MECHANICS OF MATERIALS

CHAPTER THREE TORSION

Prepared by : Dr. Mahmoud Rababah 1

3.1 INTRODUCTION

Transmission shaft

Steam turbine

Electric generator

Transmission shaft

2

3.2 - 3.3 DEFORMATION IN A CIRCULAR SHAFT

Every cross-section remains plane and undistorted

i.e. each cross-section rotates as a rigid disk.

This is only applicable for circular cross-sections.

3

: the angle of twist

it is proportional to and L

T

For small shear strain

L

L

4

maxc

The shearing strain varies linearly

with the distance from the axis of

the shaft

SHEARING STRAIN ALONG THE RADIAL

DIRECTION

Shaft radius

5

Hook’s law is applied

Thus, linear variation in shearing

strain leads to linear variation in

shearing stress.

3.4 STRESSES IN THE ELASTIC

RANGE

max max

max

max

but

thus

G

G

c

c

6

THE TORSION FORMULA

max

2ma

max

x

2

( )

but

(polar moment of inertia)

(Torsion formula)

A A

A

A

c

J

dA dAc

J

dAc

J dA

T

T

T

T

7

POLAR MOMENT OF INERTIA

8

max 120 MPa

Find

Example :

6 4 4

max

120 10 (0.03) (0.02)2 4.08 kN.m

0.03o

J

c

Solution :

T

9

1

2

max min

90 mm 0.045 m

120 mm 0.06 m

shafts and are solids of diameter

Find:

1- and in shaft .

2- diameter for 65 MPa.

BC inner

BC outer

all

d C

d C

AB CD d

BC

d

Example :

3

2max

4 4

1min max

2

20 10 0.061 86.2 MPa

(0.06) (0.045)2

64.7 MPa

BC

C

J

C

C

BCT

0

6 kN

20 kN

6 kN

AB

BC

CD

T

T

T

Solution :

M

( / 2)2 65 MPa

77.8 mm

all

AB

d

J

d

ABT

10

80 MPa

50 MPa

Find

1 for not exceeding the maximum shear stress in sleeve

2

all AB

all CD

s

CD

d

max

Example :

T

6 4 4

36

4

1

50 10 (0.08) (0.068)32 2.4 kN.m

0.04

2

2.4 10 ( / 2) 80 10 53.5 mm

32

CDCD

CD

ABAB

AB

ss

s

C

J

C

J

dd

d

Solution :

T

T

T

11

1000 N.m

56 mm and 42 mm

Determine the maximum shearing stress in both shafts.

AB CDd d

Example :

T

2500 N.mAB

AB CD

Tr r

CDAB

Solution :

TT 4

4

2500 0.02872.5 MPa

(0.028)2

1000 0.02168.7 MPa

(0.021)2

ABAB

AB

CDCD

CD

C

J

C

J

AB

CD

T

T

12

100 N.m

21 mm, 30 mm and 40 mm

Determine the maximum shearing stress in the three shafts.

AB CD EFd d d

Example :

T

1

2

4

4

1 2

240 N.m

= 600 N.m

100 0.010555 MPa

(0.010

note: 60 mm a

5)2

240 0.01545.3 MPa

(0.015)2

6

nd 30 mm

AB CD

CD EF

ABAB

AB

CDCD

CD

EFEF

E

CD CD

F

r r

r r

C

J

C

J

J

r r

C

CDABCD

CD EFEF

AB

CD

EF

Solution

TTT

T TT

T

T

T

4

00 0.0247.7 MPa

(0.02)2

13

3.5 ANGLE OF TWIST IN THE ELASTIC RANGE

maxmax

(analog

1

us to ) L

C C

L G J G

AJG E

L

P

T

T

For multi-sections

i i

i i i

L

J G

T

14

SIGN CONVENTION

and are positive when right hand thumb

is outward of the surface

T

15

Example:

/

70 1080 BC CDABA D

L LL

JG JG JG

16

4

150 N.m

130 N.m

170 N.m

(0.007)2

150 0.4 130 0.3 170 0.50.2121 radi i

A

J

L

JG JG JG JG

AC

CD

DE

Solution :

T

T

T

T

9

14 mm

80 10 GPa

d

G

Example :

17

ANGLE OF TWIST FOR VARIABLE CROSS-SECTION

0

( )

( )

L

d dxJG

xdx

J x G

T

T

18

9

max

60 mm (solid shaft), 75 10

Find

1

2 C

d G

Example :

max4

/ /

1.5

4 9 4 90

3

1800 0.031 42.4 MPa

(0.03)2

1200 2000 N.m

1200 N.m

1200 0.8 1200 2000

(0.03) 75 10 (0.03) 75 102 2

5.344 10 rad

C C B B A

C

C

C

J

x

xdx

max

AB

BC

Solution :

T

T

T

19

or apply superposition

=

+

1 2

14 9

1.

2

5

4 90

3

2

1200 2.3

(0.03) 75 102

2000

(0.03) 75 102

5.344 10 rad

C C C

C

C

C B

xdx

20

RELATIVE TWIST ANGLES

/

/

E B E B

E B

TL

JG

21

22.5 N.mAB CDr r

CDABCD

Solution :

TTT

20 mm

80 GPa

45 N.m

d

G

AB

Example :

T

22

4 9

/4 9

/

22.5 1.50.0269 rad

(0.01) 80 102

0.0134 rad

45 20.0716 rad

(0.01) 80 102

0.085 rad

CDC

C C B B

B

ABA B

A A B B

L

JG

r r

L

JG

CD

AB

T

T

ENGAGED TWIST

ANGLES RELATIONS

23

is steel

77 GPa

80 MPa

is Brass

38 GPa

50 MPa

Find and

steel

all steel

brass

all Brass

A

AB

G

CD

G

max

Example :

T

4 9 4 4 9

1005 0.3 1005 0.2

(0.02) 77 10 (0.04) (0.034) 38 102 2

0.01833 rad

AB CDA

steel steel brass brass

A

A

L L

J G J G

max maxT T

6

4

6

4 4

max

0.0280 10

(0.02)2

0.0450 10 2.4 kN

1.005 kN.

.m

(0.04) (0.03

m

4)2

all steel

all Brass

T

C

J

C

J

max

maxmax

Solution :

T

T

T

TT

24

25 mm

75 GPa

Find A

d

G

Example :90 N.mF

E

r

r

F ET T

FBD

25

3

4 9 4 9

3

/4 9

/

30 0.25 90 0.7520.86 10 rad

(0.0125) 75 10 (0.0125) 75 102 2

0.03129 rad

60 0.255.215 10 rad

(0.0125) 75 102

2.09 deg.

F

F F E E

E

A E

A E A E

r r

L

JG

Solution :

T

26

60 mm

80 mm

2 kN.m, 4 kN.m

75 GPa

Find

and

AC

EH

A C

d

d

G

1 2

Example :

T T

FBD

8 kN.mD

B

r

r D BT T

27

3

4 9

/

3

2/

4 9

/

1/

8 10 0.60.0159 rad

(0.08) 75 1032

1000.0159 0.0212 rad

75

4 10 0.60.025 rad

(0.06) 75 1032

0.025 0.0212 0.0462 rad

D EDD

ED

DB D

B

A B A B

ABA B

AB

A

C B C B

C B

T L

J G

r

r

T L

J G

T L

Solution :

3

4 9

2 10 0.90.01886 rad

(0.06) 75 1032

0.01886 0.0212 0.04 rad

BC

BC

C

J G

28

0 0 (1) x A BM T T T

/ 0

0 (2)

From Eqs. 1 and 2

and

A B

AC BC

BC AC

L L

JG JG

L L

L L

A B

A B

T T

T T T T

3.6 STATICALLY

INDETERMINATE SHAFTS

one equation and two unknowns

the compatibility condition is

29

Example:

d = 20 mm

Find TA and TB

0 800 500 0 (1) x B A

Solution :

M T T

30

/ 0

0.2 (800 ) 1.5 (300 ) 0.30 (2)

645 N.m

345 N.m

A B

JG JG JG

B B B

B

A

T T T

T

T

31

4 4 9 4 9

250 0 (1)

(2)

(0.02) (0.01) 80 10 (0.01) 36 102 2

st br

L L

st br

st br

Solution :

T T

T T

32

Example:

Tube of steel and core of brass

T = 250 N.m

Gst = 80 GPa, Gbr = 36 GPa.

max4 4 9

1

min max2

max4 9

min

From Eq. 1 and 2, we get

242.72 N.m

7.28 N.m

0.0220.6 MPa

(0.02) (0.01) 80 102

0.0120.6 MPa 10.3 MPa

0.02

0.014.63 MPa

(0.01) 36 102

0

st

st st

brbr

br

C

C

T

st

br

st

T

T

T

33

3.7 DESIGN OF TRANSMISSION SHAFTS

2P w f T T

Power in Watt (N.m/s)

Angular velocity in rad/sec

Frequency in Hz (1/s)

Example:

a solid shaft is used to transmit 3750 W at angular 175 rpm. If

allowable shear is 100 MPa, find the diameter of the shaft.

2175 18.33 rad/sec

60

204.6 N.m

10.92 mm ( 22 mm)all

w

P T w T

T CC d

J

Solution :

34

3.8 STRESS CONCENTRATION IN CIRCULAR

SHAFTS

Stress concentration occurs when variation

in the cross-section occurs

35

TCK

J

36

x

T

3.12 TORSION OF NON-CIRCULAR

MEMBERS

37

max 2

1

3

2

From elasticity theory

c ab

L

c ab G

T

T

38

3.15 THIN WALLED HOLLOW SHAFTS

0

( ) ( )

Thus,

constant (Shear flow)

A A B Bt x t x

t q

x

A B

F

F F

39

( ) ( )

but

1

2

2

ave

ave

m

ave m

d h dF h tds

h tds q hds

dA hds

tA

T

T

T 2ave

m

T

tA

40

Example:

20.035 0.057 0.002 m

1.75 MPa2

2.92 MPa2

m

A

A m

B

B m

A

t A

t A

T

T

41

END OF CHAPTER THREE

42