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FUNDATII IZOLATEFUNDATII IZOLATEFUNDATII IZOLATEFUNDATII IZOLATE
Date generale: n 4:= a 0:=
stalp: • bs 0.4:= armati cu: 8ϕϕϕϕ16 PC52 longitudinal
ls 0.5:= etrieri 10ϕϕϕϕ 100
pereti zidarie: • BCA 30 cm
Hp 3.0:=
cota terenului sistematizat:• CTA 0.4 a+( ) 0.4=:=
stratificatia terenului:•
0.00 ... -0.50: Umplutura
-0.50 ...-3.50: Argila prafoasa, cafenie,plastic consistenta
γγγγk1 18.2:= w1 22%:= Ip1 21%:= Ic1 0.70:=
ϕϕϕϕk1 12deg:= e1 0.69:= E1 12000:=
ck1 45 0.15n+( ) 45.6=:=
-3.50 ... -10.00: Nisip argilos ,galben cafeniu
γγγγk2 19:= w2 18%:= Ip2 11%:= ϕϕϕϕk2 24deg:=
e2 0.65:= E2 14000:= ck2 5 0.2n+( ) 5.8=:=
Incarcarile de calcul pentru fiecare stalp se vor determina considerand:
Incarcarea Distribuita Permanenta:•
Incarcarea Distribuita Variabila :•
pk 15 0.1n+( ) 15.4=:=
qk 8 0.1n+( ) 8.4=:=
Stalpul S1•
SLU - GEO+STR , CP3 : A1/A2+M2+R3 , conf SR EN 1997•SLEN conf SR EN 1997 (calculul tasarii prin insumare pe strate elementare)•
γγγγGf 1.0:= γγγγQf 0:= γγγγR 1.40:= γγγγc 1.25:=
γγγγGn 1.35:= γγγγQn 1.5:= γγγγϕϕϕϕ 1.25:=
1. ) PREDIMENSIONARE
Stabilirea Df :•
Df Hing 10...20( )cm+≥
Df. 0.90 0.2+ 1.1=:=
Stabilirea lui B si L ( bloc de fundare ) :•
peff pacc≤ AafS1 5 3⋅ 15=:=
pconv 408:= obtinut prin interpolare din tab de mai jos
pd pk γγγγGn⋅ 20.79=:= pacc pconv 0.4081
PakPa⋅=:=
qd qk γγγγQn⋅ 12.6=:=
NS1 AafS1 pd qd+( )⋅ 500.85=:=
Gf 0.25 NS1⋅ 125.213=:=
VdpS1 NS1 Gf+ 626.063=:=
ls
bs
1.25=L1
B1
ls
bs
= 1.25= L1 B1 1.25⋅=
VdpS1
pconv
1.534= B1 L1⋅VdpS1
pconv
= 2.181=
B1
VdpS1
1.25 pconv⋅1.108=:= L1 B1 1.25⋅ 1.385=:=
B , L multiplu de 5 cm
B , L > 40 cm
Bs1 1.50:=
alegem : Ls1 2.0:=
Stabilirea lui lc , bc ( cuzinet ) :•
lc1 0.5Ls1 1=:= alegem beton clasa C 12/15
bc1 0.5Bs1 0.75=:=
lcs1 1.10:=
alegembcs1 0.80:=
Stabilirea lui Hb :•
tan αααα( ) tan ααααadm( )≥
tan ααααadm( ) 1.52:=
clasa C 8/10
I.) tan αααα( )Hbloc
Ls1 lcs1−( )2
=
Hb1. 1.52Ls1 lcs1−( )
2⋅ 0.684=:=
II.) tan αααα( )Hbloc
Bs1 bcs1−( )2
=
Hb1.. 1.52Bs1 bcs1−( )
2⋅ 0.532=:=
Hb1 max Hb1. Hb1.., ( ) 0.684=:=
Hbloc-- multiplu de 5 cm
-- > 40 cmalegem : Hbloc1 0.70:=
Stabilirea lui hc :•
tan ββββ( ) tan ββββadm( )≥ tan ββββadm( ) 1.00:=
I.) tan ββββ( )hc
lcs1 ls−( )2
=
hc1. 1.00lcs1 ls−( )
2⋅ 0.3=:=
II.) tan ββββ( )hc
bcs1 bs−( )2
=
hc1.. 1.00bcs1 bs−( )
2⋅ 0.2=:=
hc max hc1. hc1.., ( ) 0.3=:=
hcz1-- multiplu de 5 cm
-- > 30 cmalegem : hcz1 0.30:=
VERIFICAREA LA CAPACITATE PORTANTA
( SLU - GEO )
C. P. 3•
Eforturile sectionale :
NS1 500.85=
Mx1 0.2 NS1⋅ 100.17=:= Tafx1 0.1 NS1⋅ 50.085=:=
My1 0.1 NS1⋅ 50.085=:= Tafy1 0.05 NS1⋅ 25.043=:=
hgr CTA 0.2+ 0.6=:=bgr 0.30:=
Adancimea de fundarehgr 2 bgr⋅ 0.6=:=
Df. 0.20 hcz1+ Hbloc1+ 1.2=:=Tr 5.0:=
Hz 3.0:=γγγγbet 25:=
bz 0.3:=
γγγγmed 22:= γγγγBCA 8:=Htot1 Df. CTA+ 1.6=:=
Vd Ned Npd+ Gfd+=
Npd1 γγγγGn hgr bgr⋅ Tr⋅ γγγγbet⋅ Hz bz⋅ Tr⋅ γγγγBCA⋅+( )⋅ 78.975=:=
Gfd1 γγγγGn Bs1 Ls1⋅ Htot1⋅ γγγγmed⋅( )⋅ 142.56=:=
Vd1 NS1 Npd1+ Gfd1+ 722.385=:=
e1
bs
2
bgr
2+ 0.35=:=
Mxfed1 Mx1 Tafx1 Hbloc1 hcz1+( )⋅+ Npd1 e1⋅+ 177.896=:=
Tyfed1 Tafy1 25.043=:=
Myfed1 My1 Tafy1 Hbloc1 hcz1+( )⋅+ 75.128=:=
Txfed1 Tafx1 50.085=:=
Hd1 Tyfed12Txfed1
2+
55.997=:=
Calculul presiunii active la talpa fundatiei•
pef
Vd
A1
=
pcr
Rd
A1
=
≤ A1 B1 L1⋅=
eB1
Myfed1
Vd1
0.104=:= B1 Bs1 2eB1− 1.292=:=
eL1
Mxfed1
Vd1
0.246=:= L1 Ls1 2eL1− 1.507=:=
eB1
Bs1
6<
A1 B1 L1⋅ 1.948=:=
eL1
Ls1
6<eB1
Bs1
2eL1
Ls1
2
+ 0.02=
E "VERIFICA"eB1
Bs1
2eL1
Ls1
2
+
1
9<if
"NU VERIFICA" otherwise
:=E "VERIFICA"=
pef1
Vd1
A1
370.899=:=
Calculul capacitatii portante•
-- se face in conditii drenate
αααα 0:= inclinarea bazei fundatiei
c1
ck1
γγγγc36.48=:=
ϕϕϕϕ1 9.56deg:=
Nq 2.374:=
Nc 8.155:=
Nγγγγ 0.463:=
mB
2B1
L1
+
1B1
L1
+
1.538=:=bq 1 αααα tan ϕϕϕϕ1( )2⋅−
1=:=
bγγγγ bq 1=:=
bc bq
1 bq−( )Nc tan ϕϕϕϕ1( )⋅( )
− 1=:=mL
2L1
B1
+
1L1
B1
+
1.462=:=
sq 1B1
L1
sin ϕϕϕϕ1( )⋅+ 1.142=:=
θθθθ 63.43deg:=
m mL cos θθθθ( )2⋅ mB sin θθθθ( )
2⋅+ 1.523=:=sγγγγ 1 0.3
B1
L1
⋅− 0.743=:=
sc
sq Nq⋅ 1−( )Nq 1−( )
1.246=:=Hd1 55.997=
iq 1Hd1
Vd1 A1 c1⋅1
tan ϕϕϕϕ1( )⋅+
−
m
0.895=:=
iγγγγ 1Hd1
Vd1 A1 c1⋅1
tan ϕϕϕϕ1( )⋅+
−
m 1+
0.831=:=
ic iq
1 iq−( )Nc tan ϕϕϕϕ1( )⋅( )
− 0.882=:=
q1 Df. γγγγk1⋅ 21.84=:=
pacc
Rd
A1
=
pacc1 c1 Nc⋅ bc⋅ sc⋅ ic⋅ q1 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γγγγk1⋅ B1⋅ Nγγγγ⋅ bγγγγ⋅ sγγγγ⋅ iγγγγ⋅+ 383.109=:=
pacc1 383.109=
pef1 370.899=
V "VERIFICA" pacc1 pef1≥if
"NU VERIFICA" otherwise
:=V "VERIFICA"=
VERIFICAREA LA TASARE
( SLEN )
conditia SED SCD≤
SEDtasarea pamantului sub fundatie
SCDtasarea admisibila
Calculul luiSED folosind metoda insumarii pe straturi elementare
-- se folosesc valorile caracteristice γγγγG 1:= γγγγQ 1:=
Vd Ned Npd+ Gfd+=
Npk1 γγγγG hgr bgr⋅ Tr⋅ γγγγbet⋅ Hz bz⋅ Tr⋅ γγγγBCA⋅+( )⋅ 58.5=:=
Gfk1 γγγγG Bs1 Ls1⋅ Htot1⋅ γγγγmed⋅( )⋅ 105.6=:=
NSk1 AafS1 pk qk+( )⋅ 357=:=
Vk1 NSk1 Npk1+ Gfk1+ 521.1=:=
pef.1
Vk1
Bs1 Ls1⋅173.7=:=
pneta1 pef.1 γγγγk1 Df.⋅− 151.86=:=
straturile elementare :•
h1. 0.4 Bs1⋅ 0.6=:=
alegem h1 0.40m:=h1.. 1.0:=
h1 min h1. h1.., ( )=
εεεεσσσσ
ΕΕΕΕ= S
σσσσ
ΕΕΕΕH⋅= σσσσz 0.2 σσσσgz⋅≤
σσσσz αααα0 pneta1⋅= σσσσz efortul− transmis de structura in pamant
σσσσgz γγγγ Df⋅ + ΣΣΣΣhi γγγγi⋅+= σσσσgz presiunea− geologica
SED ββββ 100⋅ ΣΣΣΣ⋅ σσσσzi.med
hi
ΕΕΕΕ⋅
⋅= ββββ 0.8:=
din calcule rezulta : SED1 0.325cm:=
SCD1 5.0cm:= pentru pamanturi necoezive
S1 "VERIFICA" SED1 SCD1≤if
"NU VERIFICA" otherwise
:=S1 "VERIFICA"=
?2= 19
?1= 18.2
Df= 1.2
Df*?1= 21.84
E1= 12000
E2= 14000
Sed1= 0.00325
Date generale
TASAREA S1
m m - - - kPa kPa kPa kPa kPa
1 0.4 0.4 0.267 1.333 0.914 138.8 145.33 29.12 25.48 5.096
2 0.4 0.8 0.533 1.333 0.708 107.5169 123.16 36.4 32.76 6.552
3 0.4 1.2 0.8 1.333 0.477 72.43722 89.977 43.68 40.04 8.008
4 0.4 1.6 1.067 1.333 0.359 54.51774 63.477 50.96 47.32 9.464
5 0.4 2 1.333 1.333 0.261 39.63546 47.077 58.24 54.6 10.92
6 0.3 2.3 1.533 1.333 0.211 32.04246 35.839 63.7 60.97 12.194
7 0.4 2.7 1.8 1.333 0.167 25.36062 28.702 71.3 67.5 13.5
8 0.4 3.1 2.067 1.333 0.132 20.04552 22.703 78.9 75.1 15.02
9 0.4 3.5 2.333 1.333 0.114 17.31204 18.679 86.5 82.7 16.54
10 0.4 3.9 2.6 1.333 0.095 14.4267 15.869 94.1 90.3 18.06
0.2*τgz med Nr strat
hi z z/B L/B α0 sup/inf τz sup/inf τz med τgz sup/inf τgz med
ARMAREA FUNDATIEI S1
Ned.c1 NS1 Npd1+ lcs1 bcs1⋅ hcz1⋅ γγγγbet⋅( ) γγγγGn⋅+ 588.735=:=
Mxed.c1 Mx1 Tafx1 hcz1⋅+ Npd1 e1⋅+ 142.837=:=
eL.
Mxed.c1
Ned.c1
0.243=:= pc1.2
Ned.c1
lc bc⋅1 6−
eL
lc
⋅+
⋅=
pc1
Ned.c1
lcs1 bcs1⋅1 6
eL.
lcs1
⋅+
⋅ 1.554 103
×=:=
pc2
Ned.c1
lcs1 bcs1⋅1 6
eL.
lcs1
⋅−
⋅ 216.335−=:=
pc2 0< admitem pc2=0
elcs1
60.183=:= c1
lcs1
2e− 0.367=:=
p12
3
Ned.c1
bcs1 c1⋅⋅ 1.338 10
3×=:= p0.1 975:=
Armatura As3•
cnom 0.05:=x1 1.345:=
Ta1 pc2
x1
2⋅ bcs1⋅ 116.388−=:=
alegem PC 52 cu : fyd345
1.15300=:=
armam constructiv cu 5 ϕϕϕϕ 10 ladistanta de 25 cm
A1.S3
Ta1
fyd
0.388=:=
A1.S3ef 3.14 104−
×:=ρρρρ3
A1.S3ef
bcs1 hcz1 cnom−( )⋅1.57 10
3−×=:=
Armatura As1•lcy1
lcs1 ls−( )2
0.3=:=
Mx.x1 bcs1 p0.1 lcy1⋅lcy1
2⋅ p1 p0.1−( )
lcy1
2⋅
2
3⋅ lcy1⋅+
⋅ 43.813=:=
cnom 0.05= ϕϕϕϕsl 0.014:= avem clasa de beton C12/15
fcd12
1.58=:=
dx hcz1 cnom−ϕϕϕϕsl
2− 0.243=:=
μμμμx μμμμlim<μμμμx
Mx.x1
bcs1 dx2⋅ fcd⋅ 1000⋅
0.116=:= μμμμlim 0.403:=
fyk 345MPa:= fctm 1.6MPa:=ωωωω 1 1 2μμμμx−− 0.124=:=
Asmin 0.26fctm
fyk
⋅ bcs1⋅ 104
⋅ dx⋅ 2.344=:=Asx ωωωω bcs1⋅ 10
4⋅ dx⋅
fcd
fyd
⋅ 6.406=:=
alegem ϕϕϕϕ 5 14 cu : As1ef 7.70cm2:=
distanta dintre bare : d = 17.5 cmA1.S1 7.70 10
4−⋅:=
ρρρρl
A1.S1
bcs1 dx⋅3.961 10
3−×=:= ρρρρmin 0.075% 7.5 10
4−×=:=
ρρρρ1 ρρρρmin>
Armatura As2•
pmed1
p1 0+( )2
669.017=:= lcx1
bcs1 bs−( )2
0.2=:=
My.y1 lcs1 pmed1 lcx1⋅lcx1
2⋅
⋅ 14.718=:=
ϕϕϕϕs2 0.010:= cnom 0.05=
dy hcz1 cnom−ϕϕϕϕs2
2− ϕϕϕϕsl− 0.231=:=
μμμμy
My.y1
lcs1 dy2⋅ fcd⋅ 10
3⋅
0.031=:= μμμμlim. 0.403:= μμμμy μμμμlim<
ωωωω 1 1 2μμμμy−− 0.032=:=Asmin. 0.26
fctm
fyk
⋅ lcs1⋅ dy⋅ 3.064 104−
×=:=
Asy ωωωω lcs1⋅ dy⋅fcd
fyd
⋅ 2.158 104−
×=:=
alegem ϕϕϕϕ 5 10 cu : As2ef 3.92cm2:=
distanta dintre bare : d = 25 cm
A1.S2 3.92 104−
⋅:=
ρρρρ2
A1.S2
lcs1 dy⋅1.543 10
3−×=:= ρρρρmin. 0.075% 7.5 10
4−×=:=
ρρρρ2 ρρρρmin<
Lungimi de ancoraj + lungimea ciocurilor•
As1
Lcioc1 15 ϕϕϕϕsl⋅ 0.21=:=
As2
Lcioc2 15 ϕϕϕϕs2⋅ 0.15=:=
As3
αααα1 1:= αααα3 1:= αααα5 1:=lbrqd 94:=
αααα2 0.7:= αααα4 0.7:=
la αααα1 αααα2⋅ αααα3⋅ αααα4⋅ αααα5⋅ lbrqd⋅ 46.06=:=
Lanc3. la 25+ 71.06=:=
Lanc3 75cm:=
ϕϕϕϕ Armatura din stalp 8 16
lbrqd. 105cm:=
la. αααα1 αααα2⋅ αααα3⋅ αααα4⋅ αααα5⋅ lbrqd.⋅ 0.514m=:=
LancS1. la. 250mm+ 0.764m=:=
LancS1 80cm:=
Stalpii S2+S3•
1.) PREDIMENSIONARE S2
Stabilirea Df :•
Df Hing 10...20( )cm+≥
Df2 1.1:=
Stabilirea lui L2 , B2 ,H2 :•
peff pacc≤ AafS2 6 2.5⋅ 15=:=
pconv2 408:= obtinut prin interpolare
pacc2 pconv2 408=:=pd. 20.79:=
qd. 12.6:=
NS2 AafS2 pd. qd.+( )⋅ 500.85=:=L2
B2
ls
bs
= r=
Gf2 0.25 NS2⋅ 125.213=:=ls
bs
1.25=VdpS2 NS2 Gf2+ 626.063=:=
B2.
VdpS2
1.25 pconv2⋅1.108=:= L2. 1.25 B2.⋅ 1.385=:=
alegem : B2 1.25:=
L2 1.75:=
pef2
VdpS2
B2 L2⋅286.2=:=
luam clasa de beton C12/15 si rezulta ca H/L = min 0.29
H2. 0.29 L2⋅ 0.507=:=
H2 0.75:=
1'.) PREDIMENSIONARE S3
Stabilirea Df :•
Df Hing 10...20( )cm+≥
Df3 1.1:=
Stabilirea lui L3 , B3 ,H3 :•
peff pacc≤ AafS3 6 5⋅ 30=:=
pconv3 408:= obtinut prin interpolare
pacc3 pconv3 408=:=pd.. 20.79:=
qd3.. 12.6:=
L3
B3
ls
bs
= r=NS3 AafS3 pd. qd.+( )⋅ 1.002 10
3×=:=
Gf3 0.25 NS3⋅ 250.425=:= ls
bs
1.25=
VdpS3 NS3 Gf3+ 1.252 103
×=:=
B3.
VdpS3
1.25 pconv3⋅1.567=:= L3. 1.25 B3.⋅ 1.959=:=
alegem : B3 1.60:=
L3 2.40:=
pef3
VdpS3
B3 L3⋅326.074=:=
luam clasa de beton C12/15 si rezulta ca H/L = min 0.30
H3. 0.30 L3⋅ 0.72=:=
alegem : H3 0.75:= !!! din acest motiv modificam si inaltimea fundatiei de sub stalpul S2 , rezultand
H2 = 0.7 m
2.) VERIFICARE LA CAPACITATE PORTANTA S2+S3
consideram ca presiunea se transmite numai
pe talpa fundatiei
Kgr
Kst
10≤
Ed1efectul actiunilor la talpa fundatiei din axul 1,
de forma unei forte verticale 1). Ed1
A1.2
Rd1
A1.2
≤
Ed2efectul actiunilor la talpa fundatiei din axul 2,
de forma unei forte verticale
A1.2aria efectiva la talpa fundatiei din axul 12).
Ed2
A1.3
Rd2
A1.3
≤
A1.3aria efectiva la talpa fundatiei din axul 2
Eforturile sectionale :
NS2 500.85=
Mx2 0.2 NS2⋅ 100.17=:= Tafx2 0.1 NS2⋅ 50.085=:=
My2 0.1 NS2⋅ 50.085=:= Tafy2 0.05 NS2⋅ 25.043=:=
NS3 1.002 103
×=
Mx3 0.2 NS3⋅ 200.34=:= Tafx3 0.1 NS3⋅ 100.17=:=
My3 0.1 NS3⋅ 100.17=:= Tafy3 0.05 NS3⋅ 50.085=:=
Gfd2. L2 B2⋅ H2⋅ γγγγbet⋅ 41.016=:=
Gfd3. L3 B3⋅ H3⋅ γγγγbet⋅ 72=:=
Grinda de fundare :•bs
20.2=
lgr. 5 B2
B3
2+
− 0.2+ 3.15=:=
hgr
bgr
1.5.....3= bgr = minim latimea stalpului ls. 0.5:=
bgr. 0.50:=
hgr. 0.75:= luam inaltimea grinzii egala cu inaltimea fundatiei
lgr. 3.15:=
Ggr bgr. hgr.⋅ lgr.⋅ γγγγbet⋅ 29.531=:=
Gfd2 Gfd2.
Ggr
2+ 55.781=:=
Gfd3 Gfd3.
Ggr
2+ 86.766=:=
Myfed2 My2 Tafy2 H2⋅+ 68.867=:=
Myfed3 My3 Tafy3 H3⋅+ 137.734=:=
ΣΣΣΣM1 0= obtinem E2
ΣΣΣΣM2 0= obtinem E1
ΣΣΣΣM1 0=
Ed2
Myfed3 NS3 4.575⋅+ Gfd3 4.575⋅+ Myfed2− NS2 0.425⋅−( )4.575
1.057 103
×=:=
ΣΣΣΣM2 0=
Ed1
Myfed2 NS2 5⋅+ Gfd2 4.575⋅+ Myfed3−( )4.575
588.105=:=
VERIFICARE : NS2 NS3+ Gfd2+ Gfd3+ Ed1− Ed2− 0=
Axul 1 :•
Mxfed2 Mx2 Tafx2 H2⋅+ 137.734=:=
eL2
Mxfed2
Ed1
0.234=:=
B1.2 B2 1.25=:= L1.2 L2 2eL2− 1.282=:=
A1.2 B1.2 L1.2⋅ 1.602=:=
Axul 2 :•
Mxfed3 Mx3 Tafx3 H3⋅+ 275.468=:=
eL3
Mxfed3
Ed2
0.261=:=
B1.3 B3 1.6=:= L1.3 L3 2eL3− 1.879=:=
A1.3 B1.3 L1.3⋅ 3.006=:=
Hd2 Tafx22Tafy2
2+ 55.997=:=
Hd3 Tafx32Tafy3
2+ 111.993=:=
presiunile
efective :pef2.
Ed1
A1.2
367.106=:= pef3.
Ed2
A1.3
351.623=:=
Calculul capacitatii portante S2•
-- se face in conditii drenate
ϕϕϕϕ1 9.56 deg⋅=αααα 0:= inclinarea bazei fundatiei
c1 36.48:=
Nq2 Nq 2.374=:=
Nc2 Nc 8.155=:=
Nγγγγ2 Nγγγγ 0.463=:=
mB2
2B1.2
L1.2
+
1B1.2
L1.2
+
1.506=:=bq2 1 αααα tan ϕϕϕϕ1( )2⋅−
1=:=
bγγγγ2 bq 1=:=
bc2 bq
1 bq−( )Nc tan ϕϕϕϕ1( )⋅( )
− 1=:=
mL2
2L1.2
B1.2
+
1L1.2
B1.2
+
1.494=:=
sq2 1B1.2
L1.2
sin ϕϕϕϕ1( )⋅+ 1.162=:=
θθθθ 63.43deg:=
m mL2 cos θθθθ( )2⋅ mB2 sin θθθθ( )
2⋅+ 1.504=:=sγγγγ2 1 0.3
B1.2
L1.2
⋅− 0.707=:=
sc2
sq2 Nq⋅ 1−( )Nq 1−( )
1.28=:=
iq2 0.91:=
iγγγγ2 0.854:=
ic2 0.84:=
q2 20.02:=
pacc
Rd
A1
=
pacc2 c1 Nc⋅ bc⋅ sc⋅ ic⋅ q2 Nq⋅ bq⋅ sq⋅ iq⋅+ 0.5 γγγγk1⋅ B1.2⋅ Nγγγγ⋅ bγγγγ⋅ sγγγγ⋅ iγγγγ⋅+ 378.584=:=
pacc2 378.584=
pef2. 367.106=
V2 "VERIFICA" pacc2 pef2.≥if
"NU VERIFICA" otherwise
:=V2 "VERIFICA"=
Calculul capacitatii portante S3•
-- se face in conditii drenate
ϕϕϕϕ1 9.56 deg⋅=αααα 0:= inclinarea bazei fundatiei
c1 36.48=
Nq3 Nq 2.374=:=
Nc3 Nc 8.155=:=
Nγγγγ3 Nγγγγ 0.463=:=
mB3
2B1.3
L1.3
+
1B1.3
L1.3
+
1.54=:=bq3 1 αααα tan ϕϕϕϕ1( )2⋅−
1=:=
bγγγγ3 bq 1=:=
bc3 bq
1 bq−( )Nc tan ϕϕϕϕ1( )⋅( )
− 1=:=
mL3
2L1.3
B1.3
+
1L1.3
B1.3
+
1.46=:=
sq3 1B1.3
L1.3
sin ϕϕϕϕ1( )⋅+ 1.141=:=
θθθθ 63.43deg:=
sγγγγ3 1 0.3B1.3
L1.3
⋅− 0.745=:=
m mL3 cos θθθθ( )2⋅ mB3 sin θθθθ( )
2⋅+ 1.524=:=
sc3
sq3 Nq⋅ 1−( )Nq 1−( )
1.244=:=
iq3 0.902:=
iγγγγ3 0.843:=
ic3 0.831:= pacc
Rd
A1
=
q3 20.02:=
pacc3. c1 Nc3⋅ bc3⋅ sc3⋅ ic3⋅ q3 Nq3⋅ bq3⋅ sq3⋅ iq3⋅+ 0.5 γγγγk1⋅ B1.3⋅ Nγγγγ3⋅ bγγγγ3⋅ sγγγγ3⋅ iγγγγ3⋅+:=
pacc3. 360.796=
pef3. 351.623=
V3 "VERIFICA" pacc3. pef3.≥if
"NU VERIFICA" otherwise
:=V3 "VERIFICA"=
ARMARE FUNDATIEI S2
p1.2
Ed1
L2 B2⋅1 6−
eL2
L2
+
⋅= eL2
L2
6≤
eL2.
L2
60.292=:=
p1.2
Ed1
L2 B2⋅1 6
eL2.
L2
+
⋅ 537.696=:=
p2.2
Ed1
L2 B2⋅1 6
eL2.
L2
−
⋅ 0=:= p02 350.51:=
Armatura As1•
lcy2 0.6:=cnom 0.05= ϕϕϕϕs2 0.016:=
lcx2 0.425:=
Mx.x2 B2 p02 lcy2⋅lcy2
2⋅ p1.2 p02−( )
lcy2
2⋅
2
3⋅ lcy2⋅+
⋅ 106.943=:=
fcd12
1.58=:=
dx2 H2 cnom−ϕϕϕϕs2
2− 0.692=:=
μμμμx μμμμlim<μμμμx2
Mx.x2
B2 dx22⋅ 10
3⋅ fcd⋅
0.022=:= μμμμlim 0.403:=
ωωωω2 1 1 2μμμμx2−− 0.023=:=Asmin 0.26
fctm
fyk
⋅ B2⋅ dx2⋅ 104
⋅ 10.43=:=
Asx2 ωωωω2 B2⋅ dx2⋅ 104
⋅fcd
fyd
⋅ 5.21=:=
alegem ϕϕϕϕ 6 16 cu : d = 23 cm
As1ef2 12.06cm2:=
A1.S2 12.06 104−
⋅:=
ρρρρ2
A1.S2
B2 dx⋅3.97 10
3−×=:= ρρρρmin 0.075% 7.5 10
4−×=:=
ρρρρ2 ρρρρmin>
Armatura As2•
pmed2
p1.2 p2.2+( )2
268.848=:=
My.y2 L2 pmed2 lcx2⋅lcx2
2⋅
⋅ 42.491=:=
L2 1.75=
dy2 H2 cnom−ϕϕϕϕs2
2− ϕϕϕϕs2− 0.676=:=
μμμμy2
My.y2
L2 dy22⋅ 10
3⋅ fcd⋅
6.642 103−
×=:= μμμμlim 0.403:=
μμμμx μμμμlim<
ωωωω2 1 1 2μμμμy2−− 6.664 103−
×=:=
Asmin 0.26fctm
fyk
⋅ L2⋅ dy2⋅ 104
⋅ 14.265=:=Asy2 ωωωω2 L2⋅ dy2⋅ 10
4⋅
fcd
fyd
⋅ 2.102=:=
A2.S2 16.94 104−
×:= alegem ϕϕϕϕ 11 14 cu : d = 16.5 cm
As1ef 16.94cm2:=
ρρρρy2
A2.S2
L2 dy2⋅1.432 10
3−×=:= ρρρρmin. 7.5 10
4−×= ρρρρ2 ρρρρmin>
ARMARE FUNDATIEI S3
p1.2
Ed2
L2 B2⋅1 6−
eL2
L2
+
⋅= eL2
L2
6≤
eL3.
B3
60.267=:=
p1.3
Ed2
L3 B3⋅1 6
eL3.
L3
+
⋅ 458.764=:=
p2.3
Ed2
L3 B3⋅1 6
eL3.
L3
−
⋅ 91.753=:= p03 172.13:=
Armatura As1•
lcy3 0.95:=cnom 0.05= ϕϕϕϕs3 0.016:=
lcx3 0.6:=
Mx.x3 B3 p03 lcy3⋅lcy3
2⋅ p1.3 p03−( )
lcy3
2⋅
2
3⋅ lcy3⋅+
⋅ 262.244=:=
dx3 H3 cnom−ϕϕϕϕs3
2− 0.692=:=
μμμμx3
Mx.x3
B3 dx32⋅ 10
3⋅ fcd⋅
0.043=:= μμμμlim 0.403:=
μμμμx μμμμlim<
ωωωω3 1 1 2μμμμx3−− 0.044=:=
Asmin 0.26fctm
fyk
⋅ B3⋅ dx3⋅ 104
⋅ =:=Asx3 ωωωω3 B3⋅ dx3⋅ 10
4⋅
fcd
fyd
⋅ 12.915=:=
A1.S3 14.07 104−
×:= alegem ϕϕϕϕ 7 16 cu : d = 25 cm
As1ef 14.07cm2:=
ρρρρx3
A1.S3
B3 dx3⋅1.271 10
3−×=:= ρρρρmin. 7.5 10
4−×= ρρρρ2 ρρρρmin>
Armatura As2•
pmed3
p1.3 p2.3+( )2
275.258=:=
My.y3 L3 pmed3 lcx3⋅lcx3
2⋅
⋅ 118.912=:=
dy3 H3 cnom−ϕϕϕϕs3
2− ϕϕϕϕs3− 0.676=:=
μμμμy3
My.y3
L3 dy32⋅ 10
3⋅ fcd⋅
0.014=:=μμμμlim 0.403:= μμμμy μμμμlim<
ωωωω3. 1 1 2μμμμy3−− 0.014=:=
Asmin 0.26fctm
fyk
⋅ L3⋅ dy3⋅ 104
⋅ 19.563=:=Asy3 ωωωω3. L3⋅ dy3⋅ 10
4⋅
fcd
fyd
⋅ 5.904=:=
alegem ϕϕϕϕ 11 16 cu : As1ef 22.11cm2:=
distanta dintre bare : d = 23 cm
A2.S3 22.11 104−
×:=
ρρρρy3
A2.S3
L3 dy3⋅1.363 10
3−×=:= ρρρρmin. 7.5 10
4−×= ρρρρ3 ρρρρmin>
ϕϕϕϕLUNGIMEA CIOCURILOR = 15
GRINDA DE ECHILIBRARE
Hgr 0.75:= Bgr 0.5:=
Vmax 500.85:= Mmax 281.37:=
armatura longitudinala :•
ϕϕϕϕsg 0.022:= cnom 0.05=
dg Hgr cnom−ϕϕϕϕsg
2− 0.689=:=
μμμμg
Mmax
Bgr dg2⋅ 10
3⋅ fcd⋅
0.148=:=μμμμlim 0.403:= μμμμy μμμμlim<
ωωωωg 1 1 2μμμμg−− 0.161=:=
Asmin 0.26fctm
fyk
⋅ Bgr⋅ dg⋅ 104
⋅ 4.154=:=Asg ωωωωg Bgr⋅ dg⋅ 10
4⋅
fcd
fyd
⋅ 14.806=:=
alegem ϕϕϕϕ 4 22 cu : As1efg 15.20cm2:=
Al.gr 15.20 104−
×:=
ρρρρg
Al.gr
Bgr dg⋅4.412 10
3−×=:= ρρρρmin. 7.5 10
4−×= ρρρρ3 ρρρρmin>
armatura transversala :•
In fundatia S2 :
cRd.c 0.12:= ρρρρl ρρρρg 4.412 103−
×=:= fck 8:=
ηηηη 1:= k 1200
dg 103
⋅+ 1.539=:=
VRd.c. cRd.c ηηηη⋅ k⋅ 100 ρρρρl⋅ fck⋅( )1
3⋅ dg⋅ Bgr⋅ 10
6⋅ 9.686 10
4×=:=
VRd.c
VRd.c.
100096.856=:=
ctgθθθθ 1.75:= z 0.9 dg⋅ 0.62=:=
Vmax VRd.c> fywd fyd 300=:=
Asw
s
Vmax
z fywd⋅ ctgθθθθ⋅≥
impunem s 0.10:= Asw
Vmax 106
⋅ s⋅
z fywd⋅ ctgθθθθ⋅ 103
⋅153.846=:=
ϕϕϕϕalegem Etrier 2 10 la distanta de 10 cm cu : Asw. 1.57 cm2⋅:=
In camp si in fundatia S3
Ved 31.474:=
VRd.c 96.856=
Ved VRd.c< ϕϕϕϕalegem constructiv Etrier 10 la distanta de 25 cm
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