Lec6 Torsion

7/28/2019 Lec6 Torsion

1/26

Torsion

Torsional Deformation of a circular shaft,

Torsion Formula , Power Transmission

1


2/26

Torsional Deformation of

a circular shaft

2

xBD

d

dx

d dx

dc

max

max

max

c

cdx

d

Length BD

when dxx


3/26

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.


4/26

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


5/26

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


6/26

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.


7/26

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


8/26

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.


9/26

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


10/26

Solve it!

10

Determine the maximum shear stress developed in the shaft atsection a-a.


11/26

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


12/26

Solve it!

12

The solid shaft has a diameter of 40 mm. Determine the absolutemaximum shear stress in the shaft.


13/26

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


14/26

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)/(


15/26

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.


16/26

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


17/26

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.


18/26

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


19/26

Angle Twist

Lecture 1 19

JG

TL

JG

TL


20/26

Sign Convention

Lecture 1 20

Right hand rule

+ve: direction of the thumb is awayfrom the part


21/26

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.


22/26

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:


23/26


24/26

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


25/26

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


26/26

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

/

Documents

Lec6 Torsion