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Chapter 6. Elasticity
6.1. Introduction ............................................................................................................ 2
Notation for stresses ................................................................................................ 4
Displacements and Strains ...................................................................................... 5
6.2. Equilibrium Conditions .......................................................................................... 7
6.3. Stress-Strain Relations .......................................................................................... 9
Young’s modulus and Poisson’s Ratio ..................................................................... 9
Microscopic Origins of Young’s Modulus and Poisson’s Ratio ................................. 9
Complete Stress-Strain Relations .......................................................................... 12
Elastic energy density ............................................................................................ 15
6.4. Membrane Stresses in Cylinders and Spheres.................................................. 16
Thin-Wall Cylinders and Spheres ........................................................................... 16
Thick-wall Cylinders ............................................................................................... 18
Spherical Shapes ................................................................................................... 21
6.5. Thermal Stresses ................................................................................................. 22
Axially Symmetric Cylindrical Geometry ................................................................. 24
Pellet Expansion .................................................................................................... 26
Pellet Cracking ....................................................................................................... 26
Thermal Stress Parameter ..................................................................................... 27
Problems ...................................................................................................................... 28
References ................................................................................................................... 29
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6.1. Introduction
Within this book’s confines to the reactor pressure vessel and the associated piping and the
fuel rods in the core, only a limited portion of the theory of elasticity needs to be addressed.
When a force, or load, is applied to a solid, the body changes shape and perhaps size.
These changes are called deformations. The objective of stress analysis is to quantitatively relate
loads and deformations. Elasticity theory deals with deformations sufficiently small to be
reversible; that is, the body returns to its original size and shape when the load is removed.
Within this range, the distances between atoms in the solid, or, alternatively, the atom-atom
bonds, are stretched or compressed, but are not broken. In order to remain in the elastic range,
the fractional changes in interatomic distances (on a microscopic scale) or in the body’s gross
dimensions (on a macroscopic scale) must be less than a percent or so.
When loads or deformations exceed the elastic range, the changes in shape of the body
are not recovered when the load is removed. This type of irreversible deformation is termed
plastic deformation. On a microscopic level, atomic bonds are broken and reformed between
different atoms than the original configuration. Plastic behavior of solids is treated in subsequent
chapters.
Instead of load (or force) and deformation, elasticity theory utilizes the related quantities
stress and strain. Stresses are forces per unit area acting on internal planes in the body and
strains are fractional deformations of the body.
Figure 6.1a shows a rod of cross sectional area A attached to a rigid lower plate and acted
upon by an upward force F at its top. All planes perpendicular to the rod’s axis experience the
same force. Imagine removal of the portion of the rod above plane a-a in the figure. The normal
stress on this plane is:
n = F/A
(6.1)
The stress is called normal if it acts in the direction perpendicular to the plane. To determine the
sign of the stress requires reference to a set of orthogonal coordinate axes such as that shown in
the figure. The surface is positive if its outward normal points in the direction of a positive
coordinate axis. The normal stress is positive if it also acts in the same direction. The load F in
Fig. 6.1a generates a positive normal stress on plane a-a. This stress tends to pull atomic planes
apart and is termed tensile. If the force F were reversed, the normal stress would act downward
on surface a-a and would tend to squeeze the atoms in the solid together. Such a stress is called
compressive. Thus, tensile stresses are positive and compressive stresses are negative.
In Fig. 6.1b, the force is in the direction perpendicular to the rod axis. The stress
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F
x
y
n
z
a
a
s
F
b 's  'n
b
(a)
Fig. 6.1
(b)
Normal and Shear Stresses
is generated in the internal surface rather than perpendicular to the surface. Such a stress is
termed a shear stress:
s = F/A
(6.2)
The sign convention for shear stresses is the same as for normal stresses; the shear stress
is positive if it acts on a surface with a positive normal in a positive coordinate direction.
Shear stresses are produced in a body loaded purely by external normal forces if the
internal surface normal is tilted by an angle  > 0 with respect to the axis of the applied stress.
Surface b-b in Fig. 6.1a represents such a surface. The area of this surface is A’ = A/cos and the
resolved component of the applied force along the plane b-b is F’ = Fcos(/2 - ) = Fsin. The
shear stress on b-b is:
σ s  F / A   (F / A) sin θ cos θ  0.5σ n sin( 2 θ )
(6.3a)
The resolved normal stress component on b-b is:
σ n  σ n cos 2 θ
(6.3b)
Equation (6.3a) shows that the shear stress due to normal loading of the rod is largest at  = 45o
and has a magnitude equal to one half of the applied normal stress. The maximum normal stress
on oblique planes naturally occurs at  = 0.
A similar decomposition of the shear force F in Fig. 6.1b on an oblique plane would
produce results analogous to those given by Eqs (6.3). These results are important in many
aspects of stress analysis of structures; whether or not the applied
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load is purely normal or purely shear relative to a coordinate axis, arbitrarily-oriented planes in
the body will experience both normal and shear stress components.
Stresses are generated in structures in a number of ways, which include:
1. Externally applied loads (Membrane stressses)
- mechanical loads represented by the discrete force F in Fig. 6.1. This type of loading is
the basis of the uniaxial tensile test used to measure many mechanical properties of
metals.
- Different pressures of a fluid (gas or liquid) on opposite faces of a structure. An example
of this type of loading is the high pressure of water or steam in the primary system of a
LWR. Components subject to this pressure loading are the primary system piping and the
reactor pressure vessel.
- Reaction forces due to connection of a particular structure to its supports and to other
components of a complex mechanical system. For example, the lower grid plate of a
LWR supports the weight of hundreds of fuel assemblies. In this case, the stresses are
induced by gravity.
2.
Thermal Stresses
Stresses are generated when the expansion of a heated body is restrained. This topic is treated
in detail in Sect. 6.6.
3. Residual Stresses
This type of stress arises from two principal sources
- fabrication of a component: Fabrication processes such as cold working (reduction in
cross-sectional area by passing between dies) introduce internal stresses in the finished
piece. These stresses remain during operation unless the piece is annealed at high
temperature prior to use.
- Welding of two components: welding involves melting of a metal, and introduces large
thermal stresses in the adjacent metal that does not melt (called the “heat-affected zone”).
These stresses persist in the cooled piece and are supplemented by additional stresses
arising from the solidification and cooling of the weld.
.
Notation for stresses
In general, any point within a solid subjected to one or more of the loads just enumerated
may possess as many as six components of stress. Three are normal and three are shear. It is
obvious that the simple designation n for normal stresses and s for shear stresses needs
refinement in order to cover complex stress patterns. The convention is as follows: stress
components are denoted with respect to the orthogonal coordinate system by which they are
described (x, y, z for Cartesian; r, , z for cylindrical; r, ,  for spherical). The stress
components are labeled i j, where i is the plane on which the stress component acts and j is its
direction.
The stress labeled n in Fig 6.1a is properly denoted by xx: it acts on the y-z plane,
which is called the x plane after the direction of its normal; the stress component acts in the same
direction, so the second subscript is also x. All normal stresses bear the generic designation ii,
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which is often shortened to i, with the understanding that the stress component acts on the i
plane in the i direction. In situations where the type of stress component is obvious, subscripts
are often entirely dispensed with. There are at most three normal components of the stress state
at a point.
Proper designation of shear stress components cannot be reduced to a single subscript
because i and j in ij are always different. The shear stress s in Fig. 6.1b should be written as xz
because the stress acts on the x plane and in the z direction. Equilibrium of the moments in a
stressed solid require that ij = ji so that there at most three shear components of the general
state of stress.
Displacements and Strains
In what follows, definitions and derivations are given for two-dimensional Cartesian
coordinates (i.e., x and y). This done in order to minimize the complexity of the theory.
Extensions to three dimensions or to other coordinate systems are stated without proof.
Displacements are changes in the position of a point in a body between the unstressed
and stressed states. Strain is a fractional displacement. In common with stresses, displacements
and strains come in two varieties, normal and shear.
Figure 6.2a shows the normal displacements u and v in the x and y directions,
respectively, of a body subjected to a normal force acting uniformly on both horizontal surfaces.
The original dimensions of the piece are Lo and wo. The solid rectangle represents the stress-free
solid and the dashed rectangle is its shape following application of the axial force. The
displacement u (positive) corresponds to the outward movement of the horizontal surfaces and
the displacement v (negative) represents the shrinkage of the body’s sides.
]
F
u
Lo
(a) Normal
v
F
(b) Shear
Lo

v
F
wo
Fig 6.2 Displacements in a stressed body
The strains in the x and y directions are defined as the fractional displacements:
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x
y
6
x = u/Lo
y = v/wo
(6.4)
The magnitudes of u and v vary linearly with the original dimensions but the strains x and y
are independent of Lo and wo. The point (or in this case, the plane) selected to follow the
displacement need not be an outer surface; displacements and strains of interior points are
defined in the same manner.
Figure 6.2b shows a shear displacement v resulting from a force F acting on the side of
the body only at a distance Lo from the fixed bottom. The shear strain is defined as the ratio of
these two perpendicular lengths:
xy = v/Lo
(6.5)
The two-digit subscript notation has been used – shear movement occurs on the plane normal to
the x axis and in the y direction. An alternative definition of shear strain is the angular distortion
of an initially rectangular section. In Fig. 6.2b, this is the angle  between the dashed slanted
line and the vertical solid line. In this triangle, tan = v/Lo, or because  is small (so that
deformations remain in the elastic range),  = v/Lo. Comparison of this result with Eq (6.5)
shows that xy = .
The strain definitions of Eqs (6.4) and (6.5) need to be generalized for use when the
displacements u and v are functions of position. Figure 6.3 shows how this is done.
C
B
A
B
dx
A
O
a) normal
x
dx 
O
D

dy
C
y
b) shear
Fig. 6.3 Diagrams of nonuniform deformations
The solid rectangles are the original shapes and the dashed figures represent the deformed
shapes.
Figure 6.3a shows a normal deformation in the x direction. The original length of the
body OB is taken to be a differential element dx. Upon application of a normal stress, the
displacement of the bottom surface is OA = u. The displacement of the upper surface is BC =
u+(u/x)dx, assuming that the body is continuous. The strain in the x direction, x, is the change
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in length, (OB+BC-OA) – OB, divided by the initial length OB. Expressing the lengths by their
equivalents in terms of u, u/x, and dx yields:
ε
x

u
x
(6.6a)
A similar analysis for the y direction gives:
ε
y

v
y
(6.6b)
Generalization of the shear strain definition is accomplished with the aid of
Fig. 6.3b. For simplicity, the lower left hand corners of the original and deformed figures are
superimposed at point O, which is equivalent to setting u = v = 0 at this point. Also, the corners
B and D of the deformed figure are taken to lie on the sides of the original shape. With these
simplifications, AB = (v/x)dx and tan   = AB/dx = v/x. Also, CD = (u/y)dy and tan
  = CD/dy = u/y. Since xy =  + ,
ε
ε
xy
yx

u v

y x
(6.7)
The analogous formulas for the normal strains in cylindrical coordinates with angular
symmetry (i.e., / = 0) are:
ε
r

 ur
r
εθ 
ur
r
ε
z

 uz
z
(6.8)
For spherical geometry with spherical symmetry (i.e., / = 0, / = 0)
ε
r

 ur
r
ε θ ε φ 
ur
r
(6.9)
In these equations, ur is the radial displacement,  is the polar angle, and  is the azimuthal
angle.
6.2. Equilibrium Conditions
The so-called equilibrium conditions of elasticity theory are consequences of Newton’s
third law: for a body subject to forces to remain stationary, the sum of the forces acting on it
must be zero. This condition applies to all volume elements in a stressed body, and provides
relations between stress components. Figure 6.4 provides
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stressed body, and provides relations between stress components. Figure 6.4 shows the basis for
deriving the x-direction force balance for a two-dimensional Cartesian body.
y
σ yx 
σ yx
dy
y
x
σ xx
σ xx 
dx
σ xx
dx
x
σ yx
Fig. 6.4 x direction forces on a volume element dx in length and dy in height
The balance of x-directed forces is:
σ


σ


net x force = σ xx  xx dx  σ xx dy  σ yx  yx dy  σ yx dx  0
x
y




or:
σ xx σ yx

0
x
y
(6.10)
Comparable equilibrium conditions apply to the y- and z-directions, and the extension to three
dimensions is straightforward (see Ref. 1, Appendix)
In axisymmetric cylindrical coordinates, the radial equilibrium condition is:
σ rr σ rr  σ θ θ σ rz


0
r
r
z
and for the z direction:
1  (rσ rz )  σ zz

0
r r
z
(6.11a)
(6.11b)
In spherical coordinates with spherical symmetry, the radial equilibrium condition is:
σ rr
σ  σ θθ
 2 rr
0
r
r
and, by symmetry:
σ θθ  σ φφ
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(6.12a)
(6.12b)
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6.3. Stress-Strain Relations
The final set of equations that forms the basis of elasticity theory relates stresses and
strains. Contrary to the strain-displacement relations (Eqs (6.6) – (6.8)) and the equilibrium
conditions (Eqs (6.10) – (6.12)), the connection between stresses and strains involves material
properties called elastic moduli. A single crystal can possess as many as 36 such parameters, but
polycrystalline solids with random grain orientations exhibit only two elastic moduli.
Young’s modulus and Poisson’s Ratio
Figure 6.2a represents the simplest of loading configurations because it produces only
one component of stress, the normal component x. The axial strain defined by the first of Eqs
(6.5) is related to x by:
x = x/E
(6.13)
where E is a material property called Young’s modulus or the modulus of elasticity. For steels, E
~ 2x105 MPa; the value of E for aluminum is about one third that for steel. The nuclear fuel UO2
has approximately the same Young’s modulus as steel, but this correspondence has little to do
with the mechanical performance in a reactor environment, as will be seen in subsequent
chapters
There is a stress limit (and consequently a strain limit) for the applicability of Eq (6.13).
For steel, the proportionality of stress and strain implied in this formula fails at a stress of about
500 MPa, which is called the yield stress. At this point the strain is 0.3%. These conditions
define the elastic limit of the material.
As suggested in Fig. 6.2a, a positive displacement (or strain) in the x direction produces a
negative displacement (or strain) in the y direction. The third transverse direction (z) experiences
the same strain as does the y direction, or z = y. The ratio of the magnitudes of the lateral
strains to the axial strain in the uniaxial tensile situation of Fig. 6.2a defines Poisson’s ratio, :
 = -y/x = -z/x
(6.14)
Like E,  is a material property, and moreover, is nearly the same for all materials.
It is not surprising that the deformed body in Fig. 6.2a shrinks in transverse dimensions
as it elongates axially. Otherwise, a substantial volume change would occur. However, the
transverse shrinkage does not quite compensate for the axial elongation, and the solid changes
volume as it deforms elastically.
Microscopic Origins of Young’s Modulus and Poisson’s Ratio
Macroscopic elastic strains are, on a microscopic scale, due to the stretching or
contraction of bonds between atoms. Figure 6.5 depicts the microscopic response to a tensile
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force on the (111) plane of the fcc crystal structure. The origin of Young’s modulus can be
explained by the increase in the distance ab between adjacent atoms a and b in the unstressed
condition to the distance a’b’ with the applied stress This
n
x
y
z
a'
a
c
d
c'
d'
b'
b
Unstressed
n
Fig. 6.5 Movement of atoms as a result of application of a tensile stress
Potential energy 
stretching of the a-b bond is accompanied by an increase in the potential energy
between the two atoms. The potential energy between two atoms is shown in Fig. 6.6 as a
function atom separation in the x direction. In the unstressed
0
ab a'b'
atom separation
Fig.6.6.
Fig. 5.6 Relation of Young's Modulus to Interatomic Potential
state, the atom separation ab lies close to the minimum on the potential curve. Increasing the
separation distance to a’b’ is accompanied by an increase in the potential energy between the two
atoms. Around its minimum, the potential energy curve can be approximated by a parabola:
1  d 2φ  2
φ φ o   2  x
2  dx  0
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(6.15)
11
where the origin of the separation distance x is defined as the location of ab in Fig. 6.6
The force between the two atoms is the negative of the gradient of the potential energy. The
force to increase the separation from ab to a’b’ is:
 d 2φ 
dφ
F
  2  (a ' b'ab)
dx
 dx  0
(6.16)
Macroscopically, the force is equal to the negative of the stress n times the projected area of an
atom, r2, where r = ½ab is the atomic radius. (the force is negative because it acts to reduce the
separation distance).
F = -n(r2)
(6.17)
The strain is the separation (displacement) of the atom centers divided by the original distance
between the centers:
n =
a ' b'ab
ab
(6.18)
Combining the above three equations and using ab = 2r yields the Young’s modulus in atomic
terms:
σn
2  d 2φ 
E

ε n π r  dx 2  0
(6.19)
Unfortunately, knowing the potential energy curve of Fig. 6.6 is the only way to estimate the
second derivative in Eq (6.19).
Poisson’s ratio is derivable (for the fcc structure) by superimposing the two diamondshaped figures in Fig. 6.5, as shown in Fig. 6.7. The normal strain in the x direction, n, and the
normal strains in the transverse (y or z) directions, t, are:
ε
n

a a'
oa
ε t 
c c'
oc
(6.20)
Now oc is the height of the equilateral triangle abc. Since all sides of the quadrilateral adbc are
equal to twice the atomic radius (see Fig. 6.5):
oc oc
3

 sin 60 
a c 2oa
2
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a'
a
60
d
d'

O
c
c'
b
b'
Fig 6.7 Diagram for calculating Poisson’s ratio in the fcc structure
Considering the angle  in Fig. 6.7 and noting that a’c’ equals twice the atomic radius:
sin θ 
cos θ 
o a' o a  a a' 1 
a a'  1
  1  ε

 1 
a ' c'
2oa
2
o a  2
n

o c' o c  c c '
3 o c  c c'
3
1  (ε t ) 



a ' c'
2oa
2
oc
2
Squaring and adding the above two equations yields:
1
1
1  ε
4
n
2  3 1  ε t 2
4
Since the strains are always << 1 (for elastic deformations), the squared parenthetical terms can
be expanded in one-term Taylor series, which yields:
-ε t
1
ν 
εn
3
(6.21)
Thus, Poisson’s ratio is not a material property in the sense that Young’s modulus is; 
depends only on the crystal structure. Although the above derivation was restricted to the closepacked plane of the fcc structure, similar results are obtained for other planes and other lattice
types. Nearly all crystalline solids exhibit Poisson’s ratios close to the theoretical value of 1/3.
Complete Stress-Strain Relations
Equation (6.13) is all that remains of the more general stress-strain relation when only
one normal stress component acts on a body. The corresponding formulation when all six stress
components are nonzero are known as the generalized Hooke’s Law. These equations are the
same in all coordinate systems; instead of the subscripts x, y, and z, the coordinate axes are
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labeled 1, 2, and 3 so as to include cylindrical and spherical geometries as well as Cartesian
coordinates. For the normal stresses and strains:
1
σ 1  ν (σ 2  σ 3 )
E
1
ε 2  σ 2  ν (σ 1  σ 3 )
E
1
ε 3  σ 3  ν (σ 1  σ 2 )
E
ε1
(6.22a)
(6.22b)
(6.22c)
and for shear stresses and strains:
ε 12  σ 12 / G
ε 13  σ 13 / G
ε
23
 σ 23 / G
(6.23)
In Eq (6.23), G is the shear modulus. G is not independent of E and , as is shown in the
following analysis. Figure 6.8 is a cross section of a square prism of side length L that is acted
upon by a shear stress S. When the applied shear stresses are resolved on the diagonal of the
square, the resultant components are both normal*. One, labeled n is tensile and in the direction
of the diagonal and the other, n’, is compressive and perpendicular to it. The triangles on each
of the four sides of the square in Fig. 6.8a
s
L
'n
s
n
s
L
s
(a) stresses resolved on diagonal
u v
(b) Strains
Fig. 6.8 Stresses (a) and strains (b) of a square prism subjected to pure applied shear
stress S
represent the resolution of the applied shear stress S into components parallel to and
perpendicular to the diagonal. The magnitude of each stress component is S/ 2 and the force
each exerts on the diagonal element is (S/ 2 )L. Since two of the resolved components of S
contribute, each of the normal forces on the diagonal is 2(S/ 2 )L. The length of the diagonal is
*
In this respect, pure applied shear stress differs from pure applied normal stress, which is resolved on a
tilted plane as a shear and a normal component (see Fig. 6.1)
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2 L , so conversion of the normal forces to normal stresses results in 2(S/ 2 )L/( 2 L) for
each, or:
n = -’n = S
(6.24)
Figure 6.8b depicts the strains resulting from the shear stresses applied to the square and
from the resulting normal stresses on the diagonal. The normal stresses
n and ’n produce a displacement u of the diagonal element. The normal strain of the diagonal
is u/( 2 L), which can also be expressed by Eq (6.22a) with 1 = n , 2 = ’n and 3 = 0.
Equating these two forms of the normal strain and using Eq (6.24) yields:
u
1
σ
 σ n  νσ 'n   S (1  ν )
E
2L E
(6.25)
From its definition (as Y in Eq (6.4)), the shear strain of the initially square figure is v/L.
Now u and v are related (approximately) by the 45o right triangle shown in black in the lower
right hand corner of Fig. 6.8b. Because v is the hypotenuse of this triangle, the two
displacements are related by v = 2 u. With the help of Eq (6.25), the shear strain of the square
is:
v
2u
2u
σ
2
εS  

2 L S (1  ν )  (1  ν )σ S
L
L
L
E
E


Comparing this result with the shear modulus G as defined by Eq (6.23) yields the desired
connection between G, E and :
E
G
(6.26)
2 (1  ν )
Since Poisson’s ratio is ~ 1/3 for most materials, this relation shows that the shear modulus is
approximately 3/8 of Young’s modulus.
Another elastic constant that describes the volume change due to the normal stress
components is called the bulk modulus, K. If the normal stresses are due to the pressure in a fluid
in which the body is immersed, 1 = 2 = 3 = -p. The quantity in a solid corresponding to the
pressure in a fluid is the mean hydrostatic stress, defined by:
1
σ h  (σ 1  σ 2  σ 3 )
3
(6.27)
The change in volume from initial value V to final value Vf is related to the normal strains by:
Vf/V = (1+1) (1+2) (1+3)  1 + (1+2+3)
or, the fractional volume change is the sum of the normal strains:
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ΔV
 1 + 2 + 3
V
(6.28)
Adding Eqs (6.22a) – (6.22c) gives:
1 + 2 + 3 =
1
(σ 1  σ 2  σ 3 )  2ν (σ 1  σ 2  σ 3 )
E
Substituting Eqs (6.27) and (6.28) into the above equation yields:
σ
Δ V 3(1  2ν )

σh  h
V
E
K
(6.29)
where the bulk modulus is+
K
E
3(1  2ν )
(6.30)
Note the similarity between Eq (6.29) and the uniaxial form of Hooke’s law, Eq (6.13).
The linear strain in the latter is replaced by the volumetric strain in the former. In place of the
axial stress in Hooke’s law, the bulk compressibility formula employs the mean of the three
orthogonal normal stresses. The bulk modulus K is the three-dimensional analog of Young’s
modulus. In fact, if Poisson’s ratio is 1/3, K and E are equal.
Elastic energy density
Work is required to produce the elongation of the rod depicted in Fig. 6.2a. No work is
expended in creating the transverse displacement because no forces act on the rod’s vertical
sides. The external work performed by the force F is converted to the increase in the internal
energy of the solid by stretching the interatomic bonds, as shown in Fig. 6.6. When divided by
the volume of the body, the internal energy stored in the solid is called the elastic energy density,
or sometimes simply the strain energy. This excess energy (over a perfect, unstressed solid) is
the source of thermodynamic instability of defects such as, dislocations, grain boundaries, and
voids.
Considering only the upper half of the deformed rod in Fig. 6.2a, the work needed to
u
produce the displacement u is
 Fdu  . This work is stored internally in the solid, and when
0
divided by LoA to convert to an internal energy density yields:
Like the thermal expansion coefficient , the bulk modulus is a thermodynamic property related to the
coefficient of compressibility  = -(1/V)(V/p)T . Comparison of this definition (in its integrated form) with
Eq (6.29) wherein h is replaced by –p shows that K = 1/
+
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σ 2x
F du  ε x
1 σx
E el  
 0 σ x dε x  0 σ x dσ x 
A Lo
E
2E
u
0
where Hooke’s law has been used to convert the variable of integration from strain to stress.
Similarly, the strain energy in pure shear can be deduced from the forces and
displacements shown in Fig. 6.8. This body is a square prism of unit depth and side length L.
Work is done only by the action of the force SL acting on the right hand face; the other three
faces do no work because they remain fixed in dimensions. The analog of the above equation for
this case is:
εS
σS
(σ SL) dv
1
σ S2



Eel 

σ Sdε S 
σ dσ 
L L
G 0 S S 2G
0
0

u


where the shear analog of Hooke’s law (Eq (6.23) ) has been used to convert strain to stress in
the integral.
For the general case in which all six components of the stress are nonzero, the above
equation is extended to:
E el 




1 2
ν
1 2
2
σ 1  σ 22  σ 32  σ 1σ 2  σ 1σ 3  σ 2σ 3  
σ 12  σ 13
 σ 223 (6.31)
2E
E
2G
This equation applies to all coordinate systems, whose orthogonal axes are designated by 1, 2
and 3.
6.4. Membrane Stresses in Cylinders and Spheres
The preceding sections dealt with the fundamentals of elasticity theory, including stresses,
strains and displacements arising from externally applied loads (membrane stresses). In this
section, the general theory is applied to membrane stresses in the simple geometries most
frequently encountered in nuclear systems. For example, fuel rod cladding, the reactor pressure
vessel and the piping of the primary coolant circuit are accurately modeled as annular cylinders
with large radius-to-wall thickness ratios and infinite in length. The hemispherical upper and
lower heads connected to the pressure vessel wall are treated in spherical geometry.
These components operate essentially isothermally, so the stresses are due only to the
difference in pressure across the wall. Thermal stresses due to temperature nonuniformities in a
component are analyzed in Sect. 6.6.
Thin-Wall Cylinders and Spheres
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If the radius of a cylinder or sphere is much larger than the thickness of the wall, it can be
treated in the so-called thin-wall approximation. This method replaces the mathematical
machinery of elasticity theory with simple force balances. Because the wall is thin, the spatial
stress distributions are replaced by average values.
Figure 6.9 shows a cross section of a thin-wall cylinder of radius R and wall thickness 
that is internally pressurized by a pressure p. The pressure acts radially on the inner surface, and
so must be resolved in the vertical direction in order to balance the azimuthal stress acting on the
midplane section of the tube wall. The middle diagram of Fig. 6.9 shows the resolved upward
force dFpress due to the pressure on an arc length Rd of the inner wall. Per unit length
perpendicular to the diagram, the upward pressure force is:

π
Fpress  pR sin θ dθ  2pR
0


top
P
P

R
z
z
Fig. 6.9 Average stresses in a thin-wall cylinder
Fpress is opposed by the azimuthal stress acting on the area  (per unit length) on both sides of the
cross section. This force is Fstress = 2δ σθ . At equilibrium, Fpress = Fstress, or:
σθ 
pR
δ
(6.32)
The average azimuthal stress is tensile.
In order to maintain a pressure in the tube, its ends must be closed. The right hand sketch
in Fig. 6.9 shows the closed top of the tube and a cross section far enough removed from the top
to avoid end effects that would invalidate the simple force-balance analysis. The force exerted on
the closed top, R2p, is balanced by the tensile force in the tube wall, 2Rδ σ z . Equating these
two yields:
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σz 
pR
2δ
(6.33)
which is tensile and one half as large as the azimuthal stress.
The radial stress on the inner wall is –p. Assuming zero external pressure, the average
radial stress is:
1
σr   p
2
(6.34)
which is small compared to σθ and σ z because the factor R/ is large in thin-wall tubing.
Example The upper head of the reactor pressure vessel in PWR is hemisphere 4 m in diameter and 24
cm thick (see Fig. I.15). It is attached to the cylindrical wall of the pressure vessel by a flange with a
number of 8 cm diameter bolts. The internal pressure in the vessel is 15 MPa. How many bolts are
required so that the stress in each does not exceed 2/3 of the yield stress (350 MPa)?
The bolts must resist the vertical component of the force due to the internal steam pressure acting
on the inside of the hemispherical upper head. The differential area on which the pressure acts is an
annular strip of width Rd and periphery 2Rcos, or an area equal to 2R2cosd. Here R is the radius
of the pressure vessel and  is the angle with respect to the vessel axis. Resolving the pressure stress in a
vertical direction by multiplying p by sin and integrating the polar angle  from 0 to /2 yields:

π /2

1
Fpress  2π R 2p cosθ sin θ dθ 2π R 2p cosθ d(cosθ )  π R 2p
0
0
The stress in N bolts of radius Rbolt resisting the pressure force is:
σ bolt 
π R 2p
Nπ R 2bolt
if bolt is not to exceed 2/3 of the yield stress Y, the required number of bolts is:
3 p  R

N
2 σ Y  R bolt
2

3 15  2 
 

  160
2 350  0.04 

or, the spacing between bolts is 2R/N = 7.8 cm.
There is no way to compute the strains or displacements by the simple force-balance
method used above for the average stresses. Full elasticity theory is needed to calculate the
strains and displacements, and is treated below.
Thick-wall Cylinders
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In order to calculate stresses and strains in internally-pressurized cylindrical components
which do not satisfy the thin-wall criterion employed above, the full set of elasticity equations
must be utilized. As long as the analysis is not applied near the ends of the cylinder, only normal
stresses and strains are developed. The starting point is the set of equations given by Eqs (6.8),
(6.11a), and (6.22). The approach is to combine these equations so as to sequentially eliminate
one variable at a time until a single equation with only one variable is obtained. The remaining
variable should be the one to which the boundary conditions on the inner and outer walls of the
cylinder apply.
First, ur is eliminated between the first two of Eqs (6.8), yielding:
dε θ ε θ  ε r

0
dr
r
(6.35)
Next, the strains in Eq (6.35) are replaced by the stresses by using Eqs (6.22a) and (6.22b):
d
σ θ  ν (σ r  σ z )  1  ν (σ θ  σ r )  0
dr
r
To proceed further, a condition reflecting the requirement that the analysis is applied far from the
cylinder ends is introduced. The condition is that during elastic deformation, what was a planar
cross section of the cylinder remains planar. This condition is equivalent to requiring that the
axial strain or axial displacement is not a function of radius. This condition is called plane strain,
and is mathematically expressed by
dz/dr = 0. When applied to Eq (6.22c), it yields:
dσ z
d
 ν (σ r  σ θ )  0
dr
dr
The next step is to eliminate dz/dr between the above two equations, which results in:
(1  ν 2 )
dσ θ
dσ
1ν
 ν (1  ν ) r 
(σ θ  σ r )  0
dr
dr
r
Finally, the equilibrium condition of Eq (6.11a), dropping the last term, is used to eliminate the
hoop stress from the above equation to produce the differential equation in the radial stress
component:
1 d  3 dσ r 
(6.36)
r
0
r 3 dr  dr 
which is readily integrated twice.
Using the boundary conditions r(Ro) = 0 and
r(R) = -p, where R and Ro are the inner and outer radii of the cylinder, respectively, the solution
is:
(R / r ) 2  1
(6.37)
σ r  p o
(R o / R ) 2  1
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The hoop stress is obtained by substituting the above equation into Eq (6.11a) (without the shear
stress term):
(R o / r ) 2  1
 = p
(6.38)
(R o / R ) 2  1
In the thin-wall limit as Ro – R =  << R, Eq (6.39) reduces to Eq (6.32).
There remains to specify the axial end conditions in order to calculate z and from this
and Eqs (6.37) and (6.38), the strains and displacements. Three possibilities are shown
schematically in Fig.6.10.
Fig. 6.10 End Conditions on a thick-wall cylinder
In case (a), the axial strain is zero. Setting 3 = z = 0 in Eq (6.22c), inserting the radial and
azimuthal stresses from Eqs (6.37) and (6.38) gives:
σ z (a )  p
2ν
(R o / R ) 2  1
(6.39a)
which is tensile, independent of r, and smaller than .
In case (b), the open end precludes establishment of stresses by gas pressure. However, an
internal radial stress can be produced by the so-called expanding mandrel method, in which a
cold, close fitting solid rod is inserted into the cylinder and allowed to heat up to ambient
temperature. Thermal expansion creates a radial stress on the cylinder inner wall, which, if
denoted by –p, produces the radial and hoop stresses given by Eqs (6.38) and (6.39). However,
there is no axial stress because there is no stress exerted on the upper end of the annulus, or:
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z(b) = 0
(6.39b)
In case (c), the axial stress is obtained by equating the force on the inner surface of the upper
end, R2p, with the counterbalancing force in the annular cross section, π (R o2  R 2 ) σ z . This
yields:
1
(6.39c)
σ z ( c)  p
(R o / R ) 2  1
As R  Ro, Eq (6.40c) reduces to Eq (6.34).
Once all the stresses have been computed, the strains follow from Eqs (6.22a) – (6.22c).
Spherical Shapes
Three spherically-shaped objects subject to mechanical loads are important in nuclear
technology: the spherical shell with different internal and external gas pressures, such as the top
head of the reactor pressure vessel analyzed by the force balance method earlier in this chapter;
spherical voids or gas bubbles in solids created as a result of irradiation; and solid spheres that
create stresses in the surrounding host solid. In this last category are atomic-size spheres such as
interstitial atoms and macroscopic objects such as precipitates. In all of these examples,
determination of the stresses (generally all normal stresses) in the spherical object or in the
medium in which the sphere is embedded is needed.
The normal stress components active in spherical geometry are the radial component r
the polar component  and the azimuthal component . The present analysis is limited to the
common case of spherical symmetry, meaning that none of the stress components are functions
of  or  but depend on r only. This does not mean that  and  are zero, only that they are
equal. The mathematical manipulations of the elasticity equations are analogous to those
employed for cylindrical geometry:
i)
ur is eliminated between Eqs (6.9)
ii)
The stress-strain relations of Eqs (6.22a) and (6.22b) with 1 = r and 2 = 3 =
 are substituted into the result of step i).
iii)
 and its derivative with respect to r are eliminated from the results of step ii)
using Eq (6.12a).
The result is the ordinary differential equation :
1 d  4 dσ r 
r
0
dr 
r 4 dr 
(6.40)
Integrating twice yields the general solution  = C1 + C2/r3, with the integration constants
depending on the particular problem. For the internally-pressurized spherical-shell of inner
radius R and outer radius Ro , the boundary conditions are r(R) = -p and r(Ro) = 0. The radial
stress component is:
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σ r  p
( R o / r )3  1
(6.41)
( R o / R )3  1
Substituting this result into the equilibrium condition of Eq (6.12a) yields the tangential stresses:
 =  = p
0.5 (R o / r )3  1
(6.42)
( R o / R )3  1
Note the strong similarity between these equations and the corresponding equations for the thickwall cylindrical annulus (Eqs (6.36) – (6.38)).
In the thin-wall limit (Ro – R =  << R), Eq (6.432 reduces to:
σθ 
pR
2δ
(6.43)
which is half of the hoop stress for the thin-wall cylinder (Eq (6.32)).
6.5. Thermal Stresses
Almost every component of a reactor core and the primary pressure system is subject to
nonuniform temperature. Thermal gradients in fuel are generated by internal heating due to
slowing-down of fission fragments. Gamma-rays heat metal structures. In the absence of internal
heat sources, spatial temperature variations are produced by unequal temperatures over the
surface of a body. It does not matter whether the thermal gradients are steady state or transient;
the thermal stresses are dependent only on the instantaneous temperature distribution.
When an unrestrained (i.e., stress-free) body is heated from temperature T to a higher
temperature T + T, it expands in each of the three orthogonal coordinate directions by an
amount TL, where L is the original length and  is the linear coefficient of thermal
expansion.+ When this expansion is impeded, either internally or externally, the stresses
generated in the body are called thermal stresses. These stresses arise in numerous situations,
such as:
-
+
Restraint of free thermal expansion during uniform heating by intimate
with other rigid structures.
interaction
 is a thermodynamic property, being 1/3 of the volumetric coefficient of thermal expansion,
vol =
1  v 
  , where v is the specific volume and p is pressure. vol is obtained from the equation of
v  T  p
state of the solid.
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Example: A bar of length L with a linear thermal expansion coefficient  is placed between two rigid
ends. At temperature T the bar is stress-free. The bar is then heated to temperature T+T but the end
pieces remain the same distance apart. First imagine that the top end
piece is removed and the bar allowed to expand freely by an amount TL, or
a thermal strain th = T. Now a stress  is applied to the top of the bar that
L
is just sufficient to return the bar to its original length and the top end piece
replaced. The elastic strain in this step is el = /E. The total strain is zero:
tot = th + el = 0, or the thermal stress in the bar is th = -ET. The
negative sign indicates that the stress is compressive, and the superscript
“th” on the stress indicates that it is due to restrained thermal expansion.
-
Another example of restrained thermal expansion is the compressive stresses generated
by uniform heating of a solid containing inclusions (precipitates) with a larger value of 
than that of the host solid.
-
Temperature nonuniformities (i.e., thermal gradients) in a body produced by temperature
differences on the boundaries+.
-
Thermal gradients created by heat generation in (and removal from) a component. The
most important example of this mechanism is the fission energy released in nuclear fuels,
which is removed as heat by an external coolant.
-
stresses in reactor components by producing temperature gradients where there were
none or by increasing the magnitude of existing temperature gradients.
Such changes include reactor power changes during startup or shutdown and the sudden
injection of cold emergency cooling water near the inner wall of the reactor pressure
vessel (see Chap. Xxx).
The thermal stresses produced by temperature gradients can be understood by a simple
physical picture. The hotter part of the body seeks to expand more than the cooler part. However,
this tendency is impeded by the adjacent cold region. Similarly, expansion of the cold portion is
enhanced by the greater expansion of the contiguous hot part. As a result of this interaction, the
hot zone develops compressive stresses and the cold zone is placed in tension. Between these
two stressed zones lies a surface of zero thermal stress.
Thermal stresses are readily incorporated into the framework of elasticity theory by
adding the linear thermal strain to the elastic strains given by Eqs (6.22):
1
σ 1  ν (σ 2  σ 3 ) + T
E
1
 σ 2  ν (σ 1  σ 3 )+ T
E
ε1
(6.44a)
ε
(6.44b)
2
+
One exception to this rule is a linear temperature profile in an unrestrained plate, in which no thermal
stresses are generated (see Ref., XX p. 388)
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ε
3

1
σ 3  ν (σ 1  σ 2 )+ T
E
(6.44c)
Axially Symmetric Cylindrical Geometry
The analysis below applies to infinitely long cylindrical annuli as well as to solid
cylinders at locations sufficiently far removed from the ends. In both cases, the temperature
distribution is assumed to be a function of radial position only. The outer boundary condition is
not a function of azimuthal angle  or axial location z.
Following the same sequence of manipulations starting from Eq (6.35) but using Eqs
(6.44a) and (6.44b) instead of Eqs (6.22a) and (6.22b) changes Eq (6.36) to:
1 d  3 dσ r 
 αE
r
  
3 dr 
r
 dr 
1 ν
 1 dT

 r dr
(6.45)
This equation is integrated for a long, thick-wall cylinder with inner radius R and outer
radius Ro. Pressure loading is not considered, so the boundary conditions are
r(R) = r(Ro) = 0
(6.46)
The result is:
σ
th
r
αE

1 ν
1  R 2 / r 2
 2
2
 R o  R
Ro
R
rT(r )dr 



r
T
(
r
)
dr


r2 R

1
r
(6.47)
Using this in Eq (6.11a), without the shear stress term, gives the hoop stress:
σ θth
αE

1 ν
1  R 2 / r 2
 2
2
 R o  R
Ro
R
rT(r )dr 



r
T
(
r
)
dr

T
(
r
)


r2 R

1
r
(6.48)
The temperature distribution T(r) can be due to different temperatures at the inner and outer
surfaces, internal heat generation in the wall with constant and equal surface temperatures, or a
transient temperature change in an initially isothermal cylinder. Note that these thermal stress
components are independent of the axial end conditions, although these end conditions affect the
axial component of the stress.
Example: A temperature difference T is maintained across the cladding of a reactor fuel element. The
cladding wall thickness  is sufficiently small so that the “thin-wall” approximation, Ro – R =  << R,
applies. For the purpose of calculating the temperature distribution, the wall can be considered a slab (i.e,
its curvature neglected) and the radial temperature profile is linear:
ΔT
T( r )  T ( R ) 
y
δ
where T = T(R) – T(Ro) and y = r – R. Although the cylindrical nature of the tube can be neglected
for the purpose of calculating the temperature distribution, the same simplification cannot be applied to
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the stress analysis (i.e., the thin wall tube is still a tube as far as stress calculation goes). Substituting the
above equation into Eq (6.48) and using the thin-wall condition where needed results in the hoop stress:
y
α EΔ T 
σ θth  
1  2 
2(1  ν ) 
δ
The thermal stress is negative (compressive) on the inner half and tensile on the outer half, as expected
from the discussion on the bottom of p. 22.
An important application of cylindrical thermal stresses is to the ceramic fuel pellets in
nuclear reactor fuel rod. Uniform volumetric heating in the solid cylinder cooled at its periphery
generates a parabolic temperature distribution (see Chap. XXX). With temperature TS at the
pellet surface and To on the axis, the distribution is:
T  TS
r2
1 2
To  TS
Ro
(6.49)
Using Eq (6.49) in Eqs (6.57) and (6.48) along with R = 0 and integrating yields:
σ rth
σ θth


α E(T0  Ts )  r 2 

1 

4(1  ν )  R 2 




α E(T0  Ts ) 
r2 

1  3 2 
4(1  ν ) 
R 


(6.50)
(6.51)
Calculation of the axial stress distribution proceeds as follows. For a stack of fuel pellets, the
unrestrained end condition (6.40b) is applicable:
σzth 
1
π R o2

Ro
2π rσ zth (r )dr  0
0
Next, Eq (6.44c) is solved for z and the result substituted into the above integral. In performing
this integral, (1 + 2) is (σ r  σ θ ), which is the sum of Eqs (6.50) and (6.51), and T is T –
TS given by Eq (6.49). The axial strain z in Eq (6.44c) is independent of r; this is the condition
of plane strain, which was needed in the derivation of Eq (6.36) and Eq (6.45). With these
conditions, the above integral is solved for z, which then permits Eq (6.44c) to be solved for the
axial thermal stress:
th
th
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

α E(T0  Ts ) 
r2 
th
σz  
2  4 2 
4(1  ν )

R 

(6.52)
The plane strain restriction means that the thermal stress solutions are applicable only near the
midplane of the solid cylinder; the ends do not satisfy the plane strain condition, and are in fact,
surfaces of plane stress (i.e., zero stress is not a function of radial position.
Pellet Expansion
The thickness of the gas-filled gap between the fuel pellet and the cladding tube strongly
affects heat transfer from, and consequently, the temperature of the fuel. The gap thickness
decreases from its initial “cold” value as a result of assuming the temperature distribution given
by Eq (6.49) during operation. The gap thickness reduction is largely due to thermal expansion
of the pellets.
According to Eq (6.8), the strain component  evaluated at r = R is the fractional
increase in the cylinder radius. This is expressed by Eq (6.44b) evaluated at r = R, where r = 0
and T = TS:
1
ε θ (R )  σ θ (R )  νσ z (R )  α (TS  Tref )
E
where Tref is the pellet temperature in the cold state. The stresses are obtained from Eqs (6.51)
and (6.52):
α E(To  TS )
σ θ (R )  σ z (R ) 
2(1  ν )
Combining the above two equations yields:
ε θ (R )  Δ R / R  α ( T  Tref )
(6.53)
where T = ½ (To + TS) is the average cylinder temperature. This simple result is valid only for
the parabolic temperature distribution of Eq (6.49).
Pellet Cracking
The thermal stresses due to heat removal from a solid cylinder may exceed the fracture
stress of the solid, particularly if the material is a brittle ceramic such as uranium dioxide.
Equations (6.51) and (6.52) are plotted in Fig. 6.11. The region of interest is near the outer edge
of the cylinder, where the azimuthal and axial stresses
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Fig. 6.11 Stress distributions in a cylinder with no end restraints
are tensile. If either stress exceeds the fracture stress, vertical and/or horizontal, radially-oriented
cracks appear in the pellet (see Fig. 6.12). The “hourglass” shape of the pellet in Fig. 6.12 is due
to the change from the plane strain condition near the midplane to the plane stress condition at
the ends. This thermal stress driven cracking and deformation phenomena have significant
effects on the behavior of the fuel during operation (Chap. Xxx).
Fig. 6.12 Cracking and “hourglassing” in fuel pellets with uniform heat generation
Thermal Stress Parameter
Modification of the scaling factor for the stress in the ordinate of Fig. 6.11 provides a
general measure of the ability of a material to withstand thermal stresses. Specifically, the
temperature difference is expressed in terms of Fourier’s law as
T = qL/, where q is the heat flux, L is the distance over which T occurs, and  is the thermal
conductivity. The first two of these parameters are operational, in the sense that they can be
changed at will. The last, however, is a material property. Substituting the Fourier-law
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expression for T into the scaling factor for the stress in the ordinate of Fig. 6.11 and neglecting
the product qL leaves a grouping of material properties intimately related to the thermal stress
resistance of the material. Finally, the stress  in the ordinate of Fig. 6.11 is replaced by the
fracture stress F, which is the stress that results in thermal stress cracking. With these
modifications, a parameter that contains the properties that govern the resistance of a solid to
failure by thermal stresses is:
σ (1  ν )κ
M F
αE
(6.54)
M is called the thermal stress parameter. Although it was developed from the solution for solid
cylinder, the concept is applicable to any geometry, with or without heat generation. The larger
M is, the more resistant to thermal stress failure is the solid. A high thermal conductivity is
beneficial because, for a fixed heat flux, it reduces T. A small coefficient of thermal expansion
is desirable because it reduces the thermal strain in Eqs (6.44). Solids show considerable
variation in M. Graphite, for example, has an M value of ~ 30,000, while for stainless steel, M ~
2000 (with the yield stress used instead of the fracture stress). Ceramic oxides such as UO2
generally exhibit even lower thermal stress parameters than metals. Quartz, because of its very
low thermal expansion coefficient, has a large M value.
Problems
6.1
Starting with Eq (6.9), prove that the thermal expansion of a solid cylinder with
the parabolic radial temperature distribution of Eq (6.28) is given by Eq (6.32) of the
notes. Here To and Ts are the centerline and surface temperatures, respectively, and
Tref is a reference temperature. The linear thermal expansion coefficient is .
6.2
Determine the radial depth of penetration of vertical thermal-stress cracks in a
fuel pellet with a parabolic temperature distribution
To = 2000oC, Ts = 500oC. Assume the following elastic properties for UO2:
f = 140 Mpa
 = 1.0x10-5 K-1
E = 1.6x105 Mpa
G = 5.9x104 Mpa
6.3
Derive the expression for the stored elastic energy per unit length of a tube of
radius R and wall thickness  that is internally pressurized at a pressure p. The material
has a Young’s modulus E and a Poisson’s ratio .
Use the thin-wall approximation.
6.4
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What error is incurred in calculating the average hoop stress   by using the thin-wall
approximation for cladding 1 cm inside diameter and 1 mm wall thickness? Hint: this is
most efficiently done by expanding the terms in Eq (6.23) in one-term Taylor series
before taking the average.
References
1. D. R. Olander, “Fundamentals of Nuclear Reactor Fuel Elements”, TID-26711-P1,
Technical Information Service (1976)
2. C. F. Bonilla, “Nuclear Engineering” , Chap. 11, McGraw Hill (1957)
3. J. H. Rust, “Nuclear Power Plant Engineering”, Chap. 8, Haralson Pub. Co. (1979)
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Nat’l