CET 1: Stress Analysis & Pressure Vessels

Lent Term 2005

Dr. Clemens Kaminski

Telephone: +44 1223 763135

E-mail: clemens_kaminski@cheng.cam.ac.uk URL: http://www.cheng.cam.ac.uk/research/groups/laser/

Synopsis

1 Introduction to Pressure Vessels and Failure Modes 1.1 Stresses in Cylinders and Spheres 1.2 Compressive failure. Euler buckling. Vacuum vessels 1.3 Tensile failure. Stress Stress Concentration & Cracking

2 3-D stress and strain 2.1 Elasticity and Strains-Young's Modulus and Poisson's Ratio 2.2 Bulk and Shear Moduli 2.3 Hoop, Longitudinal and Volumetric Strains 2.4 Strain Energy. Overfilling of Pressure Vessels

3 Thermal Effects 3.1 Coefficient of Thermal Expansion 3.2 Thermal Effect in cylindrical Pressure Vessels 3.3 Two-Material Structures

4 Torsion. 4.1 Shear Stresses in Shafts - /r = T/J = G/L 4.2 Thin Walled Shafts 4.3 Thin Walled Pressure Vessel subject to Torque

5 Two Dimensional Stress Analysis 5.1 Nomenclature and Sign Convention for Stresses 5.2 Mohr's Circle for Stresses 5.3 Worked Examples 5.4 Application of Mohr's Circle to Three Dimensional Systems

6 Bulk Failure Criteria 6.1 Tresca's Criterion. The Stress Hexagon 6.2 Von Mises' Failure Criterion. The Stress Ellipse

7 Two Dimensional Strain Analysis 7.1 Direct and Shear Strains 7.2 Mohr's Circle for Strains 7.3 Measurement of Strain - Strain Gauges 7.4 Hookes Law for Shear Stresses

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 examined in CET I (OR) Paper IV Section 1. There are consequently a large number of old Tripos questions in existence, which are of the appropriate standard. From 1999 onwards the course was taught in CET1, paper 5. Chapters 7 and 8 in Gere and Timoshenko contain a large number of example problems and questions. Nomenclature

The following symbols will be used as consistently as possible in the lectures. E Youngs modulus G Shear modulus I second moment of area J polar moment of area R radius t thickness T thermal expansivity linear strain shear strain angle Poissons ratio Normal stress Shear stress

A pressure vessel near you!

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

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 vessels only, 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 Spheres Consider a cylindrical pressure vessel

internal gauge pressure P

L

r

h

External diameter D

wall thickness, t

L

The hydrostatic pressure causes stresses in three dimensions. 1. Longitudinal stress (axial) L 2. Radial stress r3. Hoop stress h all are normal stresses.

SAPV LT 2005 CFK, MRM

1

L

r

h

L

r

h

a, The longitudinal stress L

L

P

Force equilibrium L

2

t D = P 4D

t4D P =

tensileis then 0, > P if

L

L

b, The hoop stress h

SAPV LT 2005 2CFK, MRM

t2D P =

tL 2 = P L D balance, Force

h

h

P

P

P

h

h

c, Radial stress

r

r

r o ( P )

varies from P on inner surface to 0 on the outer face

h , L P ( D2 t ).thin walled, so D >> t

so h , L >> rso neglect r

Compare terms

d, The spherical pressure vessel

h

P

P

t4D P =

tD = 4D P

h

h

2

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3

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 yield criterion, e.g.

W

bulk = WDt = 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 length L 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 = WDt =2ED3t

8L2 1Dt =

2ED28L2

or

buckle = 2 E

8 L D( )2

SAPV LT 2005 CFK, MRM

4

where L/D is a slenderness ratio. The mode of failure thus depends on the geometry:

L /D ratio

Short Long

y

stress

Euler buckling locus

Bulk yield

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5

Vacuum vessels. Cylindrical pressure vessels subject to external pressure are subject to com-pressive hoop stresses

t2D P = h

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

2L D P = t L h

If this force is large enough it will cause buckling.

length

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

SAPV LT 2005 CFK, MRM

6

Buckling occurs when Force W given by.

2

2

)D (EI4W

= ( )22

D EI4=

2L D P

12 tL =

12 tb= I

33 3

=buckle Dt

32E p

SAPV LT 2005 CFK, MRM

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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 concentration of 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 = maxmean

The values of K for 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 Kc is the material fracture toughness. Tensile stresses can thus cause failure due to bulk yielding or due to cracking.

SAPV LT 2005 CFK, MRM

8

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

Yield Stress

Elastic Limit

up to a limit, known as the yield stress (stress axis) or the elastic limit (strain axis). 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 10

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 for three mutually perpendicular stresses 1, 2, 3;

1

2

3

Giving

1 = 1E 2E

3E

2 = 1E +2E

3E

3 = 1E 2

E+ 3

E

Values of the material constants in the Data Book give orders of magnitudes of these parameters for different materials;

Material E (x109 N/m2)

Steel 210 0.30

Aluminum alloy 70 0.33

Brass 105 0.35

11

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 = 1E 1 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/mm2 For water, K = 2.2 kN/mm2 For a perfect gas, K = P (1 bara, 10-4 kN/mm2)

Shear Modulus definition = G - shear strain

12

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 = LE - h

E-

rE

= PD4tE

1 - 2( ) = LL

Hoop strain:

h = nE - L

E =

PD4tE

2 - ( ) = DD

= RR

radial strain

r = 1E r - h - L[ ] = - 3PD4ET = tt

[fractional increase in wall thickness is negative!]

13

[ONGOING EXAMPLE]:

L = 1E L - n( ) = ( )[ ]669 10 x 120 3.0 - 10 x 0610x 210 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

Lo (original)

New volume V = 4

Do + D( )2 Lo + L( ) = Lo Do

2

41 + h[ ]2 1 + L[ ]

Define volumetric strain v = VV

v = V - VoVo = 1 + h( )2

1 + L( ) - 1 = 1 + 2 h + h2( )1 + L( ) - 1 v = 2h + L + h2 + 2h L + Ln2 Magnitude inspection:

14

max steel( ) = yE = 190 x 106

210 x 109 = 0.905 x 103 small

Ignoring second order terms,

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

[ ] ( ) - 14EtPD = - - 1 = rLhh E

so ( ){ }

6 - +

6 = 3

33

o

oov D

DDD

= (1 + h)3 1 3h + 0(2) (c) General result

v = 1 + 2 + 3 ii are the strains in any three mutually perpendicular directions. {Continued example} cylinder L = 0.114 mstrain n = 0.486 mstrain rr = -0.257 mstrain v = 2n + L new volume = Vo (1 + v)

Increase in volume = D2 L

4 v = 56.55 x 1.086 x 10-3

= 61 Litres

Volume of steelo = DLt = 0.377 m3v for 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

15

Work done in expanding x dW = Fx

work done

F

x

L 0

A=area

Work done in extending to x1 w = Fdxox1 = k x dx = kx1

2

2ox1 = 1

2 F1x1

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 = Alo2

1

Alo

11 U =

112

= 122E

In a 3-D system, U = 12

[11 + 22 + 33]

Now 1 = 1E - 2

E -

3E

etc

So U = 1

2E 12 + 22 + 32 - 2 12 + 23 + 3 1( )[ ]

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

U = 3P2

2E 1 - 2( ) = P2

2K energy stored in system (per unit vol.

16

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 10-3 m3 increase in volume of pressure vessel

Increasing the pressure compresses the contents normally test with water.

V water? v water( ) = PK = 6 x 105

2.2 x 109 = - 0.273 mstrains

decrease in volume of water = -Vo (0.273) = -15.4 x 10 3 m3Thus we can add more water:

Extra space = 61 + 15.4 (L)

= 76.4 L water

p=0 p=6

extra space

17

3. Thermal Effects

3.1. Coefficient of Thermal Expansion

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

Steel: L = 11 x 10-6 K-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 109 x 11 x 10-6 T = 2.3 x 106 T y = 190 MPa: failure if T > 82.6 K

18

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

if ends are rigid, = 300 MPa failure. 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) P 83.3 = 4

= t

PDL

n = 2L = 166.7 P

Strain (volume)

T + E

- E

= lhLL

= 0.3E = 210 x 109 = 11 x 106

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

Similarly

h = 6.75 x 10-10 P + 11 10 6 T v = L + 2n = 15.08 x 10-10 P + 33 x 10-6 T = 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 10-6 K-1 v = 60 x 10-6 T 4.55 x 10-10 P

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 106 T n = 22.9 Mpa per 10C rise in Temperature

20

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)

21

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.

22

CuFe

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].

CuFe

F

F

Shearing force F compressive in Cu

Tensile in Fe

Equating strains: Fe

Fecu

cu bdEF + T =

bdEF - T

So ( ) T - = E1 +

E1

bdF

FecuFecu

23

bd = 2 x 10-5 m2

Ecu = 109 GPa cu 17 x 10-6 k-1 T = 30C F = 387 NEFe = 210 GPa Fe = 11 x 10-6 K-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

ML2

= 2

This is the principle of the bimetallic strip.

24

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 = cuSteel strain: FE = Fe T + Fe/EFe (no transvere forces) = FeT + F/EFeAFecopper strain cu = cuT F/EcuAcu

25

Strains EQUAL: ( ) TFeEAEAF dstrengthsofsum

cucuFeFe

cu =

1 + 1 -

444 3444 21

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

26

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 = E2 1 + ( )

27

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: tan Lr =

Shear stress involved L

Gr = G =

28

Torque required to cause twisting:

r dr

.T = 2 r.r r or drr 2 = T

A

2 A

dA.r = GrL

T dAr L

G 2= { }J

LG = so

rLG

JT = =

cf y

= RE =

IM

DEFN: J polar second moment of area

29

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 that J = Ixx + Iyy [perpendicular axis than] see Fenner this gives an easy way to evaluate Ixx or Iyy in symmetrical geometrics:

Ixx = Iyy = D4/64 (rod)

30

Rectangular rod:

b

d

[ ]223

3

+ 12

=

12

12dbbdJ

dbI

bdI

yy

xx

=

=

31

Example: steel rod as a drive shaft

D = 25 mm

L = 1.5 m

Failure when = y = 95 MPa G = 81 GPa

Now Nm 291 = T so 10 x 383 =

32 =

0125.010 x 95 =

484

6

max

max

=

mDJ

JT

r

From ( )

5.1 10 x 81 = 10 x 6.7 =

99

JT

LG

= 0.141 rad = 8.1

Say shaft rotates at 1450 rpm: power = T = 1450 x

602 x 291

= 45 kW

32

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 = 2 x 10 x 95

JT =

DDry

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

33

4.3. Thin walled pressure vessel subject to torque

JT =

r

now cylinder ( )[ ]44 - 2 + 32

= DtDJ

[ ]... + 24 + 8 32

223 tDtD=

tD3 4

so tD

TD 3

4 = 2

tD

T2

2 =

34

CET 1, SAPV

5. Components of Stress/ Mohrs Circle

5.1 Definitions

Scalars tensor of rank 0

Vectors tensors of rank 1

ii maF

maFmaFmaF

amF

=

===

=

:or

:hence

33

22

11

rr

Tensors of rank 2

3332321313

3232221212

3132121111

3

1

:or

3,2,1,

qTqTqTpqTqTqTp

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

34

CET 1, SAPV

vessel. However it is clear that the components of the stress tensor will be different going from one set of coordinates xi to another xi. How do we transform one set of co-ordinate axes onto another, keeping the same origin?

3332313

2322212

1312111

321

'''

aaaxaaaxaaaxxxx

... where aij are the direction cosines Forward transformation:

=

= 31

'j

jiji xax New in terms of Old

Reverse transformation:

=

= 31j

jjii 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

=

='

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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 Tij is the tensor in the old co-ordinate frame xi, how do we find the corresponding tensor Tij in the new co-ordinate frame xi, such that:

form)short (in ''''' ==j

jijjiji qTqTp

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

'' qqpp 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

000000

'

:such that

'

TT

TT

TT

ij

ijij

Consequence? Consider:

333222111 ,,

then

qTpqTpqTp

qTp jiji

===

=

The diagonal T1, T2, T3 is called the PRINCIPAL AXIS. If T1, T2, T3 are stresses, then these are called PRINCIPAL STRESSES.

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CET 1, SAPV

Mohrs 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

Fxx Fxz

The faces on this cuboid are named according to the directions of their normals. There are thus two x-faces, one facing greater values of x, as shown in Figure 1 and one facing lesser values of x (not shown in the Figure). On the x-face there will be some force Fx. Since the cuboid is of infinitesimal size, the force on the opposite side will not differ significantly. The force Fx can be divided into its components parallel to the coordinate directions, 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 ,,

40

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 ij is 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 we find this frame we can apply all our stress strain relations that we have set up in the previous lectures (which assumed there were only normal stresses 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 about zz which will result in a shear Would really like to transform into a co-ordinate frame such that all components in the xi :

'ijij So stress tensor is symmetric 2nd rank tensor. Imagine we are in the co-ordinate frame xi where we only have principal stresses:

=

3

2

1

000000

ij

Transform to a new co-ordinate frame xi by rotatoin about the x3 axis in the original co-ordinate frame (this would be, in our example, z-axis)

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CET 1, SAPV

The transformation matrix is then:

=

=

1000cossin0sincos

333231

232221

131211

aaaaaaaaa

aij

Then:

++++

=

=

=

3

22

2121

212

22

1

3

2

1

000sincossincossincos0sincossincossincos

000000

1000cossin0sincos

1000cossin0sincos

'

aa kljlikij

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CET 1, SAPV

Hence:

2sin)(21

sincossincos'

2cos)(21)(

21

cossin'

2cos)(21)(

21

sincos'

12

2112

1211

22

2122

1211

22

2111

=+=

++=+=

+=+=

44

Yield conditions. Tresca and Von Mises

Mohrs circle in three dimensions.

xz

y

normal stresses

Shear stresses

x, y plane

y,z plane

x,z plane

1

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

Failure depends on the nature of the material:

Two important criteria

(i) Trescas failure criterion: brittle materials

Cast iron: concrete: ceramics

(ii) Von Mises criterion: ductile materials

Mild 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

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

1 1 max

max = 1/2 occurs along plane at 45 to 1Similarly for tensile test.

Case (ii) 2 < 0 < 1

- 1

2

M.C. Fails when

1 - 22

= max = y =

y2

i.e., when 1 - 2 = y material will not fail.

4

5

ets do an easy example. L

6

Criterion; The stress ellipse 6.2 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 = 13

1 + 2 + 3[ ] UD =

12E

12 + 22 + 32 + 2 12 + 23 + 31( ) - 12E 3C2 + 6C2[ ]

112G

1 - 2( )2 + 2 - 3( )2 + 3 - 1( )2{ } =M.C.

1

2 3

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

213

232

221

( ) ( ) ( ){ } 2 > - + + + - 2y213232221

8

Lets do a simple example.

9

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 = PD4t' = 50N / mm2

h = PD2t' =100N / mm2

and 3 = r 0

Mohr's Circle

10050s

(100,)

(50,)

Circle construction s = 75 N/mm2 t = (252+2) The principal stresses 1,2 = s t Thus 2 may be positive (case A) or negative (case B). Case A occurs if is small.

10

10050s

(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

For Case A;

max = 12 =12 75 + 252 + 2( )[ ]

2 =135 252 = 132.7N / mm2

Giving 1 = 210N / mm2 ; 2 = 60N / mm2 Case B; We now have max as the radius of the original Mohr's circle linking our stress 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 101.98 N/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

Example More of Von Mises Failure Criterion From our second Tresca Example

h = 100N / mm2 L = 50N / mm2r 0N / mm2

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

10050s

(100,)

(50,)

Giving

1 = s + t = 75 + 252 + 22 = s t = 75 252 + 23 = 0

At failure

UD = 112G (1 2 )2 + (2 3)2 + (3 1)2{ }

13

Or

(1 2 )2 + (2 3)2 + (3 1)2 = (y )2 + (0)2 + (0 y)24t2 + (s t)2 + (s + t)2 = 2 y22s2 + 6t2 = 2y2s2 + 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

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 xx and xy by,

x

xyxy

x

xxxx l

;l

== xx is 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 1 where,

xyx

xy

xxx

xy11 ll

tan =+=

Thus in the limit as x 0, 1 xy. Similarly we can define strains yy and yx = 2 by,

y

yxyx

y

yyyy ll

== ;

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

ijij

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 xy and yx act on an element as shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is sensible to take xy as +ve when the angle ABC decreases. Thus

3

xy = +(1 + 2) Or, in general terms: )( jiijij += and since

ij = ji,

we have ij = ji.

Note that the TENSOR SHEAR STRAINS are given by the averaged sum of shear strains:

jijiijij 21)(

21)(

21

21

21 =+=+= 7.2 Mohrs Circle for Strains The strain tensor can now be written as:

=

=

332313

232212

131211

333231

232221

131211

21

21

21

21

21

21

21

21

21

21

21

21

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:

4

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

=

3

2

1

000000

: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

212

22

1

3

2

1

000sincossincossincos0sincossincossincos

000000

1000cossin0sincos

1000cossin0sincos

'

aa kljlikij

And hence:

5

2sin)(21

sincossincos'21'

2cos)(21)(

21

cossin'

2cos)(21)(

21

sincos'

12

211212

1211

22

2122

1211

22

2111

=

+==

++=

+=

+=

+=

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

6

Note, however, that on this occasion we plot half the 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 a surface, 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 its electrical resistance increases and can be readily measured. In practice, multiple wire assemblies are used in strain gauges, to measure 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

principal strain

B

Mohrs circle for strains gives

7

2

A

B

C

A B C

/2

circle, centre s, radius t so we can write

)2cos(

)2cos()2sin()902cos(

tststs

C

B

A

==++=

ts +=

3 equations in 3 unknowns Using strain gauges we can find the directions of Principal strains

/2

8

7.4 Hookes Law for Shear Stresses St. Venants Principle states that the principal axes of stress and strain are co-incident. Consider a 2-D element subject to pure shear (xy = yx = o).

y

x

oo

X

Y

PQ

oo

The Mohrs circle for stresses is

where P and Q are principal stress axes and

pp = 1 = oqq = 2 = o pq = qp = 0

Since the principal stress and strain axes are coincident,

pp = 1 = 1E

2E

= oE

1+ ( ) = = 2qq 2 E

1E

= oE

1+ ( )

and the Mohrs circle for strain is thus

X

Y

PQ

ppqq

/2 the Mohrs circle shows that

xx = 0 xy

2= o

E1 + ( )

(0, )xy

9

So pure shear causes the shear strain

oo

/2 /2 and = 2 o

E(1+ )

But by definition o = G so G =o =

E2(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 3

E

2 = 1E +2E

3E

3 = 1E 2

E+ 3

E

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