Torsion of circular shaft, Non-uniform torsion, Power transmission of
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Transcript Torsion of circular shaft, Non-uniform torsion, Power transmission of
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) 2f
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