Cezar Doca - Quasi Static Bending of Beams

QUASISTATIC BENDING OF BEAMS – SOME GENERALIZED

ANALYTICAL MODELS APPLICABLE TO NPP’S MECHANICAL

STRUCTURES

CEZAR DOCA, Ph.D.

ABSTRACT

Addressing the problem of the quasistatic bending of homogenous straight beams

(bars), the paper presents, from simple to complex, the analytical models for a total of

24 ( )2232 ×××= general situations, individualized by the combination of: (a) the

absence / presence of an elastic base (bed, support), i.e. 2 cases; (b) the absence /

presence of a compressive / tensile axial force, i.e. 3 cases; (c) beam with constant /

variable moment of inertia, i.e. 2 cases, and (d) beam with constant / variable distributed

load, i.e. 2 cases. The generality of these (mathematically both beautiful and powerful)

models enables their application in the study of the quasistatic bending of any

mechanical structure equivalent to a homogeneous straight beam (bar), examples for a

NPP being: fuel elements, pressure tubes, pipes, structural beams etc.

INTRODUCTION

It is known that, in the classical theory concerning the quasistatic bending of beams, we have

the (adapted / simplified) differential equation of second order:

(1) ( ) ( )

( )0

2

2

=+xEI

xM

dx

xwd

where: x is the axial coordinate; ( )xw is the beam’s deflection; ( )xM is the bending moment

acting on the beam; E is the beam’s elastic modulus, and ( )xI is the beam’s moment of

inertia. The product ( )xEI is the flexural stiffness.

If the bending moment ( )xM is unique determinate by the equilibrium conditions, then the

general solution of equation (1) is provided by the algorithm:

( ) ( )( )∫ ∫ ξ

ξ

ξ

ξ−⋅+=

ξx

ddEI

MxCCxw

1

1

1

2

2

210

1

where 0C and 1C are two integration constants. We only mention here that an “exact”

expression of (1) is:

( )

( )

( )( )

0

1

23

2

2

2

=+

+

xEI

xM

dx

xdw

dx

xwd

If ( )xM is not unique determinate, then we must solve the differential equation of fourth

order:

(2) ( ) ( ) ( ) 02

2

2

2

=−

xq

dx

xwdxEI

dx

d

where ( )xq is the distributed load. Knowing the functions ( )xI and ( )xq , the general solution

of equation (2) is provided by the algorithm:

( )( ) ( )

( ) ( )

( )∫ ∫∫∫

ξ

ξ

ξ

ξξξ−ξξξ

+ξ

ξ+

ξ++=

ξ

ξξ

x

ddEI

dqdq

I

C

I

CxCCxw

1

1

1

2

2

1

333

1

332

2

23

2

210

1

22

where, usually, the four integration constants 0C , 1C , 2C and 3C are determinate from the

support conditions at the beam’s ends 0=x and Lx = ; here L is the beam’s length .

BENDING OF BEAMS SUPPORTED ON ELASTIC BASE (BED, SUPPORT)

If the beam is supported on an elastic base (bed, support) characterized by the elastic constant

bk , then the equation (2) becomes:

( ) ( ) ( ) ( ) 02

2

2

2

=−+

xqxwk

dx

xwdxEI

dx

db

BENDING OF BEAMS WITH AXIAL FORCE

If a compressive or tensile axial force N ( )0>N acts on the beam, then the equation (2)

becomes:

( ) ( ) ( ) ( ) 02

2

2

2

2

2

=−±

xq

dx

xwdN

dx

xwdxEI

dx

d

where N± is substituted by N+ for compressive axial force, and by N− for the tensile axial

force.

EQUATIONS, ALGORITHMS AND GENERAL ANALYTICAL SOLUTIONS

Now, combining the above last two situations, the equation (2) acquires the general form:

(3) ( ) ( ) ( ) ( ) ( ) 02

2

2

2

2

2

=−+±

xqxwk

dx

xwdN

dx

xwdxEI

dx

db

and can be particularized, from simple to complex, as below:

Beam without elastic base (bed, support), without axial force, with constant moment of

inertia and with constant distributed load: for 0=bk , 0=N , ( ) 0IxI = and ( ) 0qxq = , the

equation (3) becomes:

( )004

4

0 =− qdx

xwdEI

and has the general solution:

( ) 4

0

03

3

2

21024

xEI

qxCxCxCCxw ++++=

Beam without elastic base (bed, support), without axial force, with constant moment of

inertia and with variable distributed load: for 0=bk , 0=N , ( ) 0IxI = and ( )xqq = , the


( ) ( ) 04

4

0 =− xqdx

xwdEI

and has the general solution provided by the algorithm:

( ) ( )∫ ∫ ∫ ∫ ξ

ξ

ξ

ξ

ζ++++=

ξ ξ ξx

ddddEI

qxCxCxCCxw

1

1

1

2

1

3

1

4

0

43

3

2

210

1 2 3

For example, in the particular cases ( ) xqqxq 10 += we have:

( ) 5

0

14

0

03

3

2

21012024

xEI

qx

EI

qxCxCxCCxw +++++=

Beam without elastic base (bed, support), without axial force, with variable moment of

inertia and with constant distributed load: for 0=bk , 0=N , ( )xII = and ( ) 0qxq = , the


( ) ( ) ( ) ( ) ( ) ( )02 02

2

2

2

3

3

4

4

=−⋅+⋅+ qdx

xwd

dx

xIdE

dx

xwd

dx

xdIE

dx

xwdxEI


( )( ) ( ) ( )∫ ∫ ξ

ξ

ξ

ξ+

ξ

ξ+

ξ++=

ξx

ddEI

q

I

C

I

CxCCxw

1

1

1

2

2

2

20

2

23

2

210

1

2

For example, in the particular case ( ) xeIxI ⋅α−= 1 we have:

( ) ( ) ( )[ ] xe

EI

EIxCCCqxxxCCxw

α

α

αα++−+α+α−++=

1

4

13230

22

102

2246

Beam without elastic base (bed, support), without axial force, with variable moment of

inertia and with variable distributed load: for 0=bk , 0=N , ( )xII = and ( )xqq = , the


( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

3

3

4

4

=−⋅+⋅+ xqdx

xwd

dx

xIdE

dx

xwd

dx

xdIE

dx

xwdxEI


( )( ) ( )

( ) ( )

( )∫ ∫∫∫

ξ

ξ

ξ

ξξξ−ξξξ

+ξ

ξ+

ξ++=

ξ

ξξ

x

ddEI

dqdq

I

C

I

CxCCxw

1

1

1

2

2

1

333

1

332

2

23

2

210

1

22

For example, in the particular case ( ) xeIxI ⋅α−= 1 we have:

( ) ( ) ( )∫ ∫ ∫∫

+−++++=

x

dddEI

pd

EI

pCCexCCxw

1

1

1

2

1

3

1

332

1

3

1

3323210

1 22

2 ξξξξξ

ξξξξ

ξξ ξξ

αξ

Forwards, for ( ) xqqxq 10 += the solution becomes:

( ) ( ){ ( )

( )[ ] }1

5

3

132

3

1

2

0

2

31

2

1010110

663

2261824

EI

eEIxCCxqxq

CEIxqxqxqqqxCCxw

x

αα+++

+α++−α++−++=α

Beam without elastic base (bed, support), with compressive axial force, with constant

moment of inertia and with constant distributed load: for 0=bk , 0>N , ( ) 0IxI = and

( ) 0qxq = , the equation (3) becomes:

( ) ( )002

2

4

4

0 =−+ qdx

xwdN

dx

xwdEI


( ) ( ) ( )[ ]xsinCxcosCxN

qxCCxw ⋅κ+⋅κ

κ−++= 322

2010

1

2

where 0EI

N=κ .

Beam without elastic base (bed, support), with compressive axial force, with constant

moment of inertia and with variable distributed load: for 0=bk , 0>N , ( ) 0IxI = and

( )xqq = , the equation (3) becomes:

( ) ( ) ( ) 02

2

4

4

0 =−+ xqdx

xwdN

dx

xwdEI


( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) 12

1

3332

1 1 1

3332

232210

2

1 2

ξ

ξ

ξξ⋅κξ

ξ⋅κ⋅κ

+ξξ⋅κξ

ξ⋅κ⋅κ−ξ⋅κ+ξ⋅κ++=

∫

∫ ∫ ∫

ξ

ξ ξ

dddcosqN

sin

dsinqN

cossinCcosCxCCxw

x

where 0EI

N=κ . For example, in the particular case ( ) xqqxq 10 += we have:

( ) ( ) ( )[ ]xsinCxcosCxN

qx

N

qxCCxw ⋅κ+⋅κ

κ−+++= 322

312010

1

62

Beam without elastic base (bed, support), with compressive axial force, with variable

moment of inertia and with constant distributed load: for 0=bk , 0>N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( )02 02

2

2

2

2

2

3

3

4

4

=−+⋅+⋅+ qdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI

but the author did not find a general analytical solution or algorithm. However, in the

particular case ( ) xeIxI ⋅α−= 1 , we have:

( )

12

1

32

02

0

1

32

02

0

1

0

1 1

32

022

010

2 322 322

1 2

2

2

2

2222

22

2

ξ

ξ

ξ

α

κ

α

κ−ξ

α

κ

α

κ

α

π

+

α

κ+

α

κ++=

∫∫

∫ ∫

ξ αξαξξ αξαξαξ

ξ αξαξ

αξαξ

dddeYeJdeJeYEI

eq

CeYeCeJexCCxw

x

where 1EI

N=κ . Here ( )zJ 0 and ( )zY0 are the Bessel functions of first and second order; we

know that ( )zJ n and ( )zYn are the solutions of the differential equation

( ) 0222 =−++ ynz'zy''yz .

Beam without elastic base (bed, support), with compressive axial force, with variable

moment of inertia and with variable distributed load: for 0=bk , 0>N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

2

2

3

3

4

4

=−+⋅+⋅+ xqdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI



( )

( )

( ) 12

1

32

032

0

1

32

032

0

1

1 1

32

022

010

2 32

2 322

1 2

2

2

2

22

22

22

2

ξ

ξ

ξ

α

κξ

α

κ

−

ξ

α

κξ

α

κ

α

π

+

α

κ+

α

κ++=

∫

∫

∫ ∫

ξ αξαξ

ξ αξαξαξ

ξ αξαξ

αξαξ

dddeYqeJ

deJqeYEI

e

CeYeCeJexCCxw

x

where 1EI

N=κ . Forwards, for ( ) xqqxq 10 += the solution becomes:

( )

( )

( ) 12

1

32

03102

0

1

32

03102

0

1

1 1

32

022

010

2 32

2 322

1 2

2

2

2

22

22

22

2

ξ

ξ

ξ

α

κξ+

α

κ

−

ξ

α

κξ+

α

κ

α

π

+

α

κ+

α

κ++=

∫

∫

∫ ∫

ξ αξαξ

ξ αξαξαξ

ξ αξαξ

αξαξ

dddeYqqeJ

deJqqeYEI

e

CeYeCeJexCCxw

x

Beam without elastic base (bed, support), with tensile axial force, with constant moment

of inertia and with constant distributed load: for 0=bk , 0<N , ( ) 0IxI = and ( ) 0qxq = ,

the equation (3) becomes:

( ) ( )002

2

4

4

0 =−− qdx

xwdN

dx

xwdEI


( ) ( )xxeCeCx

N

qxCCxw

⋅κ−⋅κ +κ

+−+= 322

2010

1

2

where 0EI

N=κ .

Beam without elastic base (bed, support), with tensile axial force, with constant moment

of inertia and with variable distributed load: for 0=bk , 0<N , ( ) 0IxI = and ( )xqq = , the


( ) ( ) ( ) 02

2

4

4

0 =−− xqdx

xwdN

dx

xwdEI


( ) [

( ) ( ) 12

1

33

1

33

1 1

3210

2

32

2

32

1

22

2ξ

ξ

ξξ−ξξ

κ

++++=

∫∫

∫ ∫

ξ

ξ⋅κξ⋅κ−

ξ

ξ⋅κ−ξ⋅κ

ξ

ξ⋅κ−ξ⋅κ

dddeqedeqeN

eCeCxCCxw

x

where 0EI

N=κ . In the particular case ( ) xqqxq 10 += we have:

( ) ( )xxeCeCx

N

qx

N

qxCCxw

⋅κ−⋅κ +κ

+−−+= 322

312010

1

62

Beam without elastic base (bed, support), with tensile axial force, with variable moment

of inertia and with constant distributed load: for 0=bk , 0<N , ( )xII = and ( ) 0qxq = ,

the equation (3) becomes:

( ) ( ) ( ) ( ) ( ) ( ) ( )02 02

2

2

2

2

2

3

3

4

4

=−−⋅+⋅+ qdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI



( )

12

1

32

02

0

1

32

02

0

1

0

1 1

203

20210

2 32

2 322

1 2

2

2

2

22

222

22

2

ξ

ξ

ξ

α

κ

α

κ

−ξ

α

κ

α

κ

α

−

α

κ+

α

κ++=

∫

∫

∫ ∫

ξ αξαξ

ξ αξαξαξ

ξ αξαξ

αξαξ

dddeKeI

deIeKEI

eq

eKeCeIeCxCCxw

x

where 1EI

N=κ . Here ( )zI0 and ( )zK0 are the Bessel functions of first order; we know that

( )zI n and ( )zKn are the solutions of the differential equation ( ) 0222 =+−+ ynz'zy"yz .

Beam without elastic base (bed, support), with tensile axial force, with variable moment

of inertia and with variable distributed load: for 0=bk , 0<N , ( )xII = and ( )xqq = , the


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

2

2

3

3

4

4

=−−⋅+⋅+ xqdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI



( )

( )

( ) 12

1

3030

1

3030

1

1 1

302010

2

32

2

32

2

1

2222

222

22

22

2

ξ

ξ

ξ

α

κξ

α

κ

−ξ

α

κξ

α

κ

α

+

α

κ+

α

κ++=

∫

∫

∫ ∫

ξ

αξαξ

ξ

αξαξαξ

ξ

αξαξαξαξ

dddeIqeK

deKqeIEI

e

CeKeCeIexCCxw

x

where 1EI

N=κ . Forwards, for ( ) xqqxq 10 += :

( )

( )

( ) 12

1

303100

1

303100

1

1 1

302010

2

32

2

32

2

1

2222

222

22

22

2

ξ

ξ

ξ

α

κξ+

α

κ

−ξ

α

κξ+

α

κ

α

+

α

κ+

α

κ++=

∫

∫

∫ ∫

ξ

αξαξ

ξ

αξαξαξ

ξ

αξαξαξαξ

dddeIqqeK

deKqqeIEI

e

CeKeCeIexCCxw

x

Beam supported on elastic base (bed, support), without axial force, with constant

moment of inertia and with constant distributed load: for 0>bk , 0=N , ( ) 0IxI = and


( ) ( ) 004

4

0 =−− qxwkdx

xwdEI b


( ) ( ) ( )b

xx

k

qeCeCxcosCxsinCxw 0

3210 −++⋅κ+⋅κ= ⋅κ−⋅κ

where 4

0EI

kb=κ .

Beam supported on elastic base (bed, support), without axial force, with constant

moment of inertia and with variable distributed load: for 0>bk , 0=N , ( ) 0IxI = and


( ) ( ) ( ) 02

2

0 =−− xqxwkdx

xwdEI b


( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

ξξ⋅κξ⋅κ−ξξ⋅κξ⋅κ

+

ξξ−ξξ

κ

+++⋅κ+⋅κ=

∫∫

∫∫ξ⋅κ

⋅κ−ξ⋅κ−

⋅κ

⋅κ−⋅κ

xx

xxxx

b

xx

dcosqxsindsinqxcos

dqee

dqee

k

eCeCxcosCxsinCxw

11

11

3210

222

where 4

0EI

kb=κ . In the particular case ( ) xqqxq 10 += we have:

( ) ( ) ( ) xk

q

k

qeCeCxcosCxsinCxw

bb

xx 103210 −−++⋅κ+⋅κ= ⋅κ−⋅κ

Beam supported on elastic base (bed, support), without axial force, with variable

moment of inertia and with constant distributed load: for 0>bk , 0=N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( ) 02 02

2

2

2

3

3

4

4

=−−⋅+⋅+ qxwkdx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b


particular case ( ) xeIxI ⋅α−= 1 and 00 =q we have:

( ) { } { } { } { }{ } { } { }{ }

{ } { }{ } { } { }{ } { } { }{ } { } { }{ }

α⋅+

α−⋅

+

α⋅+

α⋅=

⋅α⋅α

⋅α⋅α⋅α

1

43

1

42

1

41

1

40

011000110

0011221

EI

ke,,,,,,,GC

EI

ke,,,,,,GC

EI

ke,,,,,,GC

EI

ke,,,,pFqCkexw

b

x

b

x

b

x

b

x

p

x

where pFq is the generalized hyper-geometrical function:

{ }{ }( )( ) ( )( ) ( )∑

∞

=

⋅=0 1

1

11

1

k kpk

kpk

ppb...b

a...a

!kz,b,...,b,a,...,apFq

Here G is the Meijer function defined, for any real number r, as:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )∫ −

++ −−Γ−−Γ+Γ+Γ

+Γ+Γ−−Γ−−Γ

π=

ds

zrsb...rsbrsa...rsa

rsb...rsbrsa...rsa

i

r

b,...,b

a,...,azG

s

qmpn

mn

q

pmn

pq11

11

2 11

11

1

1

where ( ) ∫∞

−−=Γ0

1dtetz

tz .

Beam supported on elastic base (bed, support), without axial force, with variable

moment of inertia and with variable distributed load: for 0>bk , 0=N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

3

3

4

4

=−−⋅+⋅+ xqxwkdx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b

but the author did not find a general analytical solution or algorithm.

Beam supported on elastic base (bed, support), with compressive axial force, with

constant moment of inertia and with constant distributed load: for 0>bk , 0>N ,

( ) 0IxI = and ( ) 0qxq = , the equation (3) becomes:

( ) ( ) ( ) 002

2

4

4

0 =−−+ qxwkdx

xwdN

dx

xwdEI b


( )

xkEINNEI

xkEINNEI

xkEINNEI

xkEINNEI

p

bb

bb

eCeC

eCeCk

qxw

++−−

++−

+−−−

+−−

⋅+⋅

+⋅+⋅+−=

02

00

2

0

02

00

2

0

42

1

3

42

1

2

42

1

1

42

1

00


constant moment of inertia and with variable distributed load: for 0>bk , 0>N ,

( ) 0IxI = and ( )xqq = , the equation (3) becomes:

( ) ( ) ( ) ( ) 02

2

4

4

0 =−−+ xqxwkdx

xwdN

dx

xwdEI b


( )

( ) ( )

( ) ( )

ξξ−ξξ

++

β

+

ξξ−ξξ

+−

α

+⋅+⋅+⋅+⋅=

∫∫

∫∫

ξ⋅β⋅β−ξ⋅β−⋅β

ξ⋅α⋅α−ξ⋅α−⋅α

⋅β−⋅β⋅α−⋅α

x

x

x

x

bb

x

x

x

x

bb

xxxx

dqeedqeekEIN

N

k

dqeedqeekEIN

N

k

eCeCeCeCxw

110

2

110

2

3210

41

4

41

4

where ( )bkEINN

EI0

2

0

42

1++−=α and ( )

bkEINNEI

0

2

0

42

1+−−=β . In the

particular case ( ) xqqxq 10 += we have:

( )

xkEINNEI

xkEINNEI

xkEINNEI

xkEINNEI

bb

bb

bb

eCeC

eCeCxk

q

k

qxw

++−−

++−

+−−−

+−−

⋅+⋅

+⋅+⋅+−−=

02

00

2

0

02

00

2

0

42

1

3

42

1

2

42

1

1

42

1

010


variable moment of inertia and with constant distributed load: for 0>bk , 0>N ,

( )xII = and ( ) 0qxq = , the equation (3) becomes:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 02 02

2

2

2

2

2

3

3

4

4

=−−+⋅+⋅+ qxwkdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b



variable moment of inertia and with variable distributed load: for 0>bk , 0>N ,

( )xII = and ( )xqq = , the equation (3) becomes:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

2

2

3

3

4

4

=−−+⋅+⋅+ xqxwkdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b


Beam supported on elastic base (bed, support), with tensile axial force, with constant

moment of inertia and with constant distributed load: for 0>bk , 0<N , ( ) 0IxI = and


( ) ( ) ( ) 002

2

4

4

0 =−−− qxwkdx

xwdN

dx

xwdEI b


( )

xkEINNEI

xkEINNEI

xkEINNEI

xkEINNEI

b

bb

bb

eCeC

eCeCk

qxw

++−

++

+−−

+−

⋅+⋅

+⋅+⋅+−=

02

00

2

0

02

00

2

0

42

1

3

42

1

2

42

1

1

42

1

00

Beam supported on elastic base (bed, support), with tensile axial force, with constant

moment of inertia and with variable distributed load: for 0>bk , 0<N , ( ) 0IxI = and


( ) ( ) ( ) ( ) 02

2

4

4

0 =−−− xqxwkdx

xwdN

dx

xwdEI b


( )

( ) ( )

( ) ( )

ξξ−ξξ

+−

β

+

ξξ−ξξ

+−

α

+⋅+⋅+⋅+⋅=

∫∫

∫∫

ξ⋅β⋅β−ξ⋅β−⋅β

ξ⋅α⋅α−ξ⋅α−⋅α

⋅β−⋅β⋅α−⋅α

x

x

x

x

bb

x

x

x

x

bb

xxxx

dqeedqeekEIN

N

k

dqeedqeekEIN

N

k

eCeCeCeCxw

110

2

110

2

3210

41

4

41

4

where ( )bkEINN

EI0

2

0

42

1+−=α and ( )

bkEINNEI

0

2

0

42

1++=β . In the particular

case ( ) xqqxq 10 += we have:

( )

xkEINNEI

xkEINNEI

xkEINNEI

xkEINNEI

bb

bb

bb

eCeC

eCeCxk

q

k

qxw

++−

++

+−−

+−

⋅+⋅

+⋅+⋅+−−=

02

00

2

0

02

00

2

0

42

1

3

42

1

2

42

1

1

42

1

010

Beam supported on elastic base (bed, support), with tensile axial force, with variable

moment of inertia and with constant distributed load: for 0>bk , 0<N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 02 02

2

2

2

2

2

3

3

4

4

=−−−⋅+⋅+ qxwkdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b


Beam supported on elastic base (bed, support), with tensile axial force, with variable

moment of inertia and with variable distributed load: for 0>bk , 0<N , ( )xII = and


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 022

2

2

2

2

2

3

3

4

4

=−−−⋅+⋅+ xqxwkdx

xwdN

dx

xwd

dx

xdIE

dx

xwd

dx

xdIE

dx

xwdxEI b


CONCLUSIONS

The generality of the above (mathematically both beautiful and powerful) models enables

their application in the study of the quasistatic bending of any mechanical structure equivalent

to a homogeneous straight beam (bar), examples for a NPP being: fuel elements, pressure

tubes, pipes, structural beams etc.

REFERENCES

[1.] BUZDUGAN GH., “Strength of Materials” (in Romanian), Editura Tehnică (publishing

house), Bucharest, 1980

[2.] CASE J., CHILVER A.H., and ROSS C.T.F, “Strength of Materials and Structures”,

Arnold (publishing house), London, 1999, ISBN: 0-340-71920-6

[3.] DEUTSCH I., “Strength of Materials” (in Romanian), Editura Didactică şi Pedagogică

(publishing house), Bucharest, 1979

[4.] DOCA C., “Bending of Beams; a falsifiable Essay” (in Romanian), University from

Piteşti (publishing house), 2007, ISBN: 978-973-690-683-1

[5.] POSEA N., “Strength of Materials” (in Romanian), Editura Didactică şi Pedagogică

(publishing house), Bucharest, 1979

[6.] RADEŞ M., “Strength of Materials” (in Romanian) Vol. I, Editura Printed (publishing


[7.] RADEŞ M., “Strength of Materials” (in Romanian) Vol. II, Editura Printed (publishing


[8.] da SILVA V.D., “Mechanics and Strength of Materials”, Springer-Verlag Berlin

Heidelberg 2006, ISBN: 10 3-540-25131-6

[9.] WOLFRAM S., “The MATHEMATICA®

Book”, WOLFRAM MEDIA (publishing

house), Champaingn, IL, USA, 2003, ISBN: 1-57955-022-3

Documents

Cezar Doca - Quasi Static Bending of Beams