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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
Right pulley has exactly equal and
opposite torque:
T2 = (6750-2250)x380=1,710,000 N-mm
FV2
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
From Horizontal forces (FH) and vertical
forces (Fv), Bending moments MH & MV are
drawn separately.
Then the resultant moments at various
points on the shaft can be found from
Torque and Bending moment diagrams for the pulley
system
2
2
V
H
R M
M
M 

FH
9900N
MH
2,227,500
FV
900N 9900N
M V
2,227,500
911,250
T1 T2
Torque diag.
1,710,000 N-mm
Resultant bending moment
2,227,500
2,406,685
2
2
V
H
R M
M
M 

The section of shaft where the left pulley
is located has obviously the highest
combination of Torque (1,710,000 N-mm)
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)
3
4
16
2
)
32
/
( d
T
d
d
T
r
J
T
p
p
t 


For solid circular section:
For hollow circular section:
3
4
32
2
)
64
/
( d
M
d
d
M
r
I
M
p
p
s 


o
i
o
i
o
i d
d
where
d
T
d
d
d
T
d
d
d
T
r
J
T







 

p
p
p
t ,
)
1
(
16
)
(
16
2
)
32
/
)
(
( 4
3
0
4
4
0
4
4
0
o
i
o
i
o
i d
d
where
d
M
d
d
d
M
d
d
d
M
r
I
M







 

p
p
p
s ,
)
1
(
32
)
(
32
2
)
64
/
)
(
( 4
3
0
4
4
0
4
4
0
Principal Normal Stresses and Max Distortion Energy
Failure criterion for non-rotating shafts
2
2
2
2
2
1
2
2
2
2
t
s
s
t
s
s

















 S
and
S
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:
Max Distortion Energy theory:
2
2
1
2
2
2
1 










fs
yp
N
S
S
S
S
S
Putting values of S1 & S2 and simplifying:
2
2
2
3










fs
yp
N
S
t
s This is the design
equation for non
rotating shaft
Design of rotating shafts and fatigue consideration
The most generalized situation the rotating shaft may
have both steady and cyclic components of bending
stress (sav,sr) and torsional stress (tav,tr).
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.
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

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Design of Shaft.ppt

  • 1. 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).
  • 2. The connecting shaft is loaded primarily in torsion.
  • 3. 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.
  • 4. 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 Right pulley has exactly equal and opposite torque: T2 = (6750-2250)x380=1,710,000 N-mm FV2 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
  • 5. From Horizontal forces (FH) and vertical forces (Fv), Bending moments MH & MV are drawn separately. Then the resultant moments at various points on the shaft can be found from Torque and Bending moment diagrams for the pulley system 2 2 V H R M M M   FH 9900N MH 2,227,500 FV 900N 9900N M V 2,227,500 911,250 T1 T2 Torque diag. 1,710,000 N-mm Resultant bending moment 2,227,500 2,406,685 2 2 V H R M M M   The section of shaft where the left pulley is located has obviously the highest combination of Torque (1,710,000 N-mm) and Bending moment (2,406,685 N-mm)
  • 6. 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
  • 7. Shear (t) and bending (s) stresses on the outer surface of a shaft, for a torque (T) and bending moment (M) 3 4 16 2 ) 32 / ( d T d d T r J T p p t    For solid circular section: For hollow circular section: 3 4 32 2 ) 64 / ( d M d d M r I M p p s    o i o i o i d d where d T d d d T d d d T r J T           p p p t , ) 1 ( 16 ) ( 16 2 ) 32 / ) ( ( 4 3 0 4 4 0 4 4 0 o i o i o i d d where d M d d d M d d d M r I M           p p p s , ) 1 ( 32 ) ( 32 2 ) 64 / ) ( ( 4 3 0 4 4 0 4 4 0
  • 8. Principal Normal Stresses and Max Distortion Energy Failure criterion for non-rotating shafts 2 2 2 2 2 1 2 2 2 2 t s s t s s                   S and S 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: Max Distortion Energy theory: 2 2 1 2 2 2 1            fs yp N S S S S S Putting values of S1 & S2 and simplifying: 2 2 2 3           fs yp N S t s This is the design equation for non rotating shaft
  • 9. Design of rotating shafts and fatigue consideration The most generalized situation the rotating shaft may have both steady and cyclic components of bending stress (sav,sr) and torsional stress (tav,tr). 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. 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:
  • 10. Bearing mounting considerations and stress concentration
  • 11. 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.
  • 12. Push type retaining rings – no grooves required, less stress concentration, but less axial support
  • 13. Various types of keys for transmitting torque
  • 15. Various types of collar pins
  • 16. 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