EMA 3702 Mechanics & Materials Science (Mechanics of ...€¦ · Stresses in a Shaft under Torsion...

EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 3 Torsion

Introduction

Stress and strain in components subjected to torque T

Circular

Non-circular

Irregular shape

Elastic

Elastoplastic

Solid

Hollow

Material

Shaft design

Cross-section shape

Introduction

Component subject to twisting couples, or torques of T & T ’

• T is a vector and has two ways of representations

Example:

Transmission of torque

in shafts

Curved arrow Couple vector

(right hand rule)

Stresses in a Shaft under Torsion (1)

dF = dA Define as (local) lever arm,

i.e., the perpendicular distance

from the elemental area to the

axis (center), total torque T is:

A shaft subjected to equal

and opposite torques (or

moments of force) of T and

T’ at A and B

In a normal cross-section at

C, for an elementary area dA,

the element shearing force

dF and local stress τ satisfy:

TdF

Stresses in a Shaft under Torsion (2)

dF = dA

Complications for torsion:

• Distributions of and the resulting , i.e., how they change

over the cross-section plane is statically indeterminate.

• Unlike normal stress or simple shearing, distribution of for

torsion is NOT uniform!

Therefore,

From previous

TdF

TdA)(

Consider a small element as

illustrated. Based on previous

considerations for shearing

stress, if xz 0, axial shearing

stress

Implication: under torsion,

shearing stress exists along longitude planes, and neighboring

elements have tendency to slide against each other along axial

direction!

Axial Shearing Stress in Shaft under Torsion

y

x z zx = xz 0

zx xz

• Cross-sections remain planar

Axi-symmetric Property

of a Circular Shaft

• Cross-section remain undistorted

For circular shaft under torsion

L

Along axial direction, there is strain

(deformation) due to axial shear zx

Shearing Strain in Circular

Shaft under Torsion

Define the following terms

L Length along shaft axis

Radial distance from the shaft axis

Angle of twist for a cross-section

Shearing strain (change in angle

from 90o)

L

Within each cross-section, NO change

Far away from location of loading

& for small strain and angle

For a given L, when = c , i.e.,

radius of the shaft

Shearing Strain in Circular

Shaft under Torsion

L

Additionally, for a given L

c

max

L

c max

max

c

Shearing Stresses in Circular Shaft under

Torsion Hooke’s Law for shear

Therefore,

shearing stress

For given L and ϕ, when = c = radius of

the shaft, shearing stress reaches maximum:

From previous page,

torsion shearing strain

Distribution of shearing

stress is linear w.r.t.

radius from the axis ρ c

max

L

GcG

maxmax

The ratio between τ and τmax

L

L

G

G

Use when shaft twist angle, length,

AND materials G are known

Shearing Stresses & Torque in Circular

Shaft under Torsion

Recall relationship of T and

local shearing stress τ

max

c

For a circular shaft with fixed radius c under torque T

Recall the definition for moment of inertia:

We have Or J

Tcmax

J

T and

TdA)(

dAc

dAc

dAT 2maxmax)(

dAJ 2

c

JT max

Use when shaft geometry AND applied torque known

For a solid cylinder For a hollow cylinder

Solid & Hollow Circular Shaft under Torsion

J

T

2

1

max

min

c

c

J

Tcmax

4

2

1cJ

)(2

1 4

1

4

2 ccJ

J

T

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 3 Torsion

Example Problem for Torsion 3.1

Shaft ABCD subject to torques as

illustrated. Knowing section BC is

hollow with ID = 90 mm and OD = 120

mm. Sections AB and CD are solid.

Calculate

(a) The maximum and minimum

shearing stress in section BC;

(b) The required minimal diameter for

AB and CD if shearing stress

should not exceed 65 MPa.


Section AB,

Net internal torque TAB =

Section BC,

Net internal torque TBC =


6 kNm

(6 +14) kNm = 20 kNm



Min stress in BC occurs at the inner surface

BC

BC

BC

J

cToutermax

MPa

m

mmN

J

cT

BC

BC

innerBC 7.64

045.006.01416.32

1

045.01020

444

3

min

For section BC, TBC= 20 kNm

Max stress in BC occurs at the outer surface

“section BC is hollow with ID = 90

mm and OD = 120 mm.”

MPa

m

mmN2.86

045.006.01416.32

1

06.01020

444

3



Section AB, TAB = 6 kNm

Section BC, TBC = 20 kNm

Section CD, TCD = 6 kNm

For both AB and CD, “shearing stress

should not exceed 65 MPa”

Minimum radius for AB or CD:

Minimum diameter = 7.78 cm

MPa

c

T

J

cT65

2

1 3max

cm 89.3

/10651416.32

1

106

2

1

3/1

26

3

3/1

max

mN

mNTc


Class Example

A hollow cylinder shaft is 1 m long and has inner and

outer diameter of 20 and 40 mm. (a) What is the largest

torque that can be applied if shearing stress should

not exceed 100 MPa? (b) What is the minimum

shearing stress when maximum reaching 100 MPa?

MPaJ

Tcouter 100max

474444 1035.201.002.01416.32

1

2

1mccJ innerouter

“shearing stress should not

exceed 100 MPa”:

mNm

mNm

c

MPaJT

out

117502.0

/101001035.2100 2647

maxmin outer

inner

c

c

Largest torque that can be applied:

From geometry, moment of inertia

Min shearing stress when

max is 100 MPa MPaMPa 50100

02.0

01.0

Normal Stress in Circular Shaft under

Torsion

J

Tcmax

0max AFF BDBC

To balance, force on CD surface FCD must be normal force

0max245cos/ AFF BCCD

Since 02AACD

Normal stress

due to torsion J

Tc

A

F

CD

CD max

For element a at shaft surface FCD

For half of a at 45o, shear force along BC and BD are:

For torsion, significant

normal stress still exists

and may cause failure!

• Ductile materials are

weaker in shear and

fail with 90o fracture

(i.e., fracture surface

perpendicular to axis)

• Brittle materials are

weaker in tension and

fail with 45o fracture

(i.e., fracture surface

at 45o from axis)

Failure of Material under Torsion

Photo by Jeff Thomas, 11/1997

http://classes.mst.edu/civeng120/lessons/torsion/fracture/index.html

For circular shaft under

torsion

Brittle torsion failure

Ductile torsion failure

Angle of Twist within

Elastic Range

For torsion, based on geometry

Angle of twist

increases with

• Increasing T and L

• Decreasing J and G

Within elastic limit, Hooke’s law

L

c max

G

maxmax

J

Tcmax

JG

TcmaxMax sharing stress

due to torque

JG

TL

JG

T

L

Angle of Twist for Multiple Cross-Section

Shafts

in which Ti, Ji, and Li, Gi are

obtained and/or analyzed for each section

Angle of Twist for Variable

Cross-Section Shafts

Overall twist of A w.r.t B

i ii

ii

GJ

LT

For element dx Overall

JG

Tdxd

L

JG

Tdx

0


Class Exercise

A cylindrical shaft is 0.5 m long and has diameter

of 20 mm fixed at one end. If a torque of 76 Nm is

applied to its free end, and the measured angle

of twist at the loading end is 1.8o, please

estimate the shear modulus knowing moment of

inertia is

Angle of twist JG

TL

GPa

m

mmN

J

TLG 77

1416.3180

8.101.01416.3

2

1

5.076

44

Shear modulus G

4

2

1cJ

Example of Statically Indeterminate Shaft (1)

A cylindrical shaft AB with

two sections is fixed between

two rigid end support at A and

B. Section AC is solid with

diameter of 1.0 in;

Section CB is hollow with outside diameter of 1.0 in

and inner diameter of 0.8 in. Both sections have length of

10 in. If an external torque of 100 lbft is applied at center

C, please determine the reaction torque at A and B,

assuming the deformation is within proportional limit.

A C B

10 in 10 in

Example of Statically Indeterminate Shaft (2) From statics, balance of torque

One equation – two variables

Statically indeterminate

Consider geometry/deformation:

For section AC, angle of twist AC

For section CB, angle of twist CB

inlbTTT BA 100

CB

CB

CBB

AC

ACAAC

GJ

LT

GJ

LT

Therefore, CBAC

CB

B

AC

A

J

T

J

T

Two equations & two variables, problem could be solved

CBAC

100

Example of Statically Indeterminate Shaft (3) Therefore,

100 BA TT

Solving these two

444 4.05.02

15.0

2

1

BA TT

369625

BA TT

ftlbTB 1.37

ftlbTA 9.62

Design of Transmission Shafts (1) Two parameters in transmission shaft design:

• Power P

• Speed of rotation

For power:

is angular velocity, = 2, and has unit of radians/s

is frequency in unit Hz or s-1

has unit of Nm/s = W

As a result, for given P and f, the resulting torque will be

TP

Therefore, power

fTP 2

f

PT

2

Example of a solid circular shaft, moment of inertia:

Design of Transmission Shafts (2) Relationship between power output P, angular speed (or

frequency f), and torque T

On the other hand, max shearing stress for a given shaft

geometry and applied torque J

f

PT

2

J

Tcmax

max

T

c

J

4

2

1cJ 3

2

1c

c

J

3/1

max_

2

allowable

Tc

max

T


Class Exercise

What size of solid shaft should be used for motor of 592 hp (1 hp = 6600

lbin/s) operating at 3600 rpm (f = 60 Hz) if the shearing stress is not to

exceed 6600 psi in the shaft?

Maximum shearing stress

Power for the shaft fTP 2

J

Tcmax

3/1

max_

2

allowable

Tc

Torque for the shaft inlbs

sinlb

f

PT

10382601416.32

6600592

2 1

1

Minimum shaft radius:

4

2

1c

Tc

3

2

c

T

2/6600 inlb

in 00.1/66001416.3

1038223/1

2

inlb

inlb

Stress Concentrations in Circular Shafts

Similar to normal stress, shearing stress may concentrate

(i.e., show higher value) at certain locations

Concentration factor K

Local max stress increases with

• Sharper transition or joining

• Larger ratio of radius

J

TcKactual max_

JTcK

actual

/

max_

or

Plastic Deformation in Circular Shafts (1)

For a given shaft construction

(geometry and material), when

torque is low and within linear

elastic region, linear stress

distribution w.r.t. radius from axis

TJ

T

As torque increases further, shearing

stress would reach yield strength Y,

it will go into non-linear (e.g., ideal

elasoplastic) distribution of w.r.t.

radial from axis in certain region

TJG

TL

For angle of twist

When torque/shearing stress

is low and within linear elastic

region, increases linearly

with torque T

Plastic Deformation in Circular Shafts (2)

As torque increases further, shearing stress reaches yield

strength Y, it will go into non-linear distribution of w.r.t.

T

Torsion of Noncircular Members

A rectangular or square shaft

does not contain axi-symmetry

• The cross-sections generally do

NOT remain flat or planar

under torsion

• Shearing stress at the edge

(corner) of the shaft is zero

• Maximum shearing stress

occur in the middle of flat face

Thin-Walled Hollow Shafts

For arbitrary shaped thin walled

hollow tube subjected to torque

T,

the shearing stress can be

approximated as

τ =𝑇

2𝑡𝕒

𝕒


Homework 3.0

Read chapter 3 textbook sections 3.1 to 3.5 and give an

honor statement confirm reading.


Homework 3.1

Electric motor applies a torque of 2.8 kN•m on shaft EF at E. Knowing

each shaft section is solid. Based on the additional torques as

depicted, please determine the maximum shearing stress in (a) shaft

EF (radius cEF = 25 mm), (b) shaft FG (radius cFG=23 mm), and c) shaft

GH (radius cGH=21 mm)

E

E F G H

TF = 1.6 kN•m TG = 0.8 kN•m TH = 0.4 kN•m


Homework 3.2

A component with both solid rod EF and

hollow tube with flange GH welded onto a flat

end plate of GFI, and the rod EF and tube GH

share the same longitude axis, as illustrated.

The hollow tube GH is fixed at rigid flange of

AA’BB’ at H using bolts. The solid rod EF has

a diameter dEF = 3.81 cm and is made of steel

with allowable shearing stress of 82.7 MPa.

Tube GH is made of copper alloy with OD

d2=7.62 cm and ID d1=6.35 cm with allowable

shearing stress of 48.3 MPa. Determine the

largest torque T that can be applied at E.

E

F

H

G

T

I

A A’ B B’


Homework 3.3

Two solid shafts GH and IJ are connected by gears K and L as

illustrated. The radius for gear K is RK = 4 in, while radius for gear L is

RL= 2.5 in. Both two shafts of GH and IJ are made of same steel with

allowable shearing stress of 8 ksi. The radius for the shafts are cGH =

0.8 in and cIJ = 0.625 in. Please determine the largest torque that can

be applied at H.

G H

RK = 4.0 in

I J

RL = 2.5 in

K

L


Homework 3.4

Motor applies torque T = 500 N•m on the HGFE shaft when it is rotating

at a constant speed. Knowing shear modulus G = 27 GPa and the

torques exerted on pulleys B and C are shown. Please calculate the

angle of twist between (a) F and G, and (b) between F and H. Knowing

shaft radius cGF = 2.2 cm; cHG = 2.4 cm; cEF = 2.0 cm;

H G F E

TG = 300 N•m TF= 200 N•m Motor

T = 500 N•m

1.2 m 0.9 m


Homework 3.5

Solid aluminum rod (GAl=26 GPa) is bonded to solid copper alloy rod

(GAl=39 GPa). The radius for both rods are 10 mm. Calculate the angle

of twist with respect to base D at E point and at F point, respectively.

D E

TF = 50 N•m

0.4 m 0.2 m

F copper aluminum


Homework 3.6

A 6 foot long composite shaft consists of 0.2 inch thick copper shell (GCu

= 5.6 106 psi) bonded to 1.2 inch (GSteel = 11.2 106 psi) diameter iron

core. If the shaft is subject to 5000 lb•in torque. Please calculate the

maximum shearing stress in the steel core and the angle of twist of one

end versus the other.

T

Cu shell

Fe core

6 ft


Homework 3.7

Determine maximum shearing stress in a solid shaft of 10 mm diameter

as it transmits 2.4 kW at a frequency of (a) 30 Hz and (b) 60 Hz.


Homework 3.8

Two solid shafts EF and IJ and gears G (RG = 3 inch) and H (RH = 5

inch) are used to transmit 16 hp (1 hp = 6600 lbin) from motor at E

operating at 1200 rpm (f = 20 Hz) to machine tool at J. Knowing the

maximum allowable shearing stress is 7.5 ksi. Please calculate the

required radius for shaft EF and shaft IJ, respectively.

I J

RH = 5.0 in

F E

RG = 3.0 in

H

G

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