1. By
ANIMESH BHATTACHARYA
CLASS ROLL: Dx- 03 EXAM. ROLL: 111210003
Semester: 6th Dept.: Mechanical Engineering

2.  What is a Transmission Shaft?
- It refers to a rotating machine element,
circular in cross-section, which supports
transmission elements like gears, pulleys and
sprockets and transmits power.
- In the industries, usually ‘stepped shaft’ is
used.

3.  Stepped Shaft:

4.  Characteristics of Stepped Shaft:
 A shaft is always stepped with maximum
diameter in the middle and minimum
diameter at the two ends, where bearings are
mounted.
 The steps on the shaft provide shoulders for
positioning transmission elements like gears,
pulleys and bearings.

5.  Characteristics of Stepped Shaft:
 The rounded-off portion between two cross-
sections of different diameters is called fillet.
The fillet radius is provided to reduce the
effect of stress-concentration due to abrupt
change in the cross-section.

6.  Material of Shaft :
 Ordinary transmission shafts are made of
medium carbon steels with a carbon content
from 0.15 to 0.40% such as 30C8 or 40C8.
These steels are commonly called machinery
steels.
 For the requirement of higher strength, high
carbon steels such as 45C8 or 50C8 or alloy
steels are employed.

7.  Material of Shaft:
 Common grades of alloy steels used for
making transmission shafts are 16Mn5Cr4,
40Cr4Mo2, 16Ni5Cr2, 35Ni5Cr2 etc.
 Commercial shafts are made of low carbon
steels. They are produced by hot-rolling and
finished to size either by cold-drawing or by
turning and grinding.

8.  Shaft Design on Strength Basis:
 When the shaft is subjected to axial tensile
force, the tensile stress is given by,
σ 𝑡 =
𝑃
𝜋𝑑2
4
; or, σ 𝑡=
4𝑃
𝜋𝑑2 …………….(1)
 When the shaft is subjected to pure bending
moment, the bending stress is given by,
𝜎 𝑏=
𝑀 𝑏 𝑦
𝐼
=
𝑀 𝑏
𝑑
2
𝜋𝑑4
64
; or, 𝜎 𝑏 =
32𝑀 𝑏
𝜋𝑑3 …..(2)

9.  Shaft Design on Strength Basis:
 When the shaft is subjected to pure torsional
moment, the torsional shear stress is given
by, 𝜏 =
𝑀𝑡 𝑟
𝐽
=
𝑀𝑡
𝑑
2
𝜋𝑑4
32
; or, 𝜏 =
16𝑀𝑡
𝜋𝑑3 ……(3)
 When the shaft is subjected to combination of
loads, the principal stress and principal shear
stress are obtained by constructing Mohr’s
circle.

10.  Shaft Design on Strength Basis:
 Mohr’s Circle diagram –

11.  Shaft Design on Strength Basis:
 Let the normal stress be 𝜎 𝑥 , while the shear
stress is denoted by 𝜏 .
 So when both are there, from Mohr’s circle
principal stress 𝜎1 is given by,
𝜎1 =
𝜎 𝑥
2
+
𝜎 𝑥
2
2
+ 𝜏 2 ……….(4)
 The maximum shear stress 𝜏 𝑚𝑎𝑥 is given by,
𝜏max=
𝜎 𝑥
2
2
+ 𝜏 2 ……………..(5)

12.  Shaft Design on Strength Basis:
 Equations (1) to (5) are fundamental
equations for design of shafts.
 A shaft can be designed on the basis of
maximum principal stress theory or
maximum shear stress theory. We shall apply
these theories to transmission shafts
subjected to combined bending and torsional
moments.

13.  Maximum Principal Stress Theory:
 The theory states that the failure of the
mechanical component subjected to bi-axial
or tri-axial stresses occurs when the
maximum principal stress reaches the yield
or ultimate strength of the material. This is
called ‘Rankine’s Criterion’.
 The maximum principal stress is 𝜎1. We have,
𝜎𝑥 = 𝜎 𝑏 =
32𝑀 𝑏
𝜋𝑑3 ; 𝜏 =
16𝑀𝑡
𝜋𝑑3

14.  Maximum Principal Stress Theory:
 Now from equation (4), we get the following,
𝜎1 =
16𝑀 𝑏
𝜋𝑑3 +
16𝑀 𝑏
𝜋𝑑3
2
+
16𝑀𝑡
𝜋𝑑3
2
;
Or, 𝜎1=
16
𝜋𝑑3 𝑀 𝑏 + 𝑀 𝑏
2 + 𝑀𝑡
2 ………... (6)
 The permissible value of maximum principal
stress is given by, 𝜎1 =
𝑆 𝑦𝑡
𝑓𝑠
………………... (7)
 fs= factor of safety; 𝑆 𝑦𝑡= yield strength.

15.  Equations (6) and (7) are used to determine
the shaft diameter on the basis of principal
stress theory.
 Experimental investigations suggest that
maximum principal stress theory gives good
predictions for brittle materials, but shafts
are made of ductile material like steel and
hence, this theory is not applicable to shaft
design. We better use maximum shear stress
theory.

16.  Maximum Shear Stress Theory:
 The theory states that the failure of a
mechanical component subjected to bi-axial
or tri-axial stresses occurs when the
maximum shear stress at any point in the
component becomes equal to the maximum
shear stress in the standard specimen of the
tension test, when yielding starts. This is
called ‘Tresca’s Criterion’.
 The maximum shear stress is 𝜏 𝑚𝑎𝑥.

17.  Maximum Shear Stress Theory:
 In a similar way, as done in case of maximum
principal stress theory, here according to
maximum shear stress theory, we have-
𝜏 𝑚𝑎𝑥 =
16𝑀 𝑏
𝜋𝑑3
2
+
16𝑀𝑡
𝜋𝑑3
2
;
or, 𝜏 𝑚𝑎𝑥=
16
𝜋𝑑3 𝑀 𝑏
2 + 𝑀𝑡
2 …………. (8)

18.  According to maximum shear stress theory,
𝑆𝑠𝑦 = 0.5𝑆 𝑦𝑡 and the permissible value of
maximum shear stress is given by,
𝜏 𝑚𝑎𝑥 =
𝑆 𝑠𝑦
𝑓𝑠
=
0.5𝑆 𝑦𝑡
𝑓𝑠
……………………… (9)
 Equations (8) and (9) are used to determine the
shaft diameter on the basis of maximum shear
stress theory.
 Finally we have the following two equations,
𝜏 𝑚𝑎𝑥=
16
𝜋𝑑3 𝑀 𝑏
2 + 𝑀𝑡
2
𝜎1=
16
𝜋𝑑3 𝑀 𝑏 + 𝑀 𝑏
2 + 𝑀𝑡
2

19.  Equivalent Torsional Moment:
 The expression 𝑀 𝑏
2 + 𝑀𝑡
2 is called
‘equivalent torsional moment’. It is defined as
the torsional moment, which when acting
alone, will produce the same torsional shear
stress in the shaft as under the combined
action of bending moment (𝑀 𝑏) and torsional
moment (𝑀𝑡).

20.  Equivalent Bending Moment:
 The expression 𝑀 𝑏 + 𝑀 𝑏
2 + 𝑀𝑡
2 is
called ‘equivalent bending moment’. It is
defined as the bending moment, which when
acting alone, will produce the same bending
stresses (tensile or compressive) in the shaft
as under the combined action of bending
moment (𝑀 𝑏) and torsional moment (𝑀𝑡).

21.  The concept of equivalent torsional moment
is used in the design of shafts on the basis of
maximum shear stress theory of failure.
 The concept of equivalent bending moment is
used in the design of shafts on the basis of
maximum principal stress theory of failure.

22.  Shaft Design on Torsional Rigidity Basis:
 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.
 Similarly, the transmission shaft is said to be
rigid on the basis of lateral rigidity, if it does not
deflect too much under the action of external
forces and bending moment.
 In some applications, like machine tool spindles,
it is necessary to design the shaft on the basis of
torsional rigidity, i.e., on the basis of permissible
angle of twist per metre length of shaft.

23.  Shaft Design on Torsional Rigidity Basis:
 The angle of twist 𝜃𝑟 (in radians) is given by,
𝜃𝑟 =
𝑀𝑡 𝑙
𝐽𝐺
;
 Converting 𝜃𝑟 from radians to degrees (θ),
θ =
180
𝜋
x
𝑀𝑡 𝑙
𝐽𝐺
;
 For solid circular shaft, 𝐽 =
𝜋𝑑4
32
, we’ve-
θ =
584𝑀𝑡 𝑙
𝐺𝑑4 ……………….……. (10)

24.  Shaft Design on Torsional Rigidity Basis:
 Where,
𝜃 = angle of twist (deg.)
𝑙 = length of shaft subjected to twisting
moment (mm)
𝑀𝑡= torsional moment (N-mm)
G = modulus of rigidity (N/𝑚𝑚2 )
d = shaft diameter (mm)
 Equation (10) is used to design the shaft on the
basis of torsional rigidity. Permissible angle of
twist for machine tool applications is 0.25 𝑜
per
metre length. Modulus of rigidity for steel is
79,300 N/𝑚𝑚2 or 80kN/𝑚𝑚2(approx.).

25.  ASME Code for Shaft Design:
 According to this code, the permissible shear
stress 𝜏 𝑚𝑎𝑥 for the shaft without keyways is
taken as 30% of yield strength in tension or
18% of the ultimate strength of the material,
whichever is minimum. Therefore,
𝜏 𝑚𝑎𝑥 = 0.30𝑆 𝑦𝑡 , or, 𝜏 𝑚𝑎𝑥 = 0.18𝑆 𝑢𝑡(whichever is
minimum).
 If keyways are present, the above values are
to be reduced by 25%.

26.  ASME Code for Shaft Design:
 According to ASME code, the bending and
torsional moments are to be multiplied by
factors 𝑘 𝑏 and 𝑘 𝑡 respectively, to account for
shock and fatigue in operating condition.
 𝑘 𝑏= combined shock and fatigue factor
applied to bending moment.
 𝑘 𝑡= combined shock and fatigue factor
applied to torsional moment.

27.  Table-1: Values of shock and fatigue factors
 So, finally we have the following equations to determine
the shaft diameter to design a solid transmission shaft,
𝜏 𝑚𝑎𝑥=
16
𝜋𝑑3 𝑘 𝑏 𝑀 𝑏
2 + 𝑘 𝑡 𝑀𝑡
2
𝜎1=
16
𝜋𝑑3 𝑘 𝑏 𝑀 𝑏 + 𝑘 𝑏 𝑀 𝑏
2 + 𝑘 𝑡 𝑀𝑡
2
Serial no. Application 𝒌 𝒃 𝒌 𝒕
1. Load gradually applied 1.5 1.0
2. Load suddenly applied (minor shock) 1.5-2.0 1.0-1.5
3. Load suddenly applied (heavy shock) 2.0-3.0 1.5-3.0