Exemplu Calcul Fundatii pe Piloti

Fundatii piloti Note de calcul: Sa se proiecteze fundatia pe piloti pentru un stalp al unei conducte de transport apa. Stalpul are dimensiunile la baza 40x50. Incarcarile la partea inferioara a stalpului sunt : n := 20 Pk  := ( 400 + 0.4⋅ n) kN

= 408⋅ kN

Qk  := ( 300 + 0.2⋅ n) kN

= 304⋅ k

M xP.ek 

:= ( 0.35⋅ Pk ) ⋅ m = 142.8⋅ kN⋅

M xQ.ek  := ( 0.25⋅ Qk ) ⋅ m M yP.ek 

  TyP.ek 

= 76⋅ kN⋅

TyQ.ek  := 0.18⋅ Pk  = 73.44 73.44⋅ kN

:= ( 0.25⋅ Pk ) ⋅ m = 102⋅ kN⋅

M yQ.ek  := ( 0.20⋅ Qk ) ⋅ m

:= 0.25⋅ Pk  = 102⋅ kN

TxP.ek 

= 60.8⋅ kN⋅

:= 0.15⋅ Pk  = 61.2⋅ kN

TxQ.ek  := 0.15⋅ Qk  = 45.6⋅ kN

Stratificatia Stratificatia terenului : C.T.N ..-1.00m : mplutura

γ k1 :=  18.2

 kN 3

m

- 1.00...-!.50 : "r#ila nisipoasa nisipoasa #albuie$ plastic moale Ic2 := 0.4 + 0.00 .004  ⋅ n c'k2 :=   18kPa

= 0.48

ϕ' k2 :=  15deg

γ k2 :=  18.5

 kN 3

m

- !.50.....-15.00 : "r#ila marnoasa %inetie$ plastic %artoasa tare Ic3 :=   0.98

S3 :=   0.98

c!k3 :=   65kPa

γ k3 :=  19.5

 kN 3

ϕ !k3 :=  5deg

m

&ilotii se %or executa in %arianta executati pe loc$ de diametru diametru ϕ'!00 mm$ folosind folosind te(nolo#ia te(nolo#ia &iloti forati cu fluid. d := 600m

Verificare la capacitate portanta : Capacitatea portanta a pilotiilor se %erifica in cazul de proiectare C.&.1.)*" )+,)+4 - coeficienti partiali de si#uranta pentru actiuni "):

γ " :=  1.00

1.3 γ Q := 1.3

- %alori de calcul ale actiuniilor : Pd

:= Pk ⋅ γ " = 408⋅ kN

Qd

:= Qk ⋅ γ Q = 395.2⋅ k 

M xP.ed

:= ( 0.35⋅ Pd ) ⋅ m = 142.8⋅ kN⋅

TyP.ed

:= 0.25⋅ Pd = 102⋅ kN

1

M xQ.ed

:= ( 0.25⋅ Qd ) ⋅ m = 98.8⋅ kN⋅

TyQ.ed

M yP.ed

:= ( 0.25⋅ Pd ) ⋅ m = 102⋅ kN⋅

TxP.ed

:= 0.15⋅ Pd = 61.2⋅ kN

M yQ.ed

:= ( 0.20⋅ Qd ) ⋅ m = 79.04⋅ kN⋅

TxQ.ed

:= 0.15⋅ Qd = 59.28⋅ kN

:= 0.18⋅ Qd = 71.136⋅ kN

- coeficienti partiali de si#uranta pentru parametrii pamantului ,):

γ ϕ' :=  1.25

γ c' :=  1.25

γ c! := 1.4

γ γ  := 1

Strat 1 :

γ d1 :=

γ k1 γ γ 

= 18.2⋅

kN 3

m

Strat ) : ϕ' d2 :=

#an ( ϕ' k2)

γ ϕ'

= 12.282⋅ deg

c'k2 = 14.4⋅ kPa c'd2 := γ c'

γ d2 :=

γ k2 γ γ 

= 18.5⋅

kN 3

m

Strat / :

ϕ d3 :=

c!d3 :=

#an ( ϕ !k3)

γ ϕ' c!k3

γ c!

= 4.01⋅ deg

= 46.429⋅ kPa

γ d3 :=

γ k3 γ γ 

= 19.5⋅

kN 3

m

I. Predimensionarea fundatiei pe piloti : 1. Stabilirea numarului de piloti: 1.1 %aluarea incarcarii %erticale: $d%

:= 1.2⋅ ( Pk  + Qk ) = 854.4⋅ kN

1.) Incastrarea pilotului in T2: # 1 := 2⋅ d  = 1.2m

se ale#e:

# &

:= 3

1./ 3istanta intre piloti : fisa reala a pilotului: S % := 2⋅ d +

3 100

1.5⋅ d  = 0.9m

 := 6.5m + # & 

 ⋅  = 1.485m

S %

= 9.5m

:= 2

-distanta de la pilot la capatul radierului

2.Capacitatea portanta a pilotului la incarcari verticale ).1 Capacitatea portanta a pilotului: - strat elementar 1* intre -1m...-/m :

)

1 := 2

 *k.1 :=   15kPa

- strat elementar )* intre -/m...-5m : 2 := 2

 *k.2 :=   19.4kPa

- strat elementar /* intre -5m...-!.5m : 3 := 1.5

 *k.3 :=   21.58kPa

- strat elementar 4* intre -!.5m...-m : 4 := 1.5

 *k.4 :=   60.42kPa

- strat elementar 5* intre -m...-6.5m : 5 := 1.5

 *k.5 :=   62.92kPa

perimetrul sectiunii : +

:= π ⋅ d  = 1.885m

, *k  := +⋅ (  *k.1⋅ 1 +  *k.2⋅ 2 +  *k.3⋅ 3 +  *k.4⋅ 4 + *k.5⋅ 5)

= 539.437⋅ kN

).) Capacitatea portanta a pilotului purtator pe %arf  - aria pilotului la baza d

:= π ⋅

- &

2

4

= 0.283m2

 

- %aloarea presiunii acceptate la baza pilotului :  Nc

:= 9

γ dm :=

( 1 + 2 + 3) ⋅ γ d2 + ( 4 + 5) ⋅ γ d3 1 + 2 + 3 + 4 + 5

 &k  := Nc ⋅ c!d3

+ γ dm⋅  = 596.96⋅ kPa

,  &k  :=  &k ⋅ - &

= 168.786⋅ kN

 

= 18.853⋅

kN 3

m

)./ Capacitatea portanta a pilotului mixt *sin#ular: - coeficienti de si#uranta pentru rezistenta pentru piloti forati cu fluid:

γ  & := 1.6 , cd

:=

,  &k 

γ  &

γ * := 1.3 +

, *k 

γ *

= 520.443⋅ k 

- numarul de piloti : n 1 :=

$d% , cd

se ale#e :

= 1.642

n %

:= 4

).4 Capacitatea portanta a pilotului mixt care lucreaza in #rup:  := 2⋅ d +

 0

3 100

:= 5.5m⋅ #an

 ⋅  = 1.485m

 ϕ' d2   4

+

 ϕ d3  

3m⋅ #an  

4

= 0.348m

/

   0

, cg

= 4.273

'7

m!

:= 1

:= m! ⋅ , cd = 520.443⋅ kN

3.Stabilirea solicitarilor de calcul in piloti: /.1 Solicitarile de calcul la baza radierului:  - #reutatea radierului:  kN

γ med := 22

  := 3.8

3

/  := 3.8

  := 80c

m

"  :=  ⋅ / ⋅ 1⋅ m⋅ γ med $d

:= γ "⋅ ( Pk  +

" )

= 317.68⋅ kN

+ γ Q⋅ Qk  = 1.121 ×

3

10 ⋅ kN

- distanta de la C.8. a #rupului de piloti de la pilotul i : x := 1

y := 1

- momentele inco%oietoare la baza radierului : M xed

:= γ "⋅ ( MxP.ek  +

TyP.ek ⋅  )

+ γ Q⋅ ( MxQ.ek  + TyQ.ek ⋅  ) = 399.578⋅ kN⋅

M yed

:= γ "⋅ ( MyP.ek  + TxP.ek  ⋅  ) + γ Q⋅ ( MyQ.ek  + TxQ.ek ⋅  ) = 277.424⋅ kN⋅

- #reutatea propriea a pilotului :  %

:= 9  kN

γ  &e# := 25

3

m " %d



:= γ "⋅ π ⋅

d

2

4

 

⋅  %⋅ γ  &e# = 63.617⋅ k 

Solicitarea din pilotul i: S1d

:=

S2d

:=

S3d

S4d

:=

:=

$d n % $d n %

$d n % $d n %

+ " %d +

+ " %d −

+ " %d −

+ " %d +

M xed ⋅ y  4⋅ ( y )

 2

M xed ⋅ y  4⋅ ( y )

 2

M xed ⋅ y  4⋅ ( y )

 2

M xed ⋅ y  4⋅ ( y )

 2

+

M yed ⋅ x 4⋅ ( x)

+

M yed ⋅ x 4⋅ ( x)

−

 2

M yed ⋅ x 4⋅ ( x)

−

 2

 2

M yed ⋅ x 4⋅ ( x)

 2

= 513.088⋅ kN

= 313.299⋅ kN

= 174.587⋅ kN

= 374.376⋅ kN

4

Sdmax := max( S1d , S2d , S3d , S4d )

Sdmax < , cd

= 513.088⋅ k 

Sdmax

=1

, cd

, cd

= 520.443⋅ k 

= 0.986

4. Armarea radierului: adierul este considerat ca o consola incastrata la fata stalpului actionata de solicitarile din piloti. c1 := 0.8

c4 := c1

= 80⋅c

c2 :=   0.75

c3 := c2

= 75⋅c

:= S1d +

S

S4d

= 887.463⋅ k 

,omente de armare dupa directia 9 : M 11 :=

S⋅ c1  

⋅ m = 186.834⋅ kN⋅

M44 := M11

= 186.834⋅ kN⋅

M33 := M22

= 175.157⋅ kN⋅

,omente de armare dupa directia x : M 22 :=

S⋅ c2 / 

⋅ m = 175.157⋅ kN⋅

 "rmatura dupa directia x-x :  & := 1 a ale#erea materialului - pentru radier se ale#e beton de clasa C)0)5 :

γ *& := 1.5 N

 ck  := 20

2

mm

 cd 

 ck 

:=

γ *&

N

= 13.333⋅

2

mm

- otel st500S :

γ y :=  1.15  yk 

N

:= 500

 yd 

2

:=

mm

 yk 

γ y

N

= 434.783⋅

2

mm

b inaltimea utila - diametrul estimat a barei : cmn

:= max( φ * ,  16mm, 10mm) = 20⋅m

cnm1

:= cmn +  10mm = 30⋅m

- se ale#e :

cnm  := 40m

d x :=   − cnm  −

dy

φ * := 20m

:=   − cnm  −

φ * 2

φ * 2

− 5cm  = 0.7m − φ * − 5cm  = 0.68m

5

- se determina momentul redus :

µ c11 :=

M11  & ⋅ dx ⋅  cd 

ωc11 := 1 − - *c11 :=

= 0.029

2

1

  c11 = 0.029 − 2⋅ µ

ωc11⋅ & ⋅ dx⋅

 cd 

= 6.229⋅ cm2

 yd 

2

- *e1 :=   6.28cm

- se ale#e armatura ϕ1)1 cu  "rmatura dupa directia ;-; :

µ c22 :=

M22

= 0.028

2

 & ⋅ dy ⋅  cd 

ωc22 := 1 −

1

  c22 = 0.029 − 2⋅ µ

- *(c22 := ωc22⋅ & ⋅ dy ⋅

 cd   yd 

2

= 6.011⋅ cm

2

- *e2 :=   6.28cm

- se ale#e armatura ϕ1)1 cu  "ria de armatura maxima si minima : 2

- *max := 0.04⋅ & ⋅ d x = 280⋅ cm  c#m

:= 2.2

N 2

mm



- *mn := max 0.26⋅

 c#m  yk 

 

2

⋅ & ⋅ d x, 0.0013⋅ & ⋅ d x = 9.1⋅ cm

&rocentul de armare : - *e  := max( -*e1, - *e2) - *e  ρ1 := ⋅ 100  = 0.09  & ⋅ dx 0.15 100

 "le#

⋅ & ⋅ dx  = 10.5 cm2

= 6.28⋅ cm2

 < =min' 0.15> -armatura minima necesara 2

- *e  :=   10.58cm

ϕ1!16 - armare minima !

- *e 

ρ1 :=

 & ⋅ d x

⋅ 100  = 0.151

7 =min' 0.15>

-%om folosi frete ϕ la pas de 150mm pentru armatura trans%ersala

.Verificarea la strapun!ere : a Stalp : xd  := 1.7

x*#

:= xd  =  1.7m

yd  :=   1.65

y* #

:= yd  =  1.65m

 % d

:=

$d

   ⋅ / 

= 77.623⋅

kN 2

m

- stabilirea fortei de strapun#ere : $ed

:= Pd + Qd = 803.2⋅ kN

- inaltimea utila a radierului pe directia trans%ersala : d x =  0.7m

- inaltimea utila a radierului pe directia lon#itudinala : d y =  0.68m

- perimetrul de control de referinta d   := 0.5⋅ ( dx +  d y)   %

= 0.69m

:= 2d  =  1.38m

!  := 2⋅ ( 400mm + 500mm   + - c#

π ⋅   % ) = 10.471m

:= 500mm⋅ 400mm + 2⋅  %  ⋅ ( 400mm +

500mm)

+ π ⋅   %  2 = 8.667m2

β := 1.5 forta de strapun#ere :

 νed  := β ⋅

$ed   !⋅ d  

= 166.758⋅

kN 2

m

Capacitatea portanta la strapun#ere a placii fara armatura pentru p reluarea eforturilor de strapun#ere in lun#ul sectiunii de control considerata : η1 := 1

k

:= 1 +

ρ(x :=

 

dc 200mm   dx

- *e ⋅ 100 1m⋅  

= 1.535

= 0.132

:=

0.18 1.5

= 0.12

<)

?

ρ(y :=

- *e ⋅ 100 1m⋅  

= 0.132

ρe  := ρx⋅ ρy  = 0.132

< 0.0) 1

 νdc := dc⋅ η 1 ⋅ k ⋅ ( 100⋅ ρe ⋅ 20) 3 ⋅

2d

⋅

N

 

  % 2 mm

= 1.028⋅

N 2

mm

 νed  < νdc = 1 ezulta ca nu este necesara pre%ederea armaturii pentru preluarea eforturilor din strapun#ere   %

:= d  =  0.69 m

+ !  := 2⋅ ( 400mm + 500mm  

π ⋅   % ) = 6.135 m

β := 1.5 - c#

:= 500mm⋅ 400mm + 2⋅  %  ⋅ ( 400mm +

500mm)

+ π ⋅   %  2 = 2.938m2

forta de strapun#ere :

 νed  := β ⋅

$ed   !⋅ d  

= 284.592⋅

kN 2

m

Capacitatea portanta la strapun#ere a placii fara a rmatura pentru preluarea eforturilor de strapun#ere in lun#ul sectiunii de control considerata :

η1 := 1 dc := k

:= 1 +

ρ(x :=

ρy :=

0.18 1.5  

= 0.12

200mm   dx

- *e ⋅ 100 1m⋅   - *e ⋅ 100   1m⋅  

= 1.535

<)

= 0.132

= 0.132

ρe  := ρx⋅ ρy  = 0.132

< 0.0) 1

 νdc := dc⋅ η 1 ⋅ k ⋅ ( 100⋅ ρe ⋅ 20) 3 ⋅  νed  < νdc = 1

2d

⋅

N

 

  % 2 mm

= 2.056⋅

N 2

mm



ezulta ca nu este necesara pre%ederea armaturii pentru preluarea eforturilor din strapun#ere

".Armarea pilotiilor : - aria de armatura minima conform C)$ pentru piloti cu diametru d'!00mm: 2

< 0.5 m)

- & =  0.283m

- rezulta ca aria minima de armatura lon#itudinala este: -*

:= 0.005⋅ - & = 14.137⋅ cm2

- diametrul minima a barelor lon#itudinale este de 1! mm: - distanta libera intre bare$ masurata pe conturul exterior al pilotului trebuie sa fie mai mic d e )00mm 2

- *e  :=   16.08cm

-se ale#e armatura ϕ1! - *e  - &

> 0.005 = 1

 "rmare la strapun#ere: b &ilotii : - perimetrul de control 

:= d  =  0.69 m

- c#

:= π

d % 4

d %

:= 2⋅  

! %

:= π  d % =  4.335m

2

= 1.496m2

- stabilirea fortei de strapun#ere : $ed%

:= Sdmax = 513.088⋅ k 

forta de strapun#ere :

 νed% := β ⋅ ρx% :=

$ed%   ! % ⋅ d  

- *e ⋅ 100   1m⋅  

= 257.279⋅

kN 2

m

= 0.201

ρy% :=

- *e ⋅ 100   1m⋅  

= 0.201

ρe% := ρx%⋅ ρy%  = 0.201 1

 νdc% := dc⋅ η 1 ⋅ k ⋅ ( 100⋅ ρe%⋅ 20) 3 ⋅

2⋅ d   %

⋅

N 2

mm

= 2.364 ×

3 kN

10

⋅

2

m

 νed% < νdc% = 1 ezulta ca nu este necesara pre%ederea armaturii pentru preluarea eforturilor din strapun#ere 6