This document discusses the design of shafts that can experience twisting moments, bending moments, or a combination of both. It provides equations to determine the diameter of shafts subjected to twisting moments only based on the torque and material shear stress. Similarly, it gives equations for sizing shafts experiencing bending moments only based on the bending moment and material bending stress. For shafts with combined loads, it describes two failure theories and the resulting equivalent moment equations that can be used for design.

1. DESIGN OF SHAFT
IMSEC GHAZIABADIMSEC GHAZIABAD
PULKIT PATHAKPULKIT PATHAK
44thth
YEARYEAR

2. SHAFT INTRODUCTIONSHAFT INTRODUCTION
THE TERM SHAFT USUALLY REFERS TO A ROTATING MACHINE ELEMENTTHE TERM SHAFT USUALLY REFERS TO A ROTATING MACHINE ELEMENT
CIRCULAR IN CROSS SECTION WHICH SUPPORTS TRANSMISSIONCIRCULAR IN CROSS SECTION WHICH SUPPORTS TRANSMISSION
ELEMENTS LIKE GEARS,PULLEYS ETCELEMENTS LIKE GEARS,PULLEYS ETC

3. DIFFERENCE BETWEEN AXLE AND SHAFTDIFFERENCE BETWEEN AXLE AND SHAFT
Ideally speaking Axles are meant forIdeally speaking Axles are meant for
balancing/transferring Bendingbalancing/transferring Bending
moment and Shafts are meant formoment and Shafts are meant for
Balancing/transferring Torque.Balancing/transferring Torque.

4. The shafts may be designed
on the basis of -
Strength
Stiffness

5. Cases of designing of shafts, on the basis of
Strength
(a) Shafts subjected to Twisting
Moment only.
(b) Shafts subjected to Bending
Moment only.
(c) Shafts subjected to Combined
Twisting and Bending Moment only.

6. SHAFTS SUBJECTED TO
TWISTING MOMENT
ONLY

7. Determination of diameter of Shafts, by
using the Torsion Equation;
T/J = ς/r
Where;
T- Twisting moment acting upon shafts.
J - Polar moment of inertia of shafts.
ς – Torsional shear stress.
r – Distance from neutral axis to
outermost fibre.

8. For round solid Shafts,Polar Moment
of inertia,
J = π/32*d^4
Now Torsion equation is written as,
T / [π/32*d^4] = ς / (d/2)
Thus;
T = π/16* ς * d^3

9. For hollow Shafts, Polar Moment of inertia ,
J = π/32 [ (D^4) – (d^4) ]
Where ,
D and d – outside and inside diameter,
Therefore, torsion equation;
T / π/32[ D^4 – d^4 ] = ς / (D/2)
Thus;
T = π/16* ς * D^3 * (1 – k^4).

10. SHAFTS SUBJECTED TO
BENDING MOMENT ONLY

11. For Bending Moment only, maximum
stress is given by BENDING
EQUATION
M / I = σ / y
Where;
M→ Bending Moment .
I → Moment of inertia of cross sectional
area of shaft.
σ → Bending stress,
y → Distance from neutral axis to outer
most fibre.

12. For round solid shafts, moment of
inertia,
I = π/64* d^4
Therefore Bending equation is;
M / [π/64 * d^4 ] = σ/ (d/2)
Thus;
M = π/32* σ * d^4

13. For hollow Shafts, Moment of Inertia,
I = π / 64 [D^4 – d^4]
Putting the value, in Bending equation;
We have;
M / [π/64 { D^4 (1- k^4) }] = σ / (D/2)
Thus;
M = π/32* σ *D^3*(1 – k^4)
k → D/d

14. SHAFTS SUBJECTED TO
COMBINED TWISTING AND
BENDING MOMENT

15. In this case, the shafts must be designed on the basis of
two moments simultaneously. Various theories have been
suggested to account for the elastic failure of the materials
when they are subjected to various types of combined
stresses. Two theories are as follows -
1.Maximum Shear Stress Theory or Guest’s
theory
2.Maximum Normal Stress Theory or Rankine’s
theory

16. According to maximum Shear Stress
theory, maximum shear stress in shaft;
ςmax = (σ² + 4 ς ²)^0.5
Putting- ς = 16T/πd ³ and σ = 32M/ πd ³
In equation, on solving we get,
π/16* ςmax *d³ = [ M ² + T ² ]^0.5
The expression [ M + T ]^0.5 is known as
equivalent twisting moment
Represented by Te

17. Now according to Maximum normal shear
stress, the maximum normal stress in the
shaft;
σmax = 0.5* σ + 0.5[σ ² + 4ς² ]^0.5
Again put the values of σ, ς in above equation
We get,
σmax* d³ * π/32 = ½ [ M +{ M² + T² }½ ]
The expression ½ [ M +{ M² + T² }½ ] is known as
equivalent bending moment
Represented by Me

18. REFERENCESREFERENCES
V B BHANDARIV B BHANDARI
INTERNETINTERNET

19. THANK YOU