Pressure Vessel (Stress Analysis)
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Transcript of Pressure Vessel (Stress Analysis)
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CET 1: Stress Analysis & Pressure Vessels
Lent Term 2005
Dr. Clemens Kaminski
Telephone: +44 1223 763135
E-mail: [email protected] URL: http://www.cheng.cam.ac.uk/research/groups/laser/
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Synopsis
1 Introduction to Pressure Vessels and Failure Modes 1.1 Stresses in Cylinders and Spheres 1.2 Compressive failure. Euler buckling. Vacuum vessels 1.3 Tensile failure. Stress Stress Concentration & Cracking
2 3-D stress and strain 2.1 Elasticity and Strains-Young's Modulus and Poisson's Ratio 2.2 Bulk and Shear Moduli 2.3 Hoop, Longitudinal and Volumetric Strains 2.4 Strain Energy. Overfilling of Pressure Vessels
3 Thermal Effects 3.1 Coefficient of Thermal Expansion 3.2 Thermal Effect in cylindrical Pressure Vessels 3.3 Two-Material Structures
4 Torsion. 4.1 Shear Stresses in Shafts - /r = T/J = G/L 4.2 Thin Walled Shafts 4.3 Thin Walled Pressure Vessel subject to Torque
5 Two Dimensional Stress Analysis 5.1 Nomenclature and Sign Convention for Stresses 5.2 Mohr's Circle for Stresses 5.3 Worked Examples 5.4 Application of Mohr's Circle to Three Dimensional Systems
6 Bulk Failure Criteria 6.1 Tresca's Criterion. The Stress Hexagon 6.2 Von Mises' Failure Criterion. The Stress Ellipse
7 Two Dimensional Strain Analysis 7.1 Direct and Shear Strains 7.2 Mohr's Circle for Strains 7.3 Measurement of Strain - Strain Gauges 7.4 Hookes Law for Shear Stresses
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Supporting Materials There is one Examples paper supporting these lectures. Two good textbooks for further explanation, worked examples and exercises are
Mechanics of Materials (1997) Gere & Timoshenko, publ. ITP [ISBN 0-534-93429-3]
Mechanics of Solids (1989) Fenner, publ. Blackwell [ISBN 0-632-02018-0] This material was taught in the CET I (Old Regulations) Structures lecture unit and was examined in CET I (OR) Paper IV Section 1. There are consequently a large number of old Tripos questions in existence, which are of the appropriate standard. From 1999 onwards the course was taught in CET1, paper 5. Chapters 7 and 8 in Gere and Timoshenko contain a large number of example problems and questions. Nomenclature
The following symbols will be used as consistently as possible in the lectures. E Youngs modulus G Shear modulus I second moment of area J polar moment of area R radius t thickness T thermal expansivity linear strain shear strain angle Poissons ratio Normal stress Shear stress
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A pressure vessel near you!
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Ongoing Example
We shall refer back to this example of a typical pressure vessel on several occasions. Distillation column
2 m
18 m
P = 7 bara carbon steelt = 5 mm
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1. Introduction to Pressure Vessels and Failure Modes Pressure vessels are very often
spherical (e.g. LPG storage tanks) cylindrical (e.g. liquid storage tanks) cylindrical shells with hemispherical ends (e.g. distillation col-
umns)
Such vessels fail when the stress state somewhere in the wall material ex-ceeds some failure criterion. It is thus important to be able to be able to un-derstand and quantify (resolve) stresses in solids. This unit will con-centrate on the application of stress analysis to bulk failure in thin walled vessels only, where (i) the vessel self weight can be neglected and (ii) the thickness of the material is much smaller than the dimensions of the vessel (D t).
1.1. Stresses in Cylinders and Spheres Consider a cylindrical pressure vessel
internal gauge pressure P
L
r
h
External diameter D
wall thickness, t
L
The hydrostatic pressure causes stresses in three dimensions. 1. Longitudinal stress (axial) L 2. Radial stress r3. Hoop stress h all are normal stresses.
SAPV LT 2005 CFK, MRM
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L
r
h
L
r
h
a, The longitudinal stress L
L
P
Force equilibrium L
2
t D = P 4D
t4D P =
tensileis then 0, > P if
L
L
b, The hoop stress h
SAPV LT 2005 2CFK, MRM
t2D P =
tL 2 = P L D balance, Force
h
h
P
P
P
h
h
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c, Radial stress
r
r
r o ( P )
varies from P on inner surface to 0 on the outer face
h , L P ( D2 t ).thin walled, so D >> t
so h , L >> rso neglect r
Compare terms
d, The spherical pressure vessel
h
P
P
t4D P =
tD = 4D P
h
h
2
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1.2. Compressive Failure: Bulk Yielding & Buckling Vacuum Vessels Consider an unpressurised cylindrical column subjected to a single load W. Bulk failure will occur when the normal compressive stress exceeds a yield criterion, e.g.
W
bulk = WDt = Y Compressive stresses can cause failure due to buckling (bending instability). The critical load for the onset of buckling is given by Euler's analysis. A full explanation is given in the texts, and the basic results are summarised in the Structures Tables. A column or strut of length L supported at one end will buckle if
W = 2EI
L2
Consider a cylindrical column. I = R3t so the compressive stress required to cause buckling is
buckle = WDt =2ED3t
8L2 1Dt =
2ED28L2
or
buckle = 2 E
8 L D( )2
SAPV LT 2005 CFK, MRM
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where L/D is a slenderness ratio. The mode of failure thus depends on the geometry:
L /D ratio
Short Long
y
stress
Euler buckling locus
Bulk yield
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Vacuum vessels. Cylindrical pressure vessels subject to external pressure are subject to com-pressive hoop stresses
t2D P = h
Consider a length L of vessel , the compressive hoop force is given by,
2L D P = t L h
If this force is large enough it will cause buckling.
length
Treat the vessel as an encastered beam of length D and breadth L
SAPV LT 2005 CFK, MRM
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Buckling occurs when Force W given by.
2
2
)D (EI4W
= ( )22
D EI4=
2L D P
12 tL =
12 tb= I
33 3
=buckle Dt
32E p
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1.3. Tensile Failure: Stress Concentration & Cracking
Consider the rod in the Figure below subject to a tensile load. The stress dis-tribution across the rod a long distance away from the change in cross sec-tion (XX) will be uniform, but near XX the stress distribution is complex.
D
d
W
X X
There is a concentration of stress at the rod surface below XX and this value should thus be considered when we consider failure mechanisms. The ratio of the maximum local stress to the mean (or apparent) stress is de-scribed by a stress concentration factor K
K = maxmean
The values of K for many geometries are available in the literature, including that of cracks. The mechanism of fast fracture involves the concentration of tensile stresses at a crack root, and gives the failure criterion for a crack of length a a = Kc
where Kc is the material fracture toughness. Tensile stresses can thus cause failure due to bulk yielding or due to cracking.
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crack = Kc 1
a
length of crack. a
stress
failure locus
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2. 3-D stress and strain 2.1. Elasticity and Yield Many materials obey Hooke's law
= E applied stress (Pa) E Young's modulus (Pa) strain (-)
failure
Yield Stress
Elastic Limit
up to a limit, known as the yield stress (stress axis) or the elastic limit (strain axis). Below these limits, deformation is reversible and the material eventually returns to its original shape. Above these limits, the material behaviour depends on its nature. Consider a sample of material subjected to a tensile force F.
F F1
2
3
An increase in length (axis 1) will be accompanied by a decrease in dimensions 2 and 3.
Hooke's Law 1 = 1 F / A( )/ E 10
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The strain in the perpendicular directions 2,3 are given by
2 = 1E ;3 = 1E
where is the Poisson ratio for that material. These effects are additive, so for three mutually perpendicular stresses 1, 2, 3;
1
2
3
Giving
1 = 1E 2E
3E
2 = 1E +2E
3E
3 = 1E 2
E+ 3
E
Values of the material constants in the Data Book give orders of magnitudes of these parameters for different materials;
Material E (x109 N/m2)
Steel 210 0.30
Aluminum alloy 70 0.33
Brass 105 0.35
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2.2 Bulk and Shear Moduli These material properties describe how a material responds to an applied stress (bulk modulus, K) or shear (shear modulus, G).
The bulk modulus is defined as Puniform = Kv i.e. the volumetric strain resulting from the application of a uniform pressure. In the case of a pressure causing expansion 1 = 2 = 3 = P so
1 = 2 = 3 = 1E 1 2 3[ ]= PE (1 2)v = 1 + 2 + 3 = 3PE (1 2)K = E
3(1 2)
For steel, E = 210 kN/mm2, = 0.3, giving K = 175 kN/mm2 For water, K = 2.2 kN/mm2 For a perfect gas, K = P (1 bara, 10-4 kN/mm2)
Shear Modulus definition = G - shear strain
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2.3. Hoop, longitudinal and volumetric strains (micro or millistrain)
Fractional increase in dimension:
L length h circumference rr wall thickness (a) Cylindrical vessel:
Longitudinal strain
L = LE - h
E-
rE
= PD4tE
1 - 2( ) = LL
Hoop strain:
h = nE - L
E =
PD4tE
2 - ( ) = DD
= RR
radial strain
r = 1E r - h - L[ ] = - 3PD4ET = tt
[fractional increase in wall thickness is negative!]
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[ONGOING EXAMPLE]:
L = 1E L - n( ) = ( )[ ]669 10 x 120 3.0 - 10 x 0610x 210 1 = 1.14 x 10-4 - 0.114 millistrain h = 0.486 millistrain
r = -0.257 millistrain Thus: pressurise the vessel to 6 bar: L and D increase: t decreases
Volume expansion
Cylindrical volume: Vo = Do2
4
Lo (original)
New volume V = 4
Do + D( )2 Lo + L( ) = Lo Do
2
41 + h[ ]2 1 + L[ ]
Define volumetric strain v = VV
v = V - VoVo = 1 + h( )2
1 + L( ) - 1 = 1 + 2 h + h2( )1 + L( ) - 1 v = 2h + L + h2 + 2h L + Ln2 Magnitude inspection:
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max steel( ) = yE = 190 x 106
210 x 109 = 0.905 x 103 small
Ignoring second order terms,
v = 2 h + L (b) Spherical volume:
[ ] ( ) - 14EtPD = - - 1 = rLhh E
so ( ){ }
6 - +
6 = 3
33
o
oov D
DDD
= (1 + h)3 1 3h + 0(2) (c) General result
v = 1 + 2 + 3 ii are the strains in any three mutually perpendicular directions. {Continued example} cylinder L = 0.114 mstrain n = 0.486 mstrain rr = -0.257 mstrain v = 2n + L new volume = Vo (1 + v)
Increase in volume = D2 L
4 v = 56.55 x 1.086 x 10-3
= 61 Litres
Volume of steelo = DLt = 0.377 m3v for steel = L + h + rr = 0.343 mstrains increase in volume of steel = 0.129 L
Strain energy measure of work done
Consider an elastic material for which F = k x
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Work done in expanding x dW = Fx
work done
F
x
L 0
A=area
Work done in extending to x1 w = Fdxox1 = k x dx = kx1
2
2ox1 = 1
2 F1x1
Sample subject to stress increased from 0 to 1: Extension Force:
x1 = Lo1F1 = A1
W = ALo11
2 (no direction here)
Strain energy, U = work done per unit volume of material, U = ALo112 ALo( )
U = Alo2
1
Alo
11 U =
112
= 122E
In a 3-D system, U = 12
[11 + 22 + 33]
Now 1 = 1E - 2
E -
3E
etc
So U = 1
2E 12 + 22 + 32 - 2 12 + 23 + 3 1( )[ ]
Consider a uniform pressure applied: 1 = 2 = 3 = P
U = 3P2
2E 1 - 2( ) = P2
2K energy stored in system (per unit vol.
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For a given P, U stored is proportional to 1/K so pressure test using liquids rather than gases.
{Ongoing Example} P 6 barg . V = 61 x 10-3 m3 increase in volume of pressure vessel
Increasing the pressure compresses the contents normally test with water.
V water? v water( ) = PK = 6 x 105
2.2 x 109 = - 0.273 mstrains
decrease in volume of water = -Vo (0.273) = -15.4 x 10 3 m3Thus we can add more water:
Extra space = 61 + 15.4 (L)
= 76.4 L water
p=0 p=6
extra space
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3. Thermal Effects
3.1. Coefficient of Thermal Expansion
Definition: coefficient of thermal expansion = LT Linear Coefficient of thermal volume expansion v = T Volume
Steel: L = 11 x 10-6 K-1 T = 10oC L = 11 10-5 reactor T = 500oC L = 5.5 millistrains (!)
Consider a steel bar mounted between rigid supports which exert stress
Heat
E - T =
If rigid: = 0 so = ET (i.e., non buckling)
steel: = 210 x 109 x 11 x 10-6 T = 2.3 x 106 T y = 190 MPa: failure if T > 82.6 K
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{Example}: steam main, installed at 10oC, to contain 6 bar steam (140oC)
if ends are rigid, = 300 MPa failure. must install expansion bends.
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3.2. Temperature effects in cylindrical pressure vessels
D = 1 m . steel construction L = 3 m . full of water t = 3 mm Initially un pressurised full of water: increase temp. by T: pressure rises to Vessel P.
The Vessel Wall stresses (tensile) P 83.3 = 4
= t
PDL
n = 2L = 166.7 P
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Strain (volume)
T + E
- E
= lhLL
= 0.3E = 210 x 109 = 11 x 106
L = 1.585 x 10-10 P + 11 x 10-6 T
Similarly
h = 6.75 x 10-10 P + 11 10 6 T v = L + 2n = 15.08 x 10-10 P + 33 x 10-6 T = vessel vol. Strain vessel expands due to temp and pressure change.
The Contents, (water) Expands due to T increase
Contracts due to P increase:
v, H2O = vT P/K H2O = v = 60 x 10-6 K-1 v = 60 x 10-6 T 4.55 x 10-10 P
Since vessel remains full on increasing T: v (H20) = v (vessel) Equating P = 13750 T
pressure, rise of 1.37 bar per 10C increase in temp. Now n = 166.7 P = 2.29 x 106 T n = 22.9 Mpa per 10C rise in Temperature
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Failure does not need a large temperature increase. Very large stress changes due to temperature fluctuations.
MORAL: Always leave a space in a liquid vessel.
(v, gas = vT P/K)
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3.3. Two material structures
Beware, different materials with different thermal expansivities
can cause difficulties.
{Example} Where there is benefit. The Bimetallic strip - temperature
controllers
a= 4 mm (2 + 2 mm) b = 10 mm
Cu
Fe
L = 100mm b
a
d
Heat by T: Cu expands more than Fe so the strip will bend: it will bend in an arc as all sections are identical.
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CuFe
The different thermal expansions, set up shearing forces
in the strip, which create a bending moment. If we apply a sagging bending
moment of equal [: opposite] magnitude, we will obtain a straight beam and
can then calculate the shearing forces [and hence the BM].
CuFe
F
F
Shearing force F compressive in Cu
Tensile in Fe
Equating strains: Fe
Fecu
cu bdEF + T =
bdEF - T
So ( ) T - = E1 +
E1
bdF
FecuFecu
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bd = 2 x 10-5 m2
Ecu = 109 GPa cu 17 x 10-6 k-1 T = 30C F = 387 NEFe = 210 GPa Fe = 11 x 10-6 K-1 (significant force)
F acts through the centroid of each section so BM = F./(d/2) = 0.774 Nm
Use data book to work out deflection.
EI
ML2
= 2
This is the principle of the bimetallic strip.
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Consider a steel rod mounted in a upper tube spacer
Analysis relevant to Heat Exchangers
Cu
Fe
assembled at room temperature . increase T Data: cu > Fe : copper expands more than steel, so will generate a TENSILE stress in the steel and a compressive
stress in the copper.
Balance forces:
Tensile force in steel |FFe| = |Fcu| = F Stress in steel = F/AFe = Fe copper = F/Acu = cuSteel strain: FE = Fe T + Fe/EFe (no transvere forces) = FeT + F/EFeAFecopper strain cu = cuT F/EcuAcu
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Strains EQUAL: ( ) TFeEAEAF dstrengthsofsum
cucuFeFe
cu =
1 + 1 -
444 3444 21
So you can work out stresses and strains in a system.
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4. Torsion Twisting Shear stresses
4.1. Shear stresses in shafts /r = T/J = G /L
Consider a rod subject to twisting:
Definition : shear strain change in angle that was originally /2 Consider three points that define a right angle and more then:
Shear strain = 1 + 2 [RADIANS]
A
B C
A
BC
1
2
Hookes Law = G G shear modulus = E2 1 + ( )
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Now consider a rod subject to an applied torque, T.
2r
T
Hold one end and rotate other by angle .
B
B
L
B
B
Plane ABO was originally to the X-X axis
Plane ABO is now inclined at angle to the axis: tan Lr =
Shear stress involved L
Gr = G =
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Torque required to cause twisting:
r dr
.T = 2 r.r r or drr 2 = T
A
2 A
dA.r = GrL
T dAr L
G 2= { }J
LG = so
rLG
JT = =
cf y
= RE =
IM
DEFN: J polar second moment of area
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Consider a rod of circular section:
42
2 = .2 RdrrrJ
R
o
=
r
y
x
32
4DJ =
Now
r2 = x2 + y2
It can be shown that J = Ixx + Iyy [perpendicular axis than] see Fenner this gives an easy way to evaluate Ixx or Iyy in symmetrical geometrics:
Ixx = Iyy = D4/64 (rod)
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Rectangular rod:
b
d
[ ]223
3
+ 12
=
12
12dbbdJ
dbI
bdI
yy
xx
=
=
31
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Example: steel rod as a drive shaft
D = 25 mm
L = 1.5 m
Failure when = y = 95 MPa G = 81 GPa
Now Nm 291 = T so 10 x 383 =
32 =
0125.010 x 95 =
484
6
max
max
=
mDJ
JT
r
From ( )
5.1 10 x 81 = 10 x 6.7 =
99
JT
LG
= 0.141 rad = 8.1
Say shaft rotates at 1450 rpm: power = T = 1450 x
602 x 291
= 45 kW
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4.2. Thin walled shafts (same Eqns apply)
Consider a bracket joining two Ex. Shafts:
T = 291 Nm
0.025mD min
What is the minimum value of D for connector?
rmax = D/2
J = (/32){D4 0.0254}
( )446
max 025.0 - 32 291 = 2 x 10 x 95
JT =
DDry
D4 0.0254 = 6.24 x 10-5 D D 4.15 cm
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4.3. Thin walled pressure vessel subject to torque
JT =
r
now cylinder ( )[ ]44 - 2 + 32
= DtDJ
[ ]... + 24 + 8 32
223 tDtD=
tD3 4
so tD
TD 3
4 = 2
tD
T2
2 =
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CET 1, SAPV
5. Components of Stress/ Mohrs Circle
5.1 Definitions
Scalars tensor of rank 0
Vectors tensors of rank 1
ii maF
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
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CET 1, SAPV
vessel. However it is clear that the components of the stress tensor will be different going from one set of coordinates xi to another xi. How do we transform one set of co-ordinate axes onto another, keeping the same origin?
3332313
2322212
1312111
321
'''
aaaxaaaxaaaxxxx
... where aij are the direction cosines Forward transformation:
=
= 31
'j
jiji xax New in terms of Old
Reverse transformation:
=
= 31j
jjii xax Old in terms of New
We always have to do summations in co-ordinate transformation and it is conventional to drop the summation signs and therefore these equations are simply written as:
jjii
jiji
xax
xax
=
='
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CET 1, SAPV
Tensor transformation How will the components of a tensor change when we go from one co-ordinate system to another? I.e. if we have a situation where
form)short (in ==j
jijjiji qTqTp
where Tij is the tensor in the old co-ordinate frame xi, how do we find the corresponding tensor Tij in the new co-ordinate frame xi, such that:
form)short (in ''''' ==j
jijjiji qTqTp
We can find this from a series of sequential co-ordinate transformations:
'' qqpp Hence:
'
'
jjll
lklk
kiki
qaq
qTp
pap
=
=
=
Thus we have:
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CET 1, SAPV
''
'
''
jij
jkljlik
jjlkliki
qT
qTaa
qaTap
=
=
=
For example:
333332233113
233222222112
133112211111
33
22
11'
TaaTaaTaa
TaaTaaTaa
TaaTaaTaa
Taa
Taa
TaaT
jijiji
jijiji
jijiji
ljli
ljli
ljliij
+++
+++
++=
+
+
=
Note that there is a difference between a transformation matrix and a 2nd rank tensor: They are both matrices containing 9 elements (constants) but:
Symmetrical Tensors: Tij=Tji
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CET 1, SAPV
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CET 1, SAPV
We can always transform a second rank tensor which is symmetrical:
=
3
2
1
000000
'
:such that
'
TT
TT
TT
ij
ijij
Consequence? Consider:
333222111 ,,
then
qTpqTpqTp
qTp jiji
===
=
The diagonal T1, T2, T3 is called the PRINCIPAL AXIS. If T1, T2, T3 are stresses, then these are called PRINCIPAL STRESSES.
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CET 1, SAPV
Mohrs circle Consider an elementary cuboid with edges parallel to the coordinate directions x,y,z.
x
y
z
z face
y face
x face
Fxy
Fx
Fxx Fxz
The faces on this cuboid are named according to the directions of their normals. There are thus two x-faces, one facing greater values of x, as shown in Figure 1 and one facing lesser values of x (not shown in the Figure). On the x-face there will be some force Fx. Since the cuboid is of infinitesimal size, the force on the opposite side will not differ significantly. The force Fx can be divided into its components parallel to the coordinate directions, Fxx, Fxy, Fxz. Dividing by the area of the x-face gives the stresses on the x-plane:
xz
xy
xx
It is traditional to write normal stresses as and shear stresses as . Similarly, on the y-face: yzyyyx ,,
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CET 1, SAPV
and on the z-face we have: zzzyzx ,, There are therefore 9 components of stress;
=
zzzyzx
yzyyyx
xzxyxx
ij
Note that the first subscript refers to the face on which the stress acts and the second subscript refers to the direction in which the associated force acts.
y
x
yy
yy
xxxx
xy
yx
xy
yx
But for non accelerating bodies (or infinitesimally small cuboids): and therefore:
=
=
zzyzxz
yzyyxy
xzxyxx
zzzyzx
yzyyyx
xzxyxx
ij
Hence ij is symmetric!
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CET 1, SAPV
This means that there must be some magic co-ordinate frame in which all the stresses are normal stresses (principal stresses) and in which the off diagonal stresses (=shear stresses) are 0. So if, in a given situation we find this frame we can apply all our stress strain relations that we have set up in the previous lectures (which assumed there were only normal stresses acting). Consider a cylindrical vessel subject to shear, and normal stresses (h, l, r). We are usually interested in shears and stresses which lie in the plane defined by the vessel walls. Is there a transformation about zz which will result in a shear Would really like to transform into a co-ordinate frame such that all components in the xi :
'ijij So stress tensor is symmetric 2nd rank tensor. Imagine we are in the co-ordinate frame xi where we only have principal stresses:
=
3
2
1
000000
ij
Transform to a new co-ordinate frame xi by rotatoin about the x3 axis in the original co-ordinate frame (this would be, in our example, z-axis)
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CET 1, SAPV
The transformation matrix is then:
=
=
1000cossin0sincos
333231
232221
131211
aaaaaaaaa
aij
Then:
++++
=
=
=
3
22
2121
212
22
1
3
2
1
000sincossincossincos0sincossincossincos
000000
1000cossin0sincos
1000cossin0sincos
'
aa kljlikij
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CET 1, SAPV
Hence:
2sin)(21
sincossincos'
2cos)(21)(
21
cossin'
2cos)(21)(
21
sincos'
12
2112
1211
22
2122
1211
22
2111
=+=
++=+=
+=+=
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Yield conditions. Tresca and Von Mises
Mohrs circle in three dimensions.
xz
y
normal stresses
Shear stresses
x, y plane
y,z plane
x,z plane
1
-
We can draw Mohrs circles for each principal plane.
6. BULK FAILURE CRITERIA
Materials fail when the largest stress exceeds a critical value. Normally we
test a material in simple tension:
P P
y = Pyield
A
= =
This material fails under the stress combination (y, 0, 0) max = 0.5 y = 95 Mpa for steel We wish to establish if a material will fail if it is subject to a stress
combination (1, 2, 3) or (n, L, ) 2
-
Failure depends on the nature of the material:
Two important criteria
(i) Trescas failure criterion: brittle materials
Cast iron: concrete: ceramics
(ii) Von Mises criterion: ductile materials
Mild steel + copper
6.1. Trescas Failure Criterion; The Stress Hexagon (Brittle)
A material fails when the largest shear stress reaches a critical value, the
yield shear stress y. Case (i) Material subject to simple compression:
1 1
Principal stresses (-1, 0, 0)
-
3
-
M.C: mc passes through (1,0), (1,0) , ( 0,0)
1 1 max
max = 1/2 occurs along plane at 45 to 1Similarly for tensile test.
Case (ii) 2 < 0 < 1
- 1
2
M.C. Fails when
1 - 22
= max = y =
y2
i.e., when 1 - 2 = y material will not fail.
4
-
5
ets do an easy example. L
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6
Criterion; The stress ellipse 6.2 Von Mises Failure (ductile materials)
Trescas criterion does not work well for ductile materials. Early hypothesis
material fails when its strain energy exceeds a critical value (cant be true
ailure when strain energy due to distortion, UD, exceeds a
) due of a compressive stress C equal to
the mean of the principal stresses.
as no failure occurs under uniform compression).
Von Mises: f
critical value.
UD = difference in strain energy (U
C = 13
1 + 2 + 3[ ] UD =
12E
12 + 22 + 32 + 2 12 + 23 + 31( ) - 12E 3C2 + 6C2[ ]
112G
1 - 2( )2 + 2 - 3( )2 + 3 - 1( )2{ } =M.C.
1
2 3
-
7
Tresca failure when max (I) y) Von Mises failure when root mean
Compare with (y, , ) . simple tensile test failure
square of {a, b, c} critical value
y
Failure if
( ) ( ) ( ){ } { } + 0 + G
11 2y
22y 12 > - + + + - G12
213
232
221
( ) ( ) ( ){ } 2 > - + + + - 2y213232221
-
8
-
Lets do a simple example.
9
-
Example Tresca's Failure Criterion The same pipe as in the first example (D = 0.2 m, t' = 0.005 m) is subject to an internal pressure of 50 barg. What torque can it support?
Calculate stresses
L = PD4t' = 50N / mm2
h = PD2t' =100N / mm2
and 3 = r 0
Mohr's Circle
10050s
(100,)
(50,)
Circle construction s = 75 N/mm2 t = (252+2) The principal stresses 1,2 = s t Thus 2 may be positive (case A) or negative (case B). Case A occurs if is small.
10
-
10050s
(100,)
(50,)
123
Case A
Case B
2
10050s
(100,)
(50,)
13
2
We do not know whether the Mohr's circle for this case follows Case A or B; determine which case applies by trial and error. Case A; 'minor' principal stress is positive (2 > 0)
Thus failure when max = 12 y =105N / mm2
11
-
For Case A;
max = 12 =12 75 + 252 + 2( )[ ]
2 =135 252 = 132.7N / mm2
Giving 1 = 210N / mm2 ; 2 = 60N / mm2 Case B; We now have max as the radius of the original Mohr's circle linking our stress data. Thus
max = 252 + 2 = 105 = 101.98N / mm2 Principal stresses
1,2 = 75 105 1 = 180N / mm2 ; 2 = 30N / mm2 Thus Case B applies and the yield stress is 101.98 N/mm2. The torque required to cause failure is
T = D2t' / 2 = 32kNm Failure will occur along a plane at angle anticlockwise from the y (hoop) direction;
tan(2 ) = 102
75 2 = 76.23 ;
2 = 90- 2 = 6.9 12
-
Example More of Von Mises Failure Criterion From our second Tresca Example
h = 100N / mm2 L = 50N / mm2r 0N / mm2
What torque will cause failure if the yield stress for steel is 210 N/mm2? Mohr's Circle
10050s
(100,)
(50,)
Giving
1 = s + t = 75 + 252 + 22 = s t = 75 252 + 23 = 0
At failure
UD = 112G (1 2 )2 + (2 3)2 + (3 1)2{ }
13
-
Or
(1 2 )2 + (2 3)2 + (3 1)2 = (y )2 + (0)2 + (0 y)24t2 + (s t)2 + (s + t)2 = 2 y22s2 + 6t2 = 2y2s2 + 3t2 = y2 752 + 3(252 + 2 ) = 2102 = 110N / mm2 The tube can thus support a torque of
T = D2t' 2
= 35kNm which is larger than the value of 32 kNm given by Tresca's criterion - in this case, Tresca is more conservative.
14
-
1
7. Strains 7.1. Direct and Shear Strains
Consider a vector of length lx lying along the x-axis as shown in Figure
1. Let it be subjected to a small strain, so that, if the left hand end is fixed the right hand end will undergo a small displacement x. This need not be in the x-direction and so will have components xx in the x-direction and xy in the y-direction.
1 x xy
lx xx Figure 1
We can define strains xx and xy by,
x
xyxy
x
xxxx l
;l
== xx is the direct strain, i.e. the fractional increase in length in the direction of the original vector. xy represents rotation of the vector through the small angle 1 where,
xyx
xy
xxx
xy11 ll
tan =+=
Thus in the limit as x 0, 1 xy. Similarly we can define strains yy and yx = 2 by,
y
yxyx
y
yyyy ll
== ;
-
2
as in Figure 2.
l
lx
y
1
2
y
yy
yx
Figure 2
Or, in general terms:
3,2,1, e wher ; == jili
ijij
The ENGINEERING SHEAR STRAIN is defined as the change in an angle relative to a set of axes originally at 90. In particular xy is the change in the angle between lines which were originally in the x- and y-directions. Thus, in our example (Figure 2 above): xy = (1 + 2 ) = xy + yx or xy = (1 + 2 ) depending on sign convention.
A
B C
A'
B'C'
Figure 3a Figure 3b
xy
yx
Positive values of the shear stresses xy and yx act on an element as shown in Figure 3a and these cause distortion as in Figure 3b. Thus it is sensible to take xy as +ve when the angle ABC decreases. Thus
-
3
xy = +(1 + 2) Or, in general terms: )( jiijij += and since
ij = ji,
we have ij = ji.
Note that the TENSOR SHEAR STRAINS are given by the averaged sum of shear strains:
jijiijij 21)(
21)(
21
21
21 =+=+= 7.2 Mohrs Circle for Strains The strain tensor can now be written as:
=
=
332313
232212
131211
333231
232221
131211
21
21
21
21
21
21
21
21
21
21
21
21
yy
yy
yy
yy
yy
yy
ij
where the diagonal elements are the stretches or tensile strains and the off diagonal elements are the tensor shear strains. Thus our strain tensor is symmetrical, and:
-
4
jiij = This means there must be a co-ordinate transformation, such that:
=
3
2
1
000000
:such that
'
ij
ijij
we only have principal (=longitudinal) strains! Exactly analogous to our discussion for the transformation of the stress tensor we find this from:
++++
=
=
=
3
22
2121
212
22
1
3
2
1
000sincossincossincos0sincossincossincos
000000
1000cossin0sincos
1000cossin0sincos
'
aa kljlikij
And hence:
-
5
2sin)(21
sincossincos'21'
2cos)(21)(
21
cossin'
2cos)(21)(
21
sincos'
12
211212
1211
22
2122
1211
22
2111
=
+==
++=
+=
+=
+=
For which we can draw a Mohrs circle in the usual manner:
-
6
Note, however, that on this occasion we plot half the shear strain against the direct strain. This stems from the fact that the engineering shear strains differs from the tensorial shear strains by a factor of 2 as as discussed. 7.3 Measurement of Stress and Strain - Strain Gauges It is difficult to measure internal stresses. Strains, at least those on a surface, are easy to measure. Glue a piece of wire on to a surface Strain in wire = strain in material As the length of the wire increases, its radius decreases so its electrical resistance increases and can be readily measured. In practice, multiple wire assemblies are used in strain gauges, to measure direct strains.
strain gauge
___
12045
Strain rosettes are employed to obtain three measurements: 7.3.1 45 Strain Rosette Three direct strains are measured
1
C B A
principal strain
B
Mohrs circle for strains gives
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7
2
A
B
C
A B C
/2
circle, centre s, radius t so we can write
)2cos(
)2cos()2sin()902cos(
tststs
C
B
A
==++=
ts +=
3 equations in 3 unknowns Using strain gauges we can find the directions of Principal strains
/2
-
8
7.4 Hookes Law for Shear Stresses St. Venants Principle states that the principal axes of stress and strain are co-incident. Consider a 2-D element subject to pure shear (xy = yx = o).
y
x
oo
X
Y
PQ
oo
The Mohrs circle for stresses is
where P and Q are principal stress axes and
pp = 1 = oqq = 2 = o pq = qp = 0
Since the principal stress and strain axes are coincident,
pp = 1 = 1E
2E
= oE
1+ ( ) = = 2qq 2 E
1E
= oE
1+ ( )
and the Mohrs circle for strain is thus
X
Y
PQ
ppqq
/2 the Mohrs circle shows that
xx = 0 xy
2= o
E1 + ( )
(0, )xy
-
9
So pure shear causes the shear strain
oo
/2 /2 and = 2 o
E(1+ )
But by definition o = G so G =o =
E2(1+)
Use St Venants principal to work out principal stress values from a knowledge of principal strains. Two Mohrs circles, strain and stress.
1 = 1E 2
E 3
E
2 = 1E +2E
3E
3 = 1E 2
E+ 3
E
-
10
So using strain gauges you can work out magnitudes of principal strains.
You can then work out magnitudes of principal stresses.
Using Tresca or Von Mises you can then work out whether your
vessel is safe to operate. ie below the yield criteria