CET 1: Stress Analysis & Pressure Vessels. Lent Term Dr. Clemens Kaminski. Telephone:

74  Download (0)

Full text

(1)

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/

(2)

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

(3)

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

(4)
(5)

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 steel

t = 5 mm

(6)

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.

(7)

L r h σ σ σ L r h

a, The longitudinal stress σL

σ L P Force equilibrium L 2 t D = P 4 D π σ π t 4 D P = tensile is then 0, > P if L L σ σ

b, The hoop stress σh

SAPV LT 2005 2 CFK, MRM

t

2

D

P

=

t

L

2

=

P

L

D

balance,

Force

h h

σ

σ

P P P σ h σ h

(8)

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 ( D 2 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 t 4 D P = t D = 4 D P h h 2 σ π σ π 3

(9)

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

σ

bulk

=

W

π

Dt

=

σ

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

=

π

2

EI

L

2

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

σ

buckle

=

W

π

Dt

=

π

2

E

π

D

3

t

8

L

2

1

π

Dt

=

π

2

ED

2

8

L

2 or

σ

buckle

=

π

2

E

8

(

L D

)

2 SAPV LT 2005 CFK, MRM 4

(10)

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

(11)

Vacuum vessels.

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

t 2 D P = h ∆ σ

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

2 L 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

SAPV LT 2005 CFK, MRM

(12)

Buckling occurs when Force W given by. 2 2 ) D ( EI 4 W

π

π

=

( )

2 2 D EI 4 = 2 L D P π π ∆ 12 t L = 12 t b = I 3 3 3 = buckle D t 3 2E p ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆ 7

(13)

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 distribusec-tion 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

=

σ

max

σ

mean

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

π

a

=

K

c

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

(14)

σ

crack

=

K

c

√ π

1

a

length of crack. a stress failure locus 9

(15)

2. 3-D stress and strain

2.1. Elasticity and Yield

Many materials obey Hooke's law

σ

=

σ

applied stress (Pa)

E Young's modulus (Pa)

ε

strain (-) ε σ 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 F

1 2

3

An increase in length (axis 1) will be accompanied by a decrease in dimensions 2 and 3.

Hooke's Law

ε

1

=

(

σ

1

F

/

A

)

/

E

10

(16)

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

ε

2

= −

ν

σ

1

E

;

ε

3

= −

ν

σ

1

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

ε

1

=

σ

1

E

νσ

2

E

νσ

3

E

ε

2

= −

νσ

1

E

+

σ

2

E

νσ

3

E

ε

3

= −

νσ

1

E

νσ

2

E

+

σ

3

E

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

Material

E

(x10

9

N/m

2

)

ν

Steel

210 0.30

Aluminum

alloy 70 0.33

Brass

105 0.35

(17)

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

P

uniform

= −

K

ε

v

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

σ

1

=

σ

2

=

σ

3

= −

P

so

ε

1

=

ε

2

=

ε

3

=

1

E

[

σ

1

νσ

2

νσ

3

]

=

P

E

(1

2

ν

)

ε

v

=

ε

1

+

ε

2

+

ε

3

=

3

P

E

(1

2

ν

)

K

=

E

3(1

2

ν

)

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

=

γ

- shear strain

(18)

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 εL = σL E - υσh E - υσr E = PD 4tE

(

1 - 2υ

)

= δL L Hoop strain: εh = σn E - υσL E = PD 4tE

(

2 - υ

)

= δD D = δR R radial strain εr = 1 E

[

σr - υσh - υσL

]

= -3PDυ 4ET = δt t

(19)

[ONGOING EXAMPLE]:

ε

L = 1 E

(

σ

L -

υσ

n

)

=

[

6

( )

6

]

9 60 x 10 - 0.3 120 x 10 10 x 210 1 = 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 =

π

Do 2 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Lo (original) New volume V =

π

4

(

Do +

δ

D

)

2 Lo +

δ

L

(

)

= Lo

π

Do2 4

[

1 +

ε

h

]

2 1 +

ε

L

[

]

Define volumetric strain εv = δV V ∴

ε

v = V - Vo Vo = 1 +

(

ε

h

)

2 1 +

ε

L

(

)

- 1 = 1 + 2

(

ε

h +

ε

h2

)

(

1 +

ε

L

)

- 1

ε

v = 2

ε

h +

ε

L +

ε

h2 + 2

ε

h

ε

L +

ε

L

ε

n2 Magnitude inspection: 14

(20)

εmax

(

steel

)

= σy E =

190 x 106

210 x 109 = 0.905 x 10

−3 small Ignoring second order terms,

εv = 2εh + ε L (b) Spherical volume:

ε

[

σ

υσ

υσ

]

( )

1-

υ

4Et PD = -1 = h L r h E so

{

(

)

}

6 -+ 6 = 3 3 3 o o o v D D D D

π

δ

π

ε

= (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

D 2 L 4

ε

v = 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

(21)

Work done in expanding δx dW = Fδx work done F x L 0 A=area

Work done in extending to x1 w = ∫ox1Fdx = k x dx = kx1 2 2 ox1

∫ = 1

2 F1x1 Sample subject to stress σ increased from 0 to σ1:

Extension Force: x1 = Lo

ε

1 F1 = A

σ

1 ⎫ ⎬ ⎭ W = ALo

ε

1

σ

1

2 (no direction here)

Strain energy, U = work done per unit volume of material, U = ALoε1σ1 2 AL

(

o

)

⇒ U = Alo 2 1 Alo ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ε1σ1 U =

ε

12

σ

1 =

σ

1 2 2E In a 3-D system, U = 1 2 [ε1σ1 + ε2σ2 + ε3σ3] Now ε1 = σ1 E - υσ2 E - υσ3 E etc So U = 1 2E

σ

1 2 +

σ

22 +

σ

32 - 2

ν

(

σ

1

σ

2 +

σ

2

σ

3 +

σ

3

σ

1

)

[

]

Consider a uniform pressure applied: σ1 = σ2 = σ3 = P ∴ U = 3P

2

2E 1 - 2

(

υ

)

= P2

2K energy stored in system (per unit vol. 16

(22)

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? εv

(

water

)

= − ∆P

K = −

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

(23)

3.

Thermal Effects

3.1. Coefficient of Thermal Expansion

Definition: coefficient of thermal expansion ε = αLT 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 ε

L = 5.5 millistrains (!)

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

σ σ Heat E -T =α∆ σ ε If rigid: ε = 0 ⇒ so σ = Eα∆T

(i.e., non buckling)

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

σy = 190 MPa: failure if ∆T > 82.6 K

(24)

{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) =83.3P 4 = t PD L

σ

σn = 2σL = 166.7 P

(25)

Strain (volume) T + E -E = L h l L α ∆ νσ σ 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

(26)

∴ 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)

(27)

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.

(28)

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 cu bdE F + T = bdE F -T α ∆ ∆ α So =

(

-

)

T E 1 + E 1 bd F Fe cu Fe cu ∆ α α ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ 23

(29)

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 ML 2 = 2

δ

This is the principle of the bimetallic strip.

(30)

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

(31)

Strains EQUAL:

(

Fe

)

T E A E A F d strengths of sum cu cu Fe Fe ∆ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⇒ ∆ cu = 1 + 1

-

α

α

4 4 4 3 4 4 4 2 1

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

(32)

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 B C γ γ 1 2

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

(33)

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

L r = θ γ ≈ γ

Shear stress involved

L Gr = G = γ θ τ 28

(34)

Torque required to cause twisting:

τ r dr τ .δT = τ 2Π r.δr r or dr r 2 = T A 2

τ Π

τ A dA . r

τ

= Grθ L ⎩ ⎫ ⎬ ⎭ T r dA L G 2

θ =

{ }

J L Gθ = so r L G J T

θ

τ

= = cf y = R E = I M σ

(35)

Consider a rod of circular section

: 4 2 2 = . 2 rr dr R J R o

π

π

= r y x 32 4 D J =

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)

(36)

Rectangular rod:

b d

[

2 2

]

3 3 + 12 = 12 12 d b bd J db I bd I yy xx ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = =

(37)

Example: steel rod as a drive shaft

D = 25 mm L = 1.5 m

Failure when τ = τy = 95 MPa

G = 81 GPa Now soT=291Nm 10 x 383 = 32 = 0125 . 0 10 x 95 = 4 8 4 6 max max ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎬ ⎫ = − m D J J T r

π

τ

From ⇒

(

)

⇒ 5 . 1 10 x 81 = 10 x 6 . 7 = 9 9

θ

θ

J T L G

θ

= 0.141 rad = 8.1°

Say shaft rotates at 1450 rpm: power = T

ω

= x 1450 60 2 x 291

π

= 45 kW 32

(38)

4.2. Thin walled shafts

(same Eqns apply)

Consider a bracket joining two Ex. Shafts:

T = 291 Nm

0.025m

D min

What is the minimum value of D for connector?

rmax = D/2 J = (

π

/32){D4 – 0.0254}

(

4 4

)

6 max -0.025 32 291 = 2 x 10 x 95 J T = D D r y

π

τ

⇒ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ D4 – 0.0254 = 6.24 x 10-5 D D 4.15 cm

(39)

4.3. Thin walled pressure vessel subject to torque

J T = r τ now cylinder

[

(

)

4 4

]

-2 + 32 = D t D J

π

[

8 +24 +...

]

32 2 2 3 t D t D

π

= t D3 4

π

≈ so t D T D 3 4 = 2

π

τ

t D T 2 2 =

π

τ

34

(40)

5. Components of Stress/ Mohr’s Circle

5.1 Definitions

Scalars

tensor of rank 0

Vectors

tensors of rank 1 i i ma F ma F ma F ma F a m F = = = = = : or : hence 3 3 2 2 1 1 r r

Tensors of rank 2

3 33 2 32 1 31 3 3 23 2 22 1 21 2 3 13 2 12 1 11 1 3 1 : or 3 , 2 , 1 , q T q T q T p q T q T q T p q T q T q T p j i q T p j ij j i + + = + + = + + = = =

=

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

(41)

vessel. However it is clear that the components of the stress tensor will be different going from one set of coordinates xito another xi’.

How do we transform one set of co-ordinate axes onto another, keeping the same origin?

33 32 31 3 23 22 21 2 13 12 11 1 3 2 1 ' ' ' a a a x a a a x a a a x x x x

... where aijare the direction cosines

Forward transformation:

= = 3 1 ' j ij j i a x

x “New in terms of Old”

Reverse transformation:

= = 3 1 j ji j i a x

x “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:

j ji i j ij i x a x x a x = = ' 35

(42)

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 ij j ij j i T q T q p

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 ij j ij j i T q T q p

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

' ' p q q p ← ← ← Hence: ' ' j jl l l kl k k ik i q a q q T p p a p = = = Thus we have:

(43)

' ' ' ' ' j ij j kl jl ik j jl kl ik i q T q T a a q a T a p = = = For example: 33 3 3 32 2 3 31 1 3 23 3 2 22 2 2 21 1 2 13 3 1 12 2 1 11 1 1 3 3 2 2 1 1 ' T a a T a a T a a T a a T a a T a a T a a T a a T a a T a a T a a T a a T j i j i j i j i j i j i j i j i j i l jl i l jl i l jl i ij + + + + + + + + = + + =

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:

T

ij

=T

ji

(44)
(45)

We can always transform a second rank tensor which is symmetrical: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = → 3 2 1 0 0 0 0 0 0 ' : such that ' T T T T T T ij ij ij Consequence? Consider: 3 3 3 2 2 2 1 1 1 , , then q T p q T p q T p q T pi ij j = = = =

The diagonal T1, T2, T3 is called the PRINCIPAL AXIS.

If T1, T2, T3 are stresses, then these are called PRINCIPAL STRESSES.

(46)

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

τ

σ

It is traditional to write normal stresses as

σ

and shear stresses as

τ

.

(47)

and on the z-face we have: τzx, τzy, σzz

There are therefore 9 components of stress;

⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = zz zy zx yz yy yx xz xy xx ij

σ

τ

τ

τ

σ

τ

τ

τ

σ

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 xx xx σ σ σ σ τ τ τ τ xy yx xy yx

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

and therefore: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = zz yz xz yz yy xy xz xy xx zz zy zx yz yy yx xz xy xx ij

σ

τ

τ

τ

σ

τ

τ

τ

σ

σ

τ

τ

τ

σ

τ

τ

τ

σ

σ

Hence

σ

ij is symmetric! 41

(48)

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 (

σ

h,

σ

l,

σ

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’ :

'

ij

ij σ

σ →

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 0 0 0 0 0 0

σ

σ

σ

σ

ij

Transform to a new co-ordinate frame xi’ by rotatoin about the x3 axis in

(49)

The transformation matrix is then: ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ = 1 0 0 0 cos sin 0 sin cos 33 32 31 23 22 21 13 12 11

θ

θ

θ

θ a a a a a a a a a aij Then: ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + − + + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − = = 3 2 2 2 1 2 1 2 1 2 2 2 1 3 2 1 0 0 0 sin cos sin cos sin cos 0 sin cos sin cos sin cos 0 0 0 0 0 0 1 0 0 0 cos sin 0 sin cos 1 0 0 0 cos sin 0 sin cos '

σ

θ

σ

θ

σ

θ

θ

σ

θ

θ

σ

θ

θ

σ

θ

θ

σ

θ

σ

θ

σ

σ

σ

σ

θ

θ

θ

θ

θ

θ

σ

σ

θ θ a aik jl kl ij 43

(50)

Hence: θ θ θ 2 sin ) ( 2 1 sin cos sin cos ' 2 cos ) ( 2 1 ) ( 2 1 cos sin ' 2 cos ) ( 2 1 ) ( 2 1 sin cos ' 1 2 2 1 12 1 2 1 1 2 2 2 1 22 1 2 1 1 2 2 2 1 11

σ

σ

θ

θ

σ

θ

θ

σ

σ

σ

σ

σ

σ

θ

σ

θ

σ

σ

σ

σ

σ

σ

θ

σ

θ

σ

σ

− = + − = − + + = + = − − + = + =

(51)

Yield conditions. Tresca and Von Mises

Mohrs circle in three dimensions.

x z y normal stresses σ Shear stresses τ x, y plane y,z plane x,z plane 1

(52)

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 = σ σ = σ

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

τmax = 0.5 σy = 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)

(53)

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

(54)

M.C: mc passes through (σ1,0), (σ1,0) , ( 0,0) σ 1 σ 1 τ max τmax = σ1/2

occurs along plane at 45° to σ1

Similarly for tensile test. Case (ii) σ2 < 0 < σ1 σ τ - σ σ σ 1 2 M.C. Fails when

σ

1 -

σ

2 2 =

τ

max =

τ

y =

σ

y 2 i.e., when σ1 - σ2 = σy

material will not fail.

(55)

5

ets do an easy example. L

(56)

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 = 1 3

[

σ1 + σ2 + σ3

]

UD = 1 2E σ1 2 + σ 2 2 + σ 32 + 2υ

(

σ1σ2 + σ2σ3 + σ3σ1

)

- 2E1 3C

[

2 + 6υC2

]

⎧ ⎨ ⎩ ⎫ ⎬ ⎭ 1 12G

(

σ1 - σ2

)

2 + σ 2 - σ3

(

)

2 + σ 3 - σ1

(

)

2

{

}

= M.C. σ τ σ σ 1 2 σ 3

(57)

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 1 1 2 y 2 2 y σ σ 12 > -+ + + -G 12 2 1 3 2 3 2 2 2 1 σ σ σ σ σ σ

(

) (

) (

)

{

σ12 2 + σ23 2+ σ31 2

}

> 2σy2

(58)
(59)

Lets do a simple example.

(60)

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

=

PD

4

t

'

=

50

N

/

mm

2

σ

h

=

PD

2

t

'

=

100

N

/

mm

2 and

σ

3 =

σ

r ≈0 Mohr's Circle τ σ 100 50 s (100,τ) (50,−τ) Circle construction s = 75 N/mm2 t = √(252+τ2) The principal stresses

σ

1,2 = s ± t

Thus

σ

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

τ

is small.

(61)

τ 100 50 s (100,τ) (50,−τ) σ1 σ2 σ3 Case A Case B σ2 τ σ 100 50 s (100,τ) (50,−τ) σ1 σ3 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 (

σ

2 > 0) Thus failure when

τ

max

=

12

σ

y

=

105

N

/

mm

2

(62)

For Case A;

τ

max

=

σ

1

2

=

1 2

[

75

+

(

25

2

+

τ

2

)

]

τ

2

=

135

25

2

τ

=

132.7

N

/

mm

2 Giving

σ

1

=

210

N

/

mm

2

;

σ

2

= −

60

N

/

mm

2 Case B;

We now have

τ

max as the radius of the original Mohr's circle linking our stress data.

Thus

τ

max

=

25

2

+

τ

2

=

105

τ

=

101.98

N

/

mm

2 Principal stresses

σ

1,2

=

75

±

105

σ

1

=

180

N

/

mm

2

;

σ

2

= −

30

N

/

mm

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

T

= π

D

2

t

'

τ

/ 2

=

32

kNm

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

tan(2

λ

)

=

102

75

2

λ

=

76.23

°

;

2

β

= 90-

2

λ

β

=

6.9

°

(63)

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

σ

h

=

100

N

/

mm

2

σ

L

=

50

N

/

mm

2

σ

r

0

N

/

mm

2

What torque will cause failure if the yield stress for steel is 210 N/mm2? Mohr's Circle τ σ 100 50 s (100,τ) (50,−τ) Giving

σ

1

=

s

+

t

=

75

+

25

2

+

τ

2

σ

2

=

s

t

=

75

25

2

+

τ

2

σ

3

=

0

At failure

U

D

=

1

12

G

(

σ

1

σ

2

)

2

+

(

σ

2

σ

3

)

2

+

(

σ

3

σ

1

)

2

{

}

13

(64)

Or

(

σ

1

σ

2

)

2

+

(

σ

2

σ

3

)

2

+

(

σ

3

σ

1

)

2

=

(

σ

y

)

2

+

(0)

2

+

(0

σ

y

)

2

4

t

2

+

(

s

t

)

2

+

(

s

+

t

)

2

=

2

σ

y2

2

s

2

+

6

t

2

=

2

σ

y2

s

2

+

3

t

2

=

σ

y 2

75

2

+

3(25

2

+

τ

2

)

=

210

2

τ

= 110 N / mm

2

The tube can thus support a torque of

T

=

π

D

2

t

'

τ

2

=

35

kNm

which is larger than the value of 32 kNm given by Tresca's criterion - in this case, Tresca is more conservative.

(65)

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

δ

x. This need not be in the x-direction and so will have components

δ

xx in the x-direction and

δ

xy in the y-direction.

γ1 δx δxy

lx δxx Figure 1

We can define strains

ε

xx and

ε

xy by,

x xy xy x xx xx l ; l δ = ε δ = ε

ε

xx is the direct strain, i.e. the fractional increase in length in the direction of the original vector.

ε

xy represents rotation of the vector through the small angle

γ

1 where,

xy x xy xx x xy 1 1 l l tan ≅ δ =ε δ + δ = γ ≅ γ

Thus in the limit as

δ

x

→ 0, γ

1

ε

xy.

Similarly we can define strains

ε

yy and

ε

yx =

γ

2 by,

y yx yx y yy yy l l

δ

ε

δ

ε

= ; =

(66)

2 as in Figure 2. γ δ l lx y 1 2 y yy δ γ Figure 2

Or, in general terms:

3 , 2 , 1 , e wher ; = = i j li ij ij

δ

ε

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

γ

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):

γxy =(γ12)=εxyyx or γxy = −(γ12) depending on sign convention.

A B C A' B' C' Figure 3a Figure 3b τ τ xy yx

Positive values of the shear stresses

τ

xy and

τ

yx act on an element as shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is sensible to take

γ

xy as +ve when the angle ABC decreases. Thus

(67)

3

γxy = +(γ12) Or, in general terms:

) ( ij ji ij

ε

ε

γ

= + and since

τ

ij =

τ

ji, we have

γ

ij =

γ

ji.

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

ji ji ij ij

ε

ε

γ

γ

γ

γ

2 1 ) ( 2 1 ) ( 2 1 2 1 2 1+ = = + =

7.2 Mohr’s Circle for Strains

The strain tensor can now be written as:

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = 33 23 13 23 22 12 13 12 11 33 32 31 23 22 21 13 12 11 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1

ε

ε

ε

ε

ε

ε

ε

y y y y y y y y y y y y ij

where the diagonal elements are the stretches or tensile strains and the off diagonal elements are the tensor shear strains.

(68)

4

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

⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = → 3 2 1 0 0 0 0 0 0 : such that '

ε

ε

ε

ε

ε

ε

ij ij ij

we only have principal (=longitudinal) strains!

Exactly analogous to our discussion for the transformation of the stress tensor we find this from:

⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + − + + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ − = = 3 2 2 2 1 2 1 2 1 2 2 2 1 3 2 1 0 0 0 sin cos sin cos sin cos 0 sin cos sin cos sin cos 0 0 0 0 0 0 1 0 0 0 cos sin 0 sin cos 1 0 0 0 cos sin 0 sin cos '

ε

θ

ε

θ

ε

θ

θ

ε

θ

θ

ε

θ

θ

ε

θ

θ

ε

θ

ε

θ

ε

ε

ε

ε

θ

θ

θ

θ

θ

θ

ε

ε

θ θ a aik jl kl ij And hence:

(69)

5 θ θ θ 2 sin ) ( 2 1 sin cos sin cos ' 2 1 ' 2 cos ) ( 2 1 ) ( 2 1 cos sin ' 2 cos ) ( 2 1 ) ( 2 1 sin cos ' 1 2 2 1 12 12 1 2 1 1 2 2 2 1 22 1 2 1 1 2 2 2 1 11

ε

ε

θ

θ

ε

θ

θ

ε

γ

ε

ε

ε

ε

ε

θ

ε

θ

ε

ε

ε

ε

ε

ε

θ

ε

θ

ε

ε

− = + − = = − + + = + = − − + = + =

(70)

6

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

_ _

_ 45° 120°

Strain rosettes are employed to obtain three measurements: 7.3.1 45° Strain Rosette

Three direct strains are measured

ε 1 θ ε C ε B ε A principal strain ε B

(71)

7 2θ A B C ε A ε B ε C ε γ/2 circle, centre s, radius t so we can write ) 2 cos(

θ

ε

) 2 cos( ) 2 sin( ) 90 2 cos(

θ

ε

θ

θ

ε

t s t s t s C B A − = − = + + = t s + = 3 equations in 3 unknowns

Using strain gauges we can find the directions of Principal strains

ε

γ/2

(72)

8

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 (

τ

xy =

τ

yx =

τ

o). y x τo τo X Y P Q τo τo

The Mohr’s circle for stresses is

where P and Q are principal stress axes and

σpp1o σqq2 = −τo τpqqp =0

Since the principal stress and strain axes are coincident,

εpp= ε1= σ1 E − νσ2 E = τo E (1+ν) εqq2 = σ2 E − νσ1 E = −τo E (1+ν) and the Mohr’s circle for strain is thus

X Y P Q

ε

qq

ε

pp ε γ/2

the Mohr’s circle shows that

εxx =0 −γxy 2 = − τo E (1+ν) (0,−γ )xy

(73)

9 So pure shear causes the shear strain

γ

τo τo γ/2 γ/2 and

γ

=

2

τ

o

E

(1

+

ν

)

But by definition

τ

o = G

γ

so

G

=

τ

o

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.

ε

1

=

σ

1

E

νσ

2

E

νσ

3

E

ε

2

= −

νσ

1

E

+

σ

2

E

νσ

3

E

ε

3

= −

νσ

1

E

νσ

2

E

+

σ

3

E

(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

Figure

Updating...

References

Related subjects :