m 6000:=kN

m4

d 1.08:= m

hm 3.5 d⋅ 1.5+:= hm 5.28= m

Iπ d4

⋅

64:= I 0.07= m

4

Modulul de elasticitate al betonului din coloane C25/ 30:

Eb 33000000:=kN

m2

α5

mbc

Eb I⋅⋅:= α 0.36=

Fisa pilotului:

h 7.3:= m

hbar α h⋅:= hbar 2.59= incastrare in stinca

a0 2.33:= b0 1.59:= c0 1.69:=

Calculul si armarea consolidarii cu piloti forati

Evaluarea incarcarilor pe pilot - dispusi pe un singur rind

d 1.08:= m

bc d 1+( ):= bc 2.08= m

φu 8.7 deg⋅:= ka tan 45φu

2−

2

:=

δu2

3φu⋅:= δu 0.1=

.γu 20.68:=

kN

m3

Inaltime radier:

h1 1.20:= m

Ho 160 2.5⋅:=

Ho 400= kN

Mo Ho4.8

3⋅:=

Mo 640= kN m⋅

Evaluarea parametrilor initiali:

Coeficientul de proportionalitate:

1

A1 2.991( ) 0.9−=

A1 0.374( ) 1=

A1 x( ) linterp X Y, x,( ):=

Y A11⟨ ⟩:=

X A10⟨ ⟩:=

A1 csort A1 0,( ):=

A10 1

0.1 1

0.2 1

0.3 1

0.4 1

0.5 1

0.6 0.999

0.7 0.999

0.8 0.997

0.9 0.995

1 0.992

1.1 0.987

1.2 0.979

1.3 0.969

1.4 0.955

1.5 0.937

1.6 0.913

1.7 0.882

1.8 0.843

1.9 0.795

2 0.735

2.2 0.575

2.4 0.347

2.6 0.033

2.8 -0.385

3 -0.928

:=

Determinarea presiunilor orizontale pe fetele verticale ale elementului flexibil:

mhi 7.19=

hi1.65 Ho⋅ 2.75 Ho

2⋅ 9.4 p⋅ Ho⋅ h1⋅++

2 p⋅:=

kPap 141:=mh1 2.7:=

Determinarea adancimii de incastrare

φo 0=φoMo

α Eb⋅ I⋅c0⋅

Ho

α2Eb⋅ I⋅

b0⋅+:=

yo 0.01=yoMo

α2( ) Eb⋅ I⋅

b0⋅Ho

α3( ) Eb⋅ I⋅

a0⋅+:=

2

0 2 4 6

5

0

X-Y data

Linear interpolation

B10 0

0.1 0.1

0.2 0.2

0.3 0.3

0.4 0.4

0.5 0.5

0.6 0.6

0.7 0.7

0.8 0.799

0.9 0.899

1 0.997

1.1 1.095

1.2 1.192

1.3 1.287

1.4 1.379

1.5 1.468

1.6 1.553

1.7 1.633

1.8 1.706

1.9 1.77

2 1.823

2.2 1.887

2.4 1.874

2.6 1.755

2.8 1.49

3 1.037

:=

B1 csort B1 0,( ):=

X B10⟨ ⟩:=

Y B11⟨ ⟩:=

B1 x( ) linterp X Y, x,( ):=

B1 0.374( ) 0.37=

B1 2.991( ) 1.06=

0 2 4 6

5

0

X-Y data

Linear interpolation

3

C10 0

0.1 0.005

0.2 0.02

0.3 0.045

0.4 0.08

0.5 0.125

0.6 0.18

0.7 0.245

0.8 0.32

0.9 0.405

1 0.499

1.1 0.604

1.2 0.718

1.3 0.841

1.4 0.974

1.5 1.115

1.6 1.264

1.7 1.421

1.8 1.584

1.9 1.752

2 1.924

2.2 2.272

2.4 2.6

2.6 2.907

2.8 3.128

3 3.225

:=

C1 csort C1 0,( ):=

X C10⟨ ⟩:=

Y C11⟨ ⟩:=

C1 x( ) linterp X Y, x,( ):=

C1 0.374( ) 0.07=

C1 2.991( ) 3.22=

0 2 40

1

2

3

X-Y data

Linear interpolation

4

D10 0

0.1 0

0.2 0.001

0.3 0.005

0.4 0.011

0.5 0.021

0.6 0.036

0.7 0.057

0.8 0.085

0.9 0.121

1 0.167

1.1 0.222

1.2 0.288

1.3 0.365

1.4 0.456

1.5 0.56

1.6 0.678

1.7 0.812

1.8 0.961

1.9 1.126

2 1.308

2.2 1.72

2.4 2.195

2.6 2.724

2.8 3.288

3 3.858

:=

D1 csort D1 0,( ):=

X D10⟨ ⟩:=

Y D11⟨ ⟩:=

D1 x( ) linterp X Y, x,( ):=

D1 0.374( ) 0.01=

D1 2.991( ) 3.83=

0 2 40

2

4

X-Y data

Linear interpolation

zbar x( ) x α⋅:=

5

σz x( )m

αzbar x( )⋅ yo A1 zbar x( )( )⋅

φo B1 zbar x( )( )⋅

α−

Mo C1 zbar x( )( )⋅

α2Eb⋅ I⋅

+Ho D1 zbar x( )( )⋅

α3Eb⋅ I⋅

+

⋅:=

0 0.8 1.6 2.4 3.2 4 4.8 5.6 6.4 7.2 8

100

79

58

37

16

5

26

47

68

89

110

σz x( )

x

x 0 8..:=

x

0

1

2

3

4

5

6

7

8

= zbar x( )

0

0.36

0.71

1.07

1.42

1.78

2.13

2.49

2.84

= σz x( )

0

57.6

78.73

73.98

54.3

30.67

11.94

1.95

5.01

=

Verificarea starii limita de capacitate portanta a terenului in jurul elementului

z0.85

α:= z 2.39=

γ 21:=

φ 14 deg⋅:= c 58:=

padm 2 γ z⋅ tan φ( )⋅ c+( )⋅:=

padm 141.05=

In concluzie presiunile efective sunt mai mici decit presiunea admisibila.

6

6. Determinarea momentelor incovoietoare in elementul flexibil:

A30 0

0.1 0

0.2 -0.001

0.3 -0.005

0.4 -0.011

0.5 -0.021

0.6 -0.036

0.7 -0.057

0.8 -0.085

0.9 -0.121

1 -0.167

1.1 -0.222

1.2 -0.287

1.3 -0.365

1.4 -0.455

1.5 -0.559

1.6 -0.676

1.7 -0.808

1.8 -0.956

1.9 -1.118

2 -1.295

2.2 -1.693

2.4 -2.141

2.6 -2.621

2.8 -3.103

3 -3.541

:=

A3 csort A3 0,( ):= X A30⟨ ⟩:= Y A3

1⟨ ⟩:=

A3 x( ) linterp X Y, x,( ):=

7

0 1 24

2

0

X-Y data

Linear interpolation

Interpolated values:

A3 0.374( ) 0.01−=

A3 2.3( ) 1.92−=

B30 0

0.1 0

0.2 0

0.3 -0.001

0.4 -0.002

0.5 -0.005

0.6 -0.011

0.7 -0.02

0.8 -0.034

0.9 -0.055

1 -0.083

1.1 -0.122

1.2 -0.173

1.3 -0.238

1.4 -0.319

1.5 -0.42

1.6 -0.543

1.7 -0.691

1.8 -0.867

1.9 -1.074

2 -1.314

2.2 -1.906

2.4 -2.663

2.6 -3.6

2.8 -4.718

3 -6

:=

B3 csort B3 0,( ):= X B30⟨ ⟩:= Y B3

1⟨ ⟩:=

B3 x( ) linterp X Y, x,( ):=

8

0 1 26

4

2

0

X-Y data

Linear interpolation

Interpolated values:

B3 0.374( ) 0−=

B3 2.3( ) 2.28−=

C30 1

0.1 1

0.2 1

0.3 1

0.4 1

0.5 0.999

0.6 0.998

0.7 0.996

0.8 0.992

0.9 0.985

1 0.975

1.1 0.96

1.2 0.938

1.3 0.907

1.4 0.866

1.5 0.811

1.6 0.739

1.7 0.646

1.8 0.53

1.9 0.385

2 0.207

2.2 -0.371

2.4 -0.949

2.6 -1.988

2.8 -3.808

3 -4.988

:=

C3 csort C3 0,( ):= X C30⟨ ⟩:= Y C3

1⟨ ⟩:=

C3 x( ) linterp X Y, x,( ):=

9

0 1 2

4

2

0

X-Y data

Linear interpolation

Interpolated values:

C3 0.374( ) 1=

C3 2.3( ) 0.66−=

D30.1 0.1

0.2 0.2

0.3 0.3

0.4 0.4

0.5 0.5

0.6 0.6

0.7 0.699

0.8 0.799

0.9 0.897

1 0.994

1.1 1.09

1.2 1.183

1.3 1.273

1.4 1.358

1.5 1.437

1.6 1.507

1.7 1.566

1.8 1.612

1.9 1.64

2 1.646

2.2 1.575

2.4 1.152

2.6 0.597

2.8 0.197

3 -0.981

:=

D3 csort D3 0,( ):= X D30⟨ ⟩:= Y D3

1⟨ ⟩:=

D3 x( ) linterp X Y, x,( ):=

10

0 1 21

0

1

2

X-Y data

Linear interpolation

Interpolated values:

D3 0.374( ) 0.37=

D3 2.3( ) 1.36=

zbar y( ) y α⋅:=

mz y( ) α2Eb⋅ I⋅ yo⋅ A3 zbar y( )( )⋅ α Eb⋅ I⋅ φo⋅ B3 zbar y( )( )⋅− Mo C3 zbar y( )( )⋅+

Ho

αD3 zbar y( )( )⋅+:=

0 1 2 3 4 5 6 7 82001204040120200280360440520600680760840920100010801160124013201400

mz y( )

y

zbar z( ) α z⋅:=

z 0 8..:=

zbar z( )

0

0.36

0.71

1.07

1.42

1.78

2.13

2.49

2.84

= mz z( )

640

1014.17

1279.4

1380.12

1335.18

1176.25

955.21

446.64

178.57

=

11

m0Mmax N ead⋅+( )

Ab Rc⋅d

2⋅

:= m0 0.21=

Compresiune cu incovoiere

αc 0.22:=

Aa αc Ab⋅Rc

Ra⋅:= Aa 0.01= m

2

Aria de armatura necesara:

A 10000 Aa⋅:= A 93.82= cm2.

Se aleg 20 bare diam. 25 PC52 cu Aef=98.17 cmp

Armarea radierului- ca si grinda continua :

Ms4.1 25⋅ 2.5

2⋅

8:= Ms 80.08= kNm

Mi4.1 25⋅ 2.5

2⋅

14.3:= Mi 44.8= kNm

ho 1.2 0.05− 0.15−:= ho 1=ar 0.1:=

AsMs 10

4⋅

Ra ho ar−( )⋅:= As 3.07= cm

2

Se prevad 5 φ 16/m PC 52 cu Aef= 10 cmp la partea superioara siconstructiv 5 φ 12/m la partea inferioara.

Armarea pilotului

Mmax mz 3( ):= Mmax 1380.12= kNm

N4.1 7.5⋅ 25⋅

3

3 25⋅ π 1.082

⋅

4+:= N 324.96= kN

Rc 13500:= Ra 290000:=kN

m2

a 0.08:=

1. rad

2a−:= ra 0.46=

2. ρ ra2

d⋅:= ρ 0.85=

3. eadd

30:= ead 0.04=

4. Ab πd2

4:= Ab 0.92=

5. n0N

Ab Rc⋅:=

n0 0.03=

6.

12

Prin metoda fisiilor a rezultat valoarea de 160 kN/m, acoperitoare pentru calcul.

kN

mp 156.8=

pγ z⋅

NφD1

D1

D2

A

⋅ e

D1 D2−( ) Nφ⋅ tan φu( )⋅ tan3.14

8

φu

4+

⋅

D2

⋅ D2−

⋅:=

A Nφ0.5tan φu( )⋅ Nφ+ 1−:=

Nφ tan3.14

4

φu

2+

2

:=

kN

m3

γ 20:=

mz 4.8:=

lumina intre pilotimD2 1.42:=

distanta interaxmD1 2.50:=

Calculul presiunii pe pilot considerind ipoteza Ito-Mitsui

Armaturile din pilot vor fi prelungite 1 m in radier.

mli 0.51=li6 Mo⋅

d Rc⋅:=

Calculul lungimii de incastrare a pilotului in radier:

Se prevad 5 φ 12/m cu Aef=5.65 cmp

cm2

Az 3.24=AzMz 10

4⋅

Ra 0.45 ar−( )⋅:=

kNmMz 32.88=

Mz18

2hz2

⋅ ka⋅1

3⋅ hz⋅ cos δu( )⋅:=

mhz 1.8:=

Armarea zidului de sprijin :

13

Capacitatea portanta a unui pilot:

pv 1150:= kPa

d 1.08:= m

Apπ d2

⋅

4:= Pp π d⋅:=

Pcap 0.7 0.6 pv⋅ Ap⋅ 0.5 Pp⋅ 60⋅ 8.1⋅+( )⋅:=

Pcap 1019.61= kN.

intocmit

ing. Horea Hopirca

14