TorsionTorsion is the twisting of a straight bar when it is loaded by twisting moments or torques that tend to produce rotation about the longitudinal axes of the bar.For instance, when we turn a screw driver to produce torsion our hand applies torque ‘T’ to the handle and twists the shank of the screw driver.

Sign convention – Right Hand RuleShafts are structural members with length significantly greater than the largest cross-sectional dimension used in transmitting torque from one plane to another.Turbine: Exerts a Torque on the Shaft.Shaft: Transmit the Torque to the GeneratorGenerator: Creates an equal and opposite Torque.

Torsion: twisting of a straight bar when loaded by moments (or torques)Moments or torques tend to produce rotation about the longitudinal axis of the bar (twisting).

Couple: the pair of forces that tends to twist the bar about its longitudinal axis.

Moment of a couple:T (Nm)= Force (N) x Arm (m) or T (lb in.)= Force (lb) x Arm (in.)

Representation of the moment of a couple: a vector with a double-headed arrow

Then we can write the moments as

1 1 1T P d= 2 2 2T P d=

Net Torque Due to Internal Stresses

( )T dF dAρ ρτ= =∫ ∫

Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque,Although the net torque due to the shearing stresses is known, the distribution of the stresses is not.

Distribution of shearing stresses is statically indeterminate –must consider shaft deformations.Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed to be uniform.

Torque applied to shaft produces shearing stresses on the faces perpendicular to the axis.Conditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft.The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

Axial Shear Components

TL

φφ

∝∝

From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.

When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted then the bar is said to be under pure torsion.Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric.Cross-sections of noncircular (non-axisymmetric) shafts are distorted when subjected to torsion.

Torsional Deformations of Cylindrical Bars

Consider a cylindrical bar of circular cross section twisted by the torques T at both the ends.Since every cross section of the bar is symmetrical, we say that the bar is in pure torsion.Under action of torque T the right end of the bar will rotate through small angle φ known as the angle of twist.The angle of twist varies along the axis of the bar at intermediate cross section denoted by ( )xφ

Torsional Deformations of a Circular Bar

ddxφθ =

maxrrLφγ θ= =

|

maxbb rd rab dx

φγ θ= = =

The rate of twist or angle of twist per unit length

Shear Strain at the outer surface of the bar

For pure torsion the rate of twist is constant and equal to the total angle of twist φ divided by the length L of the bar

For Linear Elastic MaterialsFrom Hooke’s Law

G is shear modulus of elasticity and γ is shear strain

From Shear Strain equation :

Shear Stress at the outer surface of the bar :

Torsion Formula : To determine the relationship between shear stresses and torque, torsional formula is to be accomplished.

Shear Stresses in cylindrical bar with circular cross section.

max G rτ θ=

Gτ γ=

maxrrLφγ θ= =

Longitudinal and transverse shear stresses in a circular bar subjected to torsion.

Torsional Formula

Since the stresses act continously they have a resultant in the form of moment.The Moment of a small element dA located at radial distance ρ and is given by The resultant moment ( torque T ) is the summation over the entire cross sectional area of all such elemental moments.

Polar moment of inertia of circle with radius r and diameter d

Maximum Shear Stress

Distribution of stresses acting on a cross section.

4 4

2 32pr dI π π

= = max 3

16

p

Tr TI d

τπ

= =

Ar

AM Max δτρρτδδ 2==

PolarMax I

Tr

ρτρτ ==

Generalized torsion formula

2max maxp

A A

T dM dA Ir r

τ τρ= = =∫ ∫∫= APolar AI δρ 2

PolarITρτ = Generalized

torsion formula

PP GITL

LGIT

G=⇒==⇒= φρφργτγ

Circular tube in torsion.

Total angle of twist

Lφθ =

Angle of twist per unit length

Equations derived above are only for bars of circular cross sections

A solid steel bar of circular cross section has diameter d=1.5in and a length of l=54in.The bar is subjected to torques T acting at the ends if the torques have magnitude T=250 lb-ft . G=11.5x106psia) what is the maximum shear stress in the bar b) what is the angle of twist between the ends?

a) From torsional formula

b) Angle of twist 4 4

4*1.5 .497032 32pdI inπ π

= = =

max 3 3

16 16*250*12 4530*1.5

T psid

τπ π

= = =

( )( )( )( ) rad

inpsiininftlb

GITL

P

02834.04970.0105.11

541225046 =

××−

==φ

Compare the solid and hollow shafts

468.027151273

14.18.581.67

2

2

===

===

mmmm

AAth unit lengWeight per

mmmm

ddSize

solid

hollow

solid

hollow

Hollow shafts vs Solid shafts

PMax I

TR=τ

PGIT

=θ

strength

Angle of twist per unit length

Both are proportional to PI

1

Weight is proportional to the cross-sectional area (A)

Non-Uniform Torsion Bars consisting of prismatic segments with constant torque throughout each segment.

Non-Uniform Torsion Bars with continuously varying cross sections and constant torque.

Bars with continuously varying cross sections and continuously varying torque.

A shaft/gear assembly

Shaft is driven by a gear at C.

Gears at B and D are driven by the shaft. It turns freely at A and E

mNTmNTmNT

mmd

.175

.275.450

30

3

1

2

====

GPaGmmLmmL

CD

BC

80400500

===

Determine the maximum shear stress and the angle of twist between gears B and D.

mNmNmNTCD ⋅=⋅+⋅−= 175450275

( ) ( )( )

MPam

mmN

I

dT

p

CDCD 0.33

32030.0

2030.0.1752

4 ===π

τ

( )( )( ) rad

mPammN

IGLT

P

CDCDCD 0110.0

1095.710804.0.175

489 =××

=××

= −φ

mNTBC ⋅−= 275

( ) ( )MPa

m

mmN

I

dT

p

BCBC 9.51

1095.72

030.0.275248 −=

×

−== −τ

( )( )( ) rad

mPammN

IGLT

P

BCBCBC 0216.0

1095.710805.0.275

489 −=××

−=

××

= −φ

oCDBCBD rad 61.00106.0011.00216.0 −=−=+−=+= φφφ

A tapered bar in torsion

Change of diameter with length:

Evaluate the integral of this type of equation( ) ( )34 3

1bxabbxa

x+

−=+∫δ

Lddbda AB

A−

== ..........where

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−= 33

113

32BAAB ddddG

TLπ

φ

Stresses and Strains in Pure Shear

Analysis of stresses on inclined planes: (a) element in pure shear, (b) stresses acting on a triangular stress element, and (c) forces acting on the triangular stress element (Free-body diagram).

Stress elements oriented at θ = 0 and θ= 45° for pure shear.Graph of normal stresses σθ and shear stresses

τθ versus angle θ of the inclined plane.

Torsion FailureIn torsion, a ductile material will break along the plane of maximum shear, that is, a plane perpendicular to the shaft axis.A brittle material, will break along planes whose normal direction coincide with maximum tension, that is, along surfaces at 45o to the shaft.Note: Brittle materials are weaker in tension than in shear. Ductile materials generally fail in shear.

Torsion failure of a brittle material by tension cracking along a 45° helical surface.

Stress elements oriented at θ = 0 and θ = 45° for pure shear.

Strains in pure shear: (a) shear distortion of an element oriented at θ = 0, and (b) distortion of an element oriented at θ = 45°.

Strains in Pure ShearThe shear strain γ is the change in angle between two lines that were originally perpendicular to each other. The element changes its shape while the volume is maintained constant (shear distortion). For a elastic material:

( )ν

τγ

+=

=

12EG

GIf we rotate the element 45o, there will be no shear stresses and the normal stresses will be maximum. When inclined 45o, the elongation is given by:

( ) ( ) 2121 0045

45

ooo

o GE===

==

+=+= θθθ

θ

γν

τν

σε

(a) Determine the maximum shear, tensile and compressive stresses in the tube.

( )( )( ) ( )[ ]

MPamm

mmNITr

PMax 2.58

060.0080.032

080.0.400044=

−== πτ

The maximum tensile stress occurs at an inclination of 45ocw: MPa

MPa

ncompressio

tension

2.582.58−=

=σσ

Aluminum tube with G=27GPa

(b) Determine the maximum strains

radrad

radMPa

MPaGMax

Max

MaxMax

0011.02

0022.02

0022.027000

2.58

===

===

γε

τγ

Relationship Between E and G.

Before shear the distance bd is equal to: 2hLbd =After shear the distance bd is equal to: ( )Maxbd hL ε+= 12

Using the Law of Cosines: ⎟⎠⎞

⎜⎝⎛ +−+= γπ

2..2 2222 CoshhhLbd

( ) ⎟⎠⎞

⎜⎝⎛ +−=+ γπε

211 2 CosMax

( )

γγπεεε

SinCos

MaxMaxMax

−=⎟⎠⎞

⎜⎝⎛ +

++=+

2

211 22

As the ε and γ are very small, the equation is reduced to:

2

211

γε

γε

=

+=+

Max

Max

Substituting for torsion

( ) ( )ντνσε

τγ

+=+=

=

11EE

G

Max( )ν+=12EG

Transmission of Power by Circular ShaftsThe most important use of circular shafts is to transmit mechanical power from one device or machine to another. The work done by a torque of constant magnitude is equal to the product of the torque (T) and the angle through which it rotates (ψ).

ψTW = Power is the rate at which work is done (ω=rad/s): ωψ T

dtdT

dtdWP ===

)()...().....(...000,33

2

)()...()..(....60

2

hpHftlbTrpmnnTH

wattsPmNTrpmnnTP

−=

−=

π

π

Steel shaft in torsion that transmit 40hp to gear B. The allowable stress in the steel is 6000psi.

(a) Find the diameter (d) of the shaft if it operates at 500rpm.

inlbrpm

hpn

HT −=×

×== 5042

500240000,33

2000,33

ππ

ind

inpsi

inlbTddT

allowMax

62.1

280.46000

5042161616 333

=

=×

−×==⇒=

ππτπτ

The diameter of the shaft must be larger than 1.62in if the allowable shear stress is not to be exceeded.

Solid steel shaft in torsion. Motor transmits 50kW to the shaft ABC of 50mm diameter at 10Hz. The gears at B and C extract 35kW and 15kW respectively.Calculate the maximum shear stress in the shaft and the angle of twist (φAC) between the motor and the gear C. Use G=80GPa.

( )

( )

( ) mNHz

wattsf

PT

mNHz

wattsf

PT

mNHz

wattsf

PTfTP

C

B

A

.239102

150002

.557102

350002

.796102

500002

2

===

===

===⇒=

ππ

ππ

πππ

( )

( ) ( )rad

mPa

mmNGI

LT

MPam

mNdT

Polar

ABABAB

ABAB

0162.0050.0

321080

0.1.796

4.32050.0

.7961616

49

33

=⎟⎠⎞

⎜⎝⎛×

×==

=×

==

πφ

ππτ

( )

( ) ( )rad

mPa

mmN

MPam

mN

BC

BC

0058.0050.0

3280000000

2.1.239

7.9050.0

.23916

4

3

=⎟⎠⎞

⎜⎝⎛×

×=

=×

=

πφ

πτ

Maximum shear stress = 32.4MPa.

oAC

BCABAC

rad 26.1022.00058.00162.0 ==+=

+=

φ

φφφ

Statically Indeterminate Torsional Members

The equilibrium equations are not enough for determining the torque. The equilibrium equations need to be supplemented with compatibility equations pertaining to the rotational displacements.

First step : Write the equilibrium equations

21 TTT +=

Second Step : Formulate equations of compatibility. 21 φφ =

Third Step : Relate the angles of twist to the torques by torque-displacement relationships.

22

22

11

11

PP IGLT

IGLT

== φφ

Solving:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=2211

222

2211

111

PP

P

PP

P

IGIGIGTT

IGIGIGTT

Bar ABC is fixed at both ends and loaded at C by a torque TO.Find the reactive torques TA and TB., the maximum shear stresses and the angle of rotation.

Equilibrium equations: OBA TTT =+

Compatibility equations

Separate the bar from the support at the end B. The torque TO produces an angle of twist φ1. Then apply the reactive torque TB. It alone produces an angle of twist φ2.

021 =+φφ

Torque-Displacement Equations:PB

BB

PA

AB

PA

AO

GILT

GILT

GILT

−−== 21 φφ

The minus sign appears because TB produces a rotation that is opposite in direction to the positive direction of φ2

Substitute in the equation: 021 =+φφ

0=−−PB

BB

PA

AB

PA

AO

GILT

GILT

GILT

Solution:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=PBAPAB

PBAOB

PBAPAB

PABOA ILIL

ILTTILIL

ILTT

Maximum shear stresses

( ) ( )PBAPAB

BAOCB

PBAPAB

ABOAC

PB

BBCB

PA

AAAC

ILILdLT

ILILdLT

IdT

IdT

+=

+=

==

2

2

2

2

ττ

ττ

Angle of Rotation: The angle of rotation φC at section C is equal to the angle of twist of either segment of the bar.

( )PBAPAB

BAO

PB

BB

PA

AAC ILILG

LLTGI

LTGI

LT+

===φ

In the special case that the bar is prismatic LLTT

LLTT AO

BBO

A ==

( ) ( )ABP

AOCB

ABP

BOAC

P

BCB

P

AAC

LLIdLT

LLIdLT

IdT

IdT

+=

+=

==

2

2

2

2

ττ

ττ

( )ABP

BAO

P

BB

P

AAC LLGI

LLTGI

LTGI

LT+

===φ

Strain Energy in Torsion and Pure Shear

The work W done by the torque as it rotates through the angle φ is equal to the area below the torque-rotation diagram.

2φTWU ==

LGI

GILTU

GITL P

PP 22

22 φφ ===

Non-uniform torsion

( )∑∑==

==n

i iPi

iin

ii IG

LTUU1

2

1 2

The total energy of the bar is obtained by adding the strain energy of each segment

If either the cross section of the bar or the torque varies along the axis, then

( )[ ]

( )[ ]∫=

=

L

p

p

xGIdxxTU

xGIdxxTdU

0

2

2

)(2

)(2

Strain Energy Density in Pure ShearConsider an element of height h and thickness t. Then the forces V acting on each face is :

htV τ=The displacement ( δ )produced is hγδ =

2

22thU

VWU

τγ

δ

=

== Strain energy per unit volume:

2τγ

=u

2

22

2

γ

τ

Gu

Gu

=

=

Prismatic bar loaded with TA and TBsimultaneously

431052.7980

6.1.150.100

mmIGPaGmL

mNTmNT

P

B

A

×=

====

( )( ) ( ) ( )

P

B

P

BA

P

A

P

BA

P

An

i iPi

ii

GILT

GILTT

GILT

GI

LTTGI

LTIGLTU

22222

22

2

2222

1

2

++=+

+==∑=

When both loads act on the bar, the torque in segment CB is TA and in segment AC is (TA+TB)

( ) ( )( )( )

( )( )( )( )( )

( ) ( )( )( )43

2

4343

2

1052.798026.1.150

1052.798026.1.150.100

1052.798026.1.100

mmGPammN

mmGPammNmN

mmGPammNU

×+

×+

×=

JJJJU 56.441.189.126.1 =++=

Prismatic bar fixed at one end and loaded with a distributed torque of constant intensity t per unit distance.

4

6

18.17105.11

12/480

inIpsiG

ftLininlbt

P =

×=

=−=

txxT =)(

( )[ ] ( )

( ) ( )( )( ) lbin

inpsiinininlb

GILtU

GILtdxtx

GIGIdxxTU

P

P

L

P

L

p

−=×−

==

=== ∫∫

58018.17105.116

144/4806

621

2

46

3232

32

0

2

0

2

Thin-Walled Tubes under torsionStresses acting on the longitudinal faces ab and cd produce forces

dxtFdxtF cccbbb ττ ==

From equilibrium:ccbb tt ττ =

tconsftflowshear tan_ ===τ

m

L

fArdsfT

rfdsdTM

20

==

=

∫

Where LM denotes the length of the median line and AM is the area enclosed by the median line of the cross section.

MtAT

2=τ

The area enclosed by the median line is not the cross section area.

For a rectangular section:

bhtTbht

TbhA

horiz

vert

M

2

1

2

2

=

=

=

τ

τ

Area enclosed by the median line

trT

rAM

2

2

2πτ

π

=

=

Angle of Twist ( ) ( )trGTL

trGLTTUW 33

2

2222 πφ

πφ

=⇒===

Strain Energy of a Thin Walled TubeFind the strain energy of an element and then integrating throughout the volume of the bar.

dxt

dsGfdx

tds

Gttdsdx

GdU

222

2222

===ττ

Where f is the shear flow constant

∫∫∫∫∫ ==== MMM L

M

LLL

tds

GALT

tds

GLfdx

tds

GfdUU

02

2

0

2

00

2

822

Introducing a new property of cross section called the Torsion Constant

∫=

MLM

tds

AJ

0

24

GJLTU

2

2

=

For a constant thickness:M

M

LtAJ

24= 2

2

rA

rL

M

M

πFor a circular tube of thickness t

π

=

=

trJ 32π=

Comparison of circular and square tubes.

Circular Tube

rtAtrJ

rAM

ππ

π

22

1

31

21

==

=

Square Tube

rtbtA

trtbJ

rbAM

π

π

π

248

4

2

333

2

222

2

==

==

==

Both tubes have the same length, same wall thickness and same cross sectional area :

224 rbrtbt ππ =⇒=

Ratios of circular tube over square tube

79.04

42

22

1

2

2

1 ====π

π

π

ττ

r

r

AA

M

M

Stresses

Angles of twist 62.042

82

3

33

1

2

2

1 =⎟⎠⎞

⎜⎝⎛===π

π

π

φφ

tr

tr

JJ

The circular tube not only has 21% lower shear stress than does the square tube but it also has a greater stiffness against rotation.

Stress Concentration in Torsion

Stepped shaft in torsion.

The effects of stress concentration are confined to a small region around the discontinuity.

⎟⎟⎠

⎞⎜⎜⎝

⎛=== 3

1

16DTK

ITrKK

PNomMax π

ττ

stressshearalnoNom __min1 =

K = Stress Concentration Factor

=ττ

Stress-concentration factor K for a stepped shaft in torsion. (The dashed line is for a full quarter-circular fillet.)