stress analisis & pressure vessels

5/24/2018 Stress Analisis & Pressure Vessels

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CET 1:Stress Analysis &

Pressure Vessels

Lent Term 2005

Dr. Clemens Kaminski

Telephone: +44 1223 763135

E-mail: [email protected]

URL: http://www.cheng.cam.ac.uk/research/groups/laser/


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Synopsis

1 Introduction to Pressure Vessels and Failure Modes1.1 Stresses in Cylinders and Spheres1.2 Compressive failure. Euler buckling. Vacuum vessels1.3 Tensile failure. Stress Stress Concentration & Cracking

2 3-D stress and strain2.1 Elasticity and Strains-Young's Modulus and Poisson's Ratio2.2 Bulk and Shear Moduli2.3 Hoop, Longitudinal and Volumetric Strains2.4 Strain Energy. Overfilling of Pressure Vessels

3 Thermal Effects3.1 Coefficient of Thermal Expansion3.2 Thermal Effect in cylindrical Pressure Vessels3.3 Two-Material Structures

4 Torsion.4.1 Shear Stresses in Shafts - /r = T/J = G/L4.2 Thin Walled Shafts4.3 Thin Walled Pressure Vessel subject to Torque

5 Two Dimensional Stress Analysis

5.1 Nomenclature and Sign Convention for Stresses5.2 Mohr's Circle for Stresses5.3 Worked Examples5.4 Application of Mohr's Circle to Three Dimensional Systems

6 Bulk Failure Criteria6.1 Tresca's Criterion. The Stress Hexagon6.2 Von Mises' Failure Criterion. The Stress Ellipse

7 Two Dimensional Strain Analysis7.1 Direct and Shear Strains

7.2 Mohr's Circle for Strains7.3 Measurement of Strain - Strain Gauges7.4 Hookes Law for Shear Stresses


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Supporting Materials

There is one Examples paper supporting these lectures.

Two good textbooks for further explanation, worked examples and exercises are

Mechanics of Materials(1997) Gere & Timoshenko, publ. ITP [ISBN 0-534-93429-3]

Mechanics of Solids(1989) Fenner, publ. Blackwell [ISBN 0-632-02018-0]

This material was taught in the CET I (Old Regulations) Structures lecture unit and was examinedin CET I (OR) Paper IV Section 1. There are consequently a large number of old Tripos questionsin existence, which are of the appropriate standard. From 1999 onwards the course was taught inCET1, paper 5. Chapters 7 and 8 in Gere and Timoshenko contain a large number of exampleproblems and questions.

Nomenclature

The following symbols will be used as consistently as possible in the lectures.

E Youngs modulusG Shear modulus

I second moment of areaJ polar moment of areaR radius

t thicknessT

thermal expansivity linear strain shear strain angle Poissons ratio Normal stress Shear stress


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A pressure vessel near you!


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Ongoing Example

We shall refer back to this example of a typical pressure vessel on several

occasions.

Distillation column

2 m

18 m

P= 7 bara

carbon steelt= 5 mm


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1. Introduction to Pressure Vessels and Failure Modes

Pressure vessels are very often

spherical (e.g. LPG storage tanks)

cylindrical (e.g. liquid storage tanks)

cylindrical shells with hemispherical ends (e.g. distillation col-

umns)

Such vessels fail when the stress state somewhere in the wall material ex-

ceeds some failure criterion. It is thus important to be able to be able to un-derstand and quantify (resolve) stresses in solids. This unit will con-

centrate on the application of stress analysis to bulk failure in thin walled

vesselsonly, where (i) the vessel self weight can be neglected and (ii) the

thickness of the material is much smaller than the dimensions of the vessel

(D t).

1.1. Stresses in Cylinders and SpheresConsider a cylindrical pressure vessel

internal gauge pressure P

L

r

h

External diameterD

wall thickness, t

L

The hydrostatic pressure causes stresses in three dimensions.

1. Longitudinal stress (axial) L2. Radial stress r3. Hoop stress hall are normal stresses.

SAPV LT 2005CFK, MRM

1


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L

r

h

L

r

h

a, The longitudinal stress L

L

P

Force equilibrium

L

2

tD=P4

D

t4

DP=

tensileisthen0,>Pif

L

L

b, The hoop stress h

SAPV LT 2005 2

CFK, MRM

t2

DP=

tL2=PLDbalance,Force

h

h

P

P

P

h

h


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c, Radial stress

r

r

r o ( P )

varies from P on inner surface to 0 on the

outer face

h, L P (D

2 t).

thin walled, so D >> t

so h, L >> r

so neglect r

Compare terms

d, The spherical pressure vessel

h

P

P

t4

DP

=

tD=4

DP

h

h

2


3


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1.2. Compressive Failure: Bulk Yielding & Buckling Vacuum Vessels

Consider an unpressurised cylindrical column subjected to a single load W.

Bulk failure will occur when the normal compressive stress exceeds a yieldcriterion, e.g.

W

bulk =W

Dt= Y

Compressive stresses can cause failure due to buckling (bending instability).

The critical load for the onset of buckling is given by Euler's analysis. A full

explanation is given in the texts, and the basic results are summarised in the

Structures Tables. A column or strut of lengthL supported at one end will

buckle if

W=2EI

L2

Consider a cylindrical column. I = R3t so the compressive stress required

to cause buckling is

buckle =W

Dt=

2ED3t8L2

1

Dt=2ED2

8L2

or

buckle =2 E

8 L D( )2


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whereL/Dis a slenderness ratio. The mode of failure thus depends on the

geometry:

L /D ratio

Short Long

y

stress

Euler buckling locus

Bulk yield


5


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Vacuum vessels.

Cylindrical pressure vessels subject to external pressure are subject to com-

pressive hoop stresses

t2DP=h

Consider a length L of vessel , the compressive hoop force is given by,

2

LDP=tLh

If this force is large enough it will cause buckling.

length

Treat the vessel as an encastered beam of length D and breadth L


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Buckling occurs when Force W given by.

2

2

)D(

EI4W

=( )2

2

D

EI4=

2

LDP

12

tL=

12

tb=I

33 3

=buckleD

t

3

2Ep


7


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1.3. Tensile Failure: Stress Concentration & Cracking

Consider the rod in the Figure below subject to a tensile load. The stress dis-

tribution across the rod a long distance away from the change in cross sec-tion (XX) will be uniform, but near XX the stress distribution is complex.

D

d

W

X X

There is a concentrationof stress at the rod surface below XX and this value

should thus be considered when we consider failure mechanisms.

The ratio of the maximum local stress to the mean (or apparent) stress is de-

scribed by a stress concentration factor K

K= max

mean

The values of Kfor many geometries are available in the literature, including

that of cracks. The mechanism of fast fracture involves the concentration of

tensile stresses at a crack root, and gives the failure criterion for a crack of

length a

a = Kc

where Kcis the material fracture toughness. Tensile

stresses can thus cause failure due to bulk yielding or due

to cracking.


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crack =Kc

1

a

length of crack. a

stress

failure locus


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2. 3-D stress and strain

2.1. Elasticity and Yield

Many materials obey Hooke's law

= E applied stress (Pa)E Young's modulus (Pa)

strain (-)

failure

YieldStress

ElasticLimit

up to a limit, known as the yield stress (stress axis) or the elastic limit (strainaxis). Below these limits, deformation is reversible and the material eventually

returns to its original shape. Above these limits, the material behaviour depends

on its nature.

Consider a sample of material subjected to a tensile force F.

F F1

2

3

An increase in length (axis 1) will be accompanied by a decrease in dimensions

2 and 3.

Hooke's Law 1 = 1 F/A( )/E

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The strain in the perpendicular directions 2,3 are given by

2 = 1E

;3 = 1E

where is the Poisson ratio for that material. These effects are additive, so forthree mutually perpendicular stresses 1, 2, 3;

1

2

3

Giving

1 =1E

2E

3E

2 = 1E

+2E

3E

3 = 1E

2E

+3E

Values of the material constants in the Data Book give orders of magnitudes of

these parameters for different materials;

Material E

(x109N/m2)

Steel 210 0.30Aluminum alloy 70 0.33

Brass 105 0.35

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2.2 Bulk and Shear Moduli

These material properties describe how a material responds to an applied stress

(bulk modulus, K) or shear (shear modulus, G).

The bulk modulus is defined as Puniform = Kv

i.e. the volumetric strain resulting from the application of a uniform pressure. In

the case of a pressure causing expansion

1 =2 =3 = P so

1 =2 =3 =1

E1 2 3[ ]=

PE

(1 2)

v =1 + 2 +3 =3PE

(1 2)

K=E

3(1 2)

For steel, E= 210 kN/mm2,= 0.3, giving K= 175 kN/mm2For water, K= 2.2 kN/mm2

For a perfect gas, K= P(1 bara, 10-4kN/mm2)

Shear Modulus definition = G - shear strain

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2.3. Hoop, longitudinal and volumetric strains

(micro or millistrain)

Fractional increase in dimension:

L length

h circumference

rr wall thickness

(a) Cylindrical vessel:

Longitudinal strain

L =L

E -

h

E-

r

E =

PD

4tE1 - 2( ) = L

L

Hoop strain:

h =n

E -

L

E =

PD

4tE2 - ( ) =

D

D =

R

R

radial strain

r =1

Er - h - L[ ] = -

3PD

4ET =

t

[fractional increase in wall thickness is negative!]

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[ONGOING EXAMPLE]:

L =1

E

L - n( )

= ( )[ ]669

10x1203.0-10x0610x210

1

= 1.14 x 10-4 -0.114 millistrain

h= 0.486 millistrain

r = -0.257 millistrain

Thus: pressurise the vessel to 6 bar: L and D increase: t decreases

Volume expansion

Cylindrical volume: Vo =

Do2

4

L

o (original)

New volume V =

4Do + D( )

2Lo + L( )

= Lo Do

2

41 + h[ ]

21 + L[ ]

Define volumetric strain v =VV

v =V - Vo

Vo = 1 + h( )

21 + L( ) - 1

= 1 + 2h + h2( )1 + L( ) - 1

v = 2h + L + h2 + 2hL + Ln

2

Magnitude inspection:

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max steel( ) =yE

=190 x 10

6

210 x 109 = 0.905 x 10

3 small

Ignoring second order terms,

v= 2 h+ L(b) Spherical volume:

[ ] ( ) -14Et

PD=--

1=

rLhhE

so( )

6

-+

6=

3

33

o

oo

vD

DDD

= (1 +h)

3 1 3

h+ 0(2)

(c) General resultv= 1+ 2+ 3

iiare the strains in any three mutually perpendicular directions.

{Continued example} cylinder L= 0.114 mstrain

n= 0.486 mstrain

rr= -0.257 mstrainv= 2n+ L

new volume = Vo(1 + v)

Increase in volume =D

2L

4v = 56.55 x 1.086 x 10

-3

= 61 Litres

Volume of steelo= DLt = 0.377 m3

vfor steel = L+ h+ rr= 0.343 mstrains

increase in volume of steel = 0.129 L

Strain energy measure of work done

Consider an elastic material for which F = k x

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Work done in expanding x

dW = Fx

work done

F

x

L0

A=area

Work done in extending to x1 w = Fdxox1 = k x dx =

kx12

2ox1 =

1

2F1x1

Sample subject to stress increased from 0 to 1:

Extension Force:x1 = Lo1F1 = A1

W =ALo11

2 (no direction here)

Strain energy, U = work done per unit volume of material, U =ALo112 ALo( )

U =Alo

2

1

Alo

11 U =

11

2 =

12

2E

In a 3-D system, U =1

2 [11+ 22+ 33]

Now 1 =1E

-2

E -

3E

etc

So U =1

2E 1

2 + 22 +3

2 - 2 12 + 23 + 31( )[ ]

Consider a uniform pressure applied: 1= 2= 3= P

U =3P

2

2E 1 - 2( ) =

P2

2K energy stored in system (per unit vol.

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For a given P, U stored is proportional to 1/K so pressure test using liquids

rather than gases.

{Ongoing Example} P 6 barg. V = 61 x 10

-3m

3

increasein volume of pressure vessel

Increasing the pressure compresses the contents normally test with water.

V water? v water( ) = PK

= 6 x 10

5

2.2 x 109 = - 0.273mstrains

decrease in volume of water = -Vo(0.273) = -15.4 x 103

m3

Thus we can add more water:

Extra space = 61 + 15.4 (L)

= 76.4 L water

p=0p=6

extra space

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3.Thermal Effects

3.1. Coefficient of Thermal Expansion

Definition: coefficient of thermal expansion = LT LinearCoefficient of thermal volume expansion v= T Volume

Steel: L= 11 x 10-6K-1 T = 10oC L= 11 10-5

reactor T = 500oC L= 5.5 millistrains (!)

Consider a steel bar mounted between rigid supports which exert stress

Heat

E-T=

If rigid: = 0 so = ET

(i.e., non buckling)

steel: = 210 x 109x 11 x 10-6T = 2.3 x 106T

y= 190 MPa: failure if T > 82.6K

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{Example}: steam main, installed at 10oC, to contain 6 bar steam (140oC)

if ends are rigid, = 300 MPafailure.

must install expansion bends.

19

3.2. Temperature effects in cylindrical pressure vessels

D = 1 m .steel construction L = 3 m . full of water t = 3 mm

Initially un pressurised full of water: increase temp. by T: pressure rises

to Vessel P.

The Vessel Wall stresses (tensile) P83.3=4

=t

PD

L

n = 2L = 166.7 P


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Strain(volume)

T+

E

-

E

=l

hL

L

= 0.3

E = 210 x 109

= 11 x 106

L= 1.585 x 10-10P + 11 x 10-6T

Similarly

h= 6.75 x 10-10P + 11 106

T

v= L+ 2n= 15.08 x 10-10P + 33 x 10-6T = vessel vol. Strain

vessel expands due to temp and pressure change.

The Contents, (water) Expands due to T increase

Contracts due to P increase:

v, H2O = vT P/K H2O = v= 60 x 10-6K-1

v= 60 x 10-6T 4.55 x 10-10P

Since vessel remains full on increasing T:

v(H20) = v(vessel)

Equating P = 13750 T

pressure, rise of 1.37 bar per 10C increase in temp.

Now n= 166.7 P = 2.29 x 106T

n= 22.9 Mpa per 10C rise in Temperature

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Failure does not need a large temperature increase.

Very large stress changes due to temperature fluctuations.

MORAL: Always leave a space in a liquid vessel.

(v, gas = vT P/K)

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3.3. Two material structures

Beware, different materials with different thermal expansivities

can cause difficulties.

{Example} Where there is benefit. The Bimetallic strip - temperature

controllers

a= 4 mm (2 + 2 mm) b = 10 mm

Cu

Fe

L = 100mm b

a

d

Heat by T: Cu expands more than Fe so the strip will bend: it will

bend in an arc as all sections are identical.

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Cu

Fe

The different thermal expansions, set up shearing forces

in the strip, which create a bending moment. If we apply a sagging bending

moment of equal [: opposite] magnitude, we will obtain a straight beam and

can then calculate the shearing forces [and hence the BM].

Cu

Fe

F

F

Shearing force F compressive in Cu

Tensile in Fe

Equating strains:Fe

Fe

cu

cubdE

F+T=

bdE

F-T

So ( ) T-=E

1+

E

1

bd

FFecu

Fecu

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bd = 2 x 10-5m2

Ecu= 109 GPa cu17 x 10-6k-1 T = 30C F = 387 N

EFe= 210 GPa Fe= 11 x 10-6K-1 (significant force)

F acts through the centroid of each section so BM = F./(d/2)= 0.774 Nm

Use data book to work out deflection.

EI

ML

2=

2

This is the principle of the bimetallic strip.

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Consider a steel rod mounted in a upper tube spacer

Analysis relevant to Heat Exchangers

Cu

Fe

assembled at room temperature

. increase T

Data: cu> Fe : copper expands more than steel, so will generate

a TENSILE stress in the steel and a compressive

stress in the copper.

Balance forces:

Tensile force in steel |FFe|= |Fcu|= F

Stress in steel = F/AFe= Fe

copper = F/Acu= cu

Steel strain: FE= FeT + Fe/EFe (no transvere forces)

= FeT + F/EFeAFe

copper strain cu= cuT F/EcuAcu

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Strains EQUAL: ( ) TFeEAEAFd

strengthsofsum

cucuFeFe

cu

=1

+1

-

444 3444 21

So you can work out stresses and strains in a system.

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4. Torsion Twisting Shear stresses

4.1. Shear stresses in shafts /r = T/J = G /L

Consider a rod subject to twisting:

Definition : shear strain change in angle that was originally /2

Consider three points that define a right angle and more then:

Shear strain = 1+ 2 [RADIANS]

A

B C

A

BC

1

2

Hookes Law = G G shear modulus =E

2 1 + ( )

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Now consider a rod subject to an applied torque, T.

2r

T

Hold one end and rotate other by angle .

B

B

L

B

B

Plane ABO was originally to the X-X axis

Plane ABO is now inclined at angle to the axis: tanL

r=

Shear stress involvedL

Gr=G=

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Torque required to cause twisting:

r dr

.T = 2r.r r

ordrr2=TA

2 A

dA.r = GrL

T dArL

G 2

=

{ }JL

G

= so

rL

G

J

T ==

cf y=R

E

=I

M

DEFN: J polar second moment of area

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Consider a rod of circular section:

42

2=.2 RdrrrJ

R

o

=

r

y

x

32

4DJ

=

Now

r2=x2+y2

It can be shown thatJ=Ixx+Iyy [perpendicular axis than]see Fenner

this gives an easy way to evaluate Ixxor Iyyin symmetrical geometrics:

Ixx=Iyy= D4/64 (rod)

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Rectangular rod:

b

d

[ ]223

3

+12=

12

12db

bdJ

dbI

bdI

yy

xx

=

=

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Example: steel rod as a drive shaft

D = 25 mm

L = 1.5 m

Failure when = y= 95 MPa

G = 81 GPa

Now Nm291=Tso

10x383=32

=

0125.0

10x95=

484

6

max

max

=

m

DJ

J

T

r

From 5.1

10x81=10x6.7=

99

J

T

L

G = 0.141 rad= 8.1

Say shaft rotates at 1450 rpm: power = T

= 1450x60

2x291

= 45 kW

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4.2. Thin walled shafts

(same Eqns apply)

Consider a bracket joining two Ex. Shafts:T = 291

Nm

0.025mD

min

What is the minimum value of D for connector?

rmax=D/2

J= (/32){D4 0.0254}

( )446

max 025.0-

32

291

=

2x10x95

J

T

= DDr

y

D4 0.0254= 6.24 x 10-5D D 4.15 cm

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4.3. Thin walled pressure vessel subject to torque

J

T

=r

now cylinder ( )[ ]

44

-2+32= DtDJ

[ ]...+24+832

223tDtD

=

tD34

sotD

T

D 34

=2

tD

T2

2=

34


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CET 1, SAPV

5. Components of Stress/ Mohrs Circle

5.1 Definitions

Scalars

tensor of rank 0

Vectors

tensors of rank 1

ii maF

maF

maF

maF

amF

=

=

=

=

=

:or

:hence

33

22

11

r

r

Tensors of rank 2

3332321313

3232221212

3132121111

3

1

:or

3,2,1,

qTqTqTp

qTqTqTp

qTqTqTp

jiqTpj

jiji

++=

++=

++=

===

Axis transformations

The choice of axes in the description of an engineering problem is

arbitrary (as long as you choose orthogonal sets of axes!). Obviously the

physics of the problem must not depend on the choice of axis. For

example, whether a pressure vessel will explode can not depend on how

we set up our co-ordinate axes to describe the stresses acting on the

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CET 1, SAPV

vessel. However it is clear that the components of the stress tensor will

be different going from one set of coordinatesxito anotherxi.

How do we transform one set of co-ordinate axes onto another, keeping

the same origin?

3332313

2322212

1312111

321

'

'

'

aaax

aaax

aaax

xxx

... where aijare the direction cosines

Forward transformation:

=

=3

1

'j

jiji xax New in terms of Old

Reverse transformation:

==3

1jjjii xax Old in terms of New

We always have to do summations in co-ordinate transformation and it is

conventional to drop the summation signs and therefore these equations

are simply written as:

jjii

jiji

xax

xax

=

='

35


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CET 1, SAPV

Tensor transformation

How will the components of a tensor change when we go from one co-ordinate system to another? I.e. if we have a situation where

form)short(in ==j

jijjiji qTqTp

where Tijis the tensor in the old co-ordinate framexi, how do we find the

corresponding tensor Tij in the new co-ordinate framexi, such that:

form)short(in''''' == jjijjiji qTqTp

We can find this from a series of sequential co-ordinate transformations:

'' qqp

Hence:

'

'

jjll

lklk

kiki

qaq

qTp

pap

=

=

=

Thus we have:

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CET 1, SAPV

''

'

''

jij

jkljlik

jjlkliki

qT

qTaa

qaTap

=

=

=

For example:

333332233113

233222222112

133112211111

33

22

11'

TaaTaaTaa

TaaTaaTaa

TaaTaaTaa

Taa

Taa

TaaT

jijiji

jijiji

jijiji

ljli

ljli

ljliij

+++

+++

++=

+

+

=

Note that there is a difference between a transformation matrix and a 2nd

rank tensor: They are both matrices containing 9 elements (constants)

but:

Symmetrical Tensors:

Tij=Tji

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CET 1, SAPV

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CET 1, SAPV

We can always transform a second rank tensor which is symmetrical:

=

3

2

1

00

00

00

'

:such that

'

T

T

T

T

TT

ij

ijij

Consequence? Consider:

333222111 ,,

then

qTpqTpqTp

qTp jiji

===

=

The diagonal T1, T2, T3is called the PRINCIPAL AXIS.

If T1, T2, T3are stresses, then these are called PRINCIPAL STRESSES.

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CET 1, SAPV

Mohr s circle

Consider an elementary cuboid with edges parallel to the coordinate

directions x,y,z.

x

y

z

z face

y face

x face

Fxy

Fx

FxxFxz

The faces on this cuboid are named according to the directions of their

normals.

There are thus twox-faces, one facing greater values ofx, as shown in

Figure 1 and one facing lesser values ofx(not shown in the Figure).On the x-face there will be some forceFx. Since the cuboid is of

infinitesimal size, the force on the opposite side will not differ

significantly.

The forceFxcan be divided into its components parallel to the coordinatedirections,Fxx,Fxy,Fxz. Dividing by the area of the x-face gives the

stresses on the x-plane:

xz

xy

xx

It is traditional to write normal stresses as and shear stresses as .

Similarly, on the y-face: yzyyyx ,,

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CET 1, SAPV

and on the z-face we have: zzzyzx ,,

There are therefore 9 components of stress;

=

zzzyzx

yzyyyx

xzxyxx

ij

Note that the first subscript refers to the face on which the stress acts and

the second subscript refers to the direction in which the associated force

acts.

y

x

yy

yy

xxxx

xy

yx

xy

yx

But for non accelerating bodies (or infinitesimally small cuboids):

and therefore:

=

=

zzyzxz

yzyyxy

xzxyxx

zzzyzx

yzyyyx

xzxyxx

ij

Hence ijis symmetric!

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CET 1, SAPV

This means that there must be some magic co-ordinate frame in which all

the stresses are normal stresses (principal stresses) and in which the off

diagonal stresses (=shear stresses) are 0. So if, in a given situation wefind this frame we can apply all our stress strain relations that we have set

up in the previous lectures (which assumed there were only normalstresses acting).

Consider a cylindrical vessel subject to shear, and normal stresses (h, l,

r). We are usually interested in shears and stresses which lie in the

plane defined by the vessel walls.

Is there a transformation aboutzzwhich will result in a shear

Would really like to transform into a co-ordinate frame such that all

components in thexi :

'ijij

So stress tensor is symmetric 2nd

rank tensor. Imagine we are in the co-

ordinate framexiwhere we only have principal stresses:

=

3

2

1

00

0000

ij

Transform to a new co-ordinate framexiby rotatoin about thex3axis in

the original co-ordinate frame (this would be, in our example,z-axis)

42


49/74

CET 1, SAPV

The transformation matrix is then:

=

=100

0cossin

0sincos

333231

232221

131211

aaa

aaa

aaa

aij

Then:

++

++

=

=

=

3

22

2121

212

22

1

3

2

1

00

0sincossincossincos

0sincossincossincos

00

00

00

100

0cossin

0sincos

100

0cossin

0sincos

'

aa kljlikij

43


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CET 1, SAPV

Hence:

2sin)(2

1

sincossincos'

2cos)(2

1)(

2

1

cossin'

2cos)(2

1)(

2

1

sincos'

12

2112

1211

22

2122

1211

22

2111

=

+=

++=

+=

+=

+=

44


51/74

Yield conditions. Tresca and Von Mises

Mohrs circle in three dimensions.

xz

y

normalstresses

Shear stresses

x, y plane

y,z plane

x,z plane

1


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We can draw Mohrs circles for each principal plane.

6. BULK FAILURE CRITERIA

Materials fail when the largest stress exceeds a critical value. Normally we

test a material in simple tension:

P P

y =Pyield

A

= =

This material fails under the stress combination (y, 0, 0)

max= 0.5 y = 95 Mpa for steel

We wish to establish if a material will fail if it is subject to a stress

combination (1, 2, 3) or (n, L, )

2


53/74

Failure depends on the nature of the material:

Two important criteria

(i) Trescas failure criterion: brittle materialsCast iron: concrete: ceramics

(ii) Von Mises criterion: ductile materialsMild steel + copper

6.1. Trescas Failure Criterion; The Stress Hexagon (Brittle)

A material fails when the largest shear stress reaches a critical value, the

yield shear stress y.

Case (i) Material subject to simple compression:

1 1

Principal stresses (-1, 0, 0)

-

3


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M.C: mc passes through (1,0), (1,0) , ( 0,0)

1 1 max

max= 1/2

occurs along plane at 45to 1

Similarly for tensile test.

Case (ii) 2< 0 < 1

-

1

2

M.C. Fails when

1 - 22

= max

= y =y

2

i.e., when 1- 2= y

material will not fail.

4


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5

ets do an easy example.L


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6

Criterion; The stress ellipse6.2 Von Mises Failure

(ductile materials)

Trescas criterion does not work well for ductile materials. Early hypothesis

material fails when its strain energy exceeds a critical value (cant be true

ailure when strain energy due to distortion, UD, exceeds a

) due of a compressive stress C equal to

the mean of the principal stresses.

as no failure occurs under uniform compression).

Von Mises: f

critical value.

UD= difference in strain energy (U

C =1

3 1 + 2 + 3[ ]

UD =1

2E 1

2 + 2

2 + 3

2 + 2 12 + 23 + 31( ) -

1

2E 3C

2 + 6C2[ ]

1

12G 1 - 2( )

2 + 2 - 3( )

2 + 3 - 1( )

2{ =M.C.

1

23


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7

Tresca failure when max (I) y)

Von Mises failure when root mean

Compare with (y, , ).simple tensile test failure

square of {a, b, c} critical value

y

Failure if

( ) ( ) ( ){ } { }+0+G

11 2y

22y

12

>-+++-

G12

2

13

2

32

2

21

( ) ( ) ( ) 2>-+++- 2y2

13

2

32

2

21


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8


59/74

Lets do a simple example.

9


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Example Tresca's Failure Criterion

The same pipe as in the first example (D= 0.2 m, t'= 0.005 m) is subject to

an internal pressure of 50 barg. What torque can it support?

Calculate stresses

L =PD

4t'= 50N/mm2

h =PD

2t'=100N/mm2

and 3= r0

Mohr's Circle

10050

s

(100,)

(50,)

Circle construction s= 75 N/mm2

t = (252+2)

The principal stresses 1,2 = s t

Thus 2may be positive (case A) or negative (case B). Case A occurs if is

small.

10


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10050

s

(100,)

(50,)

123

Case A

Case B

2

10050s

(100,)

(50,)

13

2

We do not know whether the Mohr's circle for this case follows Case A or

B; determine which case applies by trial and error.

Case A; 'minor' principal stress is positive (2> 0)

Thus failure when max = 12 y =105N/mm2

11


62/74

For Case A;

max =1

2= 12 75 + 25

2 + 2( )[

2

=135 252

= 132.7N/mm2

Giving 1 = 210N/mm2 ; 2 = 60N/mm

2

Case B;

We now have maxas the radius of the original Mohr's circle linking ourstress data.

Thus

max = 252 +2 = 105 = 101.98N/mm2

Principal stresses

1,2

= 75 105 1

= 180N/mm2 ; 2

= 30N/mm2

Thus Case B applies and the yield stress is 101.98N/mm2. The torque

required to cause failure is

T= D2t' / 2 = 32kNm

Failure will occur along a plane at angle anticlockwise from the y (hoop)direction;

tan(2) =102

75 2= 76.23 ;

2=90- 2 = 6.9

12


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Example More of Von Mises Failure Criterion

From our second Tresca Example

h = 100N/mm2

L = 50N/mm2

r 0N/mm2

What torque will cause failure if the yield stress for steel is 210 N/mm2?

Mohr's Circle

10050

s

(100,)

(50,)

Giving

1 = s + t= 75 + 252 + 2

2 = s t= 75 252 + 2

3= 0

At failureUD=

1

12G(1 2 )

2 + (23)2 + (3 1)

2{

13


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Or

(1 2 )2 + (2 3)

2 + (3 1)2 = (y )

2 + (0)2 + (0 y)2

4t2 + (s t)2 + (s + t)2 = 2y2

2s2 + 6t2 = 2y2

s2 + 3t2 =y2

752 + 3(252 +2 ) = 2102 =110N / mm2

The tube can thus support a torque of

T=D2t'

2= 35kNm

which is larger than the value of 32 kNm given by Tresca's criterion - in this

case, Tresca is more conservative.

14


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1

7. Strains

7.1. Direct and Shear Strains

Consider a vector of length lx lying along the x-axis as shown in Figure

1. Let it be subjected to a small strain, so that, if the left hand end is fixed

the right hand end will undergo a small displacement x. This need not be

in the x-direction and so will have components xx in the x-direction and

xy in the y-direction.

1 x xy

lx xx Figure 1

We can define strains xxand xyby,

x

xy

xy

x

xxxx

l;

l

=

=

xxis the direct strain, i.e. the fractional increase in length in the direction

of the original vector. xy represents rotation of the vector through the

small angle 1where,

xy

x

xy

xxx

xy

11ll

tan =

+

=

Thus in the limit as x 0, 1xy.

Similarly we can define strains yyand yx= 2by,

y

yx

yx

y

yy

yyll

== ;


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2

as in Figure 2.

l

lx

y

1

2

y

yy

yx

Figure 2

Or, in general terms:

3,2,1,e wher; == jili

ij

ij

The ENGINEERING SHEAR STRAIN is defined as the change in an

angle relative to a set of axes originally at 90. In particular xy is the

change in the angle between lines which were originally in the x- and y-

directions. Thus, in our example (Figure 2 above):

xy= (1+2 )=xy+yx or xy= (1+2 )

depending on sign convention.

A

B C

A'

B'C'

Figure 3a Figure 3b

xy

yx

Positive values of the shear stresses xyand yx act on an element as

shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is

sensible to take xyas +ve when the angle ABC decreases. Thus


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3

xy= +(1+2)

Or, in general terms:

)( jiijij +=

and since

ij= ji,

we have

ij= ji.

Note that the TENSOR SHEAR STRAINSare given by the averaged

sum of shear strains:

jijiijij 2

1)(

2

1)(

2

1

2

121 =+=+=

7.2 Mohrs Circle for Strains

The strain tensor can now be written as:

=

=

332313

232212

131211

333231

232221

131211

2

1

2

12

1

2

12

1

2

1

2

1

2

12

1

2

12

1

2

1

yy

yy

yy

yy

yy

yy

ij

where the diagonal elements are the stretches or tensile strains and the

off diagonal elements are the tensor shear strains.

Thus our strain tensor is symmetrical, and:


68/74

4

jiij =

This means there must be a co-ordinate transformation, such that:

=

3

2

1

00

00

00

:such that

'

ij

ijij

we only have principal (=longitudinal) strains!

Exactly analogous to our discussion for the transformation of the stress

tensor we find this from:

++ ++=

=

=

3

22

2121

21

2

2

2

1

3

2

1

00

0sincossincossincos0sincossincossincos

00

00

00

100

0cossin

0sincos

100

0cossin

0sincos

'

aa kljlikij

And hence:


69/74

5

2sin)(2

1

sincossincos'2

1'

2cos)(2

1)(

2

1

cossin'

2cos)(2

1)(

2

1

sincos'

12

211212

1211

22

2122

1211

22

2111

=

+==

++=

+=

+=

+=

For which we can draw a Mohrs circle in the usual manner:


70/74

6

Note, however, that on this occasion we plot halfthe shear strain against

the direct strain. This stems from the fact that the engineering shear

strains differs from the tensorial shear strains by a factor of 2 as as

discussed.

7.3 Measurement of Stress and Strain - Strain Gauges

It is difficult to measure internal stresses. Strains, at least those on asurface, are easy to measure.

Glue a piece of wire on to a surface Strain in wire = strain in material As the length of the wire increases, its radius decreases so itselectrical resistance increases and can be readily measured.

In practice, multiple wire assemblies are used in strain gauges, tomeasure direct strains.

strain gauge

_

_

_

12045

Strain rosettes are employed to obtain three measurements:

7.3.1 45 Strain Rosette

Three direct strains are measured

1

C

B

A

rincipal strain

B

Mohrs circle for strains gives


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7

2

A

B

C

AB C

/2

circle, centre s,radiustso we can write

)2cos(

)2cos(

)2sin()902cos(

ts

tsts

C

B

A

=

=++=

ts+=

3 equations in 3 unknowns

Using strain gauges we can find the directions of Principal strains

/2


72/74

8

7.4 Hookes Law for Shear Stresses

St. Venants Principle states thatthe principal axes of stress and strain are

co-incident. Consider a 2-D element subject to pure shear (xy= yx=

o).

y

x

o

o

X

Y

PQ

o

oThe Mohrs circle for stresses is

where P and Q are principal stress axes and

pp=1=o

qq=2= opq=qp= 0

Since the principal stress and strain axes are coincident,

pp=1=1

E

2

E =

o

E1+( )

= =2

qq 2E

1

E=

oE

1+( )

and the Mohrs circle for strain is thus

X

YPQ

ppqq

/2

the Mohrs circle shows that

xx= 0

xy2

= o

E1 +( )

(0, )xy


73/74

9

So pure shear causes the shear strain

o

o/2

/2 and

= 2o

E(1+ )

But by definition o= G soG =

o =

E

2(1+ )

Use St Venants principal to work out principal stress values from a

knowledge of principal strains.

Two Mohrs circles, strain and stress.

1=1E

2

E

3E

2= 1

E

+2

E

3

E3=

1E

2

E+

3E


74/74

10

So using strain gauges you can work out magnitudes of principal strains.

You can then work out magnitudes of principal stresses.

Using Tresca or Von Mises you can then work out whether your

vessel is safe to operate. ie below the yield criteria

stress analisis & pressure vessels

Documents

d stress

stress state

stress ellipse7

stress hexagon6

dimensional stress analysis5

cylindrical pressure

typical pressure vessel

failure modespressure