89589540-Strength-of-Materials-by-S-K-Mondal-9.pdf

8/14/2019 89589540-Strength-of-Materials-by-S-K-Mondal-9.pdf

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9. Torsion

Theory at a Glance (for IES, GATE, PSU) In machinery, the general term shaft refers to a member, usually of circular cross-

section, which supports gears, sprockets, wheels, rotors, etc., and which is subjected to

torsion and to transverse or axial loads acting singly or in combination.

An axle is a rotating/non-rotating member that supports wheels, pulleys, and

carries no torque.

A spindleis a short shaft. Terms such as lineshaft, headshaft, stub shaft, transmission

shaft, countershaft, andflexible shaft are names associated with special usage.

Torsion of circular shafts

1. Equation for shafts subjected to torsion "T"

T G= =

J L

R Torsion Equation

Where J = Polar moment of inertia

= Shear stress induced due to torsion T.

G = Modulus of rigidity

= Angular deflection of shaft

R, L = Shaft radius & length respectively

Assumptions

The bar is acted upon by a pure torque.

The section under consideration is remote from the point of application of the load and from

a change in diameter.

Adjacent cross sections originally plane and parallel remain plane and parallel after

twisting, and any radial line remains straight.

The material obeys Hookes law

Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle

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Chapter-9 Torsion S K Mondals

2. Polar moment of inertia

Solid shaft J =4d

32

Hollow shaft, "J = 4 4( )32

o id d

3. The polar section modulus

Zp= J / c,where c = r = D/2

For a solid circular cross-section, Zp= D3/ 16

For a hollow circular cross-section, Zp= (Do4 - Di4) / (16Do)

Then, max = T / Zp

If design shears stress,d

is known, required polar section modulus can be calculated from:

Zp= T / d

4. Power Transmission (P)

P (in Watt ) =2

60

NT

As stated above, the polar second moment of area, J is defined as

J = 2 30

r drR

z

For a solid shaft J = 24

2

4 32

4

0

4 4

r R DR

L

NM

O

QP = = (6)

For a hollow shaft of internal radius r:

J = 2 30

r drR

z = 2 4 2 324

4 4 4 4 r

R r D d

r

R

L

NM

O

QP = = ( ) c h (7)

Where D is the external and d is the internal diameter.

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P (in hp) =2

4500

NT (1 hp = 75 Kgm/sec).

[Where N = rpm; T = Torque in N-m.]

5. Safe diameter of Shaft (d)

Stiffness consideration

=T G

J L

Shear Stress consideration

T

J R

=

We take higher value of diameter of both cases above for overall safety if other parameters are given.

6. In twisting

Solid shaft,max

=3

16T

d

Hollow shaft, max =o

4 4

16Td

( ) o id d

Diameter of a shaft to have a maximum deflection "" d = 4.9 4

TL

G

[Where T in N-mm, L in mm, G in N/mm2]

7. Comparison of solid and hollow shaft

A Hollow shaft will transmit a greater torque than a solid shaft of the same weight & same

material because the average shear stress in the hollow shaft is smaller than the average

shear stress in the solid shaft

max

max

( ) shaft 16

( ) shaft 15

=

holloow

solid

o i

If solid shaft dia = D

DHollow shaft, d = D, d =

2

Strength comparison (same weight, material, length and max )2

2

1

1

h

s

T n

T n n

+=

Externaldiameterof hollowshaftWhere, n=

Internaldiameter of hollowshaft [ONGC-2005]

Weight comparison (same Torque, material, length and max )

( )

( )

2 2/3

2/34

1

1

h

s

n nW

W n

=

Externaldiameterof hollowshaftWhere, n=

Internaldiameter of hollowshaft [WBPSC-2003]

Strain energy comparison (same weight, material, length andmax

)

2

2

1h

s

U n

U n

+=

2

11

n= +

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8. Shaft in series

1 2 = +

Torque (T) is same in all section

Electrical analogy gives torque(T) = Current (I)

9. Shaft in parallel

1 2 = and

1 2T T T= +

Electrical analogy gives torque(T) = Current (I)

10. Combined Bending and Torsion

In most practical transmission situations shafts which carry torque are also subjected to

bending, if only by virtue of the self-weight of the gears they carry. Many other practical

applications occur where bending and torsion arise simultaneously so that this type of

loading represents one of the major sources of complex stress situations.

In the case of shafts, bending gives rise to tensile stress on one surface and compressive

stress on the opposite surface while torsion gives rise to pure shear throughout the shaft.

For shafts subjected to the simultaneous application of a bending moment Mand torque Ttheprincipal stressesset up in the shaft can be shown to be equal to those produced by an

equivalent bending moment,of a certain value Meacting alone.

Figure

Maximum direct stress ( x ) & Shear stress ( ( )xy in element A

3

3

32

16

= +

=

x

xy

M P

d A

T

d

Principal normal stresses ( 1,2 ) & Maximum shearing stress ( max )Page 279 of 429


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1,2 =

2

2

2 2

+

x xxy

2

21 2max ( )

2 2

= = +

xxy

Maximum Principal Stress (max

) & Maximum shear stress (max

)

max = 2 2

3

16

+ + M M T

d

max = 2 2

3

16

+M T

d

Location of Principal plane ( )

=11 tan

2

TM

Equivalent bending moment (Me) & Equivalent torsion (Te).

2 2

2

+ +=

e

M M TM

2 2= +eT M T

Important Note

o Uses of the formulas are limited to cases in which both M & T are known. Under any

other condition Mohrs circle is used.

Safe diameter of shaft (d) on the basis of an allowable working stress.

o w in tension , d = 332 e

w

M

o w in shear , d= 316 e

w

T

11. Shaft subjected to twisting moment only

Figure

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Normal force ( nF ) & Tangential for ( tF ) on inclined plane AB

[ ]

[ ]

sin + AC cos

BC cos - AC sin

=

=

n

t

F BC

F

Normal stress ( n ) & Tangential stress (shear stress) ( t ) on inclined plane AB.

n = sin2

t = 2 cos

Maximum normal & shear stress on AB

( n )max max0 0 +

45 0

90 0

135 + 0

Important Note

Principal stresses at a point on the surface of the shaft = + , - , 0

i.e1,2 sin2 =

Principal strains

1 2 3(1 ); (1 ); 0

= + = + =

E E

Volumetric strain,

1 2 30 = + + =v

No change in volume for a shaft subjected to pure torque.

12. Torsional Stresses in Non-Circular Cross-section Members

There are some applications in machinery for non-circular cross-section members and shafts

where a regular polygonal cross-section is useful in transmitting torque to a gear or pulley

that can have an axial change in position. Because no key or keyway is needed, the

possibility of a lost key is avoided.

Saint Venant (1855) showed that max in a rectangular b c section bar occurs in the middle

of the longest side b and is of magnitude formula

max 2 2

1.83

/

T T

b cbc bc

Where b is the longer side and factor that is function of the ratio b/c.

The angle of twist is given by Page 281 of 429


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3

Tl

bc G

Where is a function of the ratio b/c

Shear stress distribution in different cross-section

Rectangular c/s Elliptical c/s Triangular c/s

13. Torsion of thin walled tube

For a thin walled tube

Shear stress,

02

= T

A t

Angle of twist,2 O

sL

A G

=

[Where S = length of mean centre line, OA = Area enclosed by mean centre line]

Special Cases

o For circular c/s

3 22 ; ; 2 = = =oJ r t A r S r

[r = radius of mean Centre line and t = wall thickness]

2

. =

2 r 2

= =

o

T T r T

t J A t

32

= = =

o

TL L TL

GJ A JG r tG

o For square c/s of length of each side b and thickness t

20

=4b

A b

S

=

o For elliptical c/s a and b are the half axis lengths.

0

3 ( )

2

A ab

S a b ab

=

+

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