Reliability Based Bridging of the Gap between System’sSafety Factors Associated with Different Failure Modes

W. Verhaeghea,∗, I. Elishakoffb

aDept. of Mechanical Engineering, KU Leuven, B3001 Heverlee, BelgiumbDept. of Ocean and Mechanical Engineering, Florida Atlantic University, Boca Raton,

FL33431, USA

Abstract

The paper presents a study of the relation between safety factors correspondingto different failure modes in a mechanical system. The failure modes consideredare displacement exceedance, yielding and buckling. A link between the safetyfactors is found using the reliability analysis. In particular, by postulating equalreliability against the considered failure modes, we derive the relation betweenthe safety factors in three typical engineering examples.

Keywords: reliability, system failure, buckling

1. Introduction

Current practice of engineering is heavily based upon safety factors. Onthe other hand there is a strong sentiment in the engineering profession thatone has to employ design that directly utilizes advances in modern uncertaintyanalyses, like probabilistic methodology that is based on the concept of relia-bility, or the fuzzy-sets based analysis that utilizes membership functions, oranti-optimization treatment that looks at worst case scenarios. Probabilisticdesign is apparently the most developed engineering tool currently. The currentstate of the art in structural reliability is discussed by several authors. Thereaders may consult with books by [10, 11, 5].

This paper deals with the central issue of the reliability based safety factors,rather than the arbitrarily assigned ones, when two failure modes are available.As an example let us review a guideline found in a Russian textbook on strengthof materials problems. Miroliubov et al [9] stress that the safety factor forstability “is always somewhat greater than the basic safety factor for strength,since during the analysis of compressed columns on stability one has to take intoaccount additional unavoidable circumstances (eccentricity of load application,initial curvature and inhomogeneity of column’s material) that facilitate the

∗Corresponding authorEmail addresses: wim.verhaeghe@mech.kuleuven.be (W. Verhaeghe), elishako@fau.edu

(I. Elishakoff)

Preprint submitted to Elsevier December 10, 2012

bending. Large eccentricities and initial curvatures are analyzed specially, butthe small ones, that are dependent on slenderness, are taken into account viaincrease of the safety factor against stability loss. Usually the following valuesare taken: for steel nst = 1.8 − 3; for cast iron nst = 5 − 5.5; for wood nst =2.8− 3.2.”

Furthermore, there appears to be no consensus on required safety factors forstability. Izkovich [6] notes that the value of the “required stability safety factordepends basically on the utilization of the designed column and its material.Thus, for steel columns they take nst = 1.7−2 in civil engineering; nst = 3.5−5in mechanical engineering. For cast iron columns in average nst = 5; for woodin average nst = 3.”

The interplay between reliability and safety factors was discussed by nu-merous authors in past decades. For the extensive summary one can consultapparently the single book on this topic [3]. This book calls for “peaceful coex-istence” between reliability concept and safety factors. Specifically, this couldbe interpreted as the reliability based assignment of safety factors without dis-carding the latter. Some recent codes practically implement such a tolerantapproach: EUROCODE introduces both safety factors and reliability implyingthat one can make a choice between these competing philosophies. Indeed, insection 3.5 of EN1990 the design engineer finds the following statements:

(a) “The requirements . . . should be achieved by partial factor (W.V.: safetyfactors) method . . . ”

(b) “As an alternative, a design directly based on probabilistic methods maybe used.”

The EUROCODE as well as the Load and Resistance Factor Design (LRFD)of the American Institute of Steel Construction introduce safety factors (partialfactors) based on a reliability assessment. In particular, the EUROCODE detailsthe background of the partial factor method in Annex 6 of EN1990. An overviewof the developments and an outlook on the next steps to undertake in the LRFDcode is given by Chen [2] and by Ellingwood [4]. Practical application of theLRFD code is explained in [1].

In this paper we deal with some academical engineering problems that ex-hibit two or more failure modes from following set: stress levels exceeding theyield stress of the material, displacement of part of the structure exceeding aprescribed maximum and buckling of the structure. Engineers usually assigndifferent safety factors to these modes. The keen reader however may not graspwhy different safety factors are assigned against these failure modes. This paperattempts to shed some light on this topic.

2. Different failure modes, different safety factors?

Let us consider class of beam and truss problems in which three failuremodes are studied. In particular we look at following conditions: the maximalstress should not exceed the yield stress σy, some displacement should not ex-ceed a maximum ∆lmax, and additionally a beam must not buckle (buckling is

2

often analysed using so called buckling curves, see i.e. [8]). The correspondingmathematical conditions are:

σ =P

A≤ σall (1)

∆l ≤ ∆lall (2)

P ≤ Nall (3)

with P and A being the load and cross-sectional area of the beam, respectively.Quantities σall, ∆lall and Nall are the allowable stress, displacement and axialforce, respectively:

σall =σyn1

(4)

∆lall =∆lmaxn2

(5)

Nall =Ncriticaln3

=1

n3

π2EI

l2(6)

with n1, n2 and n3, the associated safety factors for the stress, displacementand buckling failure mode, respectively. The study investigates the relationshipbetween these safety factors. It is shown that the concept of reliability allowsone to assign safety factors in a unified manner. This not necessarily impliesequal safety factors, but as will be shown later does not exclude that option.

3. Beam under compression load

The following study is based on an example problem due to Volmir [14]. Asimply supported steel beam of length l = 1m with rectangular cross sectionwith a height equal to three times the width (h = 3b), is axially loaded by acompression force P . Only the yield and buckling failure mode are applied tothis problem:

σ =P

A=

P

3b2≤ σall (7)

P ≤ Nall (8)

with σall and Nall, the allowable stress and axial force, respectively, defined inEq. (4) and (6).

3.1. Random dimension b

If the dimension b is considered random, then the reliability of the structure,if only the yield and buckling constraints are considered, can be calculated asfollows:

R = Prob {(σ ≤ σy) ∩ (P ≤ Ncritical)} (9)

3

where ∩ designates the operation of intersection. In terms of the random vari-able b the reliability becomes:

R = Prob

{(√P

3σy≤ b

)∩

(4

√P4l2

π2E≤ b

)}(10)

The following three cases can be encountered. In the first case the followinginequality is satisfied: √

P

3σy>

4

√P4l2

π2E(11)

then the reliability is dominated by the yield constraint. In the second case wehave:

4

√P4l2

π2E>

√P

3σy(12)

then the reliability is dominated by the buckling constraint. Finally, in the thirdcase the following equality takes place:√

P

3σy=

4

√P4l2

π2E(13)

then either of the constraints can be used to find the reliability.For the sake of simplicity let us assume a uniform distribution for b between

values g and h. (The interested reader can find an analysis with different kind ofdistribution function in Appendix) This results in a probability density function:

fb =

0 if b < g or b > h

1h−g if g ≤ b ≤ h

(14)

so that the mean equals E[b] =∫ hgb/(h− g)db = (g + h)/2, and the standard

deviation is

db =√E[(b− E[b])2] =

√∫ h

g

(b− g + h

2

)21

h− gdb =

h− g2√

3(15)

The lower and upper bound on b can be described by the mean and the standarddeviation: g = E[b] −

√3db and h= E[b] +

√3db. The reliability for the first

case is thus calculated as

R1 =

0 if√

P3σy

> h = E[b] +√

3db

1−√

P3σy−E[b]+

√3db

2√

3dbif g ≤

√P

3σy≤ h

1 if√

P3σy

< g = E[b]−√

3db

(16)

4

The reliability for the second case is evaluated as

R2 =

0 if 4

√P4l2

π2E > h = E[b] +√

3db

1−4√P4l2

π2E−E[b]+

√3db

2√

3dbif g ≤ 4

√P4l2

π2E ≤ h

1 if 4

√P4l2

π2E < g = E[b]−√

3db

(17)

The mean value of the stress is

E[Σ] =P

3E

[1

b2

]=P

3

1

(E[b])2 − 3d2b

(18)

since

E

[1

b2

]=

∫ E[b]+√

3db

E[b]−√

3db

1

b21

2√

3dbdb =

1

(E[b])2 − 3d2b

(19)

Multiplying the numerator and denominator of the middle part of Eq. (16)by√E[Σ] yields

R1 = 1−

√P

3σy

√E[Σ]− E[b]

√E[Σ] +

√3db√E[Σ]

2√

3db√E[Σ]

(20)

Defining the central safety factor for the stress condition

n1 = σy/E[Σ] (21)

and noting that the coefficient of variation of b is νb = db/E[b], Eq. (20)becomes:

R1 = 1−

√P

3n1−√

P3(1−3ν2

b )+

√P3ν2

b

3(1−3ν2b )√

P12ν2b

3(1−3ν2b )

= 1−

√1n1−√

11−3ν2

b+

√3ν2b

1−3ν2b√

12ν2b

1−3ν2b

(22)The expression for the target safety factor n1 as a function of the target relia-bility R1 = r1 is

n1 =

[(1− r1)

√12ν2

b

1− 3ν2b

+

√1

1− 3ν2b

−

√3ν2b

1− 3ν2b

]−2

(23)

On the other hand, the mean value of the critical load equals

E[Ncritical] = E

[π2Eb4

4l2

]=π2E

4l2E[b4]

=π2E

4l21

10√

3db

[10(E[b])4

√3db + 60(E[b])2

√3d3b + 18

√3d5b

](24)

5

Dividing the numerator and denominator of the middle part of Eq. (17) by4√E[Ncritical] results in

R2 = 1−4

√4l2

π2EP

E[Ncritical]− E[b]

4√E[Ncritical]

+√

3db4√E[Ncritical]

2√

3db4√E[Ncritical]

(25)

Introducing the central safety factor against buckling

n3 = E[Ncritical]/P (26)

leads to:

R2 = 1−4

√1n3− 4

√1Q + 4

√9ν4b

Q

4

√144ν4

b

Q

, Q =E[Ncritical]

(E[b])4

4l2

π2E=

1

5(5 + 30ν2

b + 9ν4b )

(27)The expression for the target safety factor n3 as a function of the target relia-bility R2 = r2 is

n3 =

(1− r2)4

√144ν4

b

Q+ 4

√1

Q− 4

√9ν4b

Q

−4

(28)

We postulate that the structure should possess the uniform reliability againsteach mode of failure, namely if r1 = r2, then the following relationship betweenthe safety factors n1 and n3 is found:

n1 =

[1√

1− 3ν2b

(4

√Q

n3− 1 + 4

√9ν4b

)+

√1

1− 3ν2b

−

√3ν2b

1− 3ν2b

]−2

(29)

or formulated differently:

n3 =

14√Q

√1− 3ν2b

n1− 1 +

√3ν2b

+ 4

√1

Q− 4

√9ν4b

Q

−4

(30)

Let us consider a numerical example. Following numerical values are adopted:F = 500kN , σy = 1100MPa, E = 210GPa. We assume the lower boundg = 0.01m and the upper bound h = 0.035m for the uniform distribution of b.Then the following values are calculated:

E[b] =0.0225m (31)

db =0.00722m (32)

νb =0.321 (33)

6

Figure 1: The safety factor n1 as a function of the required reliability r1

Figure 2: The safety factor n3 as a function of the required reliability r2

7

Figure 3: The safety factor n3 as a function of the safety factor n1 for the numerical example

Figure 1 shows the safety factor n1 as a function of target reliability r1 corre-sponding to Eq. (23). Figure 2 shows the safety factor n3 as a function of targetreliability r2 corresponding to Eq. (28). Figure 3 shows the safety factor n3 asa function of the safety factor n1 on the condition that the same reliability isrequired as expressed by Eq. (30). From the figures and the equations followingexample values are found:

• If one would demand the same reliability 0.9 for both failure modes, thenthe required safety factors would be n1 = 2.24 and n3 = 17.2.

• If one would demand the same reliability 0.99 for both failure modes, thenthe required safety factors would be n1 = 3.33 and n3 = 38.

As is seen in both cases, the different safety factors are obtained, for differentfailure modes. Without having an “intermediary” in the form of reliability, thedesign may have turned out to constitute either overdesign or underdesign.

4. Truss Problem With Three Potential Failure Modes

A structure as depicted in Figure 4 is analysed in Example Problem 7.11of the book by Vinokurov [13]. The horizontal member is loaded in tensionNAB = 1.73F and the diagonal member is loaded in compression NBC = 2F .The total structure will fail if either the tensioned member will yield, the verticaldisplacement uBy exceeds a maximum or the compressed member will buckleor both. To avoid these failures, following conditions are set forth:

8

2b

b

2b

b

30°

100

F

A

C

B

Figure 4: Problem description

σAB =NABAAB

=1.73F

2b2= 0.865

F

b2≤ σall (34)

uBy =F

EA

(−3

16l2BC

11lAB

+ 34lBC

+1

4lBC

)−1

=F

E2b2(0.7578)−1 ≤ ∆lall (35)

NBC ≤ Nall (36)

with σall, ∆lall and Nall, the allowable stress, displacement and axial force,respectively, as defined in Eq. (4), (5) and (6).

4.1. Random Dimension b

If the dimension b is considered random, then the reliability of the structurecan be calculated as follows:

R = Prob {(σ ≤ σy) ∩ (uBy ≤ ∆lmax) ∩ (NBC ≤ Ncritical)} (37)

In terms of the random variable b reliability becomes:

R = Prob

{(√0.865F

σy≤ b

)∩

(√F

1.5156E∆lmax≤ b

)∩

(4

√2F6l2BCπ2E

≤ b

)}(38)

The following four cases can be encountered:

1.√

0.865Fσy

> max

(√F

1.5156E∆lmax,

4

√2F6l2BCπ2E

), then the reliability is deter-

mined by the tension constraint.

2.√

F1.5156E∆lmax

> max

(√0.865Fσy

,4

√2F6l2BCπ2E

), then the reliability is de-

cided by the displacement constraint.

9

3.4

√2F6l2BCπ2E > max

(√0.865Fσy

,√

F1.5156E∆lmax

), then the reliability is evalu-

ated by the buckling constraint.

4.√

0.865Fσy

=√

F1.5156E∆lmax

=4

√2F6l2BCπ2E , then either constraint can be used

to find the reliability.

Let us again assume a uniform distribution for b between g and h. Thus,the mean E[b] = (g + h)/2 and the standard deviation db = (h− g)/2

√3. The

reliability against the failure in tension is calculated as

R1 =

0 if√

0.865Fσy

> h = E[b] +√

3db

1−√

0.865Fσy

−E[b]+√

3db

2√

3dbif g ≤

√0.865Fσy

≤ h

1 if√

0.865Fσy

< g = E[b]−√

3db

(39)

The against the excessive displacement is calculated as

R2 =

0 if√

F1.5156E∆lmax

> h = E[b] +√

3db

1−√

F1.5156E∆lmax

−E[b]+√

3db

2√

3dbif g ≤

√F

1.5156E∆lmax≤ h

1 if√

F1.5156E∆lmax

< g = E[b]−√

3db

(40)The reliability against buckling reads

R3 =

0 if4

√2F6l2BCπ2E > h = E[b] +

√3db

1−4

√2F6l2

BCπ2E

−E[b]+√

3db

2√

3dbif g ≤ 4

√2F6l2BCπ2E ≤ h

1 if4

√2F6l2BCπ2E < g = E[b]−

√3db

(41)

The mean of the stress is found to be

E[Σ] = 0.865FE

[1

b2

]= 0.865F

1

(E[b])2 − 3d2b

(42)

Multiplying the numerator and denominator of the middle part of Eq. (39) by√E[Σ] yields

R1 = 1−

√0.865Fσy

√E[Σ]− E[b]

√E[Σ] +

√3db√E[Σ]

2√

3db√E[Σ]

(43)

10

Defining the central safety factor as in Eq. (21) results in:

R1 = 1−

√0.865Fn1

−√

0.865F1−3ν2

b+

√0.865F3ν2

b

1−3ν2b√

0.865F12ν2b

1−3ν2b

= 1−

√1n1−√

11−3ν2

b+

√3ν2b

1−3ν2b√

12ν2b

1−3ν2b

(44)The expression for the safety factor n1 as a function of the target reliability r1

is

n1 =

[(1−R1)

√12ν2

b

1− 3ν2b

+

√1

1− 3ν2b

−

√3ν2b

1− 3ν2b

]−2

(45)

Next, noting that the mean of the vertical displacement is

E[uBy] =F

1.5156EE

[1

b2

]=

F

1.5156E

1

(E[b])2 − 3d2b

(46)

Multiplying the numerator and denominator of the middle part of Eq. (40) by√E[uBy] results in

R2 = 1−

√F

1.5156E∆lmax

√E[uBy]− E[b]

√E[uBy] +

√3db√E[uBy]

2√

3db√E[uBy]

(47)

Defining the central safety factor

n2 = ∆lmax/E[uBy] (48)

and utilizing the coefficient of variation νb results in

R2 = 1−

√F

1.5156E1n2−√

F1.5156E(1−3ν2

b )+

√F3ν2

b

1.5156E(1−3ν2b )√

F12ν2b

1.5156E(1−3ν2b )

= 1−

√1n2−√

11−3ν2

b+

√3ν2b

1−3ν2b√

12ν2b

1−3ν2b

(49)

The expression for the safety factor n2 as a function of the target reliability R2

is

n2 =

[(1− r2)

√12ν2

b

1− 3ν2b

+

√1

1− 3ν2b

−

√3ν2b

1− 3ν2b

]−2

(50)

The last two equations coincide with Eq. (44) and (45), respectively. This is aresult of both constraints depending in the same quadratic order on the randomdimension b. The different scaling coefficients 0.865F and F

1.5156E cancel out in

11

the derivation of the respective relations between the reliability and the safetyfactor.

It is instructive to search the relation between R3 and n3; we take intoaccount that the mean of the critical load is

E[Ncritical] = E

[π2Eb4

6l2BC

]=

π2E

6l2BCE[b4]

=π2E

6l2BC

1

10√

3db

[10(E[b])4

√3db + 60(E[b])2

√3d3b + 18

√3d5b

](51)

and rewrite Eq. (41) in the form

R3 = 1−4

√6l2BCπ2E

2FE[Ncritical]

− E[b]4√E[Ncritical]

+√

3db4√E[Ncritical]

2√

3db4√E[Ncritical]

(52)

Taking into account the definition (26), we arrive at:

R3 = 1−4

√1n3− 4

√1Q + 4

√9ν4b

Q

4

√144ν4

b

Q

, Q =E[Ncritical]

(E[b])4

6l2BCπ2E

=1

5(5 + 30ν2

b + 9ν4b )

(53)The expression for the safety factor n3 as a function of the target reliability r3

is

n3 =

(1− r3)4

√144ν4

b

Q+ 4

√1

Q− 4

√9ν4b

Q

−4

(54)

If the same reliability is demanded for all constraints then the followingrelationship between n1, n2 and n3 is found:

n1 = n2 =

[1√

1− 3ν2b

(4

√Q

n3− 1 + 4

√9ν4b

)+

√1

1− 3ν2b

−

√3ν2b

1− 3ν2b

]−2

(55)or, if formulated differently, we get:

n3 =

14√Q

√1− 3ν2b

n1− 1 +

√3ν2b

+ 4

√1

Q− 4

√9ν4b

Q

−4

=

14√Q

√1− 3ν2b

n2− 1 +

√3ν2b

+ 4

√1

Q− 4

√9ν4b

Q

−4

(56)

Let us consider a numerical example. Following numerical values are adopted:F = 20kN , σy = 600MPa, E = 210GPa. Assume the lower bound g = 0.005m

12

and the upper bound h = 0.009m for the uniform distribution of b. Then thefollowing values are calculated:

E[b] =0.007m (57)

db =0.001155m (58)

νb =0.165 (59)

If one would demand the same reliability 0.9 for all three failure modes, thenthe required safety factors would be n1 = n2 = 1.54 and n3 = 3.29. For a targetreliability equal to 0.99 the required safety factors would be n1 = n2 = 1.77and n3 = 4.33. In the original example in Ref. [13] the safety factors suggestedby Vinokurov are n1 = 2.5 and n3 = 4. These safety factors do not satisfy Eq.(55). For the considered random b, this safety factor n1 would result in a veryconservative design with respect to the tension constraint since the maximumsafety factor corresponding to R1 = 1 is n1 = 1.8. On the other hand, themaximum safety factor n3 corresponding to R3 = 1 is n3 = 4.47. The suggestedsafety factor n3 = 4 corresponds to an R3 = 0.964. If the same reliability R1 isrequested, then n1 = 1.7 should be taken, which is not far from the Vinokurov’s[13] values.

4.2. Safety Factors in the Case of a Random Load F

If the load F is considered random, then the reliability of the structure Eq.(37) can be expressed in terms of the random F :

R = Prob

{(F ≤ σyb

2

0.865

)∩(F ≤ 1.5156b2E∆lmax

)∩(F ≤ 1

2Ncritical

)}(60)

The following four cases can be encountered:

(a)σyb

2

0.865 < min(1.5156b2E∆lmax,

12Ncritical

), then the reliability is dominated

by the tension constraint.

(b) 1.5156b2E∆lmax < min(σyb

2

0.865 ,12Ncritical

), then the reliability is dominated

by the displacement constraint.

(c) 12Ncritical < min

(σyb

2

0.865 , 1.5156b2E∆lmax

), then the reliability is dominated

by the buckling constraint.

(d)σyb

2

0.865 = 1.5156b2E∆lmax = 12Ncritical, then either constraint can be used to

find the reliability.

Let us assume a uniform distribution for F in the similar fashion as wasdone for b in the above section. The reliability for all the cases can be writtenin a unified form

R =

0 if x < g = E[F ]−√

3dF

x−E[F ]+√

3dF2√

3dFif g ≤ x ≤ h

1 if x > h = E[F ] +√

3dF

(61)

13

where x needs to be replaced byσyb

2

0.865 , 1.5156b2E∆lmax or 12Ncritical to find the

reliability for the first, second and third cases, respectively.Taking into account that the mean value of the stress

E[Σ] =0.865

b2E[F ], (62)

and dividing the numerator and denominator of the middle part of Eq. (61) forthe first case by E[Σ] gives

R1 =

σyE[Σ]

b2

0.865 −b2

0.865 +√

3 b2

0.865νF

2√

3 b2

0.865νF(63)

Taking into account the definition of the central safety factor n1 in Eq. (21)and noting that the coefficient of variation of the force is νF = dF

E[F ] , gives the

following result:

R1 =n1 − 1 +

√3νF

2√

3νF(64)

The expression for the safety factor n1 as a function of the target reliability r1

isn1 = 2

√3νF r1 + 1−

√3νF (65)

The same calculations but considering the mean of the vertical displacementof node B E[uBy] = (1/1.5156b2E)E[F ] can be performed for the second case(dominant displacement constraint) with the central safety factor n2 defined asin Eq. (48). The result of these calculations also yield Eq. (64) and (65), butwith the variables n1, R1, r1 replaced by n2, R2, r2, respectively.

On the other hand, dividing the numerator and denominator of the middlepart of Eq. (61) for the third case by E[NBC ] = 2E[F ] gives

R3 =

Ncritical2E[NBC ] −

E[F ]2E[F ] +

√3 dF

2E[F ]

2√

3 dF2E[F ]

(66)

with the central safety factor n3 this time defined as

n3 = Ncritical/E[NBC ] (67)

we also find Eq. (64) and (65), but with n1, R1, r1 replaced by n3, R3, r3.If the same reliability is demanded for all constraints then the following

relationship between n1, n2 and n3 is found:

n1 = n2 = n3 (68)

Let us consider a numerical example. Following numerical values are adopted:b = 0.01m, σy = 600MPa, E = 210GPa. Assume the lower bound g = 30kN

14

tf

tw

d

b

x

y

Figure 5: The dimensions of the I-beam’s cross-section

and the upper bound h = 70kN for the uniform distribution of F . Then thefollowing values are calculated:

E[F ] =50kN (69)

dF =11.547kN (70)

νF =0.23094 (71)

If one would demand the same reliability 0.9 for all failure modes, then therequired safety factors would be n1 = n2 = n3 = 1.32. Instead, if one woulddemand the same reliability 0.99 for the three failure modes, then we find n1 =n2 = n3 = 1.39. In the original example in [13] the suggested safety factors aren1 = 2.5 and n3 = 4. For the considered random variable F , this would result ina very conservative design. Furthermore, these safety factors do not satisfy Eq.(68). Based on the previous probabilistic analysis, it seems that the reliabilitybased on the suggested safety factor for the buckling resistance is higher thanthe target reliability for the tension resistance, since n3 > n1.

5. Uniformly Loaded Beam with Two Potential Failure Modes

A simply supported uniformly loaded I-beam is studied in example problem12.28 by Kachurin [7]. The steel (E = 210GPa, G = 80GPa) beam has alength l = 8m and the nominal cross section is characterised by (see Figure 5):tw = 0.012m, tf = 0.0178m, b = 0.19m and d = 0.6m.

Two possible failure modes are considered: the first one being the maximumstress exceeding the allowable stress and the second one being the loading caus-ing lateral buckling of the beam. To avoid these failures following conditionsare set forth:

σmax =ql2d

16Ix≤ σall =

σyn1

(72)

q ≤ qall =qmaxn3

=1

n3γ

√EIyGJ

l3(73)

15

with q the uniform distributed load, Ix and Iy, the area moment of inertia aboutthe neutral axis parallel to x and y, respectively, J the torsional constant of thesection and γ a numerical constant. The formula for qmax and the look-up tablefor γ is found in the standard book by Timoshenko [12].

5.1. Random Dimension tfIn the following, the dimension tf is considered random. To simplify the

calculations the cross-sectional dimensions are non-dimensionalised with respectto the nominal tf defined above, as a result the whole cross section will scalewith the random tf . The reliability of the structure can thus be calculated asfollows:

R = Prob {(σmax ≤ σy) ∩ (q ≤ qmax)} (74)

in terms of the random tf and taking into account the non-dimensional quanti-ties Eq. (74) becomes:

R = Prob

{(3

√ql2

3561.1σy≤ tf

)∩

(4

√q

γ

l3

45.9√EG≤ tf

)}(75)

The following three cases can be encountered:

1. 3

√ql2

3561.1σy> 4

√qγ

l3

45.9√EG

, then the reliability is dominated by the yield

constraint.2. 4

√qγ

l3

45.9√EG

> 3

√ql2

3561.1σy, then the reliability is dominated by the lateral

buckling constraint.

3. 3

√ql2

3561.1σy= 4

√qγ

l3

45.9√EG

, then either of the constraints can be used to

find the reliability.

As above, a uniform distribution for the random variable, in this case tf , isassumed. The reliability for the first case is calculated as

R1 =

0 if 3

√ql2

3561.1σy> h = E[tf ] +

√3dtf

1−3

√ql2

3561.1σy−E[tf ]+

√3dtf

2√

3dtfif g ≤ 3

√ql2

3561.1σy≤ h

1 if 3

√ql2

3561.1σy< g = E[tf ]−

√3dtf

(76)

The reliability for the second case is calculated as

R2 =

0 if 4

√qγ

l3

45.9√EG

> h = E[tf ] +√

3dtf

1−4√

qγ

l3

45.9√EG−E[tf ]+

√3dtf

2√

3dtfif g ≤ 4

√qγ

l3

45.9√EG≤ h

1 if 4

√qγ

l3

45.9√EG

< g = E[tf ]−√

3dtf

(77)

16

The mean of the maximum stress is found to be

E[Σmax] =ql2

3561.1E

[1

t3f

]=

ql2

3561.1

1

(E[tf ])3 − 6E[tf ]d2tf

+ 9d4tf

E[tf ]

(78)

multiplying the numerator and denominator of the middle part of Eq. (76) by3√E[Σmax] gives

R1 = 1−3

√ql2

3561.1σyE[Σmax]− E[tf ] 3

√Σmax +

√3dtf

3√

Σmax

2√

3dtf3√

Σmax(79)

Taking into account the definition of the central safety factor (21), but with therandom variable of interest now Σmax instead of Σ, we get:

R1 = 1−3

√1n1− 3

√1Q +

3

√3√

3νtfQ

3

√24√

3ν3tf

Q

,

Q =

[3561.1

ql2E[Σmax](E[tf ])3

]−1

= 1− 6νtf + 9ν4tf

(80)

The expression for the safety factor n1 as a function of the target reliability r1

is

n1 =

(1− r1)3

√24√

3ν3tf

Q+ 3

√1

Q− 3

√3√

3ν3tf

Q

−3

(81)

To calculate the reliability R2 as a function of n3, first the mean of themaximum distributed load E[qmax] is calculated as

E[qmax] = 45.9γ

√EG

l3E[t4f ]

= 45.9γ

√EG

l31

10√

3

[10(E[tf ])4

√3 + 60(E[tf ])2

√3d2tf

+ 18√

3d4tf

](82)

Secondly, we rewrite Eq. (77) in the form

R2 = 1−4

√qγ

l3

45.9√EG

1E[qmax] − E[tf ] 4

√1

E[qmax] +√

3dtf4

√1

E[qmax]

2√

3dtf4

√1

E[qmax]

(83)

Taking into account following definition of the central safety factor

n3 =E[qmax]

q(84)

17

we can write

R2 = 1−4

√1n3− 4

√1P +

4

√9ν4tf

P

4

√144ν4

tf

P

,

P =E[qmax]

(E[tf ])4

l3

45.9γ√EG

=1

5(5 + 30ν2

tf+ 9ν4

tf) (85)

The expression for the safety factor n3 as a function of the target reliability r2

is

n3 =

(1− r2)4

√144ν4

tf

P+

4

√1

P−

4

√9ν4tf

P

−4

(86)

It is important to note the similarities between the above equations (85), (86)and the equations (27), (28) and (53), (54) derived in the above sections. Thesimilarity arises from the fact that the safety factors in all the three cases de-pends on the different random variables in the same way, namely with a power4. This will be discussed further in section 6.

If the same reliability is demanded for all constraints then the followingrelationship between n1 and n3 is found:

n1 =

4

√1

n3− 4

√1

P+

4

√9ν4tf

P

3

√24√

3ν3tf

Q

4

√144ν4

tf

P

+ 3

√1

Q− 3

√3√

3ν3tf

Q

−3

(87)

or equivalently

n3 =

3

√1

n1− 3

√1

Q+

3

√3√

3νtfQ

4

√144ν4

tf

P

3

√24√

3ν3tf

Q

+4

√1

P−

4

√9ν4tf

P

−4

(88)

Let us consider a numerical example. Following numerical values are adopted:σy = 800MPa, γ = 38 and q = 18000N/m. Assume the lower bound g =0.007m and the upper bound h = 0.018m for the uniform distribution of tf .Then the following values are calculated:

E[tf ] =0.0125m (89)

dtf =0.00318m (90)

νtf =0.254 (91)

If one would demand the same reliability 0.9 for both failure modes, then therequired safety factors would be n1 = 2.39 and n3 = 7.91. For a target reliabilityequal to 0.99 the required safety factors turn out to be n1 = 3.53 and n3 = 13.32.In the original example in [7] the safety factor n3 suggested by Kachurin equalsn3 = 1.7. For the considered random tf this would mean a very low reliabilityof 0.55, which is unacceptable.

18

6. Discussion

In our analyses above, we sought to bridge the gap between safety factorsliving on their respective islands, as it were, of distinct failure modes. We showedthat based on a common target reliability a clear relationship between thesedifferent safety factors can be established. The source of the randomness in theproblem together with the definitions of the central safety factors determinedthe particular kind of the relationship. In the truss problem with the randomF , all safety factors depended linearly on the random F , albeit with differentscaling factors. These scaling factors however cancel out in the expressions ofthe safety factor as a function of the reliability. In both problems where therandom b was considered, the safety factor for the tension and displacementconstraint depend both on the random b in a quadratic way, whereas the safetyfactor for the buckling constraint depends on the dimension b with a power 4.This difference results in the different values of n1, n2 on the one hand and n3 onthe other hand as expressed in Eq. (55). The already noted similarity betweenthe three sets of equations (85), (86) and (27), (28) and (53), (54) also stemsfrom the fact the safety factor n3 depends on the respective random variablesin the same way, namely with a power 4.

7. Conclusion

The assignment of safety factors related to different failure modes is stud-ied in this paper using three simple (academic) examples. It was shown that aconsistent assignment is possible through the use of the probabilistic reliabilityconcept. In particular, the origin of the randomness in the design and its re-lation to each central safety factor determines the relation between the safetyfactors. This relation varies in the studied examples from equality of safetyfactors to polynomial relations with fractional powers. Although this work isconcentrated on simple (academic) examples, it can be interesting for engineer-ing practice and teaching. However, to highlight the scientific relevance of thepresented approach and to demonstrate the field of application, more compli-cated structures have to be investigated. Appropriate analysis is underway andwill be published elsewhere.

Acknowledgment

This study was performed during the research stay of W.V. at the FloridaAtlantic University as a Fulbright scholar. W.V. would like to acknowledge thehighly appreciated support of the IWT-Flanders, the Commission for Educa-tional Exchange Between The United States of America, Belgium and Luxem-bourg (The Fulbright Fund and Vesuvius) and the Raymond Snoeys fund of theKU Leuven.The authors would like to thank the reviewers for their valuable comments.

19

Appendix

Random dimension b, quadratic type of probability density function

The procedure for linking the two safety factors in section 3.1 can be appliedfor probability distributions different from the uniform one. In this section theguidelines for dealing with a quadratic type probability density function withbounds g and h are given. Please note that towards the end of the procedurenot all expressions are explicitely reported. Let b have a probability densityfunction of the form:

fb =

0 if b < g or b > h

6(g−h)3 b

2 + −6(g+h)(g−h)3 b+ 6gh

(g−h)3 if g ≤ b ≤ h(92)

so that the mean equals E[b] =∫ hgbfb(b)db = (g + h)/2, and the standard

deviation is

db =√E[(b− E[b])2] =

√∫ h

g

(b− g + h

2

)2

fb(b)db =h− g2√

5(93)

The lower and upper bound on b can be described by the mean and the standarddeviation: g = E[b]−

√5db and h= E[b] +

√5db. Furthermore, the probability

density function Eq. (92) for values between g and h can be rewritten as functionof E[b] and db:

fb(b) =−3√

5b2

100d3b

+3E[b]

√5b

50d3b

− 3√

5((E[b])2 − 5d2b)

100d3b

(94)

The reliability for the first case is thus calculated as

R1 =

0 if√

P3σy

> h = E[b] +√

5db

1−√

5100d3

b

{−(√

P3σy

)3

+ E[b]Pσy− 3(E[b])2

√P

3σy. . .

+15d2b

√P

3σy+ (E[b])3 + 10

√5d3b − 15E[b]d2

b

}if g ≤

√P

3σy≤ h

1 if√

P3σy

< g = E[b]−√

5db

(95)

20

The reliability for the second case is evaluated as

R2 =

0 if 4

√P4l2

π2E > h = E[b] +√

5db

1−√

5100d3

b

{−(

4

√P4l2

π2E

)3

+ 3E[b] 2

√P4l2

π2E − 3(E[b])2 4

√P4l2

π2E . . .

+15d2b

4

√P4l2

π2E + (E[b])3 + 10√

5d3b − 15E[b]d2

b

}if g ≤ 4

√P4l2

π2E ≤ h

1 if 4

√P4l2

π2E < g = E[b]−√

5db

(96)The mean value of the stress is

E[Σ] =P

3E

[1

b2

]=P

3

−3

50d3b

{E[b]√

5[ln(E[b]−

√5db)− ln(E[b] +

√5db)

]+ 10db

}(97)

since

E

[1

b2

]=

∫ E[b]+√

5db

E[b]−√

5db

1

b2

(−3√

5b2

100d3b

+3E[b]

√5b

50d3b

− 3√

5[(E[b])2 − 5d2b ]

100d3b

)db

=−3

50d3b

{E[b]√

5[ln(E[b]−

√5db)− ln(E[b] +

√5db)

]+ 10db

}(98)

with the expression between brackets on the first line of the above equationbeing equal to the probability density function as reported in Eq. (94).

Multiplying the numerator and denominator of the middle part of Eq. (95)by (

√E[Σ])3 yields

R1 = 1−√

5

100d3b(√E[Σ])3

−(√

P

3σy

)3

(√E[Σ])3 +

E[b]P

σy(√E[Σ])3 . . .

− 3(E[b])2

√P

3σy(√E[Σ])3 + 15d2

b

√P

3σy(√E[Σ])3 + (E[b])3(

√E[Σ])3 . . .

+10√

5d3b(√E[Σ])3 − 15E[b]d2

b(√E[Σ])3

}(99)

Taking into account Eq. (21) and noting again that the coefficient of variation

21

of b is νb = db/E[b], the above equation becomes

R1 = 1− 10√

5ν3/2

Q3/2

−(√ P

3n1

)3

+PQ1/2

10n1ν3/2. . .

−√

3

100ν3

√P

n1Q+

√3

20ν

√P

n1Q+

1

1000ν9/2Q3/2 . . .

+

√5

100ν3/2Q3/2 − 3

200ν5/2Q3/2

](100)

with

Q = −2P

[√

5ln

(1−√

5ν

1 +√

5ν

)+ 10ν

](101)

From this equation one can solve for n1 as a function of the target reliabilityr1 leading to n1(r1) which is not written down here. The remainder of thecalculation requires to compute

E[Ncritical] = E

[π2Eb4

4l2

]=π2E

4l2E[b4]

=π2E

4l2

[(E[b])4 + 6(E[b])2d2

b +15

7d4b

](102)

When dividing the numerator and denominator of the middle part in Eq. (96)by ( 4

√E[Ncritical])

3 one finds an expression of R2 as a function of n3, with n3

defined as in Eq. (26). For space limitation reasons the expressions are nolonger reported. The expression R2(n3) can in turn be transformed to yieldn3 as a function of a target reliability r2: n3(r2). By postulating that thestructure should possess the uniform reliability against each mode of failure,namely r1 = r2, then from r1(n1) = r2(n3) one can derive the expression relatingn1 and n3.

22

References

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23