cladiri inalte model

8/12/2019 Cladiri Inalte Model

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Technical University Gh. Asachi from Iasi

Faculty of Civil Engineering and Building Services

Master of Structural Engineering

REINFORCED CONCRETE

STRUCTURES FOR TALL

BUILDINGS

Student: Coordinator:

Irimia Cristian Radu Prof. dr. ing. Gosav Ionel


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1.Project theme

It is required that each student to design one floor (nth=1) from the reinforced concrete

structure with the height of P+17E.

The used number of floor designed is: n=1

There are given the following data:

- The building is placed in city of Bucharest- The structure is made of reinforced concrete frames, the infilling is made of B.C.A.

blocks

- The foundation system is realized by a general mat with foundation beams and piles.- The buildings destination: Block of flats- The 1stfloor height is 4.8 m, current floors height is 3.45 m, and the floors 16thand

17thhave a height of 4.95 m.

- In plan dimensions are: 32.45 by 29.6 mThe materials used are:

Eb=30500 N/mm2(C25/30)

Ebg=0.6Ebfor beams and floors

Ebs=0.8Ebfor columns


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2.Structure element pre-sizing and load computation

2.1.Load computation

2.1.1 Permanent loadsa. Thermo-insulation weight:

b. Floor self weight and leveling screed: c. Infilling walls self weight:

d. Plaster self-weight: 2.1.2 Variable loadsa. Snow load:

Where:

- form factor for flat roofs - exposure factor - thermal coefficient for roofs with usual thermo-insulation

characteristic value for snow load from the soil level (CR 1-1-32005)b. Live load


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2.1.3 Exceptional loads (Seismic action)According to P100/2006 we have the following coefficients:

corner period

behaviour factor factor depending on the importance class maximum dinamic amplify factor ground acceleration factor

2.2.Element pre-sizing

2.2.1. Determining the beam dimensions (modulated to plus 50 mm)(P-E5:)

hg=max(L;T)/8=max(825; 819) mm 850 mm

bg=(0,3 0,5)hg=0.5 hg=425 mm 450 mm(E6-E11)

hg=max(L;T)/10=max(660; 655) mm 700 mm

bg=(0,3-0,5)hg= 0.5 hg=350 mm

(E12-E17)

hg=max(L;T)/12=max(550; 546) mm =550 mm

bg=(0,3-0,5)hg=0.5 hg=275mm300 mmL and T maximum opening on longitudinal and transversal direction

L=6600 mm T=6550 mm


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Eds

cd

Nh

f

2.2.2. Determining the floor height (modulated to plus 10 mm)hp=max(hp1; hp2)=max (156; 152.5) mm=156 mm 160 mm

hp1

=20mm + P/180=20+24500/180=156 mm

P=2(L0+T0)=2(6100+6150)=24500 mm

L0and T0 the maximum span on longitudinal and transversal direction

hp2= min (L0;T0)/40=min (154;152.5) mm =152.5 mm

L0=L- bg=6600-450=6150 mm

T0=T- bg=6550-450=6100 mm

2.2.3. Determining the secondary beam dimensions from the 16th and 17th floors (modulatedto plus 50 mm)

hgs=max(L;T)/15=max(6600; 6550)/15=max (440; 437) mm=440mm 450mm

bgs=(0,3-0,5)hgs=0.5 hgs=225mm 250 mm

2.2.4.

Column pre-sizing (modulated to plus 50 mm)

The columns will have 3 steps in section variation (P-E5), (E6-E11) i (E12-E17)

The maximum axial forces from the columns will be determined using the data from the

combination: 1P+0.4U+0.4Z.

In order to determine the final dimensions for the columns we first impose an initial section

of 400X400 mm, after one run with a structural program we obtain the following axial forces

and we determined the actual section for each column:

Where:

-the section area of the column- the column height- normalized axial force coefficient

2 Edbs s s s

cd

NA b h h

f


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=25/1.5=16.66 N/mm2- design compressive strength for concrete C25/30- design axial forcea. P-E5

- Marginal column N1m=3902.22 kN

- Central column N1c=5947.13 kN

b. E6-E11



c. E12-E17




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3.Static computationWith all element dimensions are determined we can now compute a final structural

analysis in order to obtain the final efforts, displacements, period of vibration and diagrams.

3.1.Static combinations

We will use the following combinations:

A. ULS:

i.

1.35P+1.5U+1.05Z

ii. 1.35P+1.05U+1.5Ziii. P+0.4U+0.4Z+1.2Sx+iv. P+0.4U+0.4Z+1.2Sx-v. P+0.4U+0.4Z+1.2Sy+

vi. P+0.4U+0.4Z+1.2Sy-B. SLS:

i. P+0.4U+0.4Z+0.72Sx+ii. P+0.4U+0.4Z+0.72Sx-

iii. P+0.4U+0.4Z+0.72Sy+iv. P+0.4U+0.4Z+0.72Sy-


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3.2.Diagrams

3.2.1. Vibration modesa. Mode 1 of vibration: T=1.410 s b. Mode 2 of vibrations: T=1.352 s


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c. Mode 3 of vibration: T=1.132 s

3.2.2. Effort diagrams for columns Fx, Fy, Fz, My, Mz (SLU1-4, envelope min-max)i. Fx:

a. Marginal column


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b. Central column

ii. Fy:a. Marginal column

b. Central column


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iii. Fz:a. Marginal column

b. Central column

iv. My:a. Marginal column


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b. Central column

v. Mz:a. Marginal column

b. Central column


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3.2.3. Bending moment diagram for beamsMy (kNm) (SLU1-4, envelope min-max)a. Beam on x direction

b. Beam on y direction


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3.2.4. Bending moment diagram for floormxx (kNm/m) (SLU1 or SLU2-E3)

3.2.5. Bending moment diagram for floormyy (kNm/m) (SLU1 or SLU2-E3)


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4.F loor design

Determining the reinforcement area for direction x

PanelMxx (kNm) b d fcd

Reinforcement

area (mm2)

Support Span mm mm N/mm2 Support Span Support Span Support Span

p1 28.34 26.69 1000 130 16.66 0.1007 0.0948 0.1063 0.0998 767.46 720.29

p2 13.27 17.39 1000 130 16.66 0.0471 0.0618 0.0483 0.0638 348.68 460.59

p3 25.96 24.49 1000 130 16.66 0.0922 0.0870 0.0969 0.0911 699.53 657.93

p4 27.49 23.93 1000 130 16.66 0.0976 0.0850 0.1029 0.0889 743.12 642.15

p5 36.11 36.27 1000 130 16.66 0.1283 0.1288 0.1377 0.1384 994.38 999.14

p6 31.74 27.95 1000 130 16.66 0.1127 0.0993 0.1199 0.1048 865.76 756.28

p7 23.04 31.24 1000 130 16.66 0.0818 0.1110 0.0855 0.1179 617.15 851.21


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p8 30.04 27.33 1000 130 16.66 0.1067 0.0971 0.1131 0.1023 816.42 738.55

p9 33.68 31.18 1000 130 16.66 0.1196 0.1107 0.1278 0.1177 922.53 849.46

p10 39.67 38.7 1000 130 16.66 0.1409 0.1375 0.1525 0.1485 1101.16 1071.88

p11 30.06 27.71 1000 130 16.66 0.1068 0.0984 0.1132 0.1038 817.00 749.41

p12 20.86 34.71 1000 130 16.66 0.0741 0.1233 0.0771 0.1320 556.31 952.89

p13 11.57 5.64 1000 130 16.66 0.0411 0.0200 0.0420 0.0202 303.03 146.09

p14 30.26 29.05 1000 130 16.66 0.1075 0.1032 0.1140 0.1091 822.78 787.86

p15 39.35 38.45 1000 130 16.66 0.1398 0.1366 0.1512 0.1474 1091.48 1064.36

p16 33.75 30.16 1000 130 16.66 0.1199 0.1071 0.1281 0.1136 924.59 819.89

p17 25.58 35.54 1000 130 16.66 0.0909 0.1262 0.0954 0.1354 688.75 977.45

p18 25.72 24.86 1000 130 16.66 0.0914 0.0883 0.0960 0.0926 692.72 668.38

p19 29.14 24.15 1000 130 16.66 0.1035 0.0858 0.1095 0.0898 790.45 648.34

p20 38.39 38.86 1000 130 16.66 0.1364 0.1380 0.1472 0.1491 1062.55 1076.70

p21 31.52 28.88 1000 130 16.66 0.1120 0.1026 0.1190 0.1085 859.35 782.97

p22 19.36 30.06 1000 130 16.66 0.0688 0.1068 0.0713 0.1132 514.76 817.00p23 32.37 26.81 1000 130 16.66 0.1150 0.0952 0.1225 0.1002 884.14 723.71

p24 30.06 35.58 1000 130 16.66 0.1068 0.1264 0.1132 0.1356 817.00 978.64

p25 37.67 37.31 1000 130 16.66 0.1338 0.1325 0.1442 0.1427 1040.94 1030.17

Determining the effective reinforcement area on x direction

Panel

Effective

reinforcement Number of bar/m

area (mm2)

Support Span Support Span

p1 1017 904 9 8

p2 565 565 5 5

p3 904 904 8 8

p4 904 791 8 7

p5 1243 1243 11 11

p6 1130 1017 10 9

p7 791 1130 7 10

p8 1017 904 9 8

p9 1130 1130 10 10

p10 1356 1356 12 12

p11 1017 1017 9 9

p12 678 1243 6 11

p13 565 565 5 5

p14 1017 1017 9 9

p15 1356 1356 12 12

p16 1130 1017 10 9

p17 904 1243 8 11

p18 904 904 8 8

p19 1017 791 9 7


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p20 1356 1356 12 12

p21 1130 1017 10 9

p22 678 1017 6 9

p23 1130 904 10 8

p24 1017 1017 9 9

p25 1356 1356 12 12

Determining the reinforcement on y direction

PanelMyy (kNm) b d fcd

Reinforcement

area (mm2)

Support Span mm mm N/mm2 Support Span Support Span Support Span

p1 19.90 12.49 1000 118 16.66 0.0858 0.0538 0.0898 0.0554 588.58 362.87

p2 13.49 7.87 1000 118 16.66 0.0582 0.0339 0.0600 0.0345 392.85 226.22

p3 14.07 7.93 1000 118 16.66 0.0607 0.0342 0.0626 0.0348 410.30 227.98

p4 17.36 7.77 1000 118 16.66 0.0748 0.0335 0.0779 0.0341 510.26 223.30

p5 22.93 17.09 1000 118 16.66 0.0988 0.0737 0.1043 0.0766 683.37 502.00

p6 32.94 31.48 1000 118 16.66 0.1420 0.1357 0.1538 0.1464 1008.04 959.51

p7 31.96 31.37 1000 118 16.66 0.1378 0.1352 0.1489 0.1459 975.42 955.87

p8 37.21 33.09 1000 118 16.66 0.1604 0.1426 0.1759 0.1546 1152.47 1013.05

p9 42.14 35.52 1000 118 16.66 0.1817 0.1531 0.2021 0.1671 1324.19 1094.85

p10 42.06 35.06 1000 118 16.66 0.1813 0.1511 0.2016 0.1647 1321.36 1079.27p11 22.83 17.46 1000 118 16.66 0.0984 0.0753 0.1038 0.0783 680.22 513.33

p12 15.26 20.28 1000 118 16.66 0.0658 0.0874 0.0681 0.0916 446.27 600.39

p13 11.23 14.97 1000 118 16.66 0.0484 0.0645 0.0496 0.0668 325.31 437.48

p14 16.35 15.65 1000 118 16.66 0.0705 0.0675 0.0732 0.0699 479.40 458.10

p15 28.38 19.81 1000 118 16.66 0.1223 0.0854 0.1309 0.0894 857.85 585.79

p16 40.52 38.97 1000 118 16.66 0.1747 0.1680 0.1934 0.1851 1267.15 1213.14

p17 39.49 38.48 1000 118 16.66 0.1702 0.1659 0.1879 0.1825 1231.20 1196.18

p18 13.64 9.45 1000 118 16.66 0.0588 0.0407 0.0606 0.0416 397.36 272.62

p19 36.18 33.31 1000 118 16.66 0.1560 0.1436 0.1705 0.1557 1117.28 1020.41

p20 37.09 35.18 1000 118 16.66 0.1599 0.1517 0.1752 0.1653 1148.36 1083.33

p21 20.07 20.80 1000 118 16.66 0.0865 0.0897 0.0906 0.0941 593.86 616.58

p22 22.09 30.13 1000 118 16.66 0.0952 0.1299 0.1003 0.1396 656.94 915.01

p23 38.98 40.28 1000 118 16.66 0.1680 0.1736 0.1852 0.1921 1213.49 1258.75

p24 39.59 39.68 1000 118 16.66 0.1707 0.1711 0.1884 0.1889 1234.68 1237.81

p25 22.43 19.88 1000 118 16.66 0.0967 0.0857 0.1019 0.0897 667.63 587.96


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Determining the effective reinforcement area on y direction

Panel

Effective

reinforcement Number of bar/m

area (mm2)

Support Span Support Span

p1 1017 904 9.00 8.00

p2 565 565 5.00 5.00

p3 904 904 8.00 8.00

p4 904 791 8.00 7.00

p5 1243 1243 11.00 11.00

p6 1130 1017 10.00 9.00

p7 791 1130 7.00 10.00

p8 1017 904 9.00 8.00

p9 1130 1130 10.00 10.00

p10 1356 1356 12.00 12.00

p11 1017 1017 9.00 9.00

p12 678 1243 6.00 11.00

p13 565 565 5.00 5.00

p14 1017 1017 9.00 9.00

p15 1356 1356 12.00 12.00

p16 1130 1017 10.00 9.00

p17 904 1243 8.00 11.00

p18 904 904 8.00 8.00

p19 1017 791 9.00 7.00p20 1356 1356 12.00 12.00

p21 1130 1017 10.00 9.00

p22 678 1017 6.00 9.00

p23 1130 904 10.00 8.00

p24 1017 1017 9.00 9.00

p25 1356 1356 12.00 12.00


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5.Beam design5.1.Determining the design bending moment on x direction

5.1.1. Determining the longitudinal reinforcement in the supports

d= hw-a-

/2 (mm)=850-30-20/2=810mm

214 13 14 13 (0.5 )

14 0.5 ( )2

cEd MAX c

q hM M R h kNm

14 13 13 14

14 ( )2

MAX SLUM M qLR kNL

2

Ed

w cd

M

b d f

(1 1 2 )x d

2

2 2 ( )w cd

w cd s yd s w

yd

x b fx b f A f A b d mm

f

0.5ctm

yk

f

f


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fcd= 16,66 (N/mm2) C25/30

fyd= fyk/ s=345/1,15=300 (N/mm2) PC52

fctm= 2,6 (N/mm2) C25/30

Mrb2=As2rfyd(d-0,5xr) (kNm)

Support A:

( ) ( )

=520

Support B:

( ) ( )

=520

2( )

s r yd

r

w cd

A fx mm

b f


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Support C:

( ) ( ) =420

Support D:

( ) ( )

=420

Support E:

( ) ( )

=420


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Support F:

( ) ( )

=420

5.1.2. Determining the longitudinal reinforcement in mid span

-if the beam connects with marginal column

-if the beam connects with central column

Span 1:

2

Ed

eff cd

M

b d f

2

1 1 20.5 ( )

eff cd

eff cd s yd s w s r

yd

x b fx b f A f A b d A mm

f

1( )

s r yd

r

f cd

A fx mm

b f

eff sb h

2eff s f b h h


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( ) ( )

=522

Span 2:

( ) ( )

=420

Span 3:

( ) ( )

=420


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Span 4:

( ) ( )

=420

Span 5:

( ) ( )

=420


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5.2.Determining the design bending moment on y direction

5.2.1. Determining the longitudinal reinforcement in the supports

d= hw-a-/2 (mm)=850-30-20/2=810mm

fcd= 16,66 (N/mm2) C25/30

fyd= fyk/ s=345/1,15=300 (N/mm2) PC52

214 13 14 13 (0.5 )14 0.5 ( )

2

cEd MAX c

q hM M R h kNm

14 13 13 14

14 ( )2

MAX SLUM M qL

R kNL

2

Ed

w cd

M

b d f

(1 1 2 )x d

2

2 2 ( )w cd

w cd s yd s w

yd

x b fx b f A f A b d mm

f

0.5 ctm

yk

f

f


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fctm= 2,6 (N/mm2) C25/30

Mrb2=As2rfyd(d-0,5xr) (kNm)

Support A:

( ) ( )

=420

Support B:

( ) ( )

=420

Support C:

( ) ( )

2( )

s r yd

r

w cd

A fx mm

b f


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=420

Support D:

( ) ( )

=420

Support E:

( ) ( )

=420

Support F:


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( ) ( )

=420

5.2.2. Determining the longitudinal reinforcement in mid span

-if the beam connects with marginal column

-if the beam connects with central column

Span 1:

( ) ( )

=520

2

Ed

eff cd

M

b d f

2

1 1 20.5 ( )

eff cd

eff cd s yd s w s r

yd

x b fx b f A f A b d A mm

f

1

( )

s r yd

rf cd

A f

x mmb f

eff sb h

2eff s f

b h h


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Span 2:

( ) ( )

=4

20

Span 3:

( ) ( )

=420

Span 4:

( ) ( )


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=420

Span 5:

( ) ( )

=420

5.3. Determining the transversal reinforcement

5.3.1. Determining the transversal reinforcement for beam on x direction

1

,max ( )( )

cw w cd

Rd Ed

b z v f

V V kN ctg tg

ctg


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cw=1coefficient that takes into account the efforts from compressed fiber

z=0.9d

v1=0,6coefficient of strength reduction for cracked concrete due to shear force

max (ctg1; ctg2)1 (if the condition is not satisfied, the section must be modified)

ctg2.5 (if results a greater value, it is taken into account ctg=2.5)

s=100 mm

fywd=0.8fywk=0.8255=204 (N/mm2) OB37

[ ]

- it is required transversal reinforcement

Propose for transversal reinforcement bar 10

, ( )SW

Rd s ywd Ed

AV z f ctg V kN

s

2

4

bwSW r

dA n


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the transversal reinforcement will be placed by construction reasons5.3.2. Determining the transversal reinforcement for beam on y direction

[ ]


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- it is required transversal reinforcementPropose for transversal reinforcement bar 10

the transversal reinforcement will be placed by construction reasons

6.Column design6.1.Marginal columns

6.1.1. Determining the design bending momentsEstablish of external bending moments is made on each direction (x and y) with the

relation:

, ,max

Rb

Ed y Rd y

Eb

MM M

M

Where:

Idirection (x or y)

- designing moment on I direction - over strength factor- maximum bending moment on column from structural analysis (seismiccombination)

- actual resisting moment on girder disposed in I direction who intersect the node- bending moment on girder disposed in I direction who intersect the node resulted fromstructural analysis


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Evaluation of slenderness is obtained according with relation:

the slenderness can be ignored.

6.1.2. Longitudinal reinforcement

a. Direction x


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the concrete doesnt crack and the column is reinforced by using the minimumreinforcement percent.

b. Direction y



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6.1.3.

Checking the biaxial moment

6.1.4. Determining the shear force design value,1 ,2db db

Ed

level

M MV

H

, , min(1, )Rb

db i Rd Rb i

Rc

MM M

M

Rdover strength factor with value 1.2;

MRb,i actual bending moments of the girder that enter the node

MRb,i actual bending moments of the column that enter the node


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Maximum shear stress level

Constructive reinforcement is needed.

6.1.5. Checking of the relative displacementsULS checking:

{ {

SLS checking:


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6.2.Central columns

6.2.1. Determining the design bending momentsEstablish of external bending moments is made on each direction( x and y) with the

relation:

, ,max

Rb

Ed y Rd y

Eb

MM M

M

slenderness can be ignored.

6.2.2. Longitudinal reinforcementa. Direction x


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b. Direction y

6.2.3. Checking the biaxial moment

6.2.4. Determining the shear force design value


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,1 ,2db db

Ed

level

M MV

H

, , min(1, )

Rb

db i Rd Rb iRc

M

M M M

Maximum shear stress level

Constructive reinforcement is needed

6.2.5. Checking of the relative displacementsULS checking:

{ {

SLS checking:


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7.Node checking7.1.Horizontal shear force design

a. Central node

Rd=1.2 over resisting factor

As1, As2the reinforcement areas from supperior and inferior part of the beams

VEdshear force from the column beneath the node

b. Marginal node

7.2. Checking the horizontal shear force

a. Central node

( )

b. Marginal node


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7.3.Checking the transversal reinforcement

a. Central node

Ash - total area of horizontal stirrups in the node

ne - the number of horizontal stirrups in the node

b.

Marginal node

7.4.Checking the longitudinal reinforcement from the node

Asv=8157+4928=6184mm2 - the vertical reinforcement that pass through the nodeAsh - the total stirrup area from the node

hjc=d-a=890-20=870 mminter-ax distance between the column reinforcement

hjw=d-a=740-20=720 mminter-ax distance between the beam reinforcement

cladiri inalte model

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