Structural reliability homework

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Reliability analysis based optimal design of monolithic RC beam-slab ... To find the reliability index of the beam and slab we examine, we write up the limit state functions. ... The type of concrete is C25/30, the reinforcing steel is S500B. The dead load is calculated from the density and the thickness of the applied materials.
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Budapesti Műszaki és Gazdaságtudományi Egyetem Hidak és Szerkezetek Tanszék

Structural reliability Reliability analysis based optimal design of monolithic RC beam-slab 2013/2014 I. félév

Lecturer: Dr. Kovács Tamás

Students: Caeiro Sénica Duarte João [L2UQSZ] Carioti Mattia [JIAUHE] Gábor Edit [JDTHH9] Kotecki Florian [WE2A3F] Kámán Csaba [RTN895] Lechard Antoine [CNN80F] Pap Zsuzsa Borbála [AGM7UG]

Structural reliability homework Reliability analysis based optimal design of monolithic RC beam-slab Supervisor: Dr. Kovács Tamás Members of the group: Caeiro João, Carioti Mattia, Gábor Edit, Kotecki Florian, Kámán Csaba, Lechard Antoine, Pap Zsuzsa Borbála

1. Outline the problem and the solution strategy

The task is to design the beam-slab of monolithic reinforced concrete building, based on reliability analysis. Aim of the optimization is to determine the minimum cost design of the structural slab. General requirement is to reach or exceed the safety level prescribed by the Eurocodes. The reliability analysis should be performed considering all the relevant design checks. Mechanical and reliability models shall be developed in accordance to Eurocodes and JCCS/EC, respectively. In our solution, first, we chose the stochastic and deterministic variables of both the action and resistance side. After this, we assume a distribution type of the stochastic variables, and calculate the parameters. We chose the deviations according to the recommendations of the JCSS. To find the reliability index of the beam and slab we examine, we write up the limit state functions. From the parameters and the failure functions we calculate the reliability index, with the help of Matlab/FERUM. We make the design checks according to the Eurocode’s partial factor method, too. After this we make a parametric study. To this we change the deviations in the program, and see, which of them has big effect on the result (rel. index). According to the parametric study and the quest for the minimum cost we can give a proposal for the optimal structural configuration. As for the cost optimization, we look for the minimum cost of the examined structure by changing the ratio of the amount of applied concrete and reinforcement. Finally, we calculate the partial factors of the proposed configuration, based on the reliability index of the structure and the sensitivity factors and distribution types of each variable. !

2. Determine the relevant design checks Given a two-storey building with an area of 10x18 meters, with 6 meter spacing in longitudinal direction and 5 meter in the other direction. The height of the structure is 2x3.30 meter. The slab is divided into six blocks by the beams running in two directions. The beams are supported by columns at their sectional points. Slab: We analyze the slab for bending at three points: the middle cross-section with positive bending moment in both x and y direction, and two cross-sections on the edges, one in x, one in y direction with negative bending moments. For the sake of simplicity we calculate only the elastic moments, no plastic moment redistribution is needed.

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Beam: We analyze the beams for bending at two and for shear failure at one cross-section: in the middle with positive bending moment, at the support with negative bending moment and shear force.

3. Define the design variables and determine their stochastic models based on literature review. We separate the variables into two parts: one for the actions and the effect of them, and one for the resistance. At the action side there are dead loads that follow normal distribution and live loads following gamma distribution. The uncertainty of the load effects can be taken into account with a parameter following lognormal distribution. We can divide the resistance side into two parts: material properties and geometry. The material strengths follow lognormal distribution as well as the variable for the uncertainty of resistance. As for the geometry, the distance between the reinforcing bars and the slab or beam bottom is a design variable following gamma distribution. The other parameters of the geometry are deterministic variables. The mean values are calculated based on both the recommendations of the EC and JCSS. The values of standard deviations and coefficients of variation are coming from the JCSS. In the Eurocode the characteristic and design value of a design variable is connected to a given quantile of the distribution function which yields a conservative solution. Our aim is to design the structure by taking into account the structural reliability index and the sensitivity factors of each variable. The type of concrete is C25/30, the reinforcing steel is S500B. The dead load is calculated from the density and the thickness of the applied materials. The live load is coming from the utilization of the structure, here the building is designed to be an office. The calculation of live load based on the recommendations of JCSS also takes into account the possible distribution of the loads acting on the slab.

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4. Complete the design checks in accordance to Eurocode partial factor method. We calculate the internal forces acting at the examined cross-sections using influence diagrams for beams and the Bares tables for the slab, then the necessary amount of reinforcement with the given geometry according to EC. The cross-sections are designed to have approximately 100% utilization but at some cross-sections this requirement cannot be satisfied because of the prescribed minimum distances between the reinforcement bars. At the shear analysis we can see that the concrete itself does not give enough load bearing capacity, in our further calculations its effect on the whole reinforced cross-section is neglected for the sake of simplicity. 5. Calculate the structural reliability index, regarding all design checks/failure modes. We calculate the beta reliability index for all failure modes using a software called FERUM (Finite Element Reliability Using Matlab). This program includes algorithms to perform FORM and simulation reliability analysis. The input data contains the design variables x from 1 to 7 and the failure functions expressed with the help of these variables. Fist we determine the reliability index for the original reinforcement area calculated based on the EC partial factor method. slab point 1x reliability index

slab point 1y

5,687

slab point 2

6,651

5,192

slab point 3

beam moment positive

5,359

beam moment negative

4,259

Beam shear

4,977

4,498

We can see that these values are greater than the prescribed 3.8 (for 50 year reference period, RC2 reliability class). Let us decrease the amount of applied reinforcement, and observe the changes in the structural reliability index:

Finding Optimum Reinforcement for Slab 7

reliability index (beta)

6 5 slab point 2

4

Beta=3,8 slab point 1 y

3

slab point 1 x

2

slab point 3

1 0 100%

90%

80%

70%

60%

relative reinforcement quantity

4

50%

40%

Finding Optimum Reinforcement for Beam 6

reliability index (beta)

5 4

Beam moment positive Beam moment negative

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Beam shear Beta=3,8

2 1 0 100%

95%

90%

85%

80%

75%

70%

65%

relative reinforcement quantity

The horizontal axis shows the relative reinforcement area where 100% means the necessary reinforcement according to Eurocode. The vertical axis shows the value of the structural reliability index. The economically most favourable case is when the applied reinforcement yields the prescribed 3.8 beta value. The optimum is different in each crosssection, the results can be seen in the chart below:

slab point 1 x slab point 1 y slab point 2 slab point 3 beam positive beam negative beam shear

original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum original reinf. (mm2/m) optimum

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314,00 168,96 314,00 126,97 628,00 414,77 628,00 383,76 1110,00 935,57 1885,00 1404,47 1210,00 1031,14

6. Complete parametric study to investigate the sensitivity of design criteria to the design variables. We observe the sensitivity of design criteria by changing the standard deviation of the design variables. From the original 100% first we decrease it by 25 then by 50%, then we increase it by the same percentages. The sensitivity factor can be regarded as a weight factor, it shows the the weight of each variable appearing in the failure function. In order to calculate it, we should derive the failure function by each variable and multiply them by their standard deviation. The ‘normalized’ value is the sensitivity factor alpha. The sum of the square of alpha values must equal to one. We analyzed the following cross-sections: slab at point 1 in direction x, and beam with negative moment and shear force. The results can be seen in the diagrams:

Variation of standard deviation (slab 1 x) 4,500

4,000 dead load

3,500

uncertainty of load effect yield strength concrete strength

3,000

uncertainty of resistance

2,500

2,000 50%

75%

100%

125%

6

150%

Variation of standard deviation (beam M negative) 4,500 4,000 dead load

3,500

uncertainty of load effect yield strength concrete strength

3,000

uncertainty of resistance

2,500 2,000 50%

75%

100%

125%

150%

Variation of standard deviation (beam x shear) 4,500 4,000 dead load

3,500

uncertainty of load effect yield strength

3,000

uncertainty of resistance

2,500 2,000 50%

75%

100%

125%

150%

The horizontal axis shows the relative value of standard deviation compared to the original value (100%) based on the recommendations of JCSS. The vertical axis contains the changes of the structural reliability index. The 100% relative standard deviation yields the prescribed 3.8 beta value. By increasing the standard deviation, the uncertainty of the values of design variables increase, too. It means that the reliability of our structure decreases. The velocity of this change can be measured by calculating the tangeant of the diagram of each variable. We can see above that the live load is missing from the variables. The explanation is that the software could not handle variables following gamma distribution during the 7

sensitivity check. Fortunately we can manage this default by observing the changes in the uncertainty of load effects. The dead load with alpha values 0.191 at slab and 0.334 at beam is not that sensitive for the changes in standard deviation. Nevertheless the uncertainty of load effect -which consist of the dead load and the live loads- has the greatest alpha value (0.636 at slab and 0.863 at beam). That means the diagram of the live load would look like and be close to the diagram of the uncertainty of load effects, showing that these variables are sensitive to the changes of standard deviation. On the resistance side, the yield strength and the uncertainty of resistance shows small sensitivity because their alpha values are between 0.161 and 0.305. The diagram of concrete strength is a horizontal line according to its alpha value which is zero. At the shear failure analysis, the concrete strength is missing from the variables, because as it was mentioned before, the load bearing capacity is mostly coming from the strength of reinforcing bars. By summarizing our observations, we can declare that the action side is much more sensitive for the changes in standard deviation than the resistance side, and within the action side, the live loads are more sensitive than the dead loads. For the sake of simplicity our structure has only dead load and live loads coming from structural and non-structural elements. By having other type of actions acting on our structure, the sensitivity might increase, because for example wind loads, snow loads etc. have more uncertainty in their behaviour then the applied loads in this case. Researches regarding the loads and the effects coming from them might decrease the uncertainty, that can result in a more safe and economical structure. 7. Give proposal for the optimal structural configuration. In this part we are about to satisfy the main goal of the project: we want to design a structure that has minimal cost besides reaching the prescribed reliability level. We optimize our structure regarding the cost functions of concrete and reinforcing steel. The given data: the cost of concrete is 100.000,- HUF per m3, the cost of steel is 300,HUF per kilogram. At the slabs and beams, their span is given so we can only modify the depth or the width and height of them and the area of reinforcement. We determine the cost functions and the failure functions regarding three failure modes: one cross-section at slab, one at beam with negative moment and shear force. The iterations are accomplished by the software FERUM.

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The results at slab:

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In the first diagram we can see the variables x8 and x9 that stand for the slab depth and the reinforcement area, accordingly. Their normalized value means that the original value of them is 1.0 and by decreasing or increasing them their multiplicators will be smaller or greater than 1.0. We can see that we reached the optimum value by 25 iteration steps, that needed about 30 minutes computational time. The green line shows that in order to decrease the costs we must use more reinforcement, and at the same time the blue line stands for the need for reducing the cross-sectional area of concrete. At the optimal configuration the reinforcement area should be ~150% and the depth of the slab should be ~70% of their original values. Of course, there can be a limit for both values. We might introduce a lower limit for the depth of the slab, and an upper limit for the amount of reinforcement. In the second diagram we can observe how the iteration converges to the optimal solution. After the 17th step we are close to the final result. The normalized cost will be ~72% of the original cost, it means we can save more than 25% at each slab. With this method at a multistorey building with large sizes, we can design a very economical building. In the third diagram the changes of the reliability index can be seen. The lower limit is the prescribed 3.8 value, that was reached at the beginning of the iteration, namely in the 5th step. After this, the reliability of our structure does not change significantly. The results at beam (shear analysis):

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Here, the design variables x7,x8,x9 are the height and the width of the beam, and the reinforcement area, accordingly. The red line shows that at the optimal configuration our structure needs ~155% shear reinforcing comparing to the original 100%. In this case, the height of the beam would increase to its ~70% and the width to its ~30% value. It yields a very tall, very slender beam, which resembles to the precast reinforced concrete beams, but they broaden at their top in order to avoid stability failures. On the other hand, at a multistorey building, having tall beams means the vertical structural elements should be taller as well, and it might result in a less economical configuration than the original one. In order to have reasonable sizes, we can limit for example the ratio of the width and the height of the beam. In the second diagram we can see that the normalized cost will be only the ~30% of the original cost. At the first iteration step, there is an increase in the cost, because the algorithm of this iterative method depends greatly on the initial data, it might find local extrema that is not favourable for the global solution. In the third diagram there are the changes of the reliability index in the function of iteration steps, and as we mentioned before, at the first iteration step there is some kind of discrepancy because of the algorithm the software is using. Summarizing our experiences, the reliability based design might take much more time than the partial factor method, but it leads to a structural configuration that is more economical and has at least the same reliability index as structures designed in accordance to Eurocode. We highly recommend to use reliability analysis based design in the future.

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8. Calculate the reliability of the proposed configuration and determine the partial factor applicable in EC design methods. The calculation of the structural reliability can be seen in chapter 7. The partial factor transforms the characteristic value of a design variable to the design value of the same variable. In the Eurocodes, these values are given as a certain quantile of the distribution functions. On the action side the characteristic value of an individual action is usually given as the 50% quantile, which must be transformed into the 95% quantile standing for the design value. Here, the partial factor is the ratio of the design value and the characteristic value. We use the partial factor in order to design more safe structures, so the partial factor must not be smaller than 1.0. On the resistance side, the characteristic value is the 5%, the design value is the 0.1% quantile of a material or geometrical property. The partial factor is the ratio of the characteristic and the design value. The variables follow three types of distribution function: normal, lognormal and gamma. The latter is approximated with Gumbel distribution because we have no data on the skewness of the distribution of our variables. In order to use the formula given for the normally distributed variables, which yields the design value of the variable, we fit normal ditstribution functions to the other distribution types and look for the replacing standard deviation. From these data we can determine the partial factor for each design variable.

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