Fully probabilistic optimization of reinforced E concrete ...½_PPT.pdfFully probabilistic...

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2020 XVI INTERNATIONAL CONFERENCE ON STRUCTURAL REPAIR AND REHABILITATION PORTO | PORTUGAL 13-15 JULY 2020 ONLINE CONFERENCE Fully probabilistic optimization of reinforced concrete elements using heuristic algorithm and neural network Venclovský J., Štěpánek P., Laníková I. Brno University of Technology, Faculty of Civil Engineering

Transcript of Fully probabilistic optimization of reinforced E concrete ...½_PPT.pdfFully probabilistic...

Page 1: Fully probabilistic optimization of reinforced E concrete ...½_PPT.pdfFully probabilistic optimization of reinforced concrete elements using heuristic algorithm and neural network

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XVI INTERNATIONAL CONFERENCE ON

STRUCTURAL REPAIR AND REHABILITATIONPORTO | PORTUGAL 13-15 JULY 2020

ONLINE CONFERENCE

Fully probabilistic optimization of reinforced concrete elements using heuristic algorithm and neural network

Venclovský J., Štěpánek P., Laníková I.

Brno University of Technology,Faculty of Civil Engineering

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Introduction

Many mathematical programming applications in civil engineering, e.g.

• structural design

• water management

• reconstruction of transportation networks

• reliability related computations

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Structural Design

Two ways of structural designing

• partial reliability factor method• all uncertainties are replaced with their characteristic values

• calculation is simple

• obtaining input data is easy

• fully probabilistic design• complex time-consuming computation

• lack of input data

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Design Optimization

Two ways to optimize structural design

• DBSO (Deterministic-Based Structural Optimization)• based on partial reliability factor method

• provides suboptimal solution

• quick

• RBSO (Reliability-Based Structural Optimization)• based on probabilistic design

• „better“ solution

• complex and time-consuming

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DBSO Definition – Objective Function

Objective is to minimize beam’s price

min(𝐶C𝑉C(𝒙) + 𝐶S𝑊S(𝒙))

𝐶C , 𝐶S – cost of concrete per m3, cost of steel per kg

𝑉C , 𝑊S – volume of concrete [m3], weight of steel [kg]

𝒙 – vector of design variables

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DBSO Definition – Design Variables

The work focuses on beams• e.g. simply supported beam with this scheme and cross section

• here then 𝒙 = (𝑏1, 𝑏2, 𝑏3, ℎ1, ℎ2, ℎ3, 𝐴𝑠1, 𝐴𝑠3)

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q

Nb2b2

b3

h1

h2

h3

h

As1

As3

b1

5 m

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DBSO Definition – ULS

First constraint is ULS – ultimate limit state

• beam has to bear the prescribed load strains in concrete and steel cannot exceed prescribed values

𝜀 ≥ 𝜀𝑐,min in cross section vertices

𝜀 ≤ 𝜀𝑠,max in reinforcing steel positions

𝜀 – actual strain value

𝜀𝑐,min – minimal allowed strain of concrete

𝜀𝑠,max – maximal allowed strain of steel

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DBSO Definition – SLS

Second constraint is SLS – serviceability limit state

• more possibilities, here it is limitation of deflection

• beam’s deflection cannot exceed prescribed maximum value

𝑤 ≤ 𝑤𝑚𝑎𝑥

𝑤 – beam’s deflection

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DBSO Definition – FEM

• Mathematical model is rather complex

• To be able to assess ULS and SLS it is necessary to proceed with matrix calculations

𝑲𝒖 = 𝑭

𝑲 – stiffness matrix; depends on 𝒙

𝑭 – vector of load; depends on load

𝒖 – vector of node deflections; strain 𝜀 and

deflection 𝑤 depend on it

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RBSO Definition

• same as DBSO, except• some variables become random variables (such as load, material characteristics of concrete and

steel, etc.)

• ULS and SLS are redefined as probabilistic constraints

• ULS

𝑃 𝜀(vertices) ≥ 𝜀𝑐,min ∧ 𝜀(steel) ≤ 𝜀𝑠,max ≥ 𝑝ULS

• SLS𝑃 𝑤 ≤ 𝑤𝑚𝑎𝑥 ≥ 𝑝SLS

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Solution

• the probability 𝑝ULS should reach the level from 1-10-4 to 1-10-6 (depending on chosenreliability class) , which is obviously troublesome to achieve

• large number of scenarios makes the problem insolvable in a reasonable time horizon

• therefore a different approach is proposed• this is based on algortihm divided into inner deterministic optimization (based on Reduced Gradiant

Method) and outer stochastic optimization (based on Regression Analysis)

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Example

• shown on the mentioned simply supported beam

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q

Nb2b2

b3

h1

h2

h3

h

As1

As3

b1

5 m

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Example

• normal force N and distributed load q are random variables with gamma distribution

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q [kN/m]

0

0.7

p

30 350

0.7

5 10N [kN]

p

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Solution – 1. Initialization

• The calculation is initialized by choosing 12 initialization scenarios

𝝃𝑘init = 𝑞𝑘

init, 𝑁𝑘init , 𝑘 = 1,… , 12

• For these scenarios, the deterministic optimization is performed, giving design variables’ values 𝒙𝑘

init and values of objective function 𝑓𝑘init

• Probabilities 𝑝U𝑘init, 𝑝S𝑘

init that configuration 𝒙𝑘init satisfies the ULS and SLS, are assessed

using Neural Network on Monte Carlo Simulation

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Solution – 2. Iteration

• Regression analysis is applied on known scenarios to approximate behavior ofobjective function values and probabilities with regards to values of q and N

• Next scenario 𝝃𝑙iter is selected based on these approximations (so that it satisfies set

probabilities and has minimal objective function value)

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Solution – 2. Iteration

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f [CZK]

q [kN/m]N [kN]

4400

380035

30 10

5

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Solution – 2. Iteration

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pU []

q [kN/m]N [kN]

1

0

35

30 10

5

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Solution – 2. Iteration

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pS []

q [kN/m]N [kN]

1

0

35

30 10

5

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Solution – 2. Iteration

• The deterministic optimization is performed for the new scenario 𝝃𝑙iter, giving design

variables’ values 𝒙𝑙iter and values of objective function 𝑓𝑙

iter

• Also, probabilities 𝑝U𝑙iter, 𝑝S𝑙

iter that configuration 𝒙𝑙iter satisfies the ULS and SLS, are

assessed using Neural Network on Monte Carlo Simulation

• If 𝑝U𝑙iter ≥ 𝑝ULS and 𝑝S𝑙

iter ≥ 𝑝SLS• than 𝒙𝑙

iter is solution, end

• else, 𝝃𝑙iter is added to regression and solution continues with next iteration

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Solution – 2. Iteration

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f [CZK]

q [kN/m]N [kN]

4400

380035

30 10

5

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Solution – 2. Iteration

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pU []

q [kN/m]N [kN]

1

0

35

30 10

5

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Solution – 2. Iteration

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pS []

q [kN/m]N [kN]

1

0

35

30 10

5

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Results

• Results• desired probabilities 𝑝ULS = 99.99277 %, 𝑝SLS = 93.31928 %

• obtained probabilities 𝑝𝑈 = 99.9928 %, 𝑝𝑆 = 99.9324 %

• solution found for scenario 𝑞 = 34.4103 kN/m, 𝑁 = 5.3090 kN

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The End

Thank you for your attention

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