Computational Modelling of Concrete Structures - Bicanic et al. (eds) ©2010 Taylor & Francis Group, London, ISBN 978-0-415-58479-1

Computation of optimal concrete reinforcement in three dimensions P.C.J. Hoogenboom Delft University of Technology Faculty of Civil Engineering and Geosciences, Delft, The Netherlands A. de Boer Ministiy of Transport, Public Works and Water Management, Centre for Public Works, Utrecht, The Netherlands

ABSTRACT: A method is proposed for determining the required reinforcement based on stresses that have been computed by the finite element method using volume elements. Included are, multiple load combinations, compression reinforcement, confinement reinforcement and crack control. The method is illustrated by several stress examples and a structural example.

1

INTRODUCTION

Many computer programs for structural analysis have post processing functionality for designing reinforcement and performing code compliance checks. For example the moments and normal forces computed with shell elements can be used to determine the required reinforcement based on the Eurocode design rules [ 1 , 2, 3]. However, for finite element models containing volume elements the codes do not provide design rules. Software companies that are developing structural analysis programs are in the process of extending their program capabilities with volume elements. Consequently, also the algorithms for computing reinforcement requirements need to be extended for use with volume elements. In 1983, Smimov pointed out the importance of this problem for design of reinforced concrete in hydroelectric power plants [4]. In 1985, Andreasen and Nielsen derived formulas for the optimal reinforcement for three-dimensional stress states [5]. They also designed a flow chart for determining which formula to use. In 1994, Kamezawa et al. proposed and tested several formulas for three-dimensional reinforcement design [6]. In 2002 and 2003, Foster, Marti and Mojsilovic published two thorough studies on the subject [7, 8]. In 2008, Hoogenboom and de Boer used analytical and numerical methods for computing three-dimensional reinforcement requirements [9]. This paper continues on this path. A n algorithm is proposed for computing the optimal reinforcement for multiple load combinations. The load combinations are related to the ultimate limit state or the serviceability limit state. Not only tension reinforcement is considered but also compression reinforcement and confinement reinforcement. A maximum crack width is imposed forthe serviceability limit state. Numerical

639

results are compared to analytical results of elementary stress states. The algorithm has been implemented in a finite element program. A structural example is included.

2

PROBLEM FORMULATION

It is assumed that reinforcing bars are present in the X, y and z direction only (Fig. 1, App. 2). The reinforcement ratios are AT, Py and p^, respectively. The smallest amount of reinforcement is obtained when the volume reinforcement ratio is minimised. Minimise

+ Py + Pz

(1)

Clearly, the reinforcement ratios need to be positive which gives a constraint on the solution.

Figure 1. Elementary part of reinforced concrete. Shown are a crack and the reinforcing bars that bridge this crack.

Av > O

3

Py>0

CONCRETE STRESSES

(2) The stresses in a structural part can be computed in a linear or non-linear finite element analysis. In this paper the stress tensor in a point is written as

Pz > O

For load combinations related to the ultimate limit state the stresses a^^^t, 0 and /s < 0 than tension reinforcement is not needed. Second, ignore the sets o f formulas that give negative reinforcement ratios. Third, c a l c u l a t e a n d 7^2 by Eqs (13), (10), (11) and ignore the sets f o r w h i c h > 0 or /c2 < 0. Fourth, select the set o f formulas f o r w h i c h ^ py Pz is smallest. W i t h the results o f step three the concrete compressive stress can be calculated and checked.

ac3 =

^/ci-y(J/ci)

- / c 2 > - / ;

(23)

2J 2.1 2X The ± signs i n Eq. ( ^ ^ 4 ) , ( > 5 ' 5 ) and ( ^ 5 - 6 ) can be replaced by the absolute value. The p r o o f f o r this is presented i n A p p e n d i x 1.

O^cl =

CTcl (PxCT^x, PyOsy,

Ö^c3 =

öTcS

W =

The barrier method is a method f o r computational optimization [ 1 4 ] . I n this method, a large cost is imposed on points that lie close to the boundary o f the feasible region. This cost is called the barrier because it makes sure that a new point is not picked outside the feasible region. + Py + Pz +

(24)

where r is a factor that is reduced i n subsequent steps and B is the barrier. A suitable barrier f u n c t i o n f o r the problem o f this paper is 0.01

0.01

+ Px

h \

+y

Example 1 to 7 have also been used by Andreasen et al. [ 5 ] . Their results and the resuhs i n this paper are the same. Example 8 and 9 have also been studied by Foster et al. [ 7 ] . I n example 8 the same results have been found. I n example 9, Foster selected = 0.75%, Py = 0^ p^ = 0.75%. Table 1 shows that the optimal reinforcement differs considerably. However, the total reinforcement is almost the same (Foster; 0.75 + 0.00 + 0.75 = 1.50%, Table 1; 0.89 + 0.00 + 0.57 = 1.46%). I t is noted that Forster et al. selected this reinforcement w i t h o u t t r y i n g to f i n d the o p t i m u m . I n fact, the o p t i m u m is an edge solution w h i c h was not considered i n their publications [7, 8 ] . Example 10 shows that i n a plane stress state several f o r m u l a sets provide the o p t i m u m reinforcement. Example 11 and 12 are included f o r comparison w i t h Example 13. Example 13 includes two load

0.01

+ Py

Pz

- c u ^ i - ^

"""^^

STRESS E X A M P L E S

Table 1 shows 13 resuhs o f the proposed algorithm. The rows contain computation examples. Columns ^xx, cfyy, ^zz, ^xy, (^xz, (^yz coutaiu the iuput strcsscs i n N / m m ^ . The reinforcement y i e l d stress is ^ = 500 N / m m ^ f o r each example. C o l u m n px, Py, pz contain the output reinforcement ratios i n %. C o l u m n del, c>c2, crc3 contain the output principal concrete stresses. C o l u m n Eq. shows the f o r m u l a number that gives the same result. Except f o r the last example all examples involve just one load combination. I n the last computation example two load combinations are included.

BARRIER METHOD

minimise

W(AT,P;;,PZ)

Note that 5 is positive f o r feasible solutions. I t goes to i n f i n i t y i f any o f the constraints is almost violated. A n advantage o f the barrier method compared to other methods o f computational optimisation is that only interior points are evaluated. Interior points are " s u f f i c i e n t reinforcement" f o r w h i c h the computation o f the crack w i d t h w converges quickly. The minimisat i o n can be performed by any unconstrained optimisat i o n algorithm such as the d o w n - h i l l simplex method or Newton's method. A good starting point f o r the Barrier method is the envelope o f the requirements f o r the individual load combinations. The required reinforcement f o r a load combination related to the serviceability l i m i t state can be quickly approximated by assuming that the reinforcement ratios are proportional to the steel stresses.

8 7

Pz^^sz)

{Px^sx. Pyfysy. Pz^sz)

+ f^j

(25)

where Uu is the number o f load combinations related to the ultimate l i m i t state and ris is the number o f load combinations related to the serviceability l i m i t state.

643

Table 1.

Stress c o m p u t a t i o n examples.

Case 1 2 3 4

Oyy

1 -5

5 6

-5 -5 1 1

7 8

1 2

9 10 11 12

-3 3 15

13

15

^xy

2 2 -6 -6 2 -2 2 -2 -7

3 3 3 -6 3 3 3 5

Oxz

-1 1 1

3 3 3

1 -1 2

3 -3 3 2 -4

-1 6 6

10 5 5

-4 5

-4 4 4

Py

Pz

1.00

1.40

2.00 1.88 1.69

-10.65 -10.31 -10.15 -10.44

-5.35 -5.89 -6.30 -6.31

2.00 1.80

-10.58

-1.42

21-8 21-11

1.80 1.40

-10.17 -9.36 -15.21

-0.64

0.57 3.00

-14.76 -10.00

21-7 21-8 21-5-1-

1.36

0.60 0.50

4

0.40 2.40 0.89

1.00 0.13 1.00 0.40

1.60 3.00 1.00 3.00 3.00

(Tel

Oc2

4 -4 -4 2 2

Eq.

Aï

1.00 0.33

-10.00

0.33

-16.67

21-10 21-421-1 -0.24

-0.79 -2.52

2 1 - 5 - , 7, 10 21-3,7,9 2 1 - 6 - , 7, 8

-1.67

The dots (.) represent zeros (0) i n order to improve readability o f the table.

combinations. The volume reinforcement ratio is ft + Py + Pz = 3.00 + 0.33 - I - 0.00 = 3.33%. A l t e r n a tively, we could have selected the envelope o f the reinforcement requirements f o r the individual load combinations, w h i c h are 11 and 12. The volume reinforcement ratio applying the envelope method is max(3.00,1.00) -I-max(0.00,1.00) = 4.00%. Consequently, the envelope method gives a safe approximation but i t overestimates the required reinforcement substantially.

1000

z 4\ 9

200

STRUCTURAL EXAMPLE

Figure 3 shows a square concrete block that is f i x e d at one face o f the block. The block is loaded by a vertical force o f 1000 k N over an area o f 0.20 x 0.20 m (25 N / m m ^ ) . Just one load is considered. Dead load (24 ItN) is neglected. Young's modulus is 30000 N / m m ^ and Poisson's ratio is 0.15. The concrete compressive strength is 35 N / m m ^ . Its tensile strength is 4 N / m m ^ . The steel y i e l d strength is 550 N / m m ^ . One load case related to the ultimate l i m i t state is considered. L o a d and resistance factors are not included. The proposed algorithm has been implemented i n a f i n i t e element program. A n eight node brick element was used. The element dimensions are 0.10 x 0.10 x 0.10 m . A linear elastic analysis is performed. The normal stressCTJ-Vis shown i n Fig. 4. The required reinforcement ratios f o r the ultimate l i m i t state are computed by the proposed algorithm (Figs. 5, 6 and 7). The results provide s u f f i c i e n t i n f o r m a t i o n f o r a structural engineer to select bar diameters and bar spacing. Subsequently, the r e i n f o r c i n g cage can be designed by applying r e i n f o r c i n g principles (hoops, hooks, hairpins, development length). Note that not

Top view

5 0 '^

I

\


50

25 N/mm^

1000

Side view 1 0 0 0 iTUn

Figure 3. Concrete b l o c k loaded by a v e r t i c a l force (dimensions in m m ) .

644

Figure 4. The normal stress a^x on the surface of the concrete block. The largest value is 6.17 N/mm^. The smallest value is -8.52 N/mm^.

Figure 7. is 1.02%.

Optmial reinforcement ratio

The largest value

only reinforcement for bending and shear are needed but also splitting reinforcement is needed for introducing the load into the concrete. The authors recommend that the reinforcement detailing is checked by mentally visualising the force flow with a strut-andtie model. This does not mean that the reinforcement needs to be quantified with a strut-and-tie model. There is no need for this time consuming task because the required amounts are already determined by the proposed algorithm.

10 CONCLUSIONS A numerical algorithm is proposed for computing the required reinforcement in solid concrete. It starts from the stresses in the integration points of a finite element model. In subsequent improvements the algorithm finds the reinforcement ratios p^, Py, p^ for which the sum is smallest. Constraints are imposed on the steel stresses and the concrete stresses for load combinations related to the ultimate limit state. A constraint is imposed on the crack widths for load combinations related to the serviceability limit state. The algorithm shows to be robust, fast and accurate.

LITERATURE [1]

Figure 6. is 0.56%.

Optimal reinforcement ratio py The largest value

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Eurocode 2, Design of concrete structures, EN 1992¬ 1-1,2004. [2] Marti P., Design of concrete slabs for transverse shear, ACI Structural Journal, Vol. 87 (1990) No 2 pp. 180-190. [3] Lourengo PB., Figueiras J.A., Solution for the design of reinforced concrete plates and shells, Journal of Structural Engineering, May 1995 pp. 815-823.

[4]

Smimov, S.B., Problems of calculating the strength of massive concrete and reinforced-concrete elements of complex hydraulic structures, Power Technology and Engineering (formerly Hydrotechnical Construction), Springer New York, Volume 17, Number 9/September, 1983, pp. 471-476. [5] Andreasen B.S., Nielsen M.P, Armiering af beton I det tredimesionale tilfaelde, Bygningsstatiske meddelelser, Vol. 56 (1985), No. 2-3, pp. 25-79 (in Danish). [6] Kamezawa Y., Hayashi N., Iwasaki I . , Tada M . , Smdy on design methods of RC structures based on FEM analysis, Proceedings of the Japan Society of Civil Engineers, Issue 502 pt 5-25, November 1994, pp. 103-112 (In Japanese). [7] Foster S.J., Marti P, Mojsilovic N . , Design of reinforced concrete solids using stress analysis, ACI StructuralJournal, Nov.-Dec. 2003, pp. 758-764. [8] Marti P, Mojsilovic N . , Foster S.J., Dimensioning of orthogonally reinforced concrete solids, structural concrete in Switzerland, The first fib-Congress, Swiss Group of fib, Osaka, Japan, Oct. 13-19, 2002, pp. 18-23. [9] Hoogenboom PCX, de Boer A., Computation of reinforcement for sohd concrete, Heron, Vol. 53 (2008), No. 4, pp. 247-271. [10] Prager W., Hodge PG., Theoiy of Perfectly Plastic Solids, New York, Wiley 1951. [11] Nielsen M.P, Limit analysis and concrete plasticity, second edition, CRC Press, 1999. [12] Vecchio EJ., Collins M.P, The modified compression-field theory for reinforced concrete elements subjected to shear, ACI Journal, Vol. 83, No. 2, March-April 1986, pp. 219-231. [13] CEB-fib Model Code 1990, Design Code, Thomas Telford, London, 1993, ISBN 0 7277 1696 4. [14] Nocedal J., Wright S., Numerical Optimization, Springer, New York, 1999, ISBN 0-387-98793-2

APPENDIX 1 The ± sign i n Eqs ( 2 1 - 4 ) , ( 2 1 - 5 ) and ( 2 1 - 6 ) can be replaced by the absolute value. Here this is proven f o r Eqs ( 2 1 , 5). Substitution o f Eqs ( 2 1 , 5) i n Eqs (10), (11) and (12) gives

T

,

h \

=

Jc2

=

^l'

CTyy +

—

+

'^yz

(^vzOxv ^

'-

646

F r o m Id

> 0 i t is concluded that the values o f cjyy

and = p ( ^ ^ | ^ — Gxz) need to have the same sign. F o r m Id < 0 it is concluded that this sign needs to be negative. Consequently, Oyy < 0 and _ cr^^^ > Q Q.E.D. APPENDIX 2 R e i n f o r c i n g bars do not need to be i n the x, y and z directions. For reinforcement i n any direction the concrete stress tensor is O'er

Cfxy