Press Vessel Compressive Stress

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/

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 - τ/r = T/J = Gθ/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 Hooke’s 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 Young’s modulus G Shear modulus I second moment of area J polar moment of area R radius t thickness T τορθυε α thermal expansivity ε linear strain γ shear strain η angle ν Poisson’s ratio σ Normal stress τ 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) σL

2. Radial stress σr

3. Hoop stress σh all are normal stresses.

SAPV LT 2005 CFK, MRM

1

L

r

h

σ

σ

σ

L

r

h

a, The longitudinal stress σL

σ L

P

Force equilibrium

L

2

t D = P 4D

σππ

t4D P =

tensileis then 0, > P if

L

L

σ

σ

b, The hoop stress σh

SAPV LT 2005 2CFK, MRM

t2D P =

tL 2 = P L D balance, Force

h

h

σ

σ

P

P

P

σ h

σ h

c, Radial stress

σ r

σ

r

σ r ≈ o ( P )

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

σ h , σ L ≈ P (D2 t

).

thin walled, so D >> t

so σ h , σ L >> σ r

so neglect σ r

Compare terms

d, The spherical pressure vessel

σ h

P

P

t4D P =

tD = 4D P

h

h

2

σ

πσπ


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

σbulk =W

πDt= σY

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 =π2EI

L2

Consider a cylindrical column. I = πR3t so the compressive stress required to cause buckling is

σbuckle =W

πDt=

π2EπD3t

8L2 ⋅1

πDt=

π2ED2

8L2

or

σbuckle =π2 E

8 L D( )2


4

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

σ

σ

L /D ratio

Short Long

y

stress

Euler buckling locus

Bulk yield


5

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

t2D P = h

∆σ

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

2L D P = t L h

∆σ

If this force is large enough it will cause buckling.

length

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


6

Buckling occurs when Force W given by.

2

2

)D (EI4W

ππ

=( )2

2

D EI4=

2L D P

ππ∆

12 tL =

12 tb= I

33 3

=buckle Dt

32E p ⎟

⎠⎞

⎜⎝⎛∆


7

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 =σmax

σmean

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 σ πa = Kc

where Kc is the material fracture toughness. Tensile stresses can thus cause failure due to bulk yielding or due to cracking.


8

σcrack =Kc

√ π⋅

1√a

length of crack. a

stress

failure locus


9

2. 3-D stress and strain 2.1. Elasticity and Yield Many materials obey Hooke's law

σ = Eε σ applied stress (Pa) E Young's modulus (Pa) ε strain (-)

ε

σfailure

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 ε1 = σ1 ≡ F / A( )/ E

10

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

ε2 = −νσ1

E;ε3 = −ν

σ1

E

where ν is the Poisson ratio for that material. These effects are additive, so for three mutually perpendicular stresses σ1, σ2, σ3;

σ

σ

σ1

2

3

Giving

ε1 = σ1

E− νσ2

E− νσ3

E

ε2 = −νσ1

E+

σ2

E−

νσ3

E

ε3 = − νσ1

E− νσ2

E+ σ3

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)

ν

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 = − Kεv

i.e. the volumetric strain resulting from the application of a uniform pressure. In the case of a pressure causing expansion σ1 = σ2 = σ3 = −P so

ε1 = ε2 = ε3 =1E

σ1 − νσ2 − νσ3[ ]=−PE

(1− 2ν)

εv = ε1 + ε2 + ε3 =−3P

E(1− 2ν)

K = E

3(1− 2ν)

For steel, E = 210 kN/mm2, ν = 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 τ = Gγ γ - shear strain

12

2.3. Hoop, longitudinal and volumetric strains (micro or millistrain)

Fractional increase in dimension:

ε L – length

ε h – circumference

ε rr – wall thickness

(a) Cylindrical vessel:

Longitudinal strain

ε L = σ LE

- υσ h

E-

υσ rE

= PD4tE

1 - 2υ( ) = δLL

Hoop strain:

εh = σ nE

- υσ L

E =

PD4tE

2 - υ( ) = δDD

= δRR

radial strain

εr = 1E

σ r - υσ h - υσ L[ ] = -3PDυ4ET

= δtt

[fractional increase in wall thickness is negative!]

13

[ONGOING EXAMPLE]:

ε L = 1E

σ L - υσ n( )

= ( )[ ]669 10 x 120 3.0 - 10 x 06

10x 2101

= 1.14 x 10-4 -≡ 0.114 millistrain

ε h = 0.486 millistrain

ε r = -0.257 millistrain

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

Volume expansion

Cylindrical volume: Vo = πDo

2

4

⎛

⎝ ⎜

⎞

⎠ ⎟ Lo (original)

New volume V = π4

Do + δD( )2Lo + δL( )

= Lo π Do

2

41 + εh[ ]2

1 + ε L[ ]

Define volumetric strain ε v = δVV

∴ εv = V - Vo

Vo = 1 + εh( )2

1 + ε L( ) - 1

= 1 + 2ε h + εh2( )1 + ε L( ) - 1

εv = 2εh + ε L + εh2 + 2εhε L + εLεn

2

Magnitude inspection:

14

ε max steel( ) = σy

E =

190 x 106

210 x 109 = 0.905 x 10−3 ∴ small

Ignoring second order terms,

εv = 2ε h + ε L

(b) Spherical volume:

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

so ( ){ }

6 - +

6 = 3

33

o

oov D

DDDπ

δπε

= (1 + εh)3 – 1 ≈ 3εh + 0(ε2)

(c) General result

εv = ε1 + ε2 + ε3

εii are the strains in any three mutually perpendicular directions.

{Continued example} – cylinder εL = 0.114 mstrain

εn = 0.486 mstrain

εrr = -0.257 mstrain

εv = 2εn + ε L

∴ new volume = Vo (1 + εv)

Increase in volume = π D2 L

4 εv = 56.55 x 1.086 x 10-3

= 61 Litres

Volume of steelo = πDLt = 0.377 m3

εv for steel = εL + εh + εrr = 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 δx

dW = Fδx

work done

F

x

L 0

A=area

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

kx12

2ox1∫ =

12

F1x1

Sample subject to stress σ increased from 0 to σ1:

Extension Force: x1 = Loε1

F1 = Aσ1

⎫ ⎬ ⎭ W =

ALoε1σ12

(no direction here)

Strain energy, U = work done per unit volume of material, U = ALoε1σ12 ALo( )

⇒ U = Alo

2

1Alo

⎛

⎝ ⎜

⎞

⎠ ⎟ ε1σ1 U =

ε1σ12

= σ1

2

2E

In a 3-D system, U = 12

[ε1σ1 + ε2σ2 + ε3σ3]

Now ε1 = σ1E

- υσ2

E -

υσ3E

etc

So U = 1

2E σ1

2 + σ22 + σ3

2 - 2ν σ1σ2 + σ 2σ3 + σ 3σ 1( )[ ] Consider a uniform pressure applied: σ1 = σ2 = σ3 = P

∴ U = 3P2

2E 1 - 2υ( ) =

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 . δV = 61 x 10-3 m3

increase in volume of pressure vessel

Increasing the pressure compresses the contents – normally test with water.

∆V water? ε 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 m3

Thus 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 ε = αL∆T Linear

Coefficient of thermal volume expansion εv = αT Volume

Steel: αL = 11 x 10-6 K-1 ∆T = 10oC ∴ εL = 11 10-5

reactor ∆T = 500oC εL = 5.5 millistrains (!)

Consider a steel bar mounted between rigid supports which exert stress σ

σ σHeat

E - T = σ

∆αε

If rigid: ε = 0 ⇒ so σ = Eα∆T

(i.e., non buckling)

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

σy = 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, σ = 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

PDLσ

σn = 2σL = 166.7 P

Strain (volume)

T + E

- E

= lhL

L ∆ανσσ

ε

υ = 0.3

E = 210 x 109

α = 11 x 10−6

⎫ ⎬ ⎪

⎭ ⎪ εL = 1.585 x 10-10 P + 11 x 10-6 ∆ T

Similarly

→ εh = 6.75 x 10-10 P + 11 10 –6 ∆T

εv = εL + 2εn = 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:

εv, H2O = αv∆T – P/K H2O = αv = 60 x 10-6 K-1

∴ εv = 60 x 10-6 ∆T – 4.55 x 10-10 P

Since vessel remains full on increasing ∆T:

εv (H20) = εv (vessel)

Equating → P = 13750 ∆T

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

Now σn = 166.7 P = 2.29 x 106 ∆T

∆σn = 22.9 Mpa per 10°C 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.

(εv, gas = αv∆T – 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 ∆α∆α

So ( ) T - = E1 +

E1

bdF

FecuFecu

∆αα⎭⎬⎫

⎩⎨⎧

23

bd = 2 x 10-5 m2

Ecu = 109 GPa αcu 17 x 10-6 k-1 ∆T = 30°C F = 387 N

EFe = 210 GPa αFe = 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

δ

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: αcu > αFe : 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 = σFe

“ “ copper = F/Acu = σcu

Steel strain: εFE = αFe ∆T + σFe/EFe (no transvere forces)

= αFe∆T + F/EFeAFe

copper strain εcu = αcu∆T – F/EcuAcu

25

Strains EQUAL: ( ) TFeEAEAF

d

strengthsofsum

cucuFeFe∆⎥

⎦

⎤⎢⎣

⎡⇒

∆

cu = 1 + 1 -

αα444 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 –τ/r = T/J = G θ/L

Consider a rod subject to twisting:

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

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

Shear strain γ = γ1 + γ2 [RADIANS]

A

B C

A

BC

γ

γ

1

2

Hooke’s Law τ = G γ G – shear modulus = E

2 1 + υ( )

27

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

2r

T

Hold one end and rotate other by angle θ .

B

B

θ γ

L

B

B

Plane ABO was originally to the X-X axis

Plane ABO is now inclined at angle γ to the axis: tan Lr = θ

γ≈γ

Shear stress involved L

Gr = G = θγτ

28

Torque required to cause twisting:

τ

r dr

τ

.δT = τ 2Π r.δr r

or drr 2 = TA

2∫ Πτ ∫ τA

dA.r τ = Grθ

L⎧ ⎨ ⎩

⎫ ⎬ ⎭

T dAr L

G 2∫θ

=

{ }J L

G θ= so

rLG

JT τθ = =

cf y

= RE =

IM σ

DEFN: J ≡ polar second moment of area

29

Consider a rod of circular section:

42

2 = .2 RdrrrJ

R

o

ππ∫=

r

y

x

32

4DJ π

=

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 = πD4/64 (rod)

30

Rectangular rod:

b

d

[ ]22

3

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 τ = τy = 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

π

τ

From ( )

⇒⇒ 5.1

10 x 81 = 10 x 6.7 = 9

9 θθJT

LG

θ = 0.141 rad = 8.1°

Say shaft rotates at 1450 rpm: power = Tω

= 1450 x 602 x 291 π

= 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 = (π/32){D4 – 0.0254}

( )44

6

max 025.0 - 32 291 = 2 x 10 x 95

JT =

DDry

πτ

⇒⎟⎠⎞

⎜⎝⎛

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

33

4.3. Thin walled pressure vessel subject to torque

JT =

rτ

now cylinder ( )[ ]44 - 2 + 32

= DtDJ π

[ ]... + 24 + 8 32

223 tDtDπ=

tD3 4

π≈

so tD

TD 3

4 = 2 π

τ

tD

T2

2 = π

τ

34

CET 1, SAPV

5. Components of Stress/ Mohr’s 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:

∑=

=3

1'

jjiji 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

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:

36

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

37

CET 1, SAPV

38

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.

39

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

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

τ

τσ

It is traditional to write normal stresses as σ and shear stresses as τ. Similarly, on the y-face: yzyyyx τστ ,,

40

CET 1, SAPV

and on the z-face we have: zzzyzx σττ ,, There are therefore 9 components of stress;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zzzyzx

yzyyyx

xzxyxx

ij

στττστττσ

σ

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

σ

σ σ

σ

τ

τ

τ

τ

xy

yx

xy

yx

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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=

zzyzxz

yzyyxy

xzxyxx

zzzyzx

yzyyyx

xzxyxx

ij

στττστττσ

στττστττσ

σ

Hence σij is symmetric!

41

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 (σh, σl, σr). 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 σσ → 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

σσ

σσ 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)

42

CET 1, SAPV

The transformation matrix is then:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

1000cossin0sincos

333231

232221

131211

θθθθ

aaaaaaaaa

aij

Then:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−++

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

=

3

22

2121

212

22

1

3

2

1

000sincossincossincos0sincossincossincos

000000

1000cossin0sincos

1000cossin0sincos

'

σθσθσθθσθθσ

θθσθθσθσθσ

σσ

σθθθ

θθθ

σσ

θθ

aa kljlikij

43

CET 1, SAPV

Hence:

θ

θ

θ

2sin)(21

sincossincos'

2cos)(21)(

21

cossin'

2cos)(21)(

21

sincos'

12

2112

1211

22

2122

1211

22

2111

σσ

θθσθθσσ

σσσσ

θσθσσ

σσσσ

θσθσσ

−=

+−=

−++=

+=

−−+=

+=

44

Yield conditions. Tresca and Von Mises

Mohrs circle in three dimensions.

xz

y

normal stresses σ

Shear stresses τ

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

σy = Pyield

A

σ

τ

σ = σσ = σ

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

τmax = 0.5 σ y = 95 Mpa for steel

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

combination (σ1, σ2, σ3) or (σn, σL, τ)

2

Failure depends on the nature of the material:

Two important criteria

(i) Tresca’s failure criterion: brittle materials

Cast iron: concrete: ceramics

(ii) Von Mises’ criterion: ductile materials

Mild steel + copper

6.1. Tresca’s Failure Criterion; The Stress Hexagon (Brittle)

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

yield shear stress τy.

Case (i) Material subject to simple compression:

σ 1 σ

1

Principal stresses (-σ1, 0, 0)

τ

- σ

3

M.C: mc passes through (σ1,0), (σ1,0) , ( 0,0)

σ 1 σ

1τ max

τmax = σ1/2

occurs along plane at 45° to σ1

Similarly for tensile test.

Case (ii) σ2 < 0 < σ1

σ

τ

- σ

σσ 1

2

M.C. Fails when

σ1 - σ22

= τ max

= τ y = σy

2

i.e., when σ1 - σ2 = σy

material will not fail.

4

5

ets do an easy example. L

6

Criterion; The stress ellipse 6.2 Von Mises’ Failure (ductile materials)

Tresca’s criterion does not work well for ductile materials. Early hypothesis

– material fails when its strain energy exceeds a critical value (can’t 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

σ1 + σ2 + σ3[ ]

UD = 1

2E σ1

2 + σ22 + σ3

2 + 2υ σ1σ2 + σ2σ3 + σ3σ1( ) - 1

2E 3C2 + 6υC2[ ]⎧ ⎨

⎩ ⎫ ⎬ ⎭

1

12G σ1 - σ2( )2

+ σ2 - σ3( )2 + σ3 - σ1( )2{ } =

M.C.

σ

τ

σσ

1

2σ

3

7

Tresca → failure when max (τI) ≥ τy)

Von Mises → failure when root mean

Compare with (σy, σ, σ) . simple tensile test – failure

square of {τa, τb, τc} ≥ critical value

σ

τ

σ y

Failure if

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

11 2y

22y σσ

12 > - + + + -

G122

132

322

21 σσσσσσ

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

213

232

221 σσσσσσσ

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

σ L = PD4t'

= 50N / mm2

σh = PD2t'

=100N / mm2 and σ3 = σr ≈0

Mohr's Circle

τ

σ10050s

(100,τ)

(50,−τ)

Circle construction s = 75 N/mm2 t = √(252+τ2) The principal stresses σ1,2 = s ± t Thus σ2 may be positive (case A) or negative (case B). Case A occurs if τ is small.

10

τ

10050s

(100,τ)

(50,−τ)

σ1σ2σ3

Case A

Case B

σ2

τ

σ10050s

(100,τ)

(50,−τ)

σ1σ3

2β

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 (σ2 > 0)

Thus failure when τmax = 12 σy =105N / mm2

11

For Case A;

τmax = σ1

2= 1

2 75 + 252 + τ 2( )[ ]⇒ τ 2 =135 − 252

⇒ τ = 132.7N / mm2

Giving σ1 = 210N / mm2 ; σ2 = −60N / mm2

Case B; We now have τmax as the radius of the original Mohr's circle linking our stress data. Thus

τmax = 252 + τ2 = 105 ⇒ τ = 101.98N / mm2

Principal stresses

σ1,2 = 75 ±105 ⇒ σ1 = 180N / mm2 ; σ2 = −30N / mm2

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

T = πD2t' τ / 2 = 32kNm Failure will occur along a plane at angle β anticlockwise from the y (hoop) direction;

tan(2λ ) =

10275

⇒ 2λ = 76.23° ;

2β = 90- 2λ ⇒ β = 6.9°

12

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

σh = 100N / mm2

σ L = 50N / mm2

σr ≈ 0N / mm2

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

τ

σ10050s

(100,τ)

(50,−τ)

Giving

σ1 = s + t = 75 + 252 + τ 2

σ2 = s − t = 75 − 252 + τ 2

σ3 = 0

At failure

UD =1

12G(σ1 − σ2 )2 + (σ2 − σ3)2 + (σ3 − σ1)2{ }

13

Or

(σ1 − σ2 )2 + (σ2 − σ3)2 + (σ3 − σ1)2 = (σy )2 + (0)2 + (0 − σ y)2

4t2 + (s − t)2 + (s + t)2 = 2σ y2

2s2 + 6t2 = 2σy2

s2 + 3t2 = σ y2

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

The tube can thus support a torque of

T =πD2t' τ

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 δx. This need not be in the x-direction and so will have components δxx in the x-direction and δxy in the y-direction.

γ1 δx δxy

lx δxx Figure 1

We can define strains εxx and εxy by,

x

xyxy

x

xxxx l

;l

δ=ε

δ=ε

εxx is the direct strain, i.e. the fractional increase in length in the direction of the original vector. εxy represents rotation of the vector through the small angle γ1 where,

xyx

xy

xxx

xy11 ll

tan ε=δ

≅δ+

δ=γ≅γ

Thus in the limit as δx→ 0, γ1 → εxy. Similarly we can define strains εyy and εyx = γ2 by,

y

yxyx

y

yyyy ll

δε

δε == ;

2

as in Figure 2.

γ

δ

δ

l

lx

y

1

2

y

yy

yx

δ

γ

Figure 2

Or, in general terms:

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

ijij

δε

The ENGINEERING SHEAR STRAIN is defined as the change in an angle relative to a set of axes originally at 90°. In particular γxy is the

change in the angle between lines which were originally in the x- and y-directions. Thus, in our example (Figure 2 above): γ xy = (γ1 + γ2 ) = εxy + εyx or γ xy = −(γ1 + γ 2 ) depending on sign convention.

A

B C

A'

B'C'

Figure 3a Figure 3b

τ

τxy

yx

Positive values of the shear stresses τxy and τyx act on an element as

shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is sensible to take γxy as +ve when the angle ABC decreases. Thus

3

γ xy = +(γ1 + γ 2) Or, in general terms: )( jiijij εεγ +=

and since

τij = τji,

we have γij = γji.

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

jijiijij γγγεεγ21)(

21)(

21

21

21 =+=+=

7.2 Mohr’s 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

ε

ε

ε

ε

ε

ε

ε

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 εε =

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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

→

3

2

1

000000

:such that

'

εε

εε

εε

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

'

εθεθεθθεθθε

θθεθθεθεθε

εε

εθθθ

θθθ

εε

θθ

aa kljlikij

And hence:

5

θ

θ

θ

2sin)(21

sincossincos'21'

2cos)(21)(

21

cossin'

2cos)(21)(

21

sincos'

12

211212

1211

22

2122

1211

22

2111

εε

θθεθθεγε

εεεε

θεθεε

εεεε

θεθεε

−=

+−==

−++=

+=

−−+=

+=

For which we can draw a Mohr’s 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

___

120°45°

Strain rosettes are employed to obtain three measurements: 7.3.1 45° Strain Rosette Three direct strains are measured

ε 1θ

ε Cε B

ε Aprincipal strain

ε B

Mohr’s circle for strains gives

7

2θ

A

B

C

ε Aε B ε C

ε

γ/2

circle, centre s, radius t so we can write

)2cos( θε

)2cos(

)2sin()902cos(θε

θθεts

tsts

C

B

A

−=−=++=

ts +=

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

ε

γ/2

8

7.4 Hooke’s Law for Shear Stresses St. Venant’s Principle states that the principal axes of stress and strain are co-incident. Consider a 2-D element subject to pure shear (τxy = τyx = τo).

y

x

τo

τo

X

Y

PQ

τo

τo

The Mohr’s circle for stresses is

where P and Q are principal stress axes and

σ pp = σ1 = τo

σqq = σ2 = −τo

τ pq = τqp = 0

Since the principal stress and strain axes are coincident,

ε pp = ε1 = σ1

E− νσ2

E

=τo

E1+ ν( )

ε = ε =σ 2

qq 2 E−

νσ1

E=

−τo

E1+ ν( )

and the Mohr’s circle for strain is thus

X

Y

PQ

ε εppqq

ε

γ/2

the Mohr’s circle shows that

ε xx = 0

−γ xy

2= −

τo

E1 + ν( )

(0,−γ )xy

9

So pure shear causes the shear strain γ

τo

τo

γ/2γ/2 and

γ = 2τ o

E(1+ν )

But by definition τo = Gγ so G = τo

γ= E

2(1+ν)

Use St Venants principal to work out principal stress values from a knowledge of principal strains. Two Mohrs circles, strain and stress.

ε1 = σ1

E− νσ2

E− νσ3

E

ε2 = −νσ1

E+

σ2

E−

νσ3

E

ε3 = − νσ1

E− νσ2

E+ σ3

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

Documents

Press Vessel Compressive Stress