lectures 1-4 hand out 2012.pdf
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MEC302 The Integrity of Materials and Components
Introduction to the Module
Two definitions of Integrityare:
1. an unimpaired condition
2. the quality of being whole and complete
In Engineering we need to design materials, components, structures, objects, systems etc.
that have no defects and that are structurally sound, i.e are unimpairedand wholeand
complete. However if they do contain defects through a material fault or due to operating
conditions then we need to know about failure so that it can be avoided. So this module
considers failure of components, how to identify it, and how to prevent it.
We will begin by studying the stress analysis of structures which have axi-symmetric
stress distributions, such as rotating shafts and pressure vessels. We will study thestresses in these components, consider how they might fail, and look at ways to strengthen
their design.
Then we will move on to how to identify defects in components using non destructive
examinationand consider how we might choose the most appropriate technique for a
particular component.
A set of lectures on fracturewill consider how to assess the integrityof a component
under load which contains a known defect.
We will then consider the collapse of cracked structuresand how to assess the failure ofsuch structures using R6 Failure Assessment Diagrams. This technique was developed in
the nuclear industry for fail safe design.
Failure due to contact loads, friction and wearwill be addressed and a study will be
undertaken of how to prevent such failure.
In week 7 a case study will be set as a learning assignmentwhich brings together stress
analysis, defect sizing and fracture mechanics assessments of a component. Students willsubmit their solutions to this problem in Week 8 and feedbackwill be given on their
understanding in Week 9.
After looking at static loading of defects we will then consider fatigue loading.
All through the course we will consider the same case studies on rotating shafts and
pressure vessels, but using different approaches and loading.
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Lectures 1-4 Axisymmetric Stress Distributions
Background reading: Mechanics of Engineering Materials, 2ndEd. P P Benham, R J
Crawford and C G Armstrong, Longman 1996, pp 367-412
Several important design problems are solved by using the results of this analysis Thick discs subjected to rotation
e.g. turbine discs , gas turbines
Thick cylinders subjected to pressure loadinge.g. gas cylinders, pressure vessels, gun barrels
Also power generating equipment, rotors, alternators, and thermal strains in pipes all
covered by this theory.
We can derive a general theory to cover all cases then modify it for specific cases to suit a
specific problem.
General Equations of Equilibrium
For an axisymmetric stress-strain system, the stresses on a typical element of material will
be as shown, the radial, hoop and axial stresses all vary with radius.
r = radial stress
= hoop stress
RB= body force in units
force/unit volume e.g. due toacceleration
Figure 1Stresses on a typical element of material under an axisymmetric stress-strain
system in the radial and hoop directions
In order to choose an appropriate material for a design we need to determine the radial
stress, r , the hoop stress, and the axial stress, Aand compare these with the properties
of the materials available. First we must consider radial equilibrium of forces.
d
2
d
2
d
r
drdr
dr
r
RBdr
x
y
r
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dr
dr
dr
r
r
A
A
Axial direction
RB
r = radial stress
= hoop stress
A= axial stress
Figure 2Stresses on a typical element of material under an axisymmetric stress-strain
system in the radial, hoop and axial directions
Radial equilibrium of forces (consider unit thickness)
02
sin2...
ddrdrdrdrRddrrdr
dr
drB
rr
Assume:22
sin dd and neglect second order terms
We also assume that the stress field is axisymmetric and stresses vary only with radius.
hence on simplifying we get
0.. rRdr
dr B
rr
(1)
From Equation (1) we can see that we have two unknowns, and r, so we need another
equation in order to solve (1).
Therefore we now consider how a typical element of material deforms such that the whole
material remains continuous, i.e., we must ensure Compatibility of Displacements. Wedo this by considering the geometry of a typical displacement or deformation and using the
elastic Hookes Lawrelationships.
Compatible deformation
We have a disc which moves radially outwards due to rotational forces
r
dru
(u + du)
dr + du
Let :u = Radial displacement at radius, r
w = axial displacement at radius, r
(out of the page)
= strain with suffix denoting
direction
= Poissons ratio
Figure 3Deformation radially
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Considering the displacement geometry we have
dr
du
dr
drdudrr
radial strain
dA
dwA axial strain
ru
rrur
2
22 hoop strain
If we want to calculate radial displacement, we cannot obtain this from radial strain, but it
can be easily obtained from hoop strain.Now we use Hookes Law to express the strains in terms of stresses:
ArrEdr
du
1 (2)
rAEr
u
1 (3)
rAAEdA
dw 1 (4)
Using the straindisplacement relationships in equations (2,3,4) gives
dr
dErrR
dr
dr ABr
...
1.1
(5)
Note: The introduction of expressions for RBand Aenables (1) and (5) to be solvedsimultaneously.
Stresses due to the rotation of thick discs of uniform thickness .
Consider an element of unit axial thickness.
dr
r
centrifugal force Let the specific weight of the material =
Centrifugal force = 2r x (volume of
element)
runitvolume
forceRB2
Figure 4 Centrifugal force due to rotation
We must now make an assumption regarding A. For a thick disc we assume that the plane
sections remain plane. This implies that the axial strain Ais constant with radius.
i.e. A= constant
0dr
d A
Thus equation (1) can now be written as:
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0... 22 rdr
dr rr
(6)
and equation (5) as
22 ..1 rdr
dr r (7)
These lead to:
22
2
22
2
18
21
2
18
23
2
rr
BA
rr
BAr
(8)
These equations (8) solve the problem once we put in boundary conditions to find A/2 and
B. The boundary conditions will depend on the component to be studied and severalexamples follow.
******************************
Note: For all problems in this course we will assume that (8) is appropriate.
In practise, if the disc is thin, then we do not assume that 0dr
d A but assume that A=0.
This leads to
222
22
2
8
31
2
8
3
2
rr
BA
rr
BAr
(8a)
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Case Study On A Rotating Shaft
Find the maximum stresses in the shaft in Figure 5
Ro
rRo
r
Ro= 0
Figure 5A long shaft of radius, Rorotating at radians/second
To find A/2 and B we must specify appropriate boundary conditions:
i.e. rat r= Rowill be zero i.e. Ro= 0.
This is the only boundary stress we can specify in this problem. But we need twoconditions since we have two unknowns.
Examine equations (8). Both these equations include
2
r
Bas a term and we have material r
= 0.
If B 0, then 2r
B, which says that we will have infinite stress at the centre of the shaft,
which is impossible. Therefore, for a solid shaftB must equal 0.
Equations (8) become
22
22
18
21
2
1823
2
rA
rAr
We now use the condition r= 0 at r= Ro
22
18
23
20 oR
A
22
1823
2 oR
A
222
222
23
21
18
23
18
23
rR
rR
o
or
(9)
At the centre of the shaft r = 0,
22
18
23or R
which are the maximum stresses in the shaft.
****************************
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Case Study On A Hollow Rotating Shaft, or Hollow Cylinder
Derive an expression for the radial and hoop stresses for the hollow rotating shaft in
Figure 6
Ro
R1
r
R1
Ro
Figure 6A long hollow shaft of outer radius, Roand inner radius R1, rotating at
radians/second
The boundary conditionshere are that at r = R1, and r = Ro, r= 0, that is:
R1 = 0, and Ro = 0
The expression for r in equations (8) becomes
22
2
2
1
2
2
1
1
18
23
20
18
23
20
o
o
Ro
R
RR
BA
RR
BA
solving these simultaneously
22
1
2
22
1
2
18
23
18
23
2
o
o
RRB
RRA
2
2
22
122
1
2
2
2
22
122
1
2
23
21
18
23
18
23
rr
RRRR
rr
RRRR
oo
oor
(10)
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Rotating Components Example 1
(a) A thinsteel disc is rotated at 6000 r.p.m. Plot the radial and hoop stresses with
increasing radius if = 8000 kg/m3, Inner radius Ri= 0.075 m, Outer radius Ro= 0.3 m
(b) If this were a long cylinder, how would the stresses differ?
(Hint: Which equations should be used? Equations (8) or (8a)?)
***************************************************************************
Rotating Components Example 2
A rotor in the form of a long hollow cylinder is shrunk onto a shaft such that the interface
radial pressure due to shrinkage is 50MN/m2.
(a) Find the speed of rotation when the shaft ceases to drive the rotor
(b) Find the maximum stress in the cylinder at this speed
= 8000 kg/m3, E = 21 x 1010N/m2, = 0.3
Inner diameter di= 100 mm, Outer diameter do= 600 mm
Note: The radial stress due to shrinkage is compressive (i.e.ve)
The radial stress due to rotation is tensile (i.e. +ve)
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Lam Equations - Thick cylinders subjected to internal and external pressure
The same theory as has been developed for rotating thick walled cylinders may be used for
thick cylinders subjected to internal and external pressure, but now the body forces are
zero (since = 0).
Equations (8) lead to:
Radial stress:222 r
BC
r
BAr
Lam Equations
Hoop stress:222 r
BC
r
BA
Axial Stress: 0A for open ends
CE
E
AA
rAA
2
for closed ends, i.e. independent of r
As before, B and C are found from boundary conditions.
Note: rand both increaseas r decreases. So the maximum stresses always occur on
the inner surfaces in this case.
We can use the Lam Equationsis terms of the diameter, provided we are consistent.
i.e.
2
2
**
**
d
BC
dBCr
and remember that the constants C* and B* will be different from C and B.
o
2max
ro
ro
r
ri
r
Figure 7Stress distribution of radial and hoop
stress in a thick walled cylinder subject to
internal pressure
Hoop stresses are positive
Radial stresses are negative
(for internal pressure conditions)
Since the problem is axisymmetric
the hoop and radial stresses are
principal stresses.
i.e. 1= 2= r
22
r21max
The maximum shear stress in
the cylinder is at the inside
maxr
2
)i()i(
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Yielding in a cylinder
Let k = ro/rifor an internally pressurized cylinder with pressure P
Using the Lam Equations
2
2
r
BC
r
BCr
with the boundary conditions:
At r = ro, r= 20or
BC
At r = ri, r= 2ir
BCP
We obtain the expressions:
1
1
1
1
2
2
2
2
k
r
rP
k
r
rP
o
o
r
12 2
2
max
k
Pkr at r = ri
Since yield will commence in the bore, using Trescas criterion
2
2
2
max
11
2
12
kP
k
Pk
y
y
************************************************************************
Pressurised Cylinders Example 1
A cylinder where ri = 0.5 rohas an internal pressure, P and closed ends. Plot the stress
distribution along the radius and find the maximum shear stress.
************************************************************************
Pressurised CylindersExample 2
A pipe of 100 mm ID is subjected to internal water pressure of 10 MN/m2. If the pipe
material has a safe tensile stress of 20 MN/m2, and a safe shear stress of 40 MN/m
2, find
the OD of the pipe.
************************************************************************
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Turbine discs of variable thickness
Turbine discs are seldom flat, but they thicker near the shaft. Optimum weight is attained
if the stresses rand are equal, and uniform at all radii.
For varying thickness, we require a new equilibrium equation, as thickness, t varies with
radius, r.
0.... 22 rttrtdr
dt
rr
may be derived from the free body diagram.
If r= f(r), the solution gives:
ro
rtt
225.0exp
and this is an ideal shape to be used in turbines.
Uniform thickness cylinders
In order to strengthen a pressurized cylinder one would automatically make the walls
thicker. However for uniform thickness cylinders, above thickness, wopt ,we get very little
reduction in ior maxfor massive increases in wall thickness.
For internal pressure, is tensile, i.e. positive. If we can induce an initial negative, i.e.
compressive hoop stress, the initial pressurisation of the cylinder would be employed in
overcoming the negative initial .
In practice to contain a higher pressure one can:
(a)shrink fitone or more cylinders around the pressure containing cylinder (e.g.hydraulic cylinders, gun barrels etc.) to form a compound cylinder to induce
compression
(b)wind wire on a cylinder under tension to induce compression at the bore.(c)overstrain once with a pressure beyond yield, to leave compressive residual stress
distribution near the bore.
imax
Cylinder wall thickness
Wopt
Thickness, t
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Shrink Fit Example 1 - Comparison of stresses in a single cylinder withthose in a compound cylinder under the same internal pressure.
(a) Single Cylinder
2r
d
BA
2d
BA
Internal Pressure = 45 MPa
d = 0.1r
= -45 -45 = A + 100B
d = 0.2 r = 0 0 = A +
25B
} B = -0.6
A = 15
2r d
6.015
2d
6.015
Stresses are max at d = 0.1:
Max. Tensile Stress, 751.0
6.015
2 MPa
Max. Shear Stress, 601.0
6.0
2 2r
MPa
d (m) )MPa(r )MPa( )MPa( 0.1 -45 75 60
0.15 -11.7 41.7 26.7
0.2 0 30 15
(b) Compound Cylinder - An outer cylinder shrunk onto an inner cylinder.(Both cylinders are made of the same material)
Shrinkage pressure between cylinders = 7 MPa
Internal pressure = 45 MPa
Resultant Stress = Stress due to shrinkage+ stress due to internalpressure.
0.1 m
0.2 m
0.2 m
0.15 m
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Due to shrinkage pressure:Inner cylinder, B100A00,1.0d r
B9
400A77,15.0d r
}6.12A
126.0B
r = -12.6 + 2d126.0
= -12.6 - 2d
126.0
d, m MPa,r MPa,
0.1 0 -25.2
0.15 -7 -18.2
Outer
Cyl inder
''r B
9
400A77,15.0d
''r 25A00,2.0d B
}
9A
36.0B
'
'
2r d
36.09
2d
36.09
d, m MPa,r MPa,
0.15 -7 25
0.2 0 18
Resultant Stress
d, mDue to
shrinkagepressure
Due to internalpressure
Resultant
r r r
InnerCylinder
0.1 0 -25.2 -45 75 60 -45 49.8 47.4
0.15 -7 -18.2 -11.7 41.7 26.7 -18.7 23.5 21.1
OuterCylinder
0.15 -7 25 -11.7 41.7 26.7 -18.7 66.7 42.7
0.2 0 18 0 30 15 0 48 24
Single
Cylinder
Compound
Cylinder
Stress Distribution
-6 0
-4 0
-2 0
0
20
40
60
80
100
0.1 0 .15 0 .2
d, mm
S
tress,
MPa
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Diametral interference required to produce a shrunk fi t
In order to produce ashrink fit, the outer diameter of the inner cylinder must be slightly
greater than the inner diameter of the outer cylinder. These dimensions need to be
accurately machined in order to obtain the required interface pressure due to shrinkage.
Ro Ri
R = initial radial difference
+ve R
-ve RFinal position o
common radius
outer Ri= change in radius ofinner cylinder
Ro= change in radius of
outer cylinder
DDDorRRR ioio for interference allowance.
General case for cylinders of different materials
I
R1 R2 R3
II
Hookes Law
rE
1 (1)
Assuming A= 0.
Hoop strain
DD
D D
DD
lengthntialcircumfereoriginal
lengthntialcircumfereinchange
Diameter change D = D (2)
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At the interface diameter, D2:
(I)cylinderinnerfor
(II)cylinderouterfor
222
22
22
2
22
I
r
IIII
II
r
IIIIIIII
E
DDD
E
DDD
Now r2I
= r2II
= r2Diametral interference allowance = D2= D2
II- D2I
This gives us the general expression:
222
22
II
II
I
I
rI
I
II
II
EEEEDD
(3)
If both cylinders are the same material this expression reduces to:
2222 IIIE
DD (4)
Shrink Fit Example 2
Look at the case in Shrink Fit Example 1
Consider the effects due to shrinkage only. What is the diametral interference allowance?
Case of sleeve on solid shaft
Let the interference pressure = -P
For the sleeve then the procedure is as before.
For the shaft:
2
2
d
BC
d
BCr
Since there is material at r = 0, then B = 0, so that we do not have r= = at r = 0.
B = 0 for the shaft
r= = C = (-P)
i.e. in the shaft we have a uniform compressive stress (-P), everywhere
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Summary of Axisymmetric Stresses
Derived expressions for radial and hoop stresses in rotating thick cylinders
22
2
22
2
18
21
2
18
23
2
rr
BA
r
r
BAr
(8)
Solution methoduse boundary conditions to find A and Bsubstitute back intoequations.
Rules:
Solid Shafts: B = 0
Hollow shafts r= 0 at inner and outer radii
Long shafts give similar solution to thin discs
Lams EquationsThick cylinders under pressure
Compound Cylindersto reduce stress at bore
Diametral Interferencehow to design to shrink fit