EMP 205 CF

Machine Design - 1
By
Prof Dr Eng \ Abdelhay M. Abdelhay

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 Chapter 2
Design
Of
Shafts
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What is a Shaft ?

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 A shaft is a rotating member
which transmits power
( i.e. Torques)
from one point to another
Through
gears, pulleys,
cams and clutches
and
is usually supported and mounted in
bearings. 4
Examples of Shafts

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Examples of Shafts

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 Examples of Shafts
(Through Belts,pulleys, gears)

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Different Types of Shafts

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 Shafts are Cylindrical in
shape.
Sometimes, however, these
may be Square in section or
Hollow cylindrical.

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 A Spindle is a short revolving
shaft which imparts motion to some
part of a machine.
Example: Drill press spindle and
lathe spindles

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 An Axle is a member resembling
shaft but it is short in length
and it does not necessarily rotate.

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Manufacturing of Shafts

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(a) Shafts up to 75 mm diameter are normal
made from cold rolled stock.

(b) For diameter range of 75 to 150 mm,

Hot Rolling is adopted.

(c) Higher than 150 mm, are manufactured

by Forging then Machined to size.

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Standard Size of Shafts

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 Shafts are Cylindrical in
shape.
Sometimes, however, these
may be Square in section or
Hollow cylindrical.

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Standard Size of Shafts

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The maximum length of a single shaft will
not exceed 7 meters because of the
difficulty of Transportation and handling.

Any desired length can be erected by joining

with detachable fastenings like coupling etc.

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 The most common SIZES
of transmission shaftings
normally manufactured and
stocked Are Listed in
Table 2-1

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 Design of
Shafts
Determine the Shape (solid Vs Hollow),Size,
Material, Load carrying Capacity, and FS

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 Shaft Loads
•Shafts subjected to twisting moment or torque (T) only
•Shafts subjected to bending moment (M) only
•Shafts subjected to combined twisting and bending moments (T and M),
•Shafts subjected to axial loads (Fa) and combined (T and M).

the nature of these loads

•A Steady load,
or
•A Fully reversing load. 20
Shaft Design

There are Different Methods for Shaft

Design:

Design for Strength

Design for Rigidity
Design for Vibration and critical Speed

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Only, Shaft Design
For
Strength is considered

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(a) Shaft’s Force
and Stress
Analyses

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 Shaft Loads
 Torsional Loads
&
 Stresses

T r

 max 
T r 16 T

J
J  d3
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 How the Torque T
is defined ?

From the Torque Diagram, Very Simple !

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 Shaft Loads
 Bending Moment
Diagrams
&
 Stresses

M r

 max 
M r 32 M
 I
I  d3
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How the Bending Moment M
is defined ?

From the BM Diagram, Very Simple !

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Be careful
If Transverse loads
are acting in
DIFFERENT Planes 

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Resolve Acting Forces in TWO
Perpendicular Planes
( Say, Vertical Plane and Horizontal
Plane)

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Determine their Resultant at
different critical Locations by:

M  M M 2
xy
2
xz

Determine: Mmax
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 (b) Shaft’s Failure
Theories
•Maximum Normal Stress Theory ( Rankine’s theory).

•Maximum shear stress theory (Guest’s or Tresca’s theory).

•Maximum Distortion Energy Theory ( Hencky and Von Mises theory).

•The ASME code for Shaft Design.

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It is Recommended to use the
following guidelines for
allowable Strengths:

These Values are Reduced by 25% if Keyway is present

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Example:
Say, A selected Shaft’s material is
having:
Sy=400 MPa and Su=780 MPa

Sall = max ( 0.6 x 400 Or 0.36 x 780)

= max (240 or 281 MPa)
= 281 MPa
Ss all = 120.5 MPa

if a Key is attached to the shaft

Sall = 211 MPa and Ssall = 105 MPa 33
 (1) Shafts Under
Torsional Load ( only)

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The principal induced is Normal stress,
which is given - as before as,

 Or for Solid Shaft:

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(2) Shafts Under
Bending Load ( only)

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The principal induced is shear stress,
which is given - as before as,

T .rmax
 max  
J all
 Or for Solid Shaft:

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 (3) Shafts Under
Combined Bending (M)
and Torsional (T) Loads

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How to Combine M and T
under Steady Loads ???

Different Failure Theories

Is going to tell us How …

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(i) Maximum Normal stress
theory (or Rankine theory)

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 M .ymax
 max  e S
I all
Me ( the Equivalent Bending moment)
is calculated as:
1 2
M  ( M  M  T2 )
e 2
Or

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(ii) Maximum SHEAR stress
theory (or Guest’s theory),

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 T . rmax
 max  e S
J s all

Te ( the Equivalent Torque) is

calculated as:
2 2
T  M T
e
Or

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(iii) Maximum Distorsion
Energy theory
(or Hencky and Von Mises theory)

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Me ( the Equivalent Bending moment)
is calculated as:

If σx ≠ 0 and σy = 0
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(iv) The ASME Code for
Transmission Shaft Design
The American Society of Mechanical Engineers has
worked out a special procedure for design of
transmission shafts.
on the assumption that both Torsional and bending
loads are present.
This code is based on the maximum shear stress
theory of failure-

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 T . rmax
 max  e S
J s all
Te ( the Equivalent Torque) is taking into
account The Shock and Fatigue
factors Km, and Kt respectively,
(See Table 2-2), Te calculated as:
2
T  ( Km . M )  ( Kt . T ) 2
e
Or
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For commercial shafting,
the ASME – code
ALSO, recommend the following
typical values for
design working stresses:

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Shaft Design Procedure

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Examples of Shafts
Shaft shown drives a gear set that is
transmitting 4.8 KW at 1750 rpm.
• Shaft is supported in self-aligning ball
bearings and gears of 20° spur gears.
Determine the torsional and bending
stresses in shaft, then Determine the Shaft
size.
d

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840
Let Learning
Continue

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