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General Structural Equations

G. 1 Beam Theory J.E. Meyer

G.2 Th e Stress Analysis of Pressure Vessels, pp 9-51 of Gill

G.3 Appendix Elasticity Theo ry, Olander

Fuel Modeling and Design

F. 1

Modeling of the Structural Be havior of Fuel Elements and Assem blies,

Chapter 21 of Olander

F.2

Fuel Rod Modelling in R eactivity-initiated Accidents at High Burnups:

Transuranus Verses Frey H. Wallin, PSI Science Report 2001, vol. 4

F.3

Extensive Programs Demonstrate ZIRLOTMCladding's performance

Benefits, A W estinghouse B rochure

F.4 Focus XS and HTS A new generation of Higher Enrichment PWR

Fuel Assemblies

A

brochure from Siemens Nuclear Power

Radiation Effects on Materials Components

R.1

Fundamental Radiation Effects on M aterials, Chapter

4

of B.M. Ma's

book

R.2 Fatigue Crack Propagation in Type s 304 and 308 Stainless Steel at

Elevated Tem peratures D.T. Raske and C. F. Cheng, Nuclear

Technology, V34, 1977

R.3 Creep-Rupture Properties of 20% Cold-Wo rked Type 16 Stainless Steel

R.4

High Fluence Neutron Irradiation R.L. Fish, Nuclear Technology, V35,

1977

R.5

Te nsile Behavior of Neu tron-Irradia ted Martensitic Ste els -R:L. Klueh ,

Nuclear Technology, v102, 199 3

R.6 Th e Evolution of Reactor Vessel PTS -A Catalyst to the Advancem ent

of Technology -T.A.'Meyer, K.R. Balkay, S.E. Yanishko, Mech. Eng.;

June 1984

R.7 Methods to Counteract radiation Effects on Reactor Pressure Vessel Steels

S.Brown, 22.3 14Term Paper Report, May 2000

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Flow Induce V ariations

V. 1

Heat Exchange Acoustics Section 12.5 of Flow-Induced Vibration of

Power and Process Plant Components M.K. Au-Yang, 2002

V.2

Summary of Important flow-Induced Vibration Relations

R. Herron, 2002

Containment Designs

C.1

Regulations and Standards for Design of Nuclear Facilities

C.2

Nonlinear analysis of Reinforced Concrete Structures Biiyiikoztiirk,

in Computers and Structures, V.7, 1977

C.3

Imaging of Concrete Structures Biiyiikoztiirk, NDT E

International, V.3 1, 1998

Pressure Vessel nalysis

P.l Structural Design Notes Topic C, Pressure Vessel Stress Analysis J.E.

Meyer, revision August 1996

P.2

Structural Design Notes Topic D, Design Rules J.E. Meyer, December

1987, revised April 1992

P.3

Reactor Pressure Vessels Notes on Elevated Temperature Effects J.E.

Meyer revision July 1994

P.4

Criteria of the ASME Boiler and Pressure Vessel Code for Design by

Analysis in Sections I11 and VIII, Division 2, ASME 1969

P.5 Subsection A Requirements for Class A Vessels Article 2

P.6 Subsection A: Requirements for Class A Vessels

P.7

Article A-2000: Analysis of Cylindrical Shells

P.8

Article A-3000: Analysis of Spherical Shells

P.9

Addendum to Topic E from ASME Cases N-47-29 and N47-32

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Fall

2006

Problem

Set

I

Due

09/19/06

1. A tensile test on 1020 steel gives the following results:

Load (kN) Diameter (mm) Length (mm)

0 12.8 50.800

22.2 50.848

28.5 (yielding begins)

50.2 12.2 56.1

51.2 (Max.) 10.4 67.3

43.6 69.8 (fracture)

(a) Calculate the elastic modulus.

(b) Calculate the maximum nominal strain.

(c) Calculate the tensile strength of this steel.

2. Calculate the maximum normal stress and the maximum shear stress in the cube shown

below.

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3. This problem illustrates some important stress tensor concepts and some calculation methods

for treating a stress tensor.

Consider three stress tensors a, b, and c defined as follows:

55 5 30

5

55

30

30

=

;

a

30

20

10 0 0

0 10 0

0

b ;=0 10

c =a +b

where the stresses are given in MPa units.

(a) What are the three principal stresses that characterize each of these stress tensors ?

(b) Draw a 3-D Mohrs circle to represent the three stress states.

(c) Also place points on your Mohrs circles that give the stresses on planes normal to each

of the original coordinate axes (x, y, z).

(d) What are the unit vectors (direction cosines) that define principal directions for the

three stress tensors?

(e) What is the Tresca Stress for each stress tensor? What is the von Mises Stress for each

stress tensor?

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Problem Set I Solution

1. Originaldimension:

D0 =

12.8

mm,

A0 =

D02

/4

=

128.68

mm2

;

and

L0 =

50.800

mm.

(a)

Stress

at

a

loadF =22.2kN:

=F/A0 =172.5MPaStrain:

= (LL0)/L0 =(50.848- 50.8)/50.8=0.009448Youngs

Modulus:

E=/=182.6GPa

(b)

Maximum

norimal

strain

is

the

strain

when

fracture

occurs:

(LmaxL0)/L0 =(69.8- 50.8)/50.8=0.374

(c) Fmax =51.2MPa.Tensilestrengthis:Fmax/A0 =397.9MPa

2.

This

problem

has

a

stress

tensor:

xx 0 0=

0 yy yz 0 zy zz

Note

that

the

shear

stresses

on

the

plane

normal

to

x

direction

are

zero.

Therefore,

we

can

derive

the

two

principal

stresses

on

the

yz

plane

using

the

solution

for

a

plane

stress

condition.yy +zz yyzz

)2 +2yzi,j (= 2 2

i=464.5MPaandj=40.5MPa.Therefore,

1 =464.5MPa,2 =440MPa,and3 =40.5MPa.Themaximumnormalstressis1 =464.5MPa.The

maximum

shear

stress

is(13)/2=212MPa.

3.

(a)

The

principal

stresses

are

the

eigenvalues

of

the

stress

tensor.

Its

solved

by

usingMATLAB(Seethecodeintheend).For

55 5 30

55a =

,530

3030 20

the

principal

stresses

are

found

to

be:

1 =80MPa,2 =60MPa,and3 =-10MPa.

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For

10 0 00 10 0

0

b = ,0 10

the

principal

stresses

are

found

to

be:

1 =

2 =

3 =

-10

MPa.

For

c =a +b,

theprincipalstressesarefoundtobe:

1 =70MPa,2 =50MPa,and3 =-20MPa.

(b) SeeFigure1andFigure2.TheMohrsCircleforb isasinglepoint(-10,0).

(c) Thepointsonthe3-DMohrscirclesthatgivethestressesonplanesnormaltoeachof

the

original

coordinate

axes

(x,

y,

z)

are:

(x, 2 +2 ),(y, 2 +2 ),and(z, 2 +2 )xy xz yx yz zx zy

as

shown

in

Figure

1

and

Figure

2.

Figure

1:

3-D

Mohrs

Circle

for

stress

tensor

a

(d)

The

unit

vectors

are

corresponding

eigenvector

of

each

eigenvalue

(principal

stress)

of

thestresstensor.Fora,theunitvectorsareasfollows:0.5774 0.7071 0.4082

u1 = ;u2 = 0.7071

0

;u3 = 0.4082 .0.57740.5474 0.8165

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Figure

2:

3-D

Mohrs

Circle

for

stress

tensorc

Forb:

u1 =

1 0 0

0 ;u2 = 1 ;u3 = 0 .0 0 1

Since

c is

the

sum

of

a and

hydrostatic

pressure

tensor

b,

the

unit

vectors

should

be

thesameasthoseofa.0.5774 0.7071 0.4082

=

0.5774 0.70710

0.4082;u2 ;u3u1 = = .0.5474 0.8165

(e)

Tresca =max{|12|,|23|,|31|}

(12)2 + (23)2 + (31)2vonMises =

2

a b c

Therefore:

Tresca 90(MPa) 0(MPa) 90(MPa)vonMises 81.85(MPa) 0(MPa) 81.85(MPa)

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The MATLAB code to solve Problem 3:

%============================================================

% 22.314 (Fall 06) Problem SetI-3.

% 09/19/2006%

%============================================================

Sigma a = [ 55 5 305

55

30

30 30 20];

Sigma b

= [10

0

0

10

0 10 00 0

10];

Sigma

c

=

Sigma

a

+

Sigma

b;

% eig returns eigenvalues and eigenvectors of a matrix

[Eigvec

a

Eigval

a]

=

eig(Sigma

a);

[Eigvec

b

Eigval

b]

= eig(Sigma b);

[Eigvec c Eigval c] = eig(Sigma c);

% Mohrs Circlefor Sigma a 20

s1

=

Eigval

a(3,3);

s2 = Eigval a(2,2);

s3

=

Eigval

a(1,1);

c1

= (s1 + s2)/2; r1 = (s1 s2)/2;c2 = (s2 + s3)/2; r2 = (s2 s3)/2;

c3

= (s1 + s3)/2;

r3

=

(s1

s3)/2;

x1

= linspace(s2, s1, 200);

y1p = sqrt( r12 (x1 c1).2); y1n = y1p;x2

= linspace(s3, s2, 200); 30

y2p = sqrt( r22 (x2 c2).2); y2n = y2p;x3

= linspace(s3, s1, 500);

y3p = sqrt( r32 (x3 c3).2); y3n = y3p;

figure; hold on;

axis([30 100 50 50]);

axis equal;

plot(x1, y1p);

plot(x1, y1n);

plot(x2, y2p); 40

plot(x2, y2n);

plot(x3, y3p);

plot(x3, y3n);

sigma x = Sigma a(1,1);

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tau x

sigma

y

tau

y

sigma z

tau

z

= sqrt(Sigma a(1,2)2 + Sigma a(1,3)2);

=

Sigma

a(2,2);

= sqrt(Sigma a(2,1)2 + Sigma a(2,3)2);

= Sigma a(3,3);

= sqrt(Sigma a(3,1)2 + Sigma a(3,2)2);

50

plot(sigma x, tau x, '+');

plot(sigma y, tau y, '+');

plot(sigma z, tau z, '+');

xlabel('\sigma (MPa)')

ylabel('\tau (MPa)')

%Mohrs

Circle

for

Sigma

c;

s1

=

Eigval

c(3,3);

s2 = Eigval c(2,2);

s3

=

Eigval

c(1,1); 60

c1

c2

c3

=

(s1

+

s2)/2; r1= (s2 + s3)/2; r2=

(s1

+

s3)/2; r3

=

(s1

s2)/2;= (s2 s3)/2;=

(s1

s3)/2;

x1

y1p

x2

y2p

x3

y3p

= linspace(s2, s1, 200);

= sqrt( r12 (x1 c1).2); y1n=

linspace(s3, s2, 200);

= sqrt( r22 (x2 c2).2); y2n

= linspace(s3, s1, 500);

=

sqrt(

r32

(x3

c3).2);

y3n

= y1p;

= y2p;

=

y3p;

70

figure; hold on;

axis([20 100 50 50]);axis equal;

plot(x1, y1p);

plot(x1, y1n);

plot(x2, y2p);

plot(x2, y2n);

plot(x3, y3p);

plot(x3, y3n);

80

sigma x

tau

x

sigma y

tau

y

sigma z

tau

z

= Sigma c(1,1);

= sqrt(Sigma c(1,2)2 + Sigma c(1,3)2);

= Sigma c(2,2);

= sqrt(Sigma c(2,1)2 + Sigma c(2,3)2);

= Sigma c(3,3);

= sqrt(Sigma c(3,1)2 + Sigma c(3,2)2);

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90plot(sigma x, tau x, '+');

plot(sigma y, tau y, '+');

plot(sigma z, tau z, '+');

xlabel('\sigma (MPa)')

ylabel('\tau (MPa)')

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22.314/1.56/2.084/13.14

Fall

2006

Problem

Set

II

Due

09/26/06

1. Consider a cylindrical vessel of inner radius R and wall thickness t with flat ends. Thepressure inside the vessel is Pi and the surrounding pressure Po. What is the relative error inestimating the maximum value of the stress intensity in the cylinder based on the thin shell

approximation for values of:

t/R= 0.03t/R= 0.10t/R= 0.15t/R= 0.30

Consider two cases:

Pi = 2PoPi = 20Po

2. A pressure vessel is constructed of a cylinder with a hemispherical head at each end. There

is no external restraint to either axial or radial displacement. Inside radius of both cylinder

and hemispheres is R. The wall thickness is uniform at a value t. The length of the cylinderis L. No flaws or stress concentrations are present. Dimensions are:

R

= 110 cmt= 11 cmL= 433 cm

Material properties:

Youngs modulus = 200 GPa

Poissons ratio = 0.3

Coefficient of thermal expansion = 12 m/mK

The vessel is pressurized to a design pressure P = 15.5 MPa.

Questions:

(a) What is the total (peak) stress as a function of radial position (z) at a junction betweencylinder and hemisphere.

(b) What is the maximum radial displacement of the vessel cylinder and sphere?

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22.314/1.56/2.084/13.14 Fall2006

ProblemSetIISolution

1.

Stress

intensity

=

max{|r|,|z|,zr|}Forthinwallapproximation:

r =Pi

+Po

(1)2

PiR2

Po(R+t)2

z =(R+t)2R2

(2)

=PiPo

(R+t

) (3)t 2

Therefore:

Sthin =r =Pi

Po

(R+t

) +Pi

+Po

(4)

t

2 2

Thickwallsolution:

Equilibrium

in

radial

direction

gives:

dr+

r= 0 (5)

dr r

Hookslaw:

1r

= (r

z) (6)E1

= (

r

z)

(7)E1

z = (zr) (8)E

duSince =u/r,r = dr

,

we

get:

d 1= (r) (9)

dr r

For thiscloseendcylinderfarfrom theend,planestresscondition isassumed, i.e.,z isconst.

Plug

Eq

6

and

Eq

5into

Eq

9,

we

get

d( +r) = 0 (10)

dr

Plug

Eq

10

into

Eq

5,

we

get

d 1 d(r2r) = 0 (11)

drrdr

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WithB.C.r(r=R) =Pi

andr(r=R+t) =Po,weget:

r

=Pi(R

r)2 +(1(

R

r)2)

Po(R+t)2 +PiR

2

(R+t)2R2

=Pi(

R

r

)2 +(1+(R

r

)2)Po(R+t)

2 +PiR2

(R

+

t)2

R2

r =2(R

r)2

(Pi

Po)(R+t)2

(R+t)2R2

Maximum

stress

intensity

is

at

the

location

of

inner

radius:

(12)

(13)

(14)

Sthick

=2(PiPo)(R+t)

2

(R+t)2R2

Theerrorinthinwallapproximationis:

|1

SthinSthick|

Resultsaretabulatedbelow:

t/RPi

=2Po

Pi =20Po

0.03

1.41%

1.31%

0.10

4.13%

4.09%

0.15

5.67%

5.88%

0.30

8.88%

10.46%

2.Weusethinshellapproximationtosolvethisproblem. Foraregionofcylinderfarfroma

junction,

stresses

are:

=

P

R

t(15)

r =P/2 (16)

x =PR

2t(17)

Radial

displacement:

uc

=PR2

2Et(2+

t

R) (18)

Forthesphere,stressesare:

= =PR

2t(19)

r

=P/2 (20)

Radial

displacement:

P R2 tus = (1+ ) (21)

2Et R

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Atthejunction,NoteCEq3033givethatattheedgeofcylinder:

u0 =uc +V0

+M0

(22)23D 22D

0 =

V0 M0(23)22D

D

at

the

edge

of

hemisphere:

2R 22u0 =us V0 + M0 (24)

Et Et

22 430

= V0

+ M0

(25)Et REt

3(12)1/4where, = sR,s = ( R2t2 ,u0 istheradialdisplacementatthejunction,0 istheslopeatthejunction.1

Due

to

the

continuity

of

the

displacement

and

slope,

the

four

unknowns

u0,

0,

M0,

and

V0

1NotethatthedirectionoftheshearforceQinNoteL4isdifferentfromthatinNoteC.Ifyouassumethe

directionofQ0S issameasQOC,thecontinuityofshearforceshouldgivethat:Q0S=Q0C.

Mo

Mo

Vo

Vo

x

z

Figure

1:

Junction

of

hemisphere

and

cylinder

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canbesolvedbyaboveequations.Itcanbefoundthat

M0 = 0 (26)

V0

=P R2/(4R+Et/3D) (27)

u

=

uc +

V0(28)2D3

P Rx

= (29)2t

=P R

+EV0

(30)t 2DR3

r

=P/2 (31)

With

mean

radiusR=innerradius+t/2=1.155m,t=0.11m,E=200GPa,=0.3,we

get

(a)

At

the

junction,

from

Eq

29Eq

31:

x=81.38MPa=122.06MPar=-7.75MPaThe

maximum

stress

is

the

hoop

stress:

122.06

MPa.

(b)

Radial

displacement

as

a

function

of

radial

postionzis:(R+z).Thus,fromEq18,

theradialdisplacementofcylinderis:

P R t(2+ )(R+z)

2Et R

From

Eq

21,

the

radial

displacement

of

hemishpere

is:

P R t(1+ )(R+z)

2Et R

From

Eq

28,

the

radial

displacement

of

junction

is:

P R t V0( (2+ ) + )(R+z)

2Et R 2DR3

Therefore:

Cylinder

Sphere

Junction

Max.

displacement

(m)

0.00089

0.00037

0.00063

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22.314/1.56/2.084/13.14

Fall

2006

Problem

Set

III

Due

10/05/06

1. For the same vessel described in Problem II-2, find out the maximum allowable pressure

of the coolant so as not to cause elastic failure under expected static loading and fatigue

conditions. State the criterion you wish to apply before proceeding to numerical evaluation.

Yield stress = 320 MPa, Ultimate tensile stress = 500 MPa.

S-N curves for both high cycle fatigue and low cycle fatigue are attached.

Expected number of cycles between cold and hot zero power conditions = 500.

Expected number of cycles between hot-zero power and hot-full power conditions =

105

Temperature profile is distributed parabolically at hot-full power conditions:

1

T=Tout T(12z/t)2

4

Where Tout is the outside wall temperature. T = -50 K is the temperature differencebetween inside and outside wall.

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22.314/1.56/2.084/13.14

Fall

2006

Problem

Set

III

Solution

We will consider the primary stress and secondary stress. Since the primary stresses are defined as

external stresses, the only primary stress for this problem is due to system pressure. The secondary

stresses are due to constraint. For this problem, the secondary stresses are junction discontinuitystress and thermal stress.

We consider the vessel subjected to static load and cyclic load respectively.

1. Under static load

Given that y = 320MPa and u = 500 MPa, from ASME code, we can get that underprimary stress:

2 1PmSmmin( y, u)=166.7 MPa

3 3For cylinder:

Pm =

|

r|

=

11P

For sphere:

Pm

=|

r|= 5.75P

Thus we have:

11P166.7 MPa

P15.6 MPa

For secondary stresses, we will consider three regions, the cylinder region far from junction,

the sphere region far from junction and junction region. The thermal stress is assumed

identical in three regions. Define:

the stresses in cylinder region: ,c, x,c,r,cthe stresses in sphere region: ,s,x,s,r,sthe stresses in junction region: ,j, x,j, r,jIt is easy to see that: ,s

< ,j

< ,c

, since the shear force in junction is in the same

direction with the system pressure for sphere, and opposite direction for cylinder. Also,

x,s

=x,j

=x,c

and r,s

=r,j

=r,c. Thus the maximum stress intensity will be in thecylinder region. From ASME code:

2 1Pm +Q3Sm3min( y, u)=500MPa

3 3Pm +Q=max(|,cr,c|,|x,cr,c|,|,cx,c|)

P R

E

P P R

E

P P R=max(

t+

1(TavgT(z)) +

2,

2t+

1(TavgT(z)) +

2,

2t)

Average temperature isTavg :

t/2

t/2T(z)2rdr

Tavg

= t/2

2rdrt/2

=Tb0.3T

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The thermal stress is:

E E 1 2z

1(TavgT(z)) =

1(

4(1

t)20.3)T

We can see that the maximum thermal stress is at the outer surface = 51.4 MPa. Therefore:

Pm

+Q=11P+51.4500 MPa

P40.8 MPa

2. Under cyclic load

The miners rule is applied.n1 n2

+ 1N1 N2

Assuming from CZP to HZP, only the pressure oscillations, n1=500.Assuming from HZP to HFP, only the temperature oscillations, n2 =10

5.

From topic D, Salt =

1

2max,Tresca. For high cycle, the alternating stress is the thermalstress, the amplitude is: Salt

=25.7MPa. Applying a safety factor of 2, and using the mostconservative curve 7 on the high cycle fatigue figure, it can be seen that at 51.4 MPa, the

cycle number is 6.5E5, i.e. N2 >6.5E5. Thus, N1n1/(1n2/N2)= 588.From low cycle fatigue figure, it can be seen that Salt 65,000 psi= 448 MPa , at lowcycle, the only stress variation is due to pressure variation and:

Salt = 21max,Tresca = 5.5P Again, applying a safety factor of 2:

5.5P448/2=224 MPa

P40.7 MPa

From above analysis results, it can be seen that the static primary stress condition in cylinder is the

limiting factor. Therefore, the system pressure:

P15.6MPa

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Fall

2006

Problem

Set

IV

Due

10/12/06

1. Consider a cantilever beam with one end fixed into a wall. The dimensions are shown in the

figure below. A force F =bhy/2is applied to the other end. y is the yield strength. Now abending moment is applied as shown in the figure. Determine the magnitude of the moment

Mat which the the beam fails. Assume idealized material law: elastic perfect plastic.

M

F

h

bL

2. The stress tensor can be written as the sum of a deviatoric stress tensor and a dilation stress

tensor and the strain can be written as the sum of volumetric strain and deviatoric strain as

follows:

1 0 0 S1 0 0 P 0 0

0 2 0

=

0 S2 0

+

0 P 0

0 0

3 0 0

S3 0 0

P

1 0 0

10 0 v/3 0 0

0 2 0=0 2 0+ 0 v/3 0 0 0 3 0 0

30 0 v/3

Where P =(1 +2 +3)/3 is the hydrostatic pressure and v = 1 +2 +3 is thevolumetric strain.

Please show that the distortion energy UDis equal to:

31

UD = Si

i2i=1

3. Consider several ceramic fuel materials with properties listed in the table below:

Property UO2 UC UN

Thermal conductivity average (W/moC) 3.6 23 21Melting point (oC

) 2800 2390 2800

Linear coefficient of expansion (/oC) 10.1E-6 11.1E-6 9.4E-6Fracture strength (MPa) 110 60

1

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(a) Calculate the theoretical maximum linear heat generation rate.

(b) Compare the the minimum linear heat generation rates at which fracture would occur

for UO2 and UC.

2

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22.314/1.56/2.084/13.14

Fall

2006

Problem

Set

IV

Solution

1. Thestressdue to force F isuniformwithavalueof y/2as showninFigure(a). ThestressduetopurebendinginelasticregimeislinearlysymmetricasshowninFigure(b)In

elasticregime,combinationofforceandmomentwillgivealinearstressdistributionwithamaximumvalueachievedatz=h/2.Thuswhenyieldingbegins,asshowninFigure(c)

=yy/2 =y/2

(a)

(c)

h/2

-h/2

z

h/2

-h/2y/2

z

y

zh/2

-h/2y

z0

-h/2

h/2z

y

(b)

(d)

Themoment forreaching the yield at a local space can becalculated. Themoment shown in

thefigureiscounterclockwise,therefore:

1 y h 2 hM1 = 2 ( )( )

3( )b

2 2 2 2

1M1 = ybh

2

12Whenfurtherincreasingthemagnitudeofmoment,theyieldingzoneinlowerbeamwill

further extend butthemaximumstressremainsy aselasticperfectplasticmaterialproperty

1

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isassumed. The upper beam become compressed and will also reach the yielding at a certain

moment. This situation continuesuntil thebeamgets totheyieldingat the wholecross

section. ThestressdistributionisshowninFigure(d). Todeterminethemoment,wefirst

needtoevaluatethepointz0 whichdividethecompressionandtension. Forcebalanceinaxialdirectionwill give:

F =ybh

= (h

2z0)(y)b+ (z0 +

h

)yb2 2

h h h=(

2

z0) + ( +z0)2 2

hz0 =

4Therefore,themomentMforactualyieldingofthewholebeamcanbecalculated:

h h h hM =y(

2

z0)( +z0)/2b+y(z0( ))( +z0)/2b2 2 2

3M =16

ybh2 islargerthanM1

2. Totalstrainenergyisthesumofdistortionenergyanddilationenergy:

UT =UD +US

UD =UTUS3

1

UT = ii

2i=1

1US = (P v)2

Therefore:

31 1

UD = ii (P v)2 2

i=1

31 1

= (SiP)(

i +v/3) (P v) (1)2 2i=1

3 3 3 31 1 1 1 1

= Si

i + Siv/3 + (P)

i + (P) v/3 (P)v2 2 2 2 2i

=1i

=1i

=1i

=1

Since

3 3

Si = (i +P) =1 +2 +3 + 3P = 0 (2)i=1 i=1

3 3

i = (iv/3)=1 +2 +3v = 0 (3)i=1 i=1

2

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Plug Eq 2 and Eq 3 into Eq 1,wecanget:

31

UD = Si

i2i=1

3. (a) From theclassnote,assuming a constant value ofthermalconductivity thelinear power

isproportionaltothethetemperaturedifferencebetweenthecenterlineandfuelsur-

face.

q = 4kT

Themaximumlinearpowerislimitedbythemeltingoffuel,i.e. whenmeltingfirstly

occursatthecenterlineofthefuel.ThusassumeTfo =700oCand plug meltingpoints

andaveragethermalconductivitiesofUO2,UCandUNintoaboveequations,wecan

getanestimationofthemaximumachievablelinearheatgenerationrates:

qmax,UO

2=

95

kW/m

qmax,UC =489 kW/m

qmax,UN =554 kW/m

(b) RefertonoteM12, thermalstressdueto aparabolictemperaturedistributionwould

give a maximumstress intensity at the outer surface of the fuel inelastic regime. Using

Trescasyieldcondition,whenfracturefirstlyoccursattheoutersurfaceofthefuel,

thelinearheatgenerationratecanberelatedtothethefracturestressasfollows:

q

=

T16k(1)

E

Fromtheclassnotes, E=175GPa, = 0.3forUO2,assumethesamepropertiesforUC.Plugmaterialpropertiesintotheaboveequation,wecanget:

q = 7.7 kW/mUO2

q =24.3 kW/mUC

Thustheq toinitiatecrackingistypicallylessthan10%ofthattoinitiatemeltingofthefuel. BothUCandUNprovideanopportunitytoincreaseq significantlywithoutviolating

thelimits.

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GPa

S T R A ~ N

MPa)

MPa,

cp)

(strain-

E,)?

(oa,J

oa A}:

MPa.

@aij) (caij)?

(oa,)

22.314/1.56/2.084/13.14 Fall 2006

Problem Set V

Due 10/19/06

This problem illustrates some "plastic strain" concepts and some calculation methods for

treating plasticity.

Uniaxial Stress-Strain Information: Consider a material that has the uniaxial stress-strain curve(actually two straight lines) shown below. The first portion of the curve, a line from the origin to

STRESS point A, has no plastic deformation. The secondportion, a line from point A to point B, has bothelastic and plastic deformation.

The elastic stress-strain behavior is characterizedby a Young's Modulus of 195 and a Poisson'sratio of 0.3.

Additional information about the uniaxial curveincludes the stress at point A (1 50 and the

0(stress, strain) at point B (260 0.54%).

Derived Properties: Using the above uniaxial stress-strain information, what is the 0.2% offset

yield stress for this material? What is the plastic stress as a function of equivalent

hardening) strain

Loadina Sequences and Questions:

A three dimensional stress is applied in a proportional manner to a solid made of the abovematerial until the following stress state is reached (note that this state is 2.5 multiplied bythe tensor of problem set

where the stresses are given in

1) What is the deviatoric stress that corresponds to What is the equivalent stressas evaluated by the following equation:

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abii),

MPa.

ab

Show that this equivalent stress is equal to the previously defined "von Mises stress."

2) What strain tensors (elastic, mechanical, and total) exist when aa is reached?

The stress on the same solid is next proportionally reduced until zero stress remains.

3) What strain tensors (elastic, mechanical, and total) exist when zero stress is reached?

The same solid is subjected to a different proportional loading to reach the stress statewhere:

where the stress is given in

4) What strain tensors (elastic, mechanical, and total) exist when is reached?

The stress is next proportionally reduced until zero stress remains.

5) What strain tensors (elastic, mechanical, and total) exist when zero stress is reached?

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22.314 Problem V Solution Fall

2006

Known

E

=195000

v

=0.3

A

= 150

oB

:=260

sB

=0.54-10

- 2

Derived

properties

EmA

=0

on

rsm

=

B

-

E

smY

:=0.002

smY - smA

ry : mA

. oB -

aA)

.

aAd

Linear

interpolation

between A anb B to get yield

stress

r

emB -

smA

aY

=

204.098

The 0.2

offset yield

stress oY

is 204.098

MPa

From the

uniaxial

stress-strain curve,

when op

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Loading

sequences

and

questions

1317.5

-12.5

75

a:=

-

12.5

137.5

75

75

75

50

1)

akk:=o

0,0

+l

a,

I

+

aa

2

,

2

ukk

=325

i:=0..2

j =0..2

ac

:=0

Sal

1

:=oa,

- --i-

j,0)-

129.167

-12.5

75

a

=12 5

29.167

75

75

75

-58.333

une

=

i

Sa)

Sao,

-)22+

Sao,2-2

(Sa

,)2

1 Sa,2)

.

2

Sa2,)]

oae

= 204.634

Now

calculate

VonMises

stress

for

comparision

eigena:=

eigenvals aa)

aVM

.[(150

- 200)

2

(150

+25)2 200+25

cVM

=204.634

We

can see cae=aVM

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2) Strain tensor

for

sa

a) Elastic

strain

i

=0 2

j

=0..2

1

selaij

=2

.[ l

v)-ai,j-

v

5 i-

j ,0 -Okk]

S4.167.10-

4

-8333-10'

5

10-4

cela

8.333.10

4.167 10

5 10

4

[510 4

5.0

- 4

-1.667-10

-4

b)

Mechanical

strain

Since

aae=204>aa=150,

mechanical

strain

exists

and

has

to

be calculated

Since

the

stress

was

applied

in

a

proportional

manner,

Saij/ce

is

constant,

thus

we

have

cm ij)= 3*Sa i,j)2*ae)*se

roota

=

roo

Lae E-

A. oB

-

A)

A],ea

Ec

= roota

sae

=

0.002

100sac

=

0 202

Mathcad

sometimes

doesn't

show

enough

significant

digits.

sae

is actually

0.00202

here

Sa

mai

2.aa

4.31810

4

-1.85110

-4

0.001

ma

-1.851-10

-

4.318-10

4

0.001

0.001

0.001

-8.637 10

Total

stress

stola

=~ela-

cma

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8.48510

4

-2.684*.t0

4

0.002

Etola

=

2.68410

8.48510

-

4

0.002

0.002

0.002

-.

OO l

3) When

stress

is

proportionally

reduced

to

zero,

sela

vanishs

while

ema remains unchanged.

stola=tma

when

zero

stress

is reached

26

0

0

ab

:=

0

00

000

akk

:=

bo,

t

bl.

I

b

2

,2

akk

260

i

=0..2

j

=0..2

Sbi,

j

abi

j

-

-5(i-

j,0 kk

173.333

Sb =

0

0

0

0

-86.667 0

0

-86.667

obe=

3.-[ Sbo

0

)

2

(Sb, 1)22+S

Sbb2)22+

Sb j 2+

Sb2)

2-2+

Sb

22

)

2

]

obe

=260

4) Strain

tensors

for

ab

a)

Elastic

strain

i

=0..2

j

=0..2

celbi

j

=

Ij

v)-obi,-

v5 i-j,

0).kk

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0 001

selb =

0

0

0 0

-4

10

-4

0

0

-4-104

b) Mechanical strain

During

the second

loading, when ae

increases

from 0

to

oae,

no

mechanical

strain is

produced.

Since

cbe=260 >

aae=204, new

mechanical

strain

will be produced

when

ae

increases

from aae

to

abe.

bshe

roothb=ro

obe-

-a aB-A)

aAse

4eB

-

zeA

she

=rootb

abe = 0.004

100-sbe= 0.407

3.Sb

i

smadd,,

:=

. sbe

-

sac

0.002 0

0

cmadd 0

0.001

0

0 0 -0.001

smadd

is

the addtional mechanical

strain

tensor that is produced

when

ve

increases

from

ose

to

abe.

Adding

smadd

to

Ema

we

can

get the total mechanical strain when

ob

is

reached.

smb := madd

sma

0.002

Emb

-1.85110

-

0.001

c) Total stress

stolb

=

celb

- smb

-1.851-10

-5.916*10

-

0.001

/ 0.004

Etolb=

1-1.851-10

o 1

-1.851 10

-

0.001

-9.91610

0.001

0.001 -f002

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5)

When

stress

is

proportionally

reduced

to zero, selb

vanishs while Emb remains unchanged.

stolb=smb

when

zero stress

is reached

Note: Some

of the

elements

in the matrices

may

not

be accurate

because

Mathcad

sometimes

doesn t show

enough

significant digits.

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22 31411 5612 084113 14Fall 2006

Problem

SetV

Due 10126106

his prob lem i l l us t ra tes a mechan ism by wh ich c lad s t resses can be

reduced du r in g up-power ramps by approaching f u l l power a t a s low er

r a t e .

1. Geometry (20C)

The c l ad ou ts id e d iameter i s 11.2 nun; and the c l ad th ickness i s 0.7 mm

2.

Clad Mater ia l Propert ies (assumed constant)

Young s Modulus = 76 GPa;

Po isson s Rat io = 0.25;

Co e f f i c i e n t o f Thermal

Expansion

=

6.7

m1m K;

and

c lad c reep ra te (s l . )

=

(1 x 10-9)ag;

9

where a has th e un i t s MPa.

9

3. Ope rat ing Con dit ion s

-

A t ho t zero power, th e c la d i s a t 280C. th e co ola nt pressure i s 15.5 MPa,

th e gas pressure ins id e the rod i s 5.4 MPa, and the c lad i s touch ing

the fue l w i t h ze ro con tac t p ressu re .

A t

ho t f u l l power the co ola nt and gas pressures are unchanged, the

average c l ad tempera ture i s 375 C, and th e r ad ia l d isplacement o f th e

ou t e r su r face o f t he f ue l has i nc reased by 35 run

Consider two cases:

- An up-power ramp from zero t o f u l l power i n 30 minutes;

- An up-power r i i p from zero t o f u l l power i n 30 hours .

4.

Ca lcu la t i ona l Bas i s

- Use a s i n g l e r i n g t o r e p r e s e nt t h e c l a d

C on sid er t h a t t h e r a d i a l d e f l e c t i o n a t t h e f u e l o u t e r s u r fa c e

and the c l ad average tempera tu re va ry l i n e a r l y w i th t ime

during each up-power ramp.

- Use on ly e l a s t i c , thermal , and c reep s t r a in s . Le t the c reep

s t ra in s equal zero a t ho t zero power.

Use zero ax ia l

f o rc e f rom fue l - c l a d con tac t.

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5.

Questions

O bta in the fo l low ing quant i t i es upon reach ing fu l l

power i n each up

power ramp:

.

c lad s t ress ;

c la d s t ra in ; and

ra d i a l d isplacem ent a t the c la d outer sur face .

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22.314/1.56/2.084/13.14 Fall 2006

Problem Set VII

Due 11/02/06

This problem set illustrates applications of beam theory to Zircaloy Follower in a BWR forcalculation of curvature caused by Zircaloy growth ; Consult Notes X on Beam Theory.

ZIRCALOY FOLLOWER

Xa) Geometry and Material properties : WY

Consider a BWR reactor core that has cruciform Zshaped control rods. When each control rod is fully

withdrawn for power operation, it is replaced in thecore by an attached Zircaloy follower to prevent Texcessive water hole peaking. The follower is also

cruciform shaped and is shown in the adjacent figure.The dimensions are :

WL = length in the z-direction = 2.4 m ; W = width or

span = 200 mm ; and T = thickness = 7 mm. T

The Zircaloy has a Youngs Modulus of 75 GPa and a Poissons

Ratio of 0.25. The growth strain in the z-direction as a function

of fast fluence is given by the following equation :

egz = C1N + C2N2 ; (1.1)

where :

- the z-direction growth strain ( egz) is given in percent;

- the fast fluence (N) is given in the units of (1021fast neutrons per cm2) with the fast flux

cutoff specified by E > 1 MeV; and

- the constants are C1= 0.013 and C2= 0.0018.

b) Notation and Support Information :

For points originally on the axial centerline (x = 0; y = 0 ), denote displacements in the x-direction, the y-direction, and the z-direction displacement, respectively, by u, v, and w.

1

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At z = 0, the follower is supported so that u, v, and w are all zero and so that no moments are

applied. At z = L, the follower is supported so that the z-direction force is zero, so that u and vare zero, and so that no moments are applied.

c) Fast Neutron Fluence:

After several refueling cycles, a follower has an accumulated fast fluence given by:

N = [ x x ] [N z ] ;N ( ) ( ) (1.2)z

Where N is the fast fluence expressed in the units of Eq 1.1; where

0.1 xN x =

W

; and where (1.3)( ) 15 1 +x

(z-(L2)

)

N z = . (1.4)( ) 1.49cos pz Le

Le is the extrapolated length of the core ( 2.54 m )

d) Questions : d.1) What is u as a function of z ? d.2) What is the value of w at z = L ?

2

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22.314/1.56/2.084/13.14 Fall 2006

Problem Set VII Solution

Solution:

x

z

y

Beam model Cross section

Geometry and material properties:

L=2.4m; W=200mm; T=7mmand

E=75 GPa and =0.25Boundary conditions:

At z=0: At z=L:

u=v=w=0 u=v=0M0=0 ML=0

Fz=0

(a) According to the beam theory:

z

z =

+

zo ; where

zo =

0.01

gzE

furthermore,

z =za kyx; where za =dw

and ky =dy =

d 2u2dz dz dz

therefore,

dw d 2uz =E( 2 x 0.01gz )dz dzAt any cross section,

My =

zxdA =M0 =0

A

In the meanwhile,

My =zxdA =E(za kyx 0.01gz )xdAA A

where u,v,w,zaand kyare only functions ofz,and

gz =C1N +C2N2 =

C1Nx (x)Nz (z) +C2 (Nx (x)Nz (z))2

1

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Therefore,

My = Eza (z)dA +Eky (z)x2dA +0.01EC1Nz (z)15

1 +0

0

.

.

1

2

x

xdA +0.01EC2Nz (z)2

Nx (x)2xdA

A A A A

where

dA=

0 for symmetryA

T w x

2dA =202x2wdx + T2x

2Tdx=4.6722 106

A 2

N (x)xdA = 15xdA +7.5x2dA =3.504 105x

A A A

Nx (x)2xdA =225xdA +225x

2dA +56.25x3dA =1.05 103

A

A A A

Then,

Eky (z) 4.6722 106 +EC1Nz (z) 3.504 10

7 +EC2Nz (z)2 1.05 105 =0

2

ky = 0.075C1Nz (z) 2.25C2Nz (z)2 = 0.001453cos

2

2

z

L

e

L

0.008991cos

2

2

z

L

e

L

2z L 2z L= 0.001453cos

2L0.004496 0.004496cos

L

e e

y (z) y (0) = 0z

ky (z)dz=0z

0.001453cos

2z

2

L

e

L

+0.004496 +0.004496cos

2z

L

e

L

dz

= 0.001453Le

sin

2z L

+sinL

0.004496z 0.004496Le

sin

2z L

+sinL

2Le 2Le 2 Le Le

= 0.001175sin(1.2368z 1.4842)

0.004496z

0.001818sin(2.4736z

2.9684)

0.001484

and let

(0) =CyAgain,

z

u(z) u(0) =y (z)dz0= 0.010

z

[0.1175sin(1.2368z1.4842) +0.4496z+0.182sin(2.4736z2.9684) +0.1484 C]dz

2= 0.00095[cos1.4842 cos(1.2368z 1.4842)]0.002248z

0.000736[cos2.9684 cos(2.4736z 2.9684)]0.001484z +Cz

=0.000736cos(2.4736z

2.9684)

+0.00095cos(1.2368z

1.4842)

0.002248z

2

0.001484z +0.0006428 +Cz

and the boundary conditions: u(0)=0 and u(L)=0

we get C=0.006879

therefore,

u(z) =0.000736cos(2.4736z 2.9684) +0.00095cos(1.2368z 1.4842) 0.002248z2

+0.0054z +0.0006428

2

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(b) To obtain w(z), we first calculate the axial strain =dw(z)

zadz

We already know

F =zdA =FzL =0zA

In the meanwhile,

Fz =zdA =

E(za kyx 0.01gz )dAA A

=Eza dA Eky (z)xdA 0.01EC1Nz (z)Nx (x)dA 0.01EC2Nz (z)2

Nx (x)2dA

A A A A

where

dA =A =2TW T2 =

0.0028A

xdA =0A

Nx (x)dA = 15dA +7.5

xdA =

0.042A A A

Nx (x)2 dA =225dA +225xdA +56.25x

2dA =0.6303A A A A

Therefore,

za =0.15C1Nz (z) +2.25107C2Nz (z)2

2

=0.002905cos

2z L

+0.008996

cos

2z L

2Le 2Le

=

0.002905cos

2z L

+

0.004498 +

0.004498cos

2z L

2Le Le

Then, we can obtain w(z) by integrating zaover zz

w(z) w(0) = za (z)dz00.002905L 2z L L 0.004498L 2z L Le e= sin +sin + sin +sin +0.004498z

2Le 2Le

2 Le Le

=0.002349sin(1.2368z 1.4842)+0.001818sin(2.4736z 2.9684)+0.004498z +0.002653

Therefore:

w(L) =w(2.4) =0.016102m

3

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22.314/1.56/2.084/13.14 Fall 2006

Problem Set VIII

Due 11/09/06

Consider the pre-stressed concrete containment example of Problem Set L.54.

Evaluate the impact of changing the longitudinal tendon pitch from 165 mm to 200 mm on the

required prestress level to prevent tensile stresses in the concrete upon pressurization.

1

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Structural Xechanics i n Nuclear Power Technoloa

(1.5653, 2.0845, 3.827, 13.143, 16.2613, 22.3143)

Problem Set L 5

This problem i l l u s t r a t e s some of the concepts i n reinforced/

pres tres sed concrete containment s t re s s calculat ionr .

1 Geometq

Consider t he c onc rete and s t e e l geometry of Fig. 1 fo r a sec t fon

of cyl ind ric al containment sh el l .

Direc t ions

- 1

depth.= 165

St ee l Radial Longitudinal

Tendons

Hoop

pi tch = 165

iameter = 57

Concrete

In sid e Radius = 20 600

Wall Thickness = 1380

Figure

1

Se cti on Normal t o the Lon gitudin al Axis

of th e Containment She ll (dimensions i n

1.

That is. th e lon gi tu di na l tendons a r e i n th e fonn of two rows of 57 p m

diameter s t e e l ba rs (one row near each surf ace) with a pi tc h of ap-

proximately L65

mm

The hoop tendons a r e i n th e form of four rows of

57 mm diameter s t e e l bar s (two rows near each sur fac e).

Harever the

p it ch of the hoop tendons i n each row is 200 mm

2

Material Properties and Operating Conditions

For th e s t e e l , Young s Modulus = 210 GPa.

For the concrete,

Young s Modulus

=

21 GPa and Poi ss on s Ratio = 0.15.

In the pressur-

ized condit ion the in ter nal pressure is 350 kPa above the external

pressure.

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Problem Set

L

54

3

Q u e s t i o n s -

(a)

What prest ;ess le ve ls (one le ve l f or a l l longi tudinal tendons,

another fo r a l l hoop tendons) ar e requi red to prevent t en si le

st resses in the concrete upon pressurizat ion?

(b)

What

is

th e mutimum te n s il e S tr es s i n the reb ars upon

pressurizat ion1

4. Calculat ional Basis

I l r sme e l as t i c behavior .

Consider that the cyl inder is long but consider the regio n near

t h e b a s e

mat

Consider the cy l inder to be bu i l t - in a t t he poa i t ion

i t

j o i n s t he

base mat (zero ra di al deflect ion. zero s lope of r ad ia l def l ec t ion) .

-Lt'

Define the f i exura l r i g id i ty on the bas i s o f the sec t io n shown i n

Fig.

1

Consider th at the tendons move in t he long itud ina l dir ec-

t i o n t h e

same a s the concrete.

C a l cu l a te r i g i d i t y on t he bas i s t ha t

plane sec t ion s remain plane and th at t ransverse Poisson's r a t i o

can e t reated the same as the concrete.

Consider &at there is no bending of the sh el l in the unpressur ized

condit ion.

Consider

the

pr es tr es si ng t o be achieved by post-tensioning .

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Problem Set VIII Solution

Solution:

Evaluate the impact of changing the longitudinal tendon pitch from 165 mm to 200 mmon the required prestress level to prevent tensile in the concrete upon pressurization.

Material properties and operating conditions:

Es =210GPa; Ec =21GPa and =0.15

Internal pressureP=350 kPa, andR=Rin+t/2= 21.19m

To calculate the required prestress level, we should go through the following steps:

At first, compute some constants that will be used:

ls =2As , where As =

dt2

=2551.8mm2

tpl 44Ass =tp

PRNl = =3.708MPa m

2

P = r

2

EcEl =(1ls ) 21

EcE =

(1

s ) 21

E* = lsEs +El

and then

=t

sEs +E(12El )

R E *

=

E

E

*

Nl +

1

t

r (1s (1 ls )

E

E

*

)

cD =E

2(

t3+lss

2t) +Eslss

2t , wheres =thickness / 2 depth =0.525m

1

12According to notes for problem set L54, we get

c =wp

=R

(P

)R

R

lc

N

t

l

Elc +

E

(1

*

ls )1

r

=

1

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The maximum stress in longitudinal direction occurs at = t

for2

c max and lc max

E clc max =12

(c max +lc max +bl ) +1

r

where

bl (z) =t d 2w

2=t2wpe

z [cosz sinz]2 dz

1

4where in turn =

4DR

Since c max and lc max reach the maximum value at z , therefore, the strain due

to bending bl =0 as z .

Now, the maximum longitudinal stress shall be offset by the tendon prestress to get

zero net concrete stress upon pressurization and therefore we get

lsprestree =1ls lc maxls

Similarly, the maximum stress in hoop direction

E cc max =1 2

(c max +lc max ) +1

r

Then, we get

1 ssprestree = c maxs

For the two cases where the longitudinal tendon pitch is equal to 165 mm and 200

mm, respectively, we just substitute corresponding numbers to the above equations, andobtain the final results:

165 mm:

lsprestree =99.5MPa

sprestree =104.36MPa

200 mm:

lsprestree =

123.7MPa

sprestree =104.41MPa

According to these result, when changing the longitudinal tendon pitch from 165mm to 200mm, the required prestress in the longitudinal direction is increased by almost

25%, however the required prestress in the hoop direction almost remains unchanged.

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Problem Set IX

Due 11/21/06

This problem set illustrates some features of seismic analysis. A response spectrum

modal (lumped mass) method is to be used.

1) Geometry, Properties, Supports: The reactor component of interest isrepresented by a pipe. The pipe is subjected to a seismic ground motion that

acts parallel to the x-direction. The axis of the pipe in the z-direction.

The pipe has an outside diameter of 210mm, a wall thickness of 7mm, and a lengthof 3m. The pipe is made of a material with Youngs modulus = 200 Gpa and density= 8500 kg/m3 . Liquid water (density = 750 kg/m3 ) is present both inside and outside

of the pipe. The added mass coefficient for the outside water is 1.1. Neglect direct

treatment of fluid friction, fluid drag, and solid friction these effects are incorporatedin the damping matrix of item 2.

The support at z = 0 is fixed (zero displacement and zero slope). The support at the

other end (z=L=3 m) is a roller (zero displacement, zero moment, and zero axialforce).

2)

Lumped Mass Treatment: Use two equal point masses (located at z = 1 mand at z = 2m). The total mass for the two points should equal the sum of the

pipe mass, the internal liquid mass, and the added mass of the external

liquid.

Massless springs should be chosen to give the same stiffness characteristics at thepoint masses as for a beam with the same supports (the resulting stiffness matrix

has no zeros).

The damping matrix should have only two non-zero elements (located on thediagonal). These elements should be equal and should give 2% critical damping for

vibraton at the system fundamental frequency.

3) Earthquake Characterization: The ground motion is characterized by the

response spectrum of Fig 8 in Note M-32 {To provide legibility, use a multiplestraight line set of segments given by:

[Sd =560mm];[Sv =1.15m /sec];[Sa =1.4g];

end points (Sq ,Sa )of (5mm,1.4g)and(0.25mm,0.33g)

and S[ =0.33g];where gis theaccerationof gravitya

4)

Questions:

a) What is the fundamental undamped frequency?

b)

What are the peak forces that act on the supports during theearthquake?

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Problem Set IX Solution

Solution:

The pipe can be modeled as in Figure 1.

Lumped mass

BmB PB

AmA PA

R

Pipe Beam with the same support

Geometry and properties:

L=3m, R=0.105m and t=0.007m

E=200GPa, =8500kg/m3and water=750kg/m3

mA =mB =(R2 Ri

2)L+(Ri

2 +1.1R2 )Lwater =133.724Kg2

The governing motion eqution for dynamic response of this pipe can be expressed as:[M ]{u&&} +[D]{u&} +[K]{u} ={F}where:

mA 0 [M]=mass matrix,M =

0 mB

[D]=damping matrix

[K]=stiffness matrix

[F]=vector of loads, earthquake load in our case, {F} = [M ] {Sa}{u}=vector of nodal displacements

First of all, we should calculate the stiffnes matrix of the pipe using beam theory:Suppose the displacements of point A (L/3) and B (2L/3) are u Aand uB, respectively.

Assuming that there is only a force PAacting on point A (at L/3), we can calculate thecorresponding displacements of points A and B, satisfying the boundary conditions:

1

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u(0) =0

u(L) =u(3) =0

(0) =du(z)

=0ydz z=0

ForL>z>L/3:dM

M =V (3 z) = y (3 z)y xdz

Solving this equation with the boundary conditionMy(L)=0, we get

My =C1(3 z)

Forz

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39C2 +

4PA 13PA 9C2 =02EI EI

3EI 2EI

23C2 = PA

27

4

C1 =

C2 +

PA =

PA27

Then, we obtain

u

C2 8 +

PA +

9C2

9C2

0.1358PA

A EI 6 EI 2EI 2EI EIu

=C1 1 9C2 8PA 13PA 9C2

= PA

B + + 0.142

EI 6 EI EI 3EI 2EI EI Similarly, assuming that there is only a force PBacting on point B (at 2L/3), we can

calculate the corresponding displacements of points A and B, satisfying the boundaryconditions:

ForL>z>2L/3:

dMM =V (3 z) = y (3 z)y x

dzSolving this equation with the boundary conditionMy(L)=0, we get

My =C1(3 z)

Forz

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C1 (3 z)3 9C2 5PB z

9C2 19PB

u(z) = ydz =

EI

C2

6

(3 z

+

)32

+

EI

PB

+2

z2

EI

+

9C

2

2

z

EI

9

C2

6EI

EI 6 2EI 2EI 2EIAt last, we have another boundary condtion, u(L)=0

So that

3

9C2 +5PB

9C2 19PB =0

2EI 2EI 2EI 6EI

13C2 = PB

27

14C1 =C2 +PB = PB

27Then, we obtain

u

C2 8 +PB +

9C2 9C2

0.142PB

A EI 6 2EI 2EI 2EI EIu =C1 1 9C2 5PB 9C2 19PB = PB B + + 0.247 EI 6 EI EI 2EI 6EI EI

So that, according to superposition when PAand PBact on the system simultaneously,

uA 1 0.1358 0.142 PA = uB

EI 0.142 0.247 PB

In the meantime,

PA uA P

=K u B B We can get

K =

EI0.1358 0.142

1

=

EI18.46232 10.614

=

4.6046 10618.46232 10.614

0.142 0.247 10.614 10.15054 10.614 10.15054

Secondly, we should compute the damping matrix:The damping matrix has only two non-zero elements, located on the diagonal. These

elements are equal and give two percent critical damping for vibration at the system

fundmental frequency.Thus, we should calculate the undamped fundmental frequency first.[ ]{u&&} +[K ]{u} =0

133.724 0 u&&A 18.46232 10.614 uA +EI =0 0 133.724u&&B 10.614 10.15054uB Solving this equation by assuming uA =Ae

t and uB =Be

t , we get

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133.7242 +18.46232EI 10.614EI A

10.614EI 133.7242 +10.15054EI B=0

(133.7242 +18.46232EI )(133.7242 +10.15054EI ) 112.657(EI )2 =0

17882.14 +

3826.226EI2 +

74.7455(EI )2 =

0

2 = 1.00126 105or

8.851105

Then, the fundamental undamped frequencys are

1 =940.8, 2 =316.43 in radius/s; note that =2fand corresponding eigenvectors are

0.826 0.56364u1 = ,u2 =

0.56364 0.826

Then, we should also calculate the critical damping matrix,Suppose that the critical damping of lumped mass is considered seperately:

C1 0 D =cr

0 C2

[M ]{u&&} +[D]{u&} +[K ]{u} =0N

and the solution is u(t) =An cosntunn=1

Multiply the above equation with ANTleftly, and also replace {u} with AN{u}, where

0.826 0.56364AN =[u1 u2 ]=

0.56364 0.826

we get

133.7242 +C+25.7EI 0 A1

1

2 =0 0 133.724 +C2+2.9EI A2 If there are nontrivial solutions for A1and A2, we obtain

133.7242 +

C1+25.7EI =0

133.7242 +

C2+2.9EI =0So that critical damping matrix:

C1 0 251592.2 0 Dcr = =

0 C2 0 84514.2

Dcr =ANDcrANT =

198513 77788 77788 137593.5

Damping matrix in this case

5031.844 0 D=0.02D =cr

0 1690.284

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At last, the force due to earthquake

{F} = [M ] {u&&g }

Natural frequency

f1 =1 =50.36 and f2 =

2 =149.72 2

According to the response spectrum of Fig 8 in note M-32, extrapolating to the calculatedfrequencies, we obtain the maximum displacements corresponding to Sa=0.33g are

Sd1 Sa

2=

0.33 92

.8=3.6538 106 m

1 940.8

Sd 2 Sa

2=

0.33 9.8=3.23105 m

2 316.432

So

133.724 0 {F} = 0 133.724

u&&g

Finally, we obtain133.724 0 u&&A 5031.844 0 u&A 25.7EI 0 uA T 133.724

&&+

&+ =AN u&

0 133.724uB 0 1690.284uB 0 2.9EI uB 133.724

u&&A 37.63 0 u&A 1 25.7EI 0 uA 0.2624u&&

+ 0 12.64u& +

133.724 0 2.9EI u = 1.39

u&&g B B B Then, maximum displacement

7.92 107

u1,max =0.2624 Sd1 u1 = 7

5.4 10

2.53105 u2,max =

1.39

Sd 2

u2 =

5 3.7110

2 2 0.5P1,max =[(Cu) +(Ku) ]

198513 77788 7.92 107 2

18.46232 10.614 7.92 107 2

0

=0.02 940.8

77788 137593.55.4 10

7

+EI

10.614 10.15054

5.4 10

7

3.75 2

93.72 2

0.5

93.795=

2.56

+ 63.95

= 64

6

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2 2 0.5P2,max =[(Cu) +(Ku) ]

=

0.02 316.43198513 77788

2.5310

5

5

2

+

EI 18.46232 10.614

2.5310

5

5

2

0.

77788 137593.53.7110 10.614 10.150543.7110

13.522

337.62 0.5 337.87

= +

=19.85 497.4

497.8

Now, given the forces of PAand PB, we calculate the forces acted on supports:

FL

BPB

APA

F0

Assuming the forces that act on the supports are F0and FL, respectively, as in the above

figure.

dM dK 4 14FL

y= =EI y = PA + PBdz dz 27 27

z=L z=L

23 13

F0 =

PA +

PB

FL =

PA +

PB27 27For P1,max:

F0 1 23 13 110.71F1 = = P1,max =

FL

27 4 14 47.08 Likewise, for P2,max:

F =F0 1 23 13

P =527.5

2 FL

=

27

4 14

2,max 308.2

2Again, F =[F1 +F22

]0.5 =

539 Newton311.8 So, the peak forces that act on the supportz=0andz=Lare 539and 311.8 Newton,

respectively.

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Department of Nuclear Engineering22.314

:

Structural Mechanics In Nuclear

PowerTechnologyQuiz No.1 Fall Term 2006

Open Book, 1.5 hours

Please state your assumptions and the definition of symbols appearing in your equations

clearly.

Note that there is an opportunity to get a score over 100 if you solve all questions

correctly.

Question #1 (45%)

For future sodium-cooled reactors, uranium carbide may be used as fuel. The fuel pin

will be of cylindrical geometry and housed in a stainless steel clad. The dimensions aswell as the physical and mechanical properties of the fuel material and the clad are given

in Table 1.

1.1 What is the maximum linear hear generation rate (in kW/m) that the fuel

pin can operate at if the fuel is not to experience any fracture for r < 0.4R,

where R is the pellet radius? You may assume the pressure at the pellet-

clad gap to be 0.3 MPa.

1.2 Does the clad design satisfy the ASME criteria for structural integrity

under static loads? The fission gases will build up within the clad until thegas plenum pressure reaches 5.0 MPa. You may assume the coolant

pressure to be 0.3 MPa and the maximum operating linear heat generation

rate is 45 kW/m.

Question #2 (45%)

A long, thin-walled cylindrical tank is used in transporting radioactive gases. While it is

being filled, the tank is subjected to radial constraint that can be approximated by a rigid

boundary that allows slip in the axial direction (see Figure 1).

2.1 What is the maximum pressure that should be allowed in the tank in order

to avoid plastic deformation of the membrane?Material Properties:

Youngs Modulus, E = 30,000 ksi (1ksi = 1000 psi)Poissons Ratio, = 0.3

Yield Stress, y = 36ksi

State any assumptions you make.

2.2 Is fatigue a factor in limiting the lifetime of the tank if the maximum gas

pressure is limited to 400 psi (see attached data in Figure 2)?

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Consider the utilization factor suggested by Soderberg for cyclical

loading: 1+N

a

y

m

Where m is the mean stress during a cycle, y is the yield stress,

a is the alternating stress, and Nis the stress intensity causing failure

after 105cycles.

Table 1

Fuel Pellet Radius = 20mm

Clad Inner Radius = 21mmClad Outer Radius = 23mm

Active Fuel Height = 1.5m

UO2 UC SteelYoungs Modulus E (GPa) 200 210 70

Poissons Ratio 0.32 0.30 0.30

Fracture Strength f( Pa) 150 300

Yield Strength y ( Pa) 330

Thermal Conductivity k(W /mC) 3 20 20

Thermal Expansion Coefficient ( m /mC) 10 10 16

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Figure by MIT OCW.

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Question #3 (20%)

Consider a beam as shown in Figure 3, subjected to an axial force, F2 and a lateral force,F1. Evaluate the value of F2 that will lead to a limit load condition for elastic behavior

of the beam when F1 = 8 MN.

Steel Data: Youngs Modulus, E = 191 GPaPoissons Ratio, = 0.28

Yield Stress, Sy = 345 Pa

Figure by MIT OCW.

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Effectiveness of the damping of a pipe during earthquakes

Prepared By: Julien Beccherle

Prepared For: 22.314

Prepared On: December 7th

, 2006

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Effectivenessofthedampingofapipeduring

earthquakes

JulienBeccherle

January10,2007

Contents

1

Introduction

2

1.1 Naturalfrequencyoftheharmonicoscillator . . . . . . . . . . . . 21.2 Forced excitation: earthquake . . . . . . . . . . . . . . . . . . . . 3

2

Decoupled

differential

equations

of

a

system

6

2.1 Lumped masses analysis method . . . . . . . . . . . . . . . . . . 62.2 Modeling the pipe . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Differentialequationgoverningthebeammotion . . . . . . . . . 82.4 Naturalfrequenciesoftheundampedsystem . . . . . . . . . . . 102.5 Uncoupled differential equation . . . . . . . . . . . . . . . . . . . 11

3

Earthquake

data

11

3.1 Hectormine,Joshuatreeearthquake,CA . . . . . . . . . . . . . 113.2

New Hampshires 1982 earthquake

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

14

4

Response

of

the

pipe

to

an

earthquake

15

4.1 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.2 CalculationsfortheJoshuaTreesearthquake . . . . . . . . . . . 16

4.2.1 = 0.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.2.2 Summary of the results . . . . . . . . . . . . . . . . . . . 17

4.3 CalculationsfortheNewHampshiresearthquake . . . . . . . . . 184.4 Interpreting the results. . . . . . . . . . . . . . . . . . . . . . . . 204.5 Conclusions and comments . . . . . . . . . . . . . . . . . . . . . 21

5

Overview

of

cost-effectiveness

design

method

21

5.1 Evaluatingtheprobabilityofoccurrenceoffailure . . . . . . . . 21

6

Conclusion

23

1

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-

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Abstract

The

paper

presents

two

methods

to

assess

the

cost

effectiveness

of

a

steam

generator

pipe.

The

first

method

is

a

time

series

analysis

with

real

earthquakedata. Thesecondmethod isaprobabilisticassessmentwhich

leads

to

the

evaluation

of

a

life-time

mean

cost.

1 Introduction

When performing a seismic design of a power plant, an engineer is faced withthe following dilemma : damping systems for pipes, vessels, steam generatorare very costly equipment but on the other hand they enable the structure towithstandalargerearthquake. Therefore,selectingtheminimumdampingisamatterofcarefulevaluation.

1.1

Natural

frequency

of

the

harmonic

oscillator

This dilemma can be trivially illustrated by a single oscillator connected to awallbyaspringandwithadampingsystemthatwewilltakeasparametric.

Figure1: Harmonicoscillator

Thefundamentalequationgoverningthissystem is:

d2u

du

m =F(t)k(u)d( ) (1)dt2 dt

Here F(t) represents the force of the earthquake on the system. Actuallythe earthquake is not really a force external to the system that is applied toit. Anearthquakeisreallyanaccelerationofthegroundwhichwassupportingthestructure. Thusequation(1)canberewrittenassumingthattheground is

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movingwithdisplacementuG(t)as :

m(

d2u d2uG) =

k(u

uG)d(du duG)m d2uG (2)dt2

dt2 dt

dt dt2

Letsdefiney(t) =u(t)uG(t). (2)becomesthesimpleequation:

md2y

+ddy

+ky=m d2uG

(3)dt2 dt dt2

Letusdefine0 =k and2Q0 =

d. Thenthehomogeneouspartof (3)m m

canbewrittensimplyas:

d2y

dy 2+ 2Q0 +0 y=0 (4)

dt2 dt

Thecharacteristicequation is

r

2 + 2Q0r+0

2 = 0

Forourproblem,wellconsiderthat0

isfixedwhereasQ istheparameter(actuallyQiscalledthequalityfactoranditisdirectlyrelatedtothedamping).

The delta of this equation is then =(2Q0)240. Thus =0 whenQ

=1 this is called critical damping, >0 ifQ >1 and

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Figure2: Amplitudeamplification

Figure3: Amplificationoftheforceduetothemotionoftheground

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FdynamicQ

Fstatic

0.01 50.00.02

25.0

0.05 10.00.1 5.00.2 2.60.5 1.2

Table1: MaximumdynamicamplificationvariationswithQ

Figure(3) isaplot fordifferentvaluesofQ : it is importanttonoticethatfor Q

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2 Decoupleddifferentialequationsofasystem

2.1

Lumped

masses

analysis

method

Theharmonicoscillatorapproachgivesinterestingresultsfortheplant,butitissomehowtoosimpletocapturethecomplexityofthepipingsysteminanuclearplant.

Theapproachbroadlyusedinseismicanalysisistheso-calledlumpedanal-ysis. The piping system of the nuclear plant is modeled as a succession ofharmonic oscillator with mass mi, coupled to one another by a spring and anabsorber. Ofcoursewehavemi

=Mtotal. Andtherelationbetweenthedifferentoscillatorsisgivenbythebeamtheory. Thestrengthofthisapproachisthatitenablestheengineertoreducetheoriginalcontinuouspipetoasuccessionoflinked masseswhichonly have one degreeof freedom(horizontal displacementif the pipe is vertical). The fundamental principle of dynamic applied to the

mass

mi yields

d2ui dui duGmi

dt2=ki,i+1(ui+1

ui)ki1,i(ui1

ui)ci(dt

dt

) (8)

Aswedidbefore,wewilldefineyi =uiuG therelativedisplacement. We

0 0y m . . .1 1

=

y2...

,M =

0 m2 0alsodefineY . . .. . ... .

yn mn...m1

k1,1

k1,2

.

.

.

k1,n

m2

.

.

,K

=

k2,1

.

.

k2,2

. .

k2,n

.

.

0 0 .

is

the

matrix

defined

as

,

m

=

..

.

kn,1

kn,2

kn,n

mn

. . .

F1F2

...Fn

the static response of the beam to a force F on each mass. So=

F =KY. Distheso-calleddampingmatrix andrepresentstheabilityofthe

systemtotransformkineticenergyintoheatandthustodissipatesomeenergy.Equation(8)canthenbewrittenas :

M Y +DY +KY =muG

(9)

Equation

(9)

is

not

trivial

to

solve

as

is.

But

it

can

be

reduced

to

a

linearlyindependentsystemofnequationsbydecomposingthehomogeneousequation

intermsofharmonicsolutions.

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2.2

Modeling

the

pipe

After

this

brief

theoretical

overview

of

the

modeling

method,

we

will

in

thissectioneffectivelymodelasteamgeneratorpipe. Wewillafterwardsstudythe

effectofanearthquakeonthepipewithdifferentdampingvalues.

Figure5: Modelingthepipe: twolumpedmasses,fixedatz=0andz=L

The pipe will be considered to be fixed between two walls. We will model

the

pipe

as

two

lumped

masses,

each

of

mass

Mtotal

at

a

distance

Ltotal

from

2 3

thewall. Seefigure(5).Some usual figures for the pipes geometry and properties are summarized

below:

Pipelength: Ltotal

= 3m

Pipeoutsidediameter21mm,materialthicknessof1.5mm

Pipecomposedofstainlesssteelgrade304: =8000kg/m3,E=193GP a

Pipefilledandsurroundedwithwaterat70M P a(water =750kg/m

3)

Virtualmasscoefficientof1.1

From

these

properties,

we

can

directly

compute

some

useful

information

onthepipe. FirstitsmassisMtotal inside+M

w WhereMpipe ==Mpipe+M

w

moved.(R2 inside =R

2 Lwater andM

w =CvirtualR

2 Lwater.outRin2 )Ltotal,Mw in moved out

ThisgivesMpipe

= 2.545kg,Mw = 0.779kg,Mw = 1.120kg. Weendinside moveduphavingatotalmassMtotal

of4.444kg. Thuseach lumpedmasswillhaveamassm= 2.222kg.

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Another parameter of importance is I

the moment of inertia of the pipe

around its x-axis. It is defined as I =A

x2dA. In our case, using polar

coordinates

we

get

x

=

r.cos

and

dA

=

rdrd.

Thus

I

=

02

RR

i

o r3cos2drd,

I=

2

cos2d

Ro r3dr. SowegetI = 6.739109 m4.0

Ri

2.3

Differential

equation

governing

the

beam

motion

Recallingequation(9)wehavetofindthematrixK.Kisrelatedtotheresponsein displacements of the beam when a force F is applied at masses locations.

We introduce the matrix A defined as Y = AF where Y =y1 and F =

y2F1

.F2

Giventhelinearityofthisequation,wecanstudytheeffectofF1

aloneand

F2 alone.ForinstanceF1

willinduceadisplacementatz=L/3andalsoatz= 2L/3proportionaltoF1.

Tofindthisrelation,onehastosolvethefundamentalequationofthebeamdynamicmotion. I isthemomentofinertiaofthepipearoundthex-axis.

4u Mtotal 2u f

z 4+

EI t2=

EI

Inourcase,weareinterestedinstaticresponse( =0)andthelinearforcest

f issimplyF1(z=L/3)where(z=L/3)istheDiracfunction.Sothedifferentialequationwehavetosolve is:

4u F1(z=L/3)

=

(10)z 4 EI

Giventhegeometryofthepipe,thereisnodisplacementattheboundaries,andthereisnoangulardeviationaswell. Wehavethefollowingboundaryconditions:

u(z=0)=0andu(z=L) = 0u

(z=0)=0and u(z=L) = 0

z zBy solving equation (10) we get figure (6). So A F

01

=yy

1

2=

0.0988F1EI

0.0679F1EI

Given the symmetry of the system we can predict that similar results will

be

found

for

F2.

Thus

putting

all

the

results

together

we

get

A.

1 0.0988 0.0679A=

EI

0.0679 0.0988

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Figure6: DisplacementresponseofthepipeatforceF1

It is importantto notethatA issymmetric,thus A1 isdefined. The wayA isdefined it isclearthatK=A1.

K=EI

19.180613.1818

13.181819.1806

(11)

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2.4

Natural

frequencies

of

the

undamped

system

The

fundamental

equation

of

the

undamped

system

with

no

excitation

is

thefollowing:

MY +KY =0 (12)

To obtain the natural frequencies,

ety1

. Equation(12)becomesy2

2My+Ky= 0

Inordertohaveanontrivialsolution(i.e. y=0) shallverify:

Det(

2M

+K)=0 (13)

Solving(13)weget

we will solve equation (12) for Y =

1 =59.3rad/sor f1 = 9.4Hz2 =137.6rad/sorf2 =21.9Hz

Correspondingtothesevalues,wegettheY1 andY2 vectorsolutions. Y1 =12

1111

This two modes represents the two fundamentals oscillation modes of the

Y2 =12

beam,andcanberepresentedonfigure(7)

Figure7: Fundamentalmodesofthebeam: for1

inblue,for2

inred.

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2.5

Uncoupled

differential

equation

Now

that

we

have

the

two

fundamental

modes

of

the

beam,

we

will

rewrite

ourinitialprobleminthiscoordinatesystem. Todoso,wedefinethemodalmatrix

associatedwiththisbase :

1 1P

= Y1 Y2 =12 1 1ThenewcoordinatevectorX issuchthatY =P X. Wealsohaveacritical

dampingmatrixDcr. SothefundamentalequationMY+DcrY +KY =muG

canberewrittenas

PtM PX +PtDcrP X +PtKP X=PtmuG (14)

We find that PtMP = M. We define a new critical damping matrix

D = PtDcrP as: cr

=d0

1

d02

. Computing K = PtKP we getcr

D

0.7802 0

3.1424K =104 . Wealsohavem =Ptm=

0 4.2091 0Soequation(14)becomessimply

MX

+D X +KX =muGcr

AndifwetakeX=x1 ,thenwecanwritethenewsetofequationasx2

theuncoupledfollowingsystemofequations:

mx1 +d x1 + 7.80103 =3.14

mx2 +d1 x2 + 4.21104

xx

1

2 = 0uG (15)

2

Wenowneedtogetvaluesford1

andd2

. Wecanfindthecriticaldampingvalues,

and

then

express

the

total

damping

of

the

system

as

a

fraction

of

the

criticaldamping. Bydefinitionofthecriticaldampingofasingleharmonicoscillatorwegetdcr,1

=

4m7.80103263.3kg/sanddcr,2

=

4m4.21104263.3kg/s.

263.3 0D =cr

0 611.6

3 Earthquakedata

The next step is to use ourmodel and to compute the displacement when thepipe is subject to an earthquake. To do this, we will use data obtained from

real

earthquake

(taken

from

[1])

and

apply

it

to

our

model.

3.1

Hector

mine,

Joshua

tree

earthquake,

CA

This earthquake occurred on October 16, 1999 in California. The data comefromtheonlinedata-baseCOSMOSStrongMotionProgram. Theseismograph

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waslocatedat48.4kmfromtheepicenter(verticalprojectionofthefocus),andtheearthquakewasrated7.1ontheRichterscale(whichgoesupto9.0).

Figure8: Joshuatree: earthquake location

The data consists of a list of 3000 acceleration recorded during 60 seconds(i.e. onerecordevery0.02s).

Also of interest is the Fourier spectrum of this signal. The acquisition frequencyis 0.

102

=50Hz. ThePowerSpectralDistribution(orFourierSpectrum)

isdefinedas |F2(f

)|2 whereF(f) isthe fouriertransformofthegroundacceleration.

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Figure9: Est-WestaccelerationofthegroundduringHectorMinesEarthquake

Figure10: FourierspectrumoftheHectorMinesseismograph

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3.2

New

Hampshires

1982

earthquake

This

earthquake

occurred

on

January

19,

1982

in

New

Hampshire.

It

was

rated4.5 on the Richter scale, so it was muchmore modest thanthe precedent one,

buteverycannotbesittingrightontheSanAnreasFaultzone! TheaccelerographwewillusewasrecordedatFranklinFallsDam,NHat10.8kmfromtheepicenter(notethatthedatawereacquiredmuchclosertotheepicenter).

Figure11: Est-WestaccelerationofthegroundduringNHsEarthquake

Even though the earthquake had some large acceleration peaks, its duration (about 20 seconds) is much shorter than the previous earthquake. Theacquisitionfrequencyis200Hz.

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Figure12: FourierspectrumofNewHampshire1982earthquakesseismograph

4 Responseofthepipetoanearthquake

Now that we both have the uncoupled equation and some sample of groundmotion we can solve (15) with the data of the two different earthquakes and

study

the

evolution

of

the

displacement

of

the

lumped

masses

and

the

force

on

thesupportcausedbythegroundmotionasafunctionofthedamping.

4.1

Methodology

Foreachearthquakewewillproceedasfollow:

Pickadampingvalue(percentofcriticaldamping)

Solve the uncoupled differential equation using a step-by-step algorithm

(weonlyknowuG

atdiscreettimes): MatLabsode45solver isused

Computethemaximumdisplacementofthe lumpedmasses

Deduce

the

forces

applied

on

the

support

for

that

particular

Incrementandstartover...

Thiswillgiveusacurveofamplificationversusdampingratio.

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4.2

Calculations

for

the

Joshua

Trees

earthquake

4.2.1

= 0.01

Firstofallwenotethatthesecondequationofequationset(15)isanunexcitedharmonic oscillator. If we assume that the pipe was originally at rest (initialdisplacement and velocity are zero) then x2(t) = 0 all along the earthquake.Thismeansthaty1(t) =y2(t). Theonlyvariableof interest isx1(t).

AMatLabsolveofthisproblemgivesthefollowingsolutionforx1(t).

Figure13: x1 asafunctionoftfor= 0.01

We also get the maximum value of x1(t) = 1.3513103 cm. We can

alsocomputethevalueofthemaximumaccelerationofthe lumpedmass. Themaximum acceleration is 65.2882103 cm/s2 and the maximum velocity is285.28103 cm/s.

WecanformulateouranswerwiththemaximumdisplacementvectorXmax

1.3513as: Xmax

=0

103 cm

0.9555ThenY =P X soYmax =P Xmax. WefindYmax =

0.9555

103 cm.

We

also

recall

that

F

=

KY

,

so

Fmax = KYmax and therefore Fmax =263.66263.66

N.

Now given the symmetry of the problem, it is clear that the force on thesupports are the same. In addition we havejust proved that F1 =F2 so by a

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trivialforcebalanceonthepipewegetthatFsupport =263.66N.

4.2.2

Summary

of

the

results

Ifweapplythesamereasoningandcalculationsfordifferentvaluesofthedampingratiowegettheresultssummarized intable

ymax103 incm Fmax inN Amplificationratio FmaxM

ax(M

uG)

0.01 0.9555 263.66 3.0000.02 0.9032 249.24 2.2840.03 0.8920 246.13 1.9040.04 0.8872 244.82 1.6160.05 0.8861 244.53 1.4370.06 0.8847 244.12 1.2470.07 0.8857 244.40 1.161

0.08

0.8926

246.30

1.070

0.09 0.8971 247.55 1.0160.10 0.9022 248.94 0.9800.20 0.9043 249.53 0.7400.50 0.8177 225.65 0.5361.00 0.7956 219.53 0.432

Table2: resultsforJoshuaTreesearthquake

Thisresultscanbesummarizedonaplot:

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Figure14: Variationoftheforcedampingwithdampingratio

4.3

Calculations

for

the

New

Hampshires

earthquake

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ymax

103 in

cm

Fmax in

N

Amplification

ratio

FmaxMax(MuG)

0.01 1.2641 348.81 1.6800.02 1.1089 305.99 1.4750.03 1.0134 279.64 1.2640.04 0.9909 273.43 1.1960.05 0.9703 267.74 1.1040.06 0.9179 253.28 1.1000.07 0.9190 253.58 0.9940.08 0.8908 245.80 0.9600.09 0.8611 237.62 0.9270.10 0.8368 230.91 0.9710.20 0.6889 190.08 0.8740.50 0.5876 162.14 0.5241.00

0.5713

157.63

0.340

Table3: ResultsforNHsearthquake

Figure15: Variationoftheforcedampingwithdampingratio

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4.4

Interpreting

the

results

The

general

idea

here

is

to

be

able

to

determine

the

force

that

applies

on

thesupportduringanearthquake. Fromtheresultsandthesketchesshowedbefore

one concludes that apart from a puzzling rise around 0.07 % for the Joshua-Treeearthquake,theamplificationofthegroundmotionisdecreasingwiththedampingratio.

WealsonotethateventhoughtheJosh