International Bulletin of Mathematical ResearchVolume 02, Issue 01, March 2015Pages 104-123, ISSN: 2394-7802

Mathematical and Computational Approach For StochasticFinite Element Method

P V Ramana1 and Vivek Singh2

Department of Civil Engineering, MNIT, Jaipur - 302017, IndiaEmail: 1pvramana.ce@mnit.ac.in, 2 2013pcs5168@mnit.ac.in

Abstract

In this paper, the effects of stochasticity in inputs on response quantity are studied in detail, using Stochastic FiniteElement Method. The comparability of the method with the techniques such as Finite Element Method and MonteCarlo Simulation is then checked. Modern structures like nuclear power plants, wind mills etc., are made of thoseengineering materials of considerable spatial statistical. For a robust design, such randomness should be consideredat the modelling and analysis phase. In the present study, the uncertainty levels in the response function are checkedwith respect to the variability in the material and geometrical properties. The method is studied and implementedfor a statistically determinate structure and a statistically indeterminate structure with material property variabilityalong the length. It can be observed that the uncertainty propagation can be tracked at every step of the analysisusing this method and comparable results are obtained. Linear dynamic equations are solved by Runge-Kutta fourthorder method and MATLAB symbolic solutions and the results are compared with analytic solutions and found tobe matching well. These problems may also be solved using SFEM.

1 Introduction

1.1 Definitions of Stochasticity

It is well known that most observable phenomena in the universe have an element of uncertainty or randomness involvedin them. And they dont follow a specific pattern or a regular fashion thereby making them totally unpredictable,giving out multiple outcomes. In other words, the occurrence of multiple outcomes without any pattern is termedas uncertainty, randomness or stochasticity. The word stochasticity comes from the Greek word stochos which meansuncertain. Concept of randomness can be explained with the help of an example as follows: if several identical specimensof a steel bar were loaded until failure in a laboratory, each specimen would fail at different values of the load. The loadcapacity of the bar is therefore, a random quantity, more technically a random variable. In general, all the parametersof interest in engineering analysis and design have some degree of uncertainty and may be considered to be randomvariables.

1.2 Definitions of Probability and Related Theorems

In the existing literature, there are three approaches for probabilitya) Classicalb) Relative Frequencyc) AxiomaticFor several centuries, the theory of probability was based on the Classical definition. This concept is used today todetermine the probabilistic data and as a working hypothesis. According to Classical definition, the probability P(A)of an event A is determined a priori without actual experimentation: It is given by the ratio

P (A) =NAN

Where N is the number of possible outcomes (which should be equally likely, mutually exclusive, and collectivelyexhaustive) and NA is the number of outcomes favourable to the event A. Here, one can define probability but using a

Received: February 20, 2015

Keywords: Stochasticity, Monte Carlo Simulation, FEM, MATLAB, Static, Dynamic

Mathematical and Computational Approach For Stochastic Finite Element Method 105

pre-conceived notion on outcomes like being equally likely which is not universally valid. For certain physical problems,experimentation is not possible. This definition may give irrational values for probability which is required to berational.

1.3 Stochastic Processes

A stochastic process or random process is a random variable that evolves in time or space or both, constituting acollection of infinite random variables, defined as follows: A random process is a function: Ω×R→ R and is denotedby X(t, ω). For a fixed value of t, X(t, ω) is a random variable. For a fixed value of ω, X(t, ω) is a function of time (arealization). For a fixed value of t and ω, X(t, ω) is a number. For variability in both t and ω, X(t, ω) is a collectionof time histories (an ensemble).

1.4 Monte Carlo Simulation

Monte Carlo simulation is a computerized mathematical technique that allows people to account for risk in quantitativeanalysis and decision making. The technique is used by professionals in such widely disparate fields as finance, projectmanagement, energy, manufacturing, engineering, research and development, insurance, oil & gas, transportation, andthe environment. Monte Carlo simulation furnishes the decision maker with a range of possible outcomes and theprobabilities they will occur for any choice of action.. It shows the extreme possibilities the outcomes of going forbroke and for the most conservative decision along with all possible consequences for middle-of-the-road decisions. Thetechnique was first used by scientists working on the atom bomb; it was named for Monte Carlo, the Monaco resorttown renowned for its casinos. Since its introduction in World War II, Monte Carlo simulation has been used to modela variety of physical and conceptual systems.

2 Stochastic Finite Element Formulation

2.1 Flexural Problem

Consider a axial bar of a length L, of material with Young’s modulus E, and of cross-section geometry giving the valueof moment of inertia as I, and subjected to static uniform vertical load over the entire span and with the governingequation EIu” = −Mx Or EIu”” = −w. Let us discretize the beam into m number of one dimensional finite elementsalong the length. Then the equation governing the displacements at each node i is P = Kd, where P is the load vector,K is the global stiffness matrix and d is the displacement matrix. l = L/m

Figure 2.1: Cantilever beam carrying udl w kN per unit length

P =

P1

M1

P2

M2

...PM+1

MM+1

; d =

U1

Θ1

U2

Θ2

...UM+1

ΘM+1

K =

EI

l3

12 6L −12 6l 06l 4l2 −6l 2l2 · · · 0−12 −6l 24 0 06l 2l2 0 8l2 0

.... . .

...12 −6l

0 0 0 0 · · ·−6l 4l2

Consider Modulus of elasticity and Cross-section size varies randomly along the length of the beam. Consider therandom field element mesh consist the same discretization locations as that for finite element discretization.

106 P V Ramana, Vivek Singh

Random field of Youngs modulus,

E(x) =[E1(x) E2(x) · · · Ei(x) Em(x)

](2.1.1)

Where Ei(x) represents the spatial average over the ith element.Mean value vector

µE =[µE1(x) µE2(x) · · · µEi(x) µIm(x)

]T(2.1.2)

Random field of Moment of Inertia,

I(x) =[I1(x) I2(x) · · · Ii(x) Im(x)

](2.1.3)

Mean value vectorµI =

[µI1(x) µI2(x) · · · µIi(x) µIm(x)

]T(2.1.4)

Random fields E(x) and I(x) are assumed to be uncorrelated. Hence

〈α(x)〉 = 〈E(x).I(x)〉 = 〈E(x)〉.〈I(x)〉 (2.1.5)

Calculation of Covariance Cov(αi, αj) can be done using the procedure proposed by E.H.Vanmarcke et al.1 for thevariance of spatial averages.The covariance between xu and x′u is expressed as :

cov(xu, x′u) =

σ2x

2[U2

0 γx(U0)− U21 γx(U1) + U2

2 γx(U2)− U23 γx(U3)] (2.1.6)

Let ∀i = 1, 2, 3....m+ 1 and j = 1, 2, 3....m+ 1 , Σ = Cov(αi, αj) represent the co-variance matrix for element flexuralrigidities αiThe variance function γx(U) describes the variation of the variance of the spatial averages with the discretization lengthor averaging interval. It has the following properties: γx(U) ≥ 0γx(0) = 1 The exact pattern of decay of the variances depends on the correlation function of the random process. i.e.

γx(U) =2

T

∫ T

0

(1− τ

T)ρ(τ)dτ (2.1.7)

Vanmarcke defined the parameter Scale of Fluctuation, Θx as the proportionality constant of variance in the asymptoticform when the averaging interval goes to infinity.

Θx = limT→∞

Uγx(U) (2.1.8)

Variance function may be approximated by the asymptotic form

γx(U) =

1, U ≤ ΘΘx

U, U ≥ Θ

(2.1.9)

And by triangulation correlation function as:

γx(U) =

1− U

3Θx, U ≤ Θ

Θx

U

[1− Θx

3U

], U ≥ Θ

(2.1.10)

Exact variation functions with simple and squared correlation functions may also be considered.Vanmarcke et al. (1983) observed that asymptotic variance function exhibit closer agreement with exact variancefunctions and thus scale of fluctuation gives simple and reasonably accurate information about the variances of localaverages of random fields.In a similar way, all the statistical parameters (mean vector µα and covariance matrix Σα can be constructed for inverserandom fields of E, I and α. Consider any element i between nodes (i-1) and (i) at any distance x from the fixed endof the beam . Local stiffness matrix will be

Mathematical and Computational Approach For Stochastic Finite Element Method 107

Figure 2.2: Finite elements and Random field elements for the flexural beam

k =1

l3

12.Ei(x).Ii(x) 6.Ei(x).Ii(x).l −12.Ei(x).Ii(x) 6.Ei(x).Ii(x).l6.Ei(x).Ii(x).l 4.Ei(x).Ii(x).l2 −6.Ei(x).Ii(x).l 2.Ei(x).Ii(x).l2

−12.Ei(x).Ii(x) −6.Ei(x).Ii(x).l 12.Ei(x).Ii(x) −6.Ei(x).Ii(x).l6.Ei(x).Ii(x).l 2.Ei(x).Ii(x).l2 −6.Ei(x).Ii(x).l 4.Ei(x).Ii(x).l2

(2.1.11)

Ei(x) represents the random variable for the ith element in the random field of Youngs modulus and Ii(x) that ofmoment of inertia of the section.

meank =µEi.µIil3

12 6l −12 6l6l 4l2 −6l 2l2

−12 −6l 12 −6l6l 2l2 −6l 4l2

(2.1.12)

The global stiffness matrix may be obtained as:

K =EI

l3

12 6L −12 6l 06l 4l2 −6l 2l2 · · · 0−12 −6l 24 0 06l l2 −6l 4l2 0

.... . .

...12 −6l

0 0 0 0 · · ·−6l 4l2

(2.1.13)

Where

E =

E1 E1 E1 E1 0E1 E1 E1 E1 · · · 0E1 E1 E1 + E2 E1 − E2 0E1 E1 E1 − E2 E1 + E2 0

.... . .

...Em Em

0 0 0 0 · · ·Em Em

(2.1.14)

I =

I1 I1 I1 I1 0I1 I1 I1 I1 · · · 0I1 I1 I1 + I2 I1 − I2 0I1 I1 I1 − I2 I1 + I2 0

.... . .

...Im Im

0 0 0 0 · · ·Im Im

(2.1.15)

Denoting the co-efficient matrix as Kc,Kbc = Ebc.(Ibc)Kcbc (2.1.16)

108 P V Ramana, Vivek Singh

where

Kcbc =1

l3

12 6L −12 6l 06l 4l2 −6l 2l2 · · · 0−12 −6l 24 0 06l l2 −6l 4l2 0

.... . .

...12 −6l

0 0 0 0 · · ·−6l 4l2

(2.1.17)

Ebc =

E1 + E2 E1 − E2 E2 E2 0E1 − E2 E1 + E2 E2 E2 · · · 0E2 E2 E2 + E3 E2 − E3 0E2 E3 E2 − E3 E2 + E3 0

.... . .

...Em Em

0 0 0 0 · · ·Em Em

(2.1.18)

Ibc =

I1 + I2 I1 − I2 I2 I2 0I1 − I2 I1 + I2 I2 I2 · · · 0I2 I2 I2 + I3 I2 − I3 0I2 I3 I2 − I3 I2 + I3 0

.... . .

...Im Im

0 0 0 0 · · ·Im Im

(2.1.19)

Also, one can have

Pbc =

P2

M2

P3

M3

...PM+1

MM+1

(2.1.20)

;

dbc =

U2

Θ2

U3

Θ3

...UM+1

ΘM+1

(2.1.21)

Then,dbc = [Kbc]

−1Pbc (2.1.22)

Mean dbc = Expected value of dbc = Expectation of ([Kbc]−1Pbc)=(Pbc)

∗ Expectation of ([Kbc]−1)

Denoting [Kbc]−1 by fbc,

fbc = ((Einv))bc · ((Iinv))bc · fcbc (2.1.23)

fcbc =fbc

Einv · Iinv(2.1.24)

Where matrices ((Einv))bc and ((Iinv))bc represents matrices constructed from inverse random fields of E (x) and I(x)respectively.Now, Expectation of

([Kbc]−1) = mean((Einv))bc ·mean((Iinv))bc · fcbc (2.1.25)

Mathematical and Computational Approach For Stochastic Finite Element Method 109

Meandbc = (mean((Einv))bc ·mean((Iinv))bc · fcbc)Pbc (2.1.26)

Vectorised Covariance matrix,

vec(Σdd) = vecCov(di, dj) = ((fcbc ⊗ fTcbc) Σ((EI)inv ((EI)inv))

(Pbc ⊗ Pbc) (2.1.27)

Where ⊗ represents the Kronecker product and represents the Hadamard product of matrices. Covariance matrix,Σddmay be obtained by re-shaping the vector into the required matrix dimensions (4m2 × 4m2)

• Derivation of expression for Vectorised Covariance matrix,vec(Σdd)

Covariance matrix for displacements = 〈(d− µd)(d− µd)T 〉

E =

U2U2 U2Θ2 U2U3 · · · U2Θm+1

Θ2U2 Θ2Θ2 Θ2U3 · · · Θ2Θm+1

U3U2 U3Θ2 U3U3 · · ·...

.... . .

...Um+1Um+1 Um+1Θm+1

Θm+1U2 Θm+1Θ2 · · ·Θm+1Um+1 Θm+1Θm+1

(2.1.28)

one can see that taking the expectation of the kronecker product of the (d − µd) vector by the (d − µd) vectorgives the vector containing all the elements of the covariance matrix.

vec(Σdd) = 〈(d− µd)⊗ (d− σµd)〉 (2.1.29)

But one can have,

dbc = (((Einv))bc · ((Iinv))bc · fcbc)Pbc (2.1.30)

Hence

vec(Σdd) = 〈[(((Einv))bc · ((Iinv))bc · fcbc)Pbc − µ((Einv))bc · µ((Iinv))bc · fcbc)Pbc]⊗ [(((Einv))bc · ((Iinv))bc · fcbc)Pbc−µ((Einv))bc · µ((Iinv))bc · fcbc)Pbc]〉

= 〈([((EIinv))bc − µ((EIinv))bc) fcbc)Pbc)⊗ ([((EIinv))bc − µ((EIinv))bc) fcbc)Pbc)〉(2.1.31)

Using mixed product rule,

vec(Σdd) = 〈([(((EIinv))bc − µ((EIinv))bc) fcbc]⊗ [(((EIinv))bc − µ((EIinv))bc) fcbc])(Pbc ⊗ Pbc)〉= 〈([fcbc ⊗ fcbc]) [(((EIinv))bc − µ((EIinv))bc)⊗ (((EIinv))bc − µ((EIinv))bc)])(Pbc ⊗ Pbc)〉

(2.1.32)

Putting

Σ(EIinvEIinv)

= 〈(((EIinv))bc − µ((EIinv))bc)⊗ (((EIinv))bc − µ((EIinv))bc)〉vec(Σdd) = ([fcbc ⊗ fcbc] Σ

(EIinvEIinv))(Pbc ⊗ Pbc)

(2.1.33)

• Construction of Σ(EIinvEIinv)

EIinvbc =

EIinv1 + EIinv2 EIinv1 − EIinv2 EIinv2 EIinv2 0EIinv1 − EIinv2 EIinv1 + EIinv2 EIinv2 EIinv2 · · · 0

EIinv2 EIinv2 EIinv2 + EIinv3 EIinv2 − EIinv3 0EIinv2 I3 EIinv2 − EIinv3 EIinv2 + EIinv3 0

.... . .

...EIinvm EIinvm

0 0 0 0 · · ·EIinvm EIinvm

(2.1.34)

Where EIinvi represents the inverse random fields (ie. The random fields expressed as reciprocal values of EIi

110 P V Ramana, Vivek Singh

as function of position varying one-dimensionally along length) . Note that the size of the matrix is 2m.

Denoting any element in EIinvbcby Ri, each element is a random variable. The covariance matrix Σ(EIinvEIinv)

will

be of the size 4m2 and will be taken as the expectation of Kronecker product of [EIinvbc − 〈EIinvbc〉] with

[EIinvbc − 〈EIinvbc〉]T .

Cov(R1, R1) = 〈(EIinv1 + EIinv2 − 〈EIinv1 + EIinv2〉)2〉= 〈(EIinv1 − EIinv2) + 〈EIinv1 − EIinv2〉)2〉

= 〈(EIinv1 − EIinv2) + 〈EIinv1 − EIinv2〉)2〉+ 2 ∗ (EIinv1 − EIinv2)(EIinv1 − EIinv2)= V ar(EIinv1) + V ar(EIinv2) + 2 ∗ Cov(EIinv1, EIinv2)

(2.1.35)

The values V ar(EIinv1), V ar(EIinv2)&Cov(EIinv1, EIinv2) are obtained from Σα. Note that α is used equivalentto EIinv . Similarly all other elements are calculated to construct the covariance matrix Σ

(EIinvEIinv).

2.1.1 Axial Problem

Consider the same beam being acted upon by axial loads as shown in figure below. The equation governing displacementin this case is AE.u

′= F

Figure 2.3: Cantilever Beam acted upon by axial body forces

• Step 1) Discretize the beam into m number of finite elements such that the discretization length = L/m . For anyelement, the random field of axial rigidity is approximated AiEi as by spatial average technique. The randomaxial rigidity vector may be written as:

αa =[A1E1 A2E2 · · · AiEi · · · AmEm

]TThe inverse random axial rigidity vector may be written as

αa =

[1

A1E1

1

A2E2· · · 1

AiEi· · · 1

AmEm

]T(2.1.36)

• Step 2) Calculate statistical parameters mean and covariance of the random field of axial rigidity vector.Mean

µaa =[〈A1E1〉 〈A2E2〉 · · · 〈AiEi〉 · · · 〈AmEm〉

]T(2.1.37)

=[µaa1 µaa2 · · · µaai · · · µaam

]T(2.1.38)

Similarly, Mean

µaa =[µaa1 µaa2 · · · µaai · · · µaam

]T(2.1.39)

Covariance matrix , Σαa = Cov(αai, αaj)Where

Cov(αai, αaj) =σ2αa

2(k − 1)2γαa

[ (k − 1)L

m

]− 2kγαa

[kLm

]+ (k + 1)2γαa

[ (k + 1)L

m] (2.1.40)

And k=[i-j]. Asymptotic or triangulation correlation function may be used for the approximate estimation ofvariance function γ , in terms of arbitrarily chosen values of scale of fluctuation to discretization length ratios.Similarly, evaluateΣαa, covariance matrix for inverse random field.

Mathematical and Computational Approach For Stochastic Finite Element Method 111

Element stiffness matrix for any element i is: kea =AiEil

[1 −1−1 1

]Global stiffness matrix will be:

ka =

A1E1 −A1E1 0 · · · 0−A1E1 A1E1 +A2E2 −A2E2 · · · 0

0 −A2E2 A2E2 +A3E3 · · ·...

.... . .

...−AmEm

0 0 · · ·−AmEm AmEm

(2.1.41)

Mean value matrix,

µ(Ka) =1

l

µαa1 −µαa1 0 · · · 0−µαa1 µαa1 + µαa2 µαa2 · · · 0

0 −µαa2 µαa2 + µαa3 · · ·...

.... . .

...−µαam

0 0 · · ·−µαam µαam

(2.1.42)

LetΣka = 〈[ka − µ(ka)⊗ [ka − µ(ka)]〉 (2.1.43)

Each element of the matrix may be computed using the values in Σαa. Applying the boundary condition thatthe displacement at node 1 (fixed end) is zero, the first row and first column of Ka is eliminated to give Kabc

Denoting [Kabc]−1 by fabc, the flexibility matrix may be expressed as:

fabc = (Einvbc Ainvbc fcabc) = (AEinv)bc fcabc (2.1.44)

The coefficient matrix may be obtained by:

fcabc = fabc /(AEinv)bc (2.1.45)

Where matrices (Einv)bc and (Ainv)bc represents matrices constructed from inverse random fields of E(x) andA(x) respectively and

(AEinv) =

1

(AE)1

1

(AE)1

1

(AE)2

1

(AE)3

1

(AE)m1

(AE)1

1

(AE)1+

1

(AE)2

1

(AE)2

1

(AE)3· · · 1

(AE)m1

(AE)2

1

(AE)2

. . ....

...

.... . .

...1

(AE)m+

1

(AE)m−1

1

(AE)m1

(AE)m

1

(AE)m· · · · · · · · ·

1

(AE)m

1

(AE)m

(2.1.46)

Now, Expectation of

Kabc]−1) = mean(AEinv)bc fcabc (2.1.47)

LetΣfa = 〈[fabc − µ(fabc)]⊗ [fabc − µ(fabc)]〉

= [fcabc ⊗ fcabc] 〈(AEinv)bc − µ((AEinv)bc)⊗ ((AEinv)bc − µ((AEinv)bc)〉(2.1.48)

Each element of the matrix may be computed using the values in Σαa

• Step 3) Calculate the load vector, Pa. Load vector for any element i′,

112 P V Ramana, Vivek Singh

pea =∫ i+1

i

1− x

lx

l

q.dx =

1− ql

2ql

2

, then pa =

1− ql

2qlql...ql

2

is the global load vector of size (m+1,1)

Applying boundary conditions, load vector becomes Pabc =

[ql ql · · · ql

2

]T• Step 4) Evaluate mean displacements and covariance between displacements. Displacement random vector, da =

[Ka]−1PaAfter applying boundary conditions, dabc = [Kabc]

−1PabcNow mean displacement vector,

µ(dabc) = 〈dabc〉 = 〈[Kabc]−1Pabc〉 = 〈[Kabc]

−1〉Pabc= (mean((AEinv)bc fcabc)Pabc

(2.1.49)

Covariance matrix,Σda = 〈(dabc − µ(dabc))2〉 It can be seen that

V ectorisedΣda = vec(Σda) = 〈(dabc − µ(dabc))⊗ (dabc − µ(dabc))〉= 〈[(AEinv)bc fcabc)Pabc − (mean(AEinv)bc fcabc)Pabc]⊗ [(AEinv)bc fcabc)Pabc − (mean((AEinv)bc fcabc)Pabc]〉

= 〈[(AEinv)bc −mean(AEinv)bc fcabc]Pabc ⊗ [(AEinv)bc −mean(AEinv)bc) fcabc]Pabc〉= 〈[(AEinv)bc −mean((AEinv)bc) fcabc]⊗ [(AEinv)bc −mean((AEinv)bc) fcabc](Pabc ⊗ Pabc)〉

= 〈[fcabc ⊗ fcabc] [(AEinv)bc − µ((AEinv)bc)⊗ (AEinv)bc − µ((AEinv)bc)](Pabc ⊗ Pabc)〉= [fcabc ⊗ fcabc] 〈(AEinv)bc − µ((AEinv)bc)⊗ (AEinv)bc − µ((AEinv)bc)〉(Pabc ⊗ Pabc)

= Σfa(Pabc ⊗ Pabc)(2.1.50)

Now, re-arrange the elements of the resulting vector to obtain the matrix.

3 Problem Statement and Results

In this paper both static and dynamic problems has been solved. In static problems both determinate and indeterminateproblems has been solved, while in dynamic problems both linear and non-linear problems has been solved. In staticdeterminate problem a cantilever beam with axial and flexure loading and in static indeterminate a simply supportedbeam is solved. In dynamic un-damped and damped free vibration problems are solved.

3.1 Statically Determinate Linear Problem

Consider the cantilever beam of steel with loading as shown in Fig.3.1 below The following parameters are considered

Figure 3.1: Cantilever beam with transverse and axial loads

for estimating the stochastic behavior of the structure: Mean value of Modulus of Elasticity, E = 2 × 105MPa, withco-efficient of variation 0.06 Mean value of Area of Cross-section, A = 4955mm2 with co-efficient of variation 0.05 Mean

Mathematical and Computational Approach For Stochastic Finite Element Method 113

value of Moment of Inertia, I = 12.5× 107mm4 with co-efficient of variation 0.05 Estimate the first and second-orderstatistics of deflection and slope at the free end.

3.1.1 Solutions

• 1)SFEM solution:Axial displacement at free end as follows: us = PL

A(ξ)E(ξ)

Vertical deflection at free end as follows: νs =PL3

3E(ξ)I(ξ)+

wL4

8E(ξ)I(ξ)

Slope at free end as follows: Θs =PL2

2E(ξ)I(ξ)+

wL3

6E(ξ)I(ξ)

• 2)MCS solution:E and I are chosen as random variables. Random numbers are generated from Normal distribution.

• 3)FEM solution:

The matrix equation, Load vector, Kd = F is solved for displacements d , after applying boundary conditions(slope and deflection at fixed end are zero). K is the global stiffness matrix for the structure.

• 4)Analytical solution:

Axial displacement as follows: uxa =Px

AE

Vertical deflection as follows: νxa =Px2(3L− x)

6EI+wx2(x2 + 6L2 − 4Lx)

24EI

Slope as follows: Θxa =Px(2L− x)

2EI+wx(x2 + 3L2 − 3Lx)

6EI

3.1.2 Results and Discussion

• A) Mean deformations:Mean values of axial displacement, slope and vertical deflection are shown in Fig.3.2(a), for different values ofdiscretization length. Axial displacement values do not show any variation when analysed for different discretiza-tion lengths. Results for slope converge as the discretization length reduces. i.e. Above four elements (m=4)discretization, the results converges. Similar results are obtained in the case of vertical deflection too. Axial andflexural combined loading gives the same results as the results obtained from axial load alone and flexural loadalone cases, as axial deformation is independent of flexural loads and flexural deformations are independent ofaxial loads.

• B) Comparison with the results from various methods:The displacement values obtained by SFEM are compared with those from analytical method, Finite ElementMethod and Monte-Carlo Simulation Technique. Fig.3.2(b) to Fig.3.2(d) shows the comparison for various valuesof finite element size, along the length of the beam for different methods. Fig.3.2(b) compares axial displacementvalues obtained from SFEM, MCS, FEM and analytical solutions. Even without discretization, SFEM resultsconverge with analytical solutions results and so do the results from other methods. Fig.3.2(c) compares slopevalues obtained from various methods. The SFEM results are found starts converging to exact solutions resultsat m=4 and converges fully at m=7. m is the no. of random field elements. Fig.3.2(d) shows the comparativestudy for vertical deflection values. Here also, it can be observed that as the discretization length reduces, SFEMresults converges with exact solutions results and the approximate results from other methods.

• C) Standard deviation results:Standard deviation =

√variance. Variance functions are estimated approximately with respect to the asymptotic

and triangulation approximations of correlation functions in terms of the parameter scale of fluctuation, Θα .Exact variance functions with respect to Type 1 (exponential squared), Type 2 and Type 3 (exponential) are alsoobtained. Covariance values for spatial averages are then calculated using the method proposed by Vanmarcke,E.H. et al. Fig.3.3 shows the variation of standard deviation values with respect to scale of fluctuation. And italso compares the results from exact and approximate variance functions. Asymptotic approximation is observed

114 P V Ramana, Vivek Singh

to be in closer agreement with the exact solutions, as observed by Vanmarcke et al. for simple shear beams.Fig.3.3(a) depicts the standard deviation values of SFEM converges after four element discretization ( m=4).i.e. for m =5,6, the results of SFEM are almost the same. While comparing SFEM and MCS results, they showalmost similarly fashioned variation (parabolic) along the length though SFEM results starts departing from MCSresults at a particular length (say 0.5 m). Variance of beam tip deflection values differ within permissible valuesbetween MCS and SFEM (Fig.3.3(c)).Similar variations are obtained for slopes and axial displacements. But thestandard deviation values for axial displacements were found to be negligible.

Mathematical and Computational Approach For Stochastic Finite Element Method 115

(a) Mean values of slope and vertical deflection (b) Mean values of axial displacement values

(c) mean slope values (d) Comparison of mean vertical deflection values

Figure 3.2: mean values

116 P V Ramana, Vivek Singh

(a) Standard deviation of vertical deflection (b) Standard deviation of the vertical deflection

(c) Beam tip deflection

(d) Probable deviation of deflection values

Figure 3.3: Standard deviation values

Mathematical and Computational Approach For Stochastic Finite Element Method 117

3.2 Statically Indeterminate Linear Problem

Consider a steel beam hinged at both the ends (statically indeterminate) with loading as shown in Fig.3.4 below The

Figure 3.4: Simply supported beam with axial and flexural loads

following parameters are considered for estimating the stochastic behavior of the structure: Mean value of Modulus ofElasticity, E = 2 × 105MPa, with co-efficient of variation 0.06 Mean value of Area of Cross-section, A = 4955mm2

with co-efficient of variation 0.05 Mean value of Moment of Inertia, I = 12.5 × 107mm4 with co-efficient of variation0.05 Estimate the first and second-order statistics of deflection and slope at the free end.

3.2.1 Solutions

• 1)SFEM solution:

Vertical deflection at mid span as follows: νa =5wL4

384E(ξ)I(ξ)+

PL3

48E(ξ)I(ξ)

Slope at free end as follows: Θs =PL2

16E(ξ)I(ξ)+

wL3

24E(ξ)I(ξ)

• 2)MCS solution:E and I are chosen as random variables. Random numbers are generated from Normal distribution.

• 3)FEM solution:

The matrix equation, Load vector, Kd = F is solved for displacements d , after applying boundary conditions(slope and deflection at fixed end are zero). K is the global stiffness matrix for the structure.

• 4)Analytical solution:Vertical deflection as follows:When x < L

2 , νxa = 1EI (wL+P12 x3 − w

24x4 − PL2

16 x−wL3

24 x)

When x > L2 , νxa = 1

EI (wL+P12 x3 − w24x

4 − PL2

16 x−wL3

24 x−p(x−L

2 )3

6 )

Slope as follows:

When x < L2 , Θxa = 1

EI (wL+P4 x2 − w6 x

3 − PL2

16 −wL3

24 )

When x > L2 , Θxa = 1

EI (wL+P4 x2 − w6 x

3 − PL2

16 −wL3

24 −P (x−L

2 )2

2 )

3.2.2 Results and Discussion

• A) Mean displacementsMean values for vertical deflection, slope and axial displacement obtained after analysis are as shown in Fig.3.5

• B) Comparison with the results from various methodsThe displacement values obtained by SFEM are compared with those from analytical method, Finite ElementMethod and Monte-Carlo Simulation Technique. Fig.3.6 shows the comparison for various values of m along thelength of the beam for different methods.

Similar variations are obtained for slopes and axial displacements. But the standard deviation values for axialdisplacements were found to be negligible

118 P V Ramana, Vivek Singh

(a) Mean values for vertical deflection, slope and axial displace-ment

(b) Mean values of vertical displacement values

(c) Mean values of vertical displacement values (d) mean rotation values

Figure 3.5: mean values

Mathematical and Computational Approach For Stochastic Finite Element Method 119

(a) Standard deviation of vertical deflection (b) Standard deviation of the vertical deflection

Figure 3.6: Standard deviation values

3.3 Linear Dynamic Problem

Governing differential equation of motion un-damped free vibration is: mu + ku = 0 The structural parameters massm is taken as 1kg and stiffness k is taken as 19.72kN/m. The initial conditions are u(0) = 0.1m and u(0) = 0.2m/s.The equation is solved using Runge-Kutta fourth order (RK4) and MATLAB (using Symbolic Math Toolbox) solutionsresults are compared with exact solutions

3.3.1 Solutions

• 1) 4th order Runge-Kutta (RK4) solution:The general differential equation given above can be transferred numerically by putting y = u. 2nd order ODEreduces to 1st order ODE my + ku = 0 and it yields to (y = −(k/m))u which can be solved by RK4.

• 2) MATLAB solution:The MATLAB solution as follows,

um =((exp((t(

√(−km)/m)((u0m+ u0

√(−km))))

((2√

(−km)))−

((u0m− u0

√(−km)))

((2exp(((t√

(−km))/m))√

(−km)))(3.3.1)

Where u0 and (u′0)are initial values (at t=0) of displacement and velocity respectively. And

um =((exp(((t(

√(−km)/m))((u0m+ u0

√(−km)(

√(−km)/m))))

((2√

(−km)))+

((u0m− u0

√(−km)(

√(−km)/m))

((2exp(((t√

(−km)/m))√

(−km)))(3.3.2)

• 3)Analytical solution:

The exact solution as follows, ua = u0cosωnt+u0ωnsinωnt

Where u0 and u0 are initial values (at t=0) of displacement and velocity respectively. And ua = −u0ωnsinωnt+u0cosωnt

120 P V Ramana, Vivek Singh

(a) Displacement and velocity values (b) Phase-Space diagram

Figure 3.7: un-damped free vibration problem

3.3.2 Results and Discussion

The general equation mu+ ku = 0 is solved in various approaches namely, RK4, MATLAB symbolic and analytically.The results are depicted as shown below and the Fig.3.7(a) shows that the displacement and velocity w.r.t time and theRK4, MATLAB symbolic compared with analytical result. It is also observed that RK4, MATLAB symbolic solutionsdisplacement and velocity matching very well with the analytical solution. Similarly, Fig.3.7(b) shows the phase spacebetween displacement and velocity with RK4, MATLAB symbolic and analytically. It is also observed that RK4,MATLAB symbolic solutions well agreed with the analytical solution. In case RK4, the step size is small the resultsshow good agreement, otherwise it gives the error solution and for this it requires more computational efforts for smallstep size. Applying the initial conditions at u(0) = a = 0.1m and u(0) = b = 0.2m/s, and the time interval is chosento vary from 0 to 20 sec. All phase portrait shows that a remarkably good match that of RK4, MATLAB symbolicand analytical solutions. The real time over which the above is achieved is expressed in terms of fundamental periodof the linear part of the system. It is achieved in about 5 sec, which means with in a real time (t) = alphaT . Wherealpha is 20/10 = 2, here T is linear fundamental period. It gives results very closely matching with the MATLAB, RK4solution up to a real time (t) as already mentioned equal. The above technique works well for all levels of nonlinearitieslike weak to strong.

3.4 Non-linear Dynamic Problem

Solving the non-linear problem with FEM and Monte-Carlo simulation method is very difficult, for solving this typeof equation Brownian motion approach should be adopted. The coefficients of stiffness are nonlinear in DH oscillatorequation also. Governing differential equation of motion in this case is: mu+cu+k(u2−1)un = F0cosωt The structuralparameters mass m is taken as 1kg. And stiffness k is taken as 19.72kN/m and c as 1.57 N-s/m. F0 = 15.78kN andω = 2πrad/s. Initial conditions are u(0) = 0.1m and u(0) = 0.2m/s Taking n=1, the equation is solved using RK4 andMATLAB ODE solver solutions and the RK4 results are compared with MATLAB ODE45 solver solutions results.

3.4.1 Solution

• 1) 4th order Runge-Kutta (RK4) solution:The general differential equation given above can be transferred numerically by putting y = u The 2nd order ODEreduces to 1st order ODE my+cy+k(u2−1)un = F0cosωt and it yields to y = (F0/m)cosωt−(c/m)y−k(u2−1)un

Mathematical and Computational Approach For Stochastic Finite Element Method 121

(a) Displacement and velocity values (b) Phase-Space diagram

Figure 3.8: DH oscillator equation problem

which can be solved by RK4.

• 2) MATLAB solution:The nonlinear differential equation my+ cy+k(u2−1)un = F0cosωt is solved in MATLAB using MATLAB ODEsolver.

• 3)Analytical solution:No exact solutions are available for the considered problem.

3.4.2 Results and Discussion

The general equation mu+ cu+ k(u2 − 1)un = F0cosωtis solved in various approaches namely, RK4, MATLAB ODEsolver. The results are depicted as shown below and the Fig.3.8(a) shows that the displacement and velocity w.r.ttime and the RK4 results are compared with MATLAB ODE solver result. It is also observed that RK4 solutionsdisplacement and velocity matching very well with the MATLAB ODE solver solutions. Similarly, Fig.8(b) shows thephase space between displacement and velocity with RK4 and MATLAB ODE solver. It is also observed that RK4solutions well agreed with MATLAB ODE solver solution. In case RK4, the step size is small the results show goodagreement, otherwise it gives the error solution and for this it requires more computational efforts for small step size.Applying the initial conditions at u(0)= a =0.1m and u(0)= b = 0.2m/s, and the time interval is chosen to vary from0 to 20 sec. All phase portrait shows that a remarkably good match that of RK4 and MATLAB ODE solver. Yetanother application of dynamic problem is shown for Duffing-Holmes equation. It is a nonlinear dynamic problem. Thecomplexity here is particularly, the influence of the initial conditions on the transient part of the response. To achieveaccuracy for much larger duration viz. for more number of cycles, one needs more number of iterative cycles in mostof the numerical methods including SFEM. However it is observed that in analytical values are close to the MATALBresponse in a few cycles only. The real time over which the above is achieved is expressed in terms of fundamentalperiod of the linear part of the system. It is achieved in about 5 sec, which means with in a real time (t) = alphaT. Where alpha is 20/7 = 2.86, here T is linear fundamental period. It gives results very closely matching with theMATLAB, RK4 solution up to a real time (t) as already mentioned equal. The above technique works well for all levelsof nonlinearities like weak to strong.

122 P V Ramana, Vivek Singh

4 Conclusions

In this paper, mainly concentrated on different types of problems in statics and dynamics are enhanced. In staticproblems, the variability has been taken in both material properties and geometry for determinate and indeterminatestructures. For static loads, structural response variability with respect to the random input quantities are studiedusing Stochastic Finite Element Method. Observations made are as follows:

4.1 CONCLUSIONS FOR STATIC PROBLEMS

1. As the discretization length reduces, the response value converges to the exact solution.

2. For axial loads, the stochasticity in axial rigidity does not result in a comparably significant randomness indisplacement with that for vertical loads.

3. The results from SFEM, is comparable with FEM, MCS and analytical solutions results.

4. SFEM gives slightly higher standard deviation values than that obtained by MCS technique.

5. Propagation of stochasticity along the length of the beam is found to be increasing towards maximum responsevalue point.

6. Asymptotic assumption of variance function was found to be in close agreement with the exact variance functions( as observed by Vanmarcke,E.H.)

4.2 CONCLUSIONS FOR DYNAMIC PROBLEMS

1. In dynamic problems, the structural response variation is estimated for linear as well as non-linear problems.Linear dynamic equations are solved by Runge-Kutta fourth order method and MATLAB symbolic solutions andthe results are compared with analytic solutions. The displacement and velocity values are observed to be foundto be matching well between the above mentioned methods. The phase space diagram is also obtained with goodcomparability between the three methods.

2. Non-linear dynamic problems of Duffing and Duffing-Holmes equations are solved by Runge-Kutta fourth ordermethod and compared with MATLAB ODE solver solutions results. In the case of displacement and velocityvalues, it is observed that the results are close to each other between these methods. The phase space diagramalso depicts close comparability.

References

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[6] Babuska, I., and Rheinboldt, W.C. (1978), Error estimates for adaptive finite element computations. SIAM Journalon Numerical Analysis, 15(4):736-754.

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[8] Sudret, B., and Der Kiureghian, A. (2000), Stochastic finite element methods and reliability - A state-of-the-artreport. Structural Engineering, Mechanics and Materials, Department of Civil and Environmental EngineeringUniversity of California, Berekely.

[9] Manohar, C.S., and Manjuprasad, M. (2007), Adaptive random field mesh refinements in stochastic finite elementreliability analysis of structures. CMES, 19(1):23-54.

[10] Ghanem, R. (1999), Stochastic finite elements with multiple random non-gaussian properties. Journal of Engineer-ing Mechanics, ASCE, 125: 26-40.

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[12] Deodatis, G. (1990), Bounds on response variability of stochastic finite element systems. Journal of EngineeringMechanics, ASCE, 116(3):565-585.

[13] Deodatis, G. (1991), The weighted integral method, I: stochastic stiffness matrix. Journal of Engineering Mechanics,ASCE, 117(8):1851-1864.

[14] Deodatis, G., and Shinozuka, M. (1991), The weighted integral method II: response variability and reliability.Journal of Engineering Mechanics, ASCE, 117(8):1865-1877.

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[18] P.V. Ramana and Vivek Singh(2013), The Magnitude Of Linear Problems, International Conference on Emerg-ing trends in Engineering & Applied Sciences-2013 (ICETEAS), published in International Journal of AdvancedEngineering & Computing Technologies, ISSN: 2249-4928.

[19] P.V. Ramana and Vivek Singh(2014), The Emerging Solution For Partial Differential Problems, SEC-2014, IITDelhi, India,(SEC14BAT), ISBN: 978-81-322-2189-0.

[20] P.V. Ramana and B K Raghu Prasad, ”Modified Adomian Decomposition Method for Van der Pol equations”,International Journal of Nonlinear Mechanics, Volume 65, October 2014, Pages 121132, 2014

[21] P.V. Ramana and B K Raghu Prasad, ”Modified Adomian decomposition method for fracture of laminated uni-directional composites”, Sadhana Vol. 37, Part 1, 33-57, 2012.

P V Ramana, working as an assistant professor at Malaviya National institute of Technology,Jaipur, India. He received his PhD from Indian institute of Science, Bangalore India, and M. Tech from Indian instituteof Technology Bombay. His research interests include computational mechanics, Non-linear structural mechanics,Fine-aggregate for concrete structures as outlet and structural Health Monitoring. He has published papers in 10International journals and in 60 National journals.

Vivek Singh is currently pursuing his Masters degree in Structural Engineering from Malaviya Nationalinstitute of Technology, Jaipur, India. He received his Bachelors Degree from Mewar University, Gangrar, India. Hisresearch interests include numerical stimulation of linear and nonlinear Structural Static Problems. He has presentedpapers in 2 National conferences and 2 International conferences.