Calcul Fundatii

Stabilirea adancimii de fundare:Construcţia este situată în Petrila, judetul Hunedoara, zonă în care adâncimea de

îngheţ este 100...110cm. Deasemenea, este o construcţie definitivă, iar terenul este supus îngheţului. Conform studiului geotehnic nu s-a gasit apa subterana, acesta efectuandu-se pe o adancime de 7 m.

Datorită sistemului structural de rezistenta – cadru spatial din beton armat – al cladirii si a terenului bun de fundare, sistemul de fundare ales este cel cu fundatii izolate, alcatuite dintr-un bloc de beton simplu si un cuzinet din beton armat.

Adâncimea de fundare stabiliă este de -5.35 m faţă de CTN.

1

-incarcarile: N=1353 kNMz=21 kNmVy=8.88 kNMy=46.55 kNmVz=16.72 kN

Alegerea materialului:

-fundatie izolata din beton armat => minim C8/10

Predimensionare:

a =0.5 mb = 0,5 m → dimensiunile stalpilor

-nisip argilos plastic consistent ⇒ pconv−−−

=290 kN/m2

⇒B =2.0 - B=L → dimensiunile blocului de beton simplu L = 2.0 m

lcL=0,5 .. . 0 ,65⇒ lc=(0,5 .. . 0 ,65 )L=(0,5 . .. 0 ,65 )×2.0=1.0 . .. 1.3

m

2

⇒ lc = bc = 1,2 m

hc≥0 ,30 mhclc

≥0 ,25⇒hc≥0 ,25 lc⇒hc≥0 ,25×1 ,20⇒hc≥0,3 m

tg β≥0 ,625⇒hcl1

≥0 ,625⇒hc≥0 ,625 l1⇒hc≥0 ,625×0 .35⇒halignl¿ c ¿¿≥0 .22¿ m⇒hc = 0,40

l1=lc−a

2=1 ,20−0,5

2=0 ,35

m

tg β≥1⇒hc

l1≥1⇒hc≥l1⇒hc≥0 ,375

m

-beton C12/15tg αmin=1,3

-pconv−−−

=290 kPa

tg α=Hl2

⇒H≥tg αmin l2=1,3×0,5=0 ,65 m ⇒ H = 0,7 m

l2=L−lc

2=2 .0−1,0

2=0,5

m

-dimensiunile: B = 2,0 mL = 2,0 m dimensiunile blocului de fundare simpluH = 0,7 m

bc = 1,0 mlc = 1,0 m dimensiunile cuzinetuluihc = 0,40 m

⇒ Df = 5.35 m

1.5 Calculul presiunii maxime acceptate de teren:

-strat de fundare : nisip argilos plastic consistent

⇒ pconv = pconv−−−

+Cb+Cd= 280+56+90.45=426.45 kN/m2 Cb=corectia de latime;pt Df>5 m Cb=0.2∗pconv=0.2∗280=56Cd=corectia de adancime;-pt Df<2m:Cd=k 2∗γ∗D f−2=1.5∗18∗(5.35−2 )=90.45

3

Calculul presiunii pe talpa fundatiei:

-solicitare excentrica pe 2 directii

pef 1,2,3,4=N+Gf

BL±M z

W x

±M y

W y

=N+Gf

BL(1±

6 e y

L±

6exB

)

Gf =BLH γ bet+bc lchc γbet=2,0×2,0×0 ,70×24+1,0×1,0×0 ,40×25=77 . 2 kN

e x=M y

N+Gf

=461353+77 . 2

=0 . 03 m

e y=M z

N +Gf

=211353+77 . 2

=0 ,01 m

pef 1=N+G f

BL(1+

6e y

L+

6 exB

)=1353+77 .22,0×2,0

×(1+ 6×0 ,012,0

+ 6×0 ,032,0

)=400 kN/m2

pef 2=N+G f

BL(1+

6e y

L−

6exB

)=1353+77 .22,0×2,0

×(1+ 6×0 ,012,0

−6×0 ,032,0

)=336 kN/m2

pef 3=N+G f

BL(1−

6e y

L+

6exB

)=1353+77 . 22,0×2,0

×(1−6×0 ,012,0

+ 6×0 ,032,0

)=356 kN/m2

pef 4=N+Gf

BL(1−

6e y

L−

6exB

)=1353+77 .22,0×2,0

×(1−6×0 ,012,0

−6×0 ,032,0

)=313 kN/m2

Calculul armaturii de rezistenta din cuzinet:

4

N01=N+G cuz=N+l c bchc γ bet=1175+1 ,20×1,20×0 ,40×25=1190 kN

M 01 x=M y+V z hc=96+50×0,4=116 kNmM 01 y=M z+V y hc=111+58×0,4=134 kNm

e x=M 01 y

N01

=1341190

=0 ,112 m

e y=M olx

N 01

=1161190

=0 ,097 m

pcx 1=N 01

lcbc(1+

6exlc

)=11901 ,20×1 ,20

×(1+ 6×0 ,1121 ,20

)=1289 kN/m2

pcx2=N 01

lcbc(1−

6exlc

)=11901,20×1,20

×(1−6×0 ,1121 ,20

)=364 kN/m2

pcy1=N 01

lcbc(1+

6e y

lc)=1190

1 ,20×1 ,20×(1+ 6×0 ,097

1 ,20)=1227

kN/m2

pcy1=N 01

lcbc(1−

6e y

lc)=1190

1 ,20×1,20×(1−6×0 ,097

1 ,20)=426

kN/m2

pcx 0=pcx 1−pcx 2

lc( lc−l x )+ pcx 2=

1289−3641 ,20

×(1 ,20−0 ,375 )+364=1000 kN/m2

pcy 0=pcy 1−pcy 2

lc( lc−l y )+ pcy 2=

1227−4261 ,20

×(1 ,20−0 ,375)+426=976 kN/m2

l x=l y=lc−a

2=1 ,20−0 ,45

2=0 ,375

m

pc ,med, x=pcx1+ pcx2

2=1289+364

2=826 .5

kN/m2

pc ,med , y=pcy1+ pcy2

2=1227+426

2=826 .5

kN/m2

M x=bc[ pcx0

lx2

2+( pcx1−pcx0 )

lx2

23lx ]=1 ,20×[1000×0 ,3752

2+(1289−1000 )×0 ,375

2×2

3×0 ,375 ]

Mx = 185 kNm

M y=lc [ pcy 0

l y2

2+( pcy 1−pcy 0 )

l y2

23l y ]=1 ,20×[ 976×0 ,3752

2+(1227−976 )×0 ,375

2×2

3×0 ,375 ]

My = 96.4 kNm

5

d=hc−(cnom+φ2

)=40−(2+ 1,82

)=37 ,1 cm

cnom=cmin+Δcdev=10+10=20 mmcmin=max {cad ;cdurab;10mm}=10 mmΔcdev=10 mmΦmax = 18 mm

h0x = d = 37,1 cmh0y = h0x-Φmax=37.1-1,1=36 cm

-armatura OB37 => Ra = 2100 kN/m2

Aax=M x

0 ,875h0 xRa

=1850 ,875×0 ,371×2100

=27 . 13 cm2 => 11Φ18 → 27.94 cm2

Aay=M y

0 ,875h0 y Ra

=96. 40 ,875×0 ,36×2100

=14 . 57 cm2 => 6Φ18 → 15.24 cm2

px%=

Aax

B∗h0 x

∗100=0. 336%

px%=

Aay

L∗h0 y

∗100=0. 178%

pmin=0.1 %

CALCULUL TASARILOR PROBABILE

Df=−1,5m;B=1,2m;

pef=N+Gf

A=1175+103

5,29=241,6[kPa]

Gf =BLH γ bet+bc lchc γbet=2,3×2,3×0 ,70×24+1 ,20×1 ,20×0 ,40×25=103 kN

σ zi=∝0 i∗pnet ;

∝0 i= f ( ziB ,LB );

σ gzi=γ∗D f+∑ hi∗γ i

hi≤0,4∗B=0,9m

s=100∗0,8∗∑ σ zmed , i∗hi

E i

≤ sadm

sadm=8 cm;

6

pnet=pef−γ∗D f ; pnet=241,6−19,65∗1,5=212,1;

σ zi=4∗Kd*pnet;

Kd=f[L/B;2zi/2];

Pentru Ϭz1;

Ϭz1=4*Kd*pn=4*0.238*212,1=202;

2*z1/B=2*0,6/2,3=0,521;

-pentru L/B=1 =>0.25….0.027 =>x=0.00210.23 0.02…..x

0.5…...0.0120.25…...x =>x=0.0072

Kd=0.23+0.0072=0.238;

Pentru Ϭz2:

Ϭz2=4*Kd*pn=4*0.1884*212,1=159.83;

2*z1/B=2*1,1/2,3=0,956;

-pentru L/B=1 =>0.25….0.031 =>x=0.02480.181 0.2…..x

7

0.5…...0.0120.25…...x =>x=0.0072

Kd=0.1812+0.0072=0.1884;

Pentru Ϭz3:

Ϭz3=4*Kd*pn=4*0.118*212,1=100.1;

2*z1/B=2*2/2,3=1.739;

-pentru L/B=1 =>0.25….0.037 =>x=0.0170.104 0.23…..x

0.5…...0.0240.25…...x =>x=0.0144

Kd=0.104+0.0144=0.118;

Pentru Ϭz4:

Ϭz4=4*Kd*pn=4*0.077*212,1=65.32;

2*z1/B=2*2,9/2,3=2.52;

-pentru L/B=1 =>1….0.039 =>x=0.0120.063 0.521…..x

0.5…...0.0230.25…...x =>x=0.0138

Kd=0.063+0.0138=0.077;

Pentru Ϭz5:

Ϭz5=4*Kd*pn=4*0.049*212,1=41.57;

2*z1/B=2*3.8/2,3=3.3;

-pentru L/B=1 =>1….0.018 =>x=0.0050.039 0.304…..x

0.5…...0.0160.25…...x =>x=0.00096Kd=0.0039+0.0096=0.049;

8

Pentru Ϭz6:

Ϭz6=4*Kd*pn=4*0.032*212,1=27.14;

2*z1/B=2*4.7/2,3=4.08;

-pentru L/B=1 =>2….0.014 =>x=0.00060.026 0.086…..x

0.5…...0.0110.25…...x =>x=0.0066

Kd=0.026+0.0066=0.032;

Pentru Ϭz7:

Ϭz7=4*Kd*pn=4*0.028*212,1=23.75;

2*z1/B=2*5.6/2,3=4.782;

-pentru L/B=1 =>2….0.014 =>x=0.00540.0215 0.782…..x

0.5…...0.0110.25…...x =>x=0.0066

Kd=0.0215+0.0066=0.028;

Pentru Ϭz8:

Ϭz8=4*Kd*pn=4*0.023*212,1=19.51;

2*z1/B=2*6.5/2,3=5.47;

-pentru L/B=1 =>2….0.014 =>x=0.01030.016 1.478…..x

2…...0.0110.25…...x =>x=0.0066

Kd=0.016+0.0066=0.023;

9

Ϭgz0=γ1*Df=19.5*1.5=29.25;

Ϭgz1=γ1*Df+ γ1*h1=19.5*1.5+19.5*0.6=41;

Ϭgz2= γ1*Df+ γ1*h1+ γ1*h2=19.5*1.5+19.5*0.6+19.5*0.5=50.75

Ϭgz3= γ1*Df+ γ1*h1+ γ1*h2+ γ1*h3=19.5*1.5+19.5*0.6+19.5*0.5+19.5*0.9=67.55

Ϭgz4= Ϭgz3+ γ2*h4=67.55+19.8*0.9=85.37

Ϭgz5= Ϭgz4+ γ2*h5=85.37+19.8*0.9=103.19

Ϭgz6= Ϭgz5+ γ2*h6=103.19+19.8*0.9=121

Ϭgz7= Ϭgz6+ γ2*h7=121+19.8*0.9=138.83

Ϭgz8= Ϭgz7+ γ2*h7=138.83+19.8*0.9=156.65

Fasia hi zi Hi γi pnet αoi Ei бzi бgzi0,2 бgzi

бgzi med s

  [m] [m] [m] [KN/mc][KN/mp]   [kPa] [KN/mp] [KN/mp] [KN/mp] [KN/mp] [cm]

0 0.6 0.600 1.500 19.5 212 0.521 14000 110 29 6    

1 0.5 1.100 2.100 19.5 212 0.956 14000 202 41 8 157 0.4472 0.9 2.000 3.000 19.5 212 0.174 32000 160 59 10 120 0.270

3 0.9 2.900 3.900 19.8 212 0.252 32000 100 77 13 45 0.102

4 0.9 3.800 4.800 19 212 0.330 32000 65 95 17 62 0.139

5 0.9 4.700 5.700 19 212 0.408 32000 41 113 20 78 0.176

6 0.9 5.600 6.500 19 212 0.478 40000 27 129 24 94 0.169

7 0.9 6.500 7.300 19 212 0.547 40000 23 145 28 109 0.196

8 0.9 7.400 7.300 19 212 0.210 45000 19 145 31 80 0.128

sadm=8 cm;

S <sadm

10

s= 1.615

2. Fronton:

2.1 Incarcari de calcul:

-incarcarile:

N=1578,05 kNMz=145,14 kNmVy=87,07 kNMy=96,68 kNmVz=56,55 kN

2.2 Alegerea materialului:

-fundatie izolata din beton armat => minim C8/10

2.3 Predimensionare:

11

a = b = 0,40 m → dimensiunile stalpilor

-roca stancoasa ⇒ pconv−−−

=1000 kN/m2

⇒ B = L = 1,50 m- B=L → dimensiunile blocului de beton simplu

lcL=0,5 .. . 0 ,65⇒ lc=(0,5 .. . 0 ,65 )L=(0,5 . .. 0 ,65 )×1 ,50=0 ,75. . .0 ,975

m⇒ lc = bc = 0,80 m

hc≥0 ,30 mhclc

≥0 ,25⇒hc≥0 ,25 lc⇒hc≥0 ,25×0 ,80⇒hc≥0 ,20 m

tg β≥0 ,65⇒hcl1

≥0 ,65⇒hc≥0 ,65 l1⇒hc≥0 ,65×0,2⇒hc≥0 ,13 m ⇒ hc = 0,50 m

l1=lc−a

2=0 ,80−0 ,40

2=0 ,20

m

tg β≥1⇒hc

l1≥1⇒hc≥l1⇒hc≥0 ,20

m

-beton C12/15tg αmin=1 ,85

-pconv−−−

=1000 kPa

12

tg α=Hl2

⇒H≥tg αmin l2=1 ,85×0 ,35=0 ,65 m ⇒ H = 1,00 m

l2=L−lc

2=1 ,50−0 ,80

2=0 ,35

m

-dimensiunile: B = 1,50 mL = 1,50 m dimensiunile blocului de fundare simpluH = 1,00 m

bc = 0,80 mlc = 0,80 m dimensiunile cuzinetuluihc = 0,50 m

⇒ Df = 3,70 m

2.4 Calculul presiunii maxime acceptate de teren:

-strat de fundare : roca stancoasa ⇒ pconv = pconv−−−

= 1000 kN/m2

pteren=βpconv=1,4×1000=1400 kN/m2 Β = 1,4 → pentru gruparea fundamentala; incarcari cu excentricitate pe ambele

directii

2.5 Calculul presiunii pe talpa fundatiei:

-solicitare excentrica pe 2 directii

pef 1,2,3,4=N+Gf

BL±M z

W x

±M y

W y

=N+Gf

BL(1±

6 e y

L±

6exB

)

Gf =BLH γ bet+bc lchc γbet=1 ,50×1 ,50×1 ,00×24+0 ,80×0 ,80×0 ,50×25=62 kN

e x=M y

N+Gf

=96 ,681578 ,05+62

=0 ,059 m

e y=M z

N +Gf

=145 ,141578 ,05+62

=0 ,089 m

pef 1=N+G f

BL(1+

6e y

L+

6 exB

)=1578 ,05+621 ,50×1 ,50

×(1+ 6×0 ,0891 ,50

+ 6×0 ,0591 ,50

)=1161 kN/m2

13

pef 2=N+G f

BL(1+

6e y

L−

6exB

)=1578 ,05+621 ,50×1 ,50

×(1+ 6×0 ,0891,50

−6×0 ,0591 ,50

)=817 kN/m2

pef 3=N+G f

BL(1−

6e y

L+

6exB

)=1578 ,05+621 ,50×1 ,50

×(1−6×0 ,0891 ,50

+ 6×0 ,0591 ,50

)=642 kN/m2

pef 4=N+Gf

BL(1−

6e y

L−

6exB

)=1578 ,05+621 ,50×1 ,50

×(1−6×0 ,0891 ,50

−6×0 ,0591,50

)=298 kN/m2

2.6 Calculul armaturii de rezistenta din cuzinet:

N01=N+G cuz=N+l c bchc γ bet=1578 ,05+0 ,80×0 ,80×0 ,50×25=1587 kN

M 01x=M y+V z hc=96 ,68+56 ,55×0,5=124 ,96 kNmM 01 y=M z+V y hc=145 ,14+87 ,07×0,5=188 ,68 kNm

e x=M 01 y

N01

=188 ,681585

=0 ,119 m

e y=M 01 z

N01

=124 ,961585

=0 ,079 m

pcx 1=N 01

lcbc(1+

6exlc

)=15870 ,80×0 ,80

×(1+ 6×0 ,1190 ,80

)=3366 kN/m2

pcx 2=N 01

lcbc(1−

6exlc

)=15870 ,80×0 ,80

×(1−6×0 ,1190 ,80

)=1595 kN/m2

14

pcy 1=N 01

lcbc(1+

6e y

lc)=1587

0 ,80×0 ,80×(1+ 6×0 ,079

0 ,80)=3068

kN/m2

pcy1=N 01

lcbc(1−

6e y

lc)=1587

0 ,80×0 ,80×(1−6×0 ,079

0 ,80)=1893

kN/m2

pcx0=pcx1−pcx2

lc( lc−l x )+ pcx2=

3366−15950 ,80

×(0 ,80−0 ,20 )+1595=2924 kN/m2

pcy 0=pcy 1−pcy 2

lc( lc−l y )+ pcy 2=

3068−18930 ,80

×(0 ,80−0 ,20 )+1893=2775 kN/m2

l x=l y=lc−a

2=0 ,80−0 ,40

2=0 ,20

m

pc ,med , x=pcx1+ pcx2

2=3366+1595

2=2481

kN/m2

pc ,med , y=pcy1+ pcy2

2=3068+1893

2=2481

kN/m2

M x=bc[ pcx0

lx2

2+( pcx1−pcx0 )

lx2

23lx ]=0 ,80×[ 2924×0 ,202

2+(3366−2924 )×0,2

2×2

3×0 ,20 ]

Mx = 51,5 kNm

M y=lc [ pcy 0

l y2

2+( pcy1−pcy0 )

l y2

23l y ]=0 ,80×[ 2775×0 ,202

2+(3068−2775 )×0,2

2×2

3×0 ,20 ]

Mx = 47,5 kNm

d=hc−(cnom+φ2

)=50−(2+ 1,42

)=47 ,3 cm

cnom=cmin+Δcdev=10+10=20 mmcmin=max {cad ;cdurab;10mm}=10 mmΔcdev=10 mmΦmax = 14 mm

h0x = d = 47,3 cmh0y = h0x-Φmax=47,3-1,4=45,9 cm

-armatura OB37 => Ra = 2100 kN/m2

15

Aax=M x

0 ,875h0 xRa

=51 ,50 ,875×0 ,473×2100

=5 ,93 cm2 => 5Φ14 → 7,70 cm2

Aay=M y

0 ,875h0 y Ra

=47 ,50 ,875×0 ,459×2100

=5 ,64 cm2 => 5Φ14 → 7,70 cm2

3.Grinda rigidizare curenta:

2.1 Incarcari de calcul:

l=5 . 00−lc2x 2=5 ,00−1 ,20

2×2=3 ,80

mggr=bgr hgr γ bet∗c=0 ,30×0 ,40×25x 1,2=3,6 kN/m

gpl=( 3 ,802

×2 )×0 ,10×25=9,5 kN/m

q pl=( 3 ,302

×2)×3 ,20=11 kN/m

gzid=hetaj γ zid=2,8×6=16 ,8 kN/mQtotal=ggr+gpl+q pl+gzid=3,6+9,5+11+16 ,8=40 ,9 kN/m

-talpa fundatiei se gaseste in stratul: praf argilos vartos

⇒ pconv = pconv−−−

+Cb+Cd= 275 kN/m2 Cb=corectia de latime;B=latimea fundatiei in m;K1=coef.care pt pamanturile coezive=0.05;

Cb=pconv−−−

*K1*(B-1)=23,25Cd=corectia de adancime;

-pt Df<2m:Cd=pconv−−−

*

Df−24 =-38,75

16

Bnec=1 ,15Qpconv

=1 ,15×41275

=0 ,167 m =17 cm < bp => bgr = 40 cm

pef=Qbgr

=410 ,40

=102 ,5 kN/m2 < pconv => armare constructiva

=> 3Φ14 jos si 3Φ14 sus - bare longitudinale => Φ10/15 - etrieri

3.Grinda rigidizare curenta:

2.1 Incarcari de calcul:

l=6 ,60−l c2

2=6 ,60−0 ,802

×2=5 ,80 m

ggr=bgr hgr γ bet=0 ,40×0 ,50×25=5 kN/m

gpl=3 ,30

2×0 ,15×25=6,2

kN/m

q pl=3 ,30

2×3 ,20=5,3

kN/mgzid=hetaj γ zid=2,8×17=47 ,6 kN/mQtotal=ggr+gpl+q pl+gzid=5+6,2+5,3+47 ,6=64 ,1 kN/m -talpa fundatiei se gaseste in stratul: nisip mare si mijlociu cu fragmente de roca

⇒ pconv−−−

=600 kN/m2

pconv=pconv−−−+CB+CD=600−36=564 kN/m2

-pamant necoeziv => k1=0,1

17

-CB=pconv−−−K1 (B−1)=600×0,1×(0,4−1 )=−36 kN/m2

-CD=K2γ (Df−2)=0

-pamant necoeziv => K2=2,5-γ = 0

Bnec=1 ,15Qpconv

=1 ,15×64 ,1564

=0 ,131 m = 13,1 cm < bp => bgr = 40 cm

pef=Qbgr

=64 ,10 ,40

=160 kN/m2 < pconv => armare constructiva

=> 3Φ14 jos si 3Φ14 sus - bare longitudinale => Φ10/15 - etrieri

18