lec6 torsion
TRANSCRIPT
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Torsion
Torsional Deformation of a circular shaft,
Torsion Formula , Power Transmission
1
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Torsional Deformation of
a circular shaft
2
xBD
d
dx
d dx
dc
max
max
max
c
cdx
d
Length BD
when dxx
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The Torsion Formula
3
Angular strain is proptional to shear stress:
max
c
Mean: highest shear stress: will be at farthestaway from center At the center point, there will be noangular strain and therefore no shearstress is developed.
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The Torsion Formula
4
A
A
A
dAc
T
dAc
T
dAT
dAdT
dAdF
2max
max)(
J
Tc
dAJ
dAc
T
A
A
max
2
2max
J
TcmaxMaximum shear stress due to torsion
Shear stress due to torsion at radius
J
T
J: Polar moment of area
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Polar Moment of Inertia
5
Polar Moment of Inertia (J)
32
2 4
4
d
cJ
Solid shaft
32
)(
)(244
44
io
io
dd
ccJ
Tubular shaft
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Solve it!
The solid circular shaft is subjected to aninternal torque of T= 5kNm. Determinethe shear stress developed at point Aand B. Represent each state of stress ona volume of element.
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Solve it!
The solid circular shaft is subjected to aninternal torque of T= 5kNm. Determinethe shear stress developed at point Aand B. Represent each state of stress ona volume of element.
MPar
TA 7.49
2/)40(
)40()10)(5000(
2/ 4
3
4
MPar
TB 3.37
2/)40(
)30()10)(5000(
2/ 4
3
4
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Example 1
The shaft shown in figure is supportedby two bearings and is subjected tothree torques. Determine the shearstress developed at points A and B,located at section a-a of the shaft.
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J
T
Point AR = 0.075 mJ = JtotalT = internal torque
Point BR = 0.015 mJ = JtotalT = internal torque
4644 )10(7.49)75(22
mmcJ
T=4.25 3.0 = 1.25 kNm
MPamm
mmNmm
J
TA 886.1
)10(7.49
)75()10(125046
3
MPa
mm
mmNmm
J
TB 377.0
)10(7.49
)15()10(125046
3
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Solve it!
10
Determine the maximum shear stress developed in the shaft atsection a-a.
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11
J
T
Maximum Torsional ShearT:2100 Nm = 2100(10)3 Nmm: 40 mmJ: tubular
464444 )10(75.2)3040(2
)(2
mmrRJ
MPaJ
Tr55.30
)10(75.2
)40()10(21006
3
max
Polar Moment of Area
Max torsional shear when r = 40mm
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Solve it!
12
The solid shaft has a diameter of 40 mm. Determine the absolutemaximum shear stress in the shaft.
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13
Maximum Torsional Shear
T:70 Nm = 70(10)3 Nmm: 20 mmJ: solid r =20 mm
MPa
r
Tr
J
Tr
57.5
20)2
(
)20()10(70
)2(
4
3
4
max
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Power Transmission
14
Power TPP: power (1 Watt = 1 Nm/s)
T: torque (Nm)
w: radian/s
)
sec
min(
60
1)(2)
min
(sec)/(
rev
radrevnrad Input n(rpm)
Input frequency of shaft rotation frad 2sec)/(
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Solve it !
The gear motor can develop 1.6kW when it turns at 450rev/min. If the shaft has a diameter of 25mm, determinethe maximum shear stress developed in the shaft.
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Calculate the T
MPa
c
Tc
J
Tc
07.11
2/)5.12(
)5.12()10(96.33
2/
4
3
4max
srad
rev
radrevrad
/12.47
)sec
min(
60
1)(2)
min(450sec)/(
NmwPT 96.33
12.471600
Maximum shear
stress
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Solve it !
The gear motor can develop 2.4kW when it turns at 150rev/min. If the allowable shear stress for the shaft is allow= 84 Mpa, determine the smallest of the shaft to nearestmultiples of 5mm that can be used.
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Solve it !
Calculate the T
3
4
3
4max
/320,97
2/)(
)()10(87.152
2/
c
c
c
c
Tc
J
Tc
srad
rev
radrevrad
/7.15
)sec
min(
60
1)(2)
min(150sec)/(
Nm
w
PT 87.152
7.15
2400
Maximum shearstress
mmd
d
c
c
all
25
21
5.10
84/320,97 3max
Calculate d
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Angle Twist
Lecture 1 19
JG
TL
JG
TL
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Sign Convention
Lecture 1 20
Right hand rule
+ve: direction of the thumb is awayfrom the part
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Solve it!
The splined ends and gears attached to the A-36 steelshaft are subjected to the torques shown. Determine theangle of twist of end B with respect to end A. The shafthas a diameter of 40 mm.
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JGLT
JGLT
JGLT
JG
TL
DBDBCDCDACAC
300Nm 500Nm 200Nm 400Nm
300Nm Tac= 300Nm
300Nm
500Nm
Tcd=200Nm
Tdb =400Nm400Nm
JGJG
LT ACACAC
)300()10)(300( 3
JGJG
LT CDCDCD
)400()10)(200(3
JGJG
LT DBDBDB
)500()10)(400(3
o
T
radJG
578.001008.0)10)(75()20)(2/(
)10)(000,190()10)(000,190(34
33
Total angle of twist:
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NF
F
TT
F
F
GF
600
)10(60)100( 3
NmNmm
FT
TT
ED
FE
90000,90
)150(600150
FF
FE
Direction of FF
Direction of FE
Equilibrium of shaft AGFB
Equilibrium at gear F and E
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NmT
T
D
D
30
090120
Direction of FFEquilibrium of shaft DCE
FE12ONm
9ONm3ONm
JG
TL
3ONm3ONm
JGJG
LT DCDCDC
)250()10)(30(3
9ONm9ONmJGJG
LT CECECE
)750()10)(90( 3
rad
JGJG
LT CECEE
3
34
33
)10(02086.0
)10(75)5.12(2/
)10(000,60)10(000,60
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rad
R
RRL
RL
F
EEF
FF
EE
03129.0
100
)150(02086.0
13
12
Length of arc 12 = length of 13
1
23
Angle of twist at end B = angle of twist at F FB : there is no torque
radB3)10(129.3
Angle of twist at end A
o
A
FAFA
rad
JG
TL
09.20365.0
)10(215.503129.0
)10)(75()5.12(2/
)250()10(60)10(129.3
)10(129.3
3
34
33
3
/