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2.6. Slab reinforcement design– SR-EN-1992-1-2004 –

2.6.1 General rules:

The design of the current slab is made according SR-EN-1992-1-2004 ( 2 way

reinforcement slab – independent bars). The design is made at ultimate limit state. The

computation is made in the following steps :

- Preliminary design of the slab (see part 2.1 Preliminary design);

- Choice of the reinforcement manner;

- Loads evaluation (see part 2.2 Loads Evaluation);

- Static analysis computation of the slab with Robot Structural Analysis;

- The computation of the reinforcement.

The design algorithm for the RC slab:

- the thickness of the slab hs is chosen after a preliminary design, hs = 130 mm;

- it is chosen the reinforcement manner on one way or two ways, function of the ratio

of the sides of the slab, in this case the way reinforcement is chosen;

- the total load on square meter is evaluated;

- the two components for each way are computed;

- the bending moments are computed for each way, in span and support;

- the following characteristics are chosen fyk , fck , a , Φ ;

- d = hs – a, is computed;

- µ is computed;

- the reinforcement are on each way in span and support is computed;

- the necessary bars number is chosen in function of the diameters, taking in account

of the constructive aspects for RC slabs;

- the input data for drawing the reinforcement slab are established: the number of bars,

the length of the bar segments.

Choice of the reinforcement manner:

- if l1/l2 > 2 – the reinforcement is made on one direction parallel with the shorter side;

- if 0.5 < l1/l2 < 2 – the slabs are reinforced on two directions;

Because 0.5 < l1/l2 < 2 the slabs are reinforced on both ways.

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For the slab reinforcement the following aspects have to be taken in account (according SR-

EN-1992-1-2004):

- the minimum diameter of the bars: 6 mm for the straight bars from the inferior part

of the plate, and for inclined bars or bars mounted at the superior part, 6 mm for

PC52 or PC60 and 8 mm for OB37;

- in the tensioned zones the minimum number of bars is of 5 bars per meter for slabs

with hs < 300 mm;

- the maximum number of bars in span and support is 12 bars/m

- the bars which take over the negative bending moments from the supports, mounted

on the superior part of the plate, are extended on each side in order to cover the entire

area of negative moments. If the computation is not performed the anchorage length

will be lc = ¼ of the maximum length of the two adjacent spans l1, l2 .

The evaluation of the loads:

- the loads have been defined in the part 2.2 Loads Evaluation. The loads have been

set in Robot Structural Analysis.

2.6.2 The static analysis :

A strip having the width 1 m is detached in the central zone on each directions. This

strip have the same behavior like in the case of beams.

The maximum and minimum bending moments in spans and supports have been

determined with Robot.

The studied floor – 4th floor

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Diagrams of bending moment on X direction:

MEd,support,x = 19.85 kNm MEd,span,x = – 6.08 kNm

Diagrams of bending moment on Y direction:

MEd,support,y = 16.30 kNm MEd,span,y = – 6.06 kNm

The reinforcement computation:

- 30015,1/345/ sykyd ff N/mm 2 ; (PC 52)

- 33,135,1/20/ bckcd ff N/mm 2 ; (C20/25)

- b = 1m; ph = 13 cm; a min = 1 cm;

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2.6.3 The computation of the reinforcement:

Establish the available heights on each direction:

1052/10201302/1 xsnomfx chd mm

952/1010201302/11 ysxsnomfy chd mm

mmmmmmc s 10;10515;10maxmin

mmcc totcmon 201010min

mmcmon 20 ;

mmca snomeff 252/10202/ ˃ ;10min mma

2.6.3.1 Design of reinforcement in span (direction x and y) :

- kNmM Edx 08.6 30015.1/345/ sykyd ff N/mm 2 ; (PC 52)

- kNmM Edy 06.6 33.135.1/20/ bckcd ff N/mm 2 ; (C20/25)

fctm = 2.2 N/mm2

- b = 1000 mm; = ∙ ∙ = 0.041= ∙ 1 − (1 − 2 ) = 4.39

-= ∙ ∙ = 195= 0.50 = 0.003188

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= ∙ ∙ = 335As1 = 335 mm2

dl =∙ = 7.80 mm

dl = 8 mm

As1r = 7∙dl2 ∙ π/4 = 352 mm2

The longitudinal reinforcement for all spans on x and y direction will be of:7 bars of Φ8

2.6.3.2 Design of reinforcement in support (direction x, direction y) :

- kNmM Edx 85.19 30015.1/345/ sykyd ff N/mm 2 ; (PC 52)

- kNmM Edy 30.16 33.135.1/20/ bckcd ff N/mm 2 ; (C20/25)

fctm = 2.2 N/mm2

- b = 1000 mm; = ∙ ∙ = 0.135= ∙ 1 − (1 − 2 ) = 15.29

-= ∙ ∙ = 679= 0.50 = 0.003188= ∙ ∙ = 335As2 = 679 mm2

dl =∙ = 9.80 mm

dl = 10 mm

As2r = 9∙dl2 ∙ π/4 = 706 mm2

The longitudinal reinforcement in all supports on each direction will be of:9 bars of Φ10