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 Lc

= 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)

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