fourth edition mechanics of materialsyunus.hacettepe.edu.tr/~boray/3_1_torsion...

MECHANICS OFFourth Edition

MECHANICS OF MATERIALS

CHAPTER

MATERIALSFerdinand P. BeerE. Russell Johnston, Jr.John T DeWolf

TorsionJohn T. DeWolf

Lecture Notes:J. Walt OlerTexas Tech University

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.


FourthEdition

Beer • Johnston • DeWolf

Contents

Introduction

T i l L d Ci l Sh ft

Statically Indeterminate Shafts

S l P bl 3 4Torsional Loads on Circular Shafts

Net Torque Due to Internal Stresses

A i l Sh C t

Sample Problem 3.4

Design of Transmission Shafts

St C t tiAxial Shear Components

Shaft Deformations

Sh i St i

Stress Concentrations

Plastic Deformations

El t l ti M t i lShearing Strain

Stresses in Elastic Range

N l St

Elastoplastic Materials

Residual Stresses

E l 3 08/3 09Normal Stresses

Torsional Failure Modes

S l P bl 3 1

Example 3.08/3.09

Torsion of Noncircular Members

Thi W ll d H ll Sh ftSample Problem 3.1

Angle of Twist in Elastic Range

Thin-Walled Hollow Shafts

Example 3.10

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 2


FourthEdition


Torsional Loads on Circular Shafts

• Interested in stresses and strains of circular shafts subjected to twistingcircular shafts subjected to twisting couples or torques

• Turbine exerts torque T on the shaft

• Shaft transmits the torque to the generator

• Turbine exerts torque T on the shaft

• Generator creates an equal and opposite torque T’

generator

opposite torque T



FourthEdition


Net Torque Due to Internal Stresses

• Net of the internal shearing stresses is an

( )∫ ∫== dAdFT τρρ

internal torque, equal and opposite to the applied torque,

( )∫ ∫== dAdFT τρρ

• Although the net torque due to the shearing stresses is known, the distribution of the stressesstresses is known, the distribution of the stresses is not.

• Distribution of shearing stresses is statically g yindeterminate – 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 uniform.


loads can not be assumed uniform.


FourthEdition


Axial Shear Components

• Torque applied to shaft produces shearing stresses on the faces perpendicular to thestresses on the faces perpendicular to the axis.

C di i f ilib i i h• 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 y g pof axial slats.

• The slats slide with respect to each other• The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.



FourthEdition


Shaft Deformations

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

T∝φ

L

T

∝

∝

φ

φ

• When subjected to torsion, every cross-sectionWhen subjected to torsion, every cross section of a circular shaft remains plane and undistorted.

• 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

bj t d t t i

y


subjected to torsion.


FourthEdition


Shearing Strain

• Consider an interior section of the shaft. As a torsional load is applied an element on thetorsional load is applied, an element on the interior cylinder deforms into a rhombus.

Si th d f th l t i l• Since the ends of the element remain planar, the shear strain is equal to angle of twist.

LL ρφγρφγ == or

• It follows that

• Shear strain is proportional to twist and radius

d ρφcmaxmax and γργφγ

cL==



FourthEdition


Stresses in Elastic Range• Multiplying the previous equation by the

shear modulus,γργ GG maxγργ G

cG =

ρFrom Hooke’s Law, γτ G= , so

421 cJ π=

maxτρτc

=

The shearing stress varies linearly with the radial position in the section

• Recall that the sum of the moments from the internal stress distribution is equal to

2 radial position in the section.

Jc

dAc

dAT max2max τρτρτ ∫ =∫ ==

qthe torque on the shaft at the section,

cc

( )41

422

1 ccJ −= πTTc ρ

• The results are known as the elastic torsion formulas,


( )2 and max J

TJ

Tc ρττ ==


FourthEdition


Normal Stresses• Elements with faces parallel and perpendicular

to the shaft axis are subjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found for other orientations.

( ) 0max0max 245cos2 ττ =°= AAF

• Consider an element at 45o to the shaft axis,

max0

0max45 2

2o ττσ ===

AA

AF

• Element a is in pure shear. • Element c is subjected to a tensile stress on

two faces and compressive stress on the other

• Note that all stresses for elements a and c have th it d

two faces and compressive stress on the other two.


the same magnitude


FourthEdition


Torsional Failure Modes

• Ductile materials generally fail in g yshear. Brittle materials are weaker in tension than shear.

• When subjected to torsion, a ductile specimen breaks along a plane ofspecimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis.

• When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction inperpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft

i


axis.


FourthEdition


Sample Problem 3.1

SOLUTION:

• Cut sections through shafts AB• Cut sections through shafts ABand BC and perform static equilibrium analyses to find t l ditorque loadings.

• Apply elastic torsion formulas to find minimum and maximum

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, • Given allowable shearing stress

find minimum and maximum stress on shaft BC.

,respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum

gand applied torque, invert the elastic torsion formula to find the required diameter.determine (a) the minimum and maximum

shearing stress in shaft BC, (b) the required diameter d of shafts AB and CDif the allowable shearing stress in these

required diameter.


if the allowable shearing stress in these shafts is 65 MPa.


FourthEdition


Sample Problem 3.1SOLUTION:• Cut sections through shafts AB and BC

and perform static equilibrium analysis p q yto find torque loadings.

( ) TM∑ kN60 ( ) ( )kN14kN60∑ TM( )

CDAB

ABx

TT

TM

=⋅=

−⋅==∑mkN6

mkN60 ( ) ( )mkN20

mkN14mkN60

⋅=

−⋅+⋅==∑

BC

BCx

T

TM



FourthEdition


Sample Problem 3.1• Apply elastic torsion formulas to

find minimum and maximum stress on shaft BC.

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter.stress on shaft BC. formula to find the required diameter.

( ) ( ) ( )[ ]46

4441

42

m109213

045.0060.022

−×=

−=−=ππ ccJ mkN665 3

24

2max

⋅===

cMPa

cTc

JTc

ππτ

m1092.13 ×=

( )( )m1092.13

m060.0mkN2046

22max

×

⋅=== −J

cTBCττm109.38 3−×=c

mm8.772 == cd

MPa2.86=

mm60mm45

MPa286min

2

1min ==τ

ττ

cc

MPa286=τ


MPa7.64

mm60MPa2.86

min

2max

=τ

τ c

MPa7.64

MPa2.86

min

max

=

=

τ

τ


FourthEdition


Angle of Twist in Elastic Range• Recall that the angle of twist and maximum

shearing strain are related,cφL

cφγ =max

• In the elastic range, the shearing strain and shear l d b H k ’ Lare related by Hooke’s Law,

JGTc

G== max

maxτγ

• Equating the expressions for shearing strain and solving for the angle of twist,

TLφJG

=φ

• If the torsional loading or shaft cross-section changes along the length, the angle of rotation is g g g , gfound as the sum of segment rotations

∑=i ii

iiGJLTφ


i ii


FourthEdition


Statically Indeterminate Shafts

• Given the shaft dimensions and the applied torque, we would like to find the torque reactions q qat A and B.

• From a free-body analysis of the shaft,flb90TT

which is not sufficient to find the end torques. Th bl i t ti ll i d t i t

ftlb90 ⋅=+ BA TT

The problem is statically indeterminate.

• Divide the shaft into two components which must have compatible deformations

ABBA T

JLJLT

GJLT

GJLT

12

21

2

2

1

121 0 ==−=+= φφφ

must have compatible deformations,

ftlb9012

21 ⋅=+ AA TJLJLT

• Substitute into the original equilibrium equation,


12


FourthEdition


Sample Problem 3.4

SOLUTION:

• Apply a static equilibrium analysis on• Apply a static equilibrium analysis on the two shafts to find a relationship between TCD and T0 .

• Apply a kinematic analysis to relate the angular rotations of the gears.

Two solid steel shafts are connected by gears. Knowing that for each shaft • Find the corresponding angle of twist

• Find the maximum allowable torque on each shaft – choose the smallest.

y g gG = 11.2 x 106 psi and that the allowable shearing stress is 8 ksi, determine (a) the largest torque T0

• Find the corresponding angle of twist for each shaft and the net angular rotation of end A.determine (a) the largest torque T0

that may be applied to the end of shaft AB, (b) the corresponding angle through which end A of shaft AB


through which end A of shaft ABrotates.


FourthEdition


Sample Problem 3.4SOLUTION:

• Apply a static equilibrium analysis on • Apply a kinematic analysis to relate pp y q ythe two shafts to find a relationship between TCD and T0 .

pp y ythe angular rotations of the gears.

( )in87500 TFM∑CCBB rr φφ =

( )( )

0

0

82

in.45.20

in.875.00

TT

TFM

TFM

CD

CDC

B

=

−==

−==

∑

∑

CB

CCB

CB r

r

φφ

φφφ

82

in.875.0in.45.2

=

==


08.2 TTCD CB φφ 8.2=

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