Torsion
Torsional Deformation of a circular shaft,
Torsion Formula , Power Transmission
1
Torsional Deformation of
a circular shaft
Length BD
when
BD    x
  d
x  dx
d
 
dx
 max  c
d
dx

d

 max 
dx
c

 

c
 max
2
The Torsion Formula
Angular strain is proptional to shear stress:
 

c
 max
Mean:
• highest shear stress: will be at farthest
away from center
• At the center point, there will be no
angular strain and therefore no shear
stress is developed.
3
The Torsion Formula
dF   dA
dT    dA
T 
T 
  dA
J 
A
T 

  ( c )
 max
c
max
dA
2

 dA
A
c
2

 dA
A
2

 dA
A
A
T 
 max
 max 
Tc
J
J: Polar moment of area
Maximum shear stress due to torsion
Shear stress due to torsion at radius 
 max
Tc

J
 
T
J
4
Polar Moment of Inertia
Polar Moment of Inertia (J)
J 

Solid shaft

Tubular shaft
J 

2
c4
d 4
32

2
(co  ci )
4
4
 (d o 4  d i 4 )
32
5
Solve it!
The solid circular shaft is subjected to an
internal torque of T= 5kNm. Determine
the shear stress developed at point A
and B. Represent each state of stress on
a volume of element.
Solve it!
The solid circular shaft is subjected to an
internal torque of T= 5kNm. Determine
the shear stress developed at point A
and B. Represent each state of stress on
a volume of element.
A
B
T
(5000)(10)3 (40)


 49.7 MPa
r 4 / 2
 (40) 4 / 2
T
(5000)(10)3 (30)


 37.3MPa
4
4
r / 2
 (40) / 2
Example 1
The shaft shown in figure is supported
by two bearings and is subjected to
three torques. Determine the shear
stress developed at points A and B,
located at section a-a of the shaft.
Point A
R = 0.075 m
J = Jtotal
T = internal torque
J 

2
c4 

2
Point B
R = 0.015 m
J = Jtotal
T = internal torque
(75) 4  49 .7(10 ) 6 mm 4
T=4.25 – 3.0 = 1.25 kNm
T
 
J
A
T
1250(103 ) Nm m(75)m m


 1.886MPa
6
4
J
49.7(10) m m
B
T
1250(103 ) Nm m(15)m m


 0.377MPa
6
4
J
49.7(10) m m
Solve it!
Determine the maximum shear stress developed in the shaft at
section a-a.
10
T
 
J
Maximum Torsional Shear
T:2100 Nm = 2100(10)3 Nmm
: 40 mm
J: tubular
Polar Moment of Area
J 

2
(R4  r 4 ) 

2
( 40 4  30 4 )  2.75(10 ) 6 mm 4
Max torsional shear when r = 40mm
 max
Tr
2100(10)3 (40)


 30.55MPa
6
J
2.75(10)
11
Solve it!
The solid shaft has a diameter of 40 mm. Determine the absolute
maximum shear stress in the shaft.
12
Maximum Torsional Shear
T:70 Nm = 70(10)3 Nmm
: 20 mm
J: solid r =20 mm
Tr
Tr

 4
J
( )r
2
70(10) 3 ( 20)
 max 

(

2
 5.57MPa
) 204
13
Power Transmission
Power
P  T
P: power (1 Watt = 1 Nm/s)
T: torque (Nm)
w: radian/s
Input n(rpm)
 ( rad / sec)  n(
Input frequency of shaft rotation
rev
rad 1 min
) 2 (
)
(
)
min
rev 60 sec
 (rad / sec)  2f
14
Solve it !
The gear motor can develop 1.6kW when it turns at 450
rev/min. If the shaft has a diameter of 25mm, determine
the maximum shear stress developed in the shaft.
 (rad / sec)  450(
Calculate the T
 47.12rad / s
T 
Maximum shear
stress
rev
rad 1 min
)2 (
)
(
)
min
rev 60 sec
P
1600

 33 .96 Nm
w 47 .12
Tc
Tc

J
c 4 / 2
33.96(10) 3 (12.5)

 (12.5) 4 / 2
 11.07MPa
 max 
Solve it !
The gear motor can develop 2.4kW when it turns at 150
rev/min. If the allowable shear stress for the shaft is allow
= 84 Mpa, determine the smallest of the shaft to nearest
multiples of 5mm that can be used.
Solve it !
Calculate the T
 (rad / sec)  150(
 15.7 rad / s
T 
Maximum shear
stress
rev
rad 1 min
)2 (
)
(
)
min
rev 60 sec
P
2400

 152 .87 Nm
w
15.7
Tc
Tc

J
c 4 / 2
152.87(10) 3 (c)

 (c ) 4 / 2
 max 
 97,320/ c 3
Calculate d
 max   all
97,320/ c 3  84
c  10.5
d  21
d  25m m
Angle Twist
TL

JG
Lecture 1
 

TL
JG
19
Right hand rule
Sign Convention
+ve: direction of the thumb is away
from the part
Lecture 1
20
Solve it!
The splined ends and gears attached to the A-36 steel
shaft are subjected to the torques shown. Determine the
angle of twist of end B with respect to end A. The shaft
has a diameter of 40 mm.
TL
JG
T L
T L
T L
 AC AC  CD CD  DB DB
JG
JG
JG
 
300Nm
300Nm
500Nm
200Nm
 AC
Tac= 300Nm
400Nm
TAC LAC
(300)(10) 3 (300)


JG
JG
500Nm
Tcd=200Nm
300Nm
400Nm
Tdb =400Nm
DB
Total angle of twist:
CD
TCD LCD
(200)(10) 3 (400)


JG
JG
TDB LDB
(400)(10) 3 (500)


JG
JG
(190,000)(10)3
(190,000)(10)3
T 

JG
( / 2)(20) 4 (75)(10)3
 0.01008rad  0.578o
Solve it !
The two shafts are made of A-36 steel. Each has a
diameter of 25mm, and they are supported by bearings at
A, B and C, which allow free rotation. If the support at D is
fixed, determine of the angle of twist of end B and A when
the toques are applied to the assembly as shown.
Equilibrium of shaft AGFB
Direction of FF
TF  TG
FF (100)  60(10) 3
FF  600N
FF
Equilibrium at gear F and E
TE  TF
Direction of FE
TD  FE 150  600(150)
 90,000Nm m  90Nm
FE
Direction of FF
Equilibrium of shaft DCE
TD  120 90  0
TD  30Nm
3ONm
 
3ONm
9ONm
12ONm

TL
JG
3ONm
DC
CE
9ONm
E
FE
9ONm
TDC LDC
(30)(10) 3 ( 250)


JG
JG
TCE LCE
(90)(10) 3 (750)


JG
JG
TCE LCE
 60,000(10)3
 60,000(10)3



 0.02086(10) 3 rad
4
3
JG
JG
 / 2(12.5) 75(10)
Length of arc 12 = length of 13
3
2
L12   E RE
L13   F RF
 F 
E RE
1
RF
0.02086(150)
100
 0.03129rad

Angle of twist at end B = angle of twist at F  FB : there is no torque
B  3.129(10) 3 rad
Angle of twist at end A
 A  F   A / F
 A  3.129(10) 3 
 3.129(10) 3 
TL
JG
60(10) 3 ( 250)
 / 2(12.5) 4 (75)(10) 3
 0.03129 5.215(10) 3
 0.0365rad  2.09o