CALCUL FUNDATII IZZOLATE
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Transcript of CALCUL FUNDATII IZZOLATE
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