week3 stress analysis fuel pin - nucmaster.web.psi.ch for reactor physics and systems behaviour...

Laboratory for Reactor Physics and Systems BehaviourNuclear Energy and Safety Department

Nuclear Fuels

Stress Analysis of Nuclear Fuel Pin

Dr. MER P. SpätigLaboratory for Nuclear Materials (LNM), PSI

Laboratory of Reactor Physics and Systems Behaviour (LRS), [email protected]


Strains Upon the application of displacements and forces F, a body deforms. So any given

segment drn changes to dRn

Strains quantify the level of deformation

Undeformed body Deformed body

drn

dRn

02.05.2016“Nuclear Fuels & Materials” - Stress analysis of nuclear fuel pin 2

F

F

drn

dRn


Normal strain: general definition

Change of length of the segment drn

22 2

2 2 2

1 12 2 2n n n n n n n

nnn n n

dS ds dR dR dr dr dRds ds ds

Definition valid for any deformation, independent of the level of deformation



Normal strain: small strain approximation

In this case only, nn represents the relative elongation of the segment drn

2 2

2 22 2 2

22

n n n n n n n nn nnn

n n n n

n n n n n

n n n

dS ds dS ds dS ds dS dsdS dsds ds ds ds

dS ds ds dS ds

ds ds ds



Shear strain: general definition

Definition valid for any deformation, independent of the level of deformation

Shear strains are associated with distortion of the body n and t are two perpendicular segments in the undeformed body, the angle changes with

deformation to become nt

1 12 2

n tnt nt

n t

dR dR cosds ds



Shear strain: small strain approximation

In this case only, nt represents half of the angle (from 90°) of the original segments t and n

1 1 1 12 2 2 2 2 2

n tnt nt nt nt

n t

dR dR cos sinds ds



Strains in function of the displacement field u

n n ndR dr du

t t tdR dr du

ndr

tdrThe image part with relationship ID rId12 was not found in the file.

tdR

u

nu du

tu du

12

n tnt

n t

dR dRds ds

2

2

1 12

nnn

n

dRds

n n

n n n n

dR dr u unds ds s s

t t

t t t t

dR dr u utds ds s s

12nt

n t n t

u u u ut ns s s s

12nn

n n n

u u uns s s



Small Strains in function of the displacement field u

Within the small strain approximation, the strain components expressed relatived to a set of n, t and s directions orthogonal and in terms of the displacement vector u are given by:

12nt

n t

u ut ns s

ttt

uts

nn

n

uns

sss

uss

12ns

n s

u us ns s

12ts

n t

u ut ns s



Small Strains

In Cartesian coordinates

xxx

ux

In cylindrical coordinates

yyy

uy

zzz

uz

12

yxxy

uuy x

normal strains

shear strains12

x zxz

u uz x

12

y zyz

u uz y

rrr

ur

1r

u ur

zzz

uz

normal strains

1 12

rr

u uur r r

12

r zrz

u uz r

12

zz

u uz r

shear strains




Types of loadsA body deforms under the action of applied loads and displacements.

1. Externally applied loads

Mechanical loads represented by discrete force F

Different pressures of a fluid (gas or liquid) on opposite faces of a structure

For instance high pressure of coolant in primary system of a LWR Components subject to this pressure loading are

Cladding Reactor pressure vessel Primary system piping

Reaction forces due to connection of a particular structure to its supports and to other components of a complex mechanical system Lower grid plate of a LWR supports weight of hundreds of fuel assemblies:

Gravity-induced stresses



Types of loads2. Thermal Stresses

Stresses are generated when expansion of a heated body is restrained

3. Residual Stresses

This type of stress arises from two principal sources: Fabrication of a component: e.g. cold working (reduction in cross-sectional area by

passing between dies) introduce internal stresses in finished piece Remain during operation unless 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 adjacent metal that does not melt (“heat affected zone”). stresses persist in cooled piece are supplemented by additional stresses arising from solidification and weld cooling


Stresses

The concept of stress is used to describe internal forces in a solid. The components of stress in a Cartesian basis are often expressed as a tensor.xx, xy, xz represent the force per unit area acting in the (x,y,z) directions on a plane which has normal in the x direction.

xy xz

yx

xx

yy y

z zx y z

z

z

normal stresses: ii shear stresses: ij



Body forces

Body forces are forces that act on all particles in a body as a result of some external body or effect not in direct contact with the body under consideration. An example of this is the gravitational force exerted on a body. This type of force is defined as a force intensity per unit mass or per unit volume at a point in the continuum.

dV

db



Equilibrium equations in Cartesian coordinatesThe linear momentum balance requires:

The angular momentum balance requiresthe symmetry of the stress tensor:

ij ji

0iji

j

bx

ij: stress conmponentsb: body force

0

0

yxxxxx xx yx yx

yxxx

2D net force in x dx dy dy dxx y

x y

3 0yxxx zxD : x y z



Equilibrium equations in cylindrical coordinates

The linear momentum balance requires:

The angular momentum balance requires the symmetry of the stress tensor:

radial direction1 0rr rrr rzrb

r r r z

2 1 0r r z br r r z

1 1 0zzr zzz

( r ) br r r z

angular direction

z direction direction

z z rz zr r r



Principal stresses (and strains)

For any stress tensor (ij) expressed in the e1, e2, e3 basis, it is always possible to find another basis m1, m2, m3 in which (ij) is diagonal. Direct consequence of the tensor symmetry.

The principal stresses are the three eigenvalues of the matrix formed by the components of (ij) in ei. By convention 1 > 2 > 3.

The three eigenvectors of (ij) in ei are used to find the unit vectors m, m2, m3 in ei. They are known as the principal stress directions, or the principal basis of the stress tensor.

The above observations can be made for the symmetric tensor of the deformation. The principal directions of the stress coincide with the principal directions of the strain (in static analysis).



Elastic strain-stress relations (1)

• By superposition of the components, we get:

• For isotropic materials, the normal components ii do not produce shear strain.Reciprocally, the shear stress ij do not produce any normal strain, then (G is theshear modulus):

1122 33 11 E

11 11 22 331 ( )E

1212 2G


• It is observed that ii = E ii where E is the Young modulus

• A tension in the x direction produces a contraction in the y and z directions.Experimentally, it was found that:


Complete elastic strain-stress relations

1xx xx yy zz( )

E

2xy

xy G

1yy yy xx zz( )

E

1zz zz yy zz( )

E

In Cartesian coordinates

2xz

xz G

2yz

yz G

In cylindrical coordinates

1rr rr zz( )

E

1rr zz( )

E

1zz zz rr( )

E

2r

r G

2rz

rz G

2z

z G



Complete elastic stress-strain relations

11 2 1xx xx yy zz

E ( )

In Cartesian coordinates In cylindrical coordinates

11 2 1yy yy xx zz

E ( )

11 2 1zz zz xx yy

E ( )

11 2 1rr rr zz

E ( )

11 2 1 rr zz

E ( )

11 2 1zz zz xx

E ( )



Isotropic elasticity

The solid has a preferred reference configuration. In the absence of loading, it has a well defined shape (this distinguishes a solid from a fluid).

The strain in the specimen depends only on the stress applied to it. It doesn’t depend on the rate of loading, or the history of loading.

For most materials, the stress is a linear function of strain.

For most, but not all, materials, the material has no characteristic orientation. Thus, if you cut a tensile specimen out of a block of material, the stress-strain curve will be independent of the orientation of the specimen relative to the block of material. If this is the case, the material is said to be isotropic.



Physical significance of the elastic constants

Young’s modulus E is the slope of the stress-strain curve in uniaxial tension. It has dimensions of stress N/m2.

Poisson’s ratio is the ratio of lateral to longitudinal strain in a uniaxial tensile stress. It is dimensionless and typically ranges from 0.2-0.49, and is around 0.3 for most metals. It is a measure of the compressibility of the solid. If =0.5, the solid is incompressible, th volume remains constant, no matter how it is deformed. If =0, then stretching a specimen causes no lateral contraction.

Thermal expansion coefficient quantifies the change in volume of a material if it is heated in the absence of stress. It has dimensions of K-1 and is usually very small. For steel, = 6x10-6 K-1.



Mathematical formulation of elasticity problemsVariables

Displacement field, ui 3

Strain field ij 9

Stress field ij 9_____

Total variables: 21

Equations

Strain-displacement 9

Equilibrium (1) 3

Equilibrium (2) 3

Strain-stress 6

_____Total equations: 21

1ij ij kk ij

( )d d dE E

ij ji

0ij

jx

12

ji k kij

j i i

uu u u( )x x x xj



Mathematical formulationOne can then find a set of 12 stress and strain conponents (2 x 6) and one set of three displacements

components that satisfies the previous equations.

In some problems, it is useful to express the equilibrium in terms of strain instead of stress, by using the

stress-strain relations, these are the Navier-Lamé equations, which can then be solved to to yield the

displacements:

with

2 3 2 0u u T

1 1 2E

2 1E



Navier-Lamé equation in Cartesian coordinates

2 3 2 0e Tuy x

u u uex y z

2 3 2 0e Tux y

2 3 2 0e Tuz z



Navier-Lamé equation in cylindrical coordinates

12 3 2 0ze Tr r z r

1 1ru v wer r r z

1 12 3 2 0r ze Tr r z r r

2 3 2 0rre Tz r r z

1 12r

w vr z

12

u wz r

12r

rv ur r



Important modes of deformation: plane strainPlane strain is defined to be a state of strain in which the strain normal to the x-y plane, z, andthe shear strain xz and yz are zero.Consider an infinitely long cylindrical (prismatic) body as shown. If the body forces and tractionson lateral boundaries are independent of the z-coordinate and have no z-component, then thedeformation field can be taken in the reduced form u=u(x,y), v=v(x,y), w=0

211yy yy xxE

10zz zz yy zz( )E

0yz xz

211xx xx yyE

2xy

xy G

21EE'

1

'

Plane strain elastic constant:



Important modes of deformation: plane stress

( , )( , )( , )

0

x x

y y

xy xy

z xz yz

x yx yx y

Plane stress is defined to be a state of stress in which the normal stress, z, and the shearstress xz and yz, dircted perpendicular to the x-y plane are zero.Two stress free planes z = h, where h is small in comparison to other dimensions in theproblem. There can be little variation in the stress components, z, xz, yz through thethickness, and thus they will be approximately zero throughout the entire domain.For a thin the region in the z-direction, at the other non-zero stresses will have little variationwith z. Thus,




Thermal stress, a simple exampleBar of length L placed between two rigid ends Has linear thermal expansion coefficient Stress-free at temperature T

Bar is now heated to temperature T+ΔT end pieces remain same distance apart

How to calculate stress? First imagine that top end piece is removed Bar allowed to expand freely by an amount αΔTL, thermal strain εth = αΔT. A stress σ is applied to top of bar sufficient to return it to original length Elastic strain in εel = σ/E. Total strain is zero

εtot = εth + εel = 0 σth = -αEΔT.

Stress is compressive Superscript “th” indicates stress is due to restrained thermal expansion

TL T+TT+TL+L

ΔTL

εel = σ/E

T+T

εth = ΔT



Thermal stresses

Restraint of free thermal expansion during uniform heating by intimate interaction with other rigid structures

Compressive stresses generated by uniform heating of a solid containing inclusions (precipitates) with larger value of than that of host solid

Temperature non-uniformities (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 Nuclear fuels

Stresses in reactor components by changing temperature gradients Reactor power changes during startup or shutdown Sudden injection of cold emergency cooling water near inner wall of RPV


02.05.2016 30

Strain-stress relation with thermal gradientAdd linear thermal strain to elastic strains

Book: Theory of thermal stresses by Bruno A. Boley, Jerome H. Weiner, (1960)

1r r z( ) + T

E

1r z( ) + T

E

1z z r( ) + T

E

“Nuclear Fuels & Materials” - Stress analysis of nuclear fuel pin


Derivation of thermal stresses in pellet and cladding (1)Problem: determine the thermal stressed in a pellet or cladding in the presence of a temperature gradient

General case: consider a tube, inner and outer radiis are R and Ro respectively.For the pellet, R=0. For symmetry reason we have: .

To fix the ideas: let’s calculate first r=r(r).

At our disposal, we have:

(1) the definition of normal strains:

(2) the strain-stress equations: with 1,2,3 = r,,z

(3) the equilibrium equation: (radial direction)

1 1 2 31 ( ) + T E

rr

ur

1r

u ur

1 0r rr rz

r r r z

zz

uz

R

oR



Derivation of thermal stresses in pellet and cladding (2)Assumptions:

• For symmetry reason, there is no dependence on , so• In addition, we consider first a long cylinder so that away form the free surfaces,

there is no gradient of vertical displacement, plane strain condition.

These two conditions allow to simplify some equations as:

(1) the definition of normal strains:

(2) the strain-stress equations: with 1,2,3 = r,,z

(3) the equilibrium equation: (radial direction)

R

oR

0/

rr

ur

rur

1 1 2 31 ( ) + T E

0rrddr r

0z



Derivation of thermal stresses in pellet and cladding (3)

(1) Eliminate the displacement ur with er and e:

(2) In previous equation, replace the strains by the stresses from strain-stress relations:

(3) We use the condition dz/dr = 0, to eliminate z:

(4) Combine (2) and (3) yields:

10r z r

d ( ) + ( )dr r

2 11 1 1 0rrd d dTE

dr dr r r dr

0rddr r

1 0zz r d d ( ) + T

dr dr E




(5) Use the radial equilibrium equation to eliminate in (4):

(6) to finally get

(7) Integrate the previous equation, cylinder with inner radius R and outer radius Ro (Exercise 3). No pressure on the cylinder faces, r(R) = r(Ro) = 0

This equation is valid for a pellet (R=0) and for a hollow cylinder (cladding tube).

rr

drdr

32

11

rdd E dTrr dr dr dr

0

22

2 2 20

1 11

R rthr

R R

RE r rT r dr r T r drR R r




(8) is deduced from the equilibrium equation:

Observations: - Stresses build up in the cylinder due to the temperature gradient only !!

T(r) can be due to: - different temperatures at the inner and outer surfaces- internal heat generation in the wall with contant and equal surface temperatures- transient temperature change in an initially isothermal cylinder

The thermal stress components are independent of axial end conditions, although these endconditions affect the axial component of the stress

rr

drdr

0

22

2 2 20

1 11

R rth

R R

RE r rT r dr r T r dr T(r)R R r




Application to fuel pellet thermal stresses (1)

Uniform volumetric heating in the solid cylinder cooled at periphery generates parabolic temperature, T0=T(r=0), Ts=T(r=R0) (see previous lecture)

Inserting into radial and azimuthal equations (exercise 4)

2

20 0

1S

S

T T rT T R

0

22

2 2 20

1 11

R rthr

R R

RE r rT r dr r T r drR R r

0

22

2 2 20

1 11

R rth

R R

RE r rT r dr r T r dr T rR R r

20

20

14 1

th Sr

E T T rR

20

20

1 34 1

th SE T T rR



Application to fuel pellet thermal stresses (2) The axial component is obtained from the plane strain condition (z and dz/dz=0), which yields:

(1) (2)

Due to plain strain, there is a normal stress z.

For a pellet with unrestrained ends, the situation is slightly different (quasi-plane strain, z = cst). The pellets have unconstrained ends, which means that for any horizontal plane wemust have:

(3) which combined with (1) yields z = T

With z = T, we get from eq.1:

1th th th thz z r( ) + T =0

E

20

20

4 2 44 1

Sthz

E T T rR

2

1 2 02

oRth thz z

o o

r ( r )drR

20

20

2 44 1

Sthz

E T T rR



Stress profiles in the pellet horizontal plane

20

20

14 1

th Sr

E T T rR

20

20

1 34 1

th SE T T rR

Region of interest is near outer edge of cylinder Azimuthal stresses is tensile but radial stress remains compressive in the pellet If either azimuthal stress exceeds fracture stress, vertical radially-oriented cracks appear in pellet



Stress profiles in the pellet vertical plane

Region of interest is again near outer edge of cylinder If either axial stress exceeds fracture stress, horizontal cracks appear in pellet



Pellet Cracking (1)



Pellet Cracking (2)

Masaomi OGUMA, “CRACKING AND RELOCATION BEHAVIOR OF NUCLEAR FUEL PELLETS DURING RISE TO POWER”, Nuclear Engineering and Design 76 (1983) 35-45



Pellet relocationConsequence of cracking as well as swelling, are important in fuel performance:

1. The fragments tend to move outward. So they reduce the pellet-clad gap size andthus reducing the fuel temperature.

2. The fragments can get in contact with the cladding that creeps down under theinfluence of the pressure difference between the coolant and the gas in the gap,reverting the direction of relocation.

3. Cracking leads to a completely different stress distribution in the intact part of thepellet.

4. Pellet relocations results in different possible fuel-cladding interactions.



Fuel pellet interactions

These interactions are critical inthe evaluation of the axialstrain/stress in the cladding.

The radial stress/strain in thepellet are derived from…



Cladding bambooing (1) The elastic stress-strain analysis explains partly the

“bambooing”.

Radially, the expansion of the pellet in plane stress and plane strain is identical but not axially (see in Book: Theory of thermal stresses by Bruno A. Boley, Jerome H. Weiner, (1960), pp. 290-291)

Nothing limits the axial expansion on the pellet ends. The compressive radial and azimutal components in the center of the pellet are responsible for the curved ends.

The elastic stress-strain analysis, restricted to plane stress and plane strain conditions, explains partly the “bambooing”.

plane stress

plane strain

We need to consider a more realistic strain/stress state the effect of temperature variation along the

radius on the elastic contants the plastic deformation in the central part of

the pellet


“Nuclear Fuels & Materials” - Stress analysis of nuclear fuel pin

Cladding bambooing- numerical simulation (2)

Veeder calculated the thermal expansion of finite cylinder for various length-to-diameter ratios with a numerical methods and showed that the radial extension on the ends is always the largest

Ref: J. Veeder, «Thermo-elastic extansion of finite cyclinders», AECL-2660 (1967), see pdf file for the details

02.05.2016 45



Elasticity

)1(2

EG

UO2 brittle at temperatures < Tm / 2 UO2 deformation is elastic up to rupture Elastic deformation depends on Young and Shear moduli: E, G Polycrystalline stoichiometric UO2 at room temperature (25oC)

E ~ 219. MPa G ~ 81.4 MPa Poisson Ratio: ~ 0.33 E and G are related: G=E/(2(1+))

Again all the calculations presented above were done in the elasticity framework withouttemperature dependence consideration on elastic contants onset of plasticity

E



Effect of temperature on elastic constants

Gd2O3 simulates FP (segregation at GB preventing plasticity by GB sliding) Little influence on elastic behavior, but extends temperature range where elastic behavior

prevails Poisson ratio increases with T from 0.32 to 0.5

(after transition from brittle to ductile behaviour)

E25=2.19x108 kN/m2 G25=8.14x107 n/m2



Brittle versus ductile deformation (1)

Poly-cristalline U02 is a brittle material under normal conditions (no irradiation) as most other ceramics

Brittle substance has no capacity for plastic deformation Elastic straining that occurs conforms to

Hooke's law mode of failure is known as brittle

fracture -> fracture stress At sufficiently high temperatures, however,

even normally brittle materials such as U02exhibit measurable amounts of unrecoverable or plastic deformation before failure -> ductile behavior



Brittle versus ductile deformation (2)

Transition from brittle to ductile behavior occurs at a temperature characteristic of the material (ductile – brittle transition or nil ductility temperature)

Mode of fracture characterized by morphology of fracture surface Ductile: Dimples Brittle: Inter- or trans-granular fracture

Ultimate tensile stress

Elastic limit

Total plastic strain



Elastic/plastic properties in the pellet

region 3

region 3

regi

on2

region 2

region 1

region 1Region 1, T > 1200 °C has

no significant strength and is ductile

Region 2 has strength and plasticity

Region 3 is completly elasticand brittle

This description of elastic/plastic properties isquite different from the simple elastic consideration



Pressurized cladding without T gradient (1)

distance fixed, completed axial restraint

p closed-end tube



Pressurized cladding without T gradient (2) Starting equation:

Integrate twice, using boundary conditions

σr(Ro) = 0 Ro: inner radius of cylinder σr(R) = -p R: outer radius of cylinder

Substitute into radial equilibrium equation to obtain hoop stress

20

20

1

1r

Rr

pR

R

0r r

r r

20

20

1

1

Rr

pR

R

32

1 0rd drr dr dr




Complete axial restraint To obtain z, set ε = εz = 0 in equation Insert r and from above

1z z rE

20

2

1z p

RR

20

20

1

1

Rr

pR

R

<

distance fixed, completed axial restraint




Closed-end tube

Obtain z, by equating gas pressure on top surface to axial stress on annulus: R2p = (R02-R2)z

20

1

1z p

RR

p closed-end tube



Thermal stress in a cladding with T gradient Temperature difference ΔT is maintained across the cladding of fuel pin Cladding wall thickness δ sufficiently small (Ro – R = δ << R) “thin-wall” approximation is valid Consider the radial temperature profile linear

With these assumptions

becomes (exercise 5)

thermal stress is negative (compressive) on inner half tensile on outer half

0

22

2 2 20

1 11

R rth

R R

RE r rT r dr r T r dr T rR R r

0

1 22 1

th E T r RR R


Lecture 3 Mechanical Solution of Fuel PinBe well informed about elasticity concept:

• Stress and strain concept, normal and shear components, principal stresses, etc…• Plane strain condition versus plane stress conditions• Know the strain-stress equations in elasticity, their origin and the significance of E and

• Types of loads, in particular explain the origin of thermal stresses

Good understanding of the cracking patterns and bambooing in the pellets:• Origin of the thermal stess in a pellet and in a cladding tube.• Qualitative description of the different thermal stress components in a pellet, with their consequences

(pellet cracking patterns, bambooing)• Discuss the limit of the elasticity theory in the determination of the stress state in pellets

1122 33 11 E

1212 2G

11 11 22 331 ( ) + TE


week3 stress analysis fuel pin - nucmaster.web.psi.ch for reactor physics and systems behaviour...

Documents