Torsion of circular shafts presentation1 - libvolume3.xyzlibvolume3.xyz/.../torsionofcircularshaftspresentation1.pdf · Overview of Mechanical Engineering 3 - 3 Torsional Loads on

Overview of Mechanical Engineering

for Non-MEs

Part 2: Mechanics of Materials

.

9 Torsion


3 - 2

Contents

Introduction

Torsional Loads on Circular Shafts

Net Torque Due to Internal Stresses

Axial Shear Components

Shaft Deformations

Shearing Strain

Stresses in Elastic Range

Normal Stresses

Torsional Failure Modes

Sample Problem 3.1

Angle of Twist in Elastic Range

Statically Indeterminate Shafts

Sample Problem 3.4

Design of Transmission Shafts

Stress Concentrations

Plastic Deformations

Elastoplastic Materials

Residual Stresses

Example 3.08/3.09

Torsion of Noncircular Members

Thin-Walled Hollow Shafts

Example 3.10


3 - 3

Torsional Loads on Circular Shafts

• Interested in stresses and strains of

circular shafts subjected to twisting

couples or torques

• Generator creates an equal and

opposite torque T’ on the shaft

(action-reaction principle)

• Shaft transmits the torque to the

generator

• Turbine exerts torque T on the shaft


3 - 4

Net Torque Due to Internal Stresses

( )∫ ∫== dAdFT τρρ

• 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.

• Unlike the normal stress due to axial loads, the

distribution of shearing stresses due to torsional

loads can not be assumed uniform.

• Distribution of shearing stresses is statically

indeterminate – must consider shaft

deformations.


3 - 5

Axial Shear Components

• Torque applied to shaft produces shearing

stresses on the faces perpendicular to the

axis.

• The existence of the axial shear components is

demonstrated by considering a shaft made up

of axial slats.

• Conditions of equilibrium require the

existence of equal stresses on the faces of the

two planes containing the axis of the shaft.

• The slats slide with respect to each other

when equal and opposite torques are applied

to the ends of the shaft.


3 - 6

• From observation, the angle of twist of the

shaft is proportional to the applied torque and

to the shaft length.

L

T

∝

∝

φ

φ

Shaft Deformations

• When subjected to torsion, every cross-section

of a circular shaft remains plane and

undistorted.

• Cross-sections of noncircular (non-

axisymmetric) shafts are distorted when

subjected to torsion.

• Cross-sections for hollow and solid circular

shafts remain plain and undistorted because a

circular shaft is axisymmetric.


3 - 7

Shearing Strain

• Consider an interior section of the shaft. As a

torsional load is applied, an element on the

interior cylinder deforms into a rhombus.

• Shear strain is proportional to twist and radius

maxmax and γρ

γφ

γcL

c==

LLAAarch

ρφγρφγ === or )( '

• It follows that

• Since the ends of the element remain planar,

the shear strain is equal to angle of twist.


3 - 8

Stresses in Elastic Range

Jc

dAc

dAT max2max τρ

τρτ ∫ =∫ ==

• Recall that the sum of the moments from

the internal stress distribution is equal to

the torque on the shaft at the section,

4

21 cJ π=

( )

Ocenterinertiaofmoment

dAJ

ccJ

=

=

−=

∫ 2

4

1

4

22

1

ρ

π

and maxJ

T

J

Tc ρττ ==

• The results are known as the elastic torsion

formulas,

• Multiplying the previous equation by the

shear modulus,

maxγρ

γ Gc

G =

maxτρ

τc

=

From Hooke’s Law, γτ G= , so

The shearing stress varies linearly with the

radial position in the section.


3 - 9

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.

( )

max0

0max45

0max0max

2

2

245cos2

o ττ

σ

ττ

===

=°=

A

A

A

F

AAF

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

• Element a is in pure shear.

• Note that all stresses for elements a and c have

the same magnitude

• Element c is subjected to a tensile stress on

two faces and compressive stress on the other

two.


3 - 10

Torsional Failure Modes

• Ductile materials generally fail in

shear. Brittle materials are weaker in

tension than shear.

• When subjected to torsion, a ductile

specimen 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 in

which tension is a maximum, i.e.,

along surfaces at 45o to the shaft

axis.


3 - 11

Shaft BC is hollow with inner and outer

diameters of 90 mm and 120 mm,

respectively. Shafts AB and CD are solid

of diameter d. For the loading shown,

determine (a) the minimum and maximum

shearing stress in shaft BC, (b) the

required diameter d of shafts AB and CD

if the allowable shearing stress in these

shafts is 65 MPa.

Sample Problem 3.1

SOLUTION:

• Cut sections through shafts AB

and BC and perform static

equilibrium analyses to find

torque loadings.

• Given allowable shearing stress

and applied torque, invert the

elastic torsion formula to find the

required diameter.

• Apply elastic torsion formulas to

find minimum and maximum

stress on shaft BC.


3 - 12

SOLUTION:

• Cut sections through shafts AB and BC

and perform static equilibrium analysis

to find torque loadings.

( )

CDAB

ABx

TT

TM

=⋅=

−⋅==∑

mkN6

mkN60 ( ) ( )

mkN20

mkN14mkN60

⋅=

−⋅+⋅==∑

BC

BCx

T

TM

Sample Problem 3.1


3 - 13

• Apply elastic torsion formulas to

find minimum and maximum

stress on shaft BC.

( ) ( ) ( )[ ]46

4441

42

m1092.13

045.0060.022

−×=

−=−=ππ

ccJ

( )( )

MPa2.86

m1092.13

m060.0mkN2046

22max

=

×

⋅===

−J

cTBCττ

MPa7.64

mm60

mm45

MPa2.86

min

min

2

1

max

min

=

==

τ

τττ

c

c

MPa7.64

MPa2.86

min

max

=

=

τ

τ

• Given allowable shearing stress and

applied torque, invert the elastic torsion

formula to find the required diameter.

m109.38

mkN665

3

3

2

4

2

max

−×=

⋅===

c

cMPa

c

Tc

J

Tc

ππτ

mm8.772 == cd

Sample Problem 3.1


3 - 14

Angle of Twist in Elastic Range

• Recall that the angle of twist and maximum

shearing strain are related,

L

cφγ =max

• In the elastic range, the shearing strain and shear

are related by Hooke’s Law,

JG

Tc

G== max

maxτ

γ

• Equating the expressions for shearing strain and

solving for the angle of twist,

JG

TL=φ

• If the torsional loading or shaft cross-section

changes along the length, the angle of rotation is

found as the sum of segment rotations

∑=i ii

ii

GJ

LTφ


3 - 15

• Given the shaft dimensions and the applied

torque, we would like to find the torque reactions

at A and B.

Statically Indeterminate Shafts

• From a free-body analysis of the shaft,

which is not sufficient to find the end torques.

The problem is statically indeterminate.

ftlb90 ⋅=+ BA TT

ftlb9012

21 ⋅=+ AA TJL

JLT

• Substitute into the original equilibrium equation,

ABBA T

JL

JLT

GJ

LT

GJ

LT

12

21

2

2

1

121 0 ==−=+= φφφ

• Divide the shaft into two components which

must have compatible deformations,


3 - 16

Sample Problem 3.4

Two solid steel shafts are connected

by gears. Knowing that for each shaft

G = 11.2 x 106 psi and that the

allowable shearing stress is 8 ksi,

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

rotates.

SOLUTION:

• Apply a static equilibrium analysis on

the two shafts to find a relationship

between TCD and T0 .

• Find the corresponding angle of twist

for each shaft and the net angular

rotation of end A.

• Find the maximum allowable torque

on each shaft – choose the smallest.

• Apply a kinematic analysis to relate

the angular rotations of the gears.


3 - 17

SOLUTION:

• Apply a static equilibrium analysis on

the two shafts to find a relationship

between TCD and T0 .

( )

( )

0

0

8.2

in.45.20

in.875.00

TT

TFM

TFM

CD

CDC

B

=

−==

−==

∑

∑

• Apply a kinematic analysis to relate

the angular rotations of the gears.

CB

CCB

CB

CCBB

r

r

rr

φφ

φφφ

φφ

8.2

in.875.0

in.45.2

=

==

=

Sample Problem 3.4


3 - 18

• Find the T0 for the maximum

allowable torque on each shaft –

choose the smallest.

( )( )

( )( )

in.lb561

in.5.0

in.5.08.28000

in.lb663

in.375.0

in.375.08000

0

4

2

0max

0

4

2

0max

⋅=

==

⋅=

==

T

Tpsi

J

cT

T

Tpsi

J

cT

CD

CD

AB

AB

π

π

τ

τ

inlb5610 ⋅=T

• Find the corresponding angle of twist for each

shaft and the net angular rotation of end A.

( )( )( ) ( )

( )( )( ) ( )

( )oo

/

oo

o

64

2

/

o

64

2

/

2.2226.8

26.895.28.28.2

95.2rad514.0

psi102.11in.5.0

.in24in.lb5618.2

2.22rad387.0

psi102.11in.375.0

.in24in.lb561

+=+=

===

==

×

⋅==

==

×

⋅==

BABA

CB

CD

CDDC

AB

ABBA

GJ

LT

GJ

LT

φφφ

φφ

φ

φ

π

π

o48.10=Aφ

Sample Problem 3.4


3 - 19

Stress Concentrations

• The derivation of the torsion formula,

assumed a circular shaft with uniform

cross-section loaded through rigid end

plates.

J

Tc=maxτ

J

TcK=maxτ

• Experimental or numerically determined

concentration factors are applied as

• The use of flange couplings, gears and

pulleys attached to shafts by keys in

keyways, and cross-section discontinuities

can cause stress concentrations


3 - 20

Thin-Walled Hollow Shafts

• Summing forces in the x-direction on AB,

shear stress varies inversely with thickness

( ) ( )

flowshear

0

====

∆−∆==∑

qttt

xtxtF

BBAA

BBAAx

τττ

ττ

( ) ( )

tA

T

qAdAqdMT

dAqpdsqdstpdFpdM

2

22

2

0

0

=

===

====

∫∫

τ

τ

• Compute the shaft torque from the integral

of the moments due to shear stress

∫=t

ds

GA

TL2

4φ

• Angle of twist (from Chapter 11)


3 - 21

Example 3.10

Extruded aluminum tubing with a rectangular

cross-section has a torque loading of 24 kip-

in. Determine the shearing stress in each of

the four walls with (a) uniform wall thickness

of 0.160 in. and wall thicknesses of (b) 0.120

in. on AB and CD and 0.200 in. on CD and

BD.

SOLUTION:

• Determine the shear flow through the

tubing walls.

• Find the corresponding shearing stress

with each wall thickness .


3 - 22

SOLUTION:

• Determine the shear flow through the

tubing walls.

( )( )

( ) in.

kip335.1

in.986.82

in.-kip24

2

in.986.8in.34.2in.84.3

2

2

===

==

A

Tq

A

• Find the corresponding shearing

stress with each wall thickness.

With a uniform wall thickness,

in.160.0

in.kip335.1==t

qτ

ksi34.8=τ

With a variable wall thickness

in.120.0

in.kip335.1== ACAB ττ

in.200.0

in.kip335.1== CDBD ττ

ksi13.11== BCAB ττ

ksi68.6== CDBC ττ

Example 3.10

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Torsion of circular shafts presentation1 - libvolume3.xyzlibvolume3.xyz/.../torsionofcircularshaftspresentation1.pdf · Overview of Mechanical Engineering 3 - 3 Torsional Loads on