Chapter 3 Torsion Introduction -- Analyzing the stresses and strains in machine parts which are subjected to torque T Circular -- Cross-section Non-circular.

Chapter 3 Torsion 

Introduction

-- Analyzing the stresses and strains in machine parts which are subjected to torque T

Circular

-- Cross-section Non-circular

Irregular shapes

-- Material (1) Elastic

(2) Elasto-plastic

-- Shaft (1) Solid

(2) Hollow

3.1 Introduction

T is a vector

Two ways of expression

-- Applications:

a. Transmission of torque in shafts, e.g. in automobiles

Assumptions in Torque Analysis:

a. Every cross section remains plane and undistorted.

b. Shearing strain varies linearly along the axis of the shaft.

3.2 Preliminary Discussion of the Stresses in a Shaft

( ) dA T

dF T

Free-body Diagram

Where = distance (torque arm)

Since dF = dA

The stress distribution is Statically Indeterminate.

-- Must rely on “deformation” to solve the problem.

Analyzing a small element:

3.3 Deformations in a Circular Shaft

= (T, L) -- the angle of twist (deformation)

Rectangular cross section warps under torsion

' 'CD C D

A circular plane remains circular plane

L (in radians)

Determination of Shear Strain

The shear strain

max

cL

max

c

max L

c

= c = radius of the shaft

L Since

G

3.4 Stresses in the Elastic Range3.4 Stresses in the Elastic Range

Hooke’s LawHooke’s Law

max

c

max

G Gc

G max max G

max

c

Therefore, Therefore, (3.6)

1

2min max

cc

max

JT

c

2maxmax

T dA dA dAc c

( ) dA T (3.1) max

c

(3.9)

But 2 dA J

Therefore, Or, max TcJ

(3.6)

Substituting Eq. (3.9) into Eq. (3.6)

JT

max TcJ

412J c

(3.10)

(3.9)

These are elastic torsion formulas.

For a solid cylinder:

For a hollow cylinder: 4 42 1

12

( ) J c c

0 02 45 2max max( )cos F A A

2 oA A 3 13. ( )EqA

Since

max 0max

0

2

2

F A

A A

(3-13)

Mohr’s Circle (Sec. 7.4)

-- Pure Shear Condition

Ductile materials fail in shear (90o fracture)

Brittle materials are weaker in tension (45o fracture)

3.5 Angle of Twist in the Elastic Range

max

cL

maxmax maxsin

Tcce

G J

TL

JG

(3.3)

max

TcJG

(3.15)

max

c TcL JG Eq. (3.3) = Eq. (3.15)

Therefore,

Hence,

i i

i i i

T J

J G

For Multiple-Section Shafts:

Shafts with a Variable Circular Cross SectionShafts with a Variable Circular Cross Section

0

LTdxJG

Tdx

dJG

3.6 Statically Indeterminate Shafts

-- Must rely on both

(1) Torque equations and

(2) Deformation equation, i.e. TLJG

0T

Example 3.05

3.7 Design of Transmission Shafts

P power T

2P f T

fP

T2

-- Two Parameters in Transmission Shafts:

a. Power P

b. Speed of rotation

where = angular velocity (radians/s) = 2

= frequency (Hz)

[N.m/s = watts (W)] (3.21)

max TcJ

fP

T2

max

J Tc

(3.21)

(3.9)

4 31 12 2

/J c and J c c

For a Solid Circular Shaft:

Therefore,

312 max

Tc

1 32

/

max

Tc

3.8 Stress Concentrations in Circular Shafts

max Tc

KJ

3.9 Plastic Deformation sin Circular Shafts

max

c

(3.4)

c = radius of the shaft

c

d

dA = 2 d

c

( )T dF dA d2

dF T (3.1)

Knowing dF = dA

2

02

cT d

Where = ()

(3.26)

UT

T cR

J

max TcJ

(3.9)

If we can determine experimentally an Ultimate Torque, TU,

then by means of Eq. (3.9), we have

RT = Modulus of Rupture in Torsion

L

3.10 Circular Shafts Made of an Elasto-Plastic Material

Y Y

JT

c

max TcJ

Case I: < Y Hooke’s Law applies, < max Case I

Case II

Case II: < Y Hooke’s Law applies, = max

TY = max elastic torque

Since

Case III: Entering Plastic Region

312 Y YT c

312

/J C c

Y Case III

0 Y:

Y

Y

Y – region within the plastic range

Y c:

(3-29)

2

02

cT d

2 2

0

3 3 3

33

3

2 2

1 2 22 3 3

2 11

3 4( )

Y

Y

cY

elastic plastic YY

Y Y Y Y Y

YY

T T T d d

c

T cc

3

3

4 11

3 4( )

Y

YT Tc

(3.26)

By evoking Eq. (3.26)

(3.31)

312 Y YT c

Case IV

Case IV -- Fully Plastic

4

3P YT T

3

3

4 11

3 4( )

Y

YT Tc

Y 0:

= Plastic Torque (3-33)

YY

L

3

3

4 1(1 )

3 4 Y

YT T

Y Y

c

Y

YLc

T A

3.11 Residual Stresses in Circular Shafts

' P

( ) 0 dA

3.12 Torsion of Noncircular Members

0 0 yx yz

0 0 zx zy

0 0 xy xz

A rectangular shaft does not axisymmetry.

21

max Tc ab

32

TL

c ab G

From Theory of Elasticity:

1 2

11 0 630

3( . / ) (for b/a = 5 only) 3.45 c c b a

3.13 Thin-Walled Hollow Shafts

0 A BF - F = 0 xF ( ) A A AF t x

0( ) ( )

A A B B

A A B B

t x t x

t t

tan q t cons t

( ) ( ) dF dA tds t ds qds

( ) ( ) OdM pdF p qds q pds

2( )odM q dA

2T q A

24

TL ds

G t A

2

TtA