### 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/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 = G*θ*/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

### 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*

**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.

L r h σ σ σ L r h

**a, The longitudinal stress **σ**L**

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 **σ**h **

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**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

### 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}

_{E}

_{π}

_{D}

3_{D}

_{t}

_{t}

### 8

*L*

2 ### ⋅

### 1

### π

*Dt*

### =

### π

2_{ED}

2
_{ED}

### 8

*L*

2
or
### σ

_{buckle}### =

### π

2*E*

### 8

### (

*L D*

### )

2 SAPV LT 2005 CFK, MRM 4where *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

**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

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### 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}

_{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

### σ

_{crack}### =

*K*

*c*

### √ 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 σ failure Yield Stress Elastic Limitup 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
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 σ_{1, }σ_{2, }σ_{3; }

σ
σ
σ_{1}
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

### 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

### =

*Gγ*

### γ

- shear strain**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

*L*

*E*- υσ

_{h}*E*- υσ

_{r}*E*=

*PD*4

*tE*

### (

1 - 2υ### )

= δ*L*

*L*Hoop strain: ε

*= σ*

_{h}*n*

*E*- υσ

_{L}*E*=

*PD*4

*tE*

### (

2 - υ### )

= δ*D*

*D*= δ

*R*

*R*radial strain ε

*= 1*

_{r}*E*

### [

σ*r*- υσ

*h*- υσ

*L*

### ]

= -3PDυ 4ET = δ*t*

*t*

[ONGOING EXAMPLE]:

### ε

*= 1*

_{L}*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 millistrainThus: pressurise the vessel to 6 bar: L and D increase: t decreases

**Volume expansion **
**Cylindrical volume**: *V _{o}* =

### π

*Do*2 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

*L*(original) New volume

_{o}*V*=

### π

4### (

*Do*+

### δ

*D*

### )

2*L*+

_{o}### δ

*L*

### (

### )

= Lo### π

Do2 4### [

1 +### ε

h### ]

2 1 +### ε

_{L}### [

### ]

Define volumetric strain ε_{v} = δV
V
∴

### ε

_{v}=

*V*-

*Vo*V

_{o}= 1 +

### (

### ε

*h*

### )

2 1 +### ε

_{L}### (

### )

- 1 = 1 + 2### (

### ε

*+*

_{h}### ε

*2*

_{h}### )

### (

1 +### ε

_{L}### )

- 1### ε

*= 2*

_{v}### ε

*+*

_{h}### ε

*+*

_{L}### ε

*2 + 2*

_{h}### ε

_{h}### ε

*+*

_{L}### ε

_{L}### ε

*2 Magnitude inspection: 14*

_{n}ε_{max}

### (

steel### )

= σy E =190 x 106

210 x 109 = 0.905 x 10

−3 _{∴}_{ small}
Ignoring second order terms,

εv** = 2**ε**h + **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**≈

**3**εh

**+ 0(**ε

**2**

_{) }**(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 **

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

Work done in extending to x1 *w* = ∫_{o}x1Fdx = k x dx = kx1
2
2
ox1

∫ = 1

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

Extension Force: *x*1 = *Lo*

### ε

1*F*

_{1}=

*A*

### σ

_{1}⎫ ⎬ ⎭ W = AL

_{o}

### 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 Al_{o}⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ε1σ1

*U*=

### ε

1_{2}

### σ

1 =### σ

1 2 2*E*In a 3-D system, U = 1 2 [ε1σ1 + ε2σ2 + ε3σ3] Now ε

_{1}= σ1 E - υσ

_{2}E - υσ

_{3}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### )

= P22K 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 . δ**V = 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

**3.**

**Thermal Effects**

### 3.1. Coefficient of Thermal Expansion

**Definition**: coefficient of thermal expansion ε** = **α**L**∆**T** **Linear **
Coefficient of thermal volume expansion ε**v = **α**T** **Volume **

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

reactor ∆T = 500o_{C } _{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}-6_{∆}_{T = 2.3 x 10}6_{∆}_{T }

σy = 190 MPa: failure if ∆T > 82.6 K

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

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**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

∴ 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)

### 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.

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 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ 23bd = 2 x 10-5_{ m}2

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.

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

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 1So you can work out stresses and strains in a system.

**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

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

**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**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*r*2

_{ = }

*2*

_{x}_{ + }

*2*

_{y}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* = π*D*4/64 (rod)

**Rectangular rod: **

b
d
### [

2 2### ]

3 3 + 12 = 12 12*d*

*b*

*bd*

*J*

*db*

*I*

*bd*

*I*

*yy*

*xx*⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = =

**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### 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){*D*4

_{ – 0.025}4

_{} }

### (

4 4### )

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

*D*

*r*

*y*

### π

### τ

⇒ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ D4_{ – 0.025}4

_{ = 6.24 x 10}-5

_{ D }

_{D }

_{≥}

_{ 4.15 cm }

### 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*

*D*3 4

### π

≈ so*t*

*D*

*T*

*D*3 4 = 2

### π

### τ

*t*

*D*

*T*2 2 =

### π

### τ

34**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

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?

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 *aij*are 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

**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:

'
'
'
'
'
*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}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*
*p _{i}*

_{ij}*= = = =*

_{j}The diagonal *T1*, *T2*, *T3* is called the PRINCIPAL AXIS.

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

**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

.and on the z-face we have: τ* _{zx}*, τ

*, σ*

_{zy}

_{zz}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

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 theplane 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

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*

*a*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 '

_{ij}### 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*

*a*

_{ik}

_{jl}

_{kl}*ij*43

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

− = + − = − + + = + = − − + = + =**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

**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
σ_{y} = 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)

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}
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.

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 = 1 3

### [

σ1 + σ2 + σ3### ]

U_{D}= 1 2E σ1 2

_{ + }

_{σ}2 2

_{ + }

_{σ}32 + 2υ

### (

σ1σ2 + σ2σ3 + σ3σ1### )

-_{2E}1 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}

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### (

### ) (

### ) (

### )

### {

σ_{1}-σ

_{2}2 + σ

_{2}+σ

_{3}2+ σ

_{3}-σ

_{1}2

### }

> 2σ_{y}2

### Lets do a simple example.

### 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

*≈0 Mohr's Circle τ σ 100 50 s (100,τ) (50,−τ) Circle construction*

_{r}*s*= 75 N/mm2

*t*= √(252+τ2) The principal stresses

### σ

_{1,2 = }

*s ± t*

Thus

### σ

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

_{2}### τ

is small.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

*> 0) Thus failure when*

_{2}### τ

_{max}

### =

1_{2}

### σ

_{y}### =

### 105

*N*

### /

*mm*

2
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

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

_{max}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

### °

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### {

### }

13Or

### (

### 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

*2*

_{y}### 2

*s*

2 ### +

### 6

*t*

2 ### =

### 2

### σ

*2*

_{y}*s*

2 _{+}

_{3}

_{t}

2 _{t}

_{=}

_{σ}

*y*2

### ⇒

### 75

2### +

### 3(25

2### +

### τ

2_{)}

_{=}

_{210}

2_{ }

_{ }

_{τ}

_{= 110 N / mm}

_{= 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.

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

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

_{x}### δ

*in the x-direction and*

_{xx }### δ

_{xy}_{ in the y-direction. }

γ1 δx δxy

lx δxx Figure 1

We can define strains

### ε

*and*

_{xx}### ε

*by,*

_{xy }x
xy
xy
x
xx
xx
l
;
l
δ
=
ε
δ
=
ε _{ }

### ε

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

_{xx}### ε

*represents rotation of the vector through the small angle*

_{xy }### 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

*and*

_{yy}### ε

_{yx = }### γ

_{2}by,

*y*
*yx*
*yx*
*y*
*yy*
*yy*
*l*
*l*

### δ

### ε

### δ

### ε

= ; =2
as in Figure 2.
γ
δ
l
l_{x}
y
1
2
y
yy
δ
γ
Figure 2

Or, in general terms:

3
,
2
,
1
,
e
wher
; =
= *i* *j*
*l _{i}*

*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

*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 }γ* _{xy}* =(γ

_{1}+γ

_{2})=ε

*+ε*

_{xy}*γ*

_{yx }or*= −(γ*

_{xy}_{1}+γ

_{2}) 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

*and*

_{xy}### τ

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

_{yx }### γ

*as +ve when the angle ABC decreases. Thus*

_{xy }3

γ* _{xy}* = +(γ

_{1}+γ

_{2}) Or, in general terms:

)
( _{ij}_{ji}*ij*

### ε

### ε

### γ

= + and since### τ

_{ij}=### τ

*, we have*

_{ji}### 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*.

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*

*a*

_{ik}

_{jl}

_{kl}*ij*And hence:

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

− = + − = = − + + = + = − − + = + =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}

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

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

*). y x τo τ*

_{o}_{o}X Y P Q τ

_{o}τo

The Mohr’s circle for stresses is

where P and Q are principal stress axes and

σ* _{pp}* =σ

_{1}=τ

*σ*

_{o}*=σ*

_{qq}_{2}= −τ

*τ*

_{o}*=τ*

_{pq}*=0*

_{qp}Since the principal stress and strain axes are coincident,

ε* _{pp}*= ε

_{1}= σ1

*E*− νσ

_{2}

*E*

*= τ*

*o*

*E*(1+ν) ε

*=ε*

_{qq}_{2}= σ2

*E*− νσ

_{1}

*E*= −τ

_{o}*E*(1+ν) and the Mohr’s circle for strain is thus

X Y P Q

### ε

_{qq}_{ε}

*ε γ/2*

_{pp}the Mohr’s circle shows that

ε* _{xx}* =0
−γ

*2 = − τ*

_{xy}

_{o}*E*(1+ν) (0,−γ )

_{xy}

9 So pure shear causes the shear strain

### γ

τo
τ_{o}
γ/2
γ/2_{ and }

### γ

### =

### 2

### τ

**o****E**

**E**

### (1

### +

### ν

### )

But by definition### τ

_{o = G}### γ

so**G**

**G**

### =

### τ

_{o}### γ

### =

**E**

**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*

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 **