Design of Shaft

• A shaft is a rotating member usually of circular cross-section (solid

or hollow), which transmits power and rotational motion.
• Machine elements such as gears, pulleys (sheaves), flywheels,
clutches, and sprockets are mounted on the shaft and are used to
transmit power from the driving device (motor or engine) through a
machine.
• Press fit, keys, dowel, pins and splines are used to attach these
machine elements on the shaft.
• The shaft rotates on rolling contact bearings or bush bearings.
• Various types of retaining rings, thrust bearings, grooves and steps
in the shaft are used to take up axial loads and locate the rotating
elements.
• Couplings are used to transmit power from drive shaft (e.g., motor)
to the driven shaft (e.g. gearbox, wheels).
The connecting shaft is loaded
primarily in torsion.
Combined bending and torsion loads on shaft:
Shaft carrying gears.

From power and rpm find the torque (T), which gives rise to shear stress.
From Torque (T) and diameter (d), find Ft = 2T/d. From Ft and pressure
angles of gears you can find Fr and Fa.
Fr and Ft are orthogonal to each other and are both transverse forces to
the shaft axis, which will give rise to normal bending stress in the shaft.
When shaft rotates, bending stress changes from tensile to compressive
and then compressive to tensile, ie, completely reversing state of stress.
Fa will give rise to normal axial stress in the shaft.
Loads on shaft due to pulleys
Pulley torque (T) = Difference in belt
tensions in the tight (t1) and slack (t2)
sides of a pulley times the radius (r), ie
T = (t1-t2)xr
Left pulley torque
T1 = (7200-2700)x380=1,710,000 N-mm
FV2
Right pulley has exactly equal and
opposite torque:
T2 = (6750-2250)x380=1,710,000 N-mm

Bending forces in vertical (Fv) and horizontal (FH) directions:

At the left pulley: FV1=900N; FH1=7200+2700 = 9900N

At the right pulley: FV2=900+6750+2250=9900N; FH2=0

Torque and Bending moment diagrams for the pulley
system 9900N

T1 T2 FH

1,710,000 N-mm 2,227,500

Torque diag. MH

900N 9900N
From Horizontal forces (FH) and vertical
forces (Fv), Bending moments MH & MV are FV
drawn separately.
Then the resultant moments at various 911,250 MV
points on the shaft can be found from
2,227,500
M R  M H2  M V2
M R  M H2  M V2
The section of shaft where the left pulley
is located has obviously the highest 2,227,500
2,406,685
combination of Torque (1,710,000 N-mm)
Resultant bending moment
and Bending moment (2,406,685 N-mm)
 Power, toque & speed
For linear motion:
Power = F.v (force x velocity)
For rotational motion
Power P = Torque x angular velocity
= T (in-lb).w (rad/sec) in-lb/sec
= T.(2 p n/60) in-lb/sec [n=rpm]
= T.(2 p n/(60*12*550)) HP [HP=550 ft-lb/sec]
= T.n/63,025 HP
or, T= 63,025HP/n (in-lb), where n = rpm
Similarly, T= 9,550,000kW/n (N-mm), where n = rpm
Shear (t) and bending (s) stresses on the outer surface of
a shaft, for a torque (T) and bending moment (M)

T T d 16T
t  r 
J (p d / 32) 2 p d 3
4

For solid circular section:

M M d 32M
s r 
I (p d 4 / 64) 2 p d 3
For hollow circular section:
T T do 16T 16T di
t r  d  where, l 
(p (d 04  d i4 ) / 32) 2 p (d 04  d i4 ) p d 03 (1  l4 )
o
J do

M M do 32M 32M di
s r  d  where, l 
(p (d 04  d i4 ) / 64) 2 p (d 04  d i4 ) p d 03 (1  l4 )
o
I do
 Principal Normal Stresses and Max Distortion Energy
Failure criterion for non-rotating shafts
The stress at a point on the shaft is normal stress
(s) in X direction and shear stress (t) in XY plane.

From Mohr Circle:

s s  s s 
2 2

S1      t 2 and S2      t 2
2 2 2 2
Max Distortion Energy theory:
2
 S yp 
S  S  S1 S 2  
2 2 
1 2 N 
 fs 
Putting values of S1 & S2 and simplifying:
2
 S yp 
s  3t  
2 2 

This is the design
 N fs  equation for non
rotating shaft
 Design of rotating shafts and fatigue consideration
The most frequently encountered stress situation for a
rotating shaft is to have completely reversed bending and
steady torsional stress. In other situations, a shaft may
have a reversed torsional stress along with reversed
bending stress.

The most generalized situation the rotating shaft may

have both steady and cyclic components of bending
stress (sav,sr) and torsional stress (tav,tr).

From Soderberg’s fatigue criterion, the equivalent

static bending and torsional stresses are:

Using these equivalent static stresses in our static

design equation, the equation for rotating shaft is:
Bearing mounting considerations
and stress concentration
Conventional retaining (or snap) rings fit
in grooves and take axial load, but groves
cause stress concentration in shaft

Retaining rings are

standardized items,
available in various
standard sizes with
various axial load
capacities.
Push type retaining rings – no grooves
required, less stress concentration, but
less axial support
Various types of keys for transmitting
torque
Other common types of keys
Various types of collar pins
Integrated splines in hubs and shafts
allow axial motion and transmits torque

All keys, pins and splines give rise to stress

concentration in the hub and shaft
Chapter 11 - Keys, Couplings and Seals

11.1 Chapter Objectives:

How attach power transmission components to shaft to prevent

rotation and axial motion?
Torque resistance: keys, splines, pins, weld, press fit, etc..
Axial positioning: retaining rings, locking collars,
shoulders machined into shaft, etc.
What is the purpose of rigid and flexible couplings in a power
transmission system?
Specify seals for shafts and other types of machine elements.
11.2 Keys

Most common for shafts up to 6.5 is thesquare and rectangular keys:

Advantages:
1. Cost effective means of locking the
2. Can replace damaged component
3. Ease of installation
4. Can use key as fuse fails in shear at some predetermined
torque to avoid damaging drive train.

Figure 11.1
Square and rectangular keys:

Step 1 
Determine
key size
based on
shaft
diameter

Step 2 
Calculate
required
length, L,
based on
torque (11.4)
Step 3 Specify appropriate shaft and bore dimensions for
keyseat:
See Figure 11.2

For 516 key

SHAFT BORE

Note, should also specify fillet radii and key

chamfers see Table 112
http:www.drivlok.com
 Other types of keys:

a. Tapered key can install

after hub (gear) is
installed over shaft.
b. Gib head key ease of
extraction
c. Pin keys low stress
concentration
d. Woodruff key light
loading offers ease of
assembly
11.4 Design of Keys stress analysis to determine
required length:

Torque being
transmitted
No load

T  F(D2) or F  T(D2) this is the

force the key must react!!!
 Bearing stress

Shear stress

Required Length based on

Required Length based on
Bearing Stress:
Shear Stress:

2T 4T
L where t d  0.5Sy / N L where s d  Sy / N
t d DW s d DH

Typical parameters for keys:

N  3, material 1020 CD (Sy  21,000 psi)
Example: Specify the complete key geometry and material for an application
requiring a gear (AISI OQT 1000) with a 4 hub to be mounted to a 3.6 diameter
shaft (AISI 1040 CD). The torque delivered through the system is 21,000 lbin.
Assume the key material is 1020 CD (Sy  21,000 psi) and N  3.

Solution (note since key is weakest material, focus analysis on key!):

See handout
11.4 Splines

Advantages:
Can carry higher torque for given diameter (vs keys) or
Lower stress on attachment (gear)
Better fit, less vibration (spline integral to shaft so no vibrating key)
May allow axial motion while reacting torque

Disadvantage:
Cost
Impractical to use as fuse
 Splines

Axial keys
machined into a shaft
Transmit torque from
shaft to another
machine element
 Advantages
• Uniform transfer of torque
• Lower loading on elements
• No relative motion between “key” and
shaft
• Axial motion can be accommodated
(can cause fretting and corrosion)
• Mating element can be indexed with a
spline
• Generally hardened to resist wear
 Spline Types
• Straight
– SAE
– 4, 6, 10 or 16 splines
• Involute
– Pressure angles of 30, 37.5, or 45
deg.
– Tend to center shafts for better
concentricity
 SAE Spline Sizes

A: Permanent Fit
B: Slide without Load
C: Slide under Load

Pg 504
 Two types of splines:

Straight Sided Involute:

Use this for
spline design 
SAE formulas
based on 1,000 psi
bearing stress
allowable!!

Use this to get

diameter. Then
table 11.4 to get W,
h, d
 Torque Capacity
• Torque capacity is based on 1000 psi
bearing stress on the sides of the
splines
T = 1000*N*R*h
N = number of splines
R = mean radius of the splines
h = depth of the splines
 Torque Capacity Cont’d
1D d D d
R  
2 2  4
1
h  (D  d)
2

Substituting R and h into torque

equation:  D  d  D  d   D 2  d2 
T  1000N     1000N  

 4  2   8 
 Torque Capacity Cont’d
• Further refinement can be done by
substituting appropriate values for N
and d.
• For 16 spline version, with C fit,
N = 16 and d = .810D
 D 2  (.810D)2 
T  1000(16)  

 8 
T  688D 2 Torque in INLBSINCH of spline

D  T / 688 Required D for given Torque

 Torque Capacity for Straight
Splines

Pg 505
Torque Capacity for Straight
Splines
Example: A chain sprocket delivers 4076 in-
lbs of torque to a shaft having a 2.50 inch
diameter. The sprocket has a 3.25 inch hub
length. Specify a suitable spline having a B
fit.
T = kD2L
T = torque capacity in in-lbs
kD2 = torque capacity per inch
(from Table 11-5)
L = length of spline in inches
 Example Continued
T 4076 in  lbs
k 2  2
 200.7  201
D L (2.50" ) (3.25" )

• From Table 11-5,

use 6 splines
 Torque Capacity for Straight
Splines

2.5

40763.25
Example: Specify straight spline for the previous problem (i.e. Torque  21,000
lbin and shaft is 3.6 in diameter.
 Taper & Screw

Expensive machining
Good concentricity
Moderate torque capacity
Can use a key too
 Couplings
• Used to connect two shafts together at
their ends to transmit torque from one to
the other.
• Two kinds of couplings:
– RIGID
– FLEXIBLE
Rigid Couplings
8T
d
pDbc N( t d )

NO relative motion between the shafts.

Precise alignment of the shafts
Bolts in carry torque in shear. N  # of bolts.
 Flexible Couplings
• Transmit torque smoothly
• Permit some axial, radial and angular
misalignment
Flexible Couplings
Flexible Couplings
Lord Corp. Products
Flexible Coupling
 Universal Joints
Large shaft misalignments permissible
Key factors in selection are Torque, Angular
Speed and the Operating Angle

Output not uniform wrt input Output IS uniform wrt input

Axial Constraint Methods

Spacers

Retaining
ring

Retaining Shoulders
ring
Retaining Rings
Locknuts