• Nie Znaleziono Wyników

TOMASZEK Henryk, ŻUREK Józef, KLIMASZEWSKI Sławomir, ZIEJA Mariusz: A probabilistic method of determining fatigue lives of some selected structural components of an aircraft to maintain the required flight safety level – an outline. Zarys probabilistyczne

N/A
N/A
Protected

Academic year: 2021

Share "TOMASZEK Henryk, ŻUREK Józef, KLIMASZEWSKI Sławomir, ZIEJA Mariusz: A probabilistic method of determining fatigue lives of some selected structural components of an aircraft to maintain the required flight safety level – an outline. Zarys probabilistyczne"

Copied!
16
0
0

Pełen tekst

(1)

A PROBABILISTIC METHOD OF DETERMINING

FATIGUE LIVES OF SOME SELECTED STRUCTURAL

COMPONENTS OF AN AIRCRAFT TO MAINTAIN THE

REQUIRED FLIGHT SAFETY LEVEL – AN OUTLINE

ZARYS PROBABILISTYCZNEJ METODY

OKREŚLENIA TRWAŁOŚCI ZMĘCZENIOWEJ

WYBRANYCH ELEMENTÓW KONSTRUKCJI

STATKU POWIETRZNEGO DLA ZACHOWANIA

WYMAGANEGO POZIOMU BEZPIECZEŃSTWA

LOTÓW

Henryk Tomaszek

1

, Józef Żurek

2

,

Sławomir Klimaszewski

3

, Mariusz Zieja

4

(1,2,3,4) Air Force Institute of Technology ul. Księcia Bolesława 6, 01-494 Warszawa, Poland E-mails: (1) henryk.tomaszek@itwl.pl, (2) jozef.zurek@itwl.pl,

(3) slawomir.klimaszewski@itwl.pl, (4) mariusz.zieja@itwl.pl

Abstract. The paper has been intended to outline a method of determining fatigue

life of a structural component of an aircraft for some assumed flight safety level. The results gained allowed of finding the density function of time (i.e. flying time) indispensable to exceed the permissible crack length. With this function determined, one could determine reliability of the component to be then used to find fatigue life of this structural component. Two solutions have been given consideration, both depending on the m coefficient in the Paris relationship, i.e. for

m = 2 and m ≠ 2.

Key words: crack, reliability, critical condition, fatigue life

Streszczenie. W artykule przedstawiony zostanie zarys probabilistycznej metody

wyznaczania trwałości zmęczeniowej elementu konstrukcji statku powietrznego wyznaczonej dla przyjętego poziomu bezpieczeństwa lotów. Otrzymane wyniki pozwoliły na wyznaczenie postaci funkcji gęstości czasu (nalotu) niezbędnego do przekroczenia dopuszczalnej długości pęknięcia. Mając wyznaczoną tą funkcję określono niezawodność elementu, którą następnie wykorzystano do wyznaczenia trwałości zmęczeniowej elementu konstrukcji. Rozpatrzono dwa przypadki rozwiązania zależne od współczynnika m występującego we wzorze Parisa tzn. dla

m=2 i m≠2.

(2)

1. Introduction

A matter discussed in the paper is a method to determine fatigue life of a structural component of an aircraft. The following assumptions have been made:

 the component’s health/maintenance status has been determined with one parameter only, i.e. the length of a crack therein. The actual value of the parameter has been denoted with l;

 any change in the crack length may only occur in the course of the system/device being operated;

 in the case given consideration the Paris formula takes the following form:

2 2 max m m m m k z l CM dN dl

 (1) where: C, m – material constants,

Nz – a variable that denotes the number of the component-affecting load cycles due

to the system’s vibration,

Mk – coefficient of the finiteness of the component’s dimensions at the crack

location,

σmax – maximum load defined with equation (2);

 the load upon the structure’s component, with the system’s vibration taken into account, is a destructive factor. Let us assume we’ve got a component-affecting-load spectrum, with account taken of vibration. The spectrum allows for the determination of:

- the total number of load cycles Nc in the course of one flight assumed

a standard cycle,

- maximum loads within thresholds in the assumed spectrum amount to

max 1

,

2max, …,

Lmax (the assumed number of thresholds in the spectrum is L),

- the number of repetitions of specific threshold values of the loading during one flight (standard load) ni, where:

L i i c

n

N

1 ;

 maximum values of loads within the assumed thresholds are found in the following way: a i sr i i

max   (2) where: max i

- maximum value of the cyclic load within the i-th threshold,

sr i

- mean value of the cyclic load within the i-th threshold,

a i

(3)

 The following frequencies of the occurrence of loads correspond to values thereof within the thresholds

1max,

2max, …,

Lmax:

1 1

P

N

n

c

, 2

P

2

N

n

c

, …, L c L

P

N

n

.

Relationship (1) may be expressed against the flying time of the aircraft. Therefore, we assume that:

t

N

z

, (3) where:

λ – the occurrence rate of load cycles upon the component, t – flying time of the aircraft.

In the case under consideration

t

 1

,

where:

∆t – the average duration of the vibration-attributable fatigue-load cycle.

The relationship (1) against the flying time takes the following form:

2 2 max m m m m k

l

CM

dt

dl

. (4) Having applied the hitherto made assumptions, one can proceed to determine the relationship that describes the dynamics of the fatigue-crack growth, i.e. of the increase in its length.

Let Ul,t denote the probability that at the time t (for the flying time equal to t) the

crack reaches the length l. With the above-shown notation used, the dynamics of the crack length increase can be described with the following difference equation:

t l l L t l l t l l t t l PU PU PU L U, 1 , 2 , ... , 2 1           , (5) where:

Pi – probability that the load

max

i

defined with equation (2) occurs, where

L

i

1

,

2

,

3

,...,

and

P

1

P

2

P

3

...

P

L

1

,

∆li – crack increment in time t for the load equal to

imax, where

i

1

,

2

,

3

,...,

L

.

The increments are to be found on the grounds of the dependence (4). Equation (5) in function notation takes the following form:

  

L i i i t t l

P

u

l

l

t

u

1 ,

,

, (6) where:

(4)

u(l,t) – the probability density function of the crack length, which depends on the

flying time of the aircraft.

The difference equation (6) can be rearranged in the following partial differential equation of the Fokker-Planck type [3]:

2 2

)

,

(

)

(

2

1

)

,

(

)

(

)

,

(

l

t

l

u

t

l

t

l

u

t

t

t

l

u

. (7)

A particular solution to equation (7) is the crack-length density function of the following form:   ) ( 2 ) ( 2

)

(

2

1

)

,

(

At t B l

e

t

A

t

l

u

 

, (8) where:

B(t) – an average crack length for the aircraft’s flying time t, A(t) – crack-length variance for the aircraft’s flying time t.

Taking eq (4) into account, two different forms of the solution can be found, depending on the value of the m coefficient:

 for m = 2, coefficients B(t) and A(t) are solutions to integrals [3]:

   t t C e l dt t t B 0 0 1 ) ( ) (

2 (9)

   t t C e C l dt t t A 0 2 2 2 0 1 2 1 ) ( ) (

2 (10) where: ] ) [( max 2 2 2  C

C ;

2 2 CMk C  ; 2 2 max 4 max

])

)

[(

(

]

)

[(

;

]

)

[(

max 2

- the second moment of structure-affecting load;

]

)

[(

max 4

- the fourth moment of structure-affecting load;

2 2

max

)

])

[(

(

- the square second moment of the structural-component-affecting load.

 for m ≠ 2 the parameters of distribution take the following forms:

0 2 2 max 2 2 2 0

[(

)

]

]

2

2

[

)

(

t

l

m

CM

t

l

B

m m m m k m

 

(11)

(5)

                  2 2 0 2 2 max 2 2 2 0 max 2 max 2 [( ) ]) 2 2 ( ] ) [( ] ) [( 2 2 ) ( m m m m m m k m m m m m k CM t l m l CM m t A

(12)

2. Finding the probability density function of the flying time until the

boundary condition of the crack length in the structural component

is reached

Using the density function of the crack length (9) dependant on the flying time of the aircraft, one can determine the probability that the actual length of the crack in the aircraft structure’s component exceeds the permissible value within the time interval (0, tN). The relationship is as follows:

 

d l d

u

l

t

dl

l

t

Q

,

,

(13) where:

ld – the permissible value of the crack length as determined for some assumed risk

of failure to the structural component.

The probability density function of the flying time up to the moment the crack exceeds the permissible value will be determined by the following equation:

 

Q

t ld

t t f ,    (14) For m = 2, the component’s unreliability will be determined with the following equation:

 

dt e e C e l l e C l t l U t Q t t C t C t C d t C d

    0 2 2 2 0 2 0 ] ) 1 ( ) ) 1 ( ( [ , 2 2 2 2

(15)

where u(ld ,t) is determined with eq (17).

From eq (14) the following is found:

] ) 1 ( ) ) 1 ( ( [ , , 2 2 2 2 2 2 2 0 2 0      t C t C t C d t C d d e e C e l l e C l t l u l t f

(16) where: ) 1 ( )) 1 ( ( 2 2 2 0 2 2 2 2 0 2 2 0 2

1

))

(

2

1

(

2

1

)

,

(

   

Ct t C d e C l e l l t C d

e

e

C

l

t

l

u

(17)

The way of finding the probability density function of time of exceeding the permissible condition (16) has been given in [4] pp. 87- 90.

(6)

 

 

                                                 ) ( 2 2 2 )) ( ( 2 2 2 2 , , 2 2 2 0 2 2 2 2 0 t A t C m l C t B l C m t C m l t l u l t f m m m d m m m d d



where:   ) ( 2 ) ( 2

)

(

2

1

)

,

(

At t B l d d

e

t

A

t

l

u

 

(19)

Relationship (18) determines the probability density function of fatigue life of the selected aircraft’s structural component under operational-conditions spectrum for the Paris formula of the m ≠ 2. Parameters A(t) and B(t) in relationships (18) and (19) have been determined with formulae (11) and (12).

3. Finding fatigue life of the structural component up to the assumed

flight-safety level

The formula for reliability of the aircraft’s structural component can be written down in the following form:

  t d dt l t f t R 0 ) , ( 1 ) ( (20)

Where probability density function f(t,ld) is determined with the formula (16) for

m = 2 and with the formula (18) for m ≠ 2.

Hence, the unreliability of the component will be given with the following equation:

 

t d dt l t f t Q 0 ) , ( (21)

The integral (21) should be rearranged to a simpler form and the problem reduced to solving an indefinite integral:

f

(

t

,

l

d

)

dt

(22) For m = 2, the following change has been made in the integrand:

" 2 " 2 2 2 0 2 0 " 1 " 2 2 2 0 2 0

)

1

(

)

)

1

(

(

)

1

(

))

1

(

(

2 2 2 2

t C d t C t C t C d

e

C

l

l

e

l

e

C

l

e

l

l

(23)

(7)

Expression “1” is to be replaced with expression “2”, and expression “2” is denoted with z: z e C l l e l t C d t C     ) 1 ( ) ) 1 ( ( 2 2 2 2 2 0 2 0

Hence, 2 2 2 2 2 4 0 2 2 2 2 0 2 0 2 2 2 0 2 0 0

)

1

(

2

]

)

1

(

[

)]

1

(

[

]

)

1

(

[

2

2 2 2 2 2 2

t C t C d t C t C t C d t C

e

C

l

e

C

l

l

e

l

e

C

l

e

C

l

l

e

l

dt

dz

2 2 2 2 2 4 0 2 2 0 2 2 2 0 2 0 2 2 3 0

)

1

(

]

)

1

(

[

2

)

1

](

)

1

(

[

2

2 2 2 2 2 2

t C t C d t C t C t C d t C

e

C

l

e

l

e

l

C

l

e

e

l

e

l

C

l

dt

dz



2 2 2 0 2 2 0 2 0 0 ) 1 ( ] ) 1 ( [ 2 ) 1 ]( ) 1 ( [ 2 2 2 2 2 2 2         t C t C d t C t C t C d t C e l e l e l e e l e l l dt dz

dz e l e l e e l e l l e l dt t C d t C t C t C d t C t C 2 2 2 2 2 2 2 2 0 2 0 0 2 2 2 0 ] ) 1 ( [ 2 ) 1 ]( ) 1 ( [ 2 ) 1 (

       

Then, the substitution is made in the indefinite integral:

     ) 1 ( )] 1 ( [( 2 2 2 2 2 2 2 0 2 0 Ct t C t C d t C e e C e l l e C l

         ] ] ) 1 ( [ 2 ) 1 ]( ) 1 ( [ 2 ) 1 ( [ 2 2 2 2 2 2 2 2 0 2 0 0 2 2 2 0 t C d t C t C t C d t C t C e l e l e e l e l l e l

dz

e

e

C

l

z t C

))

1

(

1

2 2 2 2 0

(24)

Therefore, the following is arrived at:

      ] ) 1 ( )] 1 ( [ ) 1 ( [ 2 2 2 2 2 2 2 2 0 2 2 0 t C t C t C d t C t C e e C e l l e e C l

         ] ) ) 1 ( ( 2 ) 1 ]( ) 1 ( [ 2 ) 1 ( [ 2 2 2 2 2 2 2 2 0 2 0 0 2 2 2 0 t C d t C t C t C d t C t C e l e l e e l e l l e l

dz

e

e

C

l

z t C

))

1

(

1

2 2 2 2 0

(8)

Then we get:

[

2

(

2

1

)

[

(

2

1

)]

2 2

]

2 0 2 2 0 t C t C d t C t C

e

C

e

l

l

e

e

C

l

         ] ] ) 1 ( [ 2 ) 1 ]( ) 1 ( [ 2 ) 1 ( [ 2 2 2 2 2 2 2 2 0 2 0 0 2 2 0 t C d t C t C t C d t C t C e l e l e e l e l l e l

e

dz

e

C

l

z t C

))

1

(

1

2 2 2 2 0

[

(

1

)

[

(

1

)]

]

2

1

2 2 2 2 2 0 2 0 t C t C d t C t C

e

e

l

l

e

e

l

         ] ] ) 1 ( [ ) 1 ]( ) 1 ( [ ) 1 ( [ 2 2 2 2 2 2 2 2 0 2 0 0 2 2 2 0 t C d t C t C t C d t C t C e l e l e e l e l l e C l

e

dz

e

C

l

z t C

))

1

(

1

2 2 2 2 0

[(

(

1

))

[(

(

1

)]

]

2

1

2 2 2 2 2 0 2 0 t C t C d t C t C

e

e

l

l

e

e

l

e

dz

e

l

e

l

e

e

l

e

l

l

e

C

l

z t C d t C t C t C d t C t C

]

]

)

1

(

[

)

1

](

)

1

(

[

)

1

(

[

2 2 2 2 2 2 2 2 0 2 0 0 2 2 2 0

[

(

1

)

(

(

1

))

]

2

1

2 2 2 2 2 0 2 0 t C t C d t C t C

e

e

l

l

e

e

l

dz

e

l

e

l

e

C

l

e

l

e

l

e

e

l

z z d t C t C t C d t C t C t C

1 0 2 2 2 0 2 0 2 0

[

(

1

)

]

)

1

(

)

)

1

(

(

)

1

(

1

2 2 2 2 2 2

We assume that:

))

1

(

(

)

)

1

(

(

2 2 0 0

t C d d t C

e

l

l

l

e

l

Hence,

[

(

1

)

(

(

1

)]

]

2

1

2 2 2 2 2 0 2 0 t C t C d t C t C

e

e

l

l

e

e

l

dz

e

z

e

e

l

l

e

e

l

z t C t C d t C t C

]

1

]

)

1

(

(

[

)

1

(

1

[

2 2 2 2 2 0 2 0

(9)

which means we arrive at:

e

dz

z

z

1

2

1

The indefinite integral (24) after rearrangements takes the form:

e

dz

z

z

1

2

1

(25)

Then, the second substitution has to be made in the integral (25), which should take the form:

w

z 

z dz dw 2 1  w dw dz 2  wdw dz2 (26) The dependence (26) is inserted in the integral (25). Hence, the following is arrived at:

e

wdw

e

dw

w

w w2

1

2

2

1

2

1

(27)

One more substitution:

2

2 2

y

w 

ydy

wdw 

2

dy w y dw 2  2 dy dw  (28) Hence, after inserting (28) in (27) the following integral is arrived at:

edy y 2 2 2 1

(29)

(10)

) 1 ( ) ) 1 ( ( ; ; 2 2 2 2 2 2 0 2 0 2 2       t C d t C e C l l e l z z w w y

)

1

(

)

)

1

(

(

2 2 2 2 2 0 2 0

t C d t C

e

C

l

l

e

l

w

)

1

(

)

)

1

(

(

2 2 2 2 2 0 0

t C d t C

e

C

l

l

e

l

w

2

2 2

y

w 

2 2

2w

y 

2

2w

y 

)

1

(

)

)

1

(

(

2

2

2 2 2 2 2 0 0

t C d t C

e

C

l

l

e

l

w

y

(30)

Having inserted the results gained in the equation (20) and remembering about a suitable notation of the limits of integration, the following dependence for the reliability is arrived at:

     ) ( 2 2 2 1 1 ) ( t y y dy e t R

(31)

where equation (30) should be substituted for the upper limit of the integral y(t) . The cumulative distribution function for the standard Gaussian (normal) distribution takes the form:

     x y dy e x 2 2 2 1 ) (

With the above-shown dependence taken into account, the formula for the reliability of the structure’s component is expressed with the following equation:

)

)

1

(

)

)

1

(

(

2

(

1

)

(

2 2 2 2 2 0 0

t C d t C

e

C

l

l

e

l

t

R

(32)

Hence, reliability of the structure’s component will be determined with the following dependence:

(11)

     

) 1 ( ] ) 1 ( [ 2 2 2 2 2 2 0 2 0 2

2

1

)

(

t C d t C e C l l e l y

dy

e

t

Q

(33)

Having found (assumed) the level of risk of a failure to the structure’s component, i.e. the level of exceeding the permissible value of the length of a crack in this component, we get: *

)

(

t

Q

Q

(34) Hence,

   

e dy Q y 2 * 2 2 1 (35)

For the assumed value of Q*, the value of the upper limit of the integral (for which the integral on the right side of the equation (35) takes value Q*) is to be found in the standard Gaussian distribution tables.

Hence, the following dependence is arrived at:

)

1

(

)

)

1

(

(

2

2 2 2 2 2 0 0

t C d t C

e

C

l

l

e

l

(36)

)

1

(

)

)

1

(

(

2

2 2 2 2 2 0 0

t C d t C

e

C

l

l

e

l

We assume that * 2

 Hence,

)

1

(

)

)

1

(

(

2 2 2 2 2 0 0 *

t C d t C

e

C

l

l

e

l

(37)

From (37) we can find time

t

*, for which the equality relation (34) takes place. Time t* will be the searched life of the structure’s component, i.e. it will be the aircraft’s flying time for the assumed risk of exceeding the permissible value of the crack length. We assume that

(12)

x

e

C2t

(38) Hence, ) 1 ( ) ) 1 ( ( 2 2 2 0 0 *     x C l l x l d

(39)

From (39) we can find x. With some specific value of x gained from the dependence (38), we can find t*:

x

e

C2t

x t C2 ln

2 *

ln

C

x

t

(40) Formula (40) determines fatigue life of the aircraft structure’s component t* for the assumed risk of exceeding the boundary condition Q*.

For m ≠ 2, the unreliability of the structural component can be determined with the following formula:

 

dt t A t A t B l t B t l u t Q t d d

         0 2 ( ) ) ( ' )) ( ( ) ( ' , (41) where:

  0 2 2 2 2 0

2

2

)

(

t

l

m

C

t

l

B

m m

;

    2 2 0 2 2 2 2 0

2

2

2

2

)

(

m m m m

l

t

C

m

l

C

m

t

A

;

l

t

u

d

,

- function determined with the dependence (19).

The integral (41) should be rearranged to a simpler form and the problem reduced to solving an indefinite integral:

f

(

t

,

l

d

)

dt

(42) The following change has been made in the integrand for m ≠ 2:

(13)

      " 2 " 2 " 1 " 2

)

(

2

)

)

(

(

)

(

2

))

(

(

t

A

l

t

B

t

A

t

B

l

d

d

(43)

Expression “1” is to be replaced with expression “2”, and expression “2” is denoted with z:

z

t

A

l

t

B

d

)

(

2

)

)

(

(

2 (44)

The derivative of the relationship (44) is calculated

2 2

))

(

2

(

)

(

'

2

)

)

(

(

)

(

2

)

(

'

)

)

(

(

2

t

A

t

A

l

t

B

t

A

t

B

l

t

B

dt

dz

d

d

Hence,

)

(

'

2

)

)

(

(

)

(

2

)

(

'

)

)

(

(

2

)

(

'

2

)

)

(

(

)

(

2

)

(

'

)

)

(

(

2

))

(

2

(

2 2 2

t

A

l

t

B

t

A

t

B

l

t

B

t

A

l

t

B

t

A

t

B

l

t

B

t

A

dt

d d d d

The substitution is made in the indefinite integral (42):





dz

t

A

l

t

B

t

A

t

B

l

t

B

t

A

t

A

t

A

t

B

l

t

B

e

t

A

d d d z

)]

(

'

2

)

)

(

(

)

(

2

)

(

'

)

)

(

[(

2

))

(

2

(

)

(

2

)

(

'

))

(

(

)

(

'

)

(

2

1

2 2





dz

l

t

B

t

A

l

t

B

t

A

t

B

t

A

t

A

t

A

t

B

l

t

A

t

B

e

t

A

d d d z

)

)

(

)](

(

'

)

)

(

(

)

(

)

(

'

2

[

))

(

2

(

)

(

2

)

(

'

))

(

(

)

(

)

(

'

2

)

(

2

1

2

1

2

dz e l t B t A dz l t B t A e t A z d d z  

  ) ) ( ( ) ( 2 2 1 ) ) ( ( )) ( 2 ( ) ( 2 1 2 1

For z l t B t A d 1 ) ) ( ( ) ( 2   ,

After some rearrangement we arrive at:

e

dz

z

l

t

f

d

1

z

2

1

)

,

(

(45)

(14)

Then, we introduce conversions analogous to those presented for the m=2 case. Having done suitable rearrangements, the following integral is arrived at:

edy y 2 2 2 1

where y takes value determined with the dependence (46), since

)

(

2

)

)

(

(

;

;

2

2 2 2

t

A

l

t

B

z

z

w

w

y

d

) ( 2 ) ) ( ( 2 t A l t B wd

)

(

2

)

)

(

(

t

A

l

t

B

w

d

2

2 2

y

w 

2 2

2w

y 

)

(

2

)

)

(

(

2

t

A

l

t

B

y

d (46)

Having inserted the results gained in the equation (20) and remembering about a suitable notation of the limits of integration, the following dependence for the reliability is arrived at:

     ) ( 2 2 2 1 1 ) ( t y y dy e t R

(47)

Hence, the unreliability of the structural component is determined with the relationship

   

) ( ) ) ( ( 2 2

2

1

)

(

t A l t B y d

dy

e

t

Q

(48)

Having found (assumed) the level of risk of a failure to the structure’s component, i.e. the level of exceeding the permissible value of the length of a crack in this component, we get: *

)

(

t

Q

Q

(49) Hence,

(15)

   

e dy Q y 2 * 2 2 1 (50)

For the assumed value of Q*, the value of the upper limit of the integral is to be found in the standard Gaussian distribution tables. In this way we get value of θ. Hence, we arrive at the equation that allows for the determination of the component’s life for the assumed risk level:

)

(

)

)

(

(

t

A

l

t

B

d

(51)

In the relationship (51) we look for such a value of t*, for which the left side of the equation equals the right side thereof. Solving the dependence (51) in this way, we find life of the structural component of an aircraft we have been looking for.

4. Final remarks

A probabilistic method to determine fatigue lives of some selected structural components of an aircraft has been presented for an assumed flight safety level (reliability). For the needs of the deterministic approach the physical part of the study has been based on the Paris formula. Solution to this formula depends on the value of the m coefficient because of the crack growth rate in the component. Therefore, two solutions are accepted:

 for m = 2,  for m ≠ 2.

Some random operation-induced loading in the form of a load spectrum is a fatigue-provoking destructive factor in the model of the crack growth in a structural component. An assumption has been made in the study that the sequence of load cycles, as far as values thereof are concerned, remains of no effect upon the crack growth rate. All the dependences arrived at enable specific calculations, if we have values of material constants and data on the load spectrum.

References

1. Smirnov N.N, Ickovicz A.A.: Obsłuzivanije i remont aviacjonnej techniki po

sostojaniju. Transport, 1980.

2. Kałmucki W.S.: Prognozirowanije resursov detali maszin i elementov

konstrukcji. Kisziniev,1989.

3. Tomaszek H., Klimaszewski S., Zieja M.: Zarys probabilistycznej metody

wyznaczania trwałości zmęczeniowej elementu konstrukcji z wykorzystaniem wzoru Parisa i funkcji gęstości czasu przekraczania stanu granicznego. 35th

(16)

International Scientific Congress on Powertrain and Transport Means – European KONES 2009, Zakopane 2009.

4. Tomaszek H., Żurek J., Jasztal M.: Prognozowanie uszkodzeń zagrażających

bezpieczeństwu lotów statków powietrznych. Wydawnictwo Naukowe JTE –

PIB Radom 2008.

5. Tomaszek H., Ważny M.: „Zarys metody oceny trwałości na zużycie

powierzchniowe elementów konstrukcji z wykorzystaniem rozkładu czasu przekraczania stanu granicznego”. ZEM 2008.

Prof. Henryk Tomaszek DSc. Eng, Scientific Worker of Air Force Institute of Technology. Field of science: aviation transport, reliability, safety. Specialisation: Air Armaments, Machine Building and Operation, transport, safety and reliability,

DSc. Eng. Józef Żurek, professor at Air Force Institute of Technology in Warsaw. Specialisation: Machine Building and Operation, transport, safety and reliability,

Major Mariusz Zieja PhD. Eng, Air Force Institute of Technology, graduated from Military University of Technology in 2000. M.Sc. in Mechatronics specialized in Aircraft Avionics. In 2008 achieved Ph.D. in Mechanical Engineering. He is engaged in development and implementation of IT systems to support aircraft maintenance, safety and reliability management.

Cytaty

Powiązane dokumenty

tions (see [1], [2], [4], [7]) ; from them we especially recommend results given in the Blasi’s paper [2] in which the author obtained several theorems characterizing certain

Once the pipe has been installed at the appropriate depth, finely ground unslaked lime (CaO) or cement is blown to the bottom of the pipe. At the same time, the pipe is

W reszcie ujrzał się opuszczonym przez publiczność, stał się przed­ miotem ataków ze strony m łodych, którzy napastow ali go czasem niespra­ w ied liw ie,

Badane mieszańce różniły się istotnie od form rodzicielskich pod względem większości analizowanych cech ilościowych (tab.. Rośliny były wyższe

In this paper, we propose an agent-based social simulation tool, ASSOCC, that sup- ports decision makers understand possible consequences of policy interventions, but exploring

Dział Tworzenia Zasobów Bibliografi cznych, będący częścią Centrum Usług Informacyjnych, nie zajmuje się, jak można by sądzić, katalogowaniem zbiorów, lecz

During the rollover of the bus 1 during the impact load, for example, the left wall 1 (Figure 6), the membrane of the left side or right side rack 4 and the subsequent movement of

Voor de kosten analyse van de kranen is de gelimiteerde data gebruikt uit World Cargo News.. Voor de kosten analyse van kade wanden is een promotie werk gevonden dat de vragen van