be
 x   bending
 bending
1 

2
  bending 

   zx2
 2 
2
  bending 
   zx2
 max  
 2 
2
I. Maximum shear stress theory of static failure (MSS)
  bending 
 16 M   16T 
2
   zx  
 max  
 3 
3 
 d   d 
 2 
2
 max
16
 3
d
2
2
M 2 T2
According to maximum shear stress theory, and using a
permissible value τMax of
 max 
Sy
Sy
2 F .S
16

2 F .S d 3
M 2 T2
 32 F .S
d  
  S yt

M T 

2
1
3
2
II. Distortion energy theory of static failure (DE)

 
 

1
2
2
2



 
 x   y   y   z   z   x  3  xy2   yz2   zx2
2
For plane stress (  z   yz   zx  0 ) and  y  0
    x2  3 xy2
2
 32 M 
 16T 

  
 3 3 
3 
 d 
 d 
2

32
   3 M 2  43 T 2
d
According to Von-Mises theory, and using a permissible
value σ′ of
Sy
 
F .S
Sy
32
2
2
3

M

T
4
F .S
d 3
 32 F .S
d  
  S yt
3 2
M  T 
4

1
3
2
• It is important to observe that these relations are only
valid when the stresses are truly invariable.
 ASME Code for Shaft Design
The ASME code defines a permissible shear stress which
is the smaller of the two following values:
 p  0.3S yt
or
 p  0.8 Sut
The code states that these stresses should be reduced by
25% if stress concentration possible due to shoulder fillet
or a keyway is present. Making τp equal τmax, lead to
16
p  3
d
M 2 T2
In the code, the bending moment M and the torsional
moment T are multiple by combined shock and fatigue
factors Cm and Ct respectively, depending on the condition
of particular application.
16
p  3
d
Cm M 2  CtT 2
When  p  0.3S yt
 16.96
d
 S yt
Cm M 
2
 CtT 
2



1
3
When  p  0.8 Sut
 6.36
d
 Sut
Cm M 
2
 CtT 
2



1
3
• Recommended values of Cm and Ct are listed in Table (1)
Table (1). Values of Cm and Ct for different loading type
Type of Loading
Cm
Ct
Load applied gradually
1
1
Load applied suddenly
1.5 – 2
1.5 – 2
Load applied gradually
1.5
1
Steady load
1.5
1
Load applied suddenly, minor shocks
1.5 – 2
1 – 1.5
Load applied suddenly, heavy shocks
2–3
1.5 – 3
Stationary shaft
Rotating shaft
Table (2). Typical sizes of solid shaft that are available in the market
Diameter
Increments
Up to 25 mm
0.5 mm
25 to 50 mm
1.0 mm
50 to 100 mm
2.0 mm
100 to 200 mm
5.0 mm
Example: the layout of a transmission shaft carrying two
pulleys B and C supported by bearings A and D is shown in
the figure below. Power is supplied to the shaft by means of
vertical belt on the pulley B, which is then transmitted to the
pulley C carrying a horizontal belt. The maximum tension in
the belt on the pulley B is 2.5kN. The angle of warp for both
the pulleys is 180̊ and the coefficient of friction is 0.24. The
shaft is made of plain carbon steel 30C8 Syt = 400 N/mm2
and the factor of safety is 3. Determine the shaft diameter.
600
200
200
All dimension by mm
A
D
500 φ1
B
250 φ2 C
R1
Solution:
R2
P1
2500
 e    P2  0.24  1176.673 N P3
P2
e
TB  ( P1  P2 )( D1 / 2 )
 ( 2500  1176.673 )( 500 / 2 )
TB  330831 .75 N .mm
P4
P1
P2
TB  TC  ( P3  P4 )( D2 / 2 )  ( P3  P4 )125  330831 .75 N .mm
( P3  P4 )  2646 .654 N ........( 1 )
P3  
 e  2 .125................( 2 )
P4
Substituting Eq. (1) into Eq. (2), we get
P3= 4999.235 N and P4= 2352.581 N
• Bending and torsional moments
P3+P4=7351.816 N
200
600
200
P1+P2=3676.673 N
200
C
B
5881.453
1470.363
1176290.6
600
200
C
B
735.335
2941.338
588267.6
294072.6
147067
Horizontal plane (XZ plane)
Vertical plane (XY plane)
M B  294118 2  588232 2  657675.804 N .mm
M C  1176472 2  147058 2  1185448.556 N .mm
1185448.556
330831.75
657675.804
B
C
B
Torsional moment diagram
C
Resultant bending moment diagram
This shaft will be designed based on the stress at point C at
which the bending moment is the maximum.
 32 F .S
d  
  S yt
3 2
M  Tc 
4 
1
3
2
c
 32 * 3

3
2
2

( 1185448.556 )  ( 330831.75 ) 
4
  ( 400 )

 45.3416 mm from table ( 2 ) d  46 mm
1
3
 Shaft Design on Rigidity Basis
The transverse shear V at a section of a beam in flexure
imposes a shearing deflection, which is superposed on the
bending deflection. Usually such shearing deflection is
less than 1 percent of the transverse bending deflection,
and it is seldom evaluated. However, when the shaft
length-to-diameter ratio is less than 10, the shear
component of transverse deflection merits attention.
A transmission shaft is said to be rigid on the basis of
torsional rigidity, if it does not twist too much under the
action of an external torque. The angle of twist or the
angular deflection θ (in radians) for right-circular
cylindrical solid shafts in torsion is given in is given by:
Tl

JG
Where,
G: Modulus of rigidity
 d4
J: Polar moment of inertia ( J 
)
32
By degree (θ̊),
 

180 

Tl
(
)
JG
584Tl
 
Gd 4

For a stepped shaft with individual cylinder length li and
torque Ti , the angular deflection can be estimated from
Ti li
   i  
J i Gi
For a constant torque throughout homogeneous material,
li
T
 
G
Ji
By degree (θ̊),
584T
 
G

li
 d4
i
The permissible angle of twist for machine tool applications
is 0.25̊ per meter length.
Example (2): the layout of a shaft carrying two pulleys 1
and 2, and supported on two bearings A and B is shown
in figure below. The shaft transmits 7.5kw power at
360rpm from the pulley 1 to the pulley 2. The diameter of
pulleys 1 and 2 are 250mm and 500mm respectively. The
masses of pulleys 1 and 2 are 10kg and 30kg respectively.
The belt tensions act vertically downward and the ratio of
belt tensions on the tight side to slack side for each pulley
is 2.5:1. The shaft is made of plain carbon steel 40C8
(Sy=380N/mm2) and the factor of safety is 3. Estimate
suitable diameter of shaft. If the permissible angle of twist
is 0.5̊ per meter length, calculate the shaft diameter on the
basis of torsional rigidity. Assume G=79300N/mm2.
250
500
A
250
B
1
2
Example (3): A line Shaft supporting two pulleys A and B
is shown in figure below. Power is supplied to the shaft
by means of vertical belt on the pulley A, which is then
transmitted to the pulley B carrying a horizontal belt.
The ratio of the belt tension on tight and loose sides is
3:1. The limiting value of tension in the belt is 2.7kN.
The shaft is made of plain steel 40C8 ( Sut=650N/mm2
and Sy=380N/mm2). The pulleys are keyed to the shaft.
Determine the diameter of the shaft according to the
ASME code if, load applied gradually.
450
250 φ
450
A
250
450 φ
B
H.W. Check the designed shaft according to angular
deflection.
Example (4): The layout of an intermediate shaft of a gear
box supporting two spur gears B and C is shown in the
figure below. The shaft is mounted on two bearings A
and D. The pitch circle diameters of gears B and C are
900 and 600mm respectively. The material of shaft is
steel FeE 580 (Sut=770 and Sy=580N/mm2). The factor
Cm and Ct of ASME code are 1.5 and 2 respectively.
Determine the diameter of the shaft using the ASME
code. Assume that the gears are connected to the shaft
by means of keys.
FBr
F  4421N
900
FBr  1609 N
FBt
t
B
900
900
FCt  6631.5 N
FCr  2413.67 N
φB =900 & φC =600mm
A
D
FCr
B
C
FCt